Trivia

Subject:
Re: Friday Puzzler -- Cable

Alex Y
Obviously, my memory is shorter than 11 years :-)

I was thinking of Bill's first solution, which has the advantage that it works for any number of wires (even or odd), tying the initial group of wires into a bundle of n, where n*(n+1)/2 >= total number of wires. Each other bundle is one smaller than the previous, until the last bundle is whatever is left over when you can't go one smaller.

Guess I better find some new material!

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