24
\$\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.

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  • 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
  • 48
    \$\begingroup\$ 100 rep bounty for anybody who can do this in Brainfuck. \$\endgroup\$ – Peter Olson Dec 1 '11 at 5:57
  • 3
    \$\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
  • \$\begingroup\$ @Tommy No, it does not. \$\endgroup\$ – PhiNotPi Dec 1 '11 at 21:58

56 Answers 56

28
\$\begingroup\$

Perl (no +/-, no tie-breakers, 29 chars)

s!!xx!;s!x!$"x<>!eg;say y!!!c

As a bonus, you can make the code sum more than two numbers by adding more xs to the s!!xx!.

Alternatively, here are two 21-char solutions with 1 and 3 tie-breakers respectively

say length$"x<>.$"x<>

say log exp(<>)*exp<>

Note: These solutions use the say function, available since Perl 5.10.0 with the -E command line switch or with use 5.010. See the edit history of this answer for versions that work on older perls.


How does the solution with no tie-breakers work?

  • s!!xx! is a regexp replacement operator, operating by default on the $_ variable, which replaces the empty string with the string xx. (Usually / is used as the regexp delimiter in Perl, but really almost any character can be used. I chose ! since it's not a tie-breaker.) This is just a fancy way of prepending "xx" to $_ — or, since $_ starts out empty (undefined, actually), it's really a way to write $_ = "xx" without using the equals sign (and with one character less, too).

  • s!x!$"x<>!eg is another regexp replacement, this time replacing each x in $_ with the value of the expression $" x <>. (The g switch specifies global replacement, e specifies that the replacement is to be evaluated as Perl code instead of being used as a literal string.) $" is a special variable whose default value happens to be a single space; using it instead of " " saves one char. (Any other variable known to have a one-character value, such as $& or $/, would work equally well here, except that using $/ would cost me a tie-breaker.)

    The <> line input operator, in scalar context, reads one line from standard input and returns it. The x before it is the Perl string repetition operator, and is really the core of this solution: it returns its left operand (a single space character) repeated the number of times given by its right operand (the line we just read as input).

  • y!!!c is just an obscure way to (ab)use the transliteration operator to count the characters in a string ($_ by default, again). I could've just written say length, but the obfuscated version is one character shorter. :)

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  • 3
    \$\begingroup\$ +1 - awsome and totally unreadable ;-) This seems to be the perfect answer since the character count doesn't matter in this challenge. \$\endgroup\$ – Tommy Dec 1 '11 at 22:14
  • \$\begingroup\$ @Tommy, there are other perfect answers too. Maybe we should push for character count to be the final tie-breaker. \$\endgroup\$ – Peter Taylor Dec 2 '11 at 17:04
  • \$\begingroup\$ @Peter: I sort of assumed it already was, by default. \$\endgroup\$ – Ilmari Karonen Dec 2 '11 at 17:19
  • 1
    \$\begingroup\$ if character count is the final tie breaker, and many entries exist tied for zero in the other categories, doesn't this just become code-golf with some source restrictions? \$\endgroup\$ – Sparr Sep 19 at 23:26
47
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R (24 characters)

length(sequence(scan()))

What this does:

  • scan reads input from STDIN (or a file)
  • sequence generates integer sequences starting from 1 and concatenates the sequences. For example, sequence(c(2, 3)) results in the vector 1 2 1 2 3
  • length calculates the number of elements in the concatenated vector

Example 1:

> length(sequence(scan()))
1: 123
2: 468
3:
Read 2 items
[1] 591

Example 2:

> length(sequence(scan()))
1: 702
2: 720
3:
Read 2 items
[1] 1422
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  • 1
    \$\begingroup\$ Very clever, good job. \$\endgroup\$ – Matthew Read Dec 1 '11 at 21:34
  • \$\begingroup\$ This blows my mind \$\endgroup\$ – smci Jan 21 '12 at 9:25
  • \$\begingroup\$ +/- and above mentioned tie-breakers are interesting, not characters. \$\endgroup\$ – user unknown May 30 '12 at 3:36
  • \$\begingroup\$ How does it calculate length? without using any addition? If so, I'd be surprised. \$\endgroup\$ – Tem Pora Sep 28 '13 at 10:12
  • 1
    \$\begingroup\$ @TemPora The question only restricts the code in the answer, not the operations done behind the scenes. We're not going to restrict the question so the underlying computer architecture cannot increment a register. \$\endgroup\$ – mbomb007 Jul 31 '18 at 14:57
15
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D

main(){
    int a,b;
    readf("%d %d",&a,&b);
    while(b){a^=b;b=((a^b)&b)<<1;}
    write(a);
}

bit twiddling for the win

as a bonus the compiled code doesn't contain a add operation (can't speak for the readf call though)

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12
\$\begingroup\$

Python 2, 43 bytes

print len('%s%s'%(input()*'?',input()*'!'))
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  • 3
    \$\begingroup\$ Very inventive, but you might want to change the character used in the string to something other than a tie-breaker such as "~" \$\endgroup\$ – 3Doubloons Dec 1 '11 at 5:36
  • \$\begingroup\$ Thanks for the advice Alex, I had already forgotten about the tie-breaker rule. \$\endgroup\$ – Omar Dec 2 '11 at 17:38
  • \$\begingroup\$ print sum(input(),input()) \$\endgroup\$ – razpeitia Dec 3 '11 at 2:52
  • 9
    \$\begingroup\$ razpeitia: I think sum is a "sum-like" function and thus not allowed. \$\endgroup\$ – Omar Dec 3 '11 at 12:38
6
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GolfScript

No +/- or tie-breakers:

# Twiddle some bits to get a few small integers
[]!~abs:two~abs:three!:zero~:minusone;

~[two base minusone%\two base minusone%]zip
zero:c;
{
    # Stack holds [x y] or [x] with implicit y is zero
    # Half adder using x y c: want to end up with sum on stack and carry back in c
    [~c zero]three<zero$
    ~^^\
    $~;:c;;
}%
[~c]minusone%two base

Much simpler version with two tie-breaker characters, using the same list-concatenation trick that other people are using:

~[\]{,~}%,

I'm assuming that GolfScript isn't disqualified for having ) as an increment operator, as I'm not actually using it.

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6
\$\begingroup\$

C (32bit only)

int main(int ac, char *av) {
    scanf("%d\n%d", &ac, &av);
    return printf("%d\n", &av[ac]);
}

Pointer arithmetic is just as good.
How does it match the requirements?
* No + or -
* No /, =, ., 0-9
* Only 3 pairs of parenthesis, which seems to me minimal (you need main, scanf, printf).
* One * (the pointer approach requires it).
* Four , (could save one by defining normal variables, not ac,av)

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6
\$\begingroup\$

C++ 0 +/-, 3 tie-breakers

#include <vector>
#include <iostream>

#define WAX (
#define WANE )
#define SPOT .

int main WAX WANE {
    unsigned x;
    unsigned y;
    std::cin >> x >> y;
    std::vector<int> v WAX x WANE;
    std::vector<int> u WAX y WANE;
    for WAX auto n : u WANE {
        v SPOT push_back WAX n WANE;
    }
    std::cout << v SPOT size WAX WANE;
}
\$\endgroup\$
5
\$\begingroup\$

Haskell, 0+2

import Monad
main = join $ fmap print $ fmap length $ fmap f $ fmap lines getContents
f x = join $ flip replicate [] `fmap` fmap read x

This uses no + or - characters, and only two = from the set of tie breaker characters, one of which is mandatory for binding main. The sum is done by concatenating lists of the appropriate lengths.

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4
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EDIT This was posted BEFORE the rules changed to disallow sum...

The R language: No calls to + or -... And 9 tie-breaker characters!

sum(as.numeric(readLines(n=2)))

Example:

> sum(as.numeric(readLines(n=2)))
123
456
[1] 579

The [1] 579 is the answer 579 (the [1] is to keep track of where in the result vector your are since in R all values are vectors - in this case of length 1)

Note that R has + operators just like most languages - it just so happens that it has sum too that sums up a bunch of vectors.

In this case, readLines returns a string vector of length 2. I then coerce it to numeric (doubles) and sum it up...

Just to show some other features of R:

> 11:20 # Generate a sequence
 [1] 11 12 13 14 15 16 17 18 19 20

> sum(1:10, 101:110, pi)
[1] 1113.142
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  • 1
    \$\begingroup\$ +1 For making me have to change the rules to outlaw the sum() function. \$\endgroup\$ – PhiNotPi Dec 1 '11 at 0:48
  • \$\begingroup\$ @PhiNotPi - Changing the rules?! That's cheating! :-) ...But you should probably say "sum-like functions" or I'll just use colSums instead... Maybe also outlaw "negation-like functions" while your at it... \$\endgroup\$ – Tommy Dec 1 '11 at 0:53
  • 2
    \$\begingroup\$ I'm going to take your advice. From what I can tell, everyone (including me) on this site loves to point out loopholes in the rules. \$\endgroup\$ – PhiNotPi Dec 1 '11 at 0:59
4
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The R language

New rules, new answer, same language. No calls to + or -

UPDATE Using scan, it drops to 11 tie-breaker characters (and 27 characters in all).

as.numeric(scan())%*%c(1,1)

Original: 13 tie-breaker characters!

as.numeric(readLines(n=2)) %*% c(1,1)

Example:

> as.numeric(readLines(n=2)) %*% c(1,1)
123
456
     [,1]
[1,]  579

This time the result is achieved by matrix multiplication. The answer is displayed as a 1x1 matrix.

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  • \$\begingroup\$ There's nothing that I can do to outlaw this. Perhaps R is just good at this challenge, since it is mostly mathematics-based. Or maybe this challenge is just easy. \$\endgroup\$ – PhiNotPi Dec 1 '11 at 1:14
  • \$\begingroup\$ +1 Nice. You can make this even shorter with scan() instead of readlines(n=2) \$\endgroup\$ – Andrie Dec 1 '11 at 21:53
  • \$\begingroup\$ @Andrie - yes, but then you rely on the user entering exactly two numbers... Which is OK for this challenge I guess... \$\endgroup\$ – Tommy Dec 1 '11 at 22:07
4
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Haskell, 0 +/-, 6 2 tie-breakers (=)

(does not use the string/list concatenation trick)

main = interact f
f x = show $ log $ product $ map exp $ map read $ lines x
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  • \$\begingroup\$ You can eliminate all the dots at the expense of an extra = by replacing the composition: instead of "f.g.h" write "a where a x = f$g$h x" \$\endgroup\$ – Omar Dec 7 '11 at 0:06
3
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Javascript, 56

p=prompt;alert(Array(~~p()).concat(Array(~~p())).length)

Thanks to @JiminP on the ~~ tip! I'm going for least bytes, so the 1 byte saving on the p=prompt; is still worth it. I understand your argument about tie-breaker chars, but to be honest wouldn't you rather the least bytes :-p

Version, 69

i=parseInt;p=prompt;alert(Array(i(p())).concat(Array(i(p()))).length)

Thanks to some feedback from @Ilmari and @JiminP, I've shaved 13 bytes off my original solution.

Originally, 82

i=parseInt;p=prompt;a=Array(i(p()));a.push.apply(a, Array(i(p())));alert(a.length)
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  • \$\begingroup\$ That space after the comma is completely unnecessary; removing it gets you down to 81. \$\endgroup\$ – Ilmari Karonen Dec 2 '11 at 23:53
  • \$\begingroup\$ Using concat and put calculations in alert is shorter. i=parseInt;p=prompt;alert(Array(i(p())).concat(Array(i(p()))).length) BTW, I didn't know that Array(n) returns an array with length n. The Google Chrome console gave me [] and I thought there was nothing... \$\endgroup\$ – JiminP Dec 3 '11 at 0:29
  • 1
    \$\begingroup\$ Oh, since the important thing is tie-breaker characters, p=prompt is not good. And, parseInt(x) is almost equivalent to ~~x. alert(Array(~~prompt())['concat'](Array(~~prompt()))['length']) (12 tie-breaker chars) PS. I could use this as my entry, but that just gives me feeling of stealing. \$\endgroup\$ – JiminP Dec 4 '11 at 14:14
3
\$\begingroup\$

C

b[500];i;j;main(){scanf("%d%d",&i,&j);printf("%d\n",sprintf(b,"%*s%*s",i,"",j,""));
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3
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APL (no +/-, no tie breakers, 8 or 10 characters)

This entry is similar to the other ones that concatenate sequences generated from the input and find the length... but it's in APL, which can appear confusing even for a small problem like this. I used Dyalog APL, which offers a free educational license.

Code:

⍴⊃⍪⌿⍳¨⎕⎕

From right to left:

  • Each quote-quad ( ) requests input from the user and evaluates it.
  • The each operator ( ¨ ) applies the index generator function ( ) to each of the items in the array to its right.
  • This flattens the resulting array of arrays into one array. The input array is reduced to a flat list by the reduction operator ( / ), which folds the array using the concatenation function ( , ). For the sake of this challenge, the one-dimensional reduction operator ( ) is used, along with the concatenation operator along the first axis ( ).
  • As a result of using the reduction operator, the array is enclosed, which is like placing it in the bag; all we see on the outside is a bag, not its contents. The disclose operator ( ) gives us the contents of the enclosed array (the bag).
  • Finally, the shape-of function ( ) gives us the lengths of the dimensions of an array. In this case, we have a one-dimensional array, so we obtain the number of items in the array, which is our result.

If we need to explicitly output the result, we can do so like this:

⎕←⍴⊃⍪⌿⍳¨⎕⎕

Comparable Python code, with corresponding APL symbols above:

import operator

⎕←     ⍴    ⌿        ⍪                 ¨              ⍳                ⎕        ⎕         ⊃
print (len (reduce (operator.__add__, [map (lambda n: range (1, n+1), [input(), input()])][0])))

I'd like to know if there's a shorter version possible in APL - another, simpler version I came up with that has more tie breakers (although still at 8 characters) is: ⍴(⍳⎕),⍳⎕.

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  • \$\begingroup\$ Does your index generator start at 1 or 0? \$\endgroup\$ – MrZander Aug 17 '12 at 18:33
3
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I didn't see anyone do it the Electrical Engineering way, so here's my take (in ruby):

def please_sum(a, b)
    return (a&b !=0)? please_sum( ((a&b)<<1) , a^b ):a^b
end

It is a little bit ugly, but it gets the job done. The two values are compared by a bitwise AND. If they don't have any bits in common, there is no "carry" into the next binary column, so the addition can be completed by bitwise XORing them. If there is a carry, you have to add the carry to the bitwise XOR. Here's a little ruby script I used to make sure my digital logic was not too rusty:

100.times do
    a=rand 10
    b=rand 10
    c=please_sum(a,b)
    puts "#{a}+#{b}=#{c}"
    end

Cheers!

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3
\$\begingroup\$

Seed, 3904 3846 11 bytes, 0 +/-, 10 tie breakers

4 141745954
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2
\$\begingroup\$

Shell, 52

read a
read b
(seq 1 $a;seq 1 $b)|wc|awk '{print$1}'

This is basically the same answer I gave for another problem.

\$\endgroup\$
  • \$\begingroup\$ Similar: xargs -n1 jot | wc -l which takes the same - reduction awk but I can't see how to avoid it in the xargs \$\endgroup\$ – Ben Jackson Dec 1 '11 at 20:12
  • \$\begingroup\$ +/- and above mentioned tie-breakers are interesting, not characters. \$\endgroup\$ – user unknown May 30 '12 at 3:38
2
\$\begingroup\$

C

a,b;A(int a,int b){return a&b?A(a^b,(a&b)<<1):a^b;}
main(){scanf("%d%d",&a,&b);printf("%d\n",A(a,b));}
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  • \$\begingroup\$ I count 20 tie breakers... Am I right? \$\endgroup\$ – FUZxxl Dec 1 '11 at 11:11
  • 2
    \$\begingroup\$ 22 tie breakers: 0 /*=., 7 (, 7 ), 7 ,, 1 [0-9] \$\endgroup\$ – saeedn Dec 1 '11 at 12:23
2
\$\begingroup\$

C#

It's not the shortest by any stretch:

private static int getIntFromBitArray(BitArray bitArray)
{
    int[] array = new int[1];
    bitArray.CopyTo(array, 0);
    return array[0];
}

private static BitArray getBitArrayFromInt32(Int32 a)
{
    byte[] bytes = BitConverter.GetBytes(a);
    return new BitArray(bytes);
}

static void Main(string[] args)
{
    BitArray first = getBitArrayFromInt32(int.Parse(Console.ReadLine()));
    BitArray second = getBitArrayFromInt32(int.Parse(Console.ReadLine()));
    BitArray result = new BitArray(32);

    bool carry = false;
    for (int i = 0; i < result.Length; i++)
    {
        if (first[i] && second[i] && carry)
        {
            result[i] = true;
        }
        else if (first[i] && second[i])
        {
            result[i] = false;
            carry = true;
        }
        else if (carry && (first[i] || second[i]))
        {
            result[i] = false;
            carry = true;
        }
        else
        {
            result[i] = carry || first[i] || second[i];
            carry = false;
        }
    }
    Console.WriteLine(getIntFromBitArray(result));
}
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  • \$\begingroup\$ That's exhausting, Matthew. \$\endgroup\$ – Cary Swoveland Sep 25 '13 at 2:44
2
\$\begingroup\$

J, 15 7 chars, 1 tie breaker, incomplete program

This is my J attempt. It is not a full program, because I have not yet figured out how to write one. Just put that line in a script to get the function p that can be used for adding an arbitrary amount of numbers. It is a monad and takes a list of numbers to add (such as p 1 2 3 4):

p=:#@#~

The idea is very simple. The function is written in tacit aka pointless style. Here is a pointed definition:

p=:3 :'##~y'

Read from right to left. In the tacit version, @ composes the parts of the function. (like a ∘ in mathematics [(f∘g)(x) = f(g(x)])

  • y is the parameter of p.
  • ~ makes a verb reflexive. For some verb m, m~ a is equal to a m a.
  • # (copy, a#b): Each element in a is replicated i times, where i is the element at the same index as the current element of a of b. Thus, #~ replicates an item n n times.
  • # (count, #b): Counts the number of elements in b.

Conclusion: J is awsome and less readable than Perl (that makes it even more awsome)

Edits

  • 15 -> 7 using # instead of i.. Yeah! Less chars than golfscript.

More of a program

This one queries for input, but it still isn't a full program: (13 chars, 3 breakers)

##~".1!:1<#a:
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  • \$\begingroup\$ J is definitely awesome, but believe me, you can't just leave out the parsing part of the problem when you solve challenges with it ;-) \$\endgroup\$ – J B Dec 1 '11 at 15:35
  • \$\begingroup\$ @J B Well, you can use the builtin function toJ, but I keep getting domain errors. \$\endgroup\$ – FUZxxl Dec 1 '11 at 16:54
2
\$\begingroup\$

Javascript (17 tie-breaker characters)

eval('걢갽거걲걯걭거건갨걡갽거걲걯걭거건갨걹갽걦걵걮걣건걩걯걮갨걡갩걻걣갽걮걥걷갠걕걩걮건갸걁걲걲걡걹갨걡갩갻걦걯걲갨걩갠걩걮갠걣갩걩걦갨걩갽갽걾걾걩갩걸갮거걵걳걨갨갱갩걽갬걸갽걛걝갩갩갻걹갨걡갩갻걹갨걢갩갻걡걬걥걲건갨걸갮걬걥걮걧건걨갩갻'['split']('')['map'](function(_){return String['fromCharCode'](_['charCodeAt'](~~[])^0xac00)})['join'](''))

:P ("Obfuscated" to reduce number of tie-breaker characters. Internally, it's b=prompt(a=prompt(y=function(a){c=new Uint8Array(a);for(i in c)if(i==~~i)x.push(1)},x=[]));y(a);y(b);alert(x.length); .)

\$\endgroup\$
2
\$\begingroup\$

C#,

Program works on 1 line; separated on multiple lines to avoid horizontal scrolling.

using C=System.Console;
class A{
static void Main(){
int a,b,x,y;
a=int.Parse(C.ReadLine());
b=int.Parse(C.ReadLine());
do{x=a&b;y=a^b;a=x<<1;b=y;}while(x>0);
C.WriteLine(y);
}}
\$\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\$

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\$

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\$
0
\$\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\$
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\$

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