Do I have a twin with permutated remainders?
$begingroup$
We define $R_n$ as the list of remainders of the Euclidean division of $n$ by $2$, $3$, $5$ and $7$.
Given an integer $nge0$, you have to figure out if there exists an integer $0<k<210$ such that $R_{n+k}$ is a permutation of $R_n$.
Examples
The criterion is met for $n=8$, because:
- we have $R_8=(0,2,3,1)$
- for $k=44$, we have $R_{n+k}=R_{52}=(0,1,2,3)$, which is a permutation of $R_8$
The criterion is not met for $n=48$, because:
- we have $R_{48}=(0,0,3,6)$
- the smallest integer $k>0$ such that $R_{n+k}$ is a permutation of $R_{48}$ is $k=210$ (leading to $R_{258}=(0,0,3,6)$ as well)
Rules
- You may either output a truthy value if $k$ exists and a falsy value otherwise, or two distinct and consistent values of your choice.
- This is code-golf.
Hint
Do you really need to compute $k$? Well, maybe. Or maybe not.
Test cases
Some values of $n$ for which $k$ exists:
3, 4, 5, 8, 30, 100, 200, 2019
Some values of $n$ for which $k$ does not exist:
0, 1, 2, 13, 19, 48, 210, 1999
code-golf decision-problem number-theory
asked Apr 4 at 15:36
ArnauldArnauld
81.7k797336
$endgroup$
add a comment |
$begingroup$
We define $R_n$ as the list of remainders of the Euclidean division of $n$ by $2$, $3$, $5$ and $7$.
Given an integer $nge0$, you have to figure out if there exists an integer $0<k<210$ such that $R_{n+k}$ is a permutation of $R_n$.
Examples
The criterion is met for $n=8$, because:
- we have $R_8=(0,2,3,1)$
- for $k=44$, we have $R_{n+k}=R_{52}=(0,1,2,3)$, which is a permutation of $R_8$
The criterion is not met for $n=48$, because:
- we have $R_{48}=(0,0,3,6)$
- the smallest integer $k>0$ such that $R_{n+k}$ is a permutation of $R_{48}$ is $k=210$ (leading to $R_{258}=(0,0,3,6)$ as well)
Rules
- You may either output a truthy value if $k$ exists and a falsy value otherwise, or two distinct and consistent values of your choice.
- This is code-golf.
Hint
Do you really need to compute $k$? Well, maybe. Or maybe not.
Test cases
Some values of $n$ for which $k$ exists:
3, 4, 5, 8, 30, 100, 200, 2019
Some values of $n$ for which $k$ does not exist:
0, 1, 2, 13, 19, 48, 210, 1999
code-golf decision-problem number-theory
asked Apr 4 at 15:36
ArnauldArnauld
81.7k797336
$endgroup$
add a comment |
$begingroup$
We define $R_n$ as the list of remainders of the Euclidean division of $n$ by $2$, $3$, $5$ and $7$.
Given an integer $nge0$, you have to figure out if there exists an integer $0<k<210$ such that $R_{n+k}$ is a permutation of $R_n$.
Examples
The criterion is met for $n=8$, because:
- we have $R_8=(0,2,3,1)$
- for $k=44$, we have $R_{n+k}=R_{52}=(0,1,2,3)$, which is a permutation of $R_8$
The criterion is not met for $n=48$, because:
- we have $R_{48}=(0,0,3,6)$
- the smallest integer $k>0$ such that $R_{n+k}$ is a permutation of $R_{48}$ is $k=210$ (leading to $R_{258}=(0,0,3,6)$ as well)
Rules
- You may either output a truthy value if $k$ exists and a falsy value otherwise, or two distinct and consistent values of your choice.
- This is code-golf.
Hint
Do you really need to compute $k$? Well, maybe. Or maybe not.
Test cases
Some values of $n$ for which $k$ exists:
3, 4, 5, 8, 30, 100, 200, 2019
Some values of $n$ for which $k$ does not exist:
0, 1, 2, 13, 19, 48, 210, 1999
code-golf decision-problem number-theory
asked Apr 4 at 15:36
ArnauldArnauld
81.7k797336
$endgroup$
We define $R_n$ as the list of remainders of the Euclidean division of $n$ by $2$, $3$, $5$ and $7$.
Given an integer $nge0$, you have to figure out if there exists an integer $0<k<210$ such that $R_{n+k}$ is a permutation of $R_n$.
Examples
The criterion is met for $n=8$, because:
- we have $R_8=(0,2,3,1)$
- for $k=44$, we have $R_{n+k}=R_{52}=(0,1,2,3)$, which is a permutation of $R_8$
The criterion is not met for $n=48$, because:
- we have $R_{48}=(0,0,3,6)$
- the smallest integer $k>0$ such that $R_{n+k}$ is a permutation of $R_{48}$ is $k=210$ (leading to $R_{258}=(0,0,3,6)$ as well)
Rules
- You may either output a truthy value if $k$ exists and a falsy value otherwise, or two distinct and consistent values of your choice.
- This is code-golf.
Hint
Do you really need to compute $k$? Well, maybe. Or maybe not.
Test cases
Some values of $n$ for which $k$ exists:
3, 4, 5, 8, 30, 100, 200, 2019
Some values of $n$ for which $k$ does not exist:
0, 1, 2, 13, 19, 48, 210, 1999
code-golf decision-problem number-theory
code-golf decision-problem number-theory
asked Apr 4 at 15:36
ArnauldArnauld
81.7k797336
asked Apr 4 at 15:36
ArnauldArnauld
81.7k797336
asked Apr 4 at 15:36
ArnauldArnauld
81.7k797336
asked Apr 4 at 15:36
ArnauldArnauld
81.7k797336
asked Apr 4 at 15:36
ArnauldArnauld
81.7k797336
81.7k797336
add a comment |
add a comment |
16 Answers
16
active
oldest
votes
$begingroup$
R, 63 59 bytes
s=scan()%%c(2,3,5,7);i=which(s<c(0,2,3,5));any(s[i]-s[i-1])
Try it online!
-4 bytes thanks to Giuseppe
(The explanation contains a spoiler as to how to solve the problem without computing $k$.)
Explanation:
Let $s$ be the list of remainders. Note the constraint that s[1]<2, s[2]<3, s[3]<5 and s[4]<7. By the Chinese Remainder Theorem, there exists a $k$ iff there is a permutation of $s$, distinct from $s$, which verifies the constraint. In practice, this will be verified if one of the following conditions is verified:
- s[2]<2 and s[2]!=s[1]
- s[3]<3 and s[3]!=s[2]
- s[4]<5 and s[4]!=s[3]
The code can probably be golfed further.
answered Apr 4 at 17:00
Robin RyderRobin Ryder
85112
$endgroup$
$begingroup$
Could you explain why the permutation is necessarily distinct from $s$?
$endgroup$
– dfeuer
Apr 5 at 18:41
1
$begingroup$
@dfeuer It is a consequence of the Chinese Remainder Theorem; I added a link. If two integers have the same remainders modulo 2, 3, 5 and 7 (without a permutation), then the two integers are equal modulo 2*3*5*7=210.
$endgroup$
– Robin Ryder
Apr 5 at 19:46
add a comment |
$begingroup$
Haskell, 69 bytes
Based on the Chinese remainder theorem
m=[2,3,5,7]
f x|s<-mod x<$>m=or[m!!a>b|a<-[0..2],b<-drop a s,s!!a/=b]
Try it online!
answered Apr 4 at 16:45
H.PWizH.PWiz
10.4k21353
$endgroup$
4
$begingroup$
Actually, my working title for this challenge was "Do I have a Chinese twin?" :)
$endgroup$
– Arnauld
Apr 5 at 10:22
add a comment |
$begingroup$
Perl 6, 64 61 59 43 bytes
{map($!=(*X%2,3,5,7).Bag,^209+$_+1)∋.&$!}
Try it online!
-16 thanks to @Jo King
answered Apr 4 at 16:20
VenVen
2,89011223
$endgroup$
add a comment |
$begingroup$
Python 2, 41 bytes
lambda n:n%5!=n%7<5or n%3!=n%5<3or-~n%6/4
Try it online!
Uses the same characterization as Robin Ryder. The check n%2!=n%3<2 is shortened to -~n%6/4. Writing out the three conditions turned out shorter than writing a general one:
46 bytes
lambda n:any(n%p!=n%(p+1|1)<p for p in[2,3,5])
Try it online!
answered Apr 5 at 0:34
xnorxnor
94.3k18192452
$endgroup$
add a comment |
$begingroup$
Haskell, 47 bytes
g.mod
g r|let p?q=r p/=r q&&r q<p=2?3||3?5||5?7
Try it online!
answered Apr 5 at 5:19
xnorxnor
94.3k18192452
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$begingroup$
Can you explain?
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– dfeuer
Apr 5 at 6:24
1
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@dfeuer It's using Robin Ryder's method.
$endgroup$
– Ørjan Johansen
Apr 5 at 17:39
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 125 42 38 36 bytes
n=>n%7<5&5<n%35|n%5<3&3<n%15|-~n%6>3
Direct port of @xnor's answer, which is based off of @RobinRyder's solution.
Saved 4 bytes thanks to @Ørjan Johansen!
Saved 2 more thanks to @Arnauld!
Try it online!
answered Apr 4 at 16:33
Embodiment of IgnoranceEmbodiment of Ignorance
3,024127
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1
$begingroup$
I found a variation that only ties for xnor's languages but helps for this: 38 bytes
$endgroup$
– Ørjan Johansen
Apr 5 at 6:04
1
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Isn't-~n%6/4>0just-~n%6>3?
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– Arnauld
Apr 6 at 9:55
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BTW, this is a JavaScript polyglot.
$endgroup$
– Arnauld
Apr 7 at 7:55
add a comment |
$begingroup$
Wolfram Language (Mathematica), 67 bytes
!FreeQ[Sort/@Table[R[#+k],{k,209}],Sort@R@#]&
R@n_:=n~Mod~{2,3,5,7}
Try it online!
answered Apr 4 at 15:53
J42161217J42161217
14.3k21354
$endgroup$
add a comment |
$begingroup$
Ruby, 54 bytes
->n{[2,3,5,7].each_cons(2).any?{|l,h|n%l!=n%h&&n%h<l}}
Try it online!
Uses Robin Ryder's clever solution.
answered Apr 4 at 18:59
histocrathistocrat
19.2k43173
$endgroup$
add a comment |
$begingroup$
Java (JDK), 36 bytes
n->n%7<5&5<n%35|n%5<3&3<n%15|-~n%6>3
Try it online!
Credits
- Port of xnor's solution, improved by Ørjan Johansen.
answered Apr 5 at 9:07
Olivier GrégoireOlivier Grégoire
9,47511944
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add a comment |
$begingroup$
R, 72 bytes
n=scan();b=c(2,3,5,7);for(i in n+1:209)F=F|all(sort(n%%b)==sort(i%%b));F
Try it online!
answered Apr 4 at 16:42
Aaron HaymanAaron Hayman
3516
$endgroup$
add a comment |
$begingroup$
PHP, 81 78 72 bytes
while($y<3)if($argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u)die(T);
A riff on @Robin Ryder's answer. Input is via STDIN, output is 'T' if truthy, and empty '' if falsy.
$ echo 3|php -nF euc.php
T
$ echo 5|php -nF euc.php
T
$ echo 2019|php -nF euc.php
T
$ echo 0|php -nF euc.php
$ echo 2|php -nF euc.php
$ echo 1999|php -nF euc.php
Try it online!
Or 73 bytes with 1 or 0 response
while($y<3)$r|=$argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u;echo$r;
$ echo 2019|php -nF euc.php
1
$ echo 1999|php -nF euc.php
0
Try it online (all test cases)!
Original answer, 133 127 bytes
function($n){while(++$k<210)if(($r=function($n){foreach([2,3,5,7]as$d)$o=$n%$d;sort($o);return$o;})($n+$k)==$r($n))return 1;}
Try it online!
answered Apr 4 at 16:01
gwaughgwaugh
2,4281519
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add a comment |
$begingroup$
Python 3, 69 bytes
lambda x:int('4LR2991ODO5GS2974QWH22YTLL3E3I6TDADQG87I0',36)&1<<x%210
Try it online!
Hardcoded
answered Apr 4 at 22:17
attinatattinat
60917
$endgroup$
add a comment |
$begingroup$
05AB1E, 16 bytes
Ƶ.L+ε‚ε4Åp%{}Ë}à
Try it online or verify all test cases.
Explanation:
Ƶ.L # Create a list in the range [1,209] (which is k)
+ # Add the (implicit) input to each (which is n+k)
ε # Map each value to:
‚ # Pair it with the (implicit) input
ε # Map both to:
4Åp # Get the first 4 primes: [2,3,5,7]
% # Modulo the current number by each of these four (now we have R_n and R_n+k)
{ # Sort the list
}Ë # After the inner map: check if both sorted lists are equal
}à # After the outer map: check if any are truthy by taking the maximum
# (which is output implicitly as result)
See this 05AB1E tip of mine (section How to compress large integers?) to understand why Ƶ. is 209.
answered Apr 5 at 7:29
Kevin CruijssenKevin Cruijssen
43.3k573222
$endgroup$
add a comment |
$begingroup$
J, 40 bytes
1 e.(>:+i.@209)-:&(/:~)&(2 3 5 7&|"1 0)]
Try it online!
Brute force...
answered Apr 5 at 8:28
Galen IvanovGalen Ivanov
7,54211034
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 56 bytes
Or@@(Min[s-#]>0&/@Rest@Permutations@Mod[#,s={2,3,5,7}])&
Try it online!
Finds all non-identity permutations of the remainders of the input modulo 2, 3, 5, 7, and checks if any of them are below {2,3,5,7} in each coordinate. Note that Or@@{} is False.
answered Apr 6 at 0:57
Misha LavrovMisha Lavrov
4,271424
$endgroup$
add a comment |
$begingroup$
Jelly, 15 bytes
8ÆR©PḶ+%Ṣ¥€®ċḢ$
Try it online!
I’m sure there’s a golfier answer. I’ve interpreted a truthy value as being anything that isn’t zero, so here it’s the number of possible values of k. If it needs to be two distinct values that costs me a further byte.
Explanation
8ÆR | Primes less than 8 [2,3,5,7]
© | Copy to register
P | Product [210]
Ḷ | Lowered range [0, 1, ..., 208, 209]
+ | Add to input
¥€ | For each of these 210 numbers...
% ® | Modulo 2, 3, 5, 7
Ṣ | And sort
ċḢ$ | Count how many match the first (input) number’s remainders
answered Apr 4 at 17:59
Nick KennedyNick Kennedy
1,87149
$endgroup$
1
$begingroup$
All good regarding truthy vs falsey. Using the meta agreed definition of truthy and falsey (effectively "what does the language's if-else construct do if there is one) zero is falsey and non-zeros are truthy (?is the if-else construct in Jelly; for some languages it's a harder question).
$endgroup$
– Jonathan Allan
Apr 4 at 18:08
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Oh, and you could get distinct values for no cost withḢe$if you wanted :)
$endgroup$
– Jonathan Allan
Apr 4 at 18:11
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@JonathanAllan yes of course, thanks. :)
$endgroup$
– Nick Kennedy
Apr 4 at 18:12
add a comment |
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16 Answers
16
active
oldest
votes
16 Answers
16
active
oldest
votes
active
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$begingroup$
R, 63 59 bytes
s=scan()%%c(2,3,5,7);i=which(s<c(0,2,3,5));any(s[i]-s[i-1])
Try it online!
-4 bytes thanks to Giuseppe
(The explanation contains a spoiler as to how to solve the problem without computing $k$.)
Explanation:
Let $s$ be the list of remainders. Note the constraint that s[1]<2, s[2]<3, s[3]<5 and s[4]<7. By the Chinese Remainder Theorem, there exists a $k$ iff there is a permutation of $s$, distinct from $s$, which verifies the constraint. In practice, this will be verified if one of the following conditions is verified:
- s[2]<2 and s[2]!=s[1]
- s[3]<3 and s[3]!=s[2]
- s[4]<5 and s[4]!=s[3]
The code can probably be golfed further.
answered Apr 4 at 17:00
Robin RyderRobin Ryder
85112
$endgroup$
$begingroup$
Could you explain why the permutation is necessarily distinct from $s$?
$endgroup$
– dfeuer
Apr 5 at 18:41
1
$begingroup$
@dfeuer It is a consequence of the Chinese Remainder Theorem; I added a link. If two integers have the same remainders modulo 2, 3, 5 and 7 (without a permutation), then the two integers are equal modulo 2*3*5*7=210.
$endgroup$
– Robin Ryder
Apr 5 at 19:46
add a comment |
$begingroup$
R, 63 59 bytes
s=scan()%%c(2,3,5,7);i=which(s<c(0,2,3,5));any(s[i]-s[i-1])
Try it online!
-4 bytes thanks to Giuseppe
(The explanation contains a spoiler as to how to solve the problem without computing $k$.)
Explanation:
Let $s$ be the list of remainders. Note the constraint that s[1]<2, s[2]<3, s[3]<5 and s[4]<7. By the Chinese Remainder Theorem, there exists a $k$ iff there is a permutation of $s$, distinct from $s$, which verifies the constraint. In practice, this will be verified if one of the following conditions is verified:
- s[2]<2 and s[2]!=s[1]
- s[3]<3 and s[3]!=s[2]
- s[4]<5 and s[4]!=s[3]
The code can probably be golfed further.
answered Apr 4 at 17:00
Robin RyderRobin Ryder
85112
$endgroup$
$begingroup$
Could you explain why the permutation is necessarily distinct from $s$?
$endgroup$
– dfeuer
Apr 5 at 18:41
1
$begingroup$
@dfeuer It is a consequence of the Chinese Remainder Theorem; I added a link. If two integers have the same remainders modulo 2, 3, 5 and 7 (without a permutation), then the two integers are equal modulo 2*3*5*7=210.
$endgroup$
– Robin Ryder
Apr 5 at 19:46
add a comment |
$begingroup$
R, 63 59 bytes
s=scan()%%c(2,3,5,7);i=which(s<c(0,2,3,5));any(s[i]-s[i-1])
Try it online!
-4 bytes thanks to Giuseppe
(The explanation contains a spoiler as to how to solve the problem without computing $k$.)
Explanation:
Let $s$ be the list of remainders. Note the constraint that s[1]<2, s[2]<3, s[3]<5 and s[4]<7. By the Chinese Remainder Theorem, there exists a $k$ iff there is a permutation of $s$, distinct from $s$, which verifies the constraint. In practice, this will be verified if one of the following conditions is verified:
- s[2]<2 and s[2]!=s[1]
- s[3]<3 and s[3]!=s[2]
- s[4]<5 and s[4]!=s[3]
The code can probably be golfed further.
answered Apr 4 at 17:00
Robin RyderRobin Ryder
85112
$endgroup$
R, 63 59 bytes
s=scan()%%c(2,3,5,7);i=which(s<c(0,2,3,5));any(s[i]-s[i-1])
Try it online!
-4 bytes thanks to Giuseppe
(The explanation contains a spoiler as to how to solve the problem without computing $k$.)
Explanation:
Let $s$ be the list of remainders. Note the constraint that s[1]<2, s[2]<3, s[3]<5 and s[4]<7. By the Chinese Remainder Theorem, there exists a $k$ iff there is a permutation of $s$, distinct from $s$, which verifies the constraint. In practice, this will be verified if one of the following conditions is verified:
- s[2]<2 and s[2]!=s[1]
- s[3]<3 and s[3]!=s[2]
- s[4]<5 and s[4]!=s[3]
The code can probably be golfed further.
answered Apr 4 at 17:00
Robin RyderRobin Ryder
85112
edited Apr 6 at 8:02
answered Apr 4 at 17:00
Robin RyderRobin Ryder
85112
answered Apr 4 at 17:00
Robin RyderRobin Ryder
85112
answered Apr 4 at 17:00
Robin RyderRobin Ryder
85112
85112
$begingroup$
Could you explain why the permutation is necessarily distinct from $s$?
$endgroup$
– dfeuer
Apr 5 at 18:41
1
$begingroup$
@dfeuer It is a consequence of the Chinese Remainder Theorem; I added a link. If two integers have the same remainders modulo 2, 3, 5 and 7 (without a permutation), then the two integers are equal modulo 2*3*5*7=210.
$endgroup$
– Robin Ryder
Apr 5 at 19:46
add a comment |
$begingroup$
Could you explain why the permutation is necessarily distinct from $s$?
$endgroup$
– dfeuer
Apr 5 at 18:41
1
$begingroup$
@dfeuer It is a consequence of the Chinese Remainder Theorem; I added a link. If two integers have the same remainders modulo 2, 3, 5 and 7 (without a permutation), then the two integers are equal modulo 2*3*5*7=210.
$endgroup$
– Robin Ryder
Apr 5 at 19:46
$begingroup$
Could you explain why the permutation is necessarily distinct from $s$?
$endgroup$
– dfeuer
Apr 5 at 18:41
$begingroup$
Could you explain why the permutation is necessarily distinct from $s$?
$endgroup$
– dfeuer
Apr 5 at 18:41
1
1
$begingroup$
@dfeuer It is a consequence of the Chinese Remainder Theorem; I added a link. If two integers have the same remainders modulo 2, 3, 5 and 7 (without a permutation), then the two integers are equal modulo 2*3*5*7=210.
$endgroup$
– Robin Ryder
Apr 5 at 19:46
$begingroup$
@dfeuer It is a consequence of the Chinese Remainder Theorem; I added a link. If two integers have the same remainders modulo 2, 3, 5 and 7 (without a permutation), then the two integers are equal modulo 2*3*5*7=210.
$endgroup$
– Robin Ryder
Apr 5 at 19:46
add a comment |
$begingroup$
Haskell, 69 bytes
Based on the Chinese remainder theorem
m=[2,3,5,7]
f x|s<-mod x<$>m=or[m!!a>b|a<-[0..2],b<-drop a s,s!!a/=b]
Try it online!
answered Apr 4 at 16:45
H.PWizH.PWiz
10.4k21353
$endgroup$
4
$begingroup$
Actually, my working title for this challenge was "Do I have a Chinese twin?" :)
$endgroup$
– Arnauld
Apr 5 at 10:22
add a comment |
$begingroup$
Haskell, 69 bytes
Based on the Chinese remainder theorem
m=[2,3,5,7]
f x|s<-mod x<$>m=or[m!!a>b|a<-[0..2],b<-drop a s,s!!a/=b]
Try it online!
answered Apr 4 at 16:45
H.PWizH.PWiz
10.4k21353
$endgroup$
4
$begingroup$
Actually, my working title for this challenge was "Do I have a Chinese twin?" :)
$endgroup$
– Arnauld
Apr 5 at 10:22
add a comment |
$begingroup$
Haskell, 69 bytes
Based on the Chinese remainder theorem
m=[2,3,5,7]
f x|s<-mod x<$>m=or[m!!a>b|a<-[0..2],b<-drop a s,s!!a/=b]
Try it online!
answered Apr 4 at 16:45
H.PWizH.PWiz
10.4k21353
$endgroup$
Haskell, 69 bytes
Based on the Chinese remainder theorem
m=[2,3,5,7]
f x|s<-mod x<$>m=or[m!!a>b|a<-[0..2],b<-drop a s,s!!a/=b]
Try it online!
answered Apr 4 at 16:45
H.PWizH.PWiz
10.4k21353
answered Apr 4 at 16:45
H.PWizH.PWiz
10.4k21353
answered Apr 4 at 16:45
H.PWizH.PWiz
10.4k21353
answered Apr 4 at 16:45
H.PWizH.PWiz
10.4k21353
10.4k21353
4
$begingroup$
Actually, my working title for this challenge was "Do I have a Chinese twin?" :)
$endgroup$
– Arnauld
Apr 5 at 10:22
add a comment |
4
$begingroup$
Actually, my working title for this challenge was "Do I have a Chinese twin?" :)
$endgroup$
– Arnauld
Apr 5 at 10:22
4
4
$begingroup$
Actually, my working title for this challenge was "Do I have a Chinese twin?" :)
$endgroup$
– Arnauld
Apr 5 at 10:22
$begingroup$
Actually, my working title for this challenge was "Do I have a Chinese twin?" :)
$endgroup$
– Arnauld
Apr 5 at 10:22
add a comment |
$begingroup$
Perl 6, 64 61 59 43 bytes
{map($!=(*X%2,3,5,7).Bag,^209+$_+1)∋.&$!}
Try it online!
-16 thanks to @Jo King
answered Apr 4 at 16:20
VenVen
2,89011223
$endgroup$
add a comment |
$begingroup$
Perl 6, 64 61 59 43 bytes
{map($!=(*X%2,3,5,7).Bag,^209+$_+1)∋.&$!}
Try it online!
-16 thanks to @Jo King
answered Apr 4 at 16:20
VenVen
2,89011223
$endgroup$
add a comment |
$begingroup$
Perl 6, 64 61 59 43 bytes
{map($!=(*X%2,3,5,7).Bag,^209+$_+1)∋.&$!}
Try it online!
-16 thanks to @Jo King
answered Apr 4 at 16:20
VenVen
2,89011223
$endgroup$
Perl 6, 64 61 59 43 bytes
{map($!=(*X%2,3,5,7).Bag,^209+$_+1)∋.&$!}
Try it online!
-16 thanks to @Jo King
answered Apr 4 at 16:20
VenVen
2,89011223
edited Apr 5 at 8:09
answered Apr 4 at 16:20
VenVen
2,89011223
answered Apr 4 at 16:20
VenVen
2,89011223
answered Apr 4 at 16:20
VenVen
2,89011223
2,89011223
add a comment |
add a comment |
$begingroup$
Python 2, 41 bytes
lambda n:n%5!=n%7<5or n%3!=n%5<3or-~n%6/4
Try it online!
Uses the same characterization as Robin Ryder. The check n%2!=n%3<2 is shortened to -~n%6/4. Writing out the three conditions turned out shorter than writing a general one:
46 bytes
lambda n:any(n%p!=n%(p+1|1)<p for p in[2,3,5])
Try it online!
answered Apr 5 at 0:34
xnorxnor
94.3k18192452
$endgroup$
add a comment |
$begingroup$
Python 2, 41 bytes
lambda n:n%5!=n%7<5or n%3!=n%5<3or-~n%6/4
Try it online!
Uses the same characterization as Robin Ryder. The check n%2!=n%3<2 is shortened to -~n%6/4. Writing out the three conditions turned out shorter than writing a general one:
46 bytes
lambda n:any(n%p!=n%(p+1|1)<p for p in[2,3,5])
Try it online!
answered Apr 5 at 0:34
xnorxnor
94.3k18192452
$endgroup$
add a comment |
$begingroup$
Python 2, 41 bytes
lambda n:n%5!=n%7<5or n%3!=n%5<3or-~n%6/4
Try it online!
Uses the same characterization as Robin Ryder. The check n%2!=n%3<2 is shortened to -~n%6/4. Writing out the three conditions turned out shorter than writing a general one:
46 bytes
lambda n:any(n%p!=n%(p+1|1)<p for p in[2,3,5])
Try it online!
answered Apr 5 at 0:34
xnorxnor
94.3k18192452
$endgroup$
Python 2, 41 bytes
lambda n:n%5!=n%7<5or n%3!=n%5<3or-~n%6/4
Try it online!
Uses the same characterization as Robin Ryder. The check n%2!=n%3<2 is shortened to -~n%6/4. Writing out the three conditions turned out shorter than writing a general one:
46 bytes
lambda n:any(n%p!=n%(p+1|1)<p for p in[2,3,5])
Try it online!
answered Apr 5 at 0:34
xnorxnor
94.3k18192452
edited Apr 5 at 4:16
answered Apr 5 at 0:34
xnorxnor
94.3k18192452
answered Apr 5 at 0:34
xnorxnor
94.3k18192452
answered Apr 5 at 0:34
xnorxnor
94.3k18192452
94.3k18192452
add a comment |
add a comment |
$begingroup$
Haskell, 47 bytes
g.mod
g r|let p?q=r p/=r q&&r q<p=2?3||3?5||5?7
Try it online!
answered Apr 5 at 5:19
xnorxnor
94.3k18192452
$endgroup$
$begingroup$
Can you explain?
$endgroup$
– dfeuer
Apr 5 at 6:24
1
$begingroup$
@dfeuer It's using Robin Ryder's method.
$endgroup$
– Ørjan Johansen
Apr 5 at 17:39
add a comment |
$begingroup$
Haskell, 47 bytes
g.mod
g r|let p?q=r p/=r q&&r q<p=2?3||3?5||5?7
Try it online!
answered Apr 5 at 5:19
xnorxnor
94.3k18192452
$endgroup$
$begingroup$
Can you explain?
$endgroup$
– dfeuer
Apr 5 at 6:24
1
$begingroup$
@dfeuer It's using Robin Ryder's method.
$endgroup$
– Ørjan Johansen
Apr 5 at 17:39
add a comment |
$begingroup$
Haskell, 47 bytes
g.mod
g r|let p?q=r p/=r q&&r q<p=2?3||3?5||5?7
Try it online!
answered Apr 5 at 5:19
xnorxnor
94.3k18192452
$endgroup$
Haskell, 47 bytes
g.mod
g r|let p?q=r p/=r q&&r q<p=2?3||3?5||5?7
Try it online!
answered Apr 5 at 5:19
xnorxnor
94.3k18192452
edited Apr 5 at 5:27
answered Apr 5 at 5:19
xnorxnor
94.3k18192452
answered Apr 5 at 5:19
xnorxnor
94.3k18192452
answered Apr 5 at 5:19
xnorxnor
94.3k18192452
94.3k18192452
$begingroup$
Can you explain?
$endgroup$
– dfeuer
Apr 5 at 6:24
1
$begingroup$
@dfeuer It's using Robin Ryder's method.
$endgroup$
– Ørjan Johansen
Apr 5 at 17:39
add a comment |
$begingroup$
Can you explain?
$endgroup$
– dfeuer
Apr 5 at 6:24
1
$begingroup$
@dfeuer It's using Robin Ryder's method.
$endgroup$
– Ørjan Johansen
Apr 5 at 17:39
$begingroup$
Can you explain?
$endgroup$
– dfeuer
Apr 5 at 6:24
$begingroup$
Can you explain?
$endgroup$
– dfeuer
Apr 5 at 6:24
1
1
$begingroup$
@dfeuer It's using Robin Ryder's method.
$endgroup$
– Ørjan Johansen
Apr 5 at 17:39
$begingroup$
@dfeuer It's using Robin Ryder's method.
$endgroup$
– Ørjan Johansen
Apr 5 at 17:39
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 125 42 38 36 bytes
n=>n%7<5&5<n%35|n%5<3&3<n%15|-~n%6>3
Direct port of @xnor's answer, which is based off of @RobinRyder's solution.
Saved 4 bytes thanks to @Ørjan Johansen!
Saved 2 more thanks to @Arnauld!
Try it online!
answered Apr 4 at 16:33
Embodiment of IgnoranceEmbodiment of Ignorance
3,024127
$endgroup$
1
$begingroup$
I found a variation that only ties for xnor's languages but helps for this: 38 bytes
$endgroup$
– Ørjan Johansen
Apr 5 at 6:04
1
$begingroup$
Isn't-~n%6/4>0just-~n%6>3?
$endgroup$
– Arnauld
Apr 6 at 9:55
$begingroup$
BTW, this is a JavaScript polyglot.
$endgroup$
– Arnauld
Apr 7 at 7:55
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 125 42 38 36 bytes
n=>n%7<5&5<n%35|n%5<3&3<n%15|-~n%6>3
Direct port of @xnor's answer, which is based off of @RobinRyder's solution.
Saved 4 bytes thanks to @Ørjan Johansen!
Saved 2 more thanks to @Arnauld!
Try it online!
answered Apr 4 at 16:33
Embodiment of IgnoranceEmbodiment of Ignorance
3,024127
$endgroup$
1
$begingroup$
I found a variation that only ties for xnor's languages but helps for this: 38 bytes
$endgroup$
– Ørjan Johansen
Apr 5 at 6:04
1
$begingroup$
Isn't-~n%6/4>0just-~n%6>3?
$endgroup$
– Arnauld
Apr 6 at 9:55
$begingroup$
BTW, this is a JavaScript polyglot.
$endgroup$
– Arnauld
Apr 7 at 7:55
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 125 42 38 36 bytes
n=>n%7<5&5<n%35|n%5<3&3<n%15|-~n%6>3
Direct port of @xnor's answer, which is based off of @RobinRyder's solution.
Saved 4 bytes thanks to @Ørjan Johansen!
Saved 2 more thanks to @Arnauld!
Try it online!
answered Apr 4 at 16:33
Embodiment of IgnoranceEmbodiment of Ignorance
3,024127
$endgroup$
C# (Visual C# Interactive Compiler), 125 42 38 36 bytes
n=>n%7<5&5<n%35|n%5<3&3<n%15|-~n%6>3
Direct port of @xnor's answer, which is based off of @RobinRyder's solution.
Saved 4 bytes thanks to @Ørjan Johansen!
Saved 2 more thanks to @Arnauld!
Try it online!
answered Apr 4 at 16:33
Embodiment of IgnoranceEmbodiment of Ignorance
3,024127
edited Apr 6 at 21:13
answered Apr 4 at 16:33
Embodiment of IgnoranceEmbodiment of Ignorance
3,024127
answered Apr 4 at 16:33
Embodiment of IgnoranceEmbodiment of Ignorance
3,024127
answered Apr 4 at 16:33
Embodiment of IgnoranceEmbodiment of Ignorance
3,024127
3,024127
1
$begingroup$
I found a variation that only ties for xnor's languages but helps for this: 38 bytes
$endgroup$
– Ørjan Johansen
Apr 5 at 6:04
1
$begingroup$
Isn't-~n%6/4>0just-~n%6>3?
$endgroup$
– Arnauld
Apr 6 at 9:55
$begingroup$
BTW, this is a JavaScript polyglot.
$endgroup$
– Arnauld
Apr 7 at 7:55
add a comment |
1
$begingroup$
I found a variation that only ties for xnor's languages but helps for this: 38 bytes
$endgroup$
– Ørjan Johansen
Apr 5 at 6:04
1
$begingroup$
Isn't-~n%6/4>0just-~n%6>3?
$endgroup$
– Arnauld
Apr 6 at 9:55
$begingroup$
BTW, this is a JavaScript polyglot.
$endgroup$
– Arnauld
Apr 7 at 7:55
1
1
$begingroup$
I found a variation that only ties for xnor's languages but helps for this: 38 bytes
$endgroup$
– Ørjan Johansen
Apr 5 at 6:04
$begingroup$
I found a variation that only ties for xnor's languages but helps for this: 38 bytes
$endgroup$
– Ørjan Johansen
Apr 5 at 6:04
1
1
$begingroup$
Isn't
-~n%6/4>0 just -~n%6>3?$endgroup$
– Arnauld
Apr 6 at 9:55
$begingroup$
Isn't
-~n%6/4>0 just -~n%6>3?$endgroup$
– Arnauld
Apr 6 at 9:55
$begingroup$
BTW, this is a JavaScript polyglot.
$endgroup$
– Arnauld
Apr 7 at 7:55
$begingroup$
BTW, this is a JavaScript polyglot.
$endgroup$
– Arnauld
Apr 7 at 7:55
add a comment |
$begingroup$
Wolfram Language (Mathematica), 67 bytes
!FreeQ[Sort/@Table[R[#+k],{k,209}],Sort@R@#]&
R@n_:=n~Mod~{2,3,5,7}
Try it online!
answered Apr 4 at 15:53
J42161217J42161217
14.3k21354
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 67 bytes
!FreeQ[Sort/@Table[R[#+k],{k,209}],Sort@R@#]&
R@n_:=n~Mod~{2,3,5,7}
Try it online!
answered Apr 4 at 15:53
J42161217J42161217
14.3k21354
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 67 bytes
!FreeQ[Sort/@Table[R[#+k],{k,209}],Sort@R@#]&
R@n_:=n~Mod~{2,3,5,7}
Try it online!
answered Apr 4 at 15:53
J42161217J42161217
14.3k21354
$endgroup$
Wolfram Language (Mathematica), 67 bytes
!FreeQ[Sort/@Table[R[#+k],{k,209}],Sort@R@#]&
R@n_:=n~Mod~{2,3,5,7}
Try it online!
answered Apr 4 at 15:53
J42161217J42161217
14.3k21354
edited Apr 4 at 16:00
answered Apr 4 at 15:53
J42161217J42161217
14.3k21354
answered Apr 4 at 15:53
J42161217J42161217
14.3k21354
answered Apr 4 at 15:53
J42161217J42161217
14.3k21354
14.3k21354
add a comment |
add a comment |
$begingroup$
Ruby, 54 bytes
->n{[2,3,5,7].each_cons(2).any?{|l,h|n%l!=n%h&&n%h<l}}
Try it online!
Uses Robin Ryder's clever solution.
answered Apr 4 at 18:59
histocrathistocrat
19.2k43173
$endgroup$
add a comment |
$begingroup$
Ruby, 54 bytes
->n{[2,3,5,7].each_cons(2).any?{|l,h|n%l!=n%h&&n%h<l}}
Try it online!
Uses Robin Ryder's clever solution.
answered Apr 4 at 18:59
histocrathistocrat
19.2k43173
$endgroup$
add a comment |
$begingroup$
Ruby, 54 bytes
->n{[2,3,5,7].each_cons(2).any?{|l,h|n%l!=n%h&&n%h<l}}
Try it online!
Uses Robin Ryder's clever solution.
answered Apr 4 at 18:59
histocrathistocrat
19.2k43173
$endgroup$
Ruby, 54 bytes
->n{[2,3,5,7].each_cons(2).any?{|l,h|n%l!=n%h&&n%h<l}}
Try it online!
Uses Robin Ryder's clever solution.
answered Apr 4 at 18:59
histocrathistocrat
19.2k43173
answered Apr 4 at 18:59
histocrathistocrat
19.2k43173
answered Apr 4 at 18:59
histocrathistocrat
19.2k43173
answered Apr 4 at 18:59
histocrathistocrat
19.2k43173
19.2k43173
add a comment |
add a comment |
$begingroup$
Java (JDK), 36 bytes
n->n%7<5&5<n%35|n%5<3&3<n%15|-~n%6>3
Try it online!
Credits
- Port of xnor's solution, improved by Ørjan Johansen.
answered Apr 5 at 9:07
Olivier GrégoireOlivier Grégoire
9,47511944
$endgroup$
add a comment |
$begingroup$
Java (JDK), 36 bytes
n->n%7<5&5<n%35|n%5<3&3<n%15|-~n%6>3
Try it online!
Credits
- Port of xnor's solution, improved by Ørjan Johansen.
answered Apr 5 at 9:07
Olivier GrégoireOlivier Grégoire
9,47511944
$endgroup$
add a comment |
$begingroup$
Java (JDK), 36 bytes
n->n%7<5&5<n%35|n%5<3&3<n%15|-~n%6>3
Try it online!
Credits
- Port of xnor's solution, improved by Ørjan Johansen.
answered Apr 5 at 9:07
Olivier GrégoireOlivier Grégoire
9,47511944
$endgroup$
Java (JDK), 36 bytes
n->n%7<5&5<n%35|n%5<3&3<n%15|-~n%6>3
Try it online!
Credits
- Port of xnor's solution, improved by Ørjan Johansen.
answered Apr 5 at 9:07
Olivier GrégoireOlivier Grégoire
9,47511944
edited Apr 8 at 7:43
answered Apr 5 at 9:07
Olivier GrégoireOlivier Grégoire
9,47511944
answered Apr 5 at 9:07
Olivier GrégoireOlivier Grégoire
9,47511944
answered Apr 5 at 9:07
Olivier GrégoireOlivier Grégoire
9,47511944
9,47511944
add a comment |
add a comment |
$begingroup$
R, 72 bytes
n=scan();b=c(2,3,5,7);for(i in n+1:209)F=F|all(sort(n%%b)==sort(i%%b));F
Try it online!
answered Apr 4 at 16:42
Aaron HaymanAaron Hayman
3516
$endgroup$
add a comment |
$begingroup$
R, 72 bytes
n=scan();b=c(2,3,5,7);for(i in n+1:209)F=F|all(sort(n%%b)==sort(i%%b));F
Try it online!
answered Apr 4 at 16:42
Aaron HaymanAaron Hayman
3516
$endgroup$
add a comment |
$begingroup$
R, 72 bytes
n=scan();b=c(2,3,5,7);for(i in n+1:209)F=F|all(sort(n%%b)==sort(i%%b));F
Try it online!
answered Apr 4 at 16:42
Aaron HaymanAaron Hayman
3516
$endgroup$
R, 72 bytes
n=scan();b=c(2,3,5,7);for(i in n+1:209)F=F|all(sort(n%%b)==sort(i%%b));F
Try it online!
answered Apr 4 at 16:42
Aaron HaymanAaron Hayman
3516
answered Apr 4 at 16:42
Aaron HaymanAaron Hayman
3516
answered Apr 4 at 16:42
Aaron HaymanAaron Hayman
3516
answered Apr 4 at 16:42
Aaron HaymanAaron Hayman
3516
3516
add a comment |
add a comment |
$begingroup$
PHP, 81 78 72 bytes
while($y<3)if($argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u)die(T);
A riff on @Robin Ryder's answer. Input is via STDIN, output is 'T' if truthy, and empty '' if falsy.
$ echo 3|php -nF euc.php
T
$ echo 5|php -nF euc.php
T
$ echo 2019|php -nF euc.php
T
$ echo 0|php -nF euc.php
$ echo 2|php -nF euc.php
$ echo 1999|php -nF euc.php
Try it online!
Or 73 bytes with 1 or 0 response
while($y<3)$r|=$argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u;echo$r;
$ echo 2019|php -nF euc.php
1
$ echo 1999|php -nF euc.php
0
Try it online (all test cases)!
Original answer, 133 127 bytes
function($n){while(++$k<210)if(($r=function($n){foreach([2,3,5,7]as$d)$o=$n%$d;sort($o);return$o;})($n+$k)==$r($n))return 1;}
Try it online!
answered Apr 4 at 16:01
gwaughgwaugh
2,4281519
$endgroup$
add a comment |
$begingroup$
PHP, 81 78 72 bytes
while($y<3)if($argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u)die(T);
A riff on @Robin Ryder's answer. Input is via STDIN, output is 'T' if truthy, and empty '' if falsy.
$ echo 3|php -nF euc.php
T
$ echo 5|php -nF euc.php
T
$ echo 2019|php -nF euc.php
T
$ echo 0|php -nF euc.php
$ echo 2|php -nF euc.php
$ echo 1999|php -nF euc.php
Try it online!
Or 73 bytes with 1 or 0 response
while($y<3)$r|=$argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u;echo$r;
$ echo 2019|php -nF euc.php
1
$ echo 1999|php -nF euc.php
0
Try it online (all test cases)!
Original answer, 133 127 bytes
function($n){while(++$k<210)if(($r=function($n){foreach([2,3,5,7]as$d)$o=$n%$d;sort($o);return$o;})($n+$k)==$r($n))return 1;}
Try it online!
answered Apr 4 at 16:01
gwaughgwaugh
2,4281519
$endgroup$
add a comment |
$begingroup$
PHP, 81 78 72 bytes
while($y<3)if($argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u)die(T);
A riff on @Robin Ryder's answer. Input is via STDIN, output is 'T' if truthy, and empty '' if falsy.
$ echo 3|php -nF euc.php
T
$ echo 5|php -nF euc.php
T
$ echo 2019|php -nF euc.php
T
$ echo 0|php -nF euc.php
$ echo 2|php -nF euc.php
$ echo 1999|php -nF euc.php
Try it online!
Or 73 bytes with 1 or 0 response
while($y<3)$r|=$argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u;echo$r;
$ echo 2019|php -nF euc.php
1
$ echo 1999|php -nF euc.php
0
Try it online (all test cases)!
Original answer, 133 127 bytes
function($n){while(++$k<210)if(($r=function($n){foreach([2,3,5,7]as$d)$o=$n%$d;sort($o);return$o;})($n+$k)==$r($n))return 1;}
Try it online!
answered Apr 4 at 16:01
gwaughgwaugh
2,4281519
$endgroup$
PHP, 81 78 72 bytes
while($y<3)if($argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u)die(T);
A riff on @Robin Ryder's answer. Input is via STDIN, output is 'T' if truthy, and empty '' if falsy.
$ echo 3|php -nF euc.php
T
$ echo 5|php -nF euc.php
T
$ echo 2019|php -nF euc.php
T
$ echo 0|php -nF euc.php
$ echo 2|php -nF euc.php
$ echo 1999|php -nF euc.php
Try it online!
Or 73 bytes with 1 or 0 response
while($y<3)$r|=$argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u;echo$r;
$ echo 2019|php -nF euc.php
1
$ echo 1999|php -nF euc.php
0
Try it online (all test cases)!
Original answer, 133 127 bytes
function($n){while(++$k<210)if(($r=function($n){foreach([2,3,5,7]as$d)$o=$n%$d;sort($o);return$o;})($n+$k)==$r($n))return 1;}
Try it online!
answered Apr 4 at 16:01
gwaughgwaugh
2,4281519
edited Apr 4 at 19:16
answered Apr 4 at 16:01
gwaughgwaugh
2,4281519
answered Apr 4 at 16:01
gwaughgwaugh
2,4281519
answered Apr 4 at 16:01
gwaughgwaugh
2,4281519
2,4281519
add a comment |
add a comment |
$begingroup$
Python 3, 69 bytes
lambda x:int('4LR2991ODO5GS2974QWH22YTLL3E3I6TDADQG87I0',36)&1<<x%210
Try it online!
Hardcoded
answered Apr 4 at 22:17
attinatattinat
60917
$endgroup$
add a comment |
$begingroup$
Python 3, 69 bytes
lambda x:int('4LR2991ODO5GS2974QWH22YTLL3E3I6TDADQG87I0',36)&1<<x%210
Try it online!
Hardcoded
answered Apr 4 at 22:17
attinatattinat
60917
$endgroup$
add a comment |
$begingroup$
Python 3, 69 bytes
lambda x:int('4LR2991ODO5GS2974QWH22YTLL3E3I6TDADQG87I0',36)&1<<x%210
Try it online!
Hardcoded
answered Apr 4 at 22:17
attinatattinat
60917
$endgroup$
Python 3, 69 bytes
lambda x:int('4LR2991ODO5GS2974QWH22YTLL3E3I6TDADQG87I0',36)&1<<x%210
Try it online!
Hardcoded
answered Apr 4 at 22:17
attinatattinat
60917
answered Apr 4 at 22:17
attinatattinat
60917
answered Apr 4 at 22:17
attinatattinat
60917
answered Apr 4 at 22:17
attinatattinat
60917
60917
add a comment |
add a comment |
$begingroup$
05AB1E, 16 bytes
Ƶ.L+ε‚ε4Åp%{}Ë}à
Try it online or verify all test cases.
Explanation:
Ƶ.L # Create a list in the range [1,209] (which is k)
+ # Add the (implicit) input to each (which is n+k)
ε # Map each value to:
‚ # Pair it with the (implicit) input
ε # Map both to:
4Åp # Get the first 4 primes: [2,3,5,7]
% # Modulo the current number by each of these four (now we have R_n and R_n+k)
{ # Sort the list
}Ë # After the inner map: check if both sorted lists are equal
}à # After the outer map: check if any are truthy by taking the maximum
# (which is output implicitly as result)
See this 05AB1E tip of mine (section How to compress large integers?) to understand why Ƶ. is 209.
answered Apr 5 at 7:29
Kevin CruijssenKevin Cruijssen
43.3k573222
$endgroup$
add a comment |
$begingroup$
05AB1E, 16 bytes
Ƶ.L+ε‚ε4Åp%{}Ë}à
Try it online or verify all test cases.
Explanation:
Ƶ.L # Create a list in the range [1,209] (which is k)
+ # Add the (implicit) input to each (which is n+k)
ε # Map each value to:
‚ # Pair it with the (implicit) input
ε # Map both to:
4Åp # Get the first 4 primes: [2,3,5,7]
% # Modulo the current number by each of these four (now we have R_n and R_n+k)
{ # Sort the list
}Ë # After the inner map: check if both sorted lists are equal
}à # After the outer map: check if any are truthy by taking the maximum
# (which is output implicitly as result)
See this 05AB1E tip of mine (section How to compress large integers?) to understand why Ƶ. is 209.
answered Apr 5 at 7:29
Kevin CruijssenKevin Cruijssen
43.3k573222
$endgroup$
add a comment |
$begingroup$
05AB1E, 16 bytes
Ƶ.L+ε‚ε4Åp%{}Ë}à
Try it online or verify all test cases.
Explanation:
Ƶ.L # Create a list in the range [1,209] (which is k)
+ # Add the (implicit) input to each (which is n+k)
ε # Map each value to:
‚ # Pair it with the (implicit) input
ε # Map both to:
4Åp # Get the first 4 primes: [2,3,5,7]
% # Modulo the current number by each of these four (now we have R_n and R_n+k)
{ # Sort the list
}Ë # After the inner map: check if both sorted lists are equal
}à # After the outer map: check if any are truthy by taking the maximum
# (which is output implicitly as result)
See this 05AB1E tip of mine (section How to compress large integers?) to understand why Ƶ. is 209.
answered Apr 5 at 7:29
Kevin CruijssenKevin Cruijssen
43.3k573222
$endgroup$
05AB1E, 16 bytes
Ƶ.L+ε‚ε4Åp%{}Ë}à
Try it online or verify all test cases.
Explanation:
Ƶ.L # Create a list in the range [1,209] (which is k)
+ # Add the (implicit) input to each (which is n+k)
ε # Map each value to:
‚ # Pair it with the (implicit) input
ε # Map both to:
4Åp # Get the first 4 primes: [2,3,5,7]
% # Modulo the current number by each of these four (now we have R_n and R_n+k)
{ # Sort the list
}Ë # After the inner map: check if both sorted lists are equal
}à # After the outer map: check if any are truthy by taking the maximum
# (which is output implicitly as result)
See this 05AB1E tip of mine (section How to compress large integers?) to understand why Ƶ. is 209.
answered Apr 5 at 7:29
Kevin CruijssenKevin Cruijssen
43.3k573222
answered Apr 5 at 7:29
Kevin CruijssenKevin Cruijssen
43.3k573222
answered Apr 5 at 7:29
Kevin CruijssenKevin Cruijssen
43.3k573222
answered Apr 5 at 7:29
Kevin CruijssenKevin Cruijssen
43.3k573222
43.3k573222
add a comment |
add a comment |
$begingroup$
J, 40 bytes
1 e.(>:+i.@209)-:&(/:~)&(2 3 5 7&|"1 0)]
Try it online!
Brute force...
answered Apr 5 at 8:28
Galen IvanovGalen Ivanov
7,54211034
$endgroup$
add a comment |
$begingroup$
J, 40 bytes
1 e.(>:+i.@209)-:&(/:~)&(2 3 5 7&|"1 0)]
Try it online!
Brute force...
answered Apr 5 at 8:28
Galen IvanovGalen Ivanov
7,54211034
$endgroup$
add a comment |
$begingroup$
J, 40 bytes
1 e.(>:+i.@209)-:&(/:~)&(2 3 5 7&|"1 0)]
Try it online!
Brute force...
answered Apr 5 at 8:28
Galen IvanovGalen Ivanov
7,54211034
$endgroup$
J, 40 bytes
1 e.(>:+i.@209)-:&(/:~)&(2 3 5 7&|"1 0)]
Try it online!
Brute force...
answered Apr 5 at 8:28
Galen IvanovGalen Ivanov
7,54211034
answered Apr 5 at 8:28
Galen IvanovGalen Ivanov
7,54211034
answered Apr 5 at 8:28
Galen IvanovGalen Ivanov
7,54211034
answered Apr 5 at 8:28
Galen IvanovGalen Ivanov
7,54211034
7,54211034
add a comment |
add a comment |
$begingroup$
Wolfram Language (Mathematica), 56 bytes
Or@@(Min[s-#]>0&/@Rest@Permutations@Mod[#,s={2,3,5,7}])&
Try it online!
Finds all non-identity permutations of the remainders of the input modulo 2, 3, 5, 7, and checks if any of them are below {2,3,5,7} in each coordinate. Note that Or@@{} is False.
answered Apr 6 at 0:57
Misha LavrovMisha Lavrov
4,271424
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 56 bytes
Or@@(Min[s-#]>0&/@Rest@Permutations@Mod[#,s={2,3,5,7}])&
Try it online!
Finds all non-identity permutations of the remainders of the input modulo 2, 3, 5, 7, and checks if any of them are below {2,3,5,7} in each coordinate. Note that Or@@{} is False.
answered Apr 6 at 0:57
Misha LavrovMisha Lavrov
4,271424
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 56 bytes
Or@@(Min[s-#]>0&/@Rest@Permutations@Mod[#,s={2,3,5,7}])&
Try it online!
Finds all non-identity permutations of the remainders of the input modulo 2, 3, 5, 7, and checks if any of them are below {2,3,5,7} in each coordinate. Note that Or@@{} is False.
answered Apr 6 at 0:57
Misha LavrovMisha Lavrov
4,271424
$endgroup$
Wolfram Language (Mathematica), 56 bytes
Or@@(Min[s-#]>0&/@Rest@Permutations@Mod[#,s={2,3,5,7}])&
Try it online!
Finds all non-identity permutations of the remainders of the input modulo 2, 3, 5, 7, and checks if any of them are below {2,3,5,7} in each coordinate. Note that Or@@{} is False.
answered Apr 6 at 0:57
Misha LavrovMisha Lavrov
4,271424
answered Apr 6 at 0:57
Misha LavrovMisha Lavrov
4,271424
answered Apr 6 at 0:57
Misha LavrovMisha Lavrov
4,271424
answered Apr 6 at 0:57
Misha LavrovMisha Lavrov
4,271424
4,271424
add a comment |
add a comment |
$begingroup$
Jelly, 15 bytes
8ÆR©PḶ+%Ṣ¥€®ċḢ$
Try it online!
I’m sure there’s a golfier answer. I’ve interpreted a truthy value as being anything that isn’t zero, so here it’s the number of possible values of k. If it needs to be two distinct values that costs me a further byte.
Explanation
8ÆR | Primes less than 8 [2,3,5,7]
© | Copy to register
P | Product [210]
Ḷ | Lowered range [0, 1, ..., 208, 209]
+ | Add to input
¥€ | For each of these 210 numbers...
% ® | Modulo 2, 3, 5, 7
Ṣ | And sort
ċḢ$ | Count how many match the first (input) number’s remainders
answered Apr 4 at 17:59
Nick KennedyNick Kennedy
1,87149
$endgroup$
1
$begingroup$
All good regarding truthy vs falsey. Using the meta agreed definition of truthy and falsey (effectively "what does the language's if-else construct do if there is one) zero is falsey and non-zeros are truthy (?is the if-else construct in Jelly; for some languages it's a harder question).
$endgroup$
– Jonathan Allan
Apr 4 at 18:08
$begingroup$
Oh, and you could get distinct values for no cost withḢe$if you wanted :)
$endgroup$
– Jonathan Allan
Apr 4 at 18:11
$begingroup$
@JonathanAllan yes of course, thanks. :)
$endgroup$
– Nick Kennedy
Apr 4 at 18:12
add a comment |
$begingroup$
Jelly, 15 bytes
8ÆR©PḶ+%Ṣ¥€®ċḢ$
Try it online!
I’m sure there’s a golfier answer. I’ve interpreted a truthy value as being anything that isn’t zero, so here it’s the number of possible values of k. If it needs to be two distinct values that costs me a further byte.
Explanation
8ÆR | Primes less than 8 [2,3,5,7]
© | Copy to register
P | Product [210]
Ḷ | Lowered range [0, 1, ..., 208, 209]
+ | Add to input
¥€ | For each of these 210 numbers...
% ® | Modulo 2, 3, 5, 7
Ṣ | And sort
ċḢ$ | Count how many match the first (input) number’s remainders
answered Apr 4 at 17:59
Nick KennedyNick Kennedy
1,87149
$endgroup$
1
$begingroup$
All good regarding truthy vs falsey. Using the meta agreed definition of truthy and falsey (effectively "what does the language's if-else construct do if there is one) zero is falsey and non-zeros are truthy (?is the if-else construct in Jelly; for some languages it's a harder question).
$endgroup$
– Jonathan Allan
Apr 4 at 18:08
$begingroup$
Oh, and you could get distinct values for no cost withḢe$if you wanted :)
$endgroup$
– Jonathan Allan
Apr 4 at 18:11
$begingroup$
@JonathanAllan yes of course, thanks. :)
$endgroup$
– Nick Kennedy
Apr 4 at 18:12
add a comment |
$begingroup$
Jelly, 15 bytes
8ÆR©PḶ+%Ṣ¥€®ċḢ$
Try it online!
I’m sure there’s a golfier answer. I’ve interpreted a truthy value as being anything that isn’t zero, so here it’s the number of possible values of k. If it needs to be two distinct values that costs me a further byte.
Explanation
8ÆR | Primes less than 8 [2,3,5,7]
© | Copy to register
P | Product [210]
Ḷ | Lowered range [0, 1, ..., 208, 209]
+ | Add to input
¥€ | For each of these 210 numbers...
% ® | Modulo 2, 3, 5, 7
Ṣ | And sort
ċḢ$ | Count how many match the first (input) number’s remainders
answered Apr 4 at 17:59
Nick KennedyNick Kennedy
1,87149
$endgroup$
Jelly, 15 bytes
8ÆR©PḶ+%Ṣ¥€®ċḢ$
Try it online!
I’m sure there’s a golfier answer. I’ve interpreted a truthy value as being anything that isn’t zero, so here it’s the number of possible values of k. If it needs to be two distinct values that costs me a further byte.
Explanation
8ÆR | Primes less than 8 [2,3,5,7]
© | Copy to register
P | Product [210]
Ḷ | Lowered range [0, 1, ..., 208, 209]
+ | Add to input
¥€ | For each of these 210 numbers...
% ® | Modulo 2, 3, 5, 7
Ṣ | And sort
ċḢ$ | Count how many match the first (input) number’s remainders
answered Apr 4 at 17:59
Nick KennedyNick Kennedy
1,87149
edited Apr 9 at 11:41
answered Apr 4 at 17:59
Nick KennedyNick Kennedy
1,87149
answered Apr 4 at 17:59
Nick KennedyNick Kennedy
1,87149
answered Apr 4 at 17:59
Nick KennedyNick Kennedy
1,87149
1,87149
1
$begingroup$
All good regarding truthy vs falsey. Using the meta agreed definition of truthy and falsey (effectively "what does the language's if-else construct do if there is one) zero is falsey and non-zeros are truthy (?is the if-else construct in Jelly; for some languages it's a harder question).
$endgroup$
– Jonathan Allan
Apr 4 at 18:08
$begingroup$
Oh, and you could get distinct values for no cost withḢe$if you wanted :)
$endgroup$
– Jonathan Allan
Apr 4 at 18:11
$begingroup$
@JonathanAllan yes of course, thanks. :)
$endgroup$
– Nick Kennedy
Apr 4 at 18:12
add a comment |
1
$begingroup$
All good regarding truthy vs falsey. Using the meta agreed definition of truthy and falsey (effectively "what does the language's if-else construct do if there is one) zero is falsey and non-zeros are truthy (?is the if-else construct in Jelly; for some languages it's a harder question).
$endgroup$
– Jonathan Allan
Apr 4 at 18:08
$begingroup$
Oh, and you could get distinct values for no cost withḢe$if you wanted :)
$endgroup$
– Jonathan Allan
Apr 4 at 18:11
$begingroup$
@JonathanAllan yes of course, thanks. :)
$endgroup$
– Nick Kennedy
Apr 4 at 18:12
1
1
$begingroup$
All good regarding truthy vs falsey. Using the meta agreed definition of truthy and falsey (effectively "what does the language's if-else construct do if there is one) zero is falsey and non-zeros are truthy (
? is the if-else construct in Jelly; for some languages it's a harder question).$endgroup$
– Jonathan Allan
Apr 4 at 18:08
$begingroup$
All good regarding truthy vs falsey. Using the meta agreed definition of truthy and falsey (effectively "what does the language's if-else construct do if there is one) zero is falsey and non-zeros are truthy (
? is the if-else construct in Jelly; for some languages it's a harder question).$endgroup$
– Jonathan Allan
Apr 4 at 18:08
$begingroup$
Oh, and you could get distinct values for no cost with
Ḣe$ if you wanted :)$endgroup$
– Jonathan Allan
Apr 4 at 18:11
$begingroup$
Oh, and you could get distinct values for no cost with
Ḣe$ if you wanted :)$endgroup$
– Jonathan Allan
Apr 4 at 18:11
$begingroup$
@JonathanAllan yes of course, thanks. :)
$endgroup$
– Nick Kennedy
Apr 4 at 18:12
$begingroup$
@JonathanAllan yes of course, thanks. :)
$endgroup$
– Nick Kennedy
Apr 4 at 18:12
add a comment |
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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