Permutations


I saw this post which shows some VFP code to permute a string. For example, there are 6 permutations of “abc”:

            abc, acb, bac, bca, cab, cba

 

There are n! permutations of a string of length n.

 

 

I dug up some old code that did the same thing in fewer lines.

 

 

nn=0

permute(“abcd”,0)

 

PROCEDURE permute(cstr,nLev)

      LOCAL nTrylen,i

      nTrylen= LEN(cstr)-nLev

      IF nTryLen = 0

            nn=nn+1

            ?nn,cstr

      ELSE

            FOR i = 1 to nTrylen

                  IF i>1      && swap nlev+1 and nlev+i chars

                        cstr= LEFT(cstr,nlev) + SUBSTR(cstr,nLev+i,1) +;

                              SUBSTR(cstr,nLev+2, i-2)+SUBSTR(cstr,nlev+1,1)+SUBSTR(cstr,nLev+i+1)

                  ENDIF

                  permute(cstr,nlev+1)

            ENDFOR

      ENDIF

RETURN

 

 

 

 

48662

 

Comments (2)

  1. Robert Hoogstraat says:

    Here is another possible solution with little code:

    * PERMUTATIONS

    *

    * def’n:

    * If M = (n)P(r) = P(n):(r) denotes the number of

    * permutations of (n) distinct things taken (r) at a time,

    * — M = n(n-1)(n-2)…(n-r+1) = n!/(n-r)!

    Permutations( "12345" )

    PROCEDURE Permutations ;

    ( ;

    tvThings AS CHARACTER, ;

    tvStatic AS CHARACTER, ;

    tvN AS INTEGER ;

    ) AS VOID

    LOCAL iX AS INTEGER

    IF VARTYPE(tvStatic)="L"

    * First time procedure called, preset missing parameters.

    CLEAR

    PUBLIC lnPermutation AS INTEGER

    lnPermutation = 0

    tvStatic = ""

    tvN = LEN(tvThings)

    ENDIF

    FOR iX = 1 TO tvN

    IF tvN > 2

    Permutations( RIGHT( tvThings, tvN-1 ), tvStatic+LEFT(tvThings,1), tvN-1 )

    ELSE

    lnPermutation = lnPermutation + 1

    ? lnPermutation, tvStatic+tvThings

    ENDIF

    tvThings = SUBSTR(tvThings,2) + LEFT(tvThings,1)

    ENDFOR

    ENDPROC

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