18
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Challenge

The challenge is to write a code that takes a positive integer 'n' as an input and displays all the possible ways in which the numbers from 1 - n can be written, with either positive or negative sign in between, such that their sum is equal to zero. Please remember that you may only use addition or subtraction.

For example, if the input is 3, then there are 2 ways to make the sum 0:

 1+2-3=0
-1-2+3=0

Note that, the numbers are in order, starting from 1 till n (which is 3 in this case). As it is evident from the example, the sign of the first number can also be negative, so be careful.

Now, 3 was pretty much simple. Let us list all the ways when we consider the number 7.

 1+2-3+4-5-6+7=0
 1+2-3-4+5+6-7=0
 1-2+3+4-5+6-7=0
 1-2-3-4-5+6+7=0
-1+2+3+4+5-6-7=0
-1+2-3-4+5-6+7=0
-1-2+3+4-5-6+7=0
-1-2+3-4+5+6-7=0

So here, we have got a total of 8 possible ways.


Input And Output

As stated before, the input would be a positive integer. Your output should contain all the possible ways in which the numbers give a sum of zero. In case there is no possible way to do the same, you can output anything you like.

Also, you can print the output in any format you like. But, it should be understandable. For example, you may print it as in the above example. Or, you may just print the signs of the numbers in order. Otherwise, you can also print '0's and '1's in order, where '0' would display negative sign and '1' would display positive sign (or vice versa).

For example, you can represent 1+2-3=0 using:

1+2-3=0
1+2-3
[1,2,-3]
++-
110
001    

However, I would recommend using any of the first three formats for simplicity. You can assume all the inputs to be valid.


Examples

7 ->

 1+2-3+4-5-6+7=0
 1+2-3-4+5+6-7=0
 1-2+3+4-5+6-7=0
 1-2-3-4-5+6+7=0
-1+2+3+4+5-6-7=0
-1+2-3-4+5-6+7=0
-1-2+3+4-5-6+7=0
-1-2+3-4+5+6-7=0

4 ->

 1-2-3+4=0
-1+2+3-4=0

2 -> -

8 ->

 1+2+3+4-5-6-7+8=0
 1+2+3-4+5-6+7-8=0
 1+2-3+4+5+6-7-8=0
 1+2-3-4-5-6+7+8=0
 1-2+3-4-5+6-7+8=0
 1-2-3+4+5-6-7+8=0
 1-2-3+4-5+6+7-8=0
-1+2+3-4+5-6-7+8=0
-1+2+3-4-5+6+7-8=0
-1+2-3+4+5-6+7-8=0
-1-2+3+4+5+6-7-8=0
-1-2+3-4-5-6+7+8=0
-1-2-3+4-5+6-7+8=0
-1-2-3-4+5+6+7-8=0

Scoring

This is , so the shortest code wins!

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  • \$\begingroup\$ Please note that this is not a dupe of codegolf.stackexchange.com/questions/8655/… , because this challenge is meant to take only n as input and use all the numbers 1-n in order. \$\endgroup\$ – Manish Kundu Feb 3 '18 at 15:14
  • \$\begingroup\$ May we represent + as N and - as -N, or is that taking it too far? (e.g. 3 -> [[-3,-3,3], [3,3,-3]]) \$\endgroup\$ – Jonathan Allan Feb 3 '18 at 16:05
  • \$\begingroup\$ @JonathanAllan Isn't that mentioned in the list of output formats? Or did I wrongly interpret your question? \$\endgroup\$ – Manish Kundu Feb 3 '18 at 16:07
  • \$\begingroup\$ I mean like the 0 and 1 option but using N and -N (see my edit above) \$\endgroup\$ – Jonathan Allan Feb 3 '18 at 16:09
  • 2
    \$\begingroup\$ @JonathanAllan Yes thats certainly allowed. Make sure you mention that in the answer. \$\endgroup\$ – Manish Kundu Feb 3 '18 at 16:14

21 Answers 21

8
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Haskell, 42 bytes

f n=[l|l<-mapM(\i->[i,-i])[1..n],0==sum l]

Try it online!

| improve this answer | |
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  • 1
    \$\begingroup\$ 42. \$\endgroup\$ – user202729 Feb 3 '18 at 15:40
  • 1
    \$\begingroup\$ Shouldn't it be 0== ? \$\endgroup\$ – Laikoni Feb 3 '18 at 17:48
6
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Jelly, 9 bytes

1,-ṗ×RSÐḟ

Try it online!

Exp

1,-ṗ×RSÐḟ  Main link. Input = n. Assume n=2.
1,-        Literal list [1, -1].
   ṗ       Cartesian power n. Get [[1, 1], [1, -1], [-1, 1], [-1, -1]]
    ×R     Multiply (each list) by Range 1..n.
       Ðḟ  ilter out lists with truthy (nonzero)
      S      Sum.

Jelly, 9 bytes

Jonathan Allan's suggestion, output a list of signs.

1,-ṗæ.ÐḟR

Try it online!

| improve this answer | |
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  • 1
    \$\begingroup\$ How about (ab?)using the lax output format with ,Nṗæ.ÐḟR? \$\endgroup\$ – Jonathan Allan Feb 3 '18 at 15:53
  • \$\begingroup\$ Or alternatively, this output the outputs multiplied by n. \$\endgroup\$ – user202729 Feb 3 '18 at 15:57
  • \$\begingroup\$ The N and -N output I suggested has been allowed, so that saves one byte :) (just need to mention the format in the answer) \$\endgroup\$ – Jonathan Allan Feb 3 '18 at 16:34
  • \$\begingroup\$ Not sure how recently it was added, but there's a builtin Ø+ to save a byte over 1,-. \$\endgroup\$ – Unrelated String May 25 at 19:08
5
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Python 2, 62 bytes

f=lambda n,*l:f(n-1,n,*l)+f(n-1,-n,*l)if n else[l]*(sum(l)==0)

Try it online!

Mr. Xcoder saved 4 bytes with a nifty use of starred arguments.

| improve this answer | |
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  • 1
    \$\begingroup\$ 62 bytes using *l instead of l=[] \$\endgroup\$ – Mr. Xcoder Feb 3 '18 at 22:16
3
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05AB1E, 11 bytes

®X‚¹ãʒ¹L*O_

Try it online!

The output format for e.g. input 3 is:

[[-1, -1, 1], [1, 1, -1]]

That is, -1-2+3, 1+2-3.

| improve this answer | |
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  • \$\begingroup\$ ¹L can be ā for -1 \$\endgroup\$ – Kevin Cruijssen May 25 at 15:08
3
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Perl, 37 36 bytes

perl -E 'map eval||say,glob join"{+,-}",0..<>' <<< 7
| improve this answer | |
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  • \$\begingroup\$ Nicely done. You can drop -n and <<< if you replace $_ with pop. It doesn't actually improve your score, but it makes the overall expression shorter ;) \$\endgroup\$ – Chris Feb 4 '18 at 8:08
2
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Wolfram Language (Mathematica), 36 bytes

Pick[p={1,-1}~Tuples~#,p.Range@#,0]&

Try it online!

| improve this answer | |
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2
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Husk, 10 bytes

fo¬ΣΠmSe_ḣ

Try it online!

Explanation

Not too complicated.

fo¬ΣΠmSe_ḣ  Implicit input, say n=4
         ḣ  Range: [1,2,3,4]
     m      Map over the range:
      Se     pair element with
        _    its negation.
            Result: [[1,-1],[2,-2],[3,-3],[4,-4]]
    Π       Cartesian product: [[1,2,3,4],[1,2,3,-4],..,[-1,-2,-3,-4]]
f           Keep those
   Σ        whose sum
 o¬         is falsy (equals 0): [[-1,2,3,-4],[1,-2,-3,4]]
| improve this answer | |
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2
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Python 3, 105 bytes

lambda n:[k for k in product(*[(1,-1)]*n)if sum(-~n*s for n,s in enumerate(k))==0]
from itertools import*

Try it online!

| improve this answer | |
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2
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JavaScript (V8),  69 61  58 bytes

Saved 8 bytes thanks to @Neil
Saved 3 bytes thanks to @l4m2

Prints all solutions.

f=(n,o='')=>n?f(n-1,o+'+'+n)&f(n-1,o+-n):eval(o)||print(o)

Try it online!

| improve this answer | |
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  • \$\begingroup\$ Do you need k? Something like this: f=(n,o='')=>n?['+','-'].map(c=>f(n-1,c+n+o)):eval(o)||alert(o) \$\endgroup\$ – Neil Feb 3 '18 at 23:37
  • \$\begingroup\$ @Neil I really don't... Thanks. \$\endgroup\$ – Arnauld Feb 4 '18 at 0:35
  • 1
    \$\begingroup\$ o+'-'+n => o+-n \$\endgroup\$ – l4m2 May 25 at 13:09
1
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Swift, 116 bytes

func f(n:Int){var r=[[Int]()]
for i in 1...n{r=r.flatMap{[$0+[i],$0+[-i]]}}
print(r.filter{$0.reduce(0){$0+$1}==0})}

Try it online!

Explanation

func f(n:Int){
  var r=[[Int]()]                         // Initialize r with [[]]
                                          // (list with one empty list)
  for i in 1...n{                         // For i from 1 to n:
    r=r.flatMap{[$0+[i],$0+[-i]]}         //   Replace every list in r with the list
  }                                       //   prepended with i and prepended with -i
  print(r.filter{$0.reduce(0){$0+$1}==0}) // Print all lists in r that sums to 0
}
| improve this answer | |
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1
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Python 2, 91 bytes

lambda x:[s for s in[[~j*[1,-1][i>>j&1]for j in range(x)]for i in range(2**x)]if sum(s)==0]

Try it online!

Returns a list of satisfying lists (e.g., f(3)=[[-1,-2,3], [1,2,-3]])

| improve this answer | |
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1
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APL (Dyalog), 38 bytes

{k/⍨0=+/¨k←((,o∘.,⊢)⍣(⍵-1)⊢o←¯1 1)×⊂⍳⍵}

Try it online!

| improve this answer | |
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1
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Pyth, 13 bytes

f!sT.nM*F_BMS

Try it here!

| improve this answer | |
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1
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Clean, 79 bytes

import StdEnv
$n=[k\\k<-foldr(\i l=[[p:s]\\s<-l,p<-[~i,i]])[[]][1..n]|sum k==0]

Try it online!

| improve this answer | |
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1
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Python 3 + numpy, 104 103 bytes

import itertools as I,numpy as P
lambda N:[r for r in I.product(*[[-1,1]]*N)if sum(P.arange(N)*r+r)==0]

Output is [-1, 1] corresponding to the sign.

| improve this answer | |
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  • \$\begingroup\$ You can remove the space before if for -1 byte \$\endgroup\$ – ovs Feb 3 '18 at 21:17
1
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C (gcc), 171 161 bytes

k,s;f(S,n,j)int*S;{if(s=j--)S[j]=~0,f(S,n,j),S[j]=1,f(S,n,j);else{for(k=n;k;)s+=k--*S[k];if(!s)for(puts("");k<n;)printf("%d",S[k++]+1);}}F(n){int S[n];f(S,n,n);}

Try it online!

| improve this answer | |
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  • \$\begingroup\$ @ceilingcat Thank you. \$\endgroup\$ – Jonathan Frech May 24 at 22:51
0
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Retina, 73 bytes

.+
*
_
=_$`
+0`=
-$%"+
(-(_)+|\+(_)+)+
$&=$#2=$#3=
G`(=.+)\1=
=.*

_+
$.&

Try it online! Explanation:

.+
*

Convert the input to unary.

_
=_$`

Convert the number to a list of =-prefixed numbers.

+0`=
-$%"+

Replace each = in turn with both - and +, duplicating the number of lines each time.

(-(_)+|\+(_)+)+
$&=$#2=$#3=

Separately count the number of _s after -s and +s. This sums the negative and positive numbers.

G`(=.+)\1=

Keep only those lines where the -s and +s cancel out.

=.*

Delete the counts.

_+
$.&

Convert to decimal.

| improve this answer | |
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0
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Perl 6, 43 bytes

{grep *.sum==0,[X] (1..$_ X*1,-1).rotor(2)}

Try it
Returns a sequence of lists

Expanded:

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

  grep              # only return the ones
    *.sum == 0,     # that sum to zero

    [X]             # reduce with cross meta operator

      (
          1 .. $_   # Range from 1 to the input

        X*          # cross multiplied by

          1, -1

      ).rotor(2)    # take 2 at a time (positive and negative)
}

1..$_ X* 1,-1(1, -1, 2, -2)
(…).rotor(2)((1, -1), (2, -2))
[X] …((1, 2), (1, -2), (-1, 2), (-1, -2))

| improve this answer | |
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0
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J, 35 30 bytes

-5 bytes thanks to FrownyFrog!

>:@i.(]#~0=1#.*"1)_1^2#:@i.@^]

Try it online!

Original:

J, 35 bytes

[:(#~0=+/"1)>:@i.*"1(_1^[:#:@i.2^])

How it works

I multiply the list 1..n with all possible lists of coefficients 1 / -1 and find the ones that add up to zero.

                    (             ) - the list of coefficients
                             i.     - list 0 to 
                               2^]  - 2 to the power of the input
                     _1^[:          - -1 to the power of 
                          #:@       - each binary digit of each number in 0..n-1 to 
                 *"1                - each row multiplied by
            >:@i.                   - list 1..n
  (#~      )                        - copy those rows
     0=+/"1                         - that add up to 0
[:                                  - compose   

Try it online!

As an alternative I tried an explicit verb, using the approach of cartesian product of +/-:

J, 37 bytes

3 :'(#~0=+/"1)(-y)]\;{(<"1@,.-)1+i.y'

{(<"1@,.-) finds the cartesian products for example:

{(<"1@,.-) 1 2 3
┌───────┬────────┐
│1 2 3  │1 2 _3  │
├───────┼────────┤
│1 _2 3 │1 _2 _3 │
└───────┴────────┘

┌───────┬────────┐
│_1 2 3 │_1 2 _3 │
├───────┼────────┤
│_1 _2 3│_1 _2 _3│
└───────┴────────┘

Too bad that it boxes the result, so I spent some bytes to unbox the values

Try it online!

| improve this answer | |
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  • \$\begingroup\$ @FrownyFrog Thank you, I was not happy with the right side of my code. \$\endgroup\$ – Galen Ivanov Feb 5 '18 at 7:08
0
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C (gcc), 101 100 bytes

Represents addition as 0 and subtraction as 1, but that ultimately that does not matter since for any valid sequence, the complement of the sequence will also result in zero.

i,j,s;f(n){for(i=1<<n;--i;printf(s?"\e[2K\r":"\n"))for(j=s=0;j<n;)s+=putchar(49-(i>>j++&1))&1?j:-j;}

Try it online! (Note: TIO doesn't seem to support VT100 escape sequences, and also treats carriage returns as newlines. You should probably try it in xterm or something instead)

Explanation

i,j,s;f(n){           // Declare three variables and a function.
    for ( i = 1 << n; // I use i as a bitmap that determines
                      // the negative and positive numbers.
         --i;         // Decrement, continue if non-zero.
                      // Not checking zero doesn't matter.
         printf(s?"\e[2K\r":"\n"))
                      // If sum s at the end of the inner loop is not zero,
                      // clear the line using a VT100 control sequence and a CR.
                      // Else print LF.
    for(j=s=0;j<n;)   // Inner loop from 0 to n-1
        s+=putchar(49-(i>>j++&1)) 
                      // Print 0 if jth bit set, 1 if not. Returns printed byte
                      // (or -1 on error, but we can ignore that)
           &1         // Test whether the returned byte is odd
           ? j        // ... if it is, add the next j to sum
           : -j;      // ... if it isn, subtract it instead
}
| improve this answer | |
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  • \$\begingroup\$ I do not think you need to use any prettifiers for your output, which could save some bytes. Note that printf("\n") can often be replaced by puts(""). \$\endgroup\$ – Jonathan Frech May 28 at 23:57
  • \$\begingroup\$ @JonathanFrech The line-clear would then result in an empty line for every discarded result, and I believe the standard IO rules forbid excess output like that? \$\endgroup\$ – CompilerPotato May 29 at 7:23
  • \$\begingroup\$ Ah, sorry, I misinterpreted the escape sequences' function. In that case, I think standard I/O also forbids your method since one normally thinks of bytes as a program's output, not its terminal appearance. \$\endgroup\$ – Jonathan Frech May 29 at 19:03
0
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Not the best but I tried :)

C (Linux GCC)

Entire program: 282 characters

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int n,x,i,s;int main(int c,char**z){n=atoi(z[1]);int a[n];for(x=0;x<n;x++)a[x]=x+1;for(i=0;i<pow(2, n)-1;i++){s=0;for(x=0;x<n;x++){printf("%c%d",i>>x&0x01?'-':'+',a[x]);s+=(i>>x&0x01?-1:1)*a[x];}printf(s!=0?"\r":"=%d \n",s);}}

Just as function: 196 characters

void z(int n){int a[n],x,i,s;for(x=0;x<n;x++)a[x]=x+1;for(i=0;i<pow(2,n)-1;i++){s=0;for(x=0;x<n;x++){printf("%c%d",i>>x&0x01?'-':'+',a[x]);s+=(i>>x&0x01?-1:1)*a[x];}printf(s!=0?"\r":"=%d \n",s);}}

Results

  • n = 4:
+1-2-3+4=0 
-1+2+3-4=0
  • n = 7:
+1-2-3-4-5+6+7=0 
-1+2-3-4+5-6+7=0 
-1-2+3+4-5-6+7=0 
+1+2-3+4-5-6+7=0 
-1-2+3-4+5+6-7=0 
+1+2-3-4+5+6-7=0 
+1-2+3+4-5+6-7=0 
-1+2+3+4+5-6-7=0

Abstract Printing & Just Function: 172 characters

You could shrink the print statements down much more by simply outputting only the +/-'s which are represented by 1/-1 and abstractly representing the entire sum thing, but I still don't think it would beat the other C answers anyway.

That version is:

void z(int n){int a[n],x,i,j,s;for(x=0;x<n;x++)a[x]=x+1;for(i=0;i<pow(2,n)-1;i++){s=0;for(x=0;x<n;x++){j=i>>x&0x01?-1:1;printf("%d ",j);s+=j*a[x];}printf(s!=0?"\r":"\n");}}

Results

  • n = 4:
1 -1 -1 1 
-1 1 1 -1 
  • n = 7:
1 -1 -1 -1 -1 1 1 
-1 1 -1 -1 1 -1 1  
-1 -1 1 1 -1 -1 1  
1 1 -1 1 -1 -1 1 
-1 -1 1 -1 1 1 -1 1 
1 1 -1 -1 1 1 -1 
1 -1 1 1 -1 1 -1 1 
-1 1 1 1 1 -1 -1 -1
| improve this answer | |
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