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MathGroup Archive 2004

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Re: number of switches

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47523] Re: number of switches
  • From: "Peltio" <peltio at twilight.zone>
  • Date: Thu, 15 Apr 2004 03:39:35 -0400 (EDT)
  • References: <c5j6vd$qqo$1@smc.vnet.net>
  • Reply-to: "Peltio" <peltioNOSP at Mdespammed.com.invalid>
  • Sender: owner-wri-mathgroup at wolfram.com

"fake" wrote

>Consider the lists {1,1,0,1} and {1,1,0,0},{1,0,1,0,1}.
>The first sequence (1101) switches 2 times (#2digit~#3digit,
>#3digit~#4digit}, the second (1100) 1 time, the third 10101 4 times.

>I have the following problem.
>Consider a list of binary digits. Which is the easiest way to count the
>number of switches of the list (using Mathematica commands)?

Since we're working wit 1's and 0's the difference between two values can
only be 0 (if they are the same, no switch),  +1 (0->1) or -1 (1->0).
Subtracting the list of bits from a rotated version (the last element is
adjusted in order not to give a false switch), we get a list of such
transition values. Applying Abs will turn every switch into a 1. To count
them all,
all we need to to is a simple sum:

    edgeNumber[lst_] := Plus @@ Abs[Append[Rest[lst], Last[lst]] - lst]

Some examples:

    edgeNumber[{1,1,0,1}]
        2
    edgeNumber[{1,1,0,0]
        1
    edgeNumber[{1,0,1,0,1}]
        4
    edgeNumber[{1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1}]
        8

But if you have access to a version of Mathematica that has Split built-in
(from 3.0 onward) the simplest way should be this:

    edgeNumber[lst_]=Length[Split[lst]]-1

It works for multilevel signals as well.

cheers,
Peltio
invalid address in reply-to
crafty demunging required to mail me




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