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Re^3: Algorithm for cancelling common factors between two lists of multiplicands

by jimX11 (Friar)
on Aug 09, 2005 at 01:07 UTC ( #482082=note: print w/replies, xml ) Need Help??


in reply to Re^2: Algorithm for cancelling common factors between two lists of multiplicands
in thread Algorithm for cancelling common factors between two lists of multiplicands

The original problem, framed in terms of arrays, is to reduce a fraction.

The equation above is less general.

The less general problem can be restated as reduce a fraction where the numerator and denominator are both the product of factorials.

An example of the original problem is to reduce
(9 * 8 * 8 * 8 * 6 * 4 )/(5 * 3 * 2)

An example of the factorial problem is to reduce
(9! * 8! * 8! * 8! * 6! * 4! )/(5! * 3! * 2!)

Do I understand correctly, are you are interested in the factorial type of problem?

  • Comment on Re^3: Algorithm for cancelling common factors between two lists of multiplicands

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Re^4: Algorithm for cancelling common factors between two lists of multiplicands
by BrowserUk (Pope) on Aug 09, 2005 at 01:35 UTC

    Re-frame the factorial problem thus:

    ( 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 6 * 5 * 4 * 3 * 2 * 4 * 3 * 2 ) / ( 5 * 4 * 3 * 2 * 3 * 2 * 2)

    And it reduces to

    ( 3 * 4 * 7 * 6 * 5 * 4 * 3 * 2 * 4 * 7 * 6 * 5 * 4 * 3 * 2 * 4 * 7 * 6 * 5 * 4 * 3 * 2 * 2 * 7 * 6 * 5 * 4 * 3 * 2 * 3 * 4 * 3 * 2 * 4 * 3 * 2 ) / ( 1 )

    Which eliminates factorials completely (and, in this case the need for division). More importantly and relevantly, it reduces the magnitude of the intermediate terms. In the FET, the scale of this reduction should be considerably more significant than in this example.


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