Return the Closest Prime Number
32
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Challenge
This is a simple one: Given a positive integer up to 1,000,000, return the closest prime number.
If the number itself is prime, then you should return that number; if there are two primes equally close to the provided number, return the lower of the two.
Input is in the form of a single integer, and output should be in the form of an integer as well.
I don't care how you take in the input (function, STDIN, etc.) or display the output (function, STDOUT, etc.), as long as it works.
This is code golf, so standard rules apply—the program with the least bytes wins!
Test Cases
Input => Output
------ -------
80 => 79
100 => 101
5 => 5
9 => 7
532 => 523
1 => 2
code-golf primes
edited Mar 28 at 17:17
Toby Speight
4,49711535
asked Mar 27 at 15:16
Nathan DimmerNathan Dimmer
331311
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|
show 7 more comments
32
$begingroup$
Challenge
This is a simple one: Given a positive integer up to 1,000,000, return the closest prime number.
If the number itself is prime, then you should return that number; if there are two primes equally close to the provided number, return the lower of the two.
Input is in the form of a single integer, and output should be in the form of an integer as well.
I don't care how you take in the input (function, STDIN, etc.) or display the output (function, STDOUT, etc.), as long as it works.
This is code golf, so standard rules apply—the program with the least bytes wins!
Test Cases
Input => Output
------ -------
80 => 79
100 => 101
5 => 5
9 => 7
532 => 523
1 => 2
code-golf primes
edited Mar 28 at 17:17
Toby Speight
4,49711535
asked Mar 27 at 15:16
Nathan DimmerNathan Dimmer
331311
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5
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Hi and welcome to PPCG!. To avoid down voting due to lack of quality I suggest you to post it to the sandbox first and after a couple of days post it here
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– Luis felipe De jesus Munoz
Mar 27 at 15:17
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This is one of the outputs requested in this challenge.
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– Arnauld
Mar 27 at 15:24
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Very closely related but not quite identical.
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– Giuseppe
Mar 27 at 15:35
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@Arnauld I saw that one, but I thought that they were different enough to warrant a new question.
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– Nathan Dimmer
Mar 27 at 15:37
2
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See also OEIS A051697.
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– Eric Towers
Mar 28 at 8:52
|
show 7 more comments
32
32
32
5
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Challenge
This is a simple one: Given a positive integer up to 1,000,000, return the closest prime number.
If the number itself is prime, then you should return that number; if there are two primes equally close to the provided number, return the lower of the two.
Input is in the form of a single integer, and output should be in the form of an integer as well.
I don't care how you take in the input (function, STDIN, etc.) or display the output (function, STDOUT, etc.), as long as it works.
This is code golf, so standard rules apply—the program with the least bytes wins!
Test Cases
Input => Output
------ -------
80 => 79
100 => 101
5 => 5
9 => 7
532 => 523
1 => 2
code-golf primes
edited Mar 28 at 17:17
Toby Speight
4,49711535
asked Mar 27 at 15:16
Nathan DimmerNathan Dimmer
331311
$endgroup$
Challenge
This is a simple one: Given a positive integer up to 1,000,000, return the closest prime number.
If the number itself is prime, then you should return that number; if there are two primes equally close to the provided number, return the lower of the two.
Input is in the form of a single integer, and output should be in the form of an integer as well.
I don't care how you take in the input (function, STDIN, etc.) or display the output (function, STDOUT, etc.), as long as it works.
This is code golf, so standard rules apply—the program with the least bytes wins!
Test Cases
Input => Output
------ -------
80 => 79
100 => 101
5 => 5
9 => 7
532 => 523
1 => 2
code-golf primes
code-golf primes
edited Mar 28 at 17:17
Toby Speight
4,49711535
asked Mar 27 at 15:16
Nathan DimmerNathan Dimmer
331311
edited Mar 28 at 17:17
Toby Speight
4,49711535
asked Mar 27 at 15:16
Nathan DimmerNathan Dimmer
331311
edited Mar 28 at 17:17
Toby Speight
4,49711535
edited Mar 28 at 17:17
Toby Speight
4,49711535
edited Mar 28 at 17:17
Toby Speight
4,49711535
4,49711535
asked Mar 27 at 15:16
Nathan DimmerNathan Dimmer
331311
asked Mar 27 at 15:16
Nathan DimmerNathan Dimmer
331311
asked Mar 27 at 15:16
Nathan DimmerNathan Dimmer
331311
331311
5
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Hi and welcome to PPCG!. To avoid down voting due to lack of quality I suggest you to post it to the sandbox first and after a couple of days post it here
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– Luis felipe De jesus Munoz
Mar 27 at 15:17
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This is one of the outputs requested in this challenge.
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– Arnauld
Mar 27 at 15:24
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Very closely related but not quite identical.
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– Giuseppe
Mar 27 at 15:35
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@Arnauld I saw that one, but I thought that they were different enough to warrant a new question.
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– Nathan Dimmer
Mar 27 at 15:37
2
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See also OEIS A051697.
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– Eric Towers
Mar 28 at 8:52
|
show 7 more comments
5
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Hi and welcome to PPCG!. To avoid down voting due to lack of quality I suggest you to post it to the sandbox first and after a couple of days post it here
$endgroup$
– Luis felipe De jesus Munoz
Mar 27 at 15:17
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This is one of the outputs requested in this challenge.
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– Arnauld
Mar 27 at 15:24
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Very closely related but not quite identical.
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– Giuseppe
Mar 27 at 15:35
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@Arnauld I saw that one, but I thought that they were different enough to warrant a new question.
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– Nathan Dimmer
Mar 27 at 15:37
2
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See also OEIS A051697.
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– Eric Towers
Mar 28 at 8:52
5
5
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Hi and welcome to PPCG!. To avoid down voting due to lack of quality I suggest you to post it to the sandbox first and after a couple of days post it here
$endgroup$
– Luis felipe De jesus Munoz
Mar 27 at 15:17
$begingroup$
Hi and welcome to PPCG!. To avoid down voting due to lack of quality I suggest you to post it to the sandbox first and after a couple of days post it here
$endgroup$
– Luis felipe De jesus Munoz
Mar 27 at 15:17
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This is one of the outputs requested in this challenge.
$endgroup$
– Arnauld
Mar 27 at 15:24
$begingroup$
This is one of the outputs requested in this challenge.
$endgroup$
– Arnauld
Mar 27 at 15:24
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Very closely related but not quite identical.
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– Giuseppe
Mar 27 at 15:35
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Very closely related but not quite identical.
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– Giuseppe
Mar 27 at 15:35
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@Arnauld I saw that one, but I thought that they were different enough to warrant a new question.
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– Nathan Dimmer
Mar 27 at 15:37
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@Arnauld I saw that one, but I thought that they were different enough to warrant a new question.
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– Nathan Dimmer
Mar 27 at 15:37
2
2
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See also OEIS A051697.
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– Eric Towers
Mar 28 at 8:52
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See also OEIS A051697.
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– Eric Towers
Mar 28 at 8:52
|
show 7 more comments
37 Answers
37
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JavaScript (ES6), 53 bytes
n=>(g=(o,d=N=n+o)=>N%--d?g(o,d):d-1?g(o<0?-o:~o):N)``
Try it online!
Commented
n => ( // n = input
g = ( // g = recursive function taking:
o, // o = offset
d = // d = current divisor, initialized to N
N = n + o // N = input + offset
) => //
N % --d ? // decrement d; if d is not a divisor of N:
g(o, d) // do recursive calls until it is
: // else:
d - 1 ? // if d is not equal to 1 (either N is composite or N = 1):
g( // do a recursive call with the next offset:
o < 0 ? // if o is negative:
-o // make it positive (e.g. -1 -> +1)
: // else:
~o // use -(o + 1) (e.g. +1 -> -2)
) // end of recursive call
: // else (N is prime):
N // stop recursion and return N
)`` // initial call to g with o = [''] (zero-ish)
answered Mar 27 at 16:01
ArnauldArnauld
80.6k797334
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10
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05AB1E, 5 bytes
Åps.x
Try it online!
or as a Test Suite
Inefficient for big numbers
answered Mar 27 at 15:17
EmignaEmigna
47.6k433145
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2
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Too badÅnis "In case of a tie, the higher prime is pushed" Didn't even knew we had this builtin, tbh.
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– Kevin Cruijssen
Mar 28 at 7:51
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@KevinCruijssen: Neither did I until now :)
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– Emigna
Mar 28 at 9:15
add a comment |
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Gaia, 3 bytes
ṅD⌡
Try it online!
Rather slow for large inputs, but works given enough memory/time.
I'm not sure why D⌡ implicitly pushes z again, but it makes this a remarkably short answer!
ṅ | implicit input z: push first z prime numbers, call it P
D⌡ | take the absolute difference between P and (implicit) z,
| returning the smallest value in P with the minimum absolute difference
answered Mar 27 at 20:23
GiuseppeGiuseppe
17.6k31153
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7
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Octave, 40 bytes
@(n)p([~,k]=min(abs(n-(p=primes(2*n)))))
Try it online!
This uses the fact that there is always a prime between n and 2*n (Bertrand–Chebyshev theorem).
How it works
@(n)p([~,k]=min(abs(n-(p=primes(2*n)))))
@(n) % Define anonymous function with input n
p=primes(2*n) % Vector of primes up to 2*n. Assign to p
abs(n-( )) % Absolute difference between n and each prime
[~,k]=min( ) % Index of first minimum (assign to k; not used)
p( ) % Apply that index to p
answered Mar 27 at 18:33
Luis MendoLuis Mendo
75.3k889292
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6
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Japt, 5 bytes
_j}cU
Try it or run all test cases
_j}cU :Implicit input of integer U
_ :Function taking an integer as an argument
j : Test if integer is prime
} :End function
cU :Return the first integer in [U,U-1,U+1,U-2,...] that returns true
answered Mar 27 at 15:43
ShaggyShaggy
18.9k21768
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add a comment |
5
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05AB1E, 4 bytes
z-Ån
Try it online!
answered Mar 28 at 12:48
GrimyGrimy
2,6411019
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add a comment |
4
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Pyth, 10 bytes
haDQfP_TSy
Try it online here, or verify all the test cases at once here.
haDQfP_TSyQ Implicit: Q=eval(input())
Trailing Q inferred
yQ 2 * Q
S Range from 1 to the above
f Filter keep the elements of the above, as T, where:
P_T Is T prime?
D Order the above by...
a Q ... absolute difference between each element and Q
This is a stable sort, so smaller primes will be sorted before larger ones if difference is the same
h Take the first element of the above, implicit print
answered Mar 27 at 15:46
SokSok
4,187925
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4
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Jelly, 9 7 bytes
ḤÆRạÞµḢ
Try it online!
Slow for larger input, but works ok for the requested range. Thanks to @EriktheOutgolfer for saving 2 bytes!
answered Mar 27 at 18:49
Nick KennedyNick Kennedy
1,36649
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Hey, that's clever! Save two by substituting_A¥withạ(absolute difference). Oh, andḤcan really be‘.
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– Erik the Outgolfer
Mar 27 at 19:08
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@EriktheOutgolfer thanks. Surely using‘won’t always work? It means that only primes up to n+1 will be found, while the closest might be n+2.
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– Nick Kennedy
Mar 27 at 22:39
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Hm, that's a concern.
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– Erik the Outgolfer
Mar 27 at 22:40
add a comment |
4
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Wolfram Language (Mathematica), 31 bytes
Nearest[Prime~Array~78499,#,1]&
Try it online!
& (*pure function*)
Prime~Array~78499 (*among the (ascending) first 78499 primes*)
1 (*select one*)
Nearest[ ,#, ] (*which is nearest to the argument*)
1000003 is the 78499th prime. Nearest prioritizes values which appear earlier in the list (which are lower).
answered Mar 28 at 23:46
attinatattinat
4797
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Nearest[Prime@Range@#,#,1]&for 27
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– Ben
Mar 29 at 16:10
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4
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Brachylog, 7 5 bytes
;I≜-ṗ
Try it online!
Saved 2 bytes thanks to @DLosc.
Explanation
;I≜ Label an unknown integer I (tries 0, then 1, then -1, then 2, etc.)
- Subtract I from the input
ṗ The result must be prime
answered Mar 28 at 15:11
FatalizeFatalize
28k449136
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@DLosc Mostly because I am stupid. Thanks.
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– Fatalize
Mar 29 at 8:26
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I think we just approached it from different directions. You were thinking about≜from the start, I assume, whereas I was thinking about pairing and subtracting and only later realized I'd need≜to make it work. :)
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– DLosc
Mar 29 at 21:27
add a comment |
4
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C (gcc), 87 76 74 72 bytes
Optimization of innat3's C# (Visual C# Interactive Compiler), 100 bytes
f(n,i,t,r,m){for(t=0,m=n;r-2;t++)for(r=i=1,n+=n<m?t:-t;i<n;n%++i||r++);}
Try it online!
answered Mar 29 at 11:55
Natural Number GuyNatural Number Guy
1516
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Hello and welcome to PPCG. A few tips:r!=2is equivalent tor-2,n%++i?0:r++can most likely ben%++i||r++.
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– Jonathan Frech
Mar 29 at 13:11
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I didn't immediately see that. Thanks.
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– Natural Number Guy
Mar 29 at 15:13
add a comment |
3
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Tidy, 43 bytes
{x:(prime↦splice(]x,-1,-∞],[x,∞]))@0}
Try it online!
Explanation
This is a lambda with parameter x. This works by creating the following sequence:
[x - 1, x, x - 2, x + 1, x - 3, x + 2, x - 4, x + 3, ...]
This is splicing together the two sequences ]x, -1, -∞] (left-closed, right-open) and [x, ∞] (both open).
For x = 80, this looks like:
[79, 80, 78, 81, 77, 82, 76, 83, 75, 84, 74, 85, ...]
Then, we use f↦s to select all elements from s satisfying f. In this case, we filter out all composite numbers, leaving only the prime ones. For the same x, this becomes:
[79, 83, 73, 71, 89, 67, 97, 61, 59, 101, 103, 53, ...]
Then, we use (...)@0 to select the first member of this sequence. Since the lower of the two needs to be selected, the sequence which starts with x - 1 is spliced in first.
Note: Only one of x and x - 1 can be prime, so it is okay that the spliced sequence starts with x - 1. Though the sequence could be open on both sides ([x,-1,-∞]), this would needlessly include x twice in the sequence. So, for sake of "efficiency", I chose the left-closed version (also because I like to show off Tidy).
answered Mar 27 at 21:39
Conor O'BrienConor O'Brien
30.6k264162
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3
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Python 2, 71 bytes
f=lambda n,k=1,p=1:k<n*3and min(k+n-p%k*2*n,f(n,k+1,p*k*k)-n,key=abs)+n
Try it online!
A recursive function that uses the Wilson's Theorem prime generator. The product p tracks $(k-1)!^2$, and p%k is 1 for primes and 0 for non-primes. To make it easy to compare abs(k-n) for different primes k, we store k-n and compare via abs, adding back n to get the result k.
The expression k+n-p%k*2*n is designed to give k-n on primes (where p%k=1), and otherwise a "bad" value of k+n that's always bigger in absolute value and so doesn't affect the minimum, so that non-primes are passed over.
answered Mar 28 at 0:04
xnorxnor
93.6k18190450
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3
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APL (Dyalog Extended), 20 15 bytesSBCS
Tacit prefix function inspired by Galen Ivanov's J answer.
⊢(⊃⍋⍤|⍤-⊇⊢)¯2⍭⍳
Try it online!
⍳ ɩndices one through the argument.
¯2⍭ nth primes of that
⊢(…) apply the following tacit function to that, with the original argument as left argument:
⊢ the primes
⊇ indexed by:
⍋ the ascending grade (indices which would sort ascending)
⍤ of
| the magnitude (absolute value)
⍤ of
- the differences
⊃ pick the first one (i.e. the one with smallest difference)
answered Mar 27 at 17:23
AdámAdám
28.9k276207
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3
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Perl 6, 35 bytes
{$_+=($*=-1)*$++until .is-prime;$_}
Try it online!
This uses Veitcel's technique for generating the list of 0, -1, 2, -3 but simplifies it greatly to ($*=-1)*$++ using the anonymous state variables available in P6 (I originally had -1 ** $++ * $++, but when golfed the negative loses precedence). There's a built in prime checker but unfortunately the until prevents the automagically returned value so there's an extra $_ hanging around.
answered Mar 29 at 0:15
guifaguifa
28115
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I'd usually use a sequence operator approach to something like this, but comes out to one byte longer, so nice work finding a shorter method
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– Jo King
Mar 29 at 0:39
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@JoKing good catch. The things that happen when I golf too quickly after getting a working solution. I had a similar one but the damned lack of [-1] haha
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– guifa
Mar 29 at 0:49
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3
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C, 122 121 104 bytes
p(a,i){for(i=1;++i<a;)if(a%i<1)return 0;return a>1;}c(a,b){for(b=a;;b++)if(p(--a)|p(b))return p(b)?b:a;}
Use it calling function c() and passing as argument the number; it should return the closest prime.
Thanks to Embodiment of Ignorance for 1 byte saved a big improvement.
Try it online!
answered Mar 27 at 20:24
Lince AssassinoLince Assassino
1114
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Butc()receives two parameters... Also, you can probably shorten thewhile(1)tofor(;;)(untested, since I don't get how to run your code
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– Embodiment of Ignorance
Mar 28 at 5:03
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@EmbodimentofIgnorance I wrote it and tested it all on an online c compiler, I could callc()passing only the first parameter. And you are right,for(;;)saves me a byte, only 117 left to get first place :)
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– Lince Assassino
Mar 28 at 11:30
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110 bytes:#define r return p(a,i){i=1;while(++i<a)if(a%i<1)r 0;r a>1;}c(a,b){b=a;for(;;b++){if(p(--a))r a;if(p(b))r b;}}. Here is a TIO link: tio.run/…
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– Embodiment of Ignorance
Mar 28 at 21:06
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106: tio.run/…
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– Embodiment of Ignorance
Mar 28 at 21:12
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101: tio.run/…
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– Embodiment of Ignorance
Mar 28 at 21:17
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3
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Wolfram Language (Mathematica), 52 bytes
If[PrimeQ[s=#],s,#&@@Nearest[s~NextPrime~{-1,1},s]]&
Try it online!
answered Mar 27 at 15:32
J42161217J42161217
13.9k21353
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You have an extra space that can be removed to save a byte.
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– Ben
Mar 29 at 16:03
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@Ben you are right. thanx
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– J42161217
Mar 29 at 23:08
add a comment |
2
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APL(NARS), 38 chars, 76 bytes
{⍵≤1:2⋄0π⍵:⍵⋄d←1π⍵⋄(d-⍵)≥⍵-s←¯1π⍵:s⋄d}
0π is the test for prime, ¯1π the prev prime, 1π is the next prime; test:
f←{⍵≤1:2⋄0π⍵:⍵⋄d←1π⍵⋄(d-⍵)≥⍵-s←¯1π⍵:s⋄d}
f¨80 100 5 9 532 1
79 101 5 7 523 2
answered Mar 27 at 18:05
RosLuPRosLuP
2,296514
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2
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J, 19 15 bytes
(0{]/:|@-)p:@i.
Try it online!
answered Mar 27 at 20:28
Galen IvanovGalen Ivanov
7,41211034
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2
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Perl 5, 59 bytes
$a=0;while((1x$_)=~/^.?$|^(..+?)1+$/){$_+=(-1)**$a*($a++)}
Try it online!
/^.?$|^(..+?)1+$/ is tricky regexp to check prime
(-1)**$a*($a++) generate sequence 0,-1, 2,-3 ...
answered Mar 28 at 8:35
VeitcelVeitcel
613
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2
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MathGolf, 10 bytes
∞╒g¶áÅ-±├Þ
Try it online.
Explanation:
∞ # Double the (implicit) input-integer
╒ # Create a list in the range [1, 2*n]
g¶ # Filter so only the prime numbers remain
áÅ # Sort this list using the next two character:
-± # The absolute difference with the (implicit) input-integer
├ # Push the first item of the list
# (unfortunately without popping the list itself, so:)
Þ # Discard everything from the stack except for the top
# (which is output implicitly as result)
answered Mar 28 at 8:43
Kevin CruijssenKevin Cruijssen
42.4k570217
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@JoKing Thanks! I knew Max thought about changing it, but didn't knew he actually did. The docs still state the old one.
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– Kevin Cruijssen
Mar 28 at 8:53
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Ah, I use the mathgolf.txt file as reference, since it seems to be more up to date
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– Jo King
Mar 28 at 23:21
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@JoKing Yeah, he told me yesterday about that file as well. Will use it from now on. :)
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– Kevin Cruijssen
Mar 29 at 7:23
add a comment |
2
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Factor, 91 bytes
: p ( x -- x ) [ nprimes ] keep dupd [ - abs ] curry map swap zip natural-sort first last ;
Try it online!
answered Mar 28 at 9:41
Galen IvanovGalen Ivanov
7,41211034
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2
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Python 2 (Cython), 96 bytes
l=lambda p:min(filter(lambda p:all(p%n for n in range(2,p)),range(2,p*3)),key=lambda x:abs(x-p))
Try it online!
answered Mar 27 at 16:29
Snaddyvitch DispenserSnaddyvitch Dispenser
1015
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Might be able to save a couple bytes viar=range;...
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– Skyler
Mar 28 at 20:33
1
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@Arnauld, it now works for x=1
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– Snaddyvitch Dispenser
Mar 29 at 8:20
add a comment |
2
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C# (Visual C# Interactive Compiler), 104 100 bytes
n=>{int r=0,t=0,m=n;while(r!=2){n+=(n<m)?t:-t;t++;r=0;for(int i=1;i<=n;i++)if(n%i==0)r++;}return n;}
Try it online!
Explanation:
int f(int n)
{
int r = 0; //stores the amount of factors of "n"
int t = 0; //increment used to cover all the integers surrounding "n"
int m = n; //placeholder to toggle between adding or substracting "t" to "n"
while (r != 2) //while the amount of factors found for "n" is different to 2 ("1" + itself)
{
n += (n < m) ? t : -t; //increment/decrement "n" by "t" (-0, -1, +2, -3, +4, -5,...)
t++;
r = 0;
for (int i = 1; i <= n; i++) //foreach number between "1" and "n" increment "r" if the remainder of its division with "n" is 0 (thus being a factor)
if (n % i == 0) r++;
}
return n;
}
Console.WriteLine(f(80)); //79
answered Mar 28 at 8:57
Innat3Innat3
1514
$endgroup$
add a comment |
2
$begingroup$
Java 8, 88 87 bytes
n->{for(int c=0,s=0,d,N=n;c!=2;s++)for(c=d=1,n+=n<N?s:-s;d<n;)if(n%++d<1)c++;return n;}
Port of @NaturalNumberGuy's (first) C answer, so make sure to upvote him!!
-1 byte thanks to @OlivierGrégoire.
Try it online.
Explanation:
n->{ // Method with integer as both parameter and return-type
for(int c=0, // Counter-integer, starting at 0
s=0, // Step-integer, starting at 0 as well
d, // Divisor-integer, uninitialized
N=n; // Copy of the input-integer
c!=2; // Loop as long as the counter is not exactly 2 yet:
s++) // After every iteration: increase the step-integer by 1
for(c=d=1, // (Re)set both the counter and divisor to 1
n+=n<N? // If the input is smaller than the input-copy:
s // Increase the input by the step-integer
: // Else:
-s; // Decrease the input by the step-integer
d<n;) // Inner loop as long as the divisor is smaller than the input
if(n%++d // Increase the divisor by 1 first with `++d`
<1) // And if the input is evenly divisible by the divisor:
c++; // Increase the counter-integer by 1
return n;} // Return the now modified input-integer as result
answered Mar 29 at 12:30
Kevin CruijssenKevin Cruijssen
42.4k570217
$endgroup$
add a comment |
2
$begingroup$
Java (JDK), 103 bytes
n->{int p=0,x=0,z=n,d;for(;p<1;p=p>0?z:0,z=z==n+x?n-++x:z+1)for(p=z/2,d=1;++d<z;)p=z%d<1?0:p;return p;}
Try it online!
answered Mar 29 at 11:23
Olivier GrégoireOlivier Grégoire
9,39511944
$endgroup$
$begingroup$
Umm.. I had already create a port of his answer.. ;) Although yours is 1 byte shorter, so something is different. EDIT: Ah, I have a result-integer outside the loop, and you modify the input inside the loop, hence the -1 byte for;. :) Do you want me to delete my answer?.. Feel free to copy the explanation.
$endgroup$
– Kevin Cruijssen
Mar 29 at 12:50
$begingroup$
@KevinCruijssen Oops, rollbacked!
$endgroup$
– Olivier Grégoire
Mar 29 at 12:59
$begingroup$
Sorry about that (and thanks for the -1 byte). I like your version as well, though. Already upvoted before I saw NaturalNumberGuy's answer.
$endgroup$
– Kevin Cruijssen
Mar 29 at 12:59
add a comment |
2
$begingroup$
Haskell, 79 74 bytes (thanks to Laikoni)
72 bytes as annonymus function (the initial "f=" could be removed in this case).
f=(!)(-1);n!x|x>1,all((>0).mod x)[2..x-1]=x|y<-x+n=last(-n+1:[-n-1|n>0])!y
Try it online!
original code:
f=(!)(-1);n!x|x>1&&all((>0).mod x)[2..x-1]=x|1>0=(last$(-n+1):[-n-1|n>0])!(x+n)
Try it online!
Explanation:
f x = (-1)!x
isPrime x = x > 1 && all (k -> x `mod` k /= 0)[2..x-1]
n!x | isPrime x = x -- return the first prime found
| n>0 = (-n-1)!(x+n) -- x is no prime, continue with x+n where n takes the
| otherwise = (-n+1)!(x+n) -- values -1,2,-3,4 .. in subsequent calls of (!)
answered Mar 29 at 2:38
SacheraSachera
512
$endgroup$
1
$begingroup$
Inside a guard you can use,instead of&&.(last$ ...)can belast(...), and the second guard1>0can be used for a binding to save parenthesis, e.g.y<-x+n.
$endgroup$
– Laikoni
Mar 30 at 10:57
$begingroup$
Anonymous functions are generally allowed, so the initialf=does not need to be counted. Also the parenthesis enclosing(-1+n)can be dropped.
$endgroup$
– Laikoni
Mar 30 at 11:22
$begingroup$
Thanks for the suggestions. I didn't know "," and bindings are allowed in function guards! But i don't really like the idea of an annonymous function as an answer. It doesn't feel right in my opinion.
$endgroup$
– Sachera
Mar 31 at 2:58
$begingroup$
You can find more tips in our collection of tips for golfing in Haskell. There is also a Guide to Golfing Rules in Haskell and dedicated chat room: Of Monads and Men.
$endgroup$
– Laikoni
Mar 31 at 22:13
add a comment |
2
$begingroup$
VDM-SL, 161 bytes
f(i)==(lambda p:set of nat1&let z in set p be st forall m in set p&abs(m-i)>=abs(z-i)in z)({x|x in set{1,...,9**7}&forall y in set{2,...,1003}&y<>x=>x mod y<>0})
A full program to run might look like this - it's worth noting that the bounds of the set of primes used should probably be changed if you actually want to run this, since it will take a long time to run for 1 million:
functions
f:nat1+>nat1
f(i)==(lambda p:set of nat1&let z in set p be st forall m in set p&abs(m-i)>=abs(z-i)in z)({x|x in set{1,...,9**7}&forall y in set{2,...,1003}&y<>x=>x mod y<>0})
Explanation:
f(i)== /* f is a function which takes a nat1 (natural number not including 0)*/
(lambda p:set of nat1 /* define a lambda which takes a set of nat1*/
&let z in set p be st /* which has an element z in the set such that */
forall m in set p /* for every element in the set*/
&abs(m-i) /* the difference between the element m and the input*/
>=abs(z-i) /* is greater than or equal to the difference between the element z and the input */
in z) /* and return z from the lambda */
( /* apply this lambda to... */
{ /* a set defined by comprehension as.. */
x| /* all elements x such that.. */
x in set{1,...,9**7} /* x is between 1 and 9^7 */
&forall y in set{2,...,1003} /* and for all values between 2 and 1003*/
&y<>x=>x mod y<>0 /* y is not x implies x is not divisible by y*/
}
)
answered Mar 27 at 16:31
Expired DataExpired Data
56314
$endgroup$
add a comment |
1
$begingroup$
C# (Visual C# Interactive Compiler), 112 bytes
g=>Enumerable.Range(2,2<<20).Where(x=>Enumerable.Range(1,x).Count(y=>x%y<1)<3).OrderBy(x=>Math.Abs(x-g)).First()
Try it online!
Left shifts by 20 in submission but 10 in TIO so that TIO terminates for test cases.
answered Mar 27 at 16:43
Expired DataExpired Data
56314
$endgroup$
add a comment |
1
$begingroup$
Swift, 186 bytes
func p(a:Int){let b=q(a:a,b:-1),c=q(a:a,b:1);print(a-b<=c-a ? b:c)}
func q(a:Int,b:Int)->Int{var k=max(a,2),c=2;while k>c && c != a/2{if k%c==0{k+=b;c=2}else{c=c==2 ? c+1:c+2}};return k}
Try it online!
answered Mar 27 at 18:24
onnowebonnoweb
1613
$endgroup$
add a comment |
1 2
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37 Answers
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37 Answers
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12
$begingroup$
JavaScript (ES6), 53 bytes
n=>(g=(o,d=N=n+o)=>N%--d?g(o,d):d-1?g(o<0?-o:~o):N)``
Try it online!
Commented
n => ( // n = input
g = ( // g = recursive function taking:
o, // o = offset
d = // d = current divisor, initialized to N
N = n + o // N = input + offset
) => //
N % --d ? // decrement d; if d is not a divisor of N:
g(o, d) // do recursive calls until it is
: // else:
d - 1 ? // if d is not equal to 1 (either N is composite or N = 1):
g( // do a recursive call with the next offset:
o < 0 ? // if o is negative:
-o // make it positive (e.g. -1 -> +1)
: // else:
~o // use -(o + 1) (e.g. +1 -> -2)
) // end of recursive call
: // else (N is prime):
N // stop recursion and return N
)`` // initial call to g with o = [''] (zero-ish)
answered Mar 27 at 16:01
ArnauldArnauld
80.6k797334
$endgroup$
add a comment |
12
$begingroup$
JavaScript (ES6), 53 bytes
n=>(g=(o,d=N=n+o)=>N%--d?g(o,d):d-1?g(o<0?-o:~o):N)``
Try it online!
Commented
n => ( // n = input
g = ( // g = recursive function taking:
o, // o = offset
d = // d = current divisor, initialized to N
N = n + o // N = input + offset
) => //
N % --d ? // decrement d; if d is not a divisor of N:
g(o, d) // do recursive calls until it is
: // else:
d - 1 ? // if d is not equal to 1 (either N is composite or N = 1):
g( // do a recursive call with the next offset:
o < 0 ? // if o is negative:
-o // make it positive (e.g. -1 -> +1)
: // else:
~o // use -(o + 1) (e.g. +1 -> -2)
) // end of recursive call
: // else (N is prime):
N // stop recursion and return N
)`` // initial call to g with o = [''] (zero-ish)
answered Mar 27 at 16:01
ArnauldArnauld
80.6k797334
$endgroup$
add a comment |
12
12
12
$begingroup$
JavaScript (ES6), 53 bytes
n=>(g=(o,d=N=n+o)=>N%--d?g(o,d):d-1?g(o<0?-o:~o):N)``
Try it online!
Commented
n => ( // n = input
g = ( // g = recursive function taking:
o, // o = offset
d = // d = current divisor, initialized to N
N = n + o // N = input + offset
) => //
N % --d ? // decrement d; if d is not a divisor of N:
g(o, d) // do recursive calls until it is
: // else:
d - 1 ? // if d is not equal to 1 (either N is composite or N = 1):
g( // do a recursive call with the next offset:
o < 0 ? // if o is negative:
-o // make it positive (e.g. -1 -> +1)
: // else:
~o // use -(o + 1) (e.g. +1 -> -2)
) // end of recursive call
: // else (N is prime):
N // stop recursion and return N
)`` // initial call to g with o = [''] (zero-ish)
answered Mar 27 at 16:01
ArnauldArnauld
80.6k797334
$endgroup$
JavaScript (ES6), 53 bytes
n=>(g=(o,d=N=n+o)=>N%--d?g(o,d):d-1?g(o<0?-o:~o):N)``
Try it online!
Commented
n => ( // n = input
g = ( // g = recursive function taking:
o, // o = offset
d = // d = current divisor, initialized to N
N = n + o // N = input + offset
) => //
N % --d ? // decrement d; if d is not a divisor of N:
g(o, d) // do recursive calls until it is
: // else:
d - 1 ? // if d is not equal to 1 (either N is composite or N = 1):
g( // do a recursive call with the next offset:
o < 0 ? // if o is negative:
-o // make it positive (e.g. -1 -> +1)
: // else:
~o // use -(o + 1) (e.g. +1 -> -2)
) // end of recursive call
: // else (N is prime):
N // stop recursion and return N
)`` // initial call to g with o = [''] (zero-ish)
answered Mar 27 at 16:01
ArnauldArnauld
80.6k797334
edited Mar 27 at 16:45
answered Mar 27 at 16:01
ArnauldArnauld
80.6k797334
answered Mar 27 at 16:01
ArnauldArnauld
80.6k797334
answered Mar 27 at 16:01
ArnauldArnauld
80.6k797334
80.6k797334
add a comment |
add a comment |
10
$begingroup$
05AB1E, 5 bytes
Åps.x
Try it online!
or as a Test Suite
Inefficient for big numbers
answered Mar 27 at 15:17
EmignaEmigna
47.6k433145
$endgroup$
2
$begingroup$
Too badÅnis "In case of a tie, the higher prime is pushed" Didn't even knew we had this builtin, tbh.
$endgroup$
– Kevin Cruijssen
Mar 28 at 7:51
$begingroup$
@KevinCruijssen: Neither did I until now :)
$endgroup$
– Emigna
Mar 28 at 9:15
add a comment |
10
$begingroup$
05AB1E, 5 bytes
Åps.x
Try it online!
or as a Test Suite
Inefficient for big numbers
answered Mar 27 at 15:17
EmignaEmigna
47.6k433145
$endgroup$
2
$begingroup$
Too badÅnis "In case of a tie, the higher prime is pushed" Didn't even knew we had this builtin, tbh.
$endgroup$
– Kevin Cruijssen
Mar 28 at 7:51
$begingroup$
@KevinCruijssen: Neither did I until now :)
$endgroup$
– Emigna
Mar 28 at 9:15
add a comment |
10
10
10
$begingroup$
05AB1E, 5 bytes
Åps.x
Try it online!
or as a Test Suite
Inefficient for big numbers
answered Mar 27 at 15:17
EmignaEmigna
47.6k433145
$endgroup$
05AB1E, 5 bytes
Åps.x
Try it online!
or as a Test Suite
Inefficient for big numbers
answered Mar 27 at 15:17
EmignaEmigna
47.6k433145
answered Mar 27 at 15:17
EmignaEmigna
47.6k433145
answered Mar 27 at 15:17
EmignaEmigna
47.6k433145
answered Mar 27 at 15:17
EmignaEmigna
47.6k433145
47.6k433145
2
$begingroup$
Too badÅnis "In case of a tie, the higher prime is pushed" Didn't even knew we had this builtin, tbh.
$endgroup$
– Kevin Cruijssen
Mar 28 at 7:51
$begingroup$
@KevinCruijssen: Neither did I until now :)
$endgroup$
– Emigna
Mar 28 at 9:15
add a comment |
2
$begingroup$
Too badÅnis "In case of a tie, the higher prime is pushed" Didn't even knew we had this builtin, tbh.
$endgroup$
– Kevin Cruijssen
Mar 28 at 7:51
$begingroup$
@KevinCruijssen: Neither did I until now :)
$endgroup$
– Emigna
Mar 28 at 9:15
2
2
$begingroup$
Too bad
Ån is "In case of a tie, the higher prime is pushed" Didn't even knew we had this builtin, tbh.$endgroup$
– Kevin Cruijssen
Mar 28 at 7:51
$begingroup$
Too bad
Ån is "In case of a tie, the higher prime is pushed" Didn't even knew we had this builtin, tbh.$endgroup$
– Kevin Cruijssen
Mar 28 at 7:51
$begingroup$
@KevinCruijssen: Neither did I until now :)
$endgroup$
– Emigna
Mar 28 at 9:15
$begingroup$
@KevinCruijssen: Neither did I until now :)
$endgroup$
– Emigna
Mar 28 at 9:15
add a comment |
7
$begingroup$
Gaia, 3 bytes
ṅD⌡
Try it online!
Rather slow for large inputs, but works given enough memory/time.
I'm not sure why D⌡ implicitly pushes z again, but it makes this a remarkably short answer!
ṅ | implicit input z: push first z prime numbers, call it P
D⌡ | take the absolute difference between P and (implicit) z,
| returning the smallest value in P with the minimum absolute difference
answered Mar 27 at 20:23
GiuseppeGiuseppe
17.6k31153
$endgroup$
add a comment |
7
$begingroup$
Gaia, 3 bytes
ṅD⌡
Try it online!
Rather slow for large inputs, but works given enough memory/time.
I'm not sure why D⌡ implicitly pushes z again, but it makes this a remarkably short answer!
ṅ | implicit input z: push first z prime numbers, call it P
D⌡ | take the absolute difference between P and (implicit) z,
| returning the smallest value in P with the minimum absolute difference
answered Mar 27 at 20:23
GiuseppeGiuseppe
17.6k31153
$endgroup$
add a comment |
7
7
7
$begingroup$
Gaia, 3 bytes
ṅD⌡
Try it online!
Rather slow for large inputs, but works given enough memory/time.
I'm not sure why D⌡ implicitly pushes z again, but it makes this a remarkably short answer!
ṅ | implicit input z: push first z prime numbers, call it P
D⌡ | take the absolute difference between P and (implicit) z,
| returning the smallest value in P with the minimum absolute difference
answered Mar 27 at 20:23
GiuseppeGiuseppe
17.6k31153
$endgroup$
Gaia, 3 bytes
ṅD⌡
Try it online!
Rather slow for large inputs, but works given enough memory/time.
I'm not sure why D⌡ implicitly pushes z again, but it makes this a remarkably short answer!
ṅ | implicit input z: push first z prime numbers, call it P
D⌡ | take the absolute difference between P and (implicit) z,
| returning the smallest value in P with the minimum absolute difference
answered Mar 27 at 20:23
GiuseppeGiuseppe
17.6k31153
edited Mar 27 at 20:30
answered Mar 27 at 20:23
GiuseppeGiuseppe
17.6k31153
answered Mar 27 at 20:23
GiuseppeGiuseppe
17.6k31153
answered Mar 27 at 20:23
GiuseppeGiuseppe
17.6k31153
17.6k31153
add a comment |
add a comment |
7
$begingroup$
Octave, 40 bytes
@(n)p([~,k]=min(abs(n-(p=primes(2*n)))))
Try it online!
This uses the fact that there is always a prime between n and 2*n (Bertrand–Chebyshev theorem).
How it works
@(n)p([~,k]=min(abs(n-(p=primes(2*n)))))
@(n) % Define anonymous function with input n
p=primes(2*n) % Vector of primes up to 2*n. Assign to p
abs(n-( )) % Absolute difference between n and each prime
[~,k]=min( ) % Index of first minimum (assign to k; not used)
p( ) % Apply that index to p
answered Mar 27 at 18:33
Luis MendoLuis Mendo
75.3k889292
$endgroup$
add a comment |
7
$begingroup$
Octave, 40 bytes
@(n)p([~,k]=min(abs(n-(p=primes(2*n)))))
Try it online!
This uses the fact that there is always a prime between n and 2*n (Bertrand–Chebyshev theorem).
How it works
@(n)p([~,k]=min(abs(n-(p=primes(2*n)))))
@(n) % Define anonymous function with input n
p=primes(2*n) % Vector of primes up to 2*n. Assign to p
abs(n-( )) % Absolute difference between n and each prime
[~,k]=min( ) % Index of first minimum (assign to k; not used)
p( ) % Apply that index to p
answered Mar 27 at 18:33
Luis MendoLuis Mendo
75.3k889292
$endgroup$
add a comment |
7
7
7
$begingroup$
Octave, 40 bytes
@(n)p([~,k]=min(abs(n-(p=primes(2*n)))))
Try it online!
This uses the fact that there is always a prime between n and 2*n (Bertrand–Chebyshev theorem).
How it works
@(n)p([~,k]=min(abs(n-(p=primes(2*n)))))
@(n) % Define anonymous function with input n
p=primes(2*n) % Vector of primes up to 2*n. Assign to p
abs(n-( )) % Absolute difference between n and each prime
[~,k]=min( ) % Index of first minimum (assign to k; not used)
p( ) % Apply that index to p
answered Mar 27 at 18:33
Luis MendoLuis Mendo
75.3k889292
$endgroup$
Octave, 40 bytes
@(n)p([~,k]=min(abs(n-(p=primes(2*n)))))
Try it online!
This uses the fact that there is always a prime between n and 2*n (Bertrand–Chebyshev theorem).
How it works
@(n)p([~,k]=min(abs(n-(p=primes(2*n)))))
@(n) % Define anonymous function with input n
p=primes(2*n) % Vector of primes up to 2*n. Assign to p
abs(n-( )) % Absolute difference between n and each prime
[~,k]=min( ) % Index of first minimum (assign to k; not used)
p( ) % Apply that index to p
answered Mar 27 at 18:33
Luis MendoLuis Mendo
75.3k889292
edited Mar 27 at 23:03
answered Mar 27 at 18:33
Luis MendoLuis Mendo
75.3k889292
answered Mar 27 at 18:33
Luis MendoLuis Mendo
75.3k889292
answered Mar 27 at 18:33
Luis MendoLuis Mendo
75.3k889292
75.3k889292
add a comment |
add a comment |
6
$begingroup$
Japt, 5 bytes
_j}cU
Try it or run all test cases
_j}cU :Implicit input of integer U
_ :Function taking an integer as an argument
j : Test if integer is prime
} :End function
cU :Return the first integer in [U,U-1,U+1,U-2,...] that returns true
answered Mar 27 at 15:43
ShaggyShaggy
18.9k21768
$endgroup$
add a comment |
6
$begingroup$
Japt, 5 bytes
_j}cU
Try it or run all test cases
_j}cU :Implicit input of integer U
_ :Function taking an integer as an argument
j : Test if integer is prime
} :End function
cU :Return the first integer in [U,U-1,U+1,U-2,...] that returns true
answered Mar 27 at 15:43
ShaggyShaggy
18.9k21768
$endgroup$
add a comment |
6
6
6
$begingroup$
Japt, 5 bytes
_j}cU
Try it or run all test cases
_j}cU :Implicit input of integer U
_ :Function taking an integer as an argument
j : Test if integer is prime
} :End function
cU :Return the first integer in [U,U-1,U+1,U-2,...] that returns true
answered Mar 27 at 15:43
ShaggyShaggy
18.9k21768
$endgroup$
Japt, 5 bytes
_j}cU
Try it or run all test cases
_j}cU :Implicit input of integer U
_ :Function taking an integer as an argument
j : Test if integer is prime
} :End function
cU :Return the first integer in [U,U-1,U+1,U-2,...] that returns true
answered Mar 27 at 15:43
ShaggyShaggy
18.9k21768
edited Mar 28 at 17:38
answered Mar 27 at 15:43
ShaggyShaggy
18.9k21768
answered Mar 27 at 15:43
ShaggyShaggy
18.9k21768
answered Mar 27 at 15:43
ShaggyShaggy
18.9k21768
18.9k21768
add a comment |
add a comment |
5
$begingroup$
05AB1E, 4 bytes
z-Ån
Try it online!
answered Mar 28 at 12:48
GrimyGrimy
2,6411019
$endgroup$
add a comment |
5
$begingroup$
05AB1E, 4 bytes
z-Ån
Try it online!
answered Mar 28 at 12:48
GrimyGrimy
2,6411019
$endgroup$
add a comment |
5
5
5
$begingroup$
05AB1E, 4 bytes
z-Ån
Try it online!
answered Mar 28 at 12:48
GrimyGrimy
2,6411019
$endgroup$
05AB1E, 4 bytes
z-Ån
Try it online!
answered Mar 28 at 12:48
GrimyGrimy
2,6411019
answered Mar 28 at 12:48
GrimyGrimy
2,6411019
answered Mar 28 at 12:48
GrimyGrimy
2,6411019
answered Mar 28 at 12:48
GrimyGrimy
2,6411019
2,6411019
add a comment |
add a comment |
4
$begingroup$
Pyth, 10 bytes
haDQfP_TSy
Try it online here, or verify all the test cases at once here.
haDQfP_TSyQ Implicit: Q=eval(input())
Trailing Q inferred
yQ 2 * Q
S Range from 1 to the above
f Filter keep the elements of the above, as T, where:
P_T Is T prime?
D Order the above by...
a Q ... absolute difference between each element and Q
This is a stable sort, so smaller primes will be sorted before larger ones if difference is the same
h Take the first element of the above, implicit print
answered Mar 27 at 15:46
SokSok
4,187925
$endgroup$
add a comment |
4
$begingroup$
Pyth, 10 bytes
haDQfP_TSy
Try it online here, or verify all the test cases at once here.
haDQfP_TSyQ Implicit: Q=eval(input())
Trailing Q inferred
yQ 2 * Q
S Range from 1 to the above
f Filter keep the elements of the above, as T, where:
P_T Is T prime?
D Order the above by...
a Q ... absolute difference between each element and Q
This is a stable sort, so smaller primes will be sorted before larger ones if difference is the same
h Take the first element of the above, implicit print
answered Mar 27 at 15:46
SokSok
4,187925
$endgroup$
add a comment |
4
4
4
$begingroup$
Pyth, 10 bytes
haDQfP_TSy
Try it online here, or verify all the test cases at once here.
haDQfP_TSyQ Implicit: Q=eval(input())
Trailing Q inferred
yQ 2 * Q
S Range from 1 to the above
f Filter keep the elements of the above, as T, where:
P_T Is T prime?
D Order the above by...
a Q ... absolute difference between each element and Q
This is a stable sort, so smaller primes will be sorted before larger ones if difference is the same
h Take the first element of the above, implicit print
answered Mar 27 at 15:46
SokSok
4,187925
$endgroup$
Pyth, 10 bytes
haDQfP_TSy
Try it online here, or verify all the test cases at once here.
haDQfP_TSyQ Implicit: Q=eval(input())
Trailing Q inferred
yQ 2 * Q
S Range from 1 to the above
f Filter keep the elements of the above, as T, where:
P_T Is T prime?
D Order the above by...
a Q ... absolute difference between each element and Q
This is a stable sort, so smaller primes will be sorted before larger ones if difference is the same
h Take the first element of the above, implicit print
answered Mar 27 at 15:46
SokSok
4,187925
answered Mar 27 at 15:46
SokSok
4,187925
answered Mar 27 at 15:46
SokSok
4,187925
answered Mar 27 at 15:46
SokSok
4,187925
4,187925
add a comment |
add a comment |
4
$begingroup$
Jelly, 9 7 bytes
ḤÆRạÞµḢ
Try it online!
Slow for larger input, but works ok for the requested range. Thanks to @EriktheOutgolfer for saving 2 bytes!
answered Mar 27 at 18:49
Nick KennedyNick Kennedy
1,36649
$endgroup$
$begingroup$
Hey, that's clever! Save two by substituting_A¥withạ(absolute difference). Oh, andḤcan really be‘.
$endgroup$
– Erik the Outgolfer
Mar 27 at 19:08
$begingroup$
@EriktheOutgolfer thanks. Surely using‘won’t always work? It means that only primes up to n+1 will be found, while the closest might be n+2.
$endgroup$
– Nick Kennedy
Mar 27 at 22:39
$begingroup$
Hm, that's a concern.
$endgroup$
– Erik the Outgolfer
Mar 27 at 22:40
add a comment |
4
$begingroup$
Jelly, 9 7 bytes
ḤÆRạÞµḢ
Try it online!
Slow for larger input, but works ok for the requested range. Thanks to @EriktheOutgolfer for saving 2 bytes!
answered Mar 27 at 18:49
Nick KennedyNick Kennedy
1,36649
$endgroup$
$begingroup$
Hey, that's clever! Save two by substituting_A¥withạ(absolute difference). Oh, andḤcan really be‘.
$endgroup$
– Erik the Outgolfer
Mar 27 at 19:08
$begingroup$
@EriktheOutgolfer thanks. Surely using‘won’t always work? It means that only primes up to n+1 will be found, while the closest might be n+2.
$endgroup$
– Nick Kennedy
Mar 27 at 22:39
$begingroup$
Hm, that's a concern.
$endgroup$
– Erik the Outgolfer
Mar 27 at 22:40
add a comment |
4
4
4
$begingroup$
Jelly, 9 7 bytes
ḤÆRạÞµḢ
Try it online!
Slow for larger input, but works ok for the requested range. Thanks to @EriktheOutgolfer for saving 2 bytes!
answered Mar 27 at 18:49
Nick KennedyNick Kennedy
1,36649
$endgroup$
Jelly, 9 7 bytes
ḤÆRạÞµḢ
Try it online!
Slow for larger input, but works ok for the requested range. Thanks to @EriktheOutgolfer for saving 2 bytes!
answered Mar 27 at 18:49
Nick KennedyNick Kennedy
1,36649
edited Mar 27 at 19:10
answered Mar 27 at 18:49
Nick KennedyNick Kennedy
1,36649
answered Mar 27 at 18:49
Nick KennedyNick Kennedy
1,36649
answered Mar 27 at 18:49
Nick KennedyNick Kennedy
1,36649
1,36649
$begingroup$
Hey, that's clever! Save two by substituting_A¥withạ(absolute difference). Oh, andḤcan really be‘.
$endgroup$
– Erik the Outgolfer
Mar 27 at 19:08
$begingroup$
@EriktheOutgolfer thanks. Surely using‘won’t always work? It means that only primes up to n+1 will be found, while the closest might be n+2.
$endgroup$
– Nick Kennedy
Mar 27 at 22:39
$begingroup$
Hm, that's a concern.
$endgroup$
– Erik the Outgolfer
Mar 27 at 22:40
add a comment |
$begingroup$
Hey, that's clever! Save two by substituting_A¥withạ(absolute difference). Oh, andḤcan really be‘.
$endgroup$
– Erik the Outgolfer
Mar 27 at 19:08
$begingroup$
@EriktheOutgolfer thanks. Surely using‘won’t always work? It means that only primes up to n+1 will be found, while the closest might be n+2.
$endgroup$
– Nick Kennedy
Mar 27 at 22:39
$begingroup$
Hm, that's a concern.
$endgroup$
– Erik the Outgolfer
Mar 27 at 22:40
$begingroup$
Hey, that's clever! Save two by substituting
_A¥ with ạ (absolute difference). Oh, and Ḥ can really be ‘.$endgroup$
– Erik the Outgolfer
Mar 27 at 19:08
$begingroup$
Hey, that's clever! Save two by substituting
_A¥ with ạ (absolute difference). Oh, and Ḥ can really be ‘.$endgroup$
– Erik the Outgolfer
Mar 27 at 19:08
$begingroup$
@EriktheOutgolfer thanks. Surely using
‘ won’t always work? It means that only primes up to n+1 will be found, while the closest might be n+2.$endgroup$
– Nick Kennedy
Mar 27 at 22:39
$begingroup$
@EriktheOutgolfer thanks. Surely using
‘ won’t always work? It means that only primes up to n+1 will be found, while the closest might be n+2.$endgroup$
– Nick Kennedy
Mar 27 at 22:39
$begingroup$
Hm, that's a concern.
$endgroup$
– Erik the Outgolfer
Mar 27 at 22:40
$begingroup$
Hm, that's a concern.
$endgroup$
– Erik the Outgolfer
Mar 27 at 22:40
add a comment |
4
$begingroup$
Wolfram Language (Mathematica), 31 bytes
Nearest[Prime~Array~78499,#,1]&
Try it online!
& (*pure function*)
Prime~Array~78499 (*among the (ascending) first 78499 primes*)
1 (*select one*)
Nearest[ ,#, ] (*which is nearest to the argument*)
1000003 is the 78499th prime. Nearest prioritizes values which appear earlier in the list (which are lower).
answered Mar 28 at 23:46
attinatattinat
4797
$endgroup$
4
$begingroup$
Nearest[Prime@Range@#,#,1]&for 27
$endgroup$
– Ben
Mar 29 at 16:10
add a comment |
4
$begingroup$
Wolfram Language (Mathematica), 31 bytes
Nearest[Prime~Array~78499,#,1]&
Try it online!
& (*pure function*)
Prime~Array~78499 (*among the (ascending) first 78499 primes*)
1 (*select one*)
Nearest[ ,#, ] (*which is nearest to the argument*)
1000003 is the 78499th prime. Nearest prioritizes values which appear earlier in the list (which are lower).
answered Mar 28 at 23:46
attinatattinat
4797
$endgroup$
4
$begingroup$
Nearest[Prime@Range@#,#,1]&for 27
$endgroup$
– Ben
Mar 29 at 16:10
add a comment |
4
4
4
$begingroup$
Wolfram Language (Mathematica), 31 bytes
Nearest[Prime~Array~78499,#,1]&
Try it online!
& (*pure function*)
Prime~Array~78499 (*among the (ascending) first 78499 primes*)
1 (*select one*)
Nearest[ ,#, ] (*which is nearest to the argument*)
1000003 is the 78499th prime. Nearest prioritizes values which appear earlier in the list (which are lower).
answered Mar 28 at 23:46
attinatattinat
4797
$endgroup$
Wolfram Language (Mathematica), 31 bytes
Nearest[Prime~Array~78499,#,1]&
Try it online!
& (*pure function*)
Prime~Array~78499 (*among the (ascending) first 78499 primes*)
1 (*select one*)
Nearest[ ,#, ] (*which is nearest to the argument*)
1000003 is the 78499th prime. Nearest prioritizes values which appear earlier in the list (which are lower).
answered Mar 28 at 23:46
attinatattinat
4797
edited Mar 29 at 4:19
answered Mar 28 at 23:46
attinatattinat
4797
answered Mar 28 at 23:46
attinatattinat
4797
answered Mar 28 at 23:46
attinatattinat
4797
4797
4
$begingroup$
Nearest[Prime@Range@#,#,1]&for 27
$endgroup$
– Ben
Mar 29 at 16:10
add a comment |
4
$begingroup$
Nearest[Prime@Range@#,#,1]&for 27
$endgroup$
– Ben
Mar 29 at 16:10
4
4
$begingroup$
Nearest[Prime@Range@#,#,1]& for 27$endgroup$
– Ben
Mar 29 at 16:10
$begingroup$
Nearest[Prime@Range@#,#,1]& for 27$endgroup$
– Ben
Mar 29 at 16:10
add a comment |
4
$begingroup$
Brachylog, 7 5 bytes
;I≜-ṗ
Try it online!
Saved 2 bytes thanks to @DLosc.
Explanation
;I≜ Label an unknown integer I (tries 0, then 1, then -1, then 2, etc.)
- Subtract I from the input
ṗ The result must be prime
answered Mar 28 at 15:11
FatalizeFatalize
28k449136
$endgroup$
$begingroup$
@DLosc Mostly because I am stupid. Thanks.
$endgroup$
– Fatalize
Mar 29 at 8:26
$begingroup$
I think we just approached it from different directions. You were thinking about≜from the start, I assume, whereas I was thinking about pairing and subtracting and only later realized I'd need≜to make it work. :)
$endgroup$
– DLosc
Mar 29 at 21:27
add a comment |
4
$begingroup$
Brachylog, 7 5 bytes
;I≜-ṗ
Try it online!
Saved 2 bytes thanks to @DLosc.
Explanation
;I≜ Label an unknown integer I (tries 0, then 1, then -1, then 2, etc.)
- Subtract I from the input
ṗ The result must be prime
answered Mar 28 at 15:11
FatalizeFatalize
28k449136
$endgroup$
$begingroup$
@DLosc Mostly because I am stupid. Thanks.
$endgroup$
– Fatalize
Mar 29 at 8:26
$begingroup$
I think we just approached it from different directions. You were thinking about≜from the start, I assume, whereas I was thinking about pairing and subtracting and only later realized I'd need≜to make it work. :)
$endgroup$
– DLosc
Mar 29 at 21:27
add a comment |
4
4
4
$begingroup$
Brachylog, 7 5 bytes
;I≜-ṗ
Try it online!
Saved 2 bytes thanks to @DLosc.
Explanation
;I≜ Label an unknown integer I (tries 0, then 1, then -1, then 2, etc.)
- Subtract I from the input
ṗ The result must be prime
answered Mar 28 at 15:11
FatalizeFatalize
28k449136
$endgroup$
Brachylog, 7 5 bytes
;I≜-ṗ
Try it online!
Saved 2 bytes thanks to @DLosc.
Explanation
;I≜ Label an unknown integer I (tries 0, then 1, then -1, then 2, etc.)
- Subtract I from the input
ṗ The result must be prime
answered Mar 28 at 15:11
FatalizeFatalize
28k449136
edited Mar 29 at 8:26
answered Mar 28 at 15:11
FatalizeFatalize
28k449136
answered Mar 28 at 15:11
FatalizeFatalize
28k449136
answered Mar 28 at 15:11
FatalizeFatalize
28k449136
28k449136
$begingroup$
@DLosc Mostly because I am stupid. Thanks.
$endgroup$
– Fatalize
Mar 29 at 8:26
$begingroup$
I think we just approached it from different directions. You were thinking about≜from the start, I assume, whereas I was thinking about pairing and subtracting and only later realized I'd need≜to make it work. :)
$endgroup$
– DLosc
Mar 29 at 21:27
add a comment |
$begingroup$
@DLosc Mostly because I am stupid. Thanks.
$endgroup$
– Fatalize
Mar 29 at 8:26
$begingroup$
I think we just approached it from different directions. You were thinking about≜from the start, I assume, whereas I was thinking about pairing and subtracting and only later realized I'd need≜to make it work. :)
$endgroup$
– DLosc
Mar 29 at 21:27
$begingroup$
@DLosc Mostly because I am stupid. Thanks.
$endgroup$
– Fatalize
Mar 29 at 8:26
$begingroup$
@DLosc Mostly because I am stupid. Thanks.
$endgroup$
– Fatalize
Mar 29 at 8:26
$begingroup$
I think we just approached it from different directions. You were thinking about
≜ from the start, I assume, whereas I was thinking about pairing and subtracting and only later realized I'd need ≜ to make it work. :)$endgroup$
– DLosc
Mar 29 at 21:27
$begingroup$
I think we just approached it from different directions. You were thinking about
≜ from the start, I assume, whereas I was thinking about pairing and subtracting and only later realized I'd need ≜ to make it work. :)$endgroup$
– DLosc
Mar 29 at 21:27
add a comment |
4
$begingroup$
C (gcc), 87 76 74 72 bytes
Optimization of innat3's C# (Visual C# Interactive Compiler), 100 bytes
f(n,i,t,r,m){for(t=0,m=n;r-2;t++)for(r=i=1,n+=n<m?t:-t;i<n;n%++i||r++);}
Try it online!
answered Mar 29 at 11:55
Natural Number GuyNatural Number Guy
1516
$endgroup$
$begingroup$
Hello and welcome to PPCG. A few tips:r!=2is equivalent tor-2,n%++i?0:r++can most likely ben%++i||r++.
$endgroup$
– Jonathan Frech
Mar 29 at 13:11
$begingroup$
I didn't immediately see that. Thanks.
$endgroup$
– Natural Number Guy
Mar 29 at 15:13
add a comment |
4
$begingroup$
C (gcc), 87 76 74 72 bytes
Optimization of innat3's C# (Visual C# Interactive Compiler), 100 bytes
f(n,i,t,r,m){for(t=0,m=n;r-2;t++)for(r=i=1,n+=n<m?t:-t;i<n;n%++i||r++);}
Try it online!
answered Mar 29 at 11:55
Natural Number GuyNatural Number Guy
1516
$endgroup$
$begingroup$
Hello and welcome to PPCG. A few tips:r!=2is equivalent tor-2,n%++i?0:r++can most likely ben%++i||r++.
$endgroup$
– Jonathan Frech
Mar 29 at 13:11
$begingroup$
I didn't immediately see that. Thanks.
$endgroup$
– Natural Number Guy
Mar 29 at 15:13
add a comment |
4
4
4
$begingroup$
C (gcc), 87 76 74 72 bytes
Optimization of innat3's C# (Visual C# Interactive Compiler), 100 bytes
f(n,i,t,r,m){for(t=0,m=n;r-2;t++)for(r=i=1,n+=n<m?t:-t;i<n;n%++i||r++);}
Try it online!
answered Mar 29 at 11:55
Natural Number GuyNatural Number Guy
1516
$endgroup$
C (gcc), 87 76 74 72 bytes
Optimization of innat3's C# (Visual C# Interactive Compiler), 100 bytes
f(n,i,t,r,m){for(t=0,m=n;r-2;t++)for(r=i=1,n+=n<m?t:-t;i<n;n%++i||r++);}
Try it online!
answered Mar 29 at 11:55
Natural Number GuyNatural Number Guy
1516
edited Mar 29 at 15:24
answered Mar 29 at 11:55
Natural Number GuyNatural Number Guy
1516
answered Mar 29 at 11:55
Natural Number GuyNatural Number Guy
1516
answered Mar 29 at 11:55
Natural Number GuyNatural Number Guy
1516
1516
$begingroup$
Hello and welcome to PPCG. A few tips:r!=2is equivalent tor-2,n%++i?0:r++can most likely ben%++i||r++.
$endgroup$
– Jonathan Frech
Mar 29 at 13:11
$begingroup$
I didn't immediately see that. Thanks.
$endgroup$
– Natural Number Guy
Mar 29 at 15:13
add a comment |
$begingroup$
Hello and welcome to PPCG. A few tips:r!=2is equivalent tor-2,n%++i?0:r++can most likely ben%++i||r++.
$endgroup$
– Jonathan Frech
Mar 29 at 13:11
$begingroup$
I didn't immediately see that. Thanks.
$endgroup$
– Natural Number Guy
Mar 29 at 15:13
$begingroup$
Hello and welcome to PPCG. A few tips:
r!=2 is equivalent to r-2, n%++i?0:r++ can most likely be n%++i||r++.$endgroup$
– Jonathan Frech
Mar 29 at 13:11
$begingroup$
Hello and welcome to PPCG. A few tips:
r!=2 is equivalent to r-2, n%++i?0:r++ can most likely be n%++i||r++.$endgroup$
– Jonathan Frech
Mar 29 at 13:11
$begingroup$
I didn't immediately see that. Thanks.
$endgroup$
– Natural Number Guy
Mar 29 at 15:13
$begingroup$
I didn't immediately see that. Thanks.
$endgroup$
– Natural Number Guy
Mar 29 at 15:13
add a comment |
3
$begingroup$
Tidy, 43 bytes
{x:(prime↦splice(]x,-1,-∞],[x,∞]))@0}
Try it online!
Explanation
This is a lambda with parameter x. This works by creating the following sequence:
[x - 1, x, x - 2, x + 1, x - 3, x + 2, x - 4, x + 3, ...]
This is splicing together the two sequences ]x, -1, -∞] (left-closed, right-open) and [x, ∞] (both open).
For x = 80, this looks like:
[79, 80, 78, 81, 77, 82, 76, 83, 75, 84, 74, 85, ...]
Then, we use f↦s to select all elements from s satisfying f. In this case, we filter out all composite numbers, leaving only the prime ones. For the same x, this becomes:
[79, 83, 73, 71, 89, 67, 97, 61, 59, 101, 103, 53, ...]
Then, we use (...)@0 to select the first member of this sequence. Since the lower of the two needs to be selected, the sequence which starts with x - 1 is spliced in first.
Note: Only one of x and x - 1 can be prime, so it is okay that the spliced sequence starts with x - 1. Though the sequence could be open on both sides ([x,-1,-∞]), this would needlessly include x twice in the sequence. So, for sake of "efficiency", I chose the left-closed version (also because I like to show off Tidy).
answered Mar 27 at 21:39
Conor O'BrienConor O'Brien
30.6k264162
$endgroup$
add a comment |
3
$begingroup$
Tidy, 43 bytes
{x:(prime↦splice(]x,-1,-∞],[x,∞]))@0}
Try it online!
Explanation
This is a lambda with parameter x. This works by creating the following sequence:
[x - 1, x, x - 2, x + 1, x - 3, x + 2, x - 4, x + 3, ...]
This is splicing together the two sequences ]x, -1, -∞] (left-closed, right-open) and [x, ∞] (both open).
For x = 80, this looks like:
[79, 80, 78, 81, 77, 82, 76, 83, 75, 84, 74, 85, ...]
Then, we use f↦s to select all elements from s satisfying f. In this case, we filter out all composite numbers, leaving only the prime ones. For the same x, this becomes:
[79, 83, 73, 71, 89, 67, 97, 61, 59, 101, 103, 53, ...]
Then, we use (...)@0 to select the first member of this sequence. Since the lower of the two needs to be selected, the sequence which starts with x - 1 is spliced in first.
Note: Only one of x and x - 1 can be prime, so it is okay that the spliced sequence starts with x - 1. Though the sequence could be open on both sides ([x,-1,-∞]), this would needlessly include x twice in the sequence. So, for sake of "efficiency", I chose the left-closed version (also because I like to show off Tidy).
answered Mar 27 at 21:39
Conor O'BrienConor O'Brien
30.6k264162
$endgroup$
add a comment |
3
3
3
$begingroup$
Tidy, 43 bytes
{x:(prime↦splice(]x,-1,-∞],[x,∞]))@0}
Try it online!
Explanation
This is a lambda with parameter x. This works by creating the following sequence:
[x - 1, x, x - 2, x + 1, x - 3, x + 2, x - 4, x + 3, ...]
This is splicing together the two sequences ]x, -1, -∞] (left-closed, right-open) and [x, ∞] (both open).
For x = 80, this looks like:
[79, 80, 78, 81, 77, 82, 76, 83, 75, 84, 74, 85, ...]
Then, we use f↦s to select all elements from s satisfying f. In this case, we filter out all composite numbers, leaving only the prime ones. For the same x, this becomes:
[79, 83, 73, 71, 89, 67, 97, 61, 59, 101, 103, 53, ...]
Then, we use (...)@0 to select the first member of this sequence. Since the lower of the two needs to be selected, the sequence which starts with x - 1 is spliced in first.
Note: Only one of x and x - 1 can be prime, so it is okay that the spliced sequence starts with x - 1. Though the sequence could be open on both sides ([x,-1,-∞]), this would needlessly include x twice in the sequence. So, for sake of "efficiency", I chose the left-closed version (also because I like to show off Tidy).
answered Mar 27 at 21:39
Conor O'BrienConor O'Brien
30.6k264162
$endgroup$
Tidy, 43 bytes
{x:(prime↦splice(]x,-1,-∞],[x,∞]))@0}
Try it online!
Explanation
This is a lambda with parameter x. This works by creating the following sequence:
[x - 1, x, x - 2, x + 1, x - 3, x + 2, x - 4, x + 3, ...]
This is splicing together the two sequences ]x, -1, -∞] (left-closed, right-open) and [x, ∞] (both open).
For x = 80, this looks like:
[79, 80, 78, 81, 77, 82, 76, 83, 75, 84, 74, 85, ...]
Then, we use f↦s to select all elements from s satisfying f. In this case, we filter out all composite numbers, leaving only the prime ones. For the same x, this becomes:
[79, 83, 73, 71, 89, 67, 97, 61, 59, 101, 103, 53, ...]
Then, we use (...)@0 to select the first member of this sequence. Since the lower of the two needs to be selected, the sequence which starts with x - 1 is spliced in first.
Note: Only one of x and x - 1 can be prime, so it is okay that the spliced sequence starts with x - 1. Though the sequence could be open on both sides ([x,-1,-∞]), this would needlessly include x twice in the sequence. So, for sake of "efficiency", I chose the left-closed version (also because I like to show off Tidy).
answered Mar 27 at 21:39
Conor O'BrienConor O'Brien
30.6k264162
answered Mar 27 at 21:39
Conor O'BrienConor O'Brien
30.6k264162
answered Mar 27 at 21:39
Conor O'BrienConor O'Brien
30.6k264162
answered Mar 27 at 21:39
Conor O'BrienConor O'Brien
30.6k264162
30.6k264162
add a comment |
add a comment |
3
$begingroup$
Python 2, 71 bytes
f=lambda n,k=1,p=1:k<n*3and min(k+n-p%k*2*n,f(n,k+1,p*k*k)-n,key=abs)+n
Try it online!
A recursive function that uses the Wilson's Theorem prime generator. The product p tracks $(k-1)!^2$, and p%k is 1 for primes and 0 for non-primes. To make it easy to compare abs(k-n) for different primes k, we store k-n and compare via abs, adding back n to get the result k.
The expression k+n-p%k*2*n is designed to give k-n on primes (where p%k=1), and otherwise a "bad" value of k+n that's always bigger in absolute value and so doesn't affect the minimum, so that non-primes are passed over.
answered Mar 28 at 0:04
xnorxnor
93.6k18190450
$endgroup$
add a comment |
3
$begingroup$
Python 2, 71 bytes
f=lambda n,k=1,p=1:k<n*3and min(k+n-p%k*2*n,f(n,k+1,p*k*k)-n,key=abs)+n
Try it online!
A recursive function that uses the Wilson's Theorem prime generator. The product p tracks $(k-1)!^2$, and p%k is 1 for primes and 0 for non-primes. To make it easy to compare abs(k-n) for different primes k, we store k-n and compare via abs, adding back n to get the result k.
The expression k+n-p%k*2*n is designed to give k-n on primes (where p%k=1), and otherwise a "bad" value of k+n that's always bigger in absolute value and so doesn't affect the minimum, so that non-primes are passed over.
answered Mar 28 at 0:04
xnorxnor
93.6k18190450
$endgroup$
add a comment |
3
3
3
$begingroup$
Python 2, 71 bytes
f=lambda n,k=1,p=1:k<n*3and min(k+n-p%k*2*n,f(n,k+1,p*k*k)-n,key=abs)+n
Try it online!
A recursive function that uses the Wilson's Theorem prime generator. The product p tracks $(k-1)!^2$, and p%k is 1 for primes and 0 for non-primes. To make it easy to compare abs(k-n) for different primes k, we store k-n and compare via abs, adding back n to get the result k.
The expression k+n-p%k*2*n is designed to give k-n on primes (where p%k=1), and otherwise a "bad" value of k+n that's always bigger in absolute value and so doesn't affect the minimum, so that non-primes are passed over.
answered Mar 28 at 0:04
xnorxnor
93.6k18190450
$endgroup$
Python 2, 71 bytes
f=lambda n,k=1,p=1:k<n*3and min(k+n-p%k*2*n,f(n,k+1,p*k*k)-n,key=abs)+n
Try it online!
A recursive function that uses the Wilson's Theorem prime generator. The product p tracks $(k-1)!^2$, and p%k is 1 for primes and 0 for non-primes. To make it easy to compare abs(k-n) for different primes k, we store k-n and compare via abs, adding back n to get the result k.
The expression k+n-p%k*2*n is designed to give k-n on primes (where p%k=1), and otherwise a "bad" value of k+n that's always bigger in absolute value and so doesn't affect the minimum, so that non-primes are passed over.
answered Mar 28 at 0:04
xnorxnor
93.6k18190450
answered Mar 28 at 0:04
xnorxnor
93.6k18190450
answered Mar 28 at 0:04
xnorxnor
93.6k18190450
answered Mar 28 at 0:04
xnorxnor
93.6k18190450
93.6k18190450
add a comment |
add a comment |
3
$begingroup$
APL (Dyalog Extended), 20 15 bytesSBCS
Tacit prefix function inspired by Galen Ivanov's J answer.
⊢(⊃⍋⍤|⍤-⊇⊢)¯2⍭⍳
Try it online!
⍳ ɩndices one through the argument.
¯2⍭ nth primes of that
⊢(…) apply the following tacit function to that, with the original argument as left argument:
⊢ the primes
⊇ indexed by:
⍋ the ascending grade (indices which would sort ascending)
⍤ of
| the magnitude (absolute value)
⍤ of
- the differences
⊃ pick the first one (i.e. the one with smallest difference)
answered Mar 27 at 17:23
AdámAdám
28.9k276207
$endgroup$
add a comment |
3
$begingroup$
APL (Dyalog Extended), 20 15 bytesSBCS
Tacit prefix function inspired by Galen Ivanov's J answer.
⊢(⊃⍋⍤|⍤-⊇⊢)¯2⍭⍳
Try it online!
⍳ ɩndices one through the argument.
¯2⍭ nth primes of that
⊢(…) apply the following tacit function to that, with the original argument as left argument:
⊢ the primes
⊇ indexed by:
⍋ the ascending grade (indices which would sort ascending)
⍤ of
| the magnitude (absolute value)
⍤ of
- the differences
⊃ pick the first one (i.e. the one with smallest difference)
answered Mar 27 at 17:23
AdámAdám
28.9k276207
$endgroup$
add a comment |
3
3
3
$begingroup$
APL (Dyalog Extended), 20 15 bytesSBCS
Tacit prefix function inspired by Galen Ivanov's J answer.
⊢(⊃⍋⍤|⍤-⊇⊢)¯2⍭⍳
Try it online!
⍳ ɩndices one through the argument.
¯2⍭ nth primes of that
⊢(…) apply the following tacit function to that, with the original argument as left argument:
⊢ the primes
⊇ indexed by:
⍋ the ascending grade (indices which would sort ascending)
⍤ of
| the magnitude (absolute value)
⍤ of
- the differences
⊃ pick the first one (i.e. the one with smallest difference)
answered Mar 27 at 17:23
AdámAdám
28.9k276207
$endgroup$
APL (Dyalog Extended), 20 15 bytesSBCS
Tacit prefix function inspired by Galen Ivanov's J answer.
⊢(⊃⍋⍤|⍤-⊇⊢)¯2⍭⍳
Try it online!
⍳ ɩndices one through the argument.
¯2⍭ nth primes of that
⊢(…) apply the following tacit function to that, with the original argument as left argument:
⊢ the primes
⊇ indexed by:
⍋ the ascending grade (indices which would sort ascending)
⍤ of
| the magnitude (absolute value)
⍤ of
- the differences
⊃ pick the first one (i.e. the one with smallest difference)
answered Mar 27 at 17:23
AdámAdám
28.9k276207
edited Mar 28 at 23:40
answered Mar 27 at 17:23
AdámAdám
28.9k276207
answered Mar 27 at 17:23
AdámAdám
28.9k276207
answered Mar 27 at 17:23
AdámAdám
28.9k276207
28.9k276207
add a comment |
add a comment |
3
$begingroup$
Perl 6, 35 bytes
{$_+=($*=-1)*$++until .is-prime;$_}
Try it online!
This uses Veitcel's technique for generating the list of 0, -1, 2, -3 but simplifies it greatly to ($*=-1)*$++ using the anonymous state variables available in P6 (I originally had -1 ** $++ * $++, but when golfed the negative loses precedence). There's a built in prime checker but unfortunately the until prevents the automagically returned value so there's an extra $_ hanging around.
answered Mar 29 at 0:15
guifaguifa
28115
$endgroup$
$begingroup$
I'd usually use a sequence operator approach to something like this, but comes out to one byte longer, so nice work finding a shorter method
$endgroup$
– Jo King
Mar 29 at 0:39
$begingroup$
@JoKing good catch. The things that happen when I golf too quickly after getting a working solution. I had a similar one but the damned lack of [-1] haha
$endgroup$
– guifa
Mar 29 at 0:49
add a comment |
3
$begingroup$
Perl 6, 35 bytes
{$_+=($*=-1)*$++until .is-prime;$_}
Try it online!
This uses Veitcel's technique for generating the list of 0, -1, 2, -3 but simplifies it greatly to ($*=-1)*$++ using the anonymous state variables available in P6 (I originally had -1 ** $++ * $++, but when golfed the negative loses precedence). There's a built in prime checker but unfortunately the until prevents the automagically returned value so there's an extra $_ hanging around.
answered Mar 29 at 0:15
guifaguifa
28115
$endgroup$
$begingroup$
I'd usually use a sequence operator approach to something like this, but comes out to one byte longer, so nice work finding a shorter method
$endgroup$
– Jo King
Mar 29 at 0:39
$begingroup$
@JoKing good catch. The things that happen when I golf too quickly after getting a working solution. I had a similar one but the damned lack of [-1] haha
$endgroup$
– guifa
Mar 29 at 0:49
add a comment |
3
3
3
$begingroup$
Perl 6, 35 bytes
{$_+=($*=-1)*$++until .is-prime;$_}
Try it online!
This uses Veitcel's technique for generating the list of 0, -1, 2, -3 but simplifies it greatly to ($*=-1)*$++ using the anonymous state variables available in P6 (I originally had -1 ** $++ * $++, but when golfed the negative loses precedence). There's a built in prime checker but unfortunately the until prevents the automagically returned value so there's an extra $_ hanging around.
answered Mar 29 at 0:15
guifaguifa
28115
$endgroup$
Perl 6, 35 bytes
{$_+=($*=-1)*$++until .is-prime;$_}
Try it online!
This uses Veitcel's technique for generating the list of 0, -1, 2, -3 but simplifies it greatly to ($*=-1)*$++ using the anonymous state variables available in P6 (I originally had -1 ** $++ * $++, but when golfed the negative loses precedence). There's a built in prime checker but unfortunately the until prevents the automagically returned value so there's an extra $_ hanging around.
answered Mar 29 at 0:15
guifaguifa
28115
edited Mar 29 at 0:51
answered Mar 29 at 0:15
guifaguifa
28115
answered Mar 29 at 0:15
guifaguifa
28115
answered Mar 29 at 0:15
guifaguifa
28115
28115
$begingroup$
I'd usually use a sequence operator approach to something like this, but comes out to one byte longer, so nice work finding a shorter method
$endgroup$
– Jo King
Mar 29 at 0:39
$begingroup$
@JoKing good catch. The things that happen when I golf too quickly after getting a working solution. I had a similar one but the damned lack of [-1] haha
$endgroup$
– guifa
Mar 29 at 0:49
add a comment |
$begingroup$
I'd usually use a sequence operator approach to something like this, but comes out to one byte longer, so nice work finding a shorter method
$endgroup$
– Jo King
Mar 29 at 0:39
$begingroup$
@JoKing good catch. The things that happen when I golf too quickly after getting a working solution. I had a similar one but the damned lack of [-1] haha
$endgroup$
– guifa
Mar 29 at 0:49
$begingroup$
I'd usually use a sequence operator approach to something like this, but comes out to one byte longer, so nice work finding a shorter method
$endgroup$
– Jo King
Mar 29 at 0:39
$begingroup$
I'd usually use a sequence operator approach to something like this, but comes out to one byte longer, so nice work finding a shorter method
$endgroup$
– Jo King
Mar 29 at 0:39
$begingroup$
@JoKing good catch. The things that happen when I golf too quickly after getting a working solution. I had a similar one but the damned lack of [-1] haha
$endgroup$
– guifa
Mar 29 at 0:49
$begingroup$
@JoKing good catch. The things that happen when I golf too quickly after getting a working solution. I had a similar one but the damned lack of [-1] haha
$endgroup$
– guifa
Mar 29 at 0:49
add a comment |
3
$begingroup$
C, 122 121 104 bytes
p(a,i){for(i=1;++i<a;)if(a%i<1)return 0;return a>1;}c(a,b){for(b=a;;b++)if(p(--a)|p(b))return p(b)?b:a;}
Use it calling function c() and passing as argument the number; it should return the closest prime.
Thanks to Embodiment of Ignorance for 1 byte saved a big improvement.
Try it online!
answered Mar 27 at 20:24
Lince AssassinoLince Assassino
1114
$endgroup$
$begingroup$
Butc()receives two parameters... Also, you can probably shorten thewhile(1)tofor(;;)(untested, since I don't get how to run your code
$endgroup$
– Embodiment of Ignorance
Mar 28 at 5:03
$begingroup$
@EmbodimentofIgnorance I wrote it and tested it all on an online c compiler, I could callc()passing only the first parameter. And you are right,for(;;)saves me a byte, only 117 left to get first place :)
$endgroup$
– Lince Assassino
Mar 28 at 11:30
$begingroup$
110 bytes:#define r return p(a,i){i=1;while(++i<a)if(a%i<1)r 0;r a>1;}c(a,b){b=a;for(;;b++){if(p(--a))r a;if(p(b))r b;}}. Here is a TIO link: tio.run/…
$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:06
$begingroup$
106: tio.run/…
$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:12
$begingroup$
101: tio.run/…
$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:17
|
show 2 more comments
3
$begingroup$
C, 122 121 104 bytes
p(a,i){for(i=1;++i<a;)if(a%i<1)return 0;return a>1;}c(a,b){for(b=a;;b++)if(p(--a)|p(b))return p(b)?b:a;}
Use it calling function c() and passing as argument the number; it should return the closest prime.
Thanks to Embodiment of Ignorance for 1 byte saved a big improvement.
Try it online!
answered Mar 27 at 20:24
Lince AssassinoLince Assassino
1114
$endgroup$
$begingroup$
Butc()receives two parameters... Also, you can probably shorten thewhile(1)tofor(;;)(untested, since I don't get how to run your code
$endgroup$
– Embodiment of Ignorance
Mar 28 at 5:03
$begingroup$
@EmbodimentofIgnorance I wrote it and tested it all on an online c compiler, I could callc()passing only the first parameter. And you are right,for(;;)saves me a byte, only 117 left to get first place :)
$endgroup$
– Lince Assassino
Mar 28 at 11:30
$begingroup$
110 bytes:#define r return p(a,i){i=1;while(++i<a)if(a%i<1)r 0;r a>1;}c(a,b){b=a;for(;;b++){if(p(--a))r a;if(p(b))r b;}}. Here is a TIO link: tio.run/…
$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:06
$begingroup$
106: tio.run/…
$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:12
$begingroup$
101: tio.run/…
$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:17
|
show 2 more comments
3
3
3
$begingroup$
C, 122 121 104 bytes
p(a,i){for(i=1;++i<a;)if(a%i<1)return 0;return a>1;}c(a,b){for(b=a;;b++)if(p(--a)|p(b))return p(b)?b:a;}
Use it calling function c() and passing as argument the number; it should return the closest prime.
Thanks to Embodiment of Ignorance for 1 byte saved a big improvement.
Try it online!
answered Mar 27 at 20:24
Lince AssassinoLince Assassino
1114
$endgroup$
C, 122 121 104 bytes
p(a,i){for(i=1;++i<a;)if(a%i<1)return 0;return a>1;}c(a,b){for(b=a;;b++)if(p(--a)|p(b))return p(b)?b:a;}
Use it calling function c() and passing as argument the number; it should return the closest prime.
Thanks to Embodiment of Ignorance for 1 byte saved a big improvement.
Try it online!
answered Mar 27 at 20:24
Lince AssassinoLince Assassino
1114
edited Mar 29 at 17:22
answered Mar 27 at 20:24
Lince AssassinoLince Assassino
1114
answered Mar 27 at 20:24
Lince AssassinoLince Assassino
1114
answered Mar 27 at 20:24
Lince AssassinoLince Assassino
1114
1114
$begingroup$
Butc()receives two parameters... Also, you can probably shorten thewhile(1)tofor(;;)(untested, since I don't get how to run your code
$endgroup$
– Embodiment of Ignorance
Mar 28 at 5:03
$begingroup$
@EmbodimentofIgnorance I wrote it and tested it all on an online c compiler, I could callc()passing only the first parameter. And you are right,for(;;)saves me a byte, only 117 left to get first place :)
$endgroup$
– Lince Assassino
Mar 28 at 11:30
$begingroup$
110 bytes:#define r return p(a,i){i=1;while(++i<a)if(a%i<1)r 0;r a>1;}c(a,b){b=a;for(;;b++){if(p(--a))r a;if(p(b))r b;}}. Here is a TIO link: tio.run/…
$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:06
$begingroup$
106: tio.run/…
$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:12
$begingroup$
101: tio.run/…
$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:17
|
show 2 more comments
$begingroup$
Butc()receives two parameters... Also, you can probably shorten thewhile(1)tofor(;;)(untested, since I don't get how to run your code
$endgroup$
– Embodiment of Ignorance
Mar 28 at 5:03
$begingroup$
@EmbodimentofIgnorance I wrote it and tested it all on an online c compiler, I could callc()passing only the first parameter. And you are right,for(;;)saves me a byte, only 117 left to get first place :)
$endgroup$
– Lince Assassino
Mar 28 at 11:30
$begingroup$
110 bytes:#define r return p(a,i){i=1;while(++i<a)if(a%i<1)r 0;r a>1;}c(a,b){b=a;for(;;b++){if(p(--a))r a;if(p(b))r b;}}. Here is a TIO link: tio.run/…
$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:06
$begingroup$
106: tio.run/…
$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:12
$begingroup$
101: tio.run/…
$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:17
$begingroup$
But
c() receives two parameters... Also, you can probably shorten the while(1) to for(;;) (untested, since I don't get how to run your code$endgroup$
– Embodiment of Ignorance
Mar 28 at 5:03
$begingroup$
But
c() receives two parameters... Also, you can probably shorten the while(1) to for(;;) (untested, since I don't get how to run your code$endgroup$
– Embodiment of Ignorance
Mar 28 at 5:03
$begingroup$
@EmbodimentofIgnorance I wrote it and tested it all on an online c compiler, I could call
c() passing only the first parameter. And you are right, for(;;) saves me a byte, only 117 left to get first place :)$endgroup$
– Lince Assassino
Mar 28 at 11:30
$begingroup$
@EmbodimentofIgnorance I wrote it and tested it all on an online c compiler, I could call
c() passing only the first parameter. And you are right, for(;;) saves me a byte, only 117 left to get first place :)$endgroup$
– Lince Assassino
Mar 28 at 11:30
$begingroup$
110 bytes:
#define r return p(a,i){i=1;while(++i<a)if(a%i<1)r 0;r a>1;}c(a,b){b=a;for(;;b++){if(p(--a))r a;if(p(b))r b;}}. Here is a TIO link: tio.run/…$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:06
$begingroup$
110 bytes:
#define r return p(a,i){i=1;while(++i<a)if(a%i<1)r 0;r a>1;}c(a,b){b=a;for(;;b++){if(p(--a))r a;if(p(b))r b;}}. Here is a TIO link: tio.run/…$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:06
$begingroup$
106: tio.run/…
$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:12
$begingroup$
106: tio.run/…
$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:12
$begingroup$
101: tio.run/…
$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:17
$begingroup$
101: tio.run/…
$endgroup$
– Embodiment of Ignorance
Mar 28 at 21:17
|
show 2 more comments
3
$begingroup$
Wolfram Language (Mathematica), 52 bytes
If[PrimeQ[s=#],s,#&@@Nearest[s~NextPrime~{-1,1},s]]&
Try it online!
answered Mar 27 at 15:32
J42161217J42161217
13.9k21353
$endgroup$
$begingroup$
You have an extra space that can be removed to save a byte.
$endgroup$
– Ben
Mar 29 at 16:03
$begingroup$
@Ben you are right. thanx
$endgroup$
– J42161217
Mar 29 at 23:08
add a comment |
3
$begingroup$
Wolfram Language (Mathematica), 52 bytes
If[PrimeQ[s=#],s,#&@@Nearest[s~NextPrime~{-1,1},s]]&
Try it online!
answered Mar 27 at 15:32
J42161217J42161217
13.9k21353
$endgroup$
$begingroup$
You have an extra space that can be removed to save a byte.
$endgroup$
– Ben
Mar 29 at 16:03
$begingroup$
@Ben you are right. thanx
$endgroup$
– J42161217
Mar 29 at 23:08
add a comment |
3
3
3
$begingroup$
Wolfram Language (Mathematica), 52 bytes
If[PrimeQ[s=#],s,#&@@Nearest[s~NextPrime~{-1,1},s]]&
Try it online!
answered Mar 27 at 15:32
J42161217J42161217
13.9k21353
$endgroup$
Wolfram Language (Mathematica), 52 bytes
If[PrimeQ[s=#],s,#&@@Nearest[s~NextPrime~{-1,1},s]]&
Try it online!
answered Mar 27 at 15:32
J42161217J42161217
13.9k21353
edited Mar 29 at 23:09
answered Mar 27 at 15:32
J42161217J42161217
13.9k21353
answered Mar 27 at 15:32
J42161217J42161217
13.9k21353
answered Mar 27 at 15:32
J42161217J42161217
13.9k21353
13.9k21353
$begingroup$
You have an extra space that can be removed to save a byte.
$endgroup$
– Ben
Mar 29 at 16:03
$begingroup$
@Ben you are right. thanx
$endgroup$
– J42161217
Mar 29 at 23:08
add a comment |
$begingroup$
You have an extra space that can be removed to save a byte.
$endgroup$
– Ben
Mar 29 at 16:03
$begingroup$
@Ben you are right. thanx
$endgroup$
– J42161217
Mar 29 at 23:08
$begingroup$
You have an extra space that can be removed to save a byte.
$endgroup$
– Ben
Mar 29 at 16:03
$begingroup$
You have an extra space that can be removed to save a byte.
$endgroup$
– Ben
Mar 29 at 16:03
$begingroup$
@Ben you are right. thanx
$endgroup$
– J42161217
Mar 29 at 23:08
$begingroup$
@Ben you are right. thanx
$endgroup$
– J42161217
Mar 29 at 23:08
add a comment |
2
$begingroup$
APL(NARS), 38 chars, 76 bytes
{⍵≤1:2⋄0π⍵:⍵⋄d←1π⍵⋄(d-⍵)≥⍵-s←¯1π⍵:s⋄d}
0π is the test for prime, ¯1π the prev prime, 1π is the next prime; test:
f←{⍵≤1:2⋄0π⍵:⍵⋄d←1π⍵⋄(d-⍵)≥⍵-s←¯1π⍵:s⋄d}
f¨80 100 5 9 532 1
79 101 5 7 523 2
answered Mar 27 at 18:05
RosLuPRosLuP
2,296514
$endgroup$
add a comment |
2
$begingroup$
APL(NARS), 38 chars, 76 bytes
{⍵≤1:2⋄0π⍵:⍵⋄d←1π⍵⋄(d-⍵)≥⍵-s←¯1π⍵:s⋄d}
0π is the test for prime, ¯1π the prev prime, 1π is the next prime; test:
f←{⍵≤1:2⋄0π⍵:⍵⋄d←1π⍵⋄(d-⍵)≥⍵-s←¯1π⍵:s⋄d}
f¨80 100 5 9 532 1
79 101 5 7 523 2
answered Mar 27 at 18:05
RosLuPRosLuP
2,296514
$endgroup$
add a comment |
2
2
2
$begingroup$
APL(NARS), 38 chars, 76 bytes
{⍵≤1:2⋄0π⍵:⍵⋄d←1π⍵⋄(d-⍵)≥⍵-s←¯1π⍵:s⋄d}
0π is the test for prime, ¯1π the prev prime, 1π is the next prime; test:
f←{⍵≤1:2⋄0π⍵:⍵⋄d←1π⍵⋄(d-⍵)≥⍵-s←¯1π⍵:s⋄d}
f¨80 100 5 9 532 1
79 101 5 7 523 2
answered Mar 27 at 18:05
RosLuPRosLuP
2,296514
$endgroup$
APL(NARS), 38 chars, 76 bytes
{⍵≤1:2⋄0π⍵:⍵⋄d←1π⍵⋄(d-⍵)≥⍵-s←¯1π⍵:s⋄d}
0π is the test for prime, ¯1π the prev prime, 1π is the next prime; test:
f←{⍵≤1:2⋄0π⍵:⍵⋄d←1π⍵⋄(d-⍵)≥⍵-s←¯1π⍵:s⋄d}
f¨80 100 5 9 532 1
79 101 5 7 523 2
answered Mar 27 at 18:05
RosLuPRosLuP
2,296514
answered Mar 27 at 18:05
RosLuPRosLuP
2,296514
answered Mar 27 at 18:05
RosLuPRosLuP
2,296514
answered Mar 27 at 18:05
RosLuPRosLuP
2,296514
2,296514
add a comment |
add a comment |
2
$begingroup$
J, 19 15 bytes
(0{]/:|@-)p:@i.
Try it online!
answered Mar 27 at 20:28
Galen IvanovGalen Ivanov
7,41211034
$endgroup$
add a comment |
2
$begingroup$
J, 19 15 bytes
(0{]/:|@-)p:@i.
Try it online!
answered Mar 27 at 20:28
Galen IvanovGalen Ivanov
7,41211034
$endgroup$
add a comment |
2
2
2
$begingroup$
J, 19 15 bytes
(0{]/:|@-)p:@i.
Try it online!
answered Mar 27 at 20:28
Galen IvanovGalen Ivanov
7,41211034
$endgroup$
J, 19 15 bytes
(0{]/:|@-)p:@i.
Try it online!
answered Mar 27 at 20:28
Galen IvanovGalen Ivanov
7,41211034
edited Mar 27 at 22:52
answered Mar 27 at 20:28
Galen IvanovGalen Ivanov
7,41211034
answered Mar 27 at 20:28
Galen IvanovGalen Ivanov
7,41211034
answered Mar 27 at 20:28
Galen IvanovGalen Ivanov
7,41211034
7,41211034
add a comment |
add a comment |
2
$begingroup$
Perl 5, 59 bytes
$a=0;while((1x$_)=~/^.?$|^(..+?)1+$/){$_+=(-1)**$a*($a++)}
Try it online!
/^.?$|^(..+?)1+$/ is tricky regexp to check prime
(-1)**$a*($a++) generate sequence 0,-1, 2,-3 ...
answered Mar 28 at 8:35
VeitcelVeitcel
613
$endgroup$
add a comment |
2
$begingroup$
Perl 5, 59 bytes
$a=0;while((1x$_)=~/^.?$|^(..+?)1+$/){$_+=(-1)**$a*($a++)}
Try it online!
/^.?$|^(..+?)1+$/ is tricky regexp to check prime
(-1)**$a*($a++) generate sequence 0,-1, 2,-3 ...
answered Mar 28 at 8:35
VeitcelVeitcel
613
$endgroup$
add a comment |
2
2
2
$begingroup$
Perl 5, 59 bytes
$a=0;while((1x$_)=~/^.?$|^(..+?)1+$/){$_+=(-1)**$a*($a++)}
Try it online!
/^.?$|^(..+?)1+$/ is tricky regexp to check prime
(-1)**$a*($a++) generate sequence 0,-1, 2,-3 ...
answered Mar 28 at 8:35
VeitcelVeitcel
613
$endgroup$
Perl 5, 59 bytes
$a=0;while((1x$_)=~/^.?$|^(..+?)1+$/){$_+=(-1)**$a*($a++)}
Try it online!
/^.?$|^(..+?)1+$/ is tricky regexp to check prime
(-1)**$a*($a++) generate sequence 0,-1, 2,-3 ...
answered Mar 28 at 8:35
VeitcelVeitcel
613
answered Mar 28 at 8:35
VeitcelVeitcel
613
answered Mar 28 at 8:35
VeitcelVeitcel
613
answered Mar 28 at 8:35
VeitcelVeitcel
613
613
add a comment |
add a comment |
2
$begingroup$
MathGolf, 10 bytes
∞╒g¶áÅ-±├Þ
Try it online.
Explanation:
∞ # Double the (implicit) input-integer
╒ # Create a list in the range [1, 2*n]
g¶ # Filter so only the prime numbers remain
áÅ # Sort this list using the next two character:
-± # The absolute difference with the (implicit) input-integer
├ # Push the first item of the list
# (unfortunately without popping the list itself, so:)
Þ # Discard everything from the stack except for the top
# (which is output implicitly as result)
answered Mar 28 at 8:43
Kevin CruijssenKevin Cruijssen
42.4k570217
$endgroup$
$begingroup$
@JoKing Thanks! I knew Max thought about changing it, but didn't knew he actually did. The docs still state the old one.
$endgroup$
– Kevin Cruijssen
Mar 28 at 8:53
$begingroup$
Ah, I use the mathgolf.txt file as reference, since it seems to be more up to date
$endgroup$
– Jo King
Mar 28 at 23:21
$begingroup$
@JoKing Yeah, he told me yesterday about that file as well. Will use it from now on. :)
$endgroup$
– Kevin Cruijssen
Mar 29 at 7:23
add a comment |
2
$begingroup$
MathGolf, 10 bytes
∞╒g¶áÅ-±├Þ
Try it online.
Explanation:
∞ # Double the (implicit) input-integer
╒ # Create a list in the range [1, 2*n]
g¶ # Filter so only the prime numbers remain
áÅ # Sort this list using the next two character:
-± # The absolute difference with the (implicit) input-integer
├ # Push the first item of the list
# (unfortunately without popping the list itself, so:)
Þ # Discard everything from the stack except for the top
# (which is output implicitly as result)
answered Mar 28 at 8:43
Kevin CruijssenKevin Cruijssen
42.4k570217
$endgroup$
$begingroup$
@JoKing Thanks! I knew Max thought about changing it, but didn't knew he actually did. The docs still state the old one.
$endgroup$
– Kevin Cruijssen
Mar 28 at 8:53
$begingroup$
Ah, I use the mathgolf.txt file as reference, since it seems to be more up to date
$endgroup$
– Jo King
Mar 28 at 23:21
$begingroup$
@JoKing Yeah, he told me yesterday about that file as well. Will use it from now on. :)
$endgroup$
– Kevin Cruijssen
Mar 29 at 7:23
add a comment |
2
2
2
$begingroup$
MathGolf, 10 bytes
∞╒g¶áÅ-±├Þ
Try it online.
Explanation:
∞ # Double the (implicit) input-integer
╒ # Create a list in the range [1, 2*n]
g¶ # Filter so only the prime numbers remain
áÅ # Sort this list using the next two character:
-± # The absolute difference with the (implicit) input-integer
├ # Push the first item of the list
# (unfortunately without popping the list itself, so:)
Þ # Discard everything from the stack except for the top
# (which is output implicitly as result)
answered Mar 28 at 8:43
Kevin CruijssenKevin Cruijssen
42.4k570217
$endgroup$
MathGolf, 10 bytes
∞╒g¶áÅ-±├Þ
Try it online.
Explanation:
∞ # Double the (implicit) input-integer
╒ # Create a list in the range [1, 2*n]
g¶ # Filter so only the prime numbers remain
áÅ # Sort this list using the next two character:
-± # The absolute difference with the (implicit) input-integer
├ # Push the first item of the list
# (unfortunately without popping the list itself, so:)
Þ # Discard everything from the stack except for the top
# (which is output implicitly as result)
answered Mar 28 at 8:43
Kevin CruijssenKevin Cruijssen
42.4k570217
edited Mar 28 at 8:52
answered Mar 28 at 8:43
Kevin CruijssenKevin Cruijssen
42.4k570217
answered Mar 28 at 8:43
Kevin CruijssenKevin Cruijssen
42.4k570217
answered Mar 28 at 8:43
Kevin CruijssenKevin Cruijssen
42.4k570217
42.4k570217
$begingroup$
@JoKing Thanks! I knew Max thought about changing it, but didn't knew he actually did. The docs still state the old one.
$endgroup$
– Kevin Cruijssen
Mar 28 at 8:53
$begingroup$
Ah, I use the mathgolf.txt file as reference, since it seems to be more up to date
$endgroup$
– Jo King
Mar 28 at 23:21
$begingroup$
@JoKing Yeah, he told me yesterday about that file as well. Will use it from now on. :)
$endgroup$
– Kevin Cruijssen
Mar 29 at 7:23
add a comment |
$begingroup$
@JoKing Thanks! I knew Max thought about changing it, but didn't knew he actually did. The docs still state the old one.
$endgroup$
– Kevin Cruijssen
Mar 28 at 8:53
$begingroup$
Ah, I use the mathgolf.txt file as reference, since it seems to be more up to date
$endgroup$
– Jo King
Mar 28 at 23:21
$begingroup$
@JoKing Yeah, he told me yesterday about that file as well. Will use it from now on. :)
$endgroup$
– Kevin Cruijssen
Mar 29 at 7:23
$begingroup$
@JoKing Thanks! I knew Max thought about changing it, but didn't knew he actually did. The docs still state the old one.
$endgroup$
– Kevin Cruijssen
Mar 28 at 8:53
$begingroup$
@JoKing Thanks! I knew Max thought about changing it, but didn't knew he actually did. The docs still state the old one.
$endgroup$
– Kevin Cruijssen
Mar 28 at 8:53
$begingroup$
Ah, I use the mathgolf.txt file as reference, since it seems to be more up to date
$endgroup$
– Jo King
Mar 28 at 23:21
$begingroup$
Ah, I use the mathgolf.txt file as reference, since it seems to be more up to date
$endgroup$
– Jo King
Mar 28 at 23:21
$begingroup$
@JoKing Yeah, he told me yesterday about that file as well. Will use it from now on. :)
$endgroup$
– Kevin Cruijssen
Mar 29 at 7:23
$begingroup$
@JoKing Yeah, he told me yesterday about that file as well. Will use it from now on. :)
$endgroup$
– Kevin Cruijssen
Mar 29 at 7:23
add a comment |
2
$begingroup$
Factor, 91 bytes
: p ( x -- x ) [ nprimes ] keep dupd [ - abs ] curry map swap zip natural-sort first last ;
Try it online!
answered Mar 28 at 9:41
Galen IvanovGalen Ivanov
7,41211034
$endgroup$
add a comment |
2
$begingroup$
Factor, 91 bytes
: p ( x -- x ) [ nprimes ] keep dupd [ - abs ] curry map swap zip natural-sort first last ;
Try it online!
answered Mar 28 at 9:41
Galen IvanovGalen Ivanov
7,41211034
$endgroup$
add a comment |
2
2
2
$begingroup$
Factor, 91 bytes
: p ( x -- x ) [ nprimes ] keep dupd [ - abs ] curry map swap zip natural-sort first last ;
Try it online!
answered Mar 28 at 9:41
Galen IvanovGalen Ivanov
7,41211034
$endgroup$
Factor, 91 bytes
: p ( x -- x ) [ nprimes ] keep dupd [ - abs ] curry map swap zip natural-sort first last ;
Try it online!
answered Mar 28 at 9:41
Galen IvanovGalen Ivanov
7,41211034
answered Mar 28 at 9:41
Galen IvanovGalen Ivanov
7,41211034
answered Mar 28 at 9:41
Galen IvanovGalen Ivanov
7,41211034
answered Mar 28 at 9:41
Galen IvanovGalen Ivanov
7,41211034
7,41211034
add a comment |
add a comment |
2
$begingroup$
Python 2 (Cython), 96 bytes
l=lambda p:min(filter(lambda p:all(p%n for n in range(2,p)),range(2,p*3)),key=lambda x:abs(x-p))
Try it online!
answered Mar 27 at 16:29
Snaddyvitch DispenserSnaddyvitch Dispenser
1015
$endgroup$
$begingroup$
Might be able to save a couple bytes viar=range;...
$endgroup$
– Skyler
Mar 28 at 20:33
1
$begingroup$
@Arnauld, it now works for x=1
$endgroup$
– Snaddyvitch Dispenser
Mar 29 at 8:20
add a comment |
2
$begingroup$
Python 2 (Cython), 96 bytes
l=lambda p:min(filter(lambda p:all(p%n for n in range(2,p)),range(2,p*3)),key=lambda x:abs(x-p))
Try it online!
answered Mar 27 at 16:29
Snaddyvitch DispenserSnaddyvitch Dispenser
1015
$endgroup$
$begingroup$
Might be able to save a couple bytes viar=range;...
$endgroup$
– Skyler
Mar 28 at 20:33
1
$begingroup$
@Arnauld, it now works for x=1
$endgroup$
– Snaddyvitch Dispenser
Mar 29 at 8:20
add a comment |
2
2
2
$begingroup$
Python 2 (Cython), 96 bytes
l=lambda p:min(filter(lambda p:all(p%n for n in range(2,p)),range(2,p*3)),key=lambda x:abs(x-p))
Try it online!
answered Mar 27 at 16:29
Snaddyvitch DispenserSnaddyvitch Dispenser
1015
$endgroup$
Python 2 (Cython), 96 bytes
l=lambda p:min(filter(lambda p:all(p%n for n in range(2,p)),range(2,p*3)),key=lambda x:abs(x-p))
Try it online!
answered Mar 27 at 16:29
Snaddyvitch DispenserSnaddyvitch Dispenser
1015
edited Mar 29 at 8:19
answered Mar 27 at 16:29
Snaddyvitch DispenserSnaddyvitch Dispenser
1015
answered Mar 27 at 16:29
Snaddyvitch DispenserSnaddyvitch Dispenser
1015
answered Mar 27 at 16:29
Snaddyvitch DispenserSnaddyvitch Dispenser
1015
1015
$begingroup$
Might be able to save a couple bytes viar=range;...
$endgroup$
– Skyler
Mar 28 at 20:33
1
$begingroup$
@Arnauld, it now works for x=1
$endgroup$
– Snaddyvitch Dispenser
Mar 29 at 8:20
add a comment |
$begingroup$
Might be able to save a couple bytes viar=range;...
$endgroup$
– Skyler
Mar 28 at 20:33
1
$begingroup$
@Arnauld, it now works for x=1
$endgroup$
– Snaddyvitch Dispenser
Mar 29 at 8:20
$begingroup$
Might be able to save a couple bytes via
r=range;...$endgroup$
– Skyler
Mar 28 at 20:33
$begingroup$
Might be able to save a couple bytes via
r=range;...$endgroup$
– Skyler
Mar 28 at 20:33
1
1
$begingroup$
@Arnauld, it now works for x=1
$endgroup$
– Snaddyvitch Dispenser
Mar 29 at 8:20
$begingroup$
@Arnauld, it now works for x=1
$endgroup$
– Snaddyvitch Dispenser
Mar 29 at 8:20
add a comment |
2
$begingroup$
C# (Visual C# Interactive Compiler), 104 100 bytes
n=>{int r=0,t=0,m=n;while(r!=2){n+=(n<m)?t:-t;t++;r=0;for(int i=1;i<=n;i++)if(n%i==0)r++;}return n;}
Try it online!
Explanation:
int f(int n)
{
int r = 0; //stores the amount of factors of "n"
int t = 0; //increment used to cover all the integers surrounding "n"
int m = n; //placeholder to toggle between adding or substracting "t" to "n"
while (r != 2) //while the amount of factors found for "n" is different to 2 ("1" + itself)
{
n += (n < m) ? t : -t; //increment/decrement "n" by "t" (-0, -1, +2, -3, +4, -5,...)
t++;
r = 0;
for (int i = 1; i <= n; i++) //foreach number between "1" and "n" increment "r" if the remainder of its division with "n" is 0 (thus being a factor)
if (n % i == 0) r++;
}
return n;
}
Console.WriteLine(f(80)); //79
answered Mar 28 at 8:57
Innat3Innat3
1514
$endgroup$
add a comment |
2
$begingroup$
C# (Visual C# Interactive Compiler), 104 100 bytes
n=>{int r=0,t=0,m=n;while(r!=2){n+=(n<m)?t:-t;t++;r=0;for(int i=1;i<=n;i++)if(n%i==0)r++;}return n;}
Try it online!
Explanation:
int f(int n)
{
int r = 0; //stores the amount of factors of "n"
int t = 0; //increment used to cover all the integers surrounding "n"
int m = n; //placeholder to toggle between adding or substracting "t" to "n"
while (r != 2) //while the amount of factors found for "n" is different to 2 ("1" + itself)
{
n += (n < m) ? t : -t; //increment/decrement "n" by "t" (-0, -1, +2, -3, +4, -5,...)
t++;
r = 0;
for (int i = 1; i <= n; i++) //foreach number between "1" and "n" increment "r" if the remainder of its division with "n" is 0 (thus being a factor)
if (n % i == 0) r++;
}
return n;
}
Console.WriteLine(f(80)); //79
answered Mar 28 at 8:57
Innat3Innat3
1514
$endgroup$
add a comment |
2
2
2
$begingroup$
C# (Visual C# Interactive Compiler), 104 100 bytes
n=>{int r=0,t=0,m=n;while(r!=2){n+=(n<m)?t:-t;t++;r=0;for(int i=1;i<=n;i++)if(n%i==0)r++;}return n;}
Try it online!
Explanation:
int f(int n)
{
int r = 0; //stores the amount of factors of "n"
int t = 0; //increment used to cover all the integers surrounding "n"
int m = n; //placeholder to toggle between adding or substracting "t" to "n"
while (r != 2) //while the amount of factors found for "n" is different to 2 ("1" + itself)
{
n += (n < m) ? t : -t; //increment/decrement "n" by "t" (-0, -1, +2, -3, +4, -5,...)
t++;
r = 0;
for (int i = 1; i <= n; i++) //foreach number between "1" and "n" increment "r" if the remainder of its division with "n" is 0 (thus being a factor)
if (n % i == 0) r++;
}
return n;
}
Console.WriteLine(f(80)); //79
answered Mar 28 at 8:57
Innat3Innat3
1514
$endgroup$
C# (Visual C# Interactive Compiler), 104 100 bytes
n=>{int r=0,t=0,m=n;while(r!=2){n+=(n<m)?t:-t;t++;r=0;for(int i=1;i<=n;i++)if(n%i==0)r++;}return n;}
Try it online!
Explanation:
int f(int n)
{
int r = 0; //stores the amount of factors of "n"
int t = 0; //increment used to cover all the integers surrounding "n"
int m = n; //placeholder to toggle between adding or substracting "t" to "n"
while (r != 2) //while the amount of factors found for "n" is different to 2 ("1" + itself)
{
n += (n < m) ? t : -t; //increment/decrement "n" by "t" (-0, -1, +2, -3, +4, -5,...)
t++;
r = 0;
for (int i = 1; i <= n; i++) //foreach number between "1" and "n" increment "r" if the remainder of its division with "n" is 0 (thus being a factor)
if (n % i == 0) r++;
}
return n;
}
Console.WriteLine(f(80)); //79
answered Mar 28 at 8:57
Innat3Innat3
1514
edited Mar 29 at 8:50
answered Mar 28 at 8:57
Innat3Innat3
1514
answered Mar 28 at 8:57
Innat3Innat3
1514
answered Mar 28 at 8:57
Innat3Innat3
1514
1514
add a comment |
add a comment |
2
$begingroup$
Java 8, 88 87 bytes
n->{for(int c=0,s=0,d,N=n;c!=2;s++)for(c=d=1,n+=n<N?s:-s;d<n;)if(n%++d<1)c++;return n;}
Port of @NaturalNumberGuy's (first) C answer, so make sure to upvote him!!
-1 byte thanks to @OlivierGrégoire.
Try it online.
Explanation:
n->{ // Method with integer as both parameter and return-type
for(int c=0, // Counter-integer, starting at 0
s=0, // Step-integer, starting at 0 as well
d, // Divisor-integer, uninitialized
N=n; // Copy of the input-integer
c!=2; // Loop as long as the counter is not exactly 2 yet:
s++) // After every iteration: increase the step-integer by 1
for(c=d=1, // (Re)set both the counter and divisor to 1
n+=n<N? // If the input is smaller than the input-copy:
s // Increase the input by the step-integer
: // Else:
-s; // Decrease the input by the step-integer
d<n;) // Inner loop as long as the divisor is smaller than the input
if(n%++d // Increase the divisor by 1 first with `++d`
<1) // And if the input is evenly divisible by the divisor:
c++; // Increase the counter-integer by 1
return n;} // Return the now modified input-integer as result
answered Mar 29 at 12:30
Kevin CruijssenKevin Cruijssen
42.4k570217
$endgroup$
add a comment |
2
$begingroup$
Java 8, 88 87 bytes
n->{for(int c=0,s=0,d,N=n;c!=2;s++)for(c=d=1,n+=n<N?s:-s;d<n;)if(n%++d<1)c++;return n;}
Port of @NaturalNumberGuy's (first) C answer, so make sure to upvote him!!
-1 byte thanks to @OlivierGrégoire.
Try it online.
Explanation:
n->{ // Method with integer as both parameter and return-type
for(int c=0, // Counter-integer, starting at 0
s=0, // Step-integer, starting at 0 as well
d, // Divisor-integer, uninitialized
N=n; // Copy of the input-integer
c!=2; // Loop as long as the counter is not exactly 2 yet:
s++) // After every iteration: increase the step-integer by 1
for(c=d=1, // (Re)set both the counter and divisor to 1
n+=n<N? // If the input is smaller than the input-copy:
s // Increase the input by the step-integer
: // Else:
-s; // Decrease the input by the step-integer
d<n;) // Inner loop as long as the divisor is smaller than the input
if(n%++d // Increase the divisor by 1 first with `++d`
<1) // And if the input is evenly divisible by the divisor:
c++; // Increase the counter-integer by 1
return n;} // Return the now modified input-integer as result
answered Mar 29 at 12:30
Kevin CruijssenKevin Cruijssen
42.4k570217
$endgroup$
add a comment |
2
2
2
$begingroup$
Java 8, 88 87 bytes
n->{for(int c=0,s=0,d,N=n;c!=2;s++)for(c=d=1,n+=n<N?s:-s;d<n;)if(n%++d<1)c++;return n;}
Port of @NaturalNumberGuy's (first) C answer, so make sure to upvote him!!
-1 byte thanks to @OlivierGrégoire.
Try it online.
Explanation:
n->{ // Method with integer as both parameter and return-type
for(int c=0, // Counter-integer, starting at 0
s=0, // Step-integer, starting at 0 as well
d, // Divisor-integer, uninitialized
N=n; // Copy of the input-integer
c!=2; // Loop as long as the counter is not exactly 2 yet:
s++) // After every iteration: increase the step-integer by 1
for(c=d=1, // (Re)set both the counter and divisor to 1
n+=n<N? // If the input is smaller than the input-copy:
s // Increase the input by the step-integer
: // Else:
-s; // Decrease the input by the step-integer
d<n;) // Inner loop as long as the divisor is smaller than the input
if(n%++d // Increase the divisor by 1 first with `++d`
<1) // And if the input is evenly divisible by the divisor:
c++; // Increase the counter-integer by 1
return n;} // Return the now modified input-integer as result
answered Mar 29 at 12:30
Kevin CruijssenKevin Cruijssen
42.4k570217
$endgroup$
Java 8, 88 87 bytes
n->{for(int c=0,s=0,d,N=n;c!=2;s++)for(c=d=1,n+=n<N?s:-s;d<n;)if(n%++d<1)c++;return n;}
Port of @NaturalNumberGuy's (first) C answer, so make sure to upvote him!!
-1 byte thanks to @OlivierGrégoire.
Try it online.
Explanation:
n->{ // Method with integer as both parameter and return-type
for(int c=0, // Counter-integer, starting at 0
s=0, // Step-integer, starting at 0 as well
d, // Divisor-integer, uninitialized
N=n; // Copy of the input-integer
c!=2; // Loop as long as the counter is not exactly 2 yet:
s++) // After every iteration: increase the step-integer by 1
for(c=d=1, // (Re)set both the counter and divisor to 1
n+=n<N? // If the input is smaller than the input-copy:
s // Increase the input by the step-integer
: // Else:
-s; // Decrease the input by the step-integer
d<n;) // Inner loop as long as the divisor is smaller than the input
if(n%++d // Increase the divisor by 1 first with `++d`
<1) // And if the input is evenly divisible by the divisor:
c++; // Increase the counter-integer by 1
return n;} // Return the now modified input-integer as result
answered Mar 29 at 12:30
Kevin CruijssenKevin Cruijssen
42.4k570217
edited Mar 29 at 13:06
answered Mar 29 at 12:30
Kevin CruijssenKevin Cruijssen
42.4k570217
answered Mar 29 at 12:30
Kevin CruijssenKevin Cruijssen
42.4k570217
answered Mar 29 at 12:30
Kevin CruijssenKevin Cruijssen
42.4k570217
42.4k570217
add a comment |
add a comment |
2
$begingroup$
Java (JDK), 103 bytes
n->{int p=0,x=0,z=n,d;for(;p<1;p=p>0?z:0,z=z==n+x?n-++x:z+1)for(p=z/2,d=1;++d<z;)p=z%d<1?0:p;return p;}
Try it online!
answered Mar 29 at 11:23
Olivier GrégoireOlivier Grégoire
9,39511944
$endgroup$
$begingroup$
Umm.. I had already create a port of his answer.. ;) Although yours is 1 byte shorter, so something is different. EDIT: Ah, I have a result-integer outside the loop, and you modify the input inside the loop, hence the -1 byte for;. :) Do you want me to delete my answer?.. Feel free to copy the explanation.
$endgroup$
– Kevin Cruijssen
Mar 29 at 12:50
$begingroup$
@KevinCruijssen Oops, rollbacked!
$endgroup$
– Olivier Grégoire
Mar 29 at 12:59
$begingroup$
Sorry about that (and thanks for the -1 byte). I like your version as well, though. Already upvoted before I saw NaturalNumberGuy's answer.
$endgroup$
– Kevin Cruijssen
Mar 29 at 12:59
add a comment |
2
$begingroup$
Java (JDK), 103 bytes
n->{int p=0,x=0,z=n,d;for(;p<1;p=p>0?z:0,z=z==n+x?n-++x:z+1)for(p=z/2,d=1;++d<z;)p=z%d<1?0:p;return p;}
Try it online!
answered Mar 29 at 11:23
Olivier GrégoireOlivier Grégoire
9,39511944
$endgroup$
$begingroup$
Umm.. I had already create a port of his answer.. ;) Although yours is 1 byte shorter, so something is different. EDIT: Ah, I have a result-integer outside the loop, and you modify the input inside the loop, hence the -1 byte for;. :) Do you want me to delete my answer?.. Feel free to copy the explanation.
$endgroup$
– Kevin Cruijssen
Mar 29 at 12:50
$begingroup$
@KevinCruijssen Oops, rollbacked!
$endgroup$
– Olivier Grégoire
Mar 29 at 12:59
$begingroup$
Sorry about that (and thanks for the -1 byte). I like your version as well, though. Already upvoted before I saw NaturalNumberGuy's answer.
$endgroup$
– Kevin Cruijssen
Mar 29 at 12:59
add a comment |
2
2
2
$begingroup$
Java (JDK), 103 bytes
n->{int p=0,x=0,z=n,d;for(;p<1;p=p>0?z:0,z=z==n+x?n-++x:z+1)for(p=z/2,d=1;++d<z;)p=z%d<1?0:p;return p;}
Try it online!
answered Mar 29 at 11:23
Olivier GrégoireOlivier Grégoire
9,39511944
$endgroup$
Java (JDK), 103 bytes
n->{int p=0,x=0,z=n,d;for(;p<1;p=p>0?z:0,z=z==n+x?n-++x:z+1)for(p=z/2,d=1;++d<z;)p=z%d<1?0:p;return p;}
Try it online!
answered Mar 29 at 11:23
Olivier GrégoireOlivier Grégoire
9,39511944
edited Mar 29 at 13:06
answered Mar 29 at 11:23
Olivier GrégoireOlivier Grégoire
9,39511944
answered Mar 29 at 11:23
Olivier GrégoireOlivier Grégoire
9,39511944
answered Mar 29 at 11:23
Olivier GrégoireOlivier Grégoire
9,39511944
9,39511944
$begingroup$
Umm.. I had already create a port of his answer.. ;) Although yours is 1 byte shorter, so something is different. EDIT: Ah, I have a result-integer outside the loop, and you modify the input inside the loop, hence the -1 byte for;. :) Do you want me to delete my answer?.. Feel free to copy the explanation.
$endgroup$
– Kevin Cruijssen
Mar 29 at 12:50
$begingroup$
@KevinCruijssen Oops, rollbacked!
$endgroup$
– Olivier Grégoire
Mar 29 at 12:59
$begingroup$
Sorry about that (and thanks for the -1 byte). I like your version as well, though. Already upvoted before I saw NaturalNumberGuy's answer.
$endgroup$
– Kevin Cruijssen
Mar 29 at 12:59
add a comment |
$begingroup$
Umm.. I had already create a port of his answer.. ;) Although yours is 1 byte shorter, so something is different. EDIT: Ah, I have a result-integer outside the loop, and you modify the input inside the loop, hence the -1 byte for;. :) Do you want me to delete my answer?.. Feel free to copy the explanation.
$endgroup$
– Kevin Cruijssen
Mar 29 at 12:50
$begingroup$
@KevinCruijssen Oops, rollbacked!
$endgroup$
– Olivier Grégoire
Mar 29 at 12:59
$begingroup$
Sorry about that (and thanks for the -1 byte). I like your version as well, though. Already upvoted before I saw NaturalNumberGuy's answer.
$endgroup$
– Kevin Cruijssen
Mar 29 at 12:59
$begingroup$
Umm.. I had already create a port of his answer.. ;) Although yours is 1 byte shorter, so something is different. EDIT: Ah, I have a result-integer outside the loop, and you modify the input inside the loop, hence the -1 byte for
;. :) Do you want me to delete my answer?.. Feel free to copy the explanation.$endgroup$
– Kevin Cruijssen
Mar 29 at 12:50
$begingroup$
Umm.. I had already create a port of his answer.. ;) Although yours is 1 byte shorter, so something is different. EDIT: Ah, I have a result-integer outside the loop, and you modify the input inside the loop, hence the -1 byte for
;. :) Do you want me to delete my answer?.. Feel free to copy the explanation.$endgroup$
– Kevin Cruijssen
Mar 29 at 12:50
$begingroup$
@KevinCruijssen Oops, rollbacked!
$endgroup$
– Olivier Grégoire
Mar 29 at 12:59
$begingroup$
@KevinCruijssen Oops, rollbacked!
$endgroup$
– Olivier Grégoire
Mar 29 at 12:59
$begingroup$
Sorry about that (and thanks for the -1 byte). I like your version as well, though. Already upvoted before I saw NaturalNumberGuy's answer.
$endgroup$
– Kevin Cruijssen
Mar 29 at 12:59
$begingroup$
Sorry about that (and thanks for the -1 byte). I like your version as well, though. Already upvoted before I saw NaturalNumberGuy's answer.
$endgroup$
– Kevin Cruijssen
Mar 29 at 12:59
add a comment |
2
$begingroup$
Haskell, 79 74 bytes (thanks to Laikoni)
72 bytes as annonymus function (the initial "f=" could be removed in this case).
f=(!)(-1);n!x|x>1,all((>0).mod x)[2..x-1]=x|y<-x+n=last(-n+1:[-n-1|n>0])!y
Try it online!
original code:
f=(!)(-1);n!x|x>1&&all((>0).mod x)[2..x-1]=x|1>0=(last$(-n+1):[-n-1|n>0])!(x+n)
Try it online!
Explanation:
f x = (-1)!x
isPrime x = x > 1 && all (k -> x `mod` k /= 0)[2..x-1]
n!x | isPrime x = x -- return the first prime found
| n>0 = (-n-1)!(x+n) -- x is no prime, continue with x+n where n takes the
| otherwise = (-n+1)!(x+n) -- values -1,2,-3,4 .. in subsequent calls of (!)
answered Mar 29 at 2:38
SacheraSachera
512
$endgroup$
1
$begingroup$
Inside a guard you can use,instead of&&.(last$ ...)can belast(...), and the second guard1>0can be used for a binding to save parenthesis, e.g.y<-x+n.
$endgroup$
– Laikoni
Mar 30 at 10:57
$begingroup$
Anonymous functions are generally allowed, so the initialf=does not need to be counted. Also the parenthesis enclosing(-1+n)can be dropped.
$endgroup$
– Laikoni
Mar 30 at 11:22
$begingroup$
Thanks for the suggestions. I didn't know "," and bindings are allowed in function guards! But i don't really like the idea of an annonymous function as an answer. It doesn't feel right in my opinion.
$endgroup$
– Sachera
Mar 31 at 2:58
$begingroup$
You can find more tips in our collection of tips for golfing in Haskell. There is also a Guide to Golfing Rules in Haskell and dedicated chat room: Of Monads and Men.
$endgroup$
– Laikoni
Mar 31 at 22:13
add a comment |
2
$begingroup$
Haskell, 79 74 bytes (thanks to Laikoni)
72 bytes as annonymus function (the initial "f=" could be removed in this case).
f=(!)(-1);n!x|x>1,all((>0).mod x)[2..x-1]=x|y<-x+n=last(-n+1:[-n-1|n>0])!y
Try it online!
original code:
f=(!)(-1);n!x|x>1&&all((>0).mod x)[2..x-1]=x|1>0=(last$(-n+1):[-n-1|n>0])!(x+n)
Try it online!
Explanation:
f x = (-1)!x
isPrime x = x > 1 && all (k -> x `mod` k /= 0)[2..x-1]
n!x | isPrime x = x -- return the first prime found
| n>0 = (-n-1)!(x+n) -- x is no prime, continue with x+n where n takes the
| otherwise = (-n+1)!(x+n) -- values -1,2,-3,4 .. in subsequent calls of (!)
answered Mar 29 at 2:38
SacheraSachera
512
$endgroup$
1
$begingroup$
Inside a guard you can use,instead of&&.(last$ ...)can belast(...), and the second guard1>0can be used for a binding to save parenthesis, e.g.y<-x+n.
$endgroup$
– Laikoni
Mar 30 at 10:57
$begingroup$
Anonymous functions are generally allowed, so the initialf=does not need to be counted. Also the parenthesis enclosing(-1+n)can be dropped.
$endgroup$
– Laikoni
Mar 30 at 11:22
$begingroup$
Thanks for the suggestions. I didn't know "," and bindings are allowed in function guards! But i don't really like the idea of an annonymous function as an answer. It doesn't feel right in my opinion.
$endgroup$
– Sachera
Mar 31 at 2:58
$begingroup$
You can find more tips in our collection of tips for golfing in Haskell. There is also a Guide to Golfing Rules in Haskell and dedicated chat room: Of Monads and Men.
$endgroup$
– Laikoni
Mar 31 at 22:13
add a comment |
2
2
2
$begingroup$
Haskell, 79 74 bytes (thanks to Laikoni)
72 bytes as annonymus function (the initial "f=" could be removed in this case).
f=(!)(-1);n!x|x>1,all((>0).mod x)[2..x-1]=x|y<-x+n=last(-n+1:[-n-1|n>0])!y
Try it online!
original code:
f=(!)(-1);n!x|x>1&&all((>0).mod x)[2..x-1]=x|1>0=(last$(-n+1):[-n-1|n>0])!(x+n)
Try it online!
Explanation:
f x = (-1)!x
isPrime x = x > 1 && all (k -> x `mod` k /= 0)[2..x-1]
n!x | isPrime x = x -- return the first prime found
| n>0 = (-n-1)!(x+n) -- x is no prime, continue with x+n where n takes the
| otherwise = (-n+1)!(x+n) -- values -1,2,-3,4 .. in subsequent calls of (!)
answered Mar 29 at 2:38
SacheraSachera
512
$endgroup$
Haskell, 79 74 bytes (thanks to Laikoni)
72 bytes as annonymus function (the initial "f=" could be removed in this case).
f=(!)(-1);n!x|x>1,all((>0).mod x)[2..x-1]=x|y<-x+n=last(-n+1:[-n-1|n>0])!y
Try it online!
original code:
f=(!)(-1);n!x|x>1&&all((>0).mod x)[2..x-1]=x|1>0=(last$(-n+1):[-n-1|n>0])!(x+n)
Try it online!
Explanation:
f x = (-1)!x
isPrime x = x > 1 && all (k -> x `mod` k /= 0)[2..x-1]
n!x | isPrime x = x -- return the first prime found
| n>0 = (-n-1)!(x+n) -- x is no prime, continue with x+n where n takes the
| otherwise = (-n+1)!(x+n) -- values -1,2,-3,4 .. in subsequent calls of (!)
answered Mar 29 at 2:38
SacheraSachera
512
edited Mar 31 at 2:56
answered Mar 29 at 2:38
SacheraSachera
512
answered Mar 29 at 2:38
SacheraSachera
512
answered Mar 29 at 2:38
SacheraSachera
512
512
1
$begingroup$
Inside a guard you can use,instead of&&.(last$ ...)can belast(...), and the second guard1>0can be used for a binding to save parenthesis, e.g.y<-x+n.
$endgroup$
– Laikoni
Mar 30 at 10:57
$begingroup$
Anonymous functions are generally allowed, so the initialf=does not need to be counted. Also the parenthesis enclosing(-1+n)can be dropped.
$endgroup$
– Laikoni
Mar 30 at 11:22
$begingroup$
Thanks for the suggestions. I didn't know "," and bindings are allowed in function guards! But i don't really like the idea of an annonymous function as an answer. It doesn't feel right in my opinion.
$endgroup$
– Sachera
Mar 31 at 2:58
$begingroup$
You can find more tips in our collection of tips for golfing in Haskell. There is also a Guide to Golfing Rules in Haskell and dedicated chat room: Of Monads and Men.
$endgroup$
– Laikoni
Mar 31 at 22:13
add a comment |
1
$begingroup$
Inside a guard you can use,instead of&&.(last$ ...)can belast(...), and the second guard1>0can be used for a binding to save parenthesis, e.g.y<-x+n.
$endgroup$
– Laikoni
Mar 30 at 10:57
$begingroup$
Anonymous functions are generally allowed, so the initialf=does not need to be counted. Also the parenthesis enclosing(-1+n)can be dropped.
$endgroup$
– Laikoni
Mar 30 at 11:22
$begingroup$
Thanks for the suggestions. I didn't know "," and bindings are allowed in function guards! But i don't really like the idea of an annonymous function as an answer. It doesn't feel right in my opinion.
$endgroup$
– Sachera
Mar 31 at 2:58
$begingroup$
You can find more tips in our collection of tips for golfing in Haskell. There is also a Guide to Golfing Rules in Haskell and dedicated chat room: Of Monads and Men.
$endgroup$
– Laikoni
Mar 31 at 22:13
1
1
$begingroup$
Inside a guard you can use
, instead of &&. (last$ ...) can be last(...), and the second guard 1>0 can be used for a binding to save parenthesis, e.g. y<-x+n.$endgroup$
– Laikoni
Mar 30 at 10:57
$begingroup$
Inside a guard you can use
, instead of &&. (last$ ...) can be last(...), and the second guard 1>0 can be used for a binding to save parenthesis, e.g. y<-x+n.$endgroup$
– Laikoni
Mar 30 at 10:57
$begingroup$
Anonymous functions are generally allowed, so the initial
f= does not need to be counted. Also the parenthesis enclosing (-1+n) can be dropped.$endgroup$
– Laikoni
Mar 30 at 11:22
$begingroup$
Anonymous functions are generally allowed, so the initial
f= does not need to be counted. Also the parenthesis enclosing (-1+n) can be dropped.$endgroup$
– Laikoni
Mar 30 at 11:22
$begingroup$
Thanks for the suggestions. I didn't know "," and bindings are allowed in function guards! But i don't really like the idea of an annonymous function as an answer. It doesn't feel right in my opinion.
$endgroup$
– Sachera
Mar 31 at 2:58
$begingroup$
Thanks for the suggestions. I didn't know "," and bindings are allowed in function guards! But i don't really like the idea of an annonymous function as an answer. It doesn't feel right in my opinion.
$endgroup$
– Sachera
Mar 31 at 2:58
$begingroup$
You can find more tips in our collection of tips for golfing in Haskell. There is also a Guide to Golfing Rules in Haskell and dedicated chat room: Of Monads and Men.
$endgroup$
– Laikoni
Mar 31 at 22:13
$begingroup$
You can find more tips in our collection of tips for golfing in Haskell. There is also a Guide to Golfing Rules in Haskell and dedicated chat room: Of Monads and Men.
$endgroup$
– Laikoni
Mar 31 at 22:13
add a comment |
2
$begingroup$
VDM-SL, 161 bytes
f(i)==(lambda p:set of nat1&let z in set p be st forall m in set p&abs(m-i)>=abs(z-i)in z)({x|x in set{1,...,9**7}&forall y in set{2,...,1003}&y<>x=>x mod y<>0})
A full program to run might look like this - it's worth noting that the bounds of the set of primes used should probably be changed if you actually want to run this, since it will take a long time to run for 1 million:
functions
f:nat1+>nat1
f(i)==(lambda p:set of nat1&let z in set p be st forall m in set p&abs(m-i)>=abs(z-i)in z)({x|x in set{1,...,9**7}&forall y in set{2,...,1003}&y<>x=>x mod y<>0})
Explanation:
f(i)== /* f is a function which takes a nat1 (natural number not including 0)*/
(lambda p:set of nat1 /* define a lambda which takes a set of nat1*/
&let z in set p be st /* which has an element z in the set such that */
forall m in set p /* for every element in the set*/
&abs(m-i) /* the difference between the element m and the input*/
>=abs(z-i) /* is greater than or equal to the difference between the element z and the input */
in z) /* and return z from the lambda */
( /* apply this lambda to... */
{ /* a set defined by comprehension as.. */
x| /* all elements x such that.. */
x in set{1,...,9**7} /* x is between 1 and 9^7 */
&forall y in set{2,...,1003} /* and for all values between 2 and 1003*/
&y<>x=>x mod y<>0 /* y is not x implies x is not divisible by y*/
}
)
answered Mar 27 at 16:31
Expired DataExpired Data
56314
$endgroup$
add a comment |
2
$begingroup$
VDM-SL, 161 bytes
f(i)==(lambda p:set of nat1&let z in set p be st forall m in set p&abs(m-i)>=abs(z-i)in z)({x|x in set{1,...,9**7}&forall y in set{2,...,1003}&y<>x=>x mod y<>0})
A full program to run might look like this - it's worth noting that the bounds of the set of primes used should probably be changed if you actually want to run this, since it will take a long time to run for 1 million:
functions
f:nat1+>nat1
f(i)==(lambda p:set of nat1&let z in set p be st forall m in set p&abs(m-i)>=abs(z-i)in z)({x|x in set{1,...,9**7}&forall y in set{2,...,1003}&y<>x=>x mod y<>0})
Explanation:
f(i)== /* f is a function which takes a nat1 (natural number not including 0)*/
(lambda p:set of nat1 /* define a lambda which takes a set of nat1*/
&let z in set p be st /* which has an element z in the set such that */
forall m in set p /* for every element in the set*/
&abs(m-i) /* the difference between the element m and the input*/
>=abs(z-i) /* is greater than or equal to the difference between the element z and the input */
in z) /* and return z from the lambda */
( /* apply this lambda to... */
{ /* a set defined by comprehension as.. */
x| /* all elements x such that.. */
x in set{1,...,9**7} /* x is between 1 and 9^7 */
&forall y in set{2,...,1003} /* and for all values between 2 and 1003*/
&y<>x=>x mod y<>0 /* y is not x implies x is not divisible by y*/
}
)
answered Mar 27 at 16:31
Expired DataExpired Data
56314
$endgroup$
add a comment |
2
2
2
$begingroup$
VDM-SL, 161 bytes
f(i)==(lambda p:set of nat1&let z in set p be st forall m in set p&abs(m-i)>=abs(z-i)in z)({x|x in set{1,...,9**7}&forall y in set{2,...,1003}&y<>x=>x mod y<>0})
A full program to run might look like this - it's worth noting that the bounds of the set of primes used should probably be changed if you actually want to run this, since it will take a long time to run for 1 million:
functions
f:nat1+>nat1
f(i)==(lambda p:set of nat1&let z in set p be st forall m in set p&abs(m-i)>=abs(z-i)in z)({x|x in set{1,...,9**7}&forall y in set{2,...,1003}&y<>x=>x mod y<>0})
Explanation:
f(i)== /* f is a function which takes a nat1 (natural number not including 0)*/
(lambda p:set of nat1 /* define a lambda which takes a set of nat1*/
&let z in set p be st /* which has an element z in the set such that */
forall m in set p /* for every element in the set*/
&abs(m-i) /* the difference between the element m and the input*/
>=abs(z-i) /* is greater than or equal to the difference between the element z and the input */
in z) /* and return z from the lambda */
( /* apply this lambda to... */
{ /* a set defined by comprehension as.. */
x| /* all elements x such that.. */
x in set{1,...,9**7} /* x is between 1 and 9^7 */
&forall y in set{2,...,1003} /* and for all values between 2 and 1003*/
&y<>x=>x mod y<>0 /* y is not x implies x is not divisible by y*/
}
)
answered Mar 27 at 16:31
Expired DataExpired Data
56314
$endgroup$
VDM-SL, 161 bytes
f(i)==(lambda p:set of nat1&let z in set p be st forall m in set p&abs(m-i)>=abs(z-i)in z)({x|x in set{1,...,9**7}&forall y in set{2,...,1003}&y<>x=>x mod y<>0})
A full program to run might look like this - it's worth noting that the bounds of the set of primes used should probably be changed if you actually want to run this, since it will take a long time to run for 1 million:
functions
f:nat1+>nat1
f(i)==(lambda p:set of nat1&let z in set p be st forall m in set p&abs(m-i)>=abs(z-i)in z)({x|x in set{1,...,9**7}&forall y in set{2,...,1003}&y<>x=>x mod y<>0})
Explanation:
f(i)== /* f is a function which takes a nat1 (natural number not including 0)*/
(lambda p:set of nat1 /* define a lambda which takes a set of nat1*/
&let z in set p be st /* which has an element z in the set such that */
forall m in set p /* for every element in the set*/
&abs(m-i) /* the difference between the element m and the input*/
>=abs(z-i) /* is greater than or equal to the difference between the element z and the input */
in z) /* and return z from the lambda */
( /* apply this lambda to... */
{ /* a set defined by comprehension as.. */
x| /* all elements x such that.. */
x in set{1,...,9**7} /* x is between 1 and 9^7 */
&forall y in set{2,...,1003} /* and for all values between 2 and 1003*/
&y<>x=>x mod y<>0 /* y is not x implies x is not divisible by y*/
}
)
answered Mar 27 at 16:31
Expired DataExpired Data
56314
edited Apr 1 at 15:52
answered Mar 27 at 16:31
Expired DataExpired Data
56314
answered Mar 27 at 16:31
Expired DataExpired Data
56314
answered Mar 27 at 16:31
Expired DataExpired Data
56314
56314
add a comment |
add a comment |
1
$begingroup$
C# (Visual C# Interactive Compiler), 112 bytes
g=>Enumerable.Range(2,2<<20).Where(x=>Enumerable.Range(1,x).Count(y=>x%y<1)<3).OrderBy(x=>Math.Abs(x-g)).First()
Try it online!
Left shifts by 20 in submission but 10 in TIO so that TIO terminates for test cases.
answered Mar 27 at 16:43
Expired DataExpired Data
56314
$endgroup$
add a comment |
1
$begingroup$
C# (Visual C# Interactive Compiler), 112 bytes
g=>Enumerable.Range(2,2<<20).Where(x=>Enumerable.Range(1,x).Count(y=>x%y<1)<3).OrderBy(x=>Math.Abs(x-g)).First()
Try it online!
Left shifts by 20 in submission but 10 in TIO so that TIO terminates for test cases.
answered Mar 27 at 16:43
Expired DataExpired Data
56314
$endgroup$
add a comment |
1
1
1
$begingroup$
C# (Visual C# Interactive Compiler), 112 bytes
g=>Enumerable.Range(2,2<<20).Where(x=>Enumerable.Range(1,x).Count(y=>x%y<1)<3).OrderBy(x=>Math.Abs(x-g)).First()
Try it online!
Left shifts by 20 in submission but 10 in TIO so that TIO terminates for test cases.
answered Mar 27 at 16:43
Expired DataExpired Data
56314
$endgroup$
C# (Visual C# Interactive Compiler), 112 bytes
g=>Enumerable.Range(2,2<<20).Where(x=>Enumerable.Range(1,x).Count(y=>x%y<1)<3).OrderBy(x=>Math.Abs(x-g)).First()
Try it online!
Left shifts by 20 in submission but 10 in TIO so that TIO terminates for test cases.
answered Mar 27 at 16:43
Expired DataExpired Data
56314
answered Mar 27 at 16:43
Expired DataExpired Data
56314
answered Mar 27 at 16:43
Expired DataExpired Data
56314
answered Mar 27 at 16:43
Expired DataExpired Data
56314
56314
add a comment |
add a comment |
1
$begingroup$
Swift, 186 bytes
func p(a:Int){let b=q(a:a,b:-1),c=q(a:a,b:1);print(a-b<=c-a ? b:c)}
func q(a:Int,b:Int)->Int{var k=max(a,2),c=2;while k>c && c != a/2{if k%c==0{k+=b;c=2}else{c=c==2 ? c+1:c+2}};return k}
Try it online!
answered Mar 27 at 18:24
onnowebonnoweb
1613
$endgroup$
add a comment |
1
$begingroup$
Swift, 186 bytes
func p(a:Int){let b=q(a:a,b:-1),c=q(a:a,b:1);print(a-b<=c-a ? b:c)}
func q(a:Int,b:Int)->Int{var k=max(a,2),c=2;while k>c && c != a/2{if k%c==0{k+=b;c=2}else{c=c==2 ? c+1:c+2}};return k}
Try it online!
answered Mar 27 at 18:24
onnowebonnoweb
1613
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$begingroup$
Swift, 186 bytes
func p(a:Int){let b=q(a:a,b:-1),c=q(a:a,b:1);print(a-b<=c-a ? b:c)}
func q(a:Int,b:Int)->Int{var k=max(a,2),c=2;while k>c && c != a/2{if k%c==0{k+=b;c=2}else{c=c==2 ? c+1:c+2}};return k}
Try it online!
answered Mar 27 at 18:24
onnowebonnoweb
1613
$endgroup$
Swift, 186 bytes
func p(a:Int){let b=q(a:a,b:-1),c=q(a:a,b:1);print(a-b<=c-a ? b:c)}
func q(a:Int,b:Int)->Int{var k=max(a,2),c=2;while k>c && c != a/2{if k%c==0{k+=b;c=2}else{c=c==2 ? c+1:c+2}};return k}
Try it online!
answered Mar 27 at 18:24
onnowebonnoweb
1613
answered Mar 27 at 18:24
onnowebonnoweb
1613
answered Mar 27 at 18:24
onnowebonnoweb
1613
answered Mar 27 at 18:24
onnowebonnoweb
1613
1613
add a comment |
add a comment |
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If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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Hi and welcome to PPCG!. To avoid down voting due to lack of quality I suggest you to post it to the sandbox first and after a couple of days post it here
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– Luis felipe De jesus Munoz
Mar 27 at 15:17
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This is one of the outputs requested in this challenge.
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– Arnauld
Mar 27 at 15:24
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Very closely related but not quite identical.
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– Giuseppe
Mar 27 at 15:35
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@Arnauld I saw that one, but I thought that they were different enough to warrant a new question.
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– Nathan Dimmer
Mar 27 at 15:37
2
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See also OEIS A051697.
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– Eric Towers
Mar 28 at 8:52