Can this list be balanced?

Question

To check whether a list of non-negative integers is balanced, one can imagine putting respective weights on a board and then try to balance the board on a pivot such that the summarized relative weights left and right of the pivot are the same. The relative weight is given by multiplying the weight with its distance to the pivot (see law of the lever).

^{(Source: wikipedia)}

This image corresponds to a list [100, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5]. This list is balanced because the 5 has a distance of 20 to the pivot, the 100 a distance of 1 and 5*20 = 100 = 100*1.

Examples

 3 1 5 7
#########
     ^

In this case the pivot is directly under the 5, the 3 has distance 2 and the 1 and 7 have distance 1. So both sides left and right of the pivot sum up to 7 (3*2 + 1*1 on the left and 7*1 on the right) and therefore the list [3, 1, 5, 7] is balanced.

Note, however, that the pivot does not have to be placed under one of the list elements, but might also be placed in-between two list elements:

 6 3 1
#######
  ^

In this case the distances become 0.5, 1.5, 2.5, ... and so on. This list is also balanced because 6*0.5 = 3 = 3*0.5 + 1*1.5.

The pivot can only be placed exactly below one number or exactly in the middle between two numbers, and not e.g. at two-thirds between two numbers.

Task

Given a list of non-negative integers in any reasonable format, output a truthy value if the list can be balanced and a falsy value otherwise.

You can assume that the input list contains at least two elements and that at least one element is non-zero.

This is a code-golf challenge, so the answer with the fewest amount of bytes in each language wins.

Truthy Testcases

[1, 0]
[3, 1, 5, 7]
[6, 3, 1]
[100, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5]
[10, 4, 3, 0, 2, 0, 5]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
[7, 7, 7, 7]

Falsy Testcases

[1, 2]
[3, 6, 5, 1, 12]
[0, 0, 2, 0, 1, 0]
[1, 2, 3, 4, 5, 6, 7, 8, 9]
[6, 3, 2, 4, 0, 1, 2, 3]
[4, 0, 0, 2, 3, 5, 2, 0, 1, 2, 3, 0, 0, 1, 2, 4, 3, 1, 3, 0, 0, 2]
[100, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5]

_{A lot of related challenges where found while this challenge was sand-boxed:
Is it a balanced number?, Equilibrium index of a sequence, Balance a set of weights on a seesaw, Balancing Words, Will I tip over? and Where does the pivot belong?}

Can the pivot be placed before the first number or after the last number? — Erik the Outgolfer, Commented Dec 20, 2017 at 15:45
I think this might be a dupe. Or was it sitting in the Sandbox for a while? — Shaggy, Commented Dec 20, 2017 at 16:07
related. (cc @Shaggy Maybe this was what you were thinking about) — Mr. Xcoder, Commented Dec 20, 2017 at 16:33
@Giuseppe @Steadybox I added You can assume that the input list contains at least two elements and that at least one element is non-zero. — Laikoni, Commented Dec 20, 2017 at 16:46

score 7 · Accepted Answer · 2017-12-20 15:53:08Z

7

Pyth, 12 10 bytes

!%ys*VQUQs

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Saved 2 bytes thanks to Mr. Xcoder and Erik the Outgolfer.

Explanation

!%ys*VQUQs
    *VQUQ    Multiply each input by its index.
  ys         Take twice the sum (to handle half-integer positions).
!%       sQ  Check if that's a multiple of the total weight.

edited Dec 20, 2017 at 15:53

answered Dec 20, 2017 at 15:40

user48543

\$\begingroup\$ You can use y in place of *2 \$\endgroup\$
– Mr. Xcoder
Commented Dec 20, 2017 at 15:49
\$\begingroup\$ 10 bytes: !%ys*VQUQs \$\endgroup\$
– Erik the Outgolfer
Commented Dec 20, 2017 at 15:50

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Kelly Lowder · Accepted Answer · 2017-12-20 15:48:08Z

4

Wolfram Language (Mathematica), 36 bytes

IntegerQ[2#.Range[t=Tr[1^#]]/(t-1)]&

This is a center of mass problem in a coordinate system with the origin at one of the points and then you determine if the CM falls on a lattice point where the lattice width = 1/2.

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edited Dec 20, 2017 at 15:48

answered Dec 20, 2017 at 15:42

Kelly Lowder

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Community · Accepted Answer · 2020-06-17 09:04:33Z

4

05AB1E, 6 bytes

ƶO·IOÖ

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How?

ƶO·IOÖ ~ Full program. I = input.

ƶ      ~ Lift I. Multiply each element with its 1-based index.
 O     ~ Sum.
  ·    ~ Double. 
     Ö ~ Is a multiple of?
   IO  ~ The sum of I.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 20, 2017 at 15:55

Mr. Xcoder

42.5k9 gold badges81 silver badges214 bronze badges

\$\begingroup\$ Seems to fail on [1,1] (should be truthy). It seems the implicit doubling isn't actually there. \$\endgroup\$
– Zgarb
Commented Dec 21, 2017 at 16:46
\$\begingroup\$ @Zgarb Fixed (?) \$\endgroup\$
– Mr. Xcoder
Commented Dec 21, 2017 at 17:51

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Erik the Outgolfer · Accepted Answer · 2017-12-20 16:26:37Z

2

Jelly, 6 bytes

×JSḤọS

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Well, looks like Leaky Nun pointed the pointless out.

Using Mnemonic's Pyth approach.

Returns a positive integer (truthy) or zero (falsy).

edited Dec 20, 2017 at 16:26

answered Dec 20, 2017 at 15:53

Erik the Outgolfer

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\$\begingroup\$ Would this work? \$\endgroup\$
– Leaky Nun
Commented Dec 20, 2017 at 16:20
\$\begingroup\$ @LeakyNun Not so sure, that's why I used LḶ instead (although it would succeed for all test cases). EDIT: Oooh, now that I think about it again, it seems so...(b | a ⇔ b | a + b duh) \$\endgroup\$
– Erik the Outgolfer
Commented Dec 20, 2017 at 16:22

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Giuseppe · Accepted Answer · 2017-12-20 16:52:35Z

2

R, 34 bytes

function(n)!(n%*%seq(n)*2)%%sum(n)

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Takes input as a vector. Ports mnemonic's answer. Returns a 1x1 matrix.

edited Dec 20, 2017 at 16:52

answered Dec 20, 2017 at 16:43

Giuseppe

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Community · Accepted Answer · 2020-06-17 09:04:33Z

2

Japt, 10 bytes

í* x*2 vUx

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Explanation:

 í* x*2 vUx
U            // Implicit Input                 [3, 1, 5, 7]
 í           // Pair the input with its index  [[3,0],[1,1],[5,2],[7,3]]
  *          // Multiply each item             [0,1,10,21]
    x        // Sum                            32
     *2      // Double                         64
        v    // Divisible by:
         Ux  //   Sum of Input                 16
             // Explicit Output                1

Returns 1 for truthy, 0 for falsy.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 20, 2017 at 20:07

Oliver

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Dennis · Accepted Answer · 2017-12-20 20:57:46Z

2

Python 2, 41 bytes

S=s=0
for n in input():S-=s;s-=n
1>>2*S%s

Output is via exit code, so 0 is truthy and 1 is falsy.

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edited Dec 20, 2017 at 20:57

answered Dec 20, 2017 at 18:50

Dennis

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Uriel · Accepted Answer · 2017-12-20 22:32:05Z

2

Julia, 31 27 bytes

4 bytes saved thanks to @Dennis

!n=2n[i=1:end]⋅i%sum(n)<1

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edited Dec 20, 2017 at 22:32

answered Dec 20, 2017 at 17:46

Uriel

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Tom Lazar · Accepted Answer · 2017-12-21 15:37:18Z

2

Ruby, 47 bytes

Saved 2 bytes thanks to Mr. Xcoder

->k{(k.map.with_index{|x,i|x*i*2}.sum%k.sum)<1}

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edited Dec 21, 2017 at 15:37

answered Dec 20, 2017 at 20:53

Tom Lazar

414 bronze badges

1

\$\begingroup\$ Welcome to PPCG! Nice first answer. 47 bytes \$\endgroup\$
– Mr. Xcoder
Commented Dec 20, 2017 at 20:54

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Steadybox · Accepted Answer · 2017-12-20 16:16:49Z

1

C, 140 137 bytes

float l,r;i,j,t;f(L,n)int*L;{for(i=t=-1;++i<2*n;t*=l-r)for(l=r=j=0;j<n;++j)l+=j<i/2.?L[j]*(i/2.-j):0,r+=j>i/2.?L[j]*(j-i/2.):0;return!t;}

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edited Dec 20, 2017 at 16:16

answered Dec 20, 2017 at 16:05

Steadybox

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Mr. Xcoder · Accepted Answer · 2017-12-20 16:42:13Z

1

Python 3, 51 bytes

lambda k:sum(i*e*2for i,e in enumerate(k))%sum(k)<1

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edited Dec 20, 2017 at 16:42

answered Dec 20, 2017 at 16:26

Mr. Xcoder

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Brad Gilbert b2gills · Accepted Answer · 2017-12-20 18:23:26Z

1

Perl 6, 23 bytes

{sum(1..*Z*$_)*2%%.sum}

Test it

Uses the algorithm from various other entries.

Expanded:

{  # bare block lambda with implicit parameter ｢$_｣

    sum(

        1 .. *  # Range starting from 1

      Z*        # Zip using &infix:«*»

        $_      # the input

    ) * 2

  %%            # is divisible by

    .sum        # the sum of the input (implicit method call on ｢$_｣)
}

answered Dec 20, 2017 at 18:23

Brad Gilbert b2gills

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Japt, 11 10 8 bytes

Originally inspired by Mnemonic's solution

x* vUx*½

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1 3 bytes saved thanks to ETHproductions.

Explanation

Implicit input of array U. Reduce by addition (x), multiplying each element by its 0-based index (*) in the process. Check if the result is evenly divisible (v) by the sum of the original input (Ux) with each element being multiplied by 0.5 (*½).

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 20, 2017 at 16:28

Shaggy

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\$\begingroup\$ Save a byte with m* x*2 vUx. This makes me wonder if m* x*2 can be reduced further... \$\endgroup\$
– ETHproductions
Commented Dec 21, 2017 at 18:05
\$\begingroup\$ Thanks, @ETHproductions; that's another new trick I've learned today. \$\endgroup\$
– Shaggy
Commented Dec 21, 2017 at 18:11
\$\begingroup\$ I've got it, just use x* and check if it's divisible by Ux*½ :) \$\endgroup\$
– ETHproductions
Commented Dec 21, 2017 at 18:11
\$\begingroup\$ Yes, I don't think that trick is documented anywhere... But whenever you use a binary operator as an auto-function with no second argument, it uses the index by default (like if you did XY{X*Y}) \$\endgroup\$
– ETHproductions
Commented Dec 21, 2017 at 18:12
\$\begingroup\$ Oh, now, that's just ingenious, @ETHproductions. :) \$\endgroup\$
– Shaggy
Commented Dec 21, 2017 at 18:17

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Community · Accepted Answer · 2020-06-17 09:04:33Z

C#, 71 bytes

Golfed

a=>{int i,s,S=s=i=0;while(i<a.Length){S-=s;s-=a[i++];}return 2*S%s<1;};

Ungolfed

a => {
    int
        i, s, S = s = i = 0;
    
    while( i < a.Length ) {
        S -= s;
        s -= a[ i++ ];
    }
    
    return 2 * S % s < 1;
};

Full code

using System;

namespace Namespace {
    class Program {
        static void Main( String[] args ) {
            Func<Int32[], Boolean> f = a => {
                int
                    i, s, S = s = i = 0;

                while( i < a.Length ) {
                    S -= s;
                    s -= a[ i++ ];
                }

                return 2 * S % s < 1;
            };

            List<Int32[]>
                testCases = new List<Int32[]>() {
                    new Int32[] {1, 0},
                    new Int32[] {3, 1, 5, 7},
                    new Int32[] {6, 3, 1},
                    new Int32[] {100, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5},
                    new Int32[] {10, 4, 3, 0, 2, 0, 5},
                    new Int32[] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},
                    new Int32[] {7, 7, 7, 7},

                    new Int32[] {1, 2},
                    new Int32[] {3, 6, 5, 1, 12},
                    new Int32[] {0, 0, 2, 0, 1, 0},
                    new Int32[] {1, 2, 3, 4, 5, 6, 7, 8, 9},
                    new Int32[] {6, 3, 2, 4, 0, 1, 2, 3},
                    new Int32[] {4, 0, 0, 2, 3, 5, 2, 0, 1, 2, 3, 0, 0, 1, 2, 4, 3, 1, 3, 0, 0, 2},
                    new Int32[] {100, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5},
                };

            foreach( Int32[] testCase in testCases ) {
                Console.WriteLine( $"{{ {String.Join(", ", testCase)} }}\n{f( testCase )}" );
            }

            Console.ReadLine();
        }
    }
}

Releases

v1.0 - 71 bytes - Initial solution.

Notes

I might have, or might have not, blatantly "borrowed" Dennis Python 2 solution...

naffetS · Accepted Answer · 2022-06-09 17:36:35Z

1

Vyxal, 7 bytes

ż*∑d?∑Ḋ

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How?

ż*∑d?∑Ḋ
ż*      # Multiply each by its 1-based index
  ∑     # Sum this
   d    # Double
      Ḋ # Is it divisible by...
    ?∑  # ...the sum of the input?

answered Jun 9, 2022 at 17:36

naffetS

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totallyhuman · Accepted Answer · 2017-12-20 16:22:05Z

0

Haskell, 39 bytes

f l=2*sum(zipWith(*)l[0..])`mod`sum l<1

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answered Dec 20, 2017 at 16:22

totallyhuman

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Erik the Outgolfer · Accepted Answer · 2017-12-20 16:27:45Z

0

APL (Dyalog), 15 bytes

0≡+⌿|(+⌿+⍨×⍳∘≢)

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Looks very ungolfy to me...

edited Dec 20, 2017 at 16:27

answered Dec 20, 2017 at 16:13

Erik the Outgolfer

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ovs · Accepted Answer · 2017-12-20 19:18:19Z

0

Python 2, 78 75 bytes

thanks to Mr. Xcoder for -3 bytes

lambda l:0in[sum(v*(i-y*2)for y,v in enumerate(l))for i in range(len(l)*2)]

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edited Dec 20, 2017 at 19:18

answered Dec 20, 2017 at 16:25

ovs

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2

\$\begingroup\$ No need for the space in 0 in. Also no need for the 0 in range(0,len(l)*2).. \$\endgroup\$
– Mr. Xcoder
Commented Dec 20, 2017 at 16:30

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Dennis · Accepted Answer · 2017-12-21 00:37:15Z

0

Julia 0.6, 25 bytes

~=sum
!x=~cumsum(2x)%~x<1

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answered Dec 21, 2017 at 0:37

Dennis

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Mic1780 · Accepted Answer · 2017-12-21 16:03:52Z

0

PHP, 139 128 bytes

<?php $a=explode(',',fgets(STDIN));for($i=0;$i<count($a)-.5;$i+=.5){$z=0;foreach($a as $k=>$v)$z+=($k-$i)*$v;if($z==0)die(1);}?>

Try it online!

-2 bytes for removing quotes around die thanks to manassehkatz
-9 bytes thanks to Learning the community standard

edited Dec 21, 2017 at 16:03

answered Dec 20, 2017 at 19:49

Mic1780

1214 bronze badges

1

\$\begingroup\$ Unless I misunderstand this [codegolf.meta.stackexchange.com/questions/2447/… you should be able to use die(1) and die(0) and save 4 bytes by using the exit code instead of a printed string. \$\endgroup\$
– manassehkatz-Moving 2 Codidact
Commented Dec 21, 2017 at 5:03
\$\begingroup\$ @manassehkatz If you use die without quotes on tio.run it will treat it as a status code (which it should) and not put it in the Output section. So I just added quotes to prevent people from nitpicking \$\endgroup\$
– Mic1780
Commented Dec 21, 2017 at 15:49

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Endenite · Accepted Answer · 2017-12-21 18:34:51Z

0

Swift, 76 bytes

{var i=0,t=0;print($0.reduce(0){$0+($1*i,t+=$1,i+=1).0}*2%t<1)}as([Int])->()

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answered Dec 21, 2017 at 18:34

Endenite

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Xcali · Accepted Answer · 2017-12-21 19:39:43Z

0

Perl 5, 55 + 1 (`a`) = 56 bytes

$.=0;map$.+=$p++*$_,@F;$#F+($p=$i-=.5)&&$.&&redo;say!$.

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answered Dec 21, 2017 at 19:39

Xcali

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Stack Exchange Network

Can this list be balanced?

Examples

Task

Truthy Testcases

Falsy Testcases

22 Answers 22

Pyth, 12 10 bytes

Explanation

Wolfram Language (Mathematica), 36 bytes

05AB1E, 6 bytes

How?

Jelly, 6 bytes

R, 34 bytes

Japt, 10 bytes

Explanation:

Python 2, 41 bytes

Julia, 31 27 bytes

Ruby, 47 bytes

C, 140 137 bytes

Python 3, 51 bytes

Perl 6, 23 bytes

Expanded:

Japt, 11 10 8 bytes

Explanation

C#, 71 bytes

Vyxal, 7 bytes

How?

Haskell, 39 bytes

APL (Dyalog), 15 bytes

Python 2, 78 75 bytes

Julia 0.6, 25 bytes

PHP, 139 128 bytes

Swift, 76 bytes

Perl 5, 55 + 1 (`a`) = 56 bytes

Your Answer

Not the answer you're looking for? Browse other questions tagged
code-golf
array
decision-problem
physics
or ask your own question.

Linked

Hot Network Questions

Can this list be balanced?

Examples

Task

Truthy Testcases

Falsy Testcases

22 Answers 22

Pyth, 12 10 bytes

Explanation

Wolfram Language (Mathematica), 36 bytes

05AB1E, 6 bytes

How?

Jelly, 6 bytes

R, 34 bytes

Japt, 10 bytes

Explanation:

Python 2, 41 bytes

Julia, 31 27 bytes

Ruby, 47 bytes

C, 140 137 bytes

Python 3, 51 bytes

Perl 6, 23 bytes

Expanded:

Japt, 11 10 8 bytes

Explanation

C#, 71 bytes

Vyxal, 7 bytes

How?

Haskell, 39 bytes

APL (Dyalog), 15 bytes

Python 2, 78 75 bytes

Julia 0.6, 25 bytes

PHP, 139 128 bytes

Swift, 76 bytes

Perl 5, 55 + 1 (a) = 56 bytes

Your Answer

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Not the answer you're looking for? Browse other questions tagged code-golfarraydecision-problemphysics or ask your own question.

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Hot Network Questions

Perl 5, 55 + 1 (`a`) = 56 bytes

Not the answer you're looking for? Browse other questions tagged
code-golf
array
decision-problem
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or ask your own question.