Pointer jumping
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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 Jan 23 at 17:20
BMOBMO
12k22291
$endgroup$
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 Jan 23 at 17:20
BMOBMO
12k22291
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Related: Jump the array
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– BMO
Jan 23 at 17:20
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Are we allowed to take the lengthnas additional input?
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– Kevin Cruijssen
Jan 24 at 8:55
2
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@KevinCruijssen, see this meta discussion.
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– Shaggy
Jan 24 at 12:26
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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~#&.
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– Greg Martin
Jan 25 at 8:40
add a comment |
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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 Jan 23 at 17:20
BMOBMO
12k22291
$endgroup$
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 Jan 23 at 17:20
BMOBMO
12k22291
asked Jan 23 at 17:20
BMOBMO
12k22291
edited Jan 23 at 17:55
BMO
asked Jan 23 at 17:20
BMOBMO
12k22291
asked Jan 23 at 17:20
BMOBMO
12k22291
asked Jan 23 at 17:20
BMOBMO
12k22291
12k22291
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Related: Jump the array
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– BMO
Jan 23 at 17:20
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Are we allowed to take the lengthnas additional input?
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– Kevin Cruijssen
Jan 24 at 8:55
2
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@KevinCruijssen, see this meta discussion.
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– Shaggy
Jan 24 at 12:26
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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
Jan 25 at 8:40
add a comment |
$begingroup$
Related: Jump the array
$endgroup$
– BMO
Jan 23 at 17:20
$begingroup$
Are we allowed to take the lengthnas additional input?
$endgroup$
– Kevin Cruijssen
Jan 24 at 8:55
2
$begingroup$
@KevinCruijssen, see this meta discussion.
$endgroup$
– Shaggy
Jan 24 at 12:26
$begingroup$
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
Jan 25 at 8:40
$begingroup$
Related: Jump the array
$endgroup$
– BMO
Jan 23 at 17:20
$begingroup$
Related: Jump the array
$endgroup$
– BMO
Jan 23 at 17:20
$begingroup$
Are we allowed to take the length
n as additional input?$endgroup$
– Kevin Cruijssen
Jan 24 at 8:55
$begingroup$
Are we allowed to take the length
n as additional input?$endgroup$
– Kevin Cruijssen
Jan 24 at 8:55
2
2
$begingroup$
@KevinCruijssen, see this meta discussion.
$endgroup$
– Shaggy
Jan 24 at 12:26
$begingroup$
@KevinCruijssen, see this meta discussion.
$endgroup$
– Shaggy
Jan 24 at 12:26
$begingroup$
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
Jan 25 at 8:40
$begingroup$
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
Jan 25 at 8:40
add a comment |
21 Answers
21
active
oldest
votes
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JavaScript, 36 bytes
Modifies the original input array.
a=>a.map(_=>a.map((x,y)=>a[y]=a[x]))
Try it online
answered Jan 23 at 18:45
ShaggyShaggy
19.6k21666
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add a comment |
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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 Jan 23 at 17:38
Erik the OutgolferErik the Outgolfer
31.7k429103
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add a comment |
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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 Jan 23 at 18:32
niminimi
31.7k32285
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add a comment |
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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 Jan 23 at 18:11
BierpfurzBierpfurz
1213
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add a comment |
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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 Jan 24 at 20:19
SeanSean
514
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2
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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!
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– BMO
Jan 24 at 20:48
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Thank you for the welcome and advice which I have used to update my answer.
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– Sean
Jan 24 at 22:39
add a comment |
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JavaScript (ES6), 41 bytes
f=a=>a+''==a.map((x,i)=>a[i]=a[x])?a:f(a)
Try it online!
answered Jan 23 at 17:28
ArnauldArnauld
74.8k691313
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Gah! I was waiting for this challenge to be posted so I could post the exact same solution : Damn your ninja skills! :p
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– Shaggy
Jan 23 at 17:53
2
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@shaggy 🐱👤(this is supposed to be a Ninja Cat ... but it's probably not supported everywhere)
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– Arnauld
Jan 23 at 18:10
add a comment |
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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 Jan 24 at 8:52
Kevin CruijssenKevin Cruijssen
37.3k556195
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add a comment |
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C (clang), 32-bit, 49 44 bytes
i;f(**p,n){for(i=0;i<n*n;)p[i++%n]=*p[i%n];}
Try it online!
Uses pointers.
50 45 bytes with integers:
i;f(*p,n){for(i=0;i<n*n;)p[i++%n]=p[p[i%n]];}
Try it online!
answered Jan 23 at 18:29
nwellnhofnwellnhof
6,69511126
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add a comment |
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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 Jan 23 at 18:01
Kamil DrakariKamil Drakari
3,231416
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add a comment |
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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 Jan 23 at 18:26
EmignaEmigna
46k432140
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add a comment |
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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 Jan 23 at 18:01
ShaggyShaggy
19.6k21666
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add a comment |
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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 Jan 23 at 18:16
Kirill L.Kirill L.
4,0251321
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add a comment |
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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 Jan 24 at 8:13
Galen IvanovGalen Ivanov
6,68711033
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add a comment |
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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 Jan 23 at 21:43
ngmngm
3,36924
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1
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I think you can remove the+1and simply take the 1-based input, the post states : You may choose to use 1-based indexing
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– digEmAll
Jan 24 at 7:51
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-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!
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– CriminallyVulgar
Jan 25 at 11:47
add a comment |
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Common Lisp, 59 58 bytes
(lambda(a)(dolist(j a)(map-into a(lambda(x)(elt a x))a))a)
Try it online!
answered Jan 24 at 17:06
RenzoRenzo
1,740516
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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 Jan 23 at 21:44
Ethan McCueEthan McCue
212
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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[?
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– Jonathan Frech
Jan 23 at 21:47
1
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The most recent edit should fix the test failures. Sorry for the inconvenience everybody.
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– Ethan McCue
Jan 25 at 18:32
add a comment |
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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 Jan 23 at 21:46
Nahuel FouilleulNahuel Fouilleul
2,01028
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Clojure, 88 bytes
#(reduce(fn[x i](assoc x i(get x(get x i))))%(flatten(repeat(count %)(range(count %)))))
Try it online!
answered Jan 24 at 11:40
Kirill L.Kirill L.
4,0251321
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add a comment |
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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 Jan 24 at 20:39
NeilNeil
80.2k744178
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Clean, 80 bytes
import StdEnv
limit o iterateb=foldl(l i=updateAt i(l!!(l!!i))l)b(indexList b)
Try it online!
answered Jan 25 at 1:19
ΟurousΟurous
6,69211033
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add a comment |
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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 Jan 25 at 15:48
LSM07LSM07
1213
$endgroup$
add a comment |
Your Answer
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21 Answers
21
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21 Answers
21
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$begingroup$
JavaScript, 36 bytes
Modifies the original input array.
a=>a.map(_=>a.map((x,y)=>a[y]=a[x]))
Try it online
answered Jan 23 at 18:45
ShaggyShaggy
19.6k21666
$endgroup$
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 Jan 23 at 18:45
ShaggyShaggy
19.6k21666
$endgroup$
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 Jan 23 at 18:45
ShaggyShaggy
19.6k21666
$endgroup$
JavaScript, 36 bytes
Modifies the original input array.
a=>a.map(_=>a.map((x,y)=>a[y]=a[x]))
Try it online
answered Jan 23 at 18:45
ShaggyShaggy
19.6k21666
answered Jan 23 at 18:45
ShaggyShaggy
19.6k21666
answered Jan 23 at 18:45
ShaggyShaggy
19.6k21666
answered Jan 23 at 18:45
ShaggyShaggy
19.6k21666
19.6k21666
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 Jan 23 at 17:38
Erik the OutgolferErik the Outgolfer
31.7k429103
$endgroup$
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 Jan 23 at 17:38
Erik the OutgolferErik the Outgolfer
31.7k429103
$endgroup$
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 Jan 23 at 17:38
Erik the OutgolferErik the Outgolfer
31.7k429103
$endgroup$
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 Jan 23 at 17:38
Erik the OutgolferErik the Outgolfer
31.7k429103
edited Jan 23 at 19:33
answered Jan 23 at 17:38
Erik the OutgolferErik the Outgolfer
31.7k429103
answered Jan 23 at 17:38
Erik the OutgolferErik the Outgolfer
31.7k429103
answered Jan 23 at 17:38
Erik the OutgolferErik the Outgolfer
31.7k429103
31.7k429103
add a comment |
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 Jan 23 at 18:32
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 Jan 23 at 18:32
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 Jan 23 at 18:32
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 Jan 23 at 18:32
niminimi
31.7k32285
answered Jan 23 at 18:32
niminimi
31.7k32285
answered Jan 23 at 18:32
niminimi
31.7k32285
answered Jan 23 at 18:32
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 Jan 23 at 18:11
BierpfurzBierpfurz
1213
$endgroup$
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 Jan 23 at 18:11
BierpfurzBierpfurz
1213
$endgroup$
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 Jan 23 at 18:11
BierpfurzBierpfurz
1213
$endgroup$
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 Jan 23 at 18:11
BierpfurzBierpfurz
1213
edited Jan 24 at 0:27
answered Jan 23 at 18:11
BierpfurzBierpfurz
1213
answered Jan 23 at 18:11
BierpfurzBierpfurz
1213
answered Jan 23 at 18:11
BierpfurzBierpfurz
1213
1213
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 Jan 24 at 20:19
SeanSean
514
$endgroup$
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
Jan 24 at 20:48
$begingroup$
Thank you for the welcome and advice which I have used to update my answer.
$endgroup$
– Sean
Jan 24 at 22:39
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 Jan 24 at 20:19
SeanSean
514
$endgroup$
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
Jan 24 at 20:48
$begingroup$
Thank you for the welcome and advice which I have used to update my answer.
$endgroup$
– Sean
Jan 24 at 22:39
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 Jan 24 at 20:19
SeanSean
514
$endgroup$
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 Jan 24 at 20:19
SeanSean
514
edited Jan 24 at 22:29
answered Jan 24 at 20:19
SeanSean
514
answered Jan 24 at 20:19
SeanSean
514
answered Jan 24 at 20:19
SeanSean
514
514
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
Jan 24 at 20:48
$begingroup$
Thank you for the welcome and advice which I have used to update my answer.
$endgroup$
– Sean
Jan 24 at 22:39
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
Jan 24 at 20:48
$begingroup$
Thank you for the welcome and advice which I have used to update my answer.
$endgroup$
– Sean
Jan 24 at 22:39
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
Jan 24 at 20:48
$begingroup$
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
Jan 24 at 20:48
$begingroup$
Thank you for the welcome and advice which I have used to update my answer.
$endgroup$
– Sean
Jan 24 at 22:39
$begingroup$
Thank you for the welcome and advice which I have used to update my answer.
$endgroup$
– Sean
Jan 24 at 22:39
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 Jan 23 at 17:28
ArnauldArnauld
74.8k691313
$endgroup$
$begingroup$
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
Jan 23 at 17:53
2
$begingroup$
@shaggy 🐱👤(this is supposed to be a Ninja Cat ... but it's probably not supported everywhere)
$endgroup$
– Arnauld
Jan 23 at 18:10
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 Jan 23 at 17:28
ArnauldArnauld
74.8k691313
$endgroup$
$begingroup$
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
Jan 23 at 17:53
2
$begingroup$
@shaggy 🐱👤(this is supposed to be a Ninja Cat ... but it's probably not supported everywhere)
$endgroup$
– Arnauld
Jan 23 at 18:10
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 Jan 23 at 17:28
ArnauldArnauld
74.8k691313
$endgroup$
JavaScript (ES6), 41 bytes
f=a=>a+''==a.map((x,i)=>a[i]=a[x])?a:f(a)
Try it online!
answered Jan 23 at 17:28
ArnauldArnauld
74.8k691313
answered Jan 23 at 17:28
ArnauldArnauld
74.8k691313
answered Jan 23 at 17:28
ArnauldArnauld
74.8k691313
answered Jan 23 at 17:28
ArnauldArnauld
74.8k691313
74.8k691313
$begingroup$
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
Jan 23 at 17:53
2
$begingroup$
@shaggy 🐱👤(this is supposed to be a Ninja Cat ... but it's probably not supported everywhere)
$endgroup$
– Arnauld
Jan 23 at 18:10
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
Jan 23 at 17:53
2
$begingroup$
@shaggy 🐱👤(this is supposed to be a Ninja Cat ... but it's probably not supported everywhere)
$endgroup$
– Arnauld
Jan 23 at 18:10
$begingroup$
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
Jan 23 at 17:53
$begingroup$
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
Jan 23 at 17:53
2
2
$begingroup$
@shaggy 🐱👤(this is supposed to be a Ninja Cat ... but it's probably not supported everywhere)
$endgroup$
– Arnauld
Jan 23 at 18:10
$begingroup$
@shaggy 🐱👤(this is supposed to be a Ninja Cat ... but it's probably not supported everywhere)
$endgroup$
– Arnauld
Jan 23 at 18:10
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 Jan 24 at 8:52
Kevin CruijssenKevin Cruijssen
37.3k556195
$endgroup$
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 Jan 24 at 8:52
Kevin CruijssenKevin Cruijssen
37.3k556195
$endgroup$
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 Jan 24 at 8:52
Kevin CruijssenKevin Cruijssen
37.3k556195
$endgroup$
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 Jan 24 at 8:52
Kevin CruijssenKevin Cruijssen
37.3k556195
edited Jan 24 at 9:00
answered Jan 24 at 8:52
Kevin CruijssenKevin Cruijssen
37.3k556195
answered Jan 24 at 8:52
Kevin CruijssenKevin Cruijssen
37.3k556195
answered Jan 24 at 8:52
Kevin CruijssenKevin Cruijssen
37.3k556195
37.3k556195
add a comment |
add a comment |
$begingroup$
C (clang), 32-bit, 49 44 bytes
i;f(**p,n){for(i=0;i<n*n;)p[i++%n]=*p[i%n];}
Try it online!
Uses pointers.
50 45 bytes with integers:
i;f(*p,n){for(i=0;i<n*n;)p[i++%n]=p[p[i%n]];}
Try it online!
answered Jan 23 at 18:29
nwellnhofnwellnhof
6,69511126
$endgroup$
add a comment |
$begingroup$
C (clang), 32-bit, 49 44 bytes
i;f(**p,n){for(i=0;i<n*n;)p[i++%n]=*p[i%n];}
Try it online!
Uses pointers.
50 45 bytes with integers:
i;f(*p,n){for(i=0;i<n*n;)p[i++%n]=p[p[i%n]];}
Try it online!
answered Jan 23 at 18:29
nwellnhofnwellnhof
6,69511126
$endgroup$
add a comment |
$begingroup$
C (clang), 32-bit, 49 44 bytes
i;f(**p,n){for(i=0;i<n*n;)p[i++%n]=*p[i%n];}
Try it online!
Uses pointers.
50 45 bytes with integers:
i;f(*p,n){for(i=0;i<n*n;)p[i++%n]=p[p[i%n]];}
Try it online!
answered Jan 23 at 18:29
nwellnhofnwellnhof
6,69511126
$endgroup$
C (clang), 32-bit, 49 44 bytes
i;f(**p,n){for(i=0;i<n*n;)p[i++%n]=*p[i%n];}
Try it online!
Uses pointers.
50 45 bytes with integers:
i;f(*p,n){for(i=0;i<n*n;)p[i++%n]=p[p[i%n]];}
Try it online!
answered Jan 23 at 18:29
nwellnhofnwellnhof
6,69511126
edited Jan 26 at 11:04
answered Jan 23 at 18:29
nwellnhofnwellnhof
6,69511126
answered Jan 23 at 18:29
nwellnhofnwellnhof
6,69511126
answered Jan 23 at 18:29
nwellnhofnwellnhof
6,69511126
6,69511126
add a comment |
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 Jan 23 at 18:01
Kamil DrakariKamil Drakari
3,231416
$endgroup$
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 Jan 23 at 18:01
Kamil DrakariKamil Drakari
3,231416
$endgroup$
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 Jan 23 at 18:01
Kamil DrakariKamil Drakari
3,231416
$endgroup$
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 Jan 23 at 18:01
Kamil DrakariKamil Drakari
3,231416
answered Jan 23 at 18:01
Kamil DrakariKamil Drakari
3,231416
answered Jan 23 at 18:01
Kamil DrakariKamil Drakari
3,231416
answered Jan 23 at 18:01
Kamil DrakariKamil Drakari
3,231416
3,231416
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 Jan 23 at 18:26
EmignaEmigna
46k432140
$endgroup$
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 Jan 23 at 18:26
EmignaEmigna
46k432140
$endgroup$
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 Jan 23 at 18:26
EmignaEmigna
46k432140
$endgroup$
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 Jan 23 at 18:26
EmignaEmigna
46k432140
edited Jan 23 at 18:39
answered Jan 23 at 18:26
EmignaEmigna
46k432140
answered Jan 23 at 18:26
EmignaEmigna
46k432140
answered Jan 23 at 18:26
EmignaEmigna
46k432140
46k432140
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 Jan 23 at 18:01
ShaggyShaggy
19.6k21666
$endgroup$
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 Jan 23 at 18:01
ShaggyShaggy
19.6k21666
$endgroup$
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 Jan 23 at 18:01
ShaggyShaggy
19.6k21666
$endgroup$
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 Jan 23 at 18:01
ShaggyShaggy
19.6k21666
edited Jan 23 at 20:24
answered Jan 23 at 18:01
ShaggyShaggy
19.6k21666
answered Jan 23 at 18:01
ShaggyShaggy
19.6k21666
answered Jan 23 at 18:01
ShaggyShaggy
19.6k21666
19.6k21666
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 Jan 23 at 18:16
Kirill L.Kirill L.
4,0251321
$endgroup$
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 Jan 23 at 18:16
Kirill L.Kirill L.
4,0251321
$endgroup$
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 Jan 23 at 18:16
Kirill L.Kirill L.
4,0251321
$endgroup$
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 Jan 23 at 18:16
Kirill L.Kirill L.
4,0251321
edited Jan 24 at 8:21
answered Jan 23 at 18:16
Kirill L.Kirill L.
4,0251321
answered Jan 23 at 18:16
Kirill L.Kirill L.
4,0251321
answered Jan 23 at 18:16
Kirill L.Kirill L.
4,0251321
4,0251321
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 Jan 24 at 8:13
Galen IvanovGalen Ivanov
6,68711033
$endgroup$
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 Jan 24 at 8:13
Galen IvanovGalen Ivanov
6,68711033
$endgroup$
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 Jan 24 at 8:13
Galen IvanovGalen Ivanov
6,68711033
$endgroup$
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 Jan 24 at 8:13
Galen IvanovGalen Ivanov
6,68711033
edited Jan 24 at 9:07
answered Jan 24 at 8:13
Galen IvanovGalen Ivanov
6,68711033
answered Jan 24 at 8:13
Galen IvanovGalen Ivanov
6,68711033
answered Jan 24 at 8:13
Galen IvanovGalen Ivanov
6,68711033
6,68711033
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 Jan 23 at 21:43
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
Jan 24 at 7:51
$begingroup$
-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
Jan 25 at 11:47
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 Jan 23 at 21:43
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
Jan 24 at 7:51
$begingroup$
-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
Jan 25 at 11:47
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 Jan 23 at 21:43
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 Jan 23 at 21:43
ngmngm
3,36924
edited Jan 24 at 14:46
answered Jan 23 at 21:43
ngmngm
3,36924
answered Jan 23 at 21:43
ngmngm
3,36924
answered Jan 23 at 21:43
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
Jan 24 at 7:51
$begingroup$
-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
Jan 25 at 11:47
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
Jan 24 at 7:51
$begingroup$
-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
Jan 25 at 11:47
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
Jan 24 at 7:51
$begingroup$
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
Jan 24 at 7:51
$begingroup$
-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
Jan 25 at 11:47
$begingroup$
-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
Jan 25 at 11:47
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 Jan 24 at 17:06
RenzoRenzo
1,740516
$endgroup$
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 Jan 24 at 17:06
RenzoRenzo
1,740516
$endgroup$
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 Jan 24 at 17:06
RenzoRenzo
1,740516
$endgroup$
Common Lisp, 59 58 bytes
(lambda(a)(dolist(j a)(map-into a(lambda(x)(elt a x))a))a)
Try it online!
answered Jan 24 at 17:06
RenzoRenzo
1,740516
edited Jan 24 at 17:18
answered Jan 24 at 17:06
RenzoRenzo
1,740516
answered Jan 24 at 17:06
RenzoRenzo
1,740516
answered Jan 24 at 17:06
RenzoRenzo
1,740516
1,740516
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 Jan 23 at 21:44
Ethan McCueEthan McCue
212
$endgroup$
$begingroup$
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
Jan 23 at 21:47
1
$begingroup$
The most recent edit should fix the test failures. Sorry for the inconvenience everybody.
$endgroup$
– Ethan McCue
Jan 25 at 18:32
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 Jan 23 at 21:44
Ethan McCueEthan McCue
212
$endgroup$
$begingroup$
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
Jan 23 at 21:47
1
$begingroup$
The most recent edit should fix the test failures. Sorry for the inconvenience everybody.
$endgroup$
– Ethan McCue
Jan 25 at 18:32
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 Jan 23 at 21:44
Ethan McCueEthan McCue
212
$endgroup$
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 Jan 23 at 21:44
Ethan McCueEthan McCue
212
edited Jan 25 at 18:30
answered Jan 23 at 21:44
Ethan McCueEthan McCue
212
answered Jan 23 at 21:44
Ethan McCueEthan McCue
212
answered Jan 23 at 21:44
Ethan McCueEthan McCue
212
212
$begingroup$
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
Jan 23 at 21:47
1
$begingroup$
The most recent edit should fix the test failures. Sorry for the inconvenience everybody.
$endgroup$
– Ethan McCue
Jan 25 at 18:32
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
Jan 23 at 21:47
1
$begingroup$
The most recent edit should fix the test failures. Sorry for the inconvenience everybody.
$endgroup$
– Ethan McCue
Jan 25 at 18:32
$begingroup$
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
Jan 23 at 21:47
$begingroup$
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
Jan 23 at 21:47
1
1
$begingroup$
The most recent edit should fix the test failures. Sorry for the inconvenience everybody.
$endgroup$
– Ethan McCue
Jan 25 at 18:32
$begingroup$
The most recent edit should fix the test failures. Sorry for the inconvenience everybody.
$endgroup$
– Ethan McCue
Jan 25 at 18:32
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 Jan 23 at 21:46
Nahuel FouilleulNahuel Fouilleul
2,01028
$endgroup$
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 Jan 23 at 21:46
Nahuel FouilleulNahuel Fouilleul
2,01028
$endgroup$
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 Jan 23 at 21:46
Nahuel FouilleulNahuel Fouilleul
2,01028
$endgroup$
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 Jan 23 at 21:46
Nahuel FouilleulNahuel Fouilleul
2,01028
edited Jan 24 at 8:39
answered Jan 23 at 21:46
Nahuel FouilleulNahuel Fouilleul
2,01028
answered Jan 23 at 21:46
Nahuel FouilleulNahuel Fouilleul
2,01028
answered Jan 23 at 21:46
Nahuel FouilleulNahuel Fouilleul
2,01028
2,01028
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 Jan 24 at 11:40
Kirill L.Kirill L.
4,0251321
$endgroup$
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 Jan 24 at 11:40
Kirill L.Kirill L.
4,0251321
$endgroup$
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 Jan 24 at 11:40
Kirill L.Kirill L.
4,0251321
$endgroup$
Clojure, 88 bytes
#(reduce(fn[x i](assoc x i(get x(get x i))))%(flatten(repeat(count %)(range(count %)))))
Try it online!
answered Jan 24 at 11:40
Kirill L.Kirill L.
4,0251321
answered Jan 24 at 11:40
Kirill L.Kirill L.
4,0251321
answered Jan 24 at 11:40
Kirill L.Kirill L.
4,0251321
answered Jan 24 at 11:40
Kirill L.Kirill L.
4,0251321
4,0251321
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 Jan 24 at 20:39
NeilNeil
80.2k744178
$endgroup$
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 Jan 24 at 20:39
NeilNeil
80.2k744178
$endgroup$
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 Jan 24 at 20:39
NeilNeil
80.2k744178
$endgroup$
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 Jan 24 at 20:39
NeilNeil
80.2k744178
answered Jan 24 at 20:39
NeilNeil
80.2k744178
answered Jan 24 at 20:39
NeilNeil
80.2k744178
answered Jan 24 at 20:39
NeilNeil
80.2k744178
80.2k744178
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 Jan 25 at 1:19
ΟurousΟurous
6,69211033
$endgroup$
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 Jan 25 at 1:19
ΟurousΟurous
6,69211033
$endgroup$
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 Jan 25 at 1:19
ΟurousΟurous
6,69211033
$endgroup$
Clean, 80 bytes
import StdEnv
limit o iterateb=foldl(l i=updateAt i(l!!(l!!i))l)b(indexList b)
Try it online!
answered Jan 25 at 1:19
ΟurousΟurous
6,69211033
answered Jan 25 at 1:19
ΟurousΟurous
6,69211033
answered Jan 25 at 1:19
ΟurousΟurous
6,69211033
answered Jan 25 at 1:19
ΟurousΟurous
6,69211033
6,69211033
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 Jan 25 at 15:48
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 Jan 25 at 15:48
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 Jan 25 at 15:48
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 Jan 25 at 15:48
LSM07LSM07
1213
edited Jan 25 at 16:05
answered Jan 25 at 15:48
LSM07LSM07
1213
answered Jan 25 at 15:48
LSM07LSM07
1213
answered Jan 25 at 15:48
LSM07LSM07
1213
1213
add a comment |
add a comment |
If this is an answer to a challenge…
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…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.
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More generally…
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$begingroup$
Related: Jump the array
$endgroup$
– BMO
Jan 23 at 17:20
$begingroup$
Are we allowed to take the length
nas additional input?$endgroup$
– Kevin Cruijssen
Jan 24 at 8:55
2
$begingroup$
@KevinCruijssen, see this meta discussion.
$endgroup$
– Shaggy
Jan 24 at 12:26
$begingroup$
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
Jan 25 at 8:40