# Pairs at every distance

Given a list $$\X\$$ of 2 or more integers, output whether, for all $$\n\$$ such that $$\0 \leq n < length(X)-2\$$, there exists a pair of equal integers in $$\X\$$ separated by exactly $$\n\$$ elements.

In other words: output whether, for all overlapping slices/windows of the input, there exists at least one slice/window of each length wherein the head of the slice/window equals the tail.

For example: [1, 1, 3, 2, 3, 1, 2, 1] will return truthy, because

[1, x, x, x, x, x, x, 1]
These 1s are separated by 6 elements, the most possible in an 8 element list,

[x, 1, x, x, x, x, x, 1]
These 1s are separated by 5 elements,

[1, x, x, x, x, 1, x, x]
These 1s are separated by 4 elements,

[x, 1, x, x, x, 1, x, x]
These 1s are separated by 3 elements,

[x, x, x, 2, x, x, 2, x]
These 2s are separated by 2 elements,

[x, x, 3, x, 3, 1, x, 1]
These 3s are separated by 1 element (as are the 1s, but either pair is sufficient),

[1, 1, x, x, x, x, x, x]
And these 1s are separated by 0 elements, the least possible.


But [1, 1, 2, 2, 1, 1] will return falsy, as there is no pair of equal elements separated by exactly one element. That is, there is no length 3 slice/window with a head equal to it's tail. See all length 3 slices/windows below:

[1, 1, 2]
[1, 2, 2]
[2, 2, 1]
[2, 1, 1]


Standard I/O applies, input does not have to allow negatives, or input can be a string of characters, etc.

Anything reasonable as long as you're not cheating :)

This is , so shortest code in bytes wins!

## Examples

Truthy

[1, 1]
[1, 1, 1]
[3, 3, 7, 3]
[2, 2, 1, 2, 2]
[2, 1, 2, 2, 2]
[1, 3, 1, 3, 1, 1]
[1, 1, 3, 2, 3, 1, 2, 1]
[1, 1, 1, 1, 1, 1, 2, 1]


Falsy

[1, 2]
[1, 2, 1]
[1, 2, 3, 4]
[3, 1, 3, 1, 3]
[3, 1, 1, 3, 1]
[1, 3, 1, 1, 3, 1]
[1, 1, 2, 2, 2, 2, 1, 1]
[1, 1, 1, 1, 1, 1, 1, 2]

• input does not have to allow negatives -> May the input contain 0's? Jul 1 at 13:50
• @Arnauld "etc", only if you want. EDIT: to elaborate; this isn't a challenge about integers so much as it's a challenge about matching pairs in a list. you can use church numerals for all I care :P Jul 1 at 13:52
• It seems like this problem is also interesting from the perspective of fastest code/algorithm. The best algorithm I could think of is $O(n \sqrt{n \log(n)})$ Jul 2 at 14:54
• @WheatWizard I don't understand your edit. I think it's wrong, as it seems to imply 1, 1, 3, 2, 3, 1, 2, 1 would return falsy, as 2 3 1 2 and 1 3 2 3 1 dont have the same head/tail (2 vs 1). I also don't see what's gained by using |X| in place of length(X). I'm reverting it for now. EDIT: i see that my original wording also had that mistake. This should be a tad better. Feel free to reword this version, but I still don't agree with using absolute value bars for length. Seems like a barrier for entry. Jul 3 at 15:15
• The pre-existing wording made no sense at all. It used the term "pair" to refer to nothing at all. It is really that there are n pairs of one integer. Jul 3 at 15:20

# Jelly, 118 6 bytes

Uses inverted boolean output.

IÐƤÄPẸ


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 ÐƤ     # Map over each suffix:
I       #   Increments (differences between adjacent elements)
Ä    # For each list, get cumulative sums
PẸ  # Is there a zero in each column?


This is basically: Is there a zero on each diagonal of the subtraction table?. The naive implementation would be:

_þŒDP€Ẹ


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# JavaScript (ES6), 39 bytes

Returns a Boolean value.

a=>a.every((_,d)=>a.some(v=>a[d++]==v))


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Or 38 bytes with an inverted output, assuming the input list does not contain any 0:

a=>a.some((_,d)=>a.every(v=>a[d++]^v))


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### Commented

a =>                // given an array a[] of N entries
a.every((_, d) => // for each distance d = 0 to N - 1:
a.some(v =>     //   is there some value v at position i (implied) ...
a[d++] == v   //     ... such that a[i + d] is equal to v?
//     (we increment d instead of storing i explicitly)
)               //   end of some()
)                 // end of every()


# Vyxal, 12 11 bytes

żƛ?l'ṪǏ⁼;;A


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-1 thanks to EmanresuA but also -10 rep thanks to EmanresuA so now I have to edit the post so the upvote can be returned

## Explained

żƛ?l'ṪǏ⁼;;A
żƛ           # For every number n in the range [0, len(input)]
?l         #   Overlapping windows of length n of the input
'ṪǏ⁼;    #   Get all sublists where the list is the same after appending the head of the list to the list with the tail chopped off. That is, a[:-1] + [a[0]] == a
;   # End map
A  # Are all the items truthy?

• 11 Jul 1 at 13:00
• @emanresuA that took me a bit to figure out how that worked lol Jul 1 at 13:08
• 10 bytes (link does test cases) Jul 2 at 2:30
• actually 8 Jul 2 at 2:50
• alternative 8 not using inverted booleans Jul 2 at 2:53

# J, 18 16 bytes

[:*/[:+.//.@|.e.


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-2 thanks to ovs!

Explanation slightly out of date (can't update right now), but idea is same.

Consider 3 3 7 3:

• -/~ Differences table:

0 0 _4 0
0 0 _4 0
4 4  0 4
0 0 _4 0

• |. Reversed:

0 0 _4 0
4 4  0 4
0 0 _4 0
0 0 _4 0

• 0&e./. Is there a zero in each diagonal going this way /:

1 1 1 1 1 1 1

• [:*/ Are they all 1:

1

• WTF that's incredible insight Jul 1 at 16:18
• =/~ / monadic e. can save a couple bytes
– ovs
Jul 1 at 16:19
• @thejonymyster thanks, but note ovs had the same insight as well. Jul 1 at 16:40
• @ovs Very nice, thanks! Jul 1 at 16:41
• oh. i did not understand it ^_^; thank you i will have to study further Jul 1 at 16:41

# MATL, 9 8 bytes

Saved one byte thanks to @Luis Mendo

&=T&XdaA


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Explanation:

&=        # input == input'
T       # get all diagonals from...
&Xd    # spdiags (i.e. rotate matrix 45 degrees)
a   # any applied to columns
A  # all

• @LuisMendo Ooh, I didn't realize that the comparison operators did that! That's a very handy feature. Jul 2 at 13:44
• @LuisMendo I was confused about the truthy/falsy definitions since some answers explicitly collapse to a single value at the end and some don't. I may leave it because I like the extra 'a' in 'tadaa!' Jul 2 at 13:46
• Also, a single-value result looks neater Jul 2 at 17:56

# Rust, 63 bytes

|k:&[u8]|(0..k.len()).all(|z|(z..k.len()).any(|q|k[q-z]==k[q]))

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# MATL, 9 bytes

fqG&=&f-m


Outputs a truthy or falsy value, which is allowed by default. Specifically, outputs

• a non-empty array containing only ones, which is truthy, or
• an array containing at least a zero, which is falsy.

Try it online! Or verify all test cases, including truthiness/falsihood test.

### How it works

     % Implicit input
f    % Find: gives indices of non-zeros. Since the input contains non-zero
% integers, this gives [1 2 ... n] where n is the input length
q    % Subtract 1, element-wise. Gives [0 1 ... n-1]
G    % Push input again
&=   % Matrix of pairwise equality comparisons
&f   % Two-output find: gives row and column indices of nonzero (i.e. true) entries
-    % Subtract, element-wise
m    % Ismember. This gives an array containing only true (or 1) if 0, 1,... n-1
% are all contained in the above result (note that 0 will always be contained)
% Implicit display


# Husk, 10 8 bytes

Πm▲∂↔´Ṫ=


Port of @ovs' second Jelly answer:

Is there a zero on each diagonal of the subtraction table?

-2 bytes thanks to @Steffan, by using this instead:

Is there a 1 on each diagonal of the equality table?

Outputs 1/0 for truthy/falsey respectively.

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Explanation:

     ´    # Use the given input-argument twice,
Ṫ   # to apply double-vectorized, creating a table
=  # checking for each pair if they're equal
↔     # Reverse each inner row
∂      # Now pop and take all anti-diagonals of this matrix
# (∂↔ basically takes all diagonals of the matrix†)
m        # Map over each diagonal-list:
▲       #  Maximum: check if any value in the diagonal-list is truthy
Π         # Product: check if all are truthy for each diagonal
# (after which the result is output implicitly)


† Although Husk has an anti-diagonals builtin ∂, it lacks a diagonals builtin. The reason for this is probably because the anti-diagonals can be computed on an infinite matrix, whereas this isn't the case for the diagonals, as assumed by @MartinEnder here.

• I don't actually know husk, but here's 8 bytes. Returns 0 for falsy, and any other integer for truthy. Jul 2 at 3:05
• @Steffan Thanks. I've changed the sum to a max to get a consistent 1/0 for truthy/falsey respectively. But smart to use an equality table instead of subtraction table! :) (And since this is only my 7th Husk answer, I honestly don't know Husk either ;p) Jul 2 at 9:10

# 05AB1E, 119 6 bytes

.s€αPO


Returns zero if there are pairs at every distance, otherwise returns a non-zero number.

If this output format isn't allowed, it's +1 bytes for adding _ in the end (and then it returns 1 if there are pairs at every distance and 0 otherwise). Alternatively, if you are willing to stretch the output format more you can remove the O in the end, and then it returns a list containing only zeros iff there are pairs at every distance

.s    All suffixes
€     Map each suffix to:
α    The absolute difference of each value in the suffix from the matching value in the input
P     Product of each list - it's zero iff there's at least one pair of equal elements with distance X
O     Sum - since all values are positive, this is zero iff all values in the map are zero

• Smart approach, and great use of 05AB1E's truncating unequal sized lists with that .s€α! (And here is a test suite if you want one.) Jul 2 at 9:17
• @KevinCruijssen thanks! I added the test suite Jul 2 at 9:28

# Python NumPy, 51 bytes

lambda a:all(map((a==[*zip(a)]).trace,a.argsort()))

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Takes a numpy array.

### How?

Mostly straightforward.zip and argsort are used to avoid the direct numpy import. zip does roughly the same as transpose. The comparison with a creates a boolean table of pairwise incidence. The diagonals of this table are the distance groups. Taking the trace at different offsets returns a positive (hence truthy) value if there is a pair at that distance. argsort seems to be the cheapest way to create the offsets 0,...,n-1 not necessarily in order but we don't care.

# Clojure, 137 136 bytes

(defn h[c](loop[n(count c)](if(not(contains?(set(map #(=(first %)(last %))(partition n 1 c)))true))false(if(> n 2)(recur(dec n))true))))


Ungolfed:

(defn head-eq-tail [col]
(loop [n  (count col)]
(if (not (contains? (set (map #(= (first %) (last %)) (partition n 1 col))) true))
false
(if (> n 2)
(recur (dec n))
true))))

• Here's a TIO link if you want it: Try it online!. Also, can't you remove the space after loop? Jul 2 at 15:04
• You're absolutely right. :-) Jul 2 at 16:16

# Desmos, 43 bytes

f(x)=∑_{y=1}^{x.length}abs(x[y...]-x).min


Uses 0 for truthy, and any other positive integer for falsey.

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• You can do (x[y...]-x)^2 instead of abs(x[y...]-x) for -1 byte. Jul 11 at 4:24

# Knight, 91 bytes

;=l~1;W E++'=x'=l+1l'P'1;=aT;=wF;W<=w+1w l;=bF;=s~1;W<=s+1s-l w=b|b?E+'x'sE+'x'+s w=a&a bOa


Enter each number in the input list on their own line, with a trailing newline.

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Also made a test suite cuz I'm bored lol

(Basically equivalent Python code: Try It Online!)

• Congrats on figuring out semicolon! lol Aug 9 at 15:14
• The trailing newline is only necessary in the JS interpreter. If you use the others it will be fine Aug 9 at 15:15
• @Steffan Well I still don't really understand what it does but I found a reliable way to figure out where to put semicolons through a lot of trial and error. Basically, for every "indentation level" (like in python), put a semicolon in front of every statement except for the last statement in that indentation level. Seems to work so far. Aug 10 at 2:44
• Yes, that's exactly how you use it Aug 10 at 2:45

# R, 56 bytes

\(x,n=seq(x),?=Map)any(\(i)all(\(j)x[j]-x[i+j]?n)?n-1)

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Outputs FALSE for truthy and NA for falsey.

# C (clang), 78 bytes

h;r;d;i;f(*t,n){for(r=1,d=n;--d;r&=h)for(h=i=0;i+d<n;)h|=t[i]==t[i+++d];*t=r;}


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Inputs a pointer to an array of integers and its length (because poiters in C carry no length info).
Returns, through the input pointer, $$\1\$$ if the array has pairs at every distance or $$\0\$$ otherwise.

# Charcoal, 15 bytes

⊙θ⬤θ∨‹μκ⁻λ§θ⁻μκ


Try it online! Link is to verbose version of code. Outputs an inverted Charcoal boolean, i.e. - if there is a distance with no pair, nothing if there is a pair at every distance. Explanation:

 θ              Input array
⊙               Any distance satisfies
θ            Input array
⬤             Every element satisfies
μ         Current index
‹          Is less than
κ        Current distance
∨           Logical Or
λ      Current value
⁻       Does not equal
θ    Input array
§     Indexed by
μ  Current index
⁻   Subtract
κ Current distance
Implicitly print


# Desmos, 63 bytes

k=l.length
f(l)=0^{∏_{n=2}^k∑_{i=n}^k0^{(l[i-n+1]-l[i])^2}}


Returns 0 if truthy, 1 if falsey.

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Try It On Desmos! - Prettified

# 05AB1E, 12 bytes

āsŒʒ¬Qθ}€gêQ


I have the feeling this can be shorter. EDIT: And it defintely can: see @CommandMaster's 05AB1E answer, halve the size of mine.

I do like how this challenge can be done with a lot of different approaches, though.

Explanation:

ā         # Push a list in the range [1, (implicit) input-length]
s        # Swap so the input-list is at the top
Œ       # Pop and get all its sublists
ʒ      # Filter this list of sublists by:
¬Qθ   #  Check if the first and last items are the same:
¬     #   Get the first item (without popping the list)
Q    #   Check for each item if it's equal to this first item
θ   #   Then pop and push the last check
}€     # After the filter: map over each remaining sublist:
g    #  Pop and push the length
ê   # Sorted-uniquify this list of lengths
Q  # Check if it's equal to the [1,length] list we created initially
# (after which the result is output implicitly)


# Pyth, 13 12 bytes

*FsMqVRQ>LQU


Outputs zero for false, non-zero for true.

Pretty much a port of Command Master's 05AB1E answer but double the length :/

Pyth has a prefix function (._) but no suffix function, so building the suffixes costs a few extra bytes with >LQU.

# R, 90 bytes

\(x,l=length,n=l(x),m=1:n)l(table(abs(outer(m,m,Vectorize(\(a,b)(x[a]==x[b])*(a-b))))))==n

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Just checking that all distances are represented in the matrix of distances (where elements are zeroed for non-matches).

# Julia

## 52 bytes

!X=0:length(X)-1⊆[I[1]-I[2] for I=findall(X.==X')]

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## 45 bytes - thanks to @MarcMush

!X=keys(X)⊆findall(X.==X').|>x->x[1]-x[2]+1

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• 45 bytes Jul 3 at 17:28

# x86 32-bit machine code, 18 bytes

57 89 D7 49 AF 60 89 D6 F2 A7 61 E1 F7 0F 94 C0 5F C3


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Following the fastcall calling convention, this takes the length of an array of 32-bit integers in ECX and its address in EDX, and returns 0 or 1 in AL.

In assembly:

f:  push edi      # Save EDI onto the stack.
mov edi, edx  # Set EDI to the array address.
dec ecx       # Subtract 1 from the length in ECX.
r:  scasd         # Advance EDI, also performing an unnecessary comparison.
pusha         # Push all the registers onto the stack.
mov esi, edx  # Set ESI to the array address.
# The distance between EDI and ESI increases each iteration.
repne cmpsd   # Compare values at addresses EDI and ESI and advance both,
#  repeating ECX times but stopping if they are equal.
popa          # Restore all registers' values from the stack.
loopz r       # Subtract 1 from ECX, and jump back if it's nonzero
#  and the result of the last comparison is equal.
setz al       # Set AL based on whether the result of the last comparison is equal.
pop edi       # Restore the value of EDI from the stack.
ret           # Return.


# APL (Dyalog Classic), 15 bytes

0∊1⊥⍳∘≢↓⍤¯1∘.=⍨


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Returns 0 if there are such pairs, and 1 otherwise.

# Vyxal, 7 bytes

v=ṘÞḋṠA


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Lyxal never posted my golf, so here goes

# PARI/GP, 41 bytes

a->sum(i=1,#a,prod(j=i,#a,a[j]-a[j-i+1]))

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Returns 0` for truthy, other integers for falsy.