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Tom Ogilvy Tom Ogilvy is offline
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Default how many combinations in a list equal a certain total

Quibbles.

With 479 items, there are 2^479-1 combinations to check.

Yes, don't know what I was thinking - thanks for the correction

8,640,000 secons is 100 days, not years.

Already admitted to that one

Have to think about your maximum combinations assertion - (not questioning
it) - thanks

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Tom Ogilvy

"Harlan Grove" wrote in message
...
"Tom Ogilvy" wrote...
There are 2^480 - 1 unique combinations to check. As a point of
comparison, there are only 8,640,000 seconds in 100 years (slightly

larger
then 2^23) . . .


Quibbles.

With 479 items, there are 2^479-1 combinations to check.

8,640,000 secons is 100 days, not years.

. . . In otherwords, there isn't enough time to do an exhaustive
examination without even trying to look at the results.

You could come up with a list of Unique prices and reduce the number of

rows
you need to examine. (so in your example, any solution that contains a B
could be another solution by replacing B with F)


Add to those heuristics throwing out all individual items with prices over

$100.
Also, for k = INT(100/MIN(PriceList)), k gives the largest cardinality
combinations that need to be checked. That means only 479 choose k

combinations
to check. Still, this type of problem lacks practical means to achieve a
comprehensive solution.

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