Ascending matrix

Question

The "ascending matrix" is an infinite matrix of whole numbers (0 included) in which any element is the smallest available element which has not been previously used on the respective row and column:

  | 1 2 3 4 5 6 ...
--+----------------
1 | 0 1 2 3 4 5 ...
2 | 1 0 3 2 5 4 ...
3 | 2 3 0 1 6 7 ...
4 | 3 2 1 0 7 6 ...
5 | 4 5 6 7 0 1 ...
6 | 5 4 7 6 1 0 ...
. | ...............

Your task is to write a program that will output the element found at the row and column specified by the input. (standard input and output)

Test cases:

5 3 -> 6
2 5 -> 5

Code Golf rules apply—the shortest code wins.

P.S. Even if this has an algorithmic nature the code can be very, very concise.

EDIT: I wasn't expecting to see the xor solution so early. I was really hoping to see 10 posts with an algorithmic approach and THEN the xor solution. Now, having in mind that it's not much fun to see how to write xor in different languages I recommend that you also try an algorithmic approach.

So, yes, I think no one can beat the 5 character mark now—therefore I congratulate Ilmari Karonen for the smartest and shortest solution. But there's a new challenge up ahead: write the shortest algorithmic solution.

Answer 1

GolfScript, 5 chars

~(\(^

Indeed, this task is very simple once you recognize the pattern. The only awkward bit is the 1-based indexing — if the input indices were zero-based, this 2-character solution would suffice:

~^

To explain this to readers unfamiliar with GolfScript, the ~ command evals the input, leaving the two numbers on the stack. ^ then XORs the two topmost numbers on the stack together, leaving the result for output. To deal with 1-based input, two more commands are needed: ( decrements the topmost number on the stack by one, while \ swaps the top two items on the stack.

Answer 2

Mathematica 10 44

Edit

My first response was based on a misunderstanding about the nature of the challenge, as noted by Ilmari. Here's another try.

Usage

f[n___, 1, n___] := n - 1;
j_~f~k_ := BitXor[j - 1, k - 1]

Answer 3

K, 31

{0b/:{(x|y)&~x~y}. 0b\:'-1+x,y}

Stole Ilmari Karonen's XOR logic, which I would never have spotted myself.

Answer 4

PHP, 38

Just a simple implementation of Ilmari Karonen's XOR

<?php echo --$_GET['a']^--$_GET['b']?>

Usage:

.../xor.php?a=4&b=7

will print 6

Answer 5

Python 2, 36

I figure since I'm just starting to learn Python that this would be the perfect time to submit my first answer using it (and no one has answered using Python) and perhaps I could receive some feedback.

Thank you @IlmariKaronen for the very cool shortcut.

Thank you @Gareth for the code below.

import sys
print(input()-1^input()-1)

---

Python 3, 56

The original program that I had written.

import sys
x=int(input())
y=int(input())
x-=1
y-=1
print(x^y)

IDEONE with 2 and 5

IDEONE with 3 and 3

Answer 6

MATL, 2 bytes

Z~

Try it online!

MATL post-dates the challenge by several years, but, hey, natural 1-based indexing and a bitwise xor function makes this nice and neat!

Answer 7

Haskell 174

Figured I'd make a solution that did not rely on XOR. Too lazy to golf it properly.

a 0 0=0
a b c
 |m==n=a(b-m)(c-n)
 |m>n=m+a(b-m)c
 |m<n=n+a b(c-n)
 where{g f=until(>f)(*2)1`div`2;m=g b;n=g c;}
main=do
 [x,y]<-fmap(map read.words)getLine
 print$a(x-1)(y-1)

Edit: I realized a day later that this is just calculating XOR. Thus if this counts as an algorithmic solution, so should Ilmari Karonen's.

Answer 8

Perl 5, 12 bytes

say<>-1^<>-1

Try it online!

Answer 9

Javascript 13 bytes

a=>b=>--a^--b

f=a=>b=>--a^--b

result = document.getElementById('result')

<input type="text" onkeyup="result.innerHTML = f(this.value.split(',')[0])(this.value.split(',')[1])" >
<p id="result"></p>

Answer 10

Japt, 5 bytes

UÉ^VÉ

Try it online!