29
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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.

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10
  • 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
  • 58
    \$\begingroup\$ 100 rep bounty for anybody who can do this in Brainfuck. \$\endgroup\$ 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

72 Answers 72

2
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Gol><>, 8 bytes

IRmIRmlh

Minus one byte!, I figured out that I didn't need to take both inputs first

Try it online!

Previous answer, 9 bytes

IIRfrRflh

Explanation below

II        take both inputs
  Rm      Pop and repeat that number of times, pushing -1
    r     Reverse stack (to get the first num on top)
     Rm   Pop and repeat that number of times, pushing -1
       lh Push length of stack, and output as a number halting

Basically what I did was count it all out in unary, it would only take a few more bytes to expand it to be able to handle more than two numbers

Try it online!

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2
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JavaScript (ES6), 48 bytes, 12 tie-breakers

a=>b=>"0".repeat(a).concat("0".repeat(b)).length
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2
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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!

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1
2
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JavaScript, 26 bytes

Tiebreakers: 11?

a=>b=>Math.log2(2**a*2**b)

Takes input as (a)(b). Uses the power index rule \$a^{x+y}=a^{x}a^{y}\$ and returns the \$log_2(2^{a}2^{b})\$ which is basically \$log_2(2^{a+b}) = a+b\$.

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2
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Pip, 0 +/-, 0 tiebreakers (5 bytes)

#JoXg

Try it online!

Explanation

A string-concatenation approach:

    g  List containing the command-line args
  oX   For each arg, make a string of that many 1s (o is a variable preinitialized to 1)
 J     Join the list into a single string
#      Get its length
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1
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Clojure (44 chars)

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

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

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

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

this time using array lengths

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1
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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
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1
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K, 11

{#,[!x;!y]}

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

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1
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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.

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4
1
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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.

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1
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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
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1
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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.

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1
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Ruby, 0 +/-, 0 tie-breakers, 39 bytes

And fairly unreadable.

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

Try it online!

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1
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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)
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1
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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.

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1
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Python 3, 0 +/-, 3 Tie-breaker, (40 bytes)

lambda x,y:len(f"{True<<x<<y:b}"[True:])

Try it online!

Explanation :

  • Truein python is evaluated as 1 => 1
  • shift x times then y times => 4 => 32 (for x=2 and y=3)
  • convert it to its string binary representation => "100000"
  • remove the first char of the string => "00000"
  • get the length => 5
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1
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Python 2, 0 +/-, 2 tie-breaker (144 bytes)

exec`["%c"%[ord(i)^True<<True<<True<<True][False]for i in"""xzaf| :""afx}| !":""afx}| !'':!&ja|Wdmfo|` !"""]`[True<<True::True<<True<<True|True]

Try it online!

How it works :

The general idea is to remove tie-breaker character of a non cheaty working solution. To do that I converted a solution by changing the 5th bit of each character.

Some tricks are used to avoid using tiebreaker character during the translation :

  • True and False are evaluated by python as 1 and 0. Using that I can easyly create some usefull integers like 2, 5 and 8 in my solution without using [0-9]
  • Parenthesis are bad (for this challenge at least). To replace them I used [expression][False] which return the first element of an 1-element array, using the []-priority over other operators.
  • To join an array of char, the common technique is to use "".join(my_array) but it uses 3 tie-breakers. I instead uses `my_array`[2::5] which take the representation of the array and select only the wanted character by concatenating them.
  • To get the ascii character from its interger representation, intead of using the built-in chr(i) (which uses 2 tie-breaker), I used "%c"%i wich does the same

In better understandable python, here is my solution :

exec`["%c"%(ord(i)^8)for i in""" CODED_SOLUTION """]`[2::5]

with CODED_SOLUTION once decoded :

print(2*input()*2**input()//2).bit_length()

which looks like my older solution (and which doesn't use any form of addition :p)

Older Solution (7 tiebreaker, 50 bytes)

print int.bit_length(True<<input()<<input()>>True)

Try it online!

The idea is to shift a True interpreted as 1 by python Xtimes and Ytimes then shift once the other way and take the len of the binary representation

Cheated solution (45 bytes, 6 Tie-breaker)

I also have a cheated solution in 45 bytes:

print eval("\x2binput()"*[True<<True][False])

Try it online!

which evaluates the string +input()+input() but convert the char + by its hexadecimal representation.

the [True<<True][False] is equal to 2

Other Ideas:

I had the idea to use a=abs(~a) to increment a by 1 but I didn't totaly managed to produce a working solution with that.

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0
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Python

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

37 chars with a little bit of cheating :p

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0
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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);
}
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0
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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());
    }
}
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0
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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
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3
  • \$\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
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PHP, 22 chars

echo array_sum($argv);

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

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1
  • \$\begingroup\$ +/- and above mentioned tiebreakers are interesting, not chars. \$\endgroup\$ May 30 '12 at 3:35
0
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APL (10, 3 tie-breakers)

⎕←⍴(⍳⎕),⍳⎕
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1
  • \$\begingroup\$ 10x +/-? Chars arn't interesting, just +/- and the tie-breakers. \$\endgroup\$ May 30 '12 at 3:34
0
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Common Lisp

Always surprising for such a verbose language.

((lambda(n m)(princ(length(append(make-list n)(make-list m)))))(read)(read))
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0
0
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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)
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4
  • \$\begingroup\$ +/- tiebreakers are interesting, not characters. \$\endgroup\$ 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\$ May 30 '12 at 20:41
  • \$\begingroup\$ @userunknown: Oh, right. \$\endgroup\$
    – JAB
    May 30 '12 at 20:46
0
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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());
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0
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Python 42

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

Seven tie breakers, but those are string operations

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0
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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), "")
    );
  }
}
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0
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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)

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