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Top Forums Shell Programming and Scripting All possible combinations problem Post 302961527 by wisecracker on Monday 30th of November 2015 05:13:50 PM
Old 11-30-2015
Hi LMHmedchem...

You do realise that this 23 bit word would give 8388607 combinations if all bits were set to 1?!

Is this a possibility along with just any single random bit which would not have a subset at all...

Or am I missing something?
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BITMAP_REMAP(9) 					  Basic Kernel Library Functions					   BITMAP_REMAP(9)

NAME
bitmap_remap - Apply map defined by a pair of bitmaps to another bitmap SYNOPSIS
void bitmap_remap(unsigned long * dst, const unsigned long * src, const unsigned long * old, const unsigned long * new, int bits); ARGUMENTS
dst remapped result src subset to be remapped old defines domain of map new defines range of map bits number of bits in each of these bitmaps DESCRIPTION
Let old and new define a mapping of bit positions, such that whatever position is held by the n-th set bit in old is mapped to the n-th set bit in new. In the more general case, allowing for the possibility that the weight 'w' of new is less than the weight of old, map the position of the n-th set bit in old to the position of the m-th set bit in new, where m == n % w. If either of the old and new bitmaps are empty, or if src and dst point to the same location, then this routine copies src to dst. The positions of unset bits in old are mapped to themselves (the identify map). Apply the above specified mapping to src, placing the result in dst, clearing any bits previously set in dst. For example, lets say that old has bits 4 through 7 set, and new has bits 12 through 15 set. This defines the mapping of bit position 4 to 12, 5 to 13, 6 to 14 and 7 to 15, and of all other bit positions unchanged. So if say src comes into this routine with bits 1, 5 and 7 set, then dst should leave with bits 1, 13 and 15 set. COPYRIGHT
Kernel Hackers Manual 3.10 June 2014 BITMAP_REMAP(9)
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