Well, it's simply true - we need first to look at the lowest bit of a and b. If they are both zero, then, obviously, a^b == (a^b) + (1^1) == (a+1)^(b+1).This occurs 25% of the time.

The next (interesting) case would be that both lowest bits of a and b are 1, and the second lowest bit of both, a and b, is 0. Then again, a^b == (a+1)^(b+1).

I'm not sure, if this method of counting will arrive at 33% overall, as there are some other possibilities (like a=5, b=6) which I didn't count, but 25% is close enough to 33% to mark this method of generating random numbers as dangerous.

perl -MHTTP::Daemon -MHTTP::Response -MLWP::Simple -e ' ;    # The  
$d = new HTTP::Daemon and fork and getprint $d->url and exit;#spider
($c = $d->accept())->get_request(); $c->send_response( new   #in the
HTTP::Response(200,$_,$_,qq(Just another Perl hacker\n))); ' #  web

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In reply to Re: This may be more of a math question by Corion
in thread This may be more of a math question by Fingo

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