Check my tunneling arrays

Clash Royale CLAN TAG#URR8PPP

up vote
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favorite

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Imagine you have an array of integers, whose non-negative values are pointers to other positions in the same array, only that those values represent tunnels, so if the value in position A is positive and points to position B, then the value in position B must be also positive and point to position A to represent both ends of the tunnel. So:

Challenge

• Given an array of integers, check if the array complies with the restriction to be a tunneling array and return two distinct, coherent values for truthy and falsey.


• The values in the array will be below zero for non-tunnel positions, and zero or above for tunnel positions. If your array is 1-indexed, then the zero value represents a non-tunnel position. Non-tunnel values do not need to be checked.


• If a positive value in a cell points to itself, that's a falsey. If A points to B, B to C and C to A, that's a falsey. If a positive value points beyond the limits of the array, that's a falsey.

Examples

The following examples are 0-indexed:

[-1, -1, -1, 6, -1, -1, 3, -1, -1] Truthy (position 3 points to position 6 and vice versa)
[1, 0] Truthy (position 0 points to position 1 and vice versa)
[0, 1] Falsey (positions 0 and 1 point to themselves)
[4, 2, 1, -1, 0, -1] Truthy
[2, 3, 0, 1] Truthy
[1, 2, 0] Falsey (no circular tunnels allowed)
[-1, 2, -1] Falsey (tunnel without end)
 Truthy (no tunnels, that's OK)
[-1, -2, -3] Truthy (no tunnels, that's OK)
[1, 0, 3] Falsey (tunnel goes beyond limits)
[1] Falsey (tunnel goes beyond limits)
[1, 0, 3, 7] Falsey (tunnel goes beyond limits)

This is code-golf, so may the shortest code for each language win!

code-golf array-manipulation decision-problem

share|improve this question

edited Sep 6 at 15:47

asked Sep 4 at 14:02

Charlie

6,6641976

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  what should we return for [0]?
  – ngn
  Sep 4 at 17:44
  

  
  
  
  


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  Expanding on ngn's question, are self tunnels allowed? What would the cases [0,1] and [0,-1,2] give?
  – dylnan
  Sep 4 at 17:50
  

  
  
  
  


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  @dylnan [0,1] is in the examples. "If a positive value in a cell points to itself, that's a falsey"
  – ngn
  Sep 4 at 17:51
  
  

  
  
  
  


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  suggested test: [2,3,0,1]
  – ngn
  Sep 4 at 17:55
  

  
  
  
  


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  @JonathanAllan the tunnel values are values indicating possible array positions. If your array is 0-indexed then every value below 0 is not a tunnel value. If it's 1-indexed then every value below 1 is not a tunnel value.
  – Charlie
  Sep 4 at 19:49
  

  
  
  
  

 | 
show 9 more comments

up vote
16
down vote

favorite

1

Imagine you have an array of integers, whose non-negative values are pointers to other positions in the same array, only that those values represent tunnels, so if the value in position A is positive and points to position B, then the value in position B must be also positive and point to position A to represent both ends of the tunnel. So:

Challenge

• Given an array of integers, check if the array complies with the restriction to be a tunneling array and return two distinct, coherent values for truthy and falsey.


• The values in the array will be below zero for non-tunnel positions, and zero or above for tunnel positions. If your array is 1-indexed, then the zero value represents a non-tunnel position. Non-tunnel values do not need to be checked.


• If a positive value in a cell points to itself, that's a falsey. If A points to B, B to C and C to A, that's a falsey. If a positive value points beyond the limits of the array, that's a falsey.

Examples

The following examples are 0-indexed:

[-1, -1, -1, 6, -1, -1, 3, -1, -1] Truthy (position 3 points to position 6 and vice versa)
[1, 0] Truthy (position 0 points to position 1 and vice versa)
[0, 1] Falsey (positions 0 and 1 point to themselves)
[4, 2, 1, -1, 0, -1] Truthy
[2, 3, 0, 1] Truthy
[1, 2, 0] Falsey (no circular tunnels allowed)
[-1, 2, -1] Falsey (tunnel without end)
 Truthy (no tunnels, that's OK)
[-1, -2, -3] Truthy (no tunnels, that's OK)
[1, 0, 3] Falsey (tunnel goes beyond limits)
[1] Falsey (tunnel goes beyond limits)
[1, 0, 3, 7] Falsey (tunnel goes beyond limits)

This is code-golf, so may the shortest code for each language win!

code-golf array-manipulation decision-problem

share|improve this question

edited Sep 6 at 15:47

asked Sep 4 at 14:02

Charlie

6,6641976

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  what should we return for [0]?
  – ngn
  Sep 4 at 17:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Expanding on ngn's question, are self tunnels allowed? What would the cases [0,1] and [0,-1,2] give?
  – dylnan
  Sep 4 at 17:50
  

  
  
  
  


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  @dylnan [0,1] is in the examples. "If a positive value in a cell points to itself, that's a falsey"
  – ngn
  Sep 4 at 17:51
  
  

  
  
  
  


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  suggested test: [2,3,0,1]
  – ngn
  Sep 4 at 17:55
  

  
  
  
  


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  @JonathanAllan the tunnel values are values indicating possible array positions. If your array is 0-indexed then every value below 0 is not a tunnel value. If it's 1-indexed then every value below 1 is not a tunnel value.
  – Charlie
  Sep 4 at 19:49
  

  
  
  
  

 | 
show 9 more comments

up vote
16
down vote

favorite

1

up vote
16
down vote

favorite

1

1

Imagine you have an array of integers, whose non-negative values are pointers to other positions in the same array, only that those values represent tunnels, so if the value in position A is positive and points to position B, then the value in position B must be also positive and point to position A to represent both ends of the tunnel. So:

Challenge

• Given an array of integers, check if the array complies with the restriction to be a tunneling array and return two distinct, coherent values for truthy and falsey.


• The values in the array will be below zero for non-tunnel positions, and zero or above for tunnel positions. If your array is 1-indexed, then the zero value represents a non-tunnel position. Non-tunnel values do not need to be checked.


• If a positive value in a cell points to itself, that's a falsey. If A points to B, B to C and C to A, that's a falsey. If a positive value points beyond the limits of the array, that's a falsey.

Examples

The following examples are 0-indexed:

[-1, -1, -1, 6, -1, -1, 3, -1, -1] Truthy (position 3 points to position 6 and vice versa)
[1, 0] Truthy (position 0 points to position 1 and vice versa)
[0, 1] Falsey (positions 0 and 1 point to themselves)
[4, 2, 1, -1, 0, -1] Truthy
[2, 3, 0, 1] Truthy
[1, 2, 0] Falsey (no circular tunnels allowed)
[-1, 2, -1] Falsey (tunnel without end)
 Truthy (no tunnels, that's OK)
[-1, -2, -3] Truthy (no tunnels, that's OK)
[1, 0, 3] Falsey (tunnel goes beyond limits)
[1] Falsey (tunnel goes beyond limits)
[1, 0, 3, 7] Falsey (tunnel goes beyond limits)

This is code-golf, so may the shortest code for each language win!

code-golf array-manipulation decision-problem

share|improve this question

edited Sep 6 at 15:47

asked Sep 4 at 14:02

Charlie

6,6641976

Imagine you have an array of integers, whose non-negative values are pointers to other positions in the same array, only that those values represent tunnels, so if the value in position A is positive and points to position B, then the value in position B must be also positive and point to position A to represent both ends of the tunnel. So:

Challenge

• Given an array of integers, check if the array complies with the restriction to be a tunneling array and return two distinct, coherent values for truthy and falsey.


• The values in the array will be below zero for non-tunnel positions, and zero or above for tunnel positions. If your array is 1-indexed, then the zero value represents a non-tunnel position. Non-tunnel values do not need to be checked.


• If a positive value in a cell points to itself, that's a falsey. If A points to B, B to C and C to A, that's a falsey. If a positive value points beyond the limits of the array, that's a falsey.

Examples

The following examples are 0-indexed:

[-1, -1, -1, 6, -1, -1, 3, -1, -1] Truthy (position 3 points to position 6 and vice versa)
[1, 0] Truthy (position 0 points to position 1 and vice versa)
[0, 1] Falsey (positions 0 and 1 point to themselves)
[4, 2, 1, -1, 0, -1] Truthy
[2, 3, 0, 1] Truthy
[1, 2, 0] Falsey (no circular tunnels allowed)
[-1, 2, -1] Falsey (tunnel without end)
 Truthy (no tunnels, that's OK)
[-1, -2, -3] Truthy (no tunnels, that's OK)
[1, 0, 3] Falsey (tunnel goes beyond limits)
[1] Falsey (tunnel goes beyond limits)
[1, 0, 3, 7] Falsey (tunnel goes beyond limits)

This is code-golf, so may the shortest code for each language win!

code-golf array-manipulation decision-problem

share|improve this question

edited Sep 6 at 15:47

asked Sep 4 at 14:02

Charlie

6,6641976

share|improve this question

share|improve this question

edited Sep 6 at 15:47

edited Sep 6 at 15:47

edited Sep 6 at 15:47

asked Sep 4 at 14:02

Charlie

6,6641976

asked Sep 4 at 14:02

Charlie

6,6641976

asked Sep 4 at 14:02

Charlie

6,6641976

6,6641976

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  what should we return for [0]?
  – ngn
  Sep 4 at 17:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Expanding on ngn's question, are self tunnels allowed? What would the cases [0,1] and [0,-1,2] give?
  – dylnan
  Sep 4 at 17:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @dylnan [0,1] is in the examples. "If a positive value in a cell points to itself, that's a falsey"
  – ngn
  Sep 4 at 17:51
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  suggested test: [2,3,0,1]
  – ngn
  Sep 4 at 17:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JonathanAllan the tunnel values are values indicating possible array positions. If your array is 0-indexed then every value below 0 is not a tunnel value. If it's 1-indexed then every value below 1 is not a tunnel value.
  – Charlie
  Sep 4 at 19:49
  

  
  
  
  

 | 
show 9 more comments

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  what should we return for [0]?
  – ngn
  Sep 4 at 17:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Expanding on ngn's question, are self tunnels allowed? What would the cases [0,1] and [0,-1,2] give?
  – dylnan
  Sep 4 at 17:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @dylnan [0,1] is in the examples. "If a positive value in a cell points to itself, that's a falsey"
  – ngn
  Sep 4 at 17:51
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  suggested test: [2,3,0,1]
  – ngn
  Sep 4 at 17:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JonathanAllan the tunnel values are values indicating possible array positions. If your array is 0-indexed then every value below 0 is not a tunnel value. If it's 1-indexed then every value below 1 is not a tunnel value.
  – Charlie
  Sep 4 at 19:49
  

  
  
  
  

3

3

what should we return for [0]?
– ngn
Sep 4 at 17:44

what should we return for [0]?
– ngn
Sep 4 at 17:44

Expanding on ngn's question, are self tunnels allowed? What would the cases [0,1] and [0,-1,2] give?
– dylnan
Sep 4 at 17:50

Expanding on ngn's question, are self tunnels allowed? What would the cases [0,1] and [0,-1,2] give?
– dylnan
Sep 4 at 17:50

@dylnan [0,1] is in the examples. "If a positive value in a cell points to itself, that's a falsey"
– ngn
Sep 4 at 17:51

@dylnan [0,1] is in the examples. "If a positive value in a cell points to itself, that's a falsey"
– ngn
Sep 4 at 17:51

1

1

suggested test: [2,3,0,1]
– ngn
Sep 4 at 17:55

suggested test: [2,3,0,1]
– ngn
Sep 4 at 17:55

1

1

@JonathanAllan the tunnel values are values indicating possible array positions. If your array is 0-indexed then every value below 0 is not a tunnel value. If it's 1-indexed then every value below 1 is not a tunnel value.
– Charlie
Sep 4 at 19:49

@JonathanAllan the tunnel values are values indicating possible array positions. If your array is 0-indexed then every value below 0 is not a tunnel value. If it's 1-indexed then every value below 1 is not a tunnel value.
– Charlie
Sep 4 at 19:49

 | 
show 9 more comments

24 Answers
24

active

oldest

votes

up vote
8
down vote

R, 47 bytes

function(v,a=v[v>0],b=sort(a))all(v[a]==b&a!=b)

Try it online!

---

Unrolled code and explanation :

f=
function(v) # v vector of tunnel indexes (1-based) or values <= 0

 a = v[v>0] # get the tunnel positions

 b = sort(a) # sort the tunnel positions ascending

 c1 = v[a]==b # get the values of 'v' at positions 'a'
 # and check if they're equal to the sorted positions 'b'
 # (element-wise, returns a vector of TRUE/FALSE)

 c2 = a != b # check if positions 'a' are different from sorted positions 'b' 
 # (to exclude tunnels pointing to themselves, element-wise,
 # returns a vector of TRUE/FALSE)

 all(c1 & c2) # if all logical conditions 'c1' and 'c2' are TRUE then
 # returns TRUE otherwise FALSE

share|improve this answer

edited Sep 4 at 17:26

answered Sep 4 at 15:41

digEmAll

1,73147

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  I would really appreciate an explanation for this answer. :-)
  – Charlie
  Sep 4 at 15:51
  

  
  
  
  


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  @Charlie : explanation added
  – digEmAll
  Sep 4 at 17:27
  

  
  
  
  

add a comment | 

up vote
5
down vote

Python 2, 66 61 60 bytes

lambda l:all(len(l)>v!=i==l[v]for i,v in enumerate(l)if-1<v)

Try it online!

share|improve this answer

edited Sep 4 at 14:53

answered Sep 4 at 14:37

TFeld

11.3k2833

add a comment | 

up vote
5
down vote

APL (Dyalog Unicode), 19 24 bytes

×/<∘≢⍨×≠∘⍳∘≢⍨×0∘>∨⊢=⊢⍳⍳⍨

Try it online!

Prefix anonymous lambda, returning 1 for truthy and 0 for falsy. The TIO link contains a "prettified" version of the output for the test cases.

Shoutouts to @ngn and @Adám for saving approximately a bazillion bytes.

An extra shoutout to @ngn for the help with fixing the answer for some test cases, and with making it a train.

The updated answer uses ⎕IO←0, setting the Index Origin to 0.

How:

×/<∘≢⍨×≠∘⍳∘≢⍨×0∘>∨⊢=⊢⍳⍳⍨ ⍝ Prefix lambda, argument ⍵ → 4 2 1 ¯1 0 ¯1.
 ⍳⍨ ⍝ Index of (⍳) ⍵ in ⍵. ⍵⍳⍵ → 0 1 2 3 4 3
 ⊢⍳ ⍝ Index of that in ⍵ (returns the vector length if not found). 
 ⍝ ⍵⍳⍵⍳⍵ → 4 2 1 6 0 6
 ⊢= ⍝ Compare that with ⍵. ⍵=⍵⍳⍵⍳⍵ → 1 1 1 0 1 0
 ⍝ This checks if positive indices tunnel back and forth correctly.
 ∨ ⍝ Logical OR with
 0∘> ⍝ 0>⍵ → 0 0 0 1 0 1∨1 1 1 0 1 0 → 1 1 1 1 1 1
 ⍝ Removes the zeroes generated by negative indices
 × ⍝ Multiply that vector with
 ⍨ ⍝ (using ⍵ as both arguments)
 ⍳∘≢ ⍝ Generate the range [0..length(⍵)-1]
 ≠∘ ⍝ And do ⍵≠range; this checks if any 
 ⍝ element in ⍵ is tunneling to itself.
 ⍝ ⍵≠⍳≢⍵ → 4 2 1 ¯1 0 ¯1≠0 1 2 3 4 5 → 1 1 1 1 1 1 
 × ⍝ Multiply that vector with
 ⍨ ⍝ (using ⍵ as both arguments)
 <∘≢ ⍝ ⍵ < length(⍵) → 4 2 1 ¯1 0 ¯1 < 6 → 1 1 1 1 1 1
 ⍝ This checks if any index is out of bounds
×/ ⍝ Finally, multiply and reduce.
 ⍝ ×/1 1 1 1 1 1 → 1 (truthy)

share|improve this answer

edited Sep 6 at 20:41

answered Sep 4 at 17:30

J. Sallé

1,418218

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  I think this doesn't work for (1), (3 2 1), (5 4 3 2 1).
  – nwellnhof
  Sep 4 at 23:42
  

  
  
  
  


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  0< → × I think
  – Uriel
  Sep 5 at 21:11
  

  
  
  
  

add a comment | 

up vote
4
down vote

JavaScript (ES6), 35 bytes

Saved 1 byte thanks to @Shaggy

a=>a.every((v,i)=>v<0|v!=i&a[v]==i)

Try it online!

Commented

a => // a = input array
 a.every((v, i) => // for each value v at position i in a:
 v < 0 | // force the test to succeed if v is negative (non-tunnel position)
 v != i & // make sure that this cell is not pointing to itself
 a[v] == i // check the other end of the tunnel
 ) // end of every()

share|improve this answer

edited Sep 5 at 22:02

ngn

6,13812256

answered Sep 4 at 14:32

Arnauld

63.7k580268

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  Good thing I checked the solutions before posting a port of my Japt solution, which is nearly identical to this. You can save a byte with a=>a.every((v,i)=>v<0|v!=i&a[v]==i).
  – Shaggy
  Sep 4 at 15:09
  

  
  
  
  

add a comment | 

up vote
3
down vote

Python 2, 65 bytes

lambda l:all(l[v:]>and v!=i==l[v]or v<0for i,v in enumerate(l))

Try it online!

share|improve this answer

answered Sep 4 at 14:35

ovs

17.3k21056

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  62 bytes
  – TFeld
  Sep 4 at 14:48
  

  
  
  
  


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  61 bytes
  – ÎŸurous
  Sep 5 at 0:42
  

  
  
  
  

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up vote
3
down vote

Japt, 14 bytes

eȧJªX¦Y©Y¥UgX

Try it or run all test cases

---

Explanation

 :Implicit input of array U
 È :Map each X at (0-based) index Y
 §J : X less than or equal to -1?
 ª : Logical OR
 X¦Y : X does not equal Y?
 © : Logical AND
 YÂ¥UgX : Y equals the element in U at index X?
e :All truthy?

share|improve this answer

edited Sep 4 at 14:54

answered Sep 4 at 14:45

Shaggy

16.3k21560

add a comment | 

up vote
3
down vote

Perl 6, 36 bytes

!.grep:2-set $++,$^v,.[$v]xx$v+1

Try it online!

The basic idea is to check whether the set i, a[i], a[a[i]] contains exactly two distinct elements for each index i with a[i] >= 0. If an element points to itself, the set contains only a single distinct element. If the other end doesn't point back to i, the set contains three distinct elements. If a[i] < 0, the xx factor is zero or negative, so the set is i, a[i] , also with two distinct elements.

share|improve this answer

edited Sep 4 at 17:10

answered Sep 4 at 17:00

nwellnhof

3,503714

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up vote
3
down vote

MATL, 19 18 Bytes

-1 Byte thanks to Luis

n:G=GGG0>f))7M-|hs

Try it online!, for the first one only, because I don't know how to do all of them!

Gives 0 if truthy, a non-zero integer if falsey, eg. for test case 6 gives 4.

Please remember that like MATLAB, MATL is 1-indexed so 1 must be added to the test cases!

Never golfed in an Esolang before, so advice greatly received!

Explained:

n:G=GGG0>f))7M-|hs
 Implicit - input array
n Number of values in array
 : Make array 1:n
 G Push input
 = Equality
n:G= Makes non-zero array if any of the tunnels lead to themselves
 GGG Push input 3x
 0 Push literal 0
 > Greater than
 G0> Makes array of ones where input > 0
 f Find - returns indeces of non-zero values
 Implicit - copy this matrix to clipboard
 ) Indeces - returns array of positive integers in order from input
 ) Ditto - Note, implicit non-zero any above maximum
 7M Paste from clipboard
 - Subtract
 GGG0>f))7M- Makes array of zeros if only two-ended tunnels evident
 | Absolute value (otherwise eg. [3,4,2,1] -> '0')
 h Horizontal concat (ie. joins check for self tunnels and wrong tunnels)
 s Sum; = 0 if truthy, integer otherwise 

share|improve this answer

edited Sep 5 at 23:08

answered Sep 5 at 10:36

Lui

374211

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  Is my explanation too wordy? I want to make it obvious without going totally overboard.
  – Lui
  Sep 5 at 10:39
  

  
  
  
  

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up vote
2
down vote

Python, 112 97 96 86 bytes

f=lambda l:sum(i==l[i]or len(l)<=l[i]or 0<=l[i]and i!=l[l[i]]for i in range(len(l)))<1

Try it Online!

Returns True or False.

-10 bytes thanks to @Rod and @TFeld.

share|improve this answer

edited Sep 4 at 14:37

answered Sep 4 at 14:17

DimChtz

621111

add a comment | 

up vote
2
down vote

Jelly, 16 bytes

ị=JanJ$>L<$o<1$Ạ

Try it online!

1-indexed.

share|improve this answer

answered Sep 4 at 14:42

Erik the Outgolfer

29.4k42698

add a comment | 

up vote
2
down vote

05AB1E, 16 bytes

ε©0‹®NÊI®èNQ*~}P

Try it online or verify all test cases.

Explanation:

ε } # Map each value to:
 © # Store the current item in the register (without popping)
 0‹ # If the current value is negative
 ~ # Or if:
 ®NÊ # The current value is not equal to the current index
 * # And
 I®èNQ # The current value indexed into the input is equal to the index
 P # And output whether all values are truthy after the map

share|improve this answer

edited Sep 4 at 16:51

answered Sep 4 at 15:15

Kevin Cruijssen

29.6k550162

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  My try was the same except with filter. I don't see a way to improve on this.
  – Emigna
  Sep 4 at 16:43
  

  
  
  
  

add a comment | 

up vote
2
down vote

Java (JDK 10), 95 bytes

a->int l=a.length,i=l;for(;i-->0;)if(a[i]>=0&&(a[i]>=l

Try it online!

Credits

• -3 bytes thanks to AlexRacer
  

share|improve this answer

edited Sep 4 at 19:43

answered Sep 4 at 15:32

Olivier Grégoire

7,57811741

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could have been 87 bytes if it weren't for that pesky IndexOutOfBoundsException. Maybe you see something to fix it easily?
  – Kevin Cruijssen
  Sep 4 at 15:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen I can see how to fix that for 102 bytes. Nothing shorter yet :(
  – Olivier Grégoire
  Sep 4 at 15:45
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  -3 bytes - omit r and break out of the loop analogous to here
  – AlexRacer
  Sep 4 at 19:00
  

  
  
  
  

add a comment | 

up vote
2
down vote

Haskell, 48 bytes

(all=<< u(x,y)->y<0||x/=y&&elem(y,x)u).zip[0..]

Verify all testcases!

Explanation

Let's first ungolf the code a bit. As f =<< g is the same as x -> f (g x) x, the code is equivalent to

(u->all((x,y)->y<0||x/=y&&elem(y,x)u)u).zip[0..]

which is a bit clearer.

(u -> -- given u, return
 all ((x, y) -> -- whether for all elements (x, y) of u
 y < 0 || -- either y < 0, or
 x /= y && elem (y, x) u -- (x /= y) and ((y, x) is in u)
 )
 u
) . zip [0..] -- given the array a (implicitly via point-free style),
 -- return the array augmented with indices (it's the u above)

This solution is based on a simple observation: let a be the input array, and u the list of pairs (i, a[i]) where i is an index. Then a is a valid array if and only if for every (x, y) in u with y >= 0, the pair (y, x) belongs to u as well.

share|improve this answer

edited Sep 6 at 16:32

answered Sep 5 at 16:34

Delfad0r

914

New contributor

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
1
down vote

Charcoal, 22 bytes

¬Φθ∨⁼ικ¬∨‹ι⁰∧‹ιＬθ⁼κ§θι

Try it online! Link is to verbose version of code. Outputs - for truthy and nothing for falsy. Note: Inputting an empty array seems to crash Charcoal, but for now you can enter a space instead, which is near enough. Explanation:

 θ Input array
 Φ Filter elements
 ι Current value
 ⁼ Equals
 κ Current index
 ∨ Or
 ¬ Not
 ι Current value
 ‹ ⁰ Is less than zero
 ∨ Or
 ι Current value
 ‹ Is less than
 Ｌ Length of
 θ Input array
 ∧ And
 κ Current index
 ⁼ Equals
 §θι Indexed value
¬ Logical Not (i.e. is result empty)
 Implicitly print

share|improve this answer

answered Sep 4 at 15:15

Neil

75.1k744170

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This doesn't seem to be a very Charcoalable challenge... :-)
  – Charlie
  Sep 4 at 15:24
  

  
  
  
  

add a comment | 

up vote
1
down vote

Pascal (FPC), 165 155 153 bytes

function f(a:array of int32):byte;var i:int32;begin f:=1;for i:=0to length(a)-1do if a[i]>-1then if(a[i]=i)or(a[i]>length(a))or(a[a[i]]<>i)then f:=0;end;

Try it online!

Made function this time because the input is array. Returns 1 for truthy and 0 for falsey.

share|improve this answer

edited Sep 4 at 18:38

answered Sep 4 at 18:25

AlexRacer

57928

add a comment | 

up vote
1
down vote

Clean, 60 bytes

import StdEnv
@l=and[v<0||l%(v,v)==[i]&&v<>i\v<-l&i<-[0..]]

Try it online!

Clean, 142 bytes

Vastly over-complicated monster version:

import StdEnv,Data.List,Data.Maybe
$l=and[?i(mapMaybe((!?)l)j)j\i<-l&j<-map((!?)l)l|i>=0]with?a(Just(Just c))(Just b)=a==c&&b<>c;?_ _ _=False

Try it online!

Explained:

$ l // function $ of `l` is
 = and [ // true when all elements are true
 ? // apply ? to
 i // the element `i` of `l`
 (mapMaybe // and the result of attempting to
 ((!?)l) // try gettting an element from `l`
 j) // at the potentially invalid index `j`
 j // and `j` itself, which may not exist
 \ i <- l // for every element `i` in `l`
 & j <- map // and every potential `j` in
 ((!?)l) // `l` trying to be indexed by
 l // every element in `l`
 | i >= 0 // where `i` is greater than zero
 ]
with
 ? a (Just (Just c)) (Just b) // function ? when all the arguments exist
 = a==c && b<>c // `a` equals `c` and not `b`
 ;
 ? _ _ _ = False // for all other arguments, ? is false

share|improve this answer

edited Sep 5 at 0:33

answered Sep 4 at 23:41

Οurous

5,2131931

add a comment | 

up vote
1
down vote

Ruby, 44 bytes

->aa.all?x<0

Try it online!

share|improve this answer

answered Sep 5 at 10:24

G B

6,3571324

add a comment | 

up vote
1
down vote

Pyth, 17 16 bytes

.A.e|>0b&nbkq@Qb

Try it online here, or verify all the test cases at once here.

.A.e|>0b&nbkq@QbkQ Implicit: Q=eval(input())
 Trailing k, Q inferred
 .e Q Map the input with b=element, k=index, using:
 >0b 0>b
 | OR (
 nbk b != k
 & AND
 q@Qbk Q[b] == k)
.A Check if all elements are truthy

Edit: realised that the trailing k was also unnecessary

share|improve this answer

edited Sep 5 at 12:52

answered Sep 5 at 8:19

Sok

2,851722

add a comment | 

up vote
1
down vote

Groovy, 52 bytes

o=!(i=0);it.each(it[e]==i&&i-e);i++;o

Try it online!

share|improve this answer

answered Sep 5 at 13:57

GolfIsAGoodWalkSpoilt

812

add a comment | 

up vote
1
down vote

Perl 5, 54 bytes

$i=-1;!grep$_>=0*$i++&&($_==$i

Try it online!

share|improve this answer

answered Sep 5 at 19:30

Kjetil S.

51115

add a comment | 

up vote
1
down vote

K (ngn/k), 33 bytes

(x<#x)&(~x=!#x)&x=x?x?x

Try it online!

share|improve this answer

edited Sep 5 at 21:53

answered Sep 4 at 17:34

ngn

6,13812256

add a comment | 

up vote
1
down vote

C (gcc), 95 bytes

i(_,o,O,Q,I)int*_;Q>=o

Try it online!

share|improve this answer

answered Sep 6 at 21:55

Jonathan Frech

5,60311039

add a comment | 

up vote
0
down vote

Mathematica, 42 bytes

#==||(a=0@@#)[[#]]=!=a&&a[[#]][[#]]===a&

Pure function. Takes a 1-indexed list of numbers as input and returns True or False as output. Just follows the tunnels, ensuring that 0 maps to 0, no 1-cycles exist, and all cycles are 2-cycles. (I'm not entirely sure if this fails on any edge cases, but it gives the correct results for the examples.)

share|improve this answer

answered Sep 5 at 1:36

LegionMammal978

14.8k41752

add a comment | 

up vote
0
down vote

This answer does not work. Here for illustration purposes only.

This answer passes all of the (currently) posted test cases. However, it fails (raises an error) on other valid input, such as [1, 2] or [1, 0, 3, 7].

How could it pass [1, 0, 3] and fail [1, 0, 3, 7]? Well, it proceeds through the list, just like you'd expect. When it reads an element x of the list a, it first checks whether x is less than len(a), and immediately returns False, if so. So it correctly returns False on [1, 0, 3], because 3 is not less than len(a).

But assuming that x passes that check, the code will then go on to do some other checks, and at a certain point it happens to evaluate a[a[x]]. We've already guaranteed that evaluating a[x] will be OK...but not a[a[x]], which resolves to a[7] when x is 3 in the [1, 0, 3, 7] example. At this point Python raises an IndexError, rather than returning False.

For completeness, here's the answer.

Python 2, 59 bytes

lambda a:all(x<len(a)>-1<a[x]!=x==a[a[x]]for x in a if-1<x)

Try it online!

I wanted to do x<len(a)and-1<a[x]..., but of course len(a) is always >-1, so the above is equivalent. This check is a total of 5 chained relations (<, >, <, !=, and ==), plus a separate check -1<x in the if condition.

Python (conveniently) short-circuits chained relations like this, so for example if x>=len(a) then the check returns False before it gets to a[x] (which would otherwise raise an IndexError).

share|improve this answer

edited Sep 6 at 15:43

answered Sep 5 at 20:34

mathmandan

91357

add a comment | 

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24 Answers
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up vote
8
down vote

R, 47 bytes

function(v,a=v[v>0],b=sort(a))all(v[a]==b&a!=b)

Try it online!

---

Unrolled code and explanation :

f=
function(v) # v vector of tunnel indexes (1-based) or values <= 0

 a = v[v>0] # get the tunnel positions

 b = sort(a) # sort the tunnel positions ascending

 c1 = v[a]==b # get the values of 'v' at positions 'a'
 # and check if they're equal to the sorted positions 'b'
 # (element-wise, returns a vector of TRUE/FALSE)

 c2 = a != b # check if positions 'a' are different from sorted positions 'b' 
 # (to exclude tunnels pointing to themselves, element-wise,
 # returns a vector of TRUE/FALSE)

 all(c1 & c2) # if all logical conditions 'c1' and 'c2' are TRUE then
 # returns TRUE otherwise FALSE

share|improve this answer

edited Sep 4 at 17:26

answered Sep 4 at 15:41

digEmAll

1,73147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I would really appreciate an explanation for this answer. :-)
  – Charlie
  Sep 4 at 15:51
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @Charlie : explanation added
  – digEmAll
  Sep 4 at 17:27
  

  
  
  
  

add a comment | 

up vote
8
down vote

R, 47 bytes

function(v,a=v[v>0],b=sort(a))all(v[a]==b&a!=b)

Try it online!

---

Unrolled code and explanation :

f=
function(v) # v vector of tunnel indexes (1-based) or values <= 0

 a = v[v>0] # get the tunnel positions

 b = sort(a) # sort the tunnel positions ascending

 c1 = v[a]==b # get the values of 'v' at positions 'a'
 # and check if they're equal to the sorted positions 'b'
 # (element-wise, returns a vector of TRUE/FALSE)

 c2 = a != b # check if positions 'a' are different from sorted positions 'b' 
 # (to exclude tunnels pointing to themselves, element-wise,
 # returns a vector of TRUE/FALSE)

 all(c1 & c2) # if all logical conditions 'c1' and 'c2' are TRUE then
 # returns TRUE otherwise FALSE

share|improve this answer

edited Sep 4 at 17:26

answered Sep 4 at 15:41

digEmAll

1,73147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I would really appreciate an explanation for this answer. :-)
  – Charlie
  Sep 4 at 15:51
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @Charlie : explanation added
  – digEmAll
  Sep 4 at 17:27
  

  
  
  
  

add a comment | 

up vote
8
down vote

up vote
8
down vote

R, 47 bytes

function(v,a=v[v>0],b=sort(a))all(v[a]==b&a!=b)

Try it online!

---

Unrolled code and explanation :

f=
function(v) # v vector of tunnel indexes (1-based) or values <= 0

 a = v[v>0] # get the tunnel positions

 b = sort(a) # sort the tunnel positions ascending

 c1 = v[a]==b # get the values of 'v' at positions 'a'
 # and check if they're equal to the sorted positions 'b'
 # (element-wise, returns a vector of TRUE/FALSE)

 c2 = a != b # check if positions 'a' are different from sorted positions 'b' 
 # (to exclude tunnels pointing to themselves, element-wise,
 # returns a vector of TRUE/FALSE)

 all(c1 & c2) # if all logical conditions 'c1' and 'c2' are TRUE then
 # returns TRUE otherwise FALSE

share|improve this answer

edited Sep 4 at 17:26

answered Sep 4 at 15:41

digEmAll

1,73147

R, 47 bytes

function(v,a=v[v>0],b=sort(a))all(v[a]==b&a!=b)

Try it online!

---

Unrolled code and explanation :

f=
function(v) # v vector of tunnel indexes (1-based) or values <= 0

 a = v[v>0] # get the tunnel positions

 b = sort(a) # sort the tunnel positions ascending

 c1 = v[a]==b # get the values of 'v' at positions 'a'
 # and check if they're equal to the sorted positions 'b'
 # (element-wise, returns a vector of TRUE/FALSE)

 c2 = a != b # check if positions 'a' are different from sorted positions 'b' 
 # (to exclude tunnels pointing to themselves, element-wise,
 # returns a vector of TRUE/FALSE)

 all(c1 & c2) # if all logical conditions 'c1' and 'c2' are TRUE then
 # returns TRUE otherwise FALSE

share|improve this answer

edited Sep 4 at 17:26

answered Sep 4 at 15:41

digEmAll

1,73147

share|improve this answer

share|improve this answer

edited Sep 4 at 17:26

edited Sep 4 at 17:26

edited Sep 4 at 17:26

answered Sep 4 at 15:41

digEmAll

1,73147

answered Sep 4 at 15:41

digEmAll

1,73147

answered Sep 4 at 15:41

digEmAll

1,73147

1,73147

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I would really appreciate an explanation for this answer. :-)
  – Charlie
  Sep 4 at 15:51
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @Charlie : explanation added
  – digEmAll
  Sep 4 at 17:27
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I would really appreciate an explanation for this answer. :-)
  – Charlie
  Sep 4 at 15:51
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @Charlie : explanation added
  – digEmAll
  Sep 4 at 17:27
  

  
  
  
  

I would really appreciate an explanation for this answer. :-)
– Charlie
Sep 4 at 15:51

I would really appreciate an explanation for this answer. :-)
– Charlie
Sep 4 at 15:51

3

3

@Charlie : explanation added
– digEmAll
Sep 4 at 17:27

@Charlie : explanation added
– digEmAll
Sep 4 at 17:27

add a comment | 

up vote
5
down vote

Python 2, 66 61 60 bytes

lambda l:all(len(l)>v!=i==l[v]for i,v in enumerate(l)if-1<v)

Try it online!

share|improve this answer

edited Sep 4 at 14:53

answered Sep 4 at 14:37

TFeld

11.3k2833

add a comment | 

up vote
5
down vote

Python 2, 66 61 60 bytes

lambda l:all(len(l)>v!=i==l[v]for i,v in enumerate(l)if-1<v)

Try it online!

share|improve this answer

edited Sep 4 at 14:53

answered Sep 4 at 14:37

TFeld

11.3k2833

add a comment | 

up vote
5
down vote

up vote
5
down vote

Python 2, 66 61 60 bytes

lambda l:all(len(l)>v!=i==l[v]for i,v in enumerate(l)if-1<v)

Try it online!

share|improve this answer

edited Sep 4 at 14:53

answered Sep 4 at 14:37

TFeld

11.3k2833

Python 2, 66 61 60 bytes

lambda l:all(len(l)>v!=i==l[v]for i,v in enumerate(l)if-1<v)

Try it online!

share|improve this answer

edited Sep 4 at 14:53

answered Sep 4 at 14:37

TFeld

11.3k2833

share|improve this answer

share|improve this answer

edited Sep 4 at 14:53

edited Sep 4 at 14:53

edited Sep 4 at 14:53

answered Sep 4 at 14:37

TFeld

11.3k2833

answered Sep 4 at 14:37

TFeld

11.3k2833

answered Sep 4 at 14:37

TFeld

11.3k2833

11.3k2833

add a comment | 

add a comment | 

up vote
5
down vote

APL (Dyalog Unicode), 19 24 bytes

×/<∘≢⍨×≠∘⍳∘≢⍨×0∘>∨⊢=⊢⍳⍳⍨

Try it online!

Prefix anonymous lambda, returning 1 for truthy and 0 for falsy. The TIO link contains a "prettified" version of the output for the test cases.

Shoutouts to @ngn and @Adám for saving approximately a bazillion bytes.

An extra shoutout to @ngn for the help with fixing the answer for some test cases, and with making it a train.

The updated answer uses ⎕IO←0, setting the Index Origin to 0.

How:

×/<∘≢⍨×≠∘⍳∘≢⍨×0∘>∨⊢=⊢⍳⍳⍨ ⍝ Prefix lambda, argument ⍵ → 4 2 1 ¯1 0 ¯1.
 ⍳⍨ ⍝ Index of (⍳) ⍵ in ⍵. ⍵⍳⍵ → 0 1 2 3 4 3
 ⊢⍳ ⍝ Index of that in ⍵ (returns the vector length if not found). 
 ⍝ ⍵⍳⍵⍳⍵ → 4 2 1 6 0 6
 ⊢= ⍝ Compare that with ⍵. ⍵=⍵⍳⍵⍳⍵ → 1 1 1 0 1 0
 ⍝ This checks if positive indices tunnel back and forth correctly.
 ∨ ⍝ Logical OR with
 0∘> ⍝ 0>⍵ → 0 0 0 1 0 1∨1 1 1 0 1 0 → 1 1 1 1 1 1
 ⍝ Removes the zeroes generated by negative indices
 × ⍝ Multiply that vector with
 ⍨ ⍝ (using ⍵ as both arguments)
 ⍳∘≢ ⍝ Generate the range [0..length(⍵)-1]
 ≠∘ ⍝ And do ⍵≠range; this checks if any 
 ⍝ element in ⍵ is tunneling to itself.
 ⍝ ⍵≠⍳≢⍵ → 4 2 1 ¯1 0 ¯1≠0 1 2 3 4 5 → 1 1 1 1 1 1 
 × ⍝ Multiply that vector with
 ⍨ ⍝ (using ⍵ as both arguments)
 <∘≢ ⍝ ⍵ < length(⍵) → 4 2 1 ¯1 0 ¯1 < 6 → 1 1 1 1 1 1
 ⍝ This checks if any index is out of bounds
×/ ⍝ Finally, multiply and reduce.
 ⍝ ×/1 1 1 1 1 1 → 1 (truthy)

share|improve this answer

edited Sep 6 at 20:41

answered Sep 4 at 17:30

J. Sallé

1,418218

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think this doesn't work for (1), (3 2 1), (5 4 3 2 1).
  – nwellnhof
  Sep 4 at 23:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  0< → × I think
  – Uriel
  Sep 5 at 21:11
  

  
  
  
  

add a comment | 

up vote
5
down vote

APL (Dyalog Unicode), 19 24 bytes

×/<∘≢⍨×≠∘⍳∘≢⍨×0∘>∨⊢=⊢⍳⍳⍨

Try it online!

Prefix anonymous lambda, returning 1 for truthy and 0 for falsy. The TIO link contains a "prettified" version of the output for the test cases.

Shoutouts to @ngn and @Adám for saving approximately a bazillion bytes.

An extra shoutout to @ngn for the help with fixing the answer for some test cases, and with making it a train.

The updated answer uses ⎕IO←0, setting the Index Origin to 0.

How:

×/<∘≢⍨×≠∘⍳∘≢⍨×0∘>∨⊢=⊢⍳⍳⍨ ⍝ Prefix lambda, argument ⍵ → 4 2 1 ¯1 0 ¯1.
 ⍳⍨ ⍝ Index of (⍳) ⍵ in ⍵. ⍵⍳⍵ → 0 1 2 3 4 3
 ⊢⍳ ⍝ Index of that in ⍵ (returns the vector length if not found). 
 ⍝ ⍵⍳⍵⍳⍵ → 4 2 1 6 0 6
 ⊢= ⍝ Compare that with ⍵. ⍵=⍵⍳⍵⍳⍵ → 1 1 1 0 1 0
 ⍝ This checks if positive indices tunnel back and forth correctly.
 ∨ ⍝ Logical OR with
 0∘> ⍝ 0>⍵ → 0 0 0 1 0 1∨1 1 1 0 1 0 → 1 1 1 1 1 1
 ⍝ Removes the zeroes generated by negative indices
 × ⍝ Multiply that vector with
 ⍨ ⍝ (using ⍵ as both arguments)
 ⍳∘≢ ⍝ Generate the range [0..length(⍵)-1]
 ≠∘ ⍝ And do ⍵≠range; this checks if any 
 ⍝ element in ⍵ is tunneling to itself.
 ⍝ ⍵≠⍳≢⍵ → 4 2 1 ¯1 0 ¯1≠0 1 2 3 4 5 → 1 1 1 1 1 1 
 × ⍝ Multiply that vector with
 ⍨ ⍝ (using ⍵ as both arguments)
 <∘≢ ⍝ ⍵ < length(⍵) → 4 2 1 ¯1 0 ¯1 < 6 → 1 1 1 1 1 1
 ⍝ This checks if any index is out of bounds
×/ ⍝ Finally, multiply and reduce.
 ⍝ ×/1 1 1 1 1 1 → 1 (truthy)

share|improve this answer

edited Sep 6 at 20:41

answered Sep 4 at 17:30

J. Sallé

1,418218

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think this doesn't work for (1), (3 2 1), (5 4 3 2 1).
  – nwellnhof
  Sep 4 at 23:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  0< → × I think
  – Uriel
  Sep 5 at 21:11
  

  
  
  
  

add a comment | 

up vote
5
down vote

up vote
5
down vote

APL (Dyalog Unicode), 19 24 bytes

×/<∘≢⍨×≠∘⍳∘≢⍨×0∘>∨⊢=⊢⍳⍳⍨

Try it online!

Prefix anonymous lambda, returning 1 for truthy and 0 for falsy. The TIO link contains a "prettified" version of the output for the test cases.

Shoutouts to @ngn and @Adám for saving approximately a bazillion bytes.

An extra shoutout to @ngn for the help with fixing the answer for some test cases, and with making it a train.

The updated answer uses ⎕IO←0, setting the Index Origin to 0.

How:

×/<∘≢⍨×≠∘⍳∘≢⍨×0∘>∨⊢=⊢⍳⍳⍨ ⍝ Prefix lambda, argument ⍵ → 4 2 1 ¯1 0 ¯1.
 ⍳⍨ ⍝ Index of (⍳) ⍵ in ⍵. ⍵⍳⍵ → 0 1 2 3 4 3
 ⊢⍳ ⍝ Index of that in ⍵ (returns the vector length if not found). 
 ⍝ ⍵⍳⍵⍳⍵ → 4 2 1 6 0 6
 ⊢= ⍝ Compare that with ⍵. ⍵=⍵⍳⍵⍳⍵ → 1 1 1 0 1 0
 ⍝ This checks if positive indices tunnel back and forth correctly.
 ∨ ⍝ Logical OR with
 0∘> ⍝ 0>⍵ → 0 0 0 1 0 1∨1 1 1 0 1 0 → 1 1 1 1 1 1
 ⍝ Removes the zeroes generated by negative indices
 × ⍝ Multiply that vector with
 ⍨ ⍝ (using ⍵ as both arguments)
 ⍳∘≢ ⍝ Generate the range [0..length(⍵)-1]
 ≠∘ ⍝ And do ⍵≠range; this checks if any 
 ⍝ element in ⍵ is tunneling to itself.
 ⍝ ⍵≠⍳≢⍵ → 4 2 1 ¯1 0 ¯1≠0 1 2 3 4 5 → 1 1 1 1 1 1 
 × ⍝ Multiply that vector with
 ⍨ ⍝ (using ⍵ as both arguments)
 <∘≢ ⍝ ⍵ < length(⍵) → 4 2 1 ¯1 0 ¯1 < 6 → 1 1 1 1 1 1
 ⍝ This checks if any index is out of bounds
×/ ⍝ Finally, multiply and reduce.
 ⍝ ×/1 1 1 1 1 1 → 1 (truthy)

share|improve this answer

edited Sep 6 at 20:41

answered Sep 4 at 17:30

J. Sallé

1,418218

APL (Dyalog Unicode), 19 24 bytes

×/<∘≢⍨×≠∘⍳∘≢⍨×0∘>∨⊢=⊢⍳⍳⍨

Try it online!

Prefix anonymous lambda, returning 1 for truthy and 0 for falsy. The TIO link contains a "prettified" version of the output for the test cases.

Shoutouts to @ngn and @Adám for saving approximately a bazillion bytes.

An extra shoutout to @ngn for the help with fixing the answer for some test cases, and with making it a train.

The updated answer uses ⎕IO←0, setting the Index Origin to 0.

How:

×/<∘≢⍨×≠∘⍳∘≢⍨×0∘>∨⊢=⊢⍳⍳⍨ ⍝ Prefix lambda, argument ⍵ → 4 2 1 ¯1 0 ¯1.
 ⍳⍨ ⍝ Index of (⍳) ⍵ in ⍵. ⍵⍳⍵ → 0 1 2 3 4 3
 ⊢⍳ ⍝ Index of that in ⍵ (returns the vector length if not found). 
 ⍝ ⍵⍳⍵⍳⍵ → 4 2 1 6 0 6
 ⊢= ⍝ Compare that with ⍵. ⍵=⍵⍳⍵⍳⍵ → 1 1 1 0 1 0
 ⍝ This checks if positive indices tunnel back and forth correctly.
 ∨ ⍝ Logical OR with
 0∘> ⍝ 0>⍵ → 0 0 0 1 0 1∨1 1 1 0 1 0 → 1 1 1 1 1 1
 ⍝ Removes the zeroes generated by negative indices
 × ⍝ Multiply that vector with
 ⍨ ⍝ (using ⍵ as both arguments)
 ⍳∘≢ ⍝ Generate the range [0..length(⍵)-1]
 ≠∘ ⍝ And do ⍵≠range; this checks if any 
 ⍝ element in ⍵ is tunneling to itself.
 ⍝ ⍵≠⍳≢⍵ → 4 2 1 ¯1 0 ¯1≠0 1 2 3 4 5 → 1 1 1 1 1 1 
 × ⍝ Multiply that vector with
 ⍨ ⍝ (using ⍵ as both arguments)
 <∘≢ ⍝ ⍵ < length(⍵) → 4 2 1 ¯1 0 ¯1 < 6 → 1 1 1 1 1 1
 ⍝ This checks if any index is out of bounds
×/ ⍝ Finally, multiply and reduce.
 ⍝ ×/1 1 1 1 1 1 → 1 (truthy)

share|improve this answer

edited Sep 6 at 20:41

answered Sep 4 at 17:30

J. Sallé

1,418218

share|improve this answer

share|improve this answer

edited Sep 6 at 20:41

edited Sep 6 at 20:41

edited Sep 6 at 20:41

answered Sep 4 at 17:30

J. Sallé

1,418218

answered Sep 4 at 17:30

J. Sallé

1,418218

answered Sep 4 at 17:30

J. Sallé

1,418218

1,418218

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think this doesn't work for (1), (3 2 1), (5 4 3 2 1).
  – nwellnhof
  Sep 4 at 23:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  0< → × I think
  – Uriel
  Sep 5 at 21:11
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think this doesn't work for (1), (3 2 1), (5 4 3 2 1).
  – nwellnhof
  Sep 4 at 23:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  0< → × I think
  – Uriel
  Sep 5 at 21:11
  

  
  
  
  

I think this doesn't work for (1), (3 2 1), (5 4 3 2 1).
– nwellnhof
Sep 4 at 23:42

I think this doesn't work for (1), (3 2 1), (5 4 3 2 1).
– nwellnhof
Sep 4 at 23:42

0< → × I think
– Uriel
Sep 5 at 21:11

0< → × I think
– Uriel
Sep 5 at 21:11

add a comment | 

up vote
4
down vote

JavaScript (ES6), 35 bytes

Saved 1 byte thanks to @Shaggy

a=>a.every((v,i)=>v<0|v!=i&a[v]==i)

Try it online!

Commented

a => // a = input array
 a.every((v, i) => // for each value v at position i in a:
 v < 0 | // force the test to succeed if v is negative (non-tunnel position)
 v != i & // make sure that this cell is not pointing to itself
 a[v] == i // check the other end of the tunnel
 ) // end of every()

share|improve this answer

edited Sep 5 at 22:02

ngn

6,13812256

answered Sep 4 at 14:32

Arnauld

63.7k580268

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good thing I checked the solutions before posting a port of my Japt solution, which is nearly identical to this. You can save a byte with a=>a.every((v,i)=>v<0|v!=i&a[v]==i).
  – Shaggy
  Sep 4 at 15:09
  

  
  
  
  

add a comment | 

up vote
4
down vote

JavaScript (ES6), 35 bytes

Saved 1 byte thanks to @Shaggy

a=>a.every((v,i)=>v<0|v!=i&a[v]==i)

Try it online!

Commented

a => // a = input array
 a.every((v, i) => // for each value v at position i in a:
 v < 0 | // force the test to succeed if v is negative (non-tunnel position)
 v != i & // make sure that this cell is not pointing to itself
 a[v] == i // check the other end of the tunnel
 ) // end of every()

share|improve this answer

edited Sep 5 at 22:02

ngn

6,13812256

answered Sep 4 at 14:32

Arnauld

63.7k580268

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good thing I checked the solutions before posting a port of my Japt solution, which is nearly identical to this. You can save a byte with a=>a.every((v,i)=>v<0|v!=i&a[v]==i).
  – Shaggy
  Sep 4 at 15:09
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

JavaScript (ES6), 35 bytes

Saved 1 byte thanks to @Shaggy

a=>a.every((v,i)=>v<0|v!=i&a[v]==i)

Try it online!

Commented

a => // a = input array
 a.every((v, i) => // for each value v at position i in a:
 v < 0 | // force the test to succeed if v is negative (non-tunnel position)
 v != i & // make sure that this cell is not pointing to itself
 a[v] == i // check the other end of the tunnel
 ) // end of every()

share|improve this answer

edited Sep 5 at 22:02

ngn

6,13812256

answered Sep 4 at 14:32

Arnauld

63.7k580268

JavaScript (ES6), 35 bytes

Saved 1 byte thanks to @Shaggy

a=>a.every((v,i)=>v<0|v!=i&a[v]==i)

Try it online!

Commented

a => // a = input array
 a.every((v, i) => // for each value v at position i in a:
 v < 0 | // force the test to succeed if v is negative (non-tunnel position)
 v != i & // make sure that this cell is not pointing to itself
 a[v] == i // check the other end of the tunnel
 ) // end of every()

share|improve this answer

edited Sep 5 at 22:02

ngn

6,13812256

answered Sep 4 at 14:32

Arnauld

63.7k580268

share|improve this answer

share|improve this answer

edited Sep 5 at 22:02

ngn

6,13812256

edited Sep 5 at 22:02

ngn

6,13812256

edited Sep 5 at 22:02

ngn

6,13812256

6,13812256

answered Sep 4 at 14:32

Arnauld

63.7k580268

answered Sep 4 at 14:32

Arnauld

63.7k580268

answered Sep 4 at 14:32

Arnauld

63.7k580268

63.7k580268

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good thing I checked the solutions before posting a port of my Japt solution, which is nearly identical to this. You can save a byte with a=>a.every((v,i)=>v<0|v!=i&a[v]==i).
  – Shaggy
  Sep 4 at 15:09
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good thing I checked the solutions before posting a port of my Japt solution, which is nearly identical to this. You can save a byte with a=>a.every((v,i)=>v<0|v!=i&a[v]==i).
  – Shaggy
  Sep 4 at 15:09
  

  
  
  
  

Good thing I checked the solutions before posting a port of my Japt solution, which is nearly identical to this. You can save a byte with a=>a.every((v,i)=>v<0|v!=i&a[v]==i).
– Shaggy
Sep 4 at 15:09

Good thing I checked the solutions before posting a port of my Japt solution, which is nearly identical to this. You can save a byte with a=>a.every((v,i)=>v<0|v!=i&a[v]==i).
– Shaggy
Sep 4 at 15:09

add a comment | 

up vote
3
down vote

Python 2, 65 bytes

lambda l:all(l[v:]>and v!=i==l[v]or v<0for i,v in enumerate(l))

Try it online!

share|improve this answer

answered Sep 4 at 14:35

ovs

17.3k21056

• 
  
  
  

  
  

  
  
  
  
  
  

  
  62 bytes
  – TFeld
  Sep 4 at 14:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  61 bytes
  – ÎŸurous
  Sep 5 at 0:42
  

  
  
  
  

add a comment | 

up vote
3
down vote

Python 2, 65 bytes

lambda l:all(l[v:]>and v!=i==l[v]or v<0for i,v in enumerate(l))

Try it online!

share|improve this answer

answered Sep 4 at 14:35

ovs

17.3k21056

• 
  
  
  

  
  

  
  
  
  
  
  

  
  62 bytes
  – TFeld
  Sep 4 at 14:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  61 bytes
  – ÎŸurous
  Sep 5 at 0:42
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Python 2, 65 bytes

lambda l:all(l[v:]>and v!=i==l[v]or v<0for i,v in enumerate(l))

Try it online!

share|improve this answer

answered Sep 4 at 14:35

ovs

17.3k21056

Python 2, 65 bytes

lambda l:all(l[v:]>and v!=i==l[v]or v<0for i,v in enumerate(l))

Try it online!

share|improve this answer

answered Sep 4 at 14:35

ovs

17.3k21056

share|improve this answer

share|improve this answer

answered Sep 4 at 14:35

ovs

17.3k21056

answered Sep 4 at 14:35

ovs

17.3k21056

answered Sep 4 at 14:35

ovs

17.3k21056

17.3k21056

• 
  
  
  

  
  

  
  
  
  
  
  

  
  62 bytes
  – TFeld
  Sep 4 at 14:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  61 bytes
  – ÎŸurous
  Sep 5 at 0:42
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  62 bytes
  – TFeld
  Sep 4 at 14:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  61 bytes
  – ÎŸurous
  Sep 5 at 0:42
  

  
  
  
  

62 bytes
– TFeld
Sep 4 at 14:48

62 bytes
– TFeld
Sep 4 at 14:48

61 bytes
– ÎŸurous
Sep 5 at 0:42

61 bytes
– ÎŸurous
Sep 5 at 0:42

add a comment | 

up vote
3
down vote

Japt, 14 bytes

eȧJªX¦Y©Y¥UgX

Try it or run all test cases

---

Explanation

 :Implicit input of array U
 È :Map each X at (0-based) index Y
 §J : X less than or equal to -1?
 ª : Logical OR
 X¦Y : X does not equal Y?
 © : Logical AND
 YÂ¥UgX : Y equals the element in U at index X?
e :All truthy?

share|improve this answer

edited Sep 4 at 14:54

answered Sep 4 at 14:45

Shaggy

16.3k21560

add a comment | 

up vote
3
down vote

Japt, 14 bytes

eȧJªX¦Y©Y¥UgX

Try it or run all test cases

---

Explanation

 :Implicit input of array U
 È :Map each X at (0-based) index Y
 §J : X less than or equal to -1?
 ª : Logical OR
 X¦Y : X does not equal Y?
 © : Logical AND
 YÂ¥UgX : Y equals the element in U at index X?
e :All truthy?

share|improve this answer

edited Sep 4 at 14:54

answered Sep 4 at 14:45

Shaggy

16.3k21560

add a comment | 

up vote
3
down vote

up vote
3
down vote

Japt, 14 bytes

eȧJªX¦Y©Y¥UgX

Try it or run all test cases

---

Explanation

 :Implicit input of array U
 È :Map each X at (0-based) index Y
 §J : X less than or equal to -1?
 ª : Logical OR
 X¦Y : X does not equal Y?
 © : Logical AND
 YÂ¥UgX : Y equals the element in U at index X?
e :All truthy?

share|improve this answer

edited Sep 4 at 14:54

answered Sep 4 at 14:45

Shaggy

16.3k21560

Japt, 14 bytes

eȧJªX¦Y©Y¥UgX

Try it or run all test cases

---

Explanation

 :Implicit input of array U
 È :Map each X at (0-based) index Y
 §J : X less than or equal to -1?
 ª : Logical OR
 X¦Y : X does not equal Y?
 © : Logical AND
 YÂ¥UgX : Y equals the element in U at index X?
e :All truthy?

share|improve this answer

edited Sep 4 at 14:54

answered Sep 4 at 14:45

Shaggy

16.3k21560

share|improve this answer

share|improve this answer

edited Sep 4 at 14:54

edited Sep 4 at 14:54

edited Sep 4 at 14:54

answered Sep 4 at 14:45

Shaggy

16.3k21560

answered Sep 4 at 14:45

Shaggy

16.3k21560

answered Sep 4 at 14:45

Shaggy

16.3k21560

16.3k21560

add a comment | 

add a comment | 

up vote
3
down vote

Perl 6, 36 bytes

!.grep:2-set $++,$^v,.[$v]xx$v+1

Try it online!

The basic idea is to check whether the set i, a[i], a[a[i]] contains exactly two distinct elements for each index i with a[i] >= 0. If an element points to itself, the set contains only a single distinct element. If the other end doesn't point back to i, the set contains three distinct elements. If a[i] < 0, the xx factor is zero or negative, so the set is i, a[i] , also with two distinct elements.

share|improve this answer

edited Sep 4 at 17:10

answered Sep 4 at 17:00

nwellnhof

3,503714

add a comment | 

up vote
3
down vote

Perl 6, 36 bytes

!.grep:2-set $++,$^v,.[$v]xx$v+1

Try it online!

The basic idea is to check whether the set i, a[i], a[a[i]] contains exactly two distinct elements for each index i with a[i] >= 0. If an element points to itself, the set contains only a single distinct element. If the other end doesn't point back to i, the set contains three distinct elements. If a[i] < 0, the xx factor is zero or negative, so the set is i, a[i] , also with two distinct elements.

share|improve this answer

edited Sep 4 at 17:10

answered Sep 4 at 17:00

nwellnhof

3,503714

add a comment | 

up vote
3
down vote

up vote
3
down vote

Perl 6, 36 bytes

!.grep:2-set $++,$^v,.[$v]xx$v+1

Try it online!

The basic idea is to check whether the set i, a[i], a[a[i]] contains exactly two distinct elements for each index i with a[i] >= 0. If an element points to itself, the set contains only a single distinct element. If the other end doesn't point back to i, the set contains three distinct elements. If a[i] < 0, the xx factor is zero or negative, so the set is i, a[i] , also with two distinct elements.

share|improve this answer

edited Sep 4 at 17:10

answered Sep 4 at 17:00

nwellnhof

3,503714

Perl 6, 36 bytes

!.grep:2-set $++,$^v,.[$v]xx$v+1

Try it online!

The basic idea is to check whether the set i, a[i], a[a[i]] contains exactly two distinct elements for each index i with a[i] >= 0. If an element points to itself, the set contains only a single distinct element. If the other end doesn't point back to i, the set contains three distinct elements. If a[i] < 0, the xx factor is zero or negative, so the set is i, a[i] , also with two distinct elements.

share|improve this answer

edited Sep 4 at 17:10

answered Sep 4 at 17:00

nwellnhof

3,503714

share|improve this answer

share|improve this answer

edited Sep 4 at 17:10

edited Sep 4 at 17:10

edited Sep 4 at 17:10

answered Sep 4 at 17:00

nwellnhof

3,503714

answered Sep 4 at 17:00

nwellnhof

3,503714

answered Sep 4 at 17:00

nwellnhof

3,503714

3,503714

add a comment | 

add a comment | 

up vote
3
down vote

MATL, 19 18 Bytes

-1 Byte thanks to Luis

n:G=GGG0>f))7M-|hs

Try it online!, for the first one only, because I don't know how to do all of them!

Gives 0 if truthy, a non-zero integer if falsey, eg. for test case 6 gives 4.

Please remember that like MATLAB, MATL is 1-indexed so 1 must be added to the test cases!

Never golfed in an Esolang before, so advice greatly received!

Explained:

n:G=GGG0>f))7M-|hs
 Implicit - input array
n Number of values in array
 : Make array 1:n
 G Push input
 = Equality
n:G= Makes non-zero array if any of the tunnels lead to themselves
 GGG Push input 3x
 0 Push literal 0
 > Greater than
 G0> Makes array of ones where input > 0
 f Find - returns indeces of non-zero values
 Implicit - copy this matrix to clipboard
 ) Indeces - returns array of positive integers in order from input
 ) Ditto - Note, implicit non-zero any above maximum
 7M Paste from clipboard
 - Subtract
 GGG0>f))7M- Makes array of zeros if only two-ended tunnels evident
 | Absolute value (otherwise eg. [3,4,2,1] -> '0')
 h Horizontal concat (ie. joins check for self tunnels and wrong tunnels)
 s Sum; = 0 if truthy, integer otherwise 

share|improve this answer

edited Sep 5 at 23:08

answered Sep 5 at 10:36

Lui

374211

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is my explanation too wordy? I want to make it obvious without going totally overboard.
  – Lui
  Sep 5 at 10:39
  

  
  
  
  

add a comment | 

up vote
3
down vote

MATL, 19 18 Bytes

-1 Byte thanks to Luis

n:G=GGG0>f))7M-|hs

Try it online!, for the first one only, because I don't know how to do all of them!

Gives 0 if truthy, a non-zero integer if falsey, eg. for test case 6 gives 4.

Please remember that like MATLAB, MATL is 1-indexed so 1 must be added to the test cases!

Never golfed in an Esolang before, so advice greatly received!

Explained:

n:G=GGG0>f))7M-|hs
 Implicit - input array
n Number of values in array
 : Make array 1:n
 G Push input
 = Equality
n:G= Makes non-zero array if any of the tunnels lead to themselves
 GGG Push input 3x
 0 Push literal 0
 > Greater than
 G0> Makes array of ones where input > 0
 f Find - returns indeces of non-zero values
 Implicit - copy this matrix to clipboard
 ) Indeces - returns array of positive integers in order from input
 ) Ditto - Note, implicit non-zero any above maximum
 7M Paste from clipboard
 - Subtract
 GGG0>f))7M- Makes array of zeros if only two-ended tunnels evident
 | Absolute value (otherwise eg. [3,4,2,1] -> '0')
 h Horizontal concat (ie. joins check for self tunnels and wrong tunnels)
 s Sum; = 0 if truthy, integer otherwise 

share|improve this answer

edited Sep 5 at 23:08

answered Sep 5 at 10:36

Lui

374211

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is my explanation too wordy? I want to make it obvious without going totally overboard.
  – Lui
  Sep 5 at 10:39
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

MATL, 19 18 Bytes

-1 Byte thanks to Luis

n:G=GGG0>f))7M-|hs

Try it online!, for the first one only, because I don't know how to do all of them!

Gives 0 if truthy, a non-zero integer if falsey, eg. for test case 6 gives 4.

Please remember that like MATLAB, MATL is 1-indexed so 1 must be added to the test cases!

Never golfed in an Esolang before, so advice greatly received!

Explained:

n:G=GGG0>f))7M-|hs
 Implicit - input array
n Number of values in array
 : Make array 1:n
 G Push input
 = Equality
n:G= Makes non-zero array if any of the tunnels lead to themselves
 GGG Push input 3x
 0 Push literal 0
 > Greater than
 G0> Makes array of ones where input > 0
 f Find - returns indeces of non-zero values
 Implicit - copy this matrix to clipboard
 ) Indeces - returns array of positive integers in order from input
 ) Ditto - Note, implicit non-zero any above maximum
 7M Paste from clipboard
 - Subtract
 GGG0>f))7M- Makes array of zeros if only two-ended tunnels evident
 | Absolute value (otherwise eg. [3,4,2,1] -> '0')
 h Horizontal concat (ie. joins check for self tunnels and wrong tunnels)
 s Sum; = 0 if truthy, integer otherwise 

share|improve this answer

edited Sep 5 at 23:08

answered Sep 5 at 10:36

Lui

374211

MATL, 19 18 Bytes

-1 Byte thanks to Luis

n:G=GGG0>f))7M-|hs

Try it online!, for the first one only, because I don't know how to do all of them!

Gives 0 if truthy, a non-zero integer if falsey, eg. for test case 6 gives 4.

Please remember that like MATLAB, MATL is 1-indexed so 1 must be added to the test cases!

Never golfed in an Esolang before, so advice greatly received!

Explained:

n:G=GGG0>f))7M-|hs
 Implicit - input array
n Number of values in array
 : Make array 1:n
 G Push input
 = Equality
n:G= Makes non-zero array if any of the tunnels lead to themselves
 GGG Push input 3x
 0 Push literal 0
 > Greater than
 G0> Makes array of ones where input > 0
 f Find - returns indeces of non-zero values
 Implicit - copy this matrix to clipboard
 ) Indeces - returns array of positive integers in order from input
 ) Ditto - Note, implicit non-zero any above maximum
 7M Paste from clipboard
 - Subtract
 GGG0>f))7M- Makes array of zeros if only two-ended tunnels evident
 | Absolute value (otherwise eg. [3,4,2,1] -> '0')
 h Horizontal concat (ie. joins check for self tunnels and wrong tunnels)
 s Sum; = 0 if truthy, integer otherwise 

share|improve this answer

edited Sep 5 at 23:08

answered Sep 5 at 10:36

Lui

374211

share|improve this answer

share|improve this answer

edited Sep 5 at 23:08

edited Sep 5 at 23:08

edited Sep 5 at 23:08

answered Sep 5 at 10:36

Lui

374211

answered Sep 5 at 10:36

Lui

374211

answered Sep 5 at 10:36

Lui

374211

374211

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is my explanation too wordy? I want to make it obvious without going totally overboard.
  – Lui
  Sep 5 at 10:39
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is my explanation too wordy? I want to make it obvious without going totally overboard.
  – Lui
  Sep 5 at 10:39
  

  
  
  
  

Is my explanation too wordy? I want to make it obvious without going totally overboard.
– Lui
Sep 5 at 10:39

Is my explanation too wordy? I want to make it obvious without going totally overboard.
– Lui
Sep 5 at 10:39

add a comment | 

up vote
2
down vote

Python, 112 97 96 86 bytes

f=lambda l:sum(i==l[i]or len(l)<=l[i]or 0<=l[i]and i!=l[l[i]]for i in range(len(l)))<1

Try it Online!

Returns True or False.

-10 bytes thanks to @Rod and @TFeld.

share|improve this answer

edited Sep 4 at 14:37

answered Sep 4 at 14:17

DimChtz

621111

add a comment | 

up vote
2
down vote

Python, 112 97 96 86 bytes

f=lambda l:sum(i==l[i]or len(l)<=l[i]or 0<=l[i]and i!=l[l[i]]for i in range(len(l)))<1

Try it Online!

Returns True or False.

-10 bytes thanks to @Rod and @TFeld.

share|improve this answer

edited Sep 4 at 14:37

answered Sep 4 at 14:17

DimChtz

621111

add a comment | 

up vote
2
down vote

up vote
2
down vote

Python, 112 97 96 86 bytes

f=lambda l:sum(i==l[i]or len(l)<=l[i]or 0<=l[i]and i!=l[l[i]]for i in range(len(l)))<1

Try it Online!

Returns True or False.

-10 bytes thanks to @Rod and @TFeld.

share|improve this answer

edited Sep 4 at 14:37

answered Sep 4 at 14:17

DimChtz

621111

Python, 112 97 96 86 bytes

f=lambda l:sum(i==l[i]or len(l)<=l[i]or 0<=l[i]and i!=l[l[i]]for i in range(len(l)))<1

Try it Online!

Returns True or False.

-10 bytes thanks to @Rod and @TFeld.

share|improve this answer

edited Sep 4 at 14:37

answered Sep 4 at 14:17

DimChtz

621111

share|improve this answer

share|improve this answer

edited Sep 4 at 14:37

edited Sep 4 at 14:37

edited Sep 4 at 14:37

answered Sep 4 at 14:17

DimChtz

621111

answered Sep 4 at 14:17

DimChtz

621111

answered Sep 4 at 14:17

DimChtz

621111

621111

add a comment | 

add a comment | 

up vote
2
down vote

Jelly, 16 bytes

ị=JanJ$>L<$o<1$Ạ

Try it online!

1-indexed.

share|improve this answer

answered Sep 4 at 14:42

Erik the Outgolfer

29.4k42698

add a comment | 

up vote
2
down vote

Jelly, 16 bytes

ị=JanJ$>L<$o<1$Ạ

Try it online!

1-indexed.

share|improve this answer

answered Sep 4 at 14:42

Erik the Outgolfer

29.4k42698

add a comment | 

up vote
2
down vote

up vote
2
down vote

Jelly, 16 bytes

ị=JanJ$>L<$o<1$Ạ

Try it online!

1-indexed.

share|improve this answer

answered Sep 4 at 14:42

Erik the Outgolfer

29.4k42698

Jelly, 16 bytes

ị=JanJ$>L<$o<1$Ạ

Try it online!

1-indexed.

share|improve this answer

answered Sep 4 at 14:42

Erik the Outgolfer

29.4k42698

share|improve this answer

share|improve this answer

answered Sep 4 at 14:42

Erik the Outgolfer

29.4k42698

answered Sep 4 at 14:42

Erik the Outgolfer

29.4k42698

answered Sep 4 at 14:42

Erik the Outgolfer

29.4k42698

29.4k42698

add a comment | 

add a comment | 

up vote
2
down vote

05AB1E, 16 bytes

ε©0‹®NÊI®èNQ*~}P

Try it online or verify all test cases.

Explanation:

ε } # Map each value to:
 © # Store the current item in the register (without popping)
 0‹ # If the current value is negative
 ~ # Or if:
 ®NÊ # The current value is not equal to the current index
 * # And
 I®èNQ # The current value indexed into the input is equal to the index
 P # And output whether all values are truthy after the map

share|improve this answer

edited Sep 4 at 16:51

answered Sep 4 at 15:15

Kevin Cruijssen

29.6k550162

• 
  
  
  

  
  

  
  
  
  
  
  

  
  My try was the same except with filter. I don't see a way to improve on this.
  – Emigna
  Sep 4 at 16:43
  

  
  
  
  

add a comment | 

up vote
2
down vote

05AB1E, 16 bytes

ε©0‹®NÊI®èNQ*~}P

Try it online or verify all test cases.

Explanation:

ε } # Map each value to:
 © # Store the current item in the register (without popping)
 0‹ # If the current value is negative
 ~ # Or if:
 ®NÊ # The current value is not equal to the current index
 * # And
 I®èNQ # The current value indexed into the input is equal to the index
 P # And output whether all values are truthy after the map

share|improve this answer

edited Sep 4 at 16:51

answered Sep 4 at 15:15

Kevin Cruijssen

29.6k550162

• 
  
  
  

  
  

  
  
  
  
  
  

  
  My try was the same except with filter. I don't see a way to improve on this.
  – Emigna
  Sep 4 at 16:43
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

05AB1E, 16 bytes

ε©0‹®NÊI®èNQ*~}P

Try it online or verify all test cases.

Explanation:

ε } # Map each value to:
 © # Store the current item in the register (without popping)
 0‹ # If the current value is negative
 ~ # Or if:
 ®NÊ # The current value is not equal to the current index
 * # And
 I®èNQ # The current value indexed into the input is equal to the index
 P # And output whether all values are truthy after the map

share|improve this answer

edited Sep 4 at 16:51

answered Sep 4 at 15:15

Kevin Cruijssen

29.6k550162

05AB1E, 16 bytes

ε©0‹®NÊI®èNQ*~}P

Try it online or verify all test cases.

Explanation:

ε } # Map each value to:
 © # Store the current item in the register (without popping)
 0‹ # If the current value is negative
 ~ # Or if:
 ®NÊ # The current value is not equal to the current index
 * # And
 I®èNQ # The current value indexed into the input is equal to the index
 P # And output whether all values are truthy after the map

share|improve this answer

edited Sep 4 at 16:51

answered Sep 4 at 15:15

Kevin Cruijssen

29.6k550162

share|improve this answer

share|improve this answer

edited Sep 4 at 16:51

edited Sep 4 at 16:51

edited Sep 4 at 16:51

answered Sep 4 at 15:15

Kevin Cruijssen

29.6k550162

answered Sep 4 at 15:15

Kevin Cruijssen

29.6k550162

answered Sep 4 at 15:15

Kevin Cruijssen

29.6k550162

29.6k550162

• 
  
  
  

  
  

  
  
  
  
  
  

  
  My try was the same except with filter. I don't see a way to improve on this.
  – Emigna
  Sep 4 at 16:43
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  My try was the same except with filter. I don't see a way to improve on this.
  – Emigna
  Sep 4 at 16:43
  

  
  
  
  

My try was the same except with filter. I don't see a way to improve on this.
– Emigna
Sep 4 at 16:43

My try was the same except with filter. I don't see a way to improve on this.
– Emigna
Sep 4 at 16:43

add a comment | 

up vote
2
down vote

Java (JDK 10), 95 bytes

a->int l=a.length,i=l;for(;i-->0;)if(a[i]>=0&&(a[i]>=l

Try it online!

Credits

• -3 bytes thanks to AlexRacer
  

share|improve this answer

edited Sep 4 at 19:43

answered Sep 4 at 15:32

Olivier Grégoire

7,57811741

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could have been 87 bytes if it weren't for that pesky IndexOutOfBoundsException. Maybe you see something to fix it easily?
  – Kevin Cruijssen
  Sep 4 at 15:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen I can see how to fix that for 102 bytes. Nothing shorter yet :(
  – Olivier Grégoire
  Sep 4 at 15:45
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  -3 bytes - omit r and break out of the loop analogous to here
  – AlexRacer
  Sep 4 at 19:00
  

  
  
  
  

add a comment | 

up vote
2
down vote

Java (JDK 10), 95 bytes

a->int l=a.length,i=l;for(;i-->0;)if(a[i]>=0&&(a[i]>=l

Try it online!

Credits

• -3 bytes thanks to AlexRacer
  

share|improve this answer

edited Sep 4 at 19:43

answered Sep 4 at 15:32

Olivier Grégoire

7,57811741

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could have been 87 bytes if it weren't for that pesky IndexOutOfBoundsException. Maybe you see something to fix it easily?
  – Kevin Cruijssen
  Sep 4 at 15:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen I can see how to fix that for 102 bytes. Nothing shorter yet :(
  – Olivier Grégoire
  Sep 4 at 15:45
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  -3 bytes - omit r and break out of the loop analogous to here
  – AlexRacer
  Sep 4 at 19:00
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

Java (JDK 10), 95 bytes

a->int l=a.length,i=l;for(;i-->0;)if(a[i]>=0&&(a[i]>=l

Try it online!

Credits

• -3 bytes thanks to AlexRacer
  

share|improve this answer

edited Sep 4 at 19:43

answered Sep 4 at 15:32

Olivier Grégoire

7,57811741

Java (JDK 10), 95 bytes

a->int l=a.length,i=l;for(;i-->0;)if(a[i]>=0&&(a[i]>=l

Try it online!

Credits

• -3 bytes thanks to AlexRacer
  

share|improve this answer

edited Sep 4 at 19:43

answered Sep 4 at 15:32

Olivier Grégoire

7,57811741

share|improve this answer

share|improve this answer

edited Sep 4 at 19:43

edited Sep 4 at 19:43

edited Sep 4 at 19:43

answered Sep 4 at 15:32

Olivier Grégoire

7,57811741

answered Sep 4 at 15:32

Olivier Grégoire

7,57811741

answered Sep 4 at 15:32

Olivier Grégoire

7,57811741

7,57811741

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could have been 87 bytes if it weren't for that pesky IndexOutOfBoundsException. Maybe you see something to fix it easily?
  – Kevin Cruijssen
  Sep 4 at 15:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen I can see how to fix that for 102 bytes. Nothing shorter yet :(
  – Olivier Grégoire
  Sep 4 at 15:45
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  -3 bytes - omit r and break out of the loop analogous to here
  – AlexRacer
  Sep 4 at 19:00
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Could have been 87 bytes if it weren't for that pesky IndexOutOfBoundsException. Maybe you see something to fix it easily?
  – Kevin Cruijssen
  Sep 4 at 15:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen I can see how to fix that for 102 bytes. Nothing shorter yet :(
  – Olivier Grégoire
  Sep 4 at 15:45
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  -3 bytes - omit r and break out of the loop analogous to here
  – AlexRacer
  Sep 4 at 19:00
  

  
  
  
  

Could have been 87 bytes if it weren't for that pesky IndexOutOfBoundsException. Maybe you see something to fix it easily?
– Kevin Cruijssen
Sep 4 at 15:41

Could have been 87 bytes if it weren't for that pesky IndexOutOfBoundsException. Maybe you see something to fix it easily?
– Kevin Cruijssen
Sep 4 at 15:41

@KevinCruijssen I can see how to fix that for 102 bytes. Nothing shorter yet :(
– Olivier Grégoire
Sep 4 at 15:45

@KevinCruijssen I can see how to fix that for 102 bytes. Nothing shorter yet :(
– Olivier Grégoire
Sep 4 at 15:45

1

1

-3 bytes - omit r and break out of the loop analogous to here
– AlexRacer
Sep 4 at 19:00

-3 bytes - omit r and break out of the loop analogous to here
– AlexRacer
Sep 4 at 19:00

add a comment | 

up vote
2
down vote

Haskell, 48 bytes

(all=<< u(x,y)->y<0||x/=y&&elem(y,x)u).zip[0..]

Verify all testcases!

Explanation

Let's first ungolf the code a bit. As f =<< g is the same as x -> f (g x) x, the code is equivalent to

(u->all((x,y)->y<0||x/=y&&elem(y,x)u)u).zip[0..]

which is a bit clearer.

(u -> -- given u, return
 all ((x, y) -> -- whether for all elements (x, y) of u
 y < 0 || -- either y < 0, or
 x /= y && elem (y, x) u -- (x /= y) and ((y, x) is in u)
 )
 u
) . zip [0..] -- given the array a (implicitly via point-free style),
 -- return the array augmented with indices (it's the u above)

This solution is based on a simple observation: let a be the input array, and u the list of pairs (i, a[i]) where i is an index. Then a is a valid array if and only if for every (x, y) in u with y >= 0, the pair (y, x) belongs to u as well.

share|improve this answer

edited Sep 6 at 16:32

answered Sep 5 at 16:34

Delfad0r

914

New contributor

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
2
down vote

Haskell, 48 bytes

(all=<< u(x,y)->y<0||x/=y&&elem(y,x)u).zip[0..]

Verify all testcases!

Explanation

Let's first ungolf the code a bit. As f =<< g is the same as x -> f (g x) x, the code is equivalent to

(u->all((x,y)->y<0||x/=y&&elem(y,x)u)u).zip[0..]

which is a bit clearer.

(u -> -- given u, return
 all ((x, y) -> -- whether for all elements (x, y) of u
 y < 0 || -- either y < 0, or
 x /= y && elem (y, x) u -- (x /= y) and ((y, x) is in u)
 )
 u
) . zip [0..] -- given the array a (implicitly via point-free style),
 -- return the array augmented with indices (it's the u above)

This solution is based on a simple observation: let a be the input array, and u the list of pairs (i, a[i]) where i is an index. Then a is a valid array if and only if for every (x, y) in u with y >= 0, the pair (y, x) belongs to u as well.

share|improve this answer

edited Sep 6 at 16:32

answered Sep 5 at 16:34

Delfad0r

914

New contributor

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
2
down vote

up vote
2
down vote

Haskell, 48 bytes

(all=<< u(x,y)->y<0||x/=y&&elem(y,x)u).zip[0..]

Verify all testcases!

Explanation

Let's first ungolf the code a bit. As f =<< g is the same as x -> f (g x) x, the code is equivalent to

(u->all((x,y)->y<0||x/=y&&elem(y,x)u)u).zip[0..]

which is a bit clearer.

(u -> -- given u, return
 all ((x, y) -> -- whether for all elements (x, y) of u
 y < 0 || -- either y < 0, or
 x /= y && elem (y, x) u -- (x /= y) and ((y, x) is in u)
 )
 u
) . zip [0..] -- given the array a (implicitly via point-free style),
 -- return the array augmented with indices (it's the u above)

This solution is based on a simple observation: let a be the input array, and u the list of pairs (i, a[i]) where i is an index. Then a is a valid array if and only if for every (x, y) in u with y >= 0, the pair (y, x) belongs to u as well.

share|improve this answer

edited Sep 6 at 16:32

answered Sep 5 at 16:34

Delfad0r

914

New contributor

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Haskell, 48 bytes

(all=<< u(x,y)->y<0||x/=y&&elem(y,x)u).zip[0..]

Verify all testcases!

Explanation

Let's first ungolf the code a bit. As f =<< g is the same as x -> f (g x) x, the code is equivalent to

(u->all((x,y)->y<0||x/=y&&elem(y,x)u)u).zip[0..]

which is a bit clearer.

(u -> -- given u, return
 all ((x, y) -> -- whether for all elements (x, y) of u
 y < 0 || -- either y < 0, or
 x /= y && elem (y, x) u -- (x /= y) and ((y, x) is in u)
 )
 u
) . zip [0..] -- given the array a (implicitly via point-free style),
 -- return the array augmented with indices (it's the u above)

This solution is based on a simple observation: let a be the input array, and u the list of pairs (i, a[i]) where i is an index. Then a is a valid array if and only if for every (x, y) in u with y >= 0, the pair (y, x) belongs to u as well.

share|improve this answer

edited Sep 6 at 16:32

answered Sep 5 at 16:34

Delfad0r

914

New contributor

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this answer

share|improve this answer

edited Sep 6 at 16:32

edited Sep 6 at 16:32

edited Sep 6 at 16:32

answered Sep 5 at 16:34

Delfad0r

914

New contributor

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Sep 5 at 16:34

Delfad0r

914

answered Sep 5 at 16:34

Delfad0r

914

914

New contributor

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

add a comment | 

up vote
1
down vote

Charcoal, 22 bytes

¬Φθ∨⁼ικ¬∨‹ι⁰∧‹ιＬθ⁼κ§θι

Try it online! Link is to verbose version of code. Outputs - for truthy and nothing for falsy. Note: Inputting an empty array seems to crash Charcoal, but for now you can enter a space instead, which is near enough. Explanation:

 θ Input array
 Φ Filter elements
 ι Current value
 ⁼ Equals
 κ Current index
 ∨ Or
 ¬ Not
 ι Current value
 ‹ ⁰ Is less than zero
 ∨ Or
 ι Current value
 ‹ Is less than
 Ｌ Length of
 θ Input array
 ∧ And
 κ Current index
 ⁼ Equals
 §θι Indexed value
¬ Logical Not (i.e. is result empty)
 Implicitly print

share|improve this answer

answered Sep 4 at 15:15

Neil

75.1k744170

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This doesn't seem to be a very Charcoalable challenge... :-)
  – Charlie
  Sep 4 at 15:24
  

  
  
  
  

add a comment | 

up vote
1
down vote

Charcoal, 22 bytes

¬Φθ∨⁼ικ¬∨‹ι⁰∧‹ιＬθ⁼κ§θι

Try it online! Link is to verbose version of code. Outputs - for truthy and nothing for falsy. Note: Inputting an empty array seems to crash Charcoal, but for now you can enter a space instead, which is near enough. Explanation:

 θ Input array
 Φ Filter elements
 ι Current value
 ⁼ Equals
 κ Current index
 ∨ Or
 ¬ Not
 ι Current value
 ‹ ⁰ Is less than zero
 ∨ Or
 ι Current value
 ‹ Is less than
 Ｌ Length of
 θ Input array
 ∧ And
 κ Current index
 ⁼ Equals
 §θι Indexed value
¬ Logical Not (i.e. is result empty)
 Implicitly print

share|improve this answer

answered Sep 4 at 15:15

Neil

75.1k744170

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This doesn't seem to be a very Charcoalable challenge... :-)
  – Charlie
  Sep 4 at 15:24
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Charcoal, 22 bytes

¬Φθ∨⁼ικ¬∨‹ι⁰∧‹ιＬθ⁼κ§θι

Try it online! Link is to verbose version of code. Outputs - for truthy and nothing for falsy. Note: Inputting an empty array seems to crash Charcoal, but for now you can enter a space instead, which is near enough. Explanation:

 θ Input array
 Φ Filter elements
 ι Current value
 ⁼ Equals
 κ Current index
 ∨ Or
 ¬ Not
 ι Current value
 ‹ ⁰ Is less than zero
 ∨ Or
 ι Current value
 ‹ Is less than
 Ｌ Length of
 θ Input array
 ∧ And
 κ Current index
 ⁼ Equals
 §θι Indexed value
¬ Logical Not (i.e. is result empty)
 Implicitly print

share|improve this answer

answered Sep 4 at 15:15

Neil

75.1k744170

Charcoal, 22 bytes

¬Φθ∨⁼ικ¬∨‹ι⁰∧‹ιＬθ⁼κ§θι

Try it online! Link is to verbose version of code. Outputs - for truthy and nothing for falsy. Note: Inputting an empty array seems to crash Charcoal, but for now you can enter a space instead, which is near enough. Explanation:

 θ Input array
 Φ Filter elements
 ι Current value
 ⁼ Equals
 κ Current index
 ∨ Or
 ¬ Not
 ι Current value
 ‹ ⁰ Is less than zero
 ∨ Or
 ι Current value
 ‹ Is less than
 Ｌ Length of
 θ Input array
 ∧ And
 κ Current index
 ⁼ Equals
 §θι Indexed value
¬ Logical Not (i.e. is result empty)
 Implicitly print

share|improve this answer

answered Sep 4 at 15:15

Neil

75.1k744170

share|improve this answer

share|improve this answer

answered Sep 4 at 15:15

Neil

75.1k744170

answered Sep 4 at 15:15

Neil

75.1k744170

answered Sep 4 at 15:15

Neil

75.1k744170

75.1k744170

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This doesn't seem to be a very Charcoalable challenge... :-)
  – Charlie
  Sep 4 at 15:24
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This doesn't seem to be a very Charcoalable challenge... :-)
  – Charlie
  Sep 4 at 15:24
  

  
  
  
  

This doesn't seem to be a very Charcoalable challenge... :-)
– Charlie
Sep 4 at 15:24

This doesn't seem to be a very Charcoalable challenge... :-)
– Charlie
Sep 4 at 15:24

add a comment | 

up vote
1
down vote

Pascal (FPC), 165 155 153 bytes

function f(a:array of int32):byte;var i:int32;begin f:=1;for i:=0to length(a)-1do if a[i]>-1then if(a[i]=i)or(a[i]>length(a))or(a[a[i]]<>i)then f:=0;end;

Try it online!

Made function this time because the input is array. Returns 1 for truthy and 0 for falsey.

share|improve this answer

edited Sep 4 at 18:38

answered Sep 4 at 18:25

AlexRacer

57928

add a comment | 

up vote
1
down vote

Pascal (FPC), 165 155 153 bytes

function f(a:array of int32):byte;var i:int32;begin f:=1;for i:=0to length(a)-1do if a[i]>-1then if(a[i]=i)or(a[i]>length(a))or(a[a[i]]<>i)then f:=0;end;

Try it online!

Made function this time because the input is array. Returns 1 for truthy and 0 for falsey.

share|improve this answer

edited Sep 4 at 18:38

answered Sep 4 at 18:25

AlexRacer

57928

add a comment | 

up vote
1
down vote

up vote
1
down vote

Pascal (FPC), 165 155 153 bytes

function f(a:array of int32):byte;var i:int32;begin f:=1;for i:=0to length(a)-1do if a[i]>-1then if(a[i]=i)or(a[i]>length(a))or(a[a[i]]<>i)then f:=0;end;

Try it online!

Made function this time because the input is array. Returns 1 for truthy and 0 for falsey.

share|improve this answer

edited Sep 4 at 18:38

answered Sep 4 at 18:25

AlexRacer

57928

Pascal (FPC), 165 155 153 bytes

function f(a:array of int32):byte;var i:int32;begin f:=1;for i:=0to length(a)-1do if a[i]>-1then if(a[i]=i)or(a[i]>length(a))or(a[a[i]]<>i)then f:=0;end;

Try it online!

Made function this time because the input is array. Returns 1 for truthy and 0 for falsey.

share|improve this answer

edited Sep 4 at 18:38

answered Sep 4 at 18:25

AlexRacer

57928

share|improve this answer

share|improve this answer

edited Sep 4 at 18:38

edited Sep 4 at 18:38

edited Sep 4 at 18:38

answered Sep 4 at 18:25

AlexRacer

57928

answered Sep 4 at 18:25

AlexRacer

57928

answered Sep 4 at 18:25

AlexRacer

57928

57928

add a comment | 

add a comment | 

up vote
1
down vote

Clean, 60 bytes

import StdEnv
@l=and[v<0||l%(v,v)==[i]&&v<>i\v<-l&i<-[0..]]

Try it online!

Clean, 142 bytes

Vastly over-complicated monster version:

import StdEnv,Data.List,Data.Maybe
$l=and[?i(mapMaybe((!?)l)j)j\i<-l&j<-map((!?)l)l|i>=0]with?a(Just(Just c))(Just b)=a==c&&b<>c;?_ _ _=False

Try it online!

Explained:

$ l // function $ of `l` is
 = and [ // true when all elements are true
 ? // apply ? to
 i // the element `i` of `l`
 (mapMaybe // and the result of attempting to
 ((!?)l) // try gettting an element from `l`
 j) // at the potentially invalid index `j`
 j // and `j` itself, which may not exist
 \ i <- l // for every element `i` in `l`
 & j <- map // and every potential `j` in
 ((!?)l) // `l` trying to be indexed by
 l // every element in `l`
 | i >= 0 // where `i` is greater than zero
 ]
with
 ? a (Just (Just c)) (Just b) // function ? when all the arguments exist
 = a==c && b<>c // `a` equals `c` and not `b`
 ;
 ? _ _ _ = False // for all other arguments, ? is false

share|improve this answer

edited Sep 5 at 0:33

answered Sep 4 at 23:41

Οurous

5,2131931

add a comment | 

up vote
1
down vote

Clean, 60 bytes

import StdEnv
@l=and[v<0||l%(v,v)==[i]&&v<>i\v<-l&i<-[0..]]

Try it online!

Clean, 142 bytes

Vastly over-complicated monster version:

import StdEnv,Data.List,Data.Maybe
$l=and[?i(mapMaybe((!?)l)j)j\i<-l&j<-map((!?)l)l|i>=0]with?a(Just(Just c))(Just b)=a==c&&b<>c;?_ _ _=False

Try it online!

Explained:

$ l // function $ of `l` is
 = and [ // true when all elements are true
 ? // apply ? to
 i // the element `i` of `l`
 (mapMaybe // and the result of attempting to
 ((!?)l) // try gettting an element from `l`
 j) // at the potentially invalid index `j`
 j // and `j` itself, which may not exist
 \ i <- l // for every element `i` in `l`
 & j <- map // and every potential `j` in
 ((!?)l) // `l` trying to be indexed by
 l // every element in `l`
 | i >= 0 // where `i` is greater than zero
 ]
with
 ? a (Just (Just c)) (Just b) // function ? when all the arguments exist
 = a==c && b<>c // `a` equals `c` and not `b`
 ;
 ? _ _ _ = False // for all other arguments, ? is false

share|improve this answer

edited Sep 5 at 0:33

answered Sep 4 at 23:41

Οurous

5,2131931

add a comment | 

up vote
1
down vote

up vote
1
down vote

Clean, 60 bytes

import StdEnv
@l=and[v<0||l%(v,v)==[i]&&v<>i\v<-l&i<-[0..]]

Try it online!

Clean, 142 bytes

Vastly over-complicated monster version:

import StdEnv,Data.List,Data.Maybe
$l=and[?i(mapMaybe((!?)l)j)j\i<-l&j<-map((!?)l)l|i>=0]with?a(Just(Just c))(Just b)=a==c&&b<>c;?_ _ _=False

Try it online!

Explained:

$ l // function $ of `l` is
 = and [ // true when all elements are true
 ? // apply ? to
 i // the element `i` of `l`
 (mapMaybe // and the result of attempting to
 ((!?)l) // try gettting an element from `l`
 j) // at the potentially invalid index `j`
 j // and `j` itself, which may not exist
 \ i <- l // for every element `i` in `l`
 & j <- map // and every potential `j` in
 ((!?)l) // `l` trying to be indexed by
 l // every element in `l`
 | i >= 0 // where `i` is greater than zero
 ]
with
 ? a (Just (Just c)) (Just b) // function ? when all the arguments exist
 = a==c && b<>c // `a` equals `c` and not `b`
 ;
 ? _ _ _ = False // for all other arguments, ? is false

share|improve this answer

edited Sep 5 at 0:33

answered Sep 4 at 23:41

Οurous

5,2131931

Clean, 60 bytes

import StdEnv
@l=and[v<0||l%(v,v)==[i]&&v<>i\v<-l&i<-[0..]]

Try it online!

Clean, 142 bytes

Vastly over-complicated monster version:

import StdEnv,Data.List,Data.Maybe
$l=and[?i(mapMaybe((!?)l)j)j\i<-l&j<-map((!?)l)l|i>=0]with?a(Just(Just c))(Just b)=a==c&&b<>c;?_ _ _=False

Try it online!

Explained:

$ l // function $ of `l` is
 = and [ // true when all elements are true
 ? // apply ? to
 i // the element `i` of `l`
 (mapMaybe // and the result of attempting to
 ((!?)l) // try gettting an element from `l`
 j) // at the potentially invalid index `j`
 j // and `j` itself, which may not exist
 \ i <- l // for every element `i` in `l`
 & j <- map // and every potential `j` in
 ((!?)l) // `l` trying to be indexed by
 l // every element in `l`
 | i >= 0 // where `i` is greater than zero
 ]
with
 ? a (Just (Just c)) (Just b) // function ? when all the arguments exist
 = a==c && b<>c // `a` equals `c` and not `b`
 ;
 ? _ _ _ = False // for all other arguments, ? is false

share|improve this answer

edited Sep 5 at 0:33

answered Sep 4 at 23:41

Οurous

5,2131931

share|improve this answer

share|improve this answer

edited Sep 5 at 0:33

edited Sep 5 at 0:33

edited Sep 5 at 0:33

answered Sep 4 at 23:41

Οurous

5,2131931

answered Sep 4 at 23:41

Οurous

5,2131931

answered Sep 4 at 23:41

Οurous

5,2131931

5,2131931

add a comment | 

add a comment | 

up vote
1
down vote

Ruby, 44 bytes

->aa.all?x<0

Try it online!

share|improve this answer

answered Sep 5 at 10:24

G B

6,3571324

add a comment | 

up vote
1
down vote

Ruby, 44 bytes

->aa.all?x<0

Try it online!

share|improve this answer

answered Sep 5 at 10:24

G B

6,3571324

add a comment | 

up vote
1
down vote

up vote
1
down vote

Ruby, 44 bytes

->aa.all?x<0

Try it online!

share|improve this answer

answered Sep 5 at 10:24

G B

6,3571324

Ruby, 44 bytes

->aa.all?x<0

Try it online!

share|improve this answer

answered Sep 5 at 10:24

G B

6,3571324

share|improve this answer

share|improve this answer

answered Sep 5 at 10:24

G B

6,3571324

answered Sep 5 at 10:24

G B

6,3571324

answered Sep 5 at 10:24

G B

6,3571324

6,3571324

add a comment | 

add a comment | 

up vote
1
down vote

Pyth, 17 16 bytes

.A.e|>0b&nbkq@Qb

Try it online here, or verify all the test cases at once here.

.A.e|>0b&nbkq@QbkQ Implicit: Q=eval(input())
 Trailing k, Q inferred
 .e Q Map the input with b=element, k=index, using:
 >0b 0>b
 | OR (
 nbk b != k
 & AND
 q@Qbk Q[b] == k)
.A Check if all elements are truthy

Edit: realised that the trailing k was also unnecessary

share|improve this answer

edited Sep 5 at 12:52

answered Sep 5 at 8:19

Sok

2,851722

add a comment | 

up vote
1
down vote

Pyth, 17 16 bytes

.A.e|>0b&nbkq@Qb

Try it online here, or verify all the test cases at once here.

.A.e|>0b&nbkq@QbkQ Implicit: Q=eval(input())
 Trailing k, Q inferred
 .e Q Map the input with b=element, k=index, using:
 >0b 0>b
 | OR (
 nbk b != k
 & AND
 q@Qbk Q[b] == k)
.A Check if all elements are truthy

Edit: realised that the trailing k was also unnecessary

share|improve this answer

edited Sep 5 at 12:52

answered Sep 5 at 8:19

Sok

2,851722

add a comment | 

up vote
1
down vote

up vote
1
down vote

Pyth, 17 16 bytes

.A.e|>0b&nbkq@Qb

Try it online here, or verify all the test cases at once here.

.A.e|>0b&nbkq@QbkQ Implicit: Q=eval(input())
 Trailing k, Q inferred
 .e Q Map the input with b=element, k=index, using:
 >0b 0>b
 | OR (
 nbk b != k
 & AND
 q@Qbk Q[b] == k)
.A Check if all elements are truthy

Edit: realised that the trailing k was also unnecessary

share|improve this answer

edited Sep 5 at 12:52

answered Sep 5 at 8:19

Sok

2,851722

Pyth, 17 16 bytes

.A.e|>0b&nbkq@Qb

Try it online here, or verify all the test cases at once here.

.A.e|>0b&nbkq@QbkQ Implicit: Q=eval(input())
 Trailing k, Q inferred
 .e Q Map the input with b=element, k=index, using:
 >0b 0>b
 | OR (
 nbk b != k
 & AND
 q@Qbk Q[b] == k)
.A Check if all elements are truthy

Edit: realised that the trailing k was also unnecessary

share|improve this answer

edited Sep 5 at 12:52

answered Sep 5 at 8:19

Sok

2,851722

share|improve this answer

share|improve this answer

edited Sep 5 at 12:52

edited Sep 5 at 12:52

edited Sep 5 at 12:52

answered Sep 5 at 8:19

Sok

2,851722

answered Sep 5 at 8:19

Sok

2,851722

answered Sep 5 at 8:19

Sok

2,851722

2,851722

add a comment | 

add a comment | 

up vote
1
down vote

Groovy, 52 bytes

o=!(i=0);it.each(it[e]==i&&i-e);i++;o

Try it online!

share|improve this answer

answered Sep 5 at 13:57

GolfIsAGoodWalkSpoilt

812

add a comment | 

up vote
1
down vote

Groovy, 52 bytes

o=!(i=0);it.each(it[e]==i&&i-e);i++;o

Try it online!

share|improve this answer

answered Sep 5 at 13:57

GolfIsAGoodWalkSpoilt

812

add a comment | 

up vote
1
down vote

up vote
1
down vote

Groovy, 52 bytes

o=!(i=0);it.each(it[e]==i&&i-e);i++;o

Try it online!

share|improve this answer

answered Sep 5 at 13:57

GolfIsAGoodWalkSpoilt

812

Groovy, 52 bytes

o=!(i=0);it.each(it[e]==i&&i-e);i++;o

Try it online!

share|improve this answer

answered Sep 5 at 13:57

GolfIsAGoodWalkSpoilt

812

share|improve this answer

share|improve this answer

answered Sep 5 at 13:57

GolfIsAGoodWalkSpoilt

812

answered Sep 5 at 13:57

GolfIsAGoodWalkSpoilt

812

answered Sep 5 at 13:57

GolfIsAGoodWalkSpoilt

812

812

add a comment | 

add a comment | 

up vote
1
down vote

Perl 5, 54 bytes

$i=-1;!grep$_>=0*$i++&&($_==$i

Try it online!

share|improve this answer

answered Sep 5 at 19:30

Kjetil S.

51115

add a comment | 

up vote
1
down vote

Perl 5, 54 bytes

$i=-1;!grep$_>=0*$i++&&($_==$i

Try it online!

share|improve this answer

answered Sep 5 at 19:30

Kjetil S.

51115

add a comment | 

up vote
1
down vote

up vote
1
down vote

Perl 5, 54 bytes

$i=-1;!grep$_>=0*$i++&&($_==$i

Try it online!

share|improve this answer

answered Sep 5 at 19:30

Kjetil S.

51115

Perl 5, 54 bytes

$i=-1;!grep$_>=0*$i++&&($_==$i

Try it online!

share|improve this answer

answered Sep 5 at 19:30

Kjetil S.

51115

share|improve this answer

share|improve this answer

answered Sep 5 at 19:30

Kjetil S.

51115

answered Sep 5 at 19:30

Kjetil S.

51115

answered Sep 5 at 19:30

Kjetil S.

51115

51115

add a comment | 

add a comment | 

up vote
1
down vote

K (ngn/k), 33 bytes

(x<#x)&(~x=!#x)&x=x?x?x

Try it online!

share|improve this answer

edited Sep 5 at 21:53

answered Sep 4 at 17:34

ngn

6,13812256

add a comment | 

up vote
1
down vote

K (ngn/k), 33 bytes

(x<#x)&(~x=!#x)&x=x?x?x

Try it online!

share|improve this answer

edited Sep 5 at 21:53

answered Sep 4 at 17:34

ngn

6,13812256

add a comment | 

up vote
1
down vote

up vote
1
down vote

K (ngn/k), 33 bytes

(x<#x)&(~x=!#x)&x=x?x?x

Try it online!

share|improve this answer

edited Sep 5 at 21:53

answered Sep 4 at 17:34

ngn

6,13812256

K (ngn/k), 33 bytes

(x<#x)&(~x=!#x)&x=x?x?x

Try it online!

share|improve this answer

edited Sep 5 at 21:53

answered Sep 4 at 17:34

ngn

6,13812256

share|improve this answer

share|improve this answer

edited Sep 5 at 21:53

edited Sep 5 at 21:53

edited Sep 5 at 21:53

answered Sep 4 at 17:34

ngn

6,13812256

answered Sep 4 at 17:34

ngn

6,13812256

answered Sep 4 at 17:34

ngn

6,13812256

6,13812256

add a comment | 

add a comment | 

up vote
1
down vote

C (gcc), 95 bytes

i(_,o,O,Q,I)int*_;Q>=o

Try it online!

share|improve this answer

answered Sep 6 at 21:55

Jonathan Frech

5,60311039

add a comment | 

up vote
1
down vote

C (gcc), 95 bytes

i(_,o,O,Q,I)int*_;Q>=o

Try it online!

share|improve this answer

answered Sep 6 at 21:55

Jonathan Frech

5,60311039

add a comment | 

up vote
1
down vote

up vote
1
down vote

C (gcc), 95 bytes

i(_,o,O,Q,I)int*_;Q>=o

Try it online!

share|improve this answer

answered Sep 6 at 21:55

Jonathan Frech

5,60311039

C (gcc), 95 bytes

i(_,o,O,Q,I)int*_;Q>=o

Try it online!

share|improve this answer

answered Sep 6 at 21:55

Jonathan Frech

5,60311039

share|improve this answer

share|improve this answer

answered Sep 6 at 21:55

Jonathan Frech

5,60311039

answered Sep 6 at 21:55

Jonathan Frech

5,60311039

answered Sep 6 at 21:55

Jonathan Frech

5,60311039

5,60311039

add a comment | 

add a comment | 

up vote
0
down vote

Mathematica, 42 bytes

#==||(a=0@@#)[[#]]=!=a&&a[[#]][[#]]===a&

Pure function. Takes a 1-indexed list of numbers as input and returns True or False as output. Just follows the tunnels, ensuring that 0 maps to 0, no 1-cycles exist, and all cycles are 2-cycles. (I'm not entirely sure if this fails on any edge cases, but it gives the correct results for the examples.)

share|improve this answer

answered Sep 5 at 1:36

LegionMammal978

14.8k41752

add a comment | 

up vote
0
down vote

Mathematica, 42 bytes

#==||(a=0@@#)[[#]]=!=a&&a[[#]][[#]]===a&

Pure function. Takes a 1-indexed list of numbers as input and returns True or False as output. Just follows the tunnels, ensuring that 0 maps to 0, no 1-cycles exist, and all cycles are 2-cycles. (I'm not entirely sure if this fails on any edge cases, but it gives the correct results for the examples.)

share|improve this answer

answered Sep 5 at 1:36

LegionMammal978

14.8k41752

add a comment | 

up vote
0
down vote

up vote
0
down vote

Mathematica, 42 bytes

#==||(a=0@@#)[[#]]=!=a&&a[[#]][[#]]===a&

Pure function. Takes a 1-indexed list of numbers as input and returns True or False as output. Just follows the tunnels, ensuring that 0 maps to 0, no 1-cycles exist, and all cycles are 2-cycles. (I'm not entirely sure if this fails on any edge cases, but it gives the correct results for the examples.)

share|improve this answer

answered Sep 5 at 1:36

LegionMammal978

14.8k41752

Mathematica, 42 bytes

#==||(a=0@@#)[[#]]=!=a&&a[[#]][[#]]===a&

Pure function. Takes a 1-indexed list of numbers as input and returns True or False as output. Just follows the tunnels, ensuring that 0 maps to 0, no 1-cycles exist, and all cycles are 2-cycles. (I'm not entirely sure if this fails on any edge cases, but it gives the correct results for the examples.)

share|improve this answer

answered Sep 5 at 1:36

LegionMammal978

14.8k41752

share|improve this answer

share|improve this answer

answered Sep 5 at 1:36

LegionMammal978

14.8k41752

answered Sep 5 at 1:36

LegionMammal978

14.8k41752

answered Sep 5 at 1:36

LegionMammal978

14.8k41752

14.8k41752

add a comment | 

add a comment | 

up vote
0
down vote

This answer does not work. Here for illustration purposes only.

This answer passes all of the (currently) posted test cases. However, it fails (raises an error) on other valid input, such as [1, 2] or [1, 0, 3, 7].

How could it pass [1, 0, 3] and fail [1, 0, 3, 7]? Well, it proceeds through the list, just like you'd expect. When it reads an element x of the list a, it first checks whether x is less than len(a), and immediately returns False, if so. So it correctly returns False on [1, 0, 3], because 3 is not less than len(a).

But assuming that x passes that check, the code will then go on to do some other checks, and at a certain point it happens to evaluate a[a[x]]. We've already guaranteed that evaluating a[x] will be OK...but not a[a[x]], which resolves to a[7] when x is 3 in the [1, 0, 3, 7] example. At this point Python raises an IndexError, rather than returning False.

For completeness, here's the answer.

Python 2, 59 bytes

lambda a:all(x<len(a)>-1<a[x]!=x==a[a[x]]for x in a if-1<x)

Try it online!

I wanted to do x<len(a)and-1<a[x]..., but of course len(a) is always >-1, so the above is equivalent. This check is a total of 5 chained relations (<, >, <, !=, and ==), plus a separate check -1<x in the if condition.

Python (conveniently) short-circuits chained relations like this, so for example if x>=len(a) then the check returns False before it gets to a[x] (which would otherwise raise an IndexError).

share|improve this answer

edited Sep 6 at 15:43

answered Sep 5 at 20:34

mathmandan

91357

add a comment | 

up vote
0
down vote

This answer does not work. Here for illustration purposes only.

This answer passes all of the (currently) posted test cases. However, it fails (raises an error) on other valid input, such as [1, 2] or [1, 0, 3, 7].

How could it pass [1, 0, 3] and fail [1, 0, 3, 7]? Well, it proceeds through the list, just like you'd expect. When it reads an element x of the list a, it first checks whether x is less than len(a), and immediately returns False, if so. So it correctly returns False on [1, 0, 3], because 3 is not less than len(a).

But assuming that x passes that check, the code will then go on to do some other checks, and at a certain point it happens to evaluate a[a[x]]. We've already guaranteed that evaluating a[x] will be OK...but not a[a[x]], which resolves to a[7] when x is 3 in the [1, 0, 3, 7] example. At this point Python raises an IndexError, rather than returning False.

For completeness, here's the answer.

Python 2, 59 bytes

lambda a:all(x<len(a)>-1<a[x]!=x==a[a[x]]for x in a if-1<x)

Try it online!

I wanted to do x<len(a)and-1<a[x]..., but of course len(a) is always >-1, so the above is equivalent. This check is a total of 5 chained relations (<, >, <, !=, and ==), plus a separate check -1<x in the if condition.

Python (conveniently) short-circuits chained relations like this, so for example if x>=len(a) then the check returns False before it gets to a[x] (which would otherwise raise an IndexError).

share|improve this answer

edited Sep 6 at 15:43

answered Sep 5 at 20:34

mathmandan

91357

add a comment | 

up vote
0
down vote

up vote
0
down vote

This answer does not work. Here for illustration purposes only.

This answer passes all of the (currently) posted test cases. However, it fails (raises an error) on other valid input, such as [1, 2] or [1, 0, 3, 7].

How could it pass [1, 0, 3] and fail [1, 0, 3, 7]? Well, it proceeds through the list, just like you'd expect. When it reads an element x of the list a, it first checks whether x is less than len(a), and immediately returns False, if so. So it correctly returns False on [1, 0, 3], because 3 is not less than len(a).

But assuming that x passes that check, the code will then go on to do some other checks, and at a certain point it happens to evaluate a[a[x]]. We've already guaranteed that evaluating a[x] will be OK...but not a[a[x]], which resolves to a[7] when x is 3 in the [1, 0, 3, 7] example. At this point Python raises an IndexError, rather than returning False.

For completeness, here's the answer.

Python 2, 59 bytes

lambda a:all(x<len(a)>-1<a[x]!=x==a[a[x]]for x in a if-1<x)

Try it online!

I wanted to do x<len(a)and-1<a[x]..., but of course len(a) is always >-1, so the above is equivalent. This check is a total of 5 chained relations (<, >, <, !=, and ==), plus a separate check -1<x in the if condition.

Python (conveniently) short-circuits chained relations like this, so for example if x>=len(a) then the check returns False before it gets to a[x] (which would otherwise raise an IndexError).

share|improve this answer

edited Sep 6 at 15:43

answered Sep 5 at 20:34

mathmandan

91357

This answer does not work. Here for illustration purposes only.

This answer passes all of the (currently) posted test cases. However, it fails (raises an error) on other valid input, such as [1, 2] or [1, 0, 3, 7].

How could it pass [1, 0, 3] and fail [1, 0, 3, 7]? Well, it proceeds through the list, just like you'd expect. When it reads an element x of the list a, it first checks whether x is less than len(a), and immediately returns False, if so. So it correctly returns False on [1, 0, 3], because 3 is not less than len(a).

But assuming that x passes that check, the code will then go on to do some other checks, and at a certain point it happens to evaluate a[a[x]]. We've already guaranteed that evaluating a[x] will be OK...but not a[a[x]], which resolves to a[7] when x is 3 in the [1, 0, 3, 7] example. At this point Python raises an IndexError, rather than returning False.

For completeness, here's the answer.

Python 2, 59 bytes

lambda a:all(x<len(a)>-1<a[x]!=x==a[a[x]]for x in a if-1<x)

Try it online!

I wanted to do x<len(a)and-1<a[x]..., but of course len(a) is always >-1, so the above is equivalent. This check is a total of 5 chained relations (<, >, <, !=, and ==), plus a separate check -1<x in the if condition.

Python (conveniently) short-circuits chained relations like this, so for example if x>=len(a) then the check returns False before it gets to a[x] (which would otherwise raise an IndexError).

share|improve this answer

edited Sep 6 at 15:43

answered Sep 5 at 20:34

mathmandan

91357

share|improve this answer

share|improve this answer

edited Sep 6 at 15:43

edited Sep 6 at 15:43

edited Sep 6 at 15:43

answered Sep 5 at 20:34

mathmandan

91357

answered Sep 5 at 20:34

mathmandan

91357

answered Sep 5 at 20:34

mathmandan

91357

91357

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