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Re: A mod2 Machine.

by nonsequitur (Monk)
on Jul 12, 2013 at 10:15 UTC ( #1043936=note: print w/ replies, xml ) Need Help??

in reply to A mod2 Machine.

This may be OT since it doesn't really have anything to do with Perl, but one of your original assumptions is wrong.

In an even base a integer ending in an even digit must be even because the previous digits are multiplied by a power of an even base . Leaving the last digit to decide the parity.

In an odd base the total of the digits must be even for the integer to be even because all of the digits are multiplied by a power of an odd base so that the that the parity of the integer depends on parity of all the digits.

Comment on Re: A mod2 Machine.
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Re^2: A mod2 Machine.
by code-ninja (Scribe) on Jul 12, 2013 at 12:16 UTC
    hmm... hmm... umm... I said this algorithm will work for all bases of the form 2^n and base 10. But according to you, this algorithm will work if the base is of the form 2*n. I'll just clarify right after commenting. Hold on...


    apologies, my assumption was wrong. This algorithm will work, i.e. produce correct output, if the base is of the form 2*n.

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