Adding without using a + or - sign [closed]

Question

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.

Answer 1

Python 3

import numpy as n;i=lambda:n.e**int(input());print(int(n.log(i()*i())))

Try it online!

Answer 2

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!

Answer 3

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

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

Answer 4

Zsh, 26 25 bytes, 7 5 tiebreakers

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

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

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---

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

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Answer 5

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\$.

Answer 6

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

#JoXg

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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

Answer 7

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]

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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.

Answer 8

Python 3, 0 +/-, 2 Tie-breaker, (68 bytes)

lambda x,y:A[f"{True<<x<<y:b}"[True:]]
class A:__class_getitem__=len

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class A:__class_getitem__=len allows me to replace len(x) by A[x]

Old, 3 Tie-breaker, (40 bytes)

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

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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

Answer 9

Clojure (44 chars)

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

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

Answer 10

D

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

this time using array lengths

Answer 11

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

Answer 12

K, 11

{#,[!x;!y]}

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

Answer 13

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

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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.

Answer 14

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

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

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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.

Answer 15

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

Answer 16

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

And fairly unreadable.

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

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Answer 17

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)

Answer 18

Lua, 46 bytes

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

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How It Works

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

Answer 19

brainfuck, 11 bytes, 2 +-, 3 tie breaker

,>,[-<+>]<.

Sadly i can’t think up anyway of manipulating number without +-

Try it online!

Answer 20

JavaScript, 2 tie-breakers

alert(`${``[`repeat`][`call`]`${prompt``}`}${``[`repeat`][`call`]`${prompt``}`}`[`length`])

Could I have made it shorter? Yes. But I find this version especially amusing.

Peculiarities:

• Only uses lowercase letters and the symbols $ [ ] ` { }
• Uses no mathematical operators.
• Relies heavily on especially cryptic JavaScript features

Tricks used:

• obj['prop'] instead of obj.prop to avoid using .
• tagged template literals to avoid using ( ) to call functions
  • func`string` evaluates to func(['string'])
  • func`${ expression }` evaluates to func(['', ''], expression)
• ['', ''].toString() produces ','
• ''.repeat.call(',', value) is equivalent to ','.repeat(value). This hack is used to turn the repeat function into one which takes the number of repeats as its second argument, so that we can use that function as a tag.

Edit: oops, forgot to alert the result. Uh, I'll just wrap that in an alert for now, feel free to suggest an improvement 😅

Answer 21

Desmos, 22 bytes, 0 +/-, 2 tiebreakers

lne^{\ans_0}e^{\ans_1}

Input numbers in the first two lines. I have to admit, I spent an embarrassing amount of time trying to go from the obvious 4 tiebreakers solution [\ans_0,\ans_1].total to a 2 tiebreakers solution. I considered many other interesting ideas to go from 3 to 2 (integrals, percent of, {} evaluates to 1, etc.), but at the end, it turned out to be very simple...

I'm pretty sure 2 tiebreakers is optimal for Desmos because a 0 and 1 are required to obtain input. Defining a function is even worse: You need (, ) and =, which is 3 tiebreakers.

Try It On Desmos!

Answer 22

Python

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

37 chars with a little bit of cheating :p

Answer 23

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);
}

Answer 24

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());
    }
}

Answer 25

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

Answer 26

PHP, 22 chars

echo array_sum($argv);

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

Answer 27

APL (10, 3 tie-breakers)

⎕←⍴(⍳⎕),⍳⎕

Answer 28

Common Lisp

Always surprising for such a verbose language.

((lambda(n m)(princ(length(append(make-list n)(make-list m)))))(read)(read))

Answer 29

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)

Answer 30

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());