Zero sum covers

Code Golf Asked by Zgarb on November 10, 2020

Introduction

Consider a nonempty list L of integers.
A zero-sum slice of L is a contiguous subsequence of L whose sum equals 0.
For example, [1, -3, 2] is a zero-sum slice of [-2, 4, 1, -3, 2, 2, -1, -1], but [2, 2] is not (because it doesn’t sum to 0), and neither is [4, -3, -1] (because it’s not contiguous).

A collection of zero-sum slices of L is a zero-sum cover of L if every element belongs to at least one of the slices.
For example:

L = [-2, 4, 1, -3, 2, 2, -1, -1]
A = [-2, 4, 1, -3]
B =        [1, -3, 2]
C =                  [2, -1, -1]


The three zero-sum slices A, B and C form a zero-sum cover of L.
Multiple copies of the same slice can appear in a zero-sum cover, like this:

L = [2, -1, -1, -1, 2, -1, -1]
A = [2, -1, -1]
B =        [-1, -1, 2]
C =                [2, -1, -1]


Of course, not all lists have a zero-sum cover; some examples are [2, -1] (every slice has nonzero sum) and [2, 2, -1, -1, 0, 1] (the leftmost 2 is not part of a zero-sum slice).

Your input is a nonempty integer list L, taken in any reasonable format.
Your output shall be a truthy value if L has a zero-sum cover, and a falsy value if not.

You can write a full program or a function, and the lowest byte count wins.

Test cases

[-1] -> False
[2,-1] -> False
[2,2,-1,-1,0,1] -> False
[2,-2,1,2,-2,-2,4] -> False
[3,-5,-2,0,-3,-2,-1,-2,0,-2] -> False
[-2,6,3,-3,-3,-3,1,2,2,-2,-5,1] -> False
[5,-8,2,-1,-7,-4,4,1,-8,2,-1,-3,-3,-3,5,1] -> False
[-8,-8,4,1,3,10,9,-11,4,4,10,-2,-3,4,-10,-3,-5,0,6,9,7,-5,-3,-3] -> False
[10,8,6,-4,-2,-10,1,1,-5,-11,-3,4,11,6,-3,-4,-3,-9,-11,-12,-4,7,-10,-4] -> False
[0] -> True
[4,-2,-2] -> True
[2,2,-3,1,-2,3,1] -> True
[5,-3,-1,-2,1,5,-4] -> True
[2,-1,-1,-1,2,-1,-1] -> True
[-2,4,1,-3,2,2,-1,-1] -> True
[-4,-1,-1,6,3,6,-5,1,-5,-4,5,3] -> True
[-11,8,-2,-6,2,-12,5,3,-7,4,-7,7,12,-1,-1,6,-7,-4,-5,-12,9,5,6,-3] -> True
[4,-9,12,12,-11,-11,9,-4,8,5,-10,-6,2,-9,10,-11,-9,-2,8,4,-11,7,12,-5] -> True


APL (Dyalog Unicode), 22 bytes

≢={≢∪∊↓∘⍳/¨⍸∘.=⍨0,+⍵}


Try it online!

Same byte count as Jo King's, but totally different approach.

How it works

≢={≢∪∊↓∘⍳/¨⍸∘.=⍨0,+⍵}  ⍝ Input: integer vector
{             0,+⍵}  ⍝ Cumulative sum and prepend 0
∘.=⍨        ⍝ Self outer product by equality
⍸            ⍝ Extract coords of ones; (a,b) with a>b is included
⍝ iff the interval [b+1..a] has zero sum
↓∘⍳/¨  ⍝ For each coord, [b+1..a] if a>b, empty vector otherwise
≢∪∊       ⍝ Count unique values
≢=           ⍝ Is the count same as the length of the original array?


Answered by Bubbler on November 10, 2020

Python 3, 99 bytes

lambda a,r=range:len({k for j in r(len(a)+1)for i in r(j)if sum(a[i:j])==0for k in r(i,j)})==len(a)


Try it online!

Creates a set of all indices which are part of a zero sum. Then checks if all indices are present.

Answered by Jitse on November 10, 2020

APL (Dyalog Unicode), 22 20 bytes

-2 bytes thanks to Bubbler

∧/∨⌿↑0={×⌽+⌽⍵}¨,⎕


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Explanation

                 ,⎕      ⍝ All prefixes of the input
{       }¨         ⍝ Map each to
⌽+⌽⍵           ⍝ The sum of each postfix
×                ⍝ Scan multiplied, spreading any zero sums to the other elements in the sub-array
0=                   ⍝ Map zeroes to true
↑                     ⍝ Mix the arrays into a 2D matrix
∧/∨⌿                      ⍝ Do all columns have at least one true value?


Answered by Jo King on November 10, 2020

JavaScript (ES6), 102 bytes

a=>(g=f=>a.map((_,i)=>f(i)))(i=>g(j=>j<i||(t+=a[j])||g(k=>b[k]&=k<i|k>j),t=0),b=g(_=>1))&&!/1/.test(b)


Computes the partial sums for all the elements i..j inclusive, and resets the relevant elements of b from 1 to 0 when it finds a zero sum, finally checking that there are no 1s left.

Answered by Neil on November 10, 2020

PHP, 104 bytes

Brute force and still over 99 bytes. :-(

for($p=$r=$c=$argc;$s=--$p;)for($i=$c;$s&&$k=--$i;)for($s=0;$k<$c&&($r-=!$s+=$argv[$k++])&&$s;);echo!$r;


takes input from command line arguments, 1 for truthy, empty for falsy. Run with -r.

breakdown

for($p=$r=$argc;$s=$p--;) # loop$p from $argc-1 to 0 with dummy value >0 for$s
for($i=$p;$s&&$k=$i--;) # loop$i (slice start) from $p to 1, break if sum is 0 for($s=0;                   # init sum to 0
$k<$argc                # loop $k (slice end) from$i to $argc-1 &&($r-=!$s+=$argv[$k++]) # update sum, decrement$r if sum is 0
&&$s;); # break loop if sum is 0 echo!$r;                    # $r = number of elements that are not part of a zero-sum slice  $argv[0] contains the filename; if run with -r, it will be - and evaluate to 0 for numeric ops.

Answered by Titus on November 10, 2020

Clojure, 109 bytes

#(=(count %)(count(set(flatten(for[r[(range(count %))]l r p(partition l 1 r):when(=(apply +(map % p))0)]p))))


Generates all partitions which sum to zero, checks that it has "length of input vector" distinct indices.

Answered by NikoNyrh on November 10, 2020

J, 36 35 bytes

#*/@e.[:;]({:*0=[:+/{.)@|:[email protected],.#


For every subsum I add the element's indexes and I keep the indexes iff subsum is 0 and then check whether every index is present.

Trick: 1-based indexes of a list can be generated with # i.e. length of every prefix.

Usage:

   (#*/@e.[:;]({:*0=[:+/{.)@|:[email protected],.#) 2 _1 _1 2
1
(#*/@e.[:;]({:*0=[:+/{.)@|:[email protected],.#) 2 _1
0


Try it online here.

Answered by randomra on November 10, 2020

Python, 86 Bytes

def f(l):
r=range(len(l))
if[i for i in r for j in r if sum(l[j:j+i+1])==0]:return 1


Truthy = 1 Falsy = None

import Data.Lists
g x=(1<$x)==(1<$nub(id=<<[i|(i,0)<-fmap sum.unzip<$>powerslice(zip[1..]x)]))  Usage example: g [-11,8,-2,-6,2,-12,5,3,-7,4,-7,7,12,-1,-1,6,-7,-4,-5,-12,9,5,6,-3] -> True. How it works (let's use [-1,1,5,-5] for input):  zip[1..]x -- zip each element with its index -- -> [(1,-1),(2,1),(3,5),(4,-5)] powerslice -- make a list of all continuous subsequences -- -> [[],[(1,-1)],[(1,-1),(2,1)],[(1,-1),(2,1),(3,5)],[(1,-1),(2,1),(3,5),(4,-5)],[(2,1)],[(2,1),(3,5)],[(2,1),(3,5),(4,-5)],[(3,5)],[(3,5),(4,-5)],[(4,-5)]] <$>            -- for each subsequence
unzip           --   turn the list of pairs into a pair of lists
--   -> [([],[]),([1],[-1]),([1,2],[-1,1]),([1,2,3],[-1,1,5]),([1,2,3,4],[-1,1,5,-5]),([2],[1]),([2,3],[1,5]),([2,3,4],[1,5,-5]),([3],[5]),([3,4],[5,-5]),([4],[-5])]
fmap sum         --   and sum the second element
--   -> [([],0),([1],-1),([1,2],0),([1,2,3],5),([1,2,3,4],0),([2],1),([2,3],6),([2,3,4],1),([3],5),([3,4],0),([4],-5)]
[i|(i,0)<-    ]   -- take all list of indices where the corresponding sum == 0
-- -> [[],[1,2],[1,2,3,4],[3,4]]
id=<<             -- flatten the list
-- -> [1,2,1,2,3,4,3,4]
nub                -- remove duplicates
-- -> [1,2,3,4]

(1<$x)==(1<$    )  -- check if the above list has the same length as the input list.


Answered by nimi on November 10, 2020

Python, 123 120 bytes

-3 bytes thanks to @Zgarb

Populates a list with the same size as the input with zero-sum slices and overwrites according to indexes, returning its equality to the original at the end.

def f(l):
s=len(l);n=[0]*s
for i in range(s):
for j in range(i,s+1):
if sum(l[i:j])==0:n[i:j]=l[i:j]
return n==l


Answered by dfernan on November 10, 2020

JavaScript (ES6), 109 bytes

f=([q,...a],b=[],c=[])=>1/q?f(a,[...b,0].map((x,i)=>x+q||(c=c.map((n,j)=>n|i<=j)),c.push(0)),c):c.every(x=>x)


Test snippet

f=([q,...a],b=[],c=[])=>1/q?f(a,[...b,0].map((x,i)=>x+q||(c=c.map((n,j)=>n|i<=j)),c.push(0)),c):c.every(x=>x)

I.textContent = I.textContent.replace(/.+/g,x=>x+ (\${f(eval(x.split(" ")[0]))}))
<pre id=I>[-1] -> False
[2,-1] -> False
[2,2,-1,-1,0,1] -> False
[2,-2,1,2,-2,-2,4] -> False
[3,-5,-2,0,-3,-2,-1,-2,0,-2] -> False
[-2,6,3,-3,-3,-3,1,2,2,-2,-5,1] -> False
[5,-8,2,-1,-7,-4,4,1,-8,2,-1,-3,-3,-3,5,1] -> False
[-8,-8,4,1,3,10,9,-11,4,4,10,-2,-3,4,-10,-3,-5,0,6,9,7,-5,-3,-3] -> False
[10,8,6,-4,-2,-10,1,1,-5,-11,-3,4,11,6,-3,-4,-3,-9,-11,-12,-4,7,-10,-4] -> False
[0] -> True
[4,-2,-2] -> True
[2,2,-3,1,-2,3,1] -> True
[5,-3,-1,-2,1,5,-4] -> True
[2,-1,-1,-1,2,-1,-1] -> True
[-2,4,1,-3,2,2,-1,-1] -> True
[-4,-1,-1,6,3,6,-5,1,-5,-4,5,3] -> True
[-11,8,-2,-6,2,-12,5,3,-7,4,-7,7,12,-1,-1,6,-7,-4,-5,-12,9,5,6,-3] -> True
[4,-9,12,12,-11,-11,9,-4,8,5,-10,-6,2,-9,10,-11,-9,-2,8,4,-11,7,12,-5] -> True</pre>

Answered by ETHproductions on November 10, 2020

Scala, 49 bytes

% =>(1 to%.size)flatMap(%sliding)exists(_.sum==0)


Usage:

val f:(Seq[Int]=>Boolean)= % =>(1 to%.size)flatMap(%sliding)exists(_.sum==0)
f(Seq(4, -2, -2)) //returns true


Ungolfed:

array=>(1 to array.size)
.flatMap(n => array.sliding(n))
.exists(slice => slice.sum == 0)


Explanation:

% =>            //define a anonymouns function with a parameter called %
(1 to %.size) //create a range from 1 to the size of %
flatMap(      //flatMap each number n of the range
%sliding    //to an iterator with all slices of % with length n
)exists(      //check whether there's an iterator with a sum of 0
_.sum==0
)


Answered by corvus_192 on November 10, 2020

Mathematica, 65 64 bytes

Thanks to ngenisis for saving 1 byte.

[email protected]@Cases[Subsequences[[email protected][1^#]],a_/;[email protected]#[[a]]==0]==x&


I'd rather find a pure pattern matching solution, but it's proving to be tricky (and things like {___,a__,___} are always super lengthy).

Answered by Martin Ender on November 10, 2020

Mathematica, 66 65 bytes

Saved 1 byte, and hopefully learned a new trick for the future, thanks to ngenisis!

Two equally long alternatives, both of which are unnamed functions taking a list of integers as input and returning True or False:

[email protected]@Table[0==Product[[email protected]#[[i;;j]],{i,k},{j,k,l}],{k,l=Tr[1^#]}]&

[email protected][Product[[email protected]#[[i;;j]],{i,k},{j,k,l}],{k,l=Tr[1^#]}]&


In both cases, [email protected]#[[i;;j]] computes the sum of the slice of the input from position i to position j (1-indexed). Product[...,{i,k},{j,k,l}] multiples together all these slice-sums, as i ranges over indices that are at most k and j ranges over indices that are at least k. (Note that l=Tr[1^#] defines l to be the sum of 1 to all the powers in the input list, which is simply the length of the list.) In other words, this product equals 0 if and only if the kth element belongs to a zero-sum slice.

In the first version, each of those products is compared to 0, and [email protected]@ returns True precisely when every single product equals 0. In the second version, the list of products is acted upon by the function Norm (the length of the l-dimensional vector), which equals 0 if and only if every entry equals 0.

Answered by Greg Martin on November 10, 2020

Jelly, 13 12 bytes

JẆịS¥ÐḟċþJḄẠ


Try it online!

How it works

JẆịS¥ÐḟċþJḄẠ  Main link. Argument: A (array)

J             Yield all indices of A.
Ẇ            Window; yield all slices of indices.
Ðḟ       Filter; keep slices for which the link to the left returns 0.
¥           Combine the two atoms to the left into a dyadic chain.
ị               Retrieve the elements of A at the slice's indices.
S              Take the sum.
J    Yield the indices of A.
ċþ     Count table; count how many times each index appears in each table.
Ḅ   Unbinary; convery the array of counts of each index from base 2 to
integer. This yields 0 iff an index does not appear in any slice.
Ạ  All; return 1 iff all integers are non-zero.


Answered by Dennis on November 10, 2020

Ruby, 81 bytes

Try it online

Simplistic brute-force solution; for every element of the array, try to find a zero-sum slice that contains it.

->a{(0..l=a.size).all?{|i|(0..i).any?{|j|(i..l).any?{|k|a[j..k].inject(:+)==0}}}}


Answered by Value Ink on November 10, 2020

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