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MathGroup Archive 2007

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Re: Number of Differing Digits & Another Problem (want to see different ways it can be done)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg76328] Re: [mg76310] Number of Differing Digits & Another Problem (want to see different ways it can be done)
  • From: "Patrick Scheibe" <mai99dnn at studserv.uni-leipzig.de>
  • Date: Sun, 20 May 2007 02:17:29 -0400 (EDT)
  • Organization: Universitaet Leipzig
  • References: <200705190845.EAA21281@smc.vnet.net>

Hi,

I would say that the binary form of 6 is more like

BaseForm[6, 2]=110

and so 5 and 6 differ in two digits.

BitDifference[a_, b_] := Plus @@ IntegerDigits[BitXor[a, b], 2]

Cheers
  Patrick



Am 19.05.2007, 10:45 Uhr, schrieb <VenDiddy at gmail.com>:

> I just purchased a copy of Mathematica and I've been learning it for
> about a week now. You can expect that I will be posting a lot of
> questions. One thing I've noticed is that there are so many different
> ways to do the same thing!
>
> Here is a function I came up with that calculates how many binary
> digits two numbers differ in:
>
> BitDifferences[a_, b_, n_] :=
>  Count[Equal @@@
>    Thread[{IntegerDigits[a, 2, n], IntegerDigits[b, 2, n]}], True]
>
> For example 5 = 101 differs from 6 = 111 by one digit (the middle
> digit).
>
> I want to see how you would do it so I can broaden my Mathematica
> perspective.
>
> Thanks.
>



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