27
\$\begingroup\$

There have been many "Do __ without __" challenges before, but I hope that this is one of the most challenging.

The Challenge

You are to write a program that takes two natural numbers (whole numbers > 0) from STDIN, and prints the sum of the two numbers to STDOUT. The challenge is that you must use as few + and - signs as possible. You are not allowed to use any sum-like or negation functions.

Examples

input

123
468

output

591

input

702
720

output

1422

Tie Breaker: If two programs have the same number of + and - characters, the winner is the person with fewer / * ( ) = . , and 0-9 characters.

Not Allowed: Languages in which the standard addition/subtraction and increment/decrement operators are symbols other than + or - are not allowed. This means that Whitespace the language is not allowed.

\$\endgroup\$
  • 1
    \$\begingroup\$ Perhaps this challenge was a lot easier than I thought it would be, especially in other languages, where there are sum() functions. I have to fix this. \$\endgroup\$ – PhiNotPi Dec 1 '11 at 0:45
  • 57
    \$\begingroup\$ 100 rep bounty for anybody who can do this in Brainfuck. \$\endgroup\$ – Peter Olson Dec 1 '11 at 5:57
  • 4
    \$\begingroup\$ @Peter Olson Well, I guess BF is not turing complete without either + or -... \$\endgroup\$ – FUZxxl Dec 1 '11 at 10:21
  • 3
    \$\begingroup\$ Just to clarify, this challenge does not care about code length right? Only the number of +,- and tie breaker characters? ...or do you need to change the rules again :-) \$\endgroup\$ – Tommy Dec 1 '11 at 19:39
  • 2
    \$\begingroup\$ Next challenge would be "Comparing two numbers without any of ><+-*/%&|" \$\endgroup\$ – Naruyoko Sep 19 '19 at 23:41

65 Answers 65

2
\$\begingroup\$

JavaScript (ES6), 48 bytes, 12 tie-breakers

a=>b=>"0".repeat(a).concat("0".repeat(b)).length
\$\endgroup\$
2
\$\begingroup\$

Zsh, 26 25 bytes, 7 5 tiebreakers

-1 byte, -2 tiebreakers thanks to @Nahuel Fouilleul.

set {1..$1} {1..$2}
<<<$#

Try it online!


Zsh, 62 60 bytes, 0 tiebreakers

-2 bytes because repeat foo evaluates foo in arithmetic mode.

read a b
{repeat a;printf x;repeat b;printf x}|read x
<<<$#x

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Clojure (44 chars)

(pr(#(count(concat(%)(%)))#(repeat(read)0)))

Edit: fixed to print on STDOUT instead of just returning sum.

\$\endgroup\$
  • \$\begingroup\$ +/- and above mentioned tie-breakers are interesting, not characters. \$\endgroup\$ – user unknown May 30 '12 at 3:39
1
\$\begingroup\$

D

main(){
    int a,b;
    readf("%d %d",&a,&b);
    write((new int[a]~new int[b]).length);
}

this time using array lengths

\$\endgroup\$
1
\$\begingroup\$

Scala

  • score:
    • +- : 0
    • (). : 5+5+3=13

Code:

(List.fill (readInt) (1) ::: List.fill (readInt) (2)).size
  • List.fill (4)(7) produces List (7, 7, 7, 7)
  • a ::: b concatenates 2 Lists into one
  • The rest should be obvious
\$\endgroup\$
1
\$\begingroup\$

K, 11

{#,[!x;!y]}

Same concatenation trick as the R solution. Reading right to left: Enumerate the two input variables, concatenate and then count.

\$\endgroup\$
1
\$\begingroup\$

PowerShell, 27 42 bytes, 0 +-, 4 1 secondary

Thanks to mazzy for saving a + and 4 secondaries

$args|%{[int[]]$_*$_}|measure|select count

Try it online! or Pretty Table for an extra 3 bytes

-Or- adding four secondaries to save 19 bytes:

32 23 bytes, 1 0 +-, 12 5 secondaries

-9 bytes thanks to mazzy

($args|%{,$_*$_}).count

Try it online!

For each argument, we push n array elements (consisting of [n] but that's not important) to the pipeline which are grouped by the parens and then counted.

\$\endgroup\$
1
\$\begingroup\$

Ruby -na0, no +/-, no tiebreakers, 60 bytes

$F<<?a until~%r[
]&&$F::join[%r[a{#$`}a{#$'}]]
p$F::count ?a

Try it online!

Actually harder than I thought, I'm almost certainly missing a trick.

Reads in the input, then adds a characters to the argument array until the array, when joined into a string, matches a regular expression that uses the two input numbers as quantifiers, then counts how many it added.

Syntax tricks include using %r[] instead of // for a regexp, and :: instead . for a method call.

\$\endgroup\$
1
\$\begingroup\$

Keg (SBCS on Keg wiki)

Basically a port of the R answer.

¿¿Ï_"Ï_!.

Explanation

¿¿#        Take 2 integer inputs
  Ï_"Ï_#   Generate 2 arrays the length of the integer inputs
       !.# Output the length of the stack
\$\endgroup\$
1
\$\begingroup\$

C++, () only for main

#include <iostream>

int main()
{
    unsigned int a, b;
    std::cin >> a;
    std::cin >> b;


    unsigned int p = a ^ b;
    unsigned int g = a & b;

    unsigned int gg {g | p & g << 1};
    unsigned int pp {p & p << 1};

    unsigned int ggg {gg | pp & gg << 2};
    unsigned int ppp {pp | pp << 2};

    unsigned int gggg {ggg | ppp & ggg << 4};
    unsigned int pppp {ppp & ppp << 4};

    unsigned int ggggg {gggg | pppp & gggg << 8};
    unsigned int ppppp {pppp | pppp << 8};

    unsigned int gggggg {ggggg | ppppp & ggggg << 16};

    unsigned int result {a ^ b ^ gggggg << 1};
    std::cout << result;
}

Assumes 32-bit unsigned ints. Very ugly code to remove tiebreak characters. Emulates a Kogge-Stone adder.

\$\endgroup\$
1
\$\begingroup\$

Ruby, 0 +/-, 0 tie-breakers, 39 bytes

And fairly unreadable.

p"#{"%#{"%ss%%%ss"%$F}"%%w[o o]}"::size

Try it online!

\$\endgroup\$
1
\$\begingroup\$

MathGolf, 3 bytes, 0 +-, 1 tiebreaker

~{)

Input as a list. If taking the two inputs separated is mandatory, a trailing ê should be added (read entire STDIN input as integer-array).

Try it online.

Explanation:

~    # Push the value of the (implicit) input-list to the stack
 {   # Loop the top value amount of times
  )  #  And increase the other value once every iteration
     # (after which the entire stack joined together is output implicitly as result)
\$\endgroup\$
1
\$\begingroup\$

Lua, 46 bytes

print(load'return io.read"n"\x2Bio.read"n"'())

Try it online!

How It Works

load loads a string and returns a function, we pass a hexadecimal to it and call the function.

\$\endgroup\$
0
\$\begingroup\$

Python

i=input
print eval('%d\x2b%d'%(i(),i()))

37 chars with a little bit of cheating :p

\$\endgroup\$
0
\$\begingroup\$

A very yucky C version

Has no +, -, / or any digits. One use of * for a pointer type. The others (()=,) are needed a lot in the language itself.

#include <malloc.h>
#include <string.h>
#include <stdio.h>
typedef unsigned char *ptr;

char
  y;

const size_t
  one = sizeof y;

unsigned char
  zero='a'^'a',
  a [] = "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
  z [] = "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
  x [] = "xxxxxxxxxxxxxxxxx";

size_t plus_one (size_t i)
{
  ptr b=&a[one];
  memset (a,'x',sizeof a);
  b[i]=zero;
  return strlen(a);
}

size_t minus_one (size_t i)
{
  ptr b=&a[one];
  memset (a,'x',sizeof a);
  a[i]=zero;
  return strlen(b);
}

size_t minus_zero (size_t i)
{
  size_t p;
  for (p = zero ; z[p] ; p = plus_one (p))
  {
    i = minus_one(i);
  }
  return i;
}

unsigned char to_zero (unsigned char i)
{
  size_t p;
  for (p = one>>one ; x[p] ; p = plus_one (p))
  {
    i = (unsigned char) minus_one(i);
  }
  return i;
}

main (int argc, ptr argv [])
{
  ptr
    a = argv [argc>>one],
    b = argv [argc>>one<<one];

  size_t
    len_a = strlen (a),
    len_b = strlen (b),
    len = len_a > len_b ? len_a : len_b,
    o = plus_one(len),
    i,
    c=one>>one;

  ptr
    res=(ptr)malloc(plus_one(o));

  res[o]=zero;

  for (i = zero ; i < len ; i=plus_one(i))
  {
    size_t va;
    unsigned char r;

    va = (len_a > zero ? minus_zero (a[minus_one(len_a)]) <<one<<one<<one<<one : zero) | (len_b > zero ? minus_zero (b[minus_one (len_b)]) : zero);
    r = (c?"bcdefghijAuhjsadcdefghijABsadlkfdefghijABCvsxwerefghijABCDDNIWjsfghijABCDEOQJdccghijABCDEFIndDjlhijABCDEFGCnjndwijABCDEFGHsadXSkjABCDEFGHICSjshdABCDEFGHIJcbaCBA":"abcdefghijaswhHDbcdefghijAQWuciucdefghijABCOasdpdefghijABCCnjaskefghijABCDCIUdasfghijABCDEuoiDSAghijABCDEFCsalkjhijABCDEFGcapCPcijABCDEFGHCWEoerjABCDEFGHIkjIjIj") [va];
    c=r<'a';
    r&=~(one<<one<<one<<one<<one<<one);
    o=minus_one(o);
    res[o]=to_zero(r);

    if (len_a) len_a = minus_one (len_a);
    if (len_b) len_b = minus_one (len_b);
  }
  if(c) 
  {
    o=minus_one(o);
    res[o]=to_zero('B');
  }
  printf("%s\n",&res[o]);
  free (res);
}
\$\endgroup\$
0
\$\begingroup\$

C #, 26 tiebreakers

+/- : 0
=   : 2
.   : 6 
()  : 18
using System;
using System.Linq;
class P
{
    static void Main()
    {
        Func<int[]> f=()=>new int[int.Parse(Console.ReadLine())];
        Console.Write(f().Concat(f()).Count());
    }
}
\$\endgroup\$
0
\$\begingroup\$

VBA - No +/-, 7 tie-breakers

+,-,/,*,,,.,0-9 = 0
( = 3 (3 total pairs)
) = 3
= = 1

Function a(b)
    For Each c In b
        a = a & Space(c)
    Next
    MsgBox Len(a)
End Function

Takes an array of natural numbers as an argument. This will actually add more than just 2 values, if provided, as well as return the value of a single argument.

Older versions


+,-,/,*,=,.,0-9 = 0
( = 4 (4 total pairs)
) = 4
, = 1

Function a(b, c)
MsgBox Len(Space(b) & Space(c))
End Function

+,-,/,*,=,.,0-9 = 0
( = 4 (4 total pairs)
) = 4
, = 3

Sub a(b,c)
MsgBox Len(String(b," ") & String(c," "))
End Sub

+,-,/,*,. = 0
= = 4
( = 2 (2 total pairs)
) = 2
0-9 = 2
, = 1

Sub d(e,f)
For g=1 To e:h=h & " ":Next
For g=1 To f:h=h & " ":Next
MsgBox Len(h)
End Sub
\$\endgroup\$
  • \$\begingroup\$ Character count technically doesn't matter in this challenge. You have no -+, and you have 11 tie breakers. This puts you above any solution that has a -+ character, and below any solution with no -+ with fewer tie breakers. \$\endgroup\$ – PhiNotPi Apr 12 '12 at 15:43
  • \$\begingroup\$ So, I am tied with (as one example) Uri Goren's answer, correct? He has no +/-, but 11 tie-breakers. \$\endgroup\$ – Gaffi Apr 12 '12 at 15:45
  • \$\begingroup\$ Yes, you are tied with him. \$\endgroup\$ – PhiNotPi Apr 12 '12 at 15:50
0
\$\begingroup\$

PHP, 22 chars

echo array_sum($argv);

Documentation: array_sum() and $argv
Usage: php -r 'echo array_sum($argv);' 5 6 will output 11.

\$\endgroup\$
  • \$\begingroup\$ +/- and above mentioned tiebreakers are interesting, not chars. \$\endgroup\$ – user unknown May 30 '12 at 3:35
0
\$\begingroup\$

APL (10, 3 tie-breakers)

⎕←⍴(⍳⎕),⍳⎕
\$\endgroup\$
  • \$\begingroup\$ 10x +/-? Chars arn't interesting, just +/- and the tie-breakers. \$\endgroup\$ – user unknown May 30 '12 at 3:34
0
\$\begingroup\$

Common Lisp

Always surprising for such a verbose language.

((lambda(n m)(princ(length(append(make-list n)(make-list m)))))(read)(read))
\$\endgroup\$
0
\$\begingroup\$

Python 3 - 159 characters | ().,*/= count: 39

A little long, but I felt like doing it HDL-style.

a,b,c,d=int(input()),int(input()),0,0;l=max(a,b,key=int.bit_length)
for i in range(len(bin(l<<1)[2:])):
 e=a>>i&1;f=b>>i&1;c|=(e^f^d)<<i;d=e&f|d&(e^f)
print(c)
\$\endgroup\$
  • \$\begingroup\$ +/- tiebreakers are interesting, not characters. \$\endgroup\$ – user unknown May 30 '12 at 3:33
  • \$\begingroup\$ @userunknown: What's wrong with characters? \$\endgroup\$ – JAB May 30 '12 at 16:55
  • \$\begingroup\$ Characters are counted in CodeGolf tagged questions. If a reader doesn't read the question again, he will get the impression that character count is important and probably make a misguided vote based on that assumption. So the rules say: No (+|-). In Tiebreak digits, ().,*/= are important. You should include that number in your answer, so that not every visitor has to count them himself. \$\endgroup\$ – user unknown May 30 '12 at 20:41
  • \$\begingroup\$ @userunknown: Oh, right. \$\endgroup\$ – JAB May 30 '12 at 20:46
0
\$\begingroup\$

C# 141 characters 27 tiebreakers

+/- : 0
=   : 0
.   : 9 
()  : 18


Console.WriteLine(Enumerable.Range(1,int.Parse(Console.ReadLine()))
.Concat(Enumerable.Range(1,int.Parse(Console.ReadLine())))
.Count());
\$\endgroup\$
0
\$\begingroup\$

Python 42

a,b=input()
print eval("~-"*a+"~-"*b+"0")

Seven tie breakers, but those are string operations

\$\endgroup\$
0
\$\begingroup\$

C#, 12 tiebreakers (), 3 tiebreakers .

using System;
using System.Data;

class Program {
  static string ReadLine { get { return Console.ReadLine(); } }
  static void Main(string[] args) {
    Console.Out.Write(
     new DataTable().Compute(string.Join("\u002b", 
     ReadLine,
     ReadLine), "")
    );
  }
}
\$\endgroup\$
0
\$\begingroup\$

JavaScript 69

My attempt, using bitwise operators

for(x=(z=prompt().split(" "))[0],y=z[1];y;)x^=y,y=(y&x^y)<<1;alert(x)

\$\endgroup\$
0
\$\begingroup\$

R, 44 bytes

nchar(paste(strrep(" ",scan()),collapse=""))

Try it online!

Converts the two inputs to unary and "sums" via concatenation. Check out the other R answers:

\$\endgroup\$
0
\$\begingroup\$

C, score 0 with 8 tie-breakers

f(a,b) { return b ? f(a^b, a<<1 & b<<1) : a; }
 ^ ^ ^               ^   ^    ^      ^^

I've highlighted the tie-break chars.

Negative numbers are assumed to be 2s-complement.

\$\endgroup\$
0
\$\begingroup\$

Javascript (ES6) - 8 Tie-breaker characters

add: {
    let a = prompt``|false;
    let b = prompt``|false;
    while (a & b) {
        a = a ^ b;
        b = [b & ~a][false|false] << true
    }
    alert(a | b);
}
  • 2 * (
  • 2 * )
  • 4 * =
\$\endgroup\$
0
\$\begingroup\$

Runic Enchantments, (no +/-, 1 tie-breaker, 11 bytes)

'V2,k!$wi|;

Try it online!

Not allowed to have + characters? Fine, I'll create them myself with math and reflection. V has the byte value 86, which gets divided by 2 (the , and only tie-breaker character) converted to a character, and finally, reflectively written into the program under the instruction pointer. It then reads input, hits a reflector (|), reads input again, then reaches the space originally containing a w and adds them together before printing. The IP then performs an illegal action at , and is silently terminated.

Is this cheating? Probably. But the challenge says "don't use + characters" and the source code contains 0 of them.

Alternate method, (no +/-, 3 tie-breakers, 16 bytes)

"ab"1,i*}i*qul$;

Try it online!

Makes two strings, one of length x and one of length y, concatenates them together, then computes the length of the string. Does this count as a "sum-like function"? Perhaps. + on two strings does do the same thing.

\$\endgroup\$
0
\$\begingroup\$

APL (Dyalog Unicode), 4 bytesSBCS, 0 +/-, 0 tie breakers

Can sum any number of non-negative integers.

≢⎕⌿#

Try it online!

# reference to the root object

⎕/ replicate by the number(s) from STDIN
for each number, we get that many copies of the corresponding element on the right, but since there is only one, all numbers get paired up with that one, forming a single list of N1+N2+…+Nn references

 tally

Output is implicitly to STDOUT.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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