Author: Leen Ammeraal
Date: 22:54:45 08/28/00
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On August 28, 2000 at 22:36:26, Larry Griffiths wrote: ... >Excuse me Bob, but I have not done powers in quite a while and I was thinking >2^depth amounted to shifting the binary value 2 left depth positions. Maybe I ... >Larry. To compute 2^depth, shift 1, not 2, 'depth' positions left. (Incidentally, as Bob pointed out, you had better use depth^2, which is depth * depth, for your purpose.) Leen Ammeraal
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