Solved on March 5, 2001.

Hello again followers of the Challenge

Here I send the solution Judge of Challenge 0014:

 a =  ( a^=b, b = (a^b)*( (a^b)<b ) + b*( (a^b)>=b ), a^b ); //solution JUDGE

Where the only thing that stands out, I think, is that instead of using the bucket method, as all the participants did, the XOR method (exclusive OR) was used to exchange the variables a and b, without an auxiliary variable.

a = a^b;
b = a^b;
a = a^b;

But this solution is adaptable to the other method, this is as follows:

 a =  ( a+=b, b = (a-b)*( (a-b)<b ) + b*( (a-b)>=b ), a-b ); // alternative solution

Regards.

PS: I am sending the complete program of the solution as an attachment.

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