Pointer jumping
$begingroup$
Suppose we have an array $texttt{ps}$ of length $n$ with pointers pointing to some location in the array: The process of "pointer jumping" will set every pointer to the location the pointer it points to points to.
For the purpose of this challenge a pointer is the (zero-based) index of an element of the array, this implies that every element in the array will be greater or equal to $0$ and less than $n$. Using this notation the process can be formulated as follows:
for i = 0..(n-1) {
ps[i] = ps[ps[i]]
}
This means (for this challenge) that the pointers are updated in-place in sequential order (ie. lower indices first).
Example
Let's work through an example, $texttt{ps = [2,1,4,1,3,2]}$:
$$
texttt{i = 0}: text{the element at position }texttt{ps[0] = 2}text{ points to }texttt{4} \ to texttt{ps = [4,1,4,1,3,2]} \
texttt{i = 1}: text{the element at position }texttt{ps[1] = 1}text{ points to }texttt{1} \ to texttt{ps = [4,1,4,1,3,2]} \
texttt{i = 2}: text{the element at position }texttt{ps[2] = 4}text{ points to }texttt{3} \ to texttt{ps = [4,1,3,1,3,2]} \
texttt{i = 3}: text{the element at position }texttt{ps[3] = 1}text{ points to }texttt{1} \ to texttt{ps = [4,1,3,1,3,2]} \
texttt{i = 4}: text{the element at position }texttt{ps[4] = 3}text{ points to }texttt{1} \ to texttt{ps = [4,1,3,1,1,2]} \
texttt{i = 5}: text{the element at position }texttt{ps[5] = 2}text{ points to }texttt{3} \ to texttt{ps = [4,1,3,1,1,3]}
$$
So after one iteration of "pointer jumping" we get the array $texttt{[4,1,3,1,1,3]}$.
Challenge
Given an array with indices output the array obtained by iterating the above described pointer jumping until the array does not change anymore.
Rules
Your program/function will take and return/output the same type, a list/vector/array etc. which
- is guaranteed to be non-empty and
- is guaranteed to only contain entries $0 leq p < n$.
Variants: You may choose
- to use 1-based indexing or
- use actual pointers,
however you should mention this in your submission.
Test cases
[0] → [0]
[1,0] → [0,0]
[1,2,3,4,0] → [2,2,2,2,2]
[0,1,1,1,0,3] → [0,1,1,1,0,1]
[4,1,3,0,3,2] → [3,1,3,3,3,3]
[5,1,2,0,4,5,6] → [5,1,2,5,4,5,6]
[9,9,9,2,5,4,4,5,8,1,0,0] → [1,1,1,1,4,4,4,4,8,1,1,1]
code-golf array-manipulation graph-theory
asked 2 days ago
BMOBMO
12k22290
$endgroup$
add a comment |
$begingroup$
Suppose we have an array $texttt{ps}$ of length $n$ with pointers pointing to some location in the array: The process of "pointer jumping" will set every pointer to the location the pointer it points to points to.
For the purpose of this challenge a pointer is the (zero-based) index of an element of the array, this implies that every element in the array will be greater or equal to $0$ and less than $n$. Using this notation the process can be formulated as follows:
for i = 0..(n-1) {
ps[i] = ps[ps[i]]
}
This means (for this challenge) that the pointers are updated in-place in sequential order (ie. lower indices first).
Example
Let's work through an example, $texttt{ps = [2,1,4,1,3,2]}$:
$$
texttt{i = 0}: text{the element at position }texttt{ps[0] = 2}text{ points to }texttt{4} \ to texttt{ps = [4,1,4,1,3,2]} \
texttt{i = 1}: text{the element at position }texttt{ps[1] = 1}text{ points to }texttt{1} \ to texttt{ps = [4,1,4,1,3,2]} \
texttt{i = 2}: text{the element at position }texttt{ps[2] = 4}text{ points to }texttt{3} \ to texttt{ps = [4,1,3,1,3,2]} \
texttt{i = 3}: text{the element at position }texttt{ps[3] = 1}text{ points to }texttt{1} \ to texttt{ps = [4,1,3,1,3,2]} \
texttt{i = 4}: text{the element at position }texttt{ps[4] = 3}text{ points to }texttt{1} \ to texttt{ps = [4,1,3,1,1,2]} \
texttt{i = 5}: text{the element at position }texttt{ps[5] = 2}text{ points to }texttt{3} \ to texttt{ps = [4,1,3,1,1,3]}
$$
So after one iteration of "pointer jumping" we get the array $texttt{[4,1,3,1,1,3]}$.
Challenge
Given an array with indices output the array obtained by iterating the above described pointer jumping until the array does not change anymore.
Rules
Your program/function will take and return/output the same type, a list/vector/array etc. which
- is guaranteed to be non-empty and
- is guaranteed to only contain entries $0 leq p < n$.
Variants: You may choose
- to use 1-based indexing or
- use actual pointers,
however you should mention this in your submission.
Test cases
[0] → [0]
[1,0] → [0,0]
[1,2,3,4,0] → [2,2,2,2,2]
[0,1,1,1,0,3] → [0,1,1,1,0,1]
[4,1,3,0,3,2] → [3,1,3,3,3,3]
[5,1,2,0,4,5,6] → [5,1,2,5,4,5,6]
[9,9,9,2,5,4,4,5,8,1,0,0] → [1,1,1,1,4,4,4,4,8,1,1,1]
code-golf array-manipulation graph-theory
asked 2 days ago
BMOBMO
12k22290
$endgroup$
$begingroup$
Related: Jump the array
$endgroup$
– BMO
2 days ago
$begingroup$
Are we allowed to take the lengthnas additional input?
$endgroup$
– Kevin Cruijssen
yesterday
2
$begingroup$
@KevinCruijssen, see this meta discussion.
$endgroup$
– Shaggy
yesterday
$begingroup$
It's too bad the entries need to be updated sequentially; if they could be updated simultaneously, Mathematica would have the 21-character solution#[[#]]&~FixedPoint~#&.
$endgroup$
– Greg Martin
17 hours ago
add a comment |
$begingroup$
Suppose we have an array $texttt{ps}$ of length $n$ with pointers pointing to some location in the array: The process of "pointer jumping" will set every pointer to the location the pointer it points to points to.
For the purpose of this challenge a pointer is the (zero-based) index of an element of the array, this implies that every element in the array will be greater or equal to $0$ and less than $n$. Using this notation the process can be formulated as follows:
for i = 0..(n-1) {
ps[i] = ps[ps[i]]
}
This means (for this challenge) that the pointers are updated in-place in sequential order (ie. lower indices first).
Example
Let's work through an example, $texttt{ps = [2,1,4,1,3,2]}$:
$$
texttt{i = 0}: text{the element at position }texttt{ps[0] = 2}text{ points to }texttt{4} \ to texttt{ps = [4,1,4,1,3,2]} \
texttt{i = 1}: text{the element at position }texttt{ps[1] = 1}text{ points to }texttt{1} \ to texttt{ps = [4,1,4,1,3,2]} \
texttt{i = 2}: text{the element at position }texttt{ps[2] = 4}text{ points to }texttt{3} \ to texttt{ps = [4,1,3,1,3,2]} \
texttt{i = 3}: text{the element at position }texttt{ps[3] = 1}text{ points to }texttt{1} \ to texttt{ps = [4,1,3,1,3,2]} \
texttt{i = 4}: text{the element at position }texttt{ps[4] = 3}text{ points to }texttt{1} \ to texttt{ps = [4,1,3,1,1,2]} \
texttt{i = 5}: text{the element at position }texttt{ps[5] = 2}text{ points to }texttt{3} \ to texttt{ps = [4,1,3,1,1,3]}
$$
So after one iteration of "pointer jumping" we get the array $texttt{[4,1,3,1,1,3]}$.
Challenge
Given an array with indices output the array obtained by iterating the above described pointer jumping until the array does not change anymore.
Rules
Your program/function will take and return/output the same type, a list/vector/array etc. which
- is guaranteed to be non-empty and
- is guaranteed to only contain entries $0 leq p < n$.
Variants: You may choose
- to use 1-based indexing or
- use actual pointers,
however you should mention this in your submission.
Test cases
[0] → [0]
[1,0] → [0,0]
[1,2,3,4,0] → [2,2,2,2,2]
[0,1,1,1,0,3] → [0,1,1,1,0,1]
[4,1,3,0,3,2] → [3,1,3,3,3,3]
[5,1,2,0,4,5,6] → [5,1,2,5,4,5,6]
[9,9,9,2,5,4,4,5,8,1,0,0] → [1,1,1,1,4,4,4,4,8,1,1,1]
code-golf array-manipulation graph-theory
asked 2 days ago
BMOBMO
12k22290
$endgroup$
Suppose we have an array $texttt{ps}$ of length $n$ with pointers pointing to some location in the array: The process of "pointer jumping" will set every pointer to the location the pointer it points to points to.
For the purpose of this challenge a pointer is the (zero-based) index of an element of the array, this implies that every element in the array will be greater or equal to $0$ and less than $n$. Using this notation the process can be formulated as follows:
for i = 0..(n-1) {
ps[i] = ps[ps[i]]
}
This means (for this challenge) that the pointers are updated in-place in sequential order (ie. lower indices first).
Example
Let's work through an example, $texttt{ps = [2,1,4,1,3,2]}$:
$$
texttt{i = 0}: text{the element at position }texttt{ps[0] = 2}text{ points to }texttt{4} \ to texttt{ps = [4,1,4,1,3,2]} \
texttt{i = 1}: text{the element at position }texttt{ps[1] = 1}text{ points to }texttt{1} \ to texttt{ps = [4,1,4,1,3,2]} \
texttt{i = 2}: text{the element at position }texttt{ps[2] = 4}text{ points to }texttt{3} \ to texttt{ps = [4,1,3,1,3,2]} \
texttt{i = 3}: text{the element at position }texttt{ps[3] = 1}text{ points to }texttt{1} \ to texttt{ps = [4,1,3,1,3,2]} \
texttt{i = 4}: text{the element at position }texttt{ps[4] = 3}text{ points to }texttt{1} \ to texttt{ps = [4,1,3,1,1,2]} \
texttt{i = 5}: text{the element at position }texttt{ps[5] = 2}text{ points to }texttt{3} \ to texttt{ps = [4,1,3,1,1,3]}
$$
So after one iteration of "pointer jumping" we get the array $texttt{[4,1,3,1,1,3]}$.
Challenge
Given an array with indices output the array obtained by iterating the above described pointer jumping until the array does not change anymore.
Rules
Your program/function will take and return/output the same type, a list/vector/array etc. which
- is guaranteed to be non-empty and
- is guaranteed to only contain entries $0 leq p < n$.
Variants: You may choose
- to use 1-based indexing or
- use actual pointers,
however you should mention this in your submission.
Test cases
[0] → [0]
[1,0] → [0,0]
[1,2,3,4,0] → [2,2,2,2,2]
[0,1,1,1,0,3] → [0,1,1,1,0,1]
[4,1,3,0,3,2] → [3,1,3,3,3,3]
[5,1,2,0,4,5,6] → [5,1,2,5,4,5,6]
[9,9,9,2,5,4,4,5,8,1,0,0] → [1,1,1,1,4,4,4,4,8,1,1,1]
code-golf array-manipulation graph-theory
code-golf array-manipulation graph-theory
asked 2 days ago
BMOBMO
12k22290
asked 2 days ago
BMOBMO
12k22290
edited 2 days ago
BMO
asked 2 days ago
BMOBMO
12k22290
asked 2 days ago
BMOBMO
12k22290
asked 2 days ago
BMOBMO
12k22290
12k22290
$begingroup$
Related: Jump the array
$endgroup$
– BMO
2 days ago
$begingroup$
Are we allowed to take the lengthnas additional input?
$endgroup$
– Kevin Cruijssen
yesterday
2
$begingroup$
@KevinCruijssen, see this meta discussion.
$endgroup$
– Shaggy
yesterday
$begingroup$
It's too bad the entries need to be updated sequentially; if they could be updated simultaneously, Mathematica would have the 21-character solution#[[#]]&~FixedPoint~#&.
$endgroup$
– Greg Martin
17 hours ago
add a comment |
$begingroup$
Related: Jump the array
$endgroup$
– BMO
2 days ago
$begingroup$
Are we allowed to take the lengthnas additional input?
$endgroup$
– Kevin Cruijssen
yesterday
2
$begingroup$
@KevinCruijssen, see this meta discussion.
$endgroup$
– Shaggy
yesterday
$begingroup$
It's too bad the entries need to be updated sequentially; if they could be updated simultaneously, Mathematica would have the 21-character solution#[[#]]&~FixedPoint~#&.
$endgroup$
– Greg Martin
17 hours ago
$begingroup$
Related: Jump the array
$endgroup$
– BMO
2 days ago
$begingroup$
Related: Jump the array
$endgroup$
– BMO
2 days ago
$begingroup$
Are we allowed to take the length
n as additional input?$endgroup$
– Kevin Cruijssen
yesterday
$begingroup$
Are we allowed to take the length
n as additional input?$endgroup$
– Kevin Cruijssen
yesterday
2
2
$begingroup$
@KevinCruijssen, see this meta discussion.
$endgroup$
– Shaggy
yesterday
$begingroup$
@KevinCruijssen, see this meta discussion.
$endgroup$
– Shaggy
yesterday
$begingroup$
It's too bad the entries need to be updated sequentially; if they could be updated simultaneously, Mathematica would have the 21-character solution
#[[#]]&~FixedPoint~#&.$endgroup$
– Greg Martin
17 hours ago
$begingroup$
It's too bad the entries need to be updated sequentially; if they could be updated simultaneously, Mathematica would have the 21-character solution
#[[#]]&~FixedPoint~#&.$endgroup$
– Greg Martin
17 hours ago
add a comment |
21 Answers
21
active
oldest
votes
$begingroup$
JavaScript, 36 bytes
Modifies the original input array.
a=>a.map(_=>a.map((x,y)=>a[y]=a[x]))
Try it online
answered 2 days ago
ShaggyShaggy
19.5k21666
$endgroup$
add a comment |
$begingroup$
Python 2, 53 bytes
def f(l):L=len(l);i=0;exec'l[i]=l[l[i]];i=-~i%L;'*L*L
Try it online!
-6 thanks to HyperNeutrino.
Alters l to the result in place.
answered 2 days ago
Erik the OutgolferErik the Outgolfer
31.6k429103
$endgroup$
add a comment |
$begingroup$
JavaScript (ES6), 41 bytes
f=a=>a+''==a.map((x,i)=>a[i]=a[x])?a:f(a)
Try it online!
answered 2 days ago
ArnauldArnauld
74k690310
$endgroup$
$begingroup$
Gah! I was waiting for this challenge to be posted so I could post the exact same solution : Damn your ninja skills! :p
$endgroup$
– Shaggy
2 days ago
2
$begingroup$
@shaggy 🐱👤(this is supposed to be a Ninja Cat ... but it's probably not supported everywhere)
$endgroup$
– Arnauld
2 days ago
add a comment |
$begingroup$
Haskell, 56 bytes
foldr(_->(#))=<<id
x#a@(b:c)=(x++[(x++a)!!b])#c
x#_=x
Haskell and in-place updates are a bad match.
Try it online!
answered 2 days ago
niminimi
31.7k32285
$endgroup$
add a comment |
$begingroup$
C++14 (gcc), 61 bytes
As unnamed generic lambda. Requires sequential containers like std::vector.
(auto&c){auto d=c;do{d=c;for(auto&x:c)x=c[x];}while(d!=c);}
Try it online!
answered 2 days ago
BierpfurzBierpfurz
1113
$endgroup$
add a comment |
$begingroup$
Swift, 68 53 bytes
{a in for _ in a{var i=0;a.forEach{a[i]=a[$0];i+=1}}}
Try it online!
-15 thanks to BMO
answered yesterday
SeanSean
414
New contributor
Sean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Welcome to PPCG! I don't know Swift, but on codegolf.SE the default is to accept typed lambda-functions which I guess a closure would count as. So this can be 53 bytes (you don't need to countf=). Enjoy your stay here!
$endgroup$
– BMO
yesterday
$begingroup$
Thank you for the welcome and advice which I have used to update my answer.
$endgroup$
– Sean
yesterday
add a comment |
$begingroup$
Japt, 17 bytes
®
£hYUgX
eV ?U:ß
Try all test cases
This feels like it should be shorter, but unfortunately my initial thought of UmgU doesn't work because each g accesses the original U rather than modifying it at each step. Preserving different components appropriately cost a handful of bytes as well.
Explanation:
#Blank line preserves input in U long enough for the next line
® #Copy U into V to preserve its original value
£hY #Modify U in-place by replacing each element X with...
UgX #The value from the current U at the index X
eV ?U #If the modified U is identical to the copy V, output it
:ß #Otherwise start again with the modified U as the new input
answered 2 days ago
Kamil DrakariKamil Drakari
3,191416
$endgroup$
add a comment |
$begingroup$
05AB1E (legacy), 8 bytes
ΔDvDyèNǝ
Try it online!
Explanation
Δ # apply until the top of the stack stops changing
D # duplicate current list
v # for each (element y, index N) in the list
Dyè # get the element at index y
Nǝ # and insert it at index N
05AB1E, 14 bytes
[D©vDyèNǝ}D®Q#
Try it online!
answered 2 days ago
EmignaEmigna
45.8k432139
$endgroup$
add a comment |
$begingroup$
Java 8, 105 54 bytes
a->{for(int l=a.length,i=0;i<l*l;)a[i%l]=a[a[i++%l]];}
Modifies the input-array instead of returning a new one to save bytes.
-51 bytes by just modifying the array $length^2$ times, instead of until it no longer changes.
Try it online.
Explanation:
a->{ // Method with integer-array parameter and no return-type
int l=a.length, // Length of the input-array
i=0;i<l*l;) // Loop `i` in the range [0, length squared):
a[i%l]= // Set the (`i` modulo-length)'th item in the array to:
a[ // The `p`'th value of the input-array,
a[i++%l]];} // where `p` is the (`i` modulo-length)'th value of the array
answered yesterday
Kevin CruijssenKevin Cruijssen
36.8k555193
$endgroup$
add a comment |
$begingroup$
C (gcc), 32-bit, 49 bytes
f(p,n,i)int**p;{for(i=0;i<n*n;)p[i++%n]=*p[i%n];}
Try it online!
Uses pointers.
50 bytes with integers:
f(p,n,i)int*p;{for(i=0;i<n*n;)p[i++%n]=p[p[i%n]];}
Try it online!
answered 2 days ago
nwellnhofnwellnhof
6,66511126
$endgroup$
add a comment |
$begingroup$
Japt, 15 13 7 bytes
Modifies the original input array.
££hYXgU
Try it (additional bytes are to write the modified input to the console)
££hYXgU
£ :Map
£ : Map each X at index Y
hY : Replace the element at index Y
XgU : With the element at index X
answered 2 days ago
ShaggyShaggy
19.5k21666
$endgroup$
add a comment |
$begingroup$
Ruby, 37 34 bytes
->a{a.size.times{a.map!{|x|a[x]}}}
Try it online!
Returns by modifying the input array in-place.
answered 2 days ago
Kirill L.Kirill L.
3,9451319
$endgroup$
add a comment |
$begingroup$
Red, 63 bytes
func[b][loop(l: length? b)* l[repeat i l[b/:i: b/(1 + b/:i)]]b]
Try it online!
Modifies the array in place
answered yesterday
Galen IvanovGalen Ivanov
6,64711032
$endgroup$
add a comment |
$begingroup$
R, 60 58 bytes
-2 bytes thanks to @digEmAll for reading the rules.
function(x,n=sum(x|1)){for(i in rep(1:n,n))x[i]=x[x[i]];x}
Try it online!
1-indexed.
n is the length of the input array.
rep(1:n,n) replicates 1:n n times (e.g. n=3 => 1,2,3,1,2,3,1,2,3)
Loop through the array n times. Steady state will be acheived by then for sure, in fact by the end of the n-1st time through I think. The proof is left to the reader.
answered 2 days ago
ngmngm
3,36924
$endgroup$
1
$begingroup$
I think you can remove the+1and simply take the 1-based input, the post states : You may choose to use 1-based indexing
$endgroup$
– digEmAll
yesterday
$begingroup$
-4 by switching toscan()for input. I always feel like myscan()solutions are suboptimal, so keep an eye out a shorter way to assignxandntogether:n=length(x<-scan());for(i in rep(1:n,n))x[i]=x[x[i]];xTry it online!
$endgroup$
– CriminallyVulgar
13 hours ago
add a comment |
$begingroup$
Common Lisp, 59 58 bytes
(lambda(a)(dolist(j a)(map-into a(lambda(x)(elt a x))a))a)
Try it online!
answered yesterday
RenzoRenzo
1,730516
$endgroup$
add a comment |
$begingroup$
Clojure, 136 bytes
(defn j[a](let[f(fn[a](loop[i 0 a a](if(= i(count a))a(recur(inc i)(assoc a i(a(a i)))))))](loop[p nil a a](if(= p a)a(recur a(f a))))))
Try it online!
answered 2 days ago
Ethan McCueEthan McCue
212
New contributor
Ethan McCue is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Hello and welcome to PPCG. Would it be possible for you to provide a link to an online interpreter such that one could easily verify your solution? Furthermore, canloop [not becomeloop[?
$endgroup$
– Jonathan Frech
2 days ago
1
$begingroup$
The most recent edit should fix the test failures. Sorry for the inconvenience everybody.
$endgroup$
– Ethan McCue
7 hours ago
add a comment |
$begingroup$
Perl 5, 35 34 26 bytes
using the fact that convergence is reach at most for the size number of iterations
$_=$F[$_]for@F x@F;$_="@F"
26 bytes
$_=$F[$_]for@F;/@F/ or$_="@F",redo
34 bytes
$_=$F[$_]for@F;$_="@F",redo if!/@F/
35 bytes
answered 2 days ago
Nahuel FouilleulNahuel Fouilleul
1,84028
$endgroup$
add a comment |
$begingroup$
Clojure, 88 bytes
#(reduce(fn[x i](assoc x i(get x(get x i))))%(flatten(repeat(count %)(range(count %)))))
Try it online!
answered yesterday
Kirill L.Kirill L.
3,9451319
$endgroup$
add a comment |
$begingroup$
Charcoal, 16 bytes
ＦθＦＬθ§≔θκ§θ§θκＩθ
Try it online! Link is to verbose version of code. Sadly all the usual mapping functions only operate on a copy of the array with the result being that they just permute the elements rather than jumping them, so the code has to do everything manually. Explanation:
Ｆθ
Repeat the inner loop once for each element. This just ensures that the result stabilises.
ＦＬθ
Loop over the array indices.
§≔θκ§θ§θκ
Get the array element at the current index, use that to index into the array, and replace the current element with that value.
Ｉθ
Cast the elements to string and implicitly print each on their own line.
answered yesterday
NeilNeil
80k744178
$endgroup$
add a comment |
$begingroup$
Clean, 80 bytes
import StdEnv
limit o iterateb=foldl(l i=updateAt i(l!!(l!!i))l)b(indexList b)
Try it online!
answered yesterday
ΟurousΟurous
6,63211033
$endgroup$
add a comment |
$begingroup$
F#, 74 73 bytes
fun(c:'a)->let l=c.Length in(for i in 0..l*l do c.[i%l]<-c.[c.[i%l]]);c
Nothing special. Uses the modulus idea seen in other answers.
answered 9 hours ago
LSM07LSM07
1213
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "200"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f179050%2fpointer-jumping%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
21 Answers
21
active
oldest
votes
21 Answers
21
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
JavaScript, 36 bytes
Modifies the original input array.
a=>a.map(_=>a.map((x,y)=>a[y]=a[x]))
Try it online
answered 2 days ago
ShaggyShaggy
19.5k21666
$endgroup$
add a comment |
$begingroup$
JavaScript, 36 bytes
Modifies the original input array.
a=>a.map(_=>a.map((x,y)=>a[y]=a[x]))
Try it online
answered 2 days ago
ShaggyShaggy
19.5k21666
$endgroup$
add a comment |
$begingroup$
JavaScript, 36 bytes
Modifies the original input array.
a=>a.map(_=>a.map((x,y)=>a[y]=a[x]))
Try it online
answered 2 days ago
ShaggyShaggy
19.5k21666
$endgroup$
JavaScript, 36 bytes
Modifies the original input array.
a=>a.map(_=>a.map((x,y)=>a[y]=a[x]))
Try it online
answered 2 days ago
ShaggyShaggy
19.5k21666
answered 2 days ago
ShaggyShaggy
19.5k21666
answered 2 days ago
ShaggyShaggy
19.5k21666
answered 2 days ago
ShaggyShaggy
19.5k21666
19.5k21666
add a comment |
add a comment |
$begingroup$
Python 2, 53 bytes
def f(l):L=len(l);i=0;exec'l[i]=l[l[i]];i=-~i%L;'*L*L
Try it online!
-6 thanks to HyperNeutrino.
Alters l to the result in place.
answered 2 days ago
Erik the OutgolferErik the Outgolfer
31.6k429103
$endgroup$
add a comment |
$begingroup$
Python 2, 53 bytes
def f(l):L=len(l);i=0;exec'l[i]=l[l[i]];i=-~i%L;'*L*L
Try it online!
-6 thanks to HyperNeutrino.
Alters l to the result in place.
answered 2 days ago
Erik the OutgolferErik the Outgolfer
31.6k429103
$endgroup$
add a comment |
$begingroup$
Python 2, 53 bytes
def f(l):L=len(l);i=0;exec'l[i]=l[l[i]];i=-~i%L;'*L*L
Try it online!
-6 thanks to HyperNeutrino.
Alters l to the result in place.
answered 2 days ago
Erik the OutgolferErik the Outgolfer
31.6k429103
$endgroup$
Python 2, 53 bytes
def f(l):L=len(l);i=0;exec'l[i]=l[l[i]];i=-~i%L;'*L*L
Try it online!
-6 thanks to HyperNeutrino.
Alters l to the result in place.
answered 2 days ago
Erik the OutgolferErik the Outgolfer
31.6k429103
edited 2 days ago
answered 2 days ago
Erik the OutgolferErik the Outgolfer
31.6k429103
answered 2 days ago
Erik the OutgolferErik the Outgolfer
31.6k429103
answered 2 days ago
Erik the OutgolferErik the Outgolfer
31.6k429103
31.6k429103
add a comment |
add a comment |
$begingroup$
JavaScript (ES6), 41 bytes
f=a=>a+''==a.map((x,i)=>a[i]=a[x])?a:f(a)
Try it online!
answered 2 days ago
ArnauldArnauld
74k690310
$endgroup$
$begingroup$
Gah! I was waiting for this challenge to be posted so I could post the exact same solution : Damn your ninja skills! :p
$endgroup$
– Shaggy
2 days ago
2
$begingroup$
@shaggy 🐱👤(this is supposed to be a Ninja Cat ... but it's probably not supported everywhere)
$endgroup$
– Arnauld
2 days ago
add a comment |
$begingroup$
JavaScript (ES6), 41 bytes
f=a=>a+''==a.map((x,i)=>a[i]=a[x])?a:f(a)
Try it online!
answered 2 days ago
ArnauldArnauld
74k690310
$endgroup$
$begingroup$
Gah! I was waiting for this challenge to be posted so I could post the exact same solution : Damn your ninja skills! :p
$endgroup$
– Shaggy
2 days ago
2
$begingroup$
@shaggy 🐱👤(this is supposed to be a Ninja Cat ... but it's probably not supported everywhere)
$endgroup$
– Arnauld
2 days ago
add a comment |
$begingroup$
JavaScript (ES6), 41 bytes
f=a=>a+''==a.map((x,i)=>a[i]=a[x])?a:f(a)
Try it online!
answered 2 days ago
ArnauldArnauld
74k690310
$endgroup$
JavaScript (ES6), 41 bytes
f=a=>a+''==a.map((x,i)=>a[i]=a[x])?a:f(a)
Try it online!
answered 2 days ago
ArnauldArnauld
74k690310
answered 2 days ago
ArnauldArnauld
74k690310
answered 2 days ago
ArnauldArnauld
74k690310
answered 2 days ago
ArnauldArnauld
74k690310
74k690310
$begingroup$
Gah! I was waiting for this challenge to be posted so I could post the exact same solution : Damn your ninja skills! :p
$endgroup$
– Shaggy
2 days ago
2
$begingroup$
@shaggy 🐱👤(this is supposed to be a Ninja Cat ... but it's probably not supported everywhere)
$endgroup$
– Arnauld
2 days ago
add a comment |
$begingroup$
Gah! I was waiting for this challenge to be posted so I could post the exact same solution : Damn your ninja skills! :p
$endgroup$
– Shaggy
2 days ago
2
$begingroup$
@shaggy 🐱👤(this is supposed to be a Ninja Cat ... but it's probably not supported everywhere)
$endgroup$
– Arnauld
2 days ago
$begingroup$
Gah! I was waiting for this challenge to be posted so I could post the exact same solution : Damn your ninja skills! :p
$endgroup$
– Shaggy
2 days ago
$begingroup$
Gah! I was waiting for this challenge to be posted so I could post the exact same solution : Damn your ninja skills! :p
$endgroup$
– Shaggy
2 days ago
2
2
$begingroup$
@shaggy 🐱👤(this is supposed to be a Ninja Cat ... but it's probably not supported everywhere)
$endgroup$
– Arnauld
2 days ago
$begingroup$
@shaggy 🐱👤(this is supposed to be a Ninja Cat ... but it's probably not supported everywhere)
$endgroup$
– Arnauld
2 days ago
add a comment |
$begingroup$
Haskell, 56 bytes
foldr(_->(#))=<<id
x#a@(b:c)=(x++[(x++a)!!b])#c
x#_=x
Haskell and in-place updates are a bad match.
Try it online!
answered 2 days ago
niminimi
31.7k32285
$endgroup$
add a comment |
$begingroup$
Haskell, 56 bytes
foldr(_->(#))=<<id
x#a@(b:c)=(x++[(x++a)!!b])#c
x#_=x
Haskell and in-place updates are a bad match.
Try it online!
answered 2 days ago
niminimi
31.7k32285
$endgroup$
add a comment |
$begingroup$
Haskell, 56 bytes
foldr(_->(#))=<<id
x#a@(b:c)=(x++[(x++a)!!b])#c
x#_=x
Haskell and in-place updates are a bad match.
Try it online!
answered 2 days ago
niminimi
31.7k32285
$endgroup$
Haskell, 56 bytes
foldr(_->(#))=<<id
x#a@(b:c)=(x++[(x++a)!!b])#c
x#_=x
Haskell and in-place updates are a bad match.
Try it online!
answered 2 days ago
niminimi
31.7k32285
answered 2 days ago
niminimi
31.7k32285
answered 2 days ago
niminimi
31.7k32285
answered 2 days ago
niminimi
31.7k32285
31.7k32285
add a comment |
add a comment |
$begingroup$
C++14 (gcc), 61 bytes
As unnamed generic lambda. Requires sequential containers like std::vector.
(auto&c){auto d=c;do{d=c;for(auto&x:c)x=c[x];}while(d!=c);}
Try it online!
answered 2 days ago
BierpfurzBierpfurz
1113
$endgroup$
add a comment |
$begingroup$
C++14 (gcc), 61 bytes
As unnamed generic lambda. Requires sequential containers like std::vector.
(auto&c){auto d=c;do{d=c;for(auto&x:c)x=c[x];}while(d!=c);}
Try it online!
answered 2 days ago
BierpfurzBierpfurz
1113
$endgroup$
add a comment |
$begingroup$
C++14 (gcc), 61 bytes
As unnamed generic lambda. Requires sequential containers like std::vector.
(auto&c){auto d=c;do{d=c;for(auto&x:c)x=c[x];}while(d!=c);}
Try it online!
answered 2 days ago
BierpfurzBierpfurz
1113
$endgroup$
C++14 (gcc), 61 bytes
As unnamed generic lambda. Requires sequential containers like std::vector.
(auto&c){auto d=c;do{d=c;for(auto&x:c)x=c[x];}while(d!=c);}
Try it online!
answered 2 days ago
BierpfurzBierpfurz
1113
edited 2 days ago
answered 2 days ago
BierpfurzBierpfurz
1113
answered 2 days ago
BierpfurzBierpfurz
1113
answered 2 days ago
BierpfurzBierpfurz
1113
1113
add a comment |
add a comment |
$begingroup$
Swift, 68 53 bytes
{a in for _ in a{var i=0;a.forEach{a[i]=a[$0];i+=1}}}
Try it online!
-15 thanks to BMO
answered yesterday
SeanSean
414
New contributor
Sean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Welcome to PPCG! I don't know Swift, but on codegolf.SE the default is to accept typed lambda-functions which I guess a closure would count as. So this can be 53 bytes (you don't need to countf=). Enjoy your stay here!
$endgroup$
– BMO
yesterday
$begingroup$
Thank you for the welcome and advice which I have used to update my answer.
$endgroup$
– Sean
yesterday
add a comment |
$begingroup$
Swift, 68 53 bytes
{a in for _ in a{var i=0;a.forEach{a[i]=a[$0];i+=1}}}
Try it online!
-15 thanks to BMO
answered yesterday
SeanSean
414
New contributor
Sean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Welcome to PPCG! I don't know Swift, but on codegolf.SE the default is to accept typed lambda-functions which I guess a closure would count as. So this can be 53 bytes (you don't need to countf=). Enjoy your stay here!
$endgroup$
– BMO
yesterday
$begingroup$
Thank you for the welcome and advice which I have used to update my answer.
$endgroup$
– Sean
yesterday
add a comment |
$begingroup$
Swift, 68 53 bytes
{a in for _ in a{var i=0;a.forEach{a[i]=a[$0];i+=1}}}
Try it online!
-15 thanks to BMO
answered yesterday
SeanSean
414
New contributor
Sean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Swift, 68 53 bytes
{a in for _ in a{var i=0;a.forEach{a[i]=a[$0];i+=1}}}
Try it online!
-15 thanks to BMO
answered yesterday
SeanSean
414
New contributor
Sean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
answered yesterday
SeanSean
414
New contributor
Sean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
SeanSean
414
answered yesterday
SeanSean
414
414
New contributor
Sean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Welcome to PPCG! I don't know Swift, but on codegolf.SE the default is to accept typed lambda-functions which I guess a closure would count as. So this can be 53 bytes (you don't need to countf=). Enjoy your stay here!
$endgroup$
– BMO
yesterday
$begingroup$
Thank you for the welcome and advice which I have used to update my answer.
$endgroup$
– Sean
yesterday
add a comment |
2
$begingroup$
Welcome to PPCG! I don't know Swift, but on codegolf.SE the default is to accept typed lambda-functions which I guess a closure would count as. So this can be 53 bytes (you don't need to countf=). Enjoy your stay here!
$endgroup$
– BMO
yesterday
$begingroup$
Thank you for the welcome and advice which I have used to update my answer.
$endgroup$
– Sean
yesterday
2
2
$begingroup$
Welcome to PPCG! I don't know Swift, but on codegolf.SE the default is to accept typed lambda-functions which I guess a closure would count as. So this can be 53 bytes (you don't need to count
f=). Enjoy your stay here!$endgroup$
– BMO
yesterday
$begingroup$
Welcome to PPCG! I don't know Swift, but on codegolf.SE the default is to accept typed lambda-functions which I guess a closure would count as. So this can be 53 bytes (you don't need to count
f=). Enjoy your stay here!$endgroup$
– BMO
yesterday
$begingroup$
Thank you for the welcome and advice which I have used to update my answer.
$endgroup$
– Sean
yesterday
$begingroup$
Thank you for the welcome and advice which I have used to update my answer.
$endgroup$
– Sean
yesterday
add a comment |
$begingroup$
Japt, 17 bytes
®
£hYUgX
eV ?U:ß
Try all test cases
This feels like it should be shorter, but unfortunately my initial thought of UmgU doesn't work because each g accesses the original U rather than modifying it at each step. Preserving different components appropriately cost a handful of bytes as well.
Explanation:
#Blank line preserves input in U long enough for the next line
® #Copy U into V to preserve its original value
£hY #Modify U in-place by replacing each element X with...
UgX #The value from the current U at the index X
eV ?U #If the modified U is identical to the copy V, output it
:ß #Otherwise start again with the modified U as the new input
answered 2 days ago
Kamil DrakariKamil Drakari
3,191416
$endgroup$
add a comment |
$begingroup$
Japt, 17 bytes
®
£hYUgX
eV ?U:ß
Try all test cases
This feels like it should be shorter, but unfortunately my initial thought of UmgU doesn't work because each g accesses the original U rather than modifying it at each step. Preserving different components appropriately cost a handful of bytes as well.
Explanation:
#Blank line preserves input in U long enough for the next line
® #Copy U into V to preserve its original value
£hY #Modify U in-place by replacing each element X with...
UgX #The value from the current U at the index X
eV ?U #If the modified U is identical to the copy V, output it
:ß #Otherwise start again with the modified U as the new input
answered 2 days ago
Kamil DrakariKamil Drakari
3,191416
$endgroup$
add a comment |
$begingroup$
Japt, 17 bytes
®
£hYUgX
eV ?U:ß
Try all test cases
This feels like it should be shorter, but unfortunately my initial thought of UmgU doesn't work because each g accesses the original U rather than modifying it at each step. Preserving different components appropriately cost a handful of bytes as well.
Explanation:
#Blank line preserves input in U long enough for the next line
® #Copy U into V to preserve its original value
£hY #Modify U in-place by replacing each element X with...
UgX #The value from the current U at the index X
eV ?U #If the modified U is identical to the copy V, output it
:ß #Otherwise start again with the modified U as the new input
answered 2 days ago
Kamil DrakariKamil Drakari
3,191416
$endgroup$
Japt, 17 bytes
®
£hYUgX
eV ?U:ß
Try all test cases
This feels like it should be shorter, but unfortunately my initial thought of UmgU doesn't work because each g accesses the original U rather than modifying it at each step. Preserving different components appropriately cost a handful of bytes as well.
Explanation:
#Blank line preserves input in U long enough for the next line
® #Copy U into V to preserve its original value
£hY #Modify U in-place by replacing each element X with...
UgX #The value from the current U at the index X
eV ?U #If the modified U is identical to the copy V, output it
:ß #Otherwise start again with the modified U as the new input
answered 2 days ago
Kamil DrakariKamil Drakari
3,191416
answered 2 days ago
Kamil DrakariKamil Drakari
3,191416
answered 2 days ago
Kamil DrakariKamil Drakari
3,191416
answered 2 days ago
Kamil DrakariKamil Drakari
3,191416
3,191416
add a comment |
add a comment |
$begingroup$
05AB1E (legacy), 8 bytes
ΔDvDyèNǝ
Try it online!
Explanation
Δ # apply until the top of the stack stops changing
D # duplicate current list
v # for each (element y, index N) in the list
Dyè # get the element at index y
Nǝ # and insert it at index N
05AB1E, 14 bytes
[D©vDyèNǝ}D®Q#
Try it online!
answered 2 days ago
EmignaEmigna
45.8k432139
$endgroup$
add a comment |
$begingroup$
05AB1E (legacy), 8 bytes
ΔDvDyèNǝ
Try it online!
Explanation
Δ # apply until the top of the stack stops changing
D # duplicate current list
v # for each (element y, index N) in the list
Dyè # get the element at index y
Nǝ # and insert it at index N
05AB1E, 14 bytes
[D©vDyèNǝ}D®Q#
Try it online!
answered 2 days ago
EmignaEmigna
45.8k432139
$endgroup$
add a comment |
$begingroup$
05AB1E (legacy), 8 bytes
ΔDvDyèNǝ
Try it online!
Explanation
Δ # apply until the top of the stack stops changing
D # duplicate current list
v # for each (element y, index N) in the list
Dyè # get the element at index y
Nǝ # and insert it at index N
05AB1E, 14 bytes
[D©vDyèNǝ}D®Q#
Try it online!
answered 2 days ago
EmignaEmigna
45.8k432139
$endgroup$
05AB1E (legacy), 8 bytes
ΔDvDyèNǝ
Try it online!
Explanation
Δ # apply until the top of the stack stops changing
D # duplicate current list
v # for each (element y, index N) in the list
Dyè # get the element at index y
Nǝ # and insert it at index N
05AB1E, 14 bytes
[D©vDyèNǝ}D®Q#
Try it online!
answered 2 days ago
EmignaEmigna
45.8k432139
edited 2 days ago
answered 2 days ago
EmignaEmigna
45.8k432139
answered 2 days ago
EmignaEmigna
45.8k432139
answered 2 days ago
EmignaEmigna
45.8k432139
45.8k432139
add a comment |
add a comment |
$begingroup$
Java 8, 105 54 bytes
a->{for(int l=a.length,i=0;i<l*l;)a[i%l]=a[a[i++%l]];}
Modifies the input-array instead of returning a new one to save bytes.
-51 bytes by just modifying the array $length^2$ times, instead of until it no longer changes.
Try it online.
Explanation:
a->{ // Method with integer-array parameter and no return-type
int l=a.length, // Length of the input-array
i=0;i<l*l;) // Loop `i` in the range [0, length squared):
a[i%l]= // Set the (`i` modulo-length)'th item in the array to:
a[ // The `p`'th value of the input-array,
a[i++%l]];} // where `p` is the (`i` modulo-length)'th value of the array
answered yesterday
Kevin CruijssenKevin Cruijssen
36.8k555193
$endgroup$
add a comment |
$begingroup$
Java 8, 105 54 bytes
a->{for(int l=a.length,i=0;i<l*l;)a[i%l]=a[a[i++%l]];}
Modifies the input-array instead of returning a new one to save bytes.
-51 bytes by just modifying the array $length^2$ times, instead of until it no longer changes.
Try it online.
Explanation:
a->{ // Method with integer-array parameter and no return-type
int l=a.length, // Length of the input-array
i=0;i<l*l;) // Loop `i` in the range [0, length squared):
a[i%l]= // Set the (`i` modulo-length)'th item in the array to:
a[ // The `p`'th value of the input-array,
a[i++%l]];} // where `p` is the (`i` modulo-length)'th value of the array
answered yesterday
Kevin CruijssenKevin Cruijssen
36.8k555193
$endgroup$
add a comment |
$begingroup$
Java 8, 105 54 bytes
a->{for(int l=a.length,i=0;i<l*l;)a[i%l]=a[a[i++%l]];}
Modifies the input-array instead of returning a new one to save bytes.
-51 bytes by just modifying the array $length^2$ times, instead of until it no longer changes.
Try it online.
Explanation:
a->{ // Method with integer-array parameter and no return-type
int l=a.length, // Length of the input-array
i=0;i<l*l;) // Loop `i` in the range [0, length squared):
a[i%l]= // Set the (`i` modulo-length)'th item in the array to:
a[ // The `p`'th value of the input-array,
a[i++%l]];} // where `p` is the (`i` modulo-length)'th value of the array
answered yesterday
Kevin CruijssenKevin Cruijssen
36.8k555193
$endgroup$
Java 8, 105 54 bytes
a->{for(int l=a.length,i=0;i<l*l;)a[i%l]=a[a[i++%l]];}
Modifies the input-array instead of returning a new one to save bytes.
-51 bytes by just modifying the array $length^2$ times, instead of until it no longer changes.
Try it online.
Explanation:
a->{ // Method with integer-array parameter and no return-type
int l=a.length, // Length of the input-array
i=0;i<l*l;) // Loop `i` in the range [0, length squared):
a[i%l]= // Set the (`i` modulo-length)'th item in the array to:
a[ // The `p`'th value of the input-array,
a[i++%l]];} // where `p` is the (`i` modulo-length)'th value of the array
answered yesterday
Kevin CruijssenKevin Cruijssen
36.8k555193
edited yesterday
answered yesterday
Kevin CruijssenKevin Cruijssen
36.8k555193
answered yesterday
Kevin CruijssenKevin Cruijssen
36.8k555193
answered yesterday
Kevin CruijssenKevin Cruijssen
36.8k555193
36.8k555193
add a comment |
add a comment |
$begingroup$
C (gcc), 32-bit, 49 bytes
f(p,n,i)int**p;{for(i=0;i<n*n;)p[i++%n]=*p[i%n];}
Try it online!
Uses pointers.
50 bytes with integers:
f(p,n,i)int*p;{for(i=0;i<n*n;)p[i++%n]=p[p[i%n]];}
Try it online!
answered 2 days ago
nwellnhofnwellnhof
6,66511126
$endgroup$
add a comment |
$begingroup$
C (gcc), 32-bit, 49 bytes
f(p,n,i)int**p;{for(i=0;i<n*n;)p[i++%n]=*p[i%n];}
Try it online!
Uses pointers.
50 bytes with integers:
f(p,n,i)int*p;{for(i=0;i<n*n;)p[i++%n]=p[p[i%n]];}
Try it online!
answered 2 days ago
nwellnhofnwellnhof
6,66511126
$endgroup$
add a comment |
$begingroup$
C (gcc), 32-bit, 49 bytes
f(p,n,i)int**p;{for(i=0;i<n*n;)p[i++%n]=*p[i%n];}
Try it online!
Uses pointers.
50 bytes with integers:
f(p,n,i)int*p;{for(i=0;i<n*n;)p[i++%n]=p[p[i%n]];}
Try it online!
answered 2 days ago
nwellnhofnwellnhof
6,66511126
$endgroup$
C (gcc), 32-bit, 49 bytes
f(p,n,i)int**p;{for(i=0;i<n*n;)p[i++%n]=*p[i%n];}
Try it online!
Uses pointers.
50 bytes with integers:
f(p,n,i)int*p;{for(i=0;i<n*n;)p[i++%n]=p[p[i%n]];}
Try it online!
answered 2 days ago
nwellnhofnwellnhof
6,66511126
edited 2 days ago
answered 2 days ago
nwellnhofnwellnhof
6,66511126
answered 2 days ago
nwellnhofnwellnhof
6,66511126
answered 2 days ago
nwellnhofnwellnhof
6,66511126
6,66511126
add a comment |
add a comment |
$begingroup$
Japt, 15 13 7 bytes
Modifies the original input array.
££hYXgU
Try it (additional bytes are to write the modified input to the console)
££hYXgU
£ :Map
£ : Map each X at index Y
hY : Replace the element at index Y
XgU : With the element at index X
answered 2 days ago
ShaggyShaggy
19.5k21666
$endgroup$
add a comment |
$begingroup$
Japt, 15 13 7 bytes
Modifies the original input array.
££hYXgU
Try it (additional bytes are to write the modified input to the console)
££hYXgU
£ :Map
£ : Map each X at index Y
hY : Replace the element at index Y
XgU : With the element at index X
answered 2 days ago
ShaggyShaggy
19.5k21666
$endgroup$
add a comment |
$begingroup$
Japt, 15 13 7 bytes
Modifies the original input array.
££hYXgU
Try it (additional bytes are to write the modified input to the console)
££hYXgU
£ :Map
£ : Map each X at index Y
hY : Replace the element at index Y
XgU : With the element at index X
answered 2 days ago
ShaggyShaggy
19.5k21666
$endgroup$
Japt, 15 13 7 bytes
Modifies the original input array.
££hYXgU
Try it (additional bytes are to write the modified input to the console)
££hYXgU
£ :Map
£ : Map each X at index Y
hY : Replace the element at index Y
XgU : With the element at index X
answered 2 days ago
ShaggyShaggy
19.5k21666
edited 2 days ago
answered 2 days ago
ShaggyShaggy
19.5k21666
answered 2 days ago
ShaggyShaggy
19.5k21666
answered 2 days ago
ShaggyShaggy
19.5k21666
19.5k21666
add a comment |
add a comment |
$begingroup$
Ruby, 37 34 bytes
->a{a.size.times{a.map!{|x|a[x]}}}
Try it online!
Returns by modifying the input array in-place.
answered 2 days ago
Kirill L.Kirill L.
3,9451319
$endgroup$
add a comment |
$begingroup$
Ruby, 37 34 bytes
->a{a.size.times{a.map!{|x|a[x]}}}
Try it online!
Returns by modifying the input array in-place.
answered 2 days ago
Kirill L.Kirill L.
3,9451319
$endgroup$
add a comment |
$begingroup$
Ruby, 37 34 bytes
->a{a.size.times{a.map!{|x|a[x]}}}
Try it online!
Returns by modifying the input array in-place.
answered 2 days ago
Kirill L.Kirill L.
3,9451319
$endgroup$
Ruby, 37 34 bytes
->a{a.size.times{a.map!{|x|a[x]}}}
Try it online!
Returns by modifying the input array in-place.
answered 2 days ago
Kirill L.Kirill L.
3,9451319
edited yesterday
answered 2 days ago
Kirill L.Kirill L.
3,9451319
answered 2 days ago
Kirill L.Kirill L.
3,9451319
answered 2 days ago
Kirill L.Kirill L.
3,9451319
3,9451319
add a comment |
add a comment |
$begingroup$
Red, 63 bytes
func[b][loop(l: length? b)* l[repeat i l[b/:i: b/(1 + b/:i)]]b]
Try it online!
Modifies the array in place
answered yesterday
Galen IvanovGalen Ivanov
6,64711032
$endgroup$
add a comment |
$begingroup$
Red, 63 bytes
func[b][loop(l: length? b)* l[repeat i l[b/:i: b/(1 + b/:i)]]b]
Try it online!
Modifies the array in place
answered yesterday
Galen IvanovGalen Ivanov
6,64711032
$endgroup$
add a comment |
$begingroup$
Red, 63 bytes
func[b][loop(l: length? b)* l[repeat i l[b/:i: b/(1 + b/:i)]]b]
Try it online!
Modifies the array in place
answered yesterday
Galen IvanovGalen Ivanov
6,64711032
$endgroup$
Red, 63 bytes
func[b][loop(l: length? b)* l[repeat i l[b/:i: b/(1 + b/:i)]]b]
Try it online!
Modifies the array in place
answered yesterday
Galen IvanovGalen Ivanov
6,64711032
edited yesterday
answered yesterday
Galen IvanovGalen Ivanov
6,64711032
answered yesterday
Galen IvanovGalen Ivanov
6,64711032
answered yesterday
Galen IvanovGalen Ivanov
6,64711032
6,64711032
add a comment |
add a comment |
$begingroup$
R, 60 58 bytes
-2 bytes thanks to @digEmAll for reading the rules.
function(x,n=sum(x|1)){for(i in rep(1:n,n))x[i]=x[x[i]];x}
Try it online!
1-indexed.
n is the length of the input array.
rep(1:n,n) replicates 1:n n times (e.g. n=3 => 1,2,3,1,2,3,1,2,3)
Loop through the array n times. Steady state will be acheived by then for sure, in fact by the end of the n-1st time through I think. The proof is left to the reader.
answered 2 days ago
ngmngm
3,36924
$endgroup$
1
$begingroup$
I think you can remove the+1and simply take the 1-based input, the post states : You may choose to use 1-based indexing
$endgroup$
– digEmAll
yesterday
$begingroup$
-4 by switching toscan()for input. I always feel like myscan()solutions are suboptimal, so keep an eye out a shorter way to assignxandntogether:n=length(x<-scan());for(i in rep(1:n,n))x[i]=x[x[i]];xTry it online!
$endgroup$
– CriminallyVulgar
13 hours ago
add a comment |
$begingroup$
R, 60 58 bytes
-2 bytes thanks to @digEmAll for reading the rules.
function(x,n=sum(x|1)){for(i in rep(1:n,n))x[i]=x[x[i]];x}
Try it online!
1-indexed.
n is the length of the input array.
rep(1:n,n) replicates 1:n n times (e.g. n=3 => 1,2,3,1,2,3,1,2,3)
Loop through the array n times. Steady state will be acheived by then for sure, in fact by the end of the n-1st time through I think. The proof is left to the reader.
answered 2 days ago
ngmngm
3,36924
$endgroup$
1
$begingroup$
I think you can remove the+1and simply take the 1-based input, the post states : You may choose to use 1-based indexing
$endgroup$
– digEmAll
yesterday
$begingroup$
-4 by switching toscan()for input. I always feel like myscan()solutions are suboptimal, so keep an eye out a shorter way to assignxandntogether:n=length(x<-scan());for(i in rep(1:n,n))x[i]=x[x[i]];xTry it online!
$endgroup$
– CriminallyVulgar
13 hours ago
add a comment |
$begingroup$
R, 60 58 bytes
-2 bytes thanks to @digEmAll for reading the rules.
function(x,n=sum(x|1)){for(i in rep(1:n,n))x[i]=x[x[i]];x}
Try it online!
1-indexed.
n is the length of the input array.
rep(1:n,n) replicates 1:n n times (e.g. n=3 => 1,2,3,1,2,3,1,2,3)
Loop through the array n times. Steady state will be acheived by then for sure, in fact by the end of the n-1st time through I think. The proof is left to the reader.
answered 2 days ago
ngmngm
3,36924
$endgroup$
R, 60 58 bytes
-2 bytes thanks to @digEmAll for reading the rules.
function(x,n=sum(x|1)){for(i in rep(1:n,n))x[i]=x[x[i]];x}
Try it online!
1-indexed.
n is the length of the input array.
rep(1:n,n) replicates 1:n n times (e.g. n=3 => 1,2,3,1,2,3,1,2,3)
Loop through the array n times. Steady state will be acheived by then for sure, in fact by the end of the n-1st time through I think. The proof is left to the reader.
answered 2 days ago
ngmngm
3,36924
edited yesterday
answered 2 days ago
ngmngm
3,36924
answered 2 days ago
ngmngm
3,36924
answered 2 days ago
ngmngm
3,36924
3,36924
1
$begingroup$
I think you can remove the+1and simply take the 1-based input, the post states : You may choose to use 1-based indexing
$endgroup$
– digEmAll
yesterday
$begingroup$
-4 by switching toscan()for input. I always feel like myscan()solutions are suboptimal, so keep an eye out a shorter way to assignxandntogether:n=length(x<-scan());for(i in rep(1:n,n))x[i]=x[x[i]];xTry it online!
$endgroup$
– CriminallyVulgar
13 hours ago
add a comment |
1
$begingroup$
I think you can remove the+1and simply take the 1-based input, the post states : You may choose to use 1-based indexing
$endgroup$
– digEmAll
yesterday
$begingroup$
-4 by switching toscan()for input. I always feel like myscan()solutions are suboptimal, so keep an eye out a shorter way to assignxandntogether:n=length(x<-scan());for(i in rep(1:n,n))x[i]=x[x[i]];xTry it online!
$endgroup$
– CriminallyVulgar
13 hours ago
1
1
$begingroup$
I think you can remove the
+1 and simply take the 1-based input, the post states : You may choose to use 1-based indexing$endgroup$
– digEmAll
yesterday
$begingroup$
I think you can remove the
+1 and simply take the 1-based input, the post states : You may choose to use 1-based indexing$endgroup$
– digEmAll
yesterday
$begingroup$
-4 by switching to
scan() for input. I always feel like my scan() solutions are suboptimal, so keep an eye out a shorter way to assign x and n together: n=length(x<-scan());for(i in rep(1:n,n))x[i]=x[x[i]];x Try it online!$endgroup$
– CriminallyVulgar
13 hours ago
$begingroup$
-4 by switching to
scan() for input. I always feel like my scan() solutions are suboptimal, so keep an eye out a shorter way to assign x and n together: n=length(x<-scan());for(i in rep(1:n,n))x[i]=x[x[i]];x Try it online!$endgroup$
– CriminallyVulgar
13 hours ago
add a comment |
$begingroup$
Common Lisp, 59 58 bytes
(lambda(a)(dolist(j a)(map-into a(lambda(x)(elt a x))a))a)
Try it online!
answered yesterday
RenzoRenzo
1,730516
$endgroup$
add a comment |
$begingroup$
Common Lisp, 59 58 bytes
(lambda(a)(dolist(j a)(map-into a(lambda(x)(elt a x))a))a)
Try it online!
answered yesterday
RenzoRenzo
1,730516
$endgroup$
add a comment |
$begingroup$
Common Lisp, 59 58 bytes
(lambda(a)(dolist(j a)(map-into a(lambda(x)(elt a x))a))a)
Try it online!
answered yesterday
RenzoRenzo
1,730516
$endgroup$
Common Lisp, 59 58 bytes
(lambda(a)(dolist(j a)(map-into a(lambda(x)(elt a x))a))a)
Try it online!
answered yesterday
RenzoRenzo
1,730516
edited yesterday
answered yesterday
RenzoRenzo
1,730516
answered yesterday
RenzoRenzo
1,730516
answered yesterday
RenzoRenzo
1,730516
1,730516
add a comment |
add a comment |
$begingroup$
Clojure, 136 bytes
(defn j[a](let[f(fn[a](loop[i 0 a a](if(= i(count a))a(recur(inc i)(assoc a i(a(a i)))))))](loop[p nil a a](if(= p a)a(recur a(f a))))))
Try it online!
answered 2 days ago
Ethan McCueEthan McCue
212
New contributor
Ethan McCue is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Hello and welcome to PPCG. Would it be possible for you to provide a link to an online interpreter such that one could easily verify your solution? Furthermore, canloop [not becomeloop[?
$endgroup$
– Jonathan Frech
2 days ago
1
$begingroup$
The most recent edit should fix the test failures. Sorry for the inconvenience everybody.
$endgroup$
– Ethan McCue
7 hours ago
add a comment |
$begingroup$
Clojure, 136 bytes
(defn j[a](let[f(fn[a](loop[i 0 a a](if(= i(count a))a(recur(inc i)(assoc a i(a(a i)))))))](loop[p nil a a](if(= p a)a(recur a(f a))))))
Try it online!
answered 2 days ago
Ethan McCueEthan McCue
212
New contributor
Ethan McCue is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Hello and welcome to PPCG. Would it be possible for you to provide a link to an online interpreter such that one could easily verify your solution? Furthermore, canloop [not becomeloop[?
$endgroup$
– Jonathan Frech
2 days ago
1
$begingroup$
The most recent edit should fix the test failures. Sorry for the inconvenience everybody.
$endgroup$
– Ethan McCue
7 hours ago
add a comment |
$begingroup$
Clojure, 136 bytes
(defn j[a](let[f(fn[a](loop[i 0 a a](if(= i(count a))a(recur(inc i)(assoc a i(a(a i)))))))](loop[p nil a a](if(= p a)a(recur a(f a))))))
Try it online!
answered 2 days ago
Ethan McCueEthan McCue
212
New contributor
Ethan McCue is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Clojure, 136 bytes
(defn j[a](let[f(fn[a](loop[i 0 a a](if(= i(count a))a(recur(inc i)(assoc a i(a(a i)))))))](loop[p nil a a](if(= p a)a(recur a(f a))))))
Try it online!
answered 2 days ago
Ethan McCueEthan McCue
212
New contributor
Ethan McCue is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 hours ago
answered 2 days ago
Ethan McCueEthan McCue
212
New contributor
Ethan McCue is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Ethan McCueEthan McCue
212
answered 2 days ago
Ethan McCueEthan McCue
212
212
New contributor
Ethan McCue is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ethan McCue is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ethan McCue is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Hello and welcome to PPCG. Would it be possible for you to provide a link to an online interpreter such that one could easily verify your solution? Furthermore, canloop [not becomeloop[?
$endgroup$
– Jonathan Frech
2 days ago
1
$begingroup$
The most recent edit should fix the test failures. Sorry for the inconvenience everybody.
$endgroup$
– Ethan McCue
7 hours ago
add a comment |
$begingroup$
Hello and welcome to PPCG. Would it be possible for you to provide a link to an online interpreter such that one could easily verify your solution? Furthermore, canloop [not becomeloop[?
$endgroup$
– Jonathan Frech
2 days ago
1
$begingroup$
The most recent edit should fix the test failures. Sorry for the inconvenience everybody.
$endgroup$
– Ethan McCue
7 hours ago
$begingroup$
Hello and welcome to PPCG. Would it be possible for you to provide a link to an online interpreter such that one could easily verify your solution? Furthermore, can
loop [ not become loop[?$endgroup$
– Jonathan Frech
2 days ago
$begingroup$
Hello and welcome to PPCG. Would it be possible for you to provide a link to an online interpreter such that one could easily verify your solution? Furthermore, can
loop [ not become loop[?$endgroup$
– Jonathan Frech
2 days ago
1
1
$begingroup$
The most recent edit should fix the test failures. Sorry for the inconvenience everybody.
$endgroup$
– Ethan McCue
7 hours ago
$begingroup$
The most recent edit should fix the test failures. Sorry for the inconvenience everybody.
$endgroup$
– Ethan McCue
7 hours ago
add a comment |
$begingroup$
Perl 5, 35 34 26 bytes
using the fact that convergence is reach at most for the size number of iterations
$_=$F[$_]for@F x@F;$_="@F"
26 bytes
$_=$F[$_]for@F;/@F/ or$_="@F",redo
34 bytes
$_=$F[$_]for@F;$_="@F",redo if!/@F/
35 bytes
answered 2 days ago
Nahuel FouilleulNahuel Fouilleul
1,84028
$endgroup$
add a comment |
$begingroup$
Perl 5, 35 34 26 bytes
using the fact that convergence is reach at most for the size number of iterations
$_=$F[$_]for@F x@F;$_="@F"
26 bytes
$_=$F[$_]for@F;/@F/ or$_="@F",redo
34 bytes
$_=$F[$_]for@F;$_="@F",redo if!/@F/
35 bytes
answered 2 days ago
Nahuel FouilleulNahuel Fouilleul
1,84028
$endgroup$
add a comment |
$begingroup$
Perl 5, 35 34 26 bytes
using the fact that convergence is reach at most for the size number of iterations
$_=$F[$_]for@F x@F;$_="@F"
26 bytes
$_=$F[$_]for@F;/@F/ or$_="@F",redo
34 bytes
$_=$F[$_]for@F;$_="@F",redo if!/@F/
35 bytes
answered 2 days ago
Nahuel FouilleulNahuel Fouilleul
1,84028
$endgroup$
Perl 5, 35 34 26 bytes
using the fact that convergence is reach at most for the size number of iterations
$_=$F[$_]for@F x@F;$_="@F"
26 bytes
$_=$F[$_]for@F;/@F/ or$_="@F",redo
34 bytes
$_=$F[$_]for@F;$_="@F",redo if!/@F/
35 bytes
answered 2 days ago
Nahuel FouilleulNahuel Fouilleul
1,84028
edited yesterday
answered 2 days ago
Nahuel FouilleulNahuel Fouilleul
1,84028
answered 2 days ago
Nahuel FouilleulNahuel Fouilleul
1,84028
answered 2 days ago
Nahuel FouilleulNahuel Fouilleul
1,84028
1,84028
add a comment |
add a comment |
$begingroup$
Clojure, 88 bytes
#(reduce(fn[x i](assoc x i(get x(get x i))))%(flatten(repeat(count %)(range(count %)))))
Try it online!
answered yesterday
Kirill L.Kirill L.
3,9451319
$endgroup$
add a comment |
$begingroup$
Clojure, 88 bytes
#(reduce(fn[x i](assoc x i(get x(get x i))))%(flatten(repeat(count %)(range(count %)))))
Try it online!
answered yesterday
Kirill L.Kirill L.
3,9451319
$endgroup$
add a comment |
$begingroup$
Clojure, 88 bytes
#(reduce(fn[x i](assoc x i(get x(get x i))))%(flatten(repeat(count %)(range(count %)))))
Try it online!
answered yesterday
Kirill L.Kirill L.
3,9451319
$endgroup$
Clojure, 88 bytes
#(reduce(fn[x i](assoc x i(get x(get x i))))%(flatten(repeat(count %)(range(count %)))))
Try it online!
answered yesterday
Kirill L.Kirill L.
3,9451319
answered yesterday
Kirill L.Kirill L.
3,9451319
answered yesterday
Kirill L.Kirill L.
3,9451319
answered yesterday
Kirill L.Kirill L.
3,9451319
3,9451319
add a comment |
add a comment |
$begingroup$
Charcoal, 16 bytes
ＦθＦＬθ§≔θκ§θ§θκＩθ
Try it online! Link is to verbose version of code. Sadly all the usual mapping functions only operate on a copy of the array with the result being that they just permute the elements rather than jumping them, so the code has to do everything manually. Explanation:
Ｆθ
Repeat the inner loop once for each element. This just ensures that the result stabilises.
ＦＬθ
Loop over the array indices.
§≔θκ§θ§θκ
Get the array element at the current index, use that to index into the array, and replace the current element with that value.
Ｉθ
Cast the elements to string and implicitly print each on their own line.
answered yesterday
NeilNeil
80k744178
$endgroup$
add a comment |
$begingroup$
Charcoal, 16 bytes
ＦθＦＬθ§≔θκ§θ§θκＩθ
Try it online! Link is to verbose version of code. Sadly all the usual mapping functions only operate on a copy of the array with the result being that they just permute the elements rather than jumping them, so the code has to do everything manually. Explanation:
Ｆθ
Repeat the inner loop once for each element. This just ensures that the result stabilises.
ＦＬθ
Loop over the array indices.
§≔θκ§θ§θκ
Get the array element at the current index, use that to index into the array, and replace the current element with that value.
Ｉθ
Cast the elements to string and implicitly print each on their own line.
answered yesterday
NeilNeil
80k744178
$endgroup$
add a comment |
$begingroup$
Charcoal, 16 bytes
ＦθＦＬθ§≔θκ§θ§θκＩθ
Try it online! Link is to verbose version of code. Sadly all the usual mapping functions only operate on a copy of the array with the result being that they just permute the elements rather than jumping them, so the code has to do everything manually. Explanation:
Ｆθ
Repeat the inner loop once for each element. This just ensures that the result stabilises.
ＦＬθ
Loop over the array indices.
§≔θκ§θ§θκ
Get the array element at the current index, use that to index into the array, and replace the current element with that value.
Ｉθ
Cast the elements to string and implicitly print each on their own line.
answered yesterday
NeilNeil
80k744178
$endgroup$
Charcoal, 16 bytes
ＦθＦＬθ§≔θκ§θ§θκＩθ
Try it online! Link is to verbose version of code. Sadly all the usual mapping functions only operate on a copy of the array with the result being that they just permute the elements rather than jumping them, so the code has to do everything manually. Explanation:
Ｆθ
Repeat the inner loop once for each element. This just ensures that the result stabilises.
ＦＬθ
Loop over the array indices.
§≔θκ§θ§θκ
Get the array element at the current index, use that to index into the array, and replace the current element with that value.
Ｉθ
Cast the elements to string and implicitly print each on their own line.
answered yesterday
NeilNeil
80k744178
answered yesterday
NeilNeil
80k744178
answered yesterday
NeilNeil
80k744178
answered yesterday
NeilNeil
80k744178
80k744178
add a comment |
add a comment |
$begingroup$
Clean, 80 bytes
import StdEnv
limit o iterateb=foldl(l i=updateAt i(l!!(l!!i))l)b(indexList b)
Try it online!
answered yesterday
ΟurousΟurous
6,63211033
$endgroup$
add a comment |
$begingroup$
Clean, 80 bytes
import StdEnv
limit o iterateb=foldl(l i=updateAt i(l!!(l!!i))l)b(indexList b)
Try it online!
answered yesterday
ΟurousΟurous
6,63211033
$endgroup$
add a comment |
$begingroup$
Clean, 80 bytes
import StdEnv
limit o iterateb=foldl(l i=updateAt i(l!!(l!!i))l)b(indexList b)
Try it online!
answered yesterday
ΟurousΟurous
6,63211033
$endgroup$
Clean, 80 bytes
import StdEnv
limit o iterateb=foldl(l i=updateAt i(l!!(l!!i))l)b(indexList b)
Try it online!
answered yesterday
ΟurousΟurous
6,63211033
answered yesterday
ΟurousΟurous
6,63211033
answered yesterday
ΟurousΟurous
6,63211033
answered yesterday
ΟurousΟurous
6,63211033
6,63211033
add a comment |
add a comment |
$begingroup$
F#, 74 73 bytes
fun(c:'a)->let l=c.Length in(for i in 0..l*l do c.[i%l]<-c.[c.[i%l]]);c
Nothing special. Uses the modulus idea seen in other answers.
answered 9 hours ago
LSM07LSM07
1213
$endgroup$
add a comment |
$begingroup$
F#, 74 73 bytes
fun(c:'a)->let l=c.Length in(for i in 0..l*l do c.[i%l]<-c.[c.[i%l]]);c
Nothing special. Uses the modulus idea seen in other answers.
answered 9 hours ago
LSM07LSM07
1213
$endgroup$
add a comment |
$begingroup$
F#, 74 73 bytes
fun(c:'a)->let l=c.Length in(for i in 0..l*l do c.[i%l]<-c.[c.[i%l]]);c
Nothing special. Uses the modulus idea seen in other answers.
answered 9 hours ago
LSM07LSM07
1213
$endgroup$
F#, 74 73 bytes
fun(c:'a)->let l=c.Length in(for i in 0..l*l do c.[i%l]<-c.[c.[i%l]]);c
Nothing special. Uses the modulus idea seen in other answers.
answered 9 hours ago
LSM07LSM07
1213
edited 9 hours ago
answered 9 hours ago
LSM07LSM07
1213
answered 9 hours ago
LSM07LSM07
1213
answered 9 hours ago
LSM07LSM07
1213
1213
add a comment |
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).
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f179050%2fpointer-jumping%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Related: Jump the array
$endgroup$
– BMO
2 days ago
$begingroup$
Are we allowed to take the length
nas additional input?$endgroup$
– Kevin Cruijssen
yesterday
2
$begingroup$
@KevinCruijssen, see this meta discussion.
$endgroup$
– Shaggy
yesterday
$begingroup$
It's too bad the entries need to be updated sequentially; if they could be updated simultaneously, Mathematica would have the 21-character solution
#[[#]]&~FixedPoint~#&.$endgroup$
– Greg Martin
17 hours ago