[gwu78]

Here is a spitbol/snobol4 solution. Assumes the number of items in the set is not greater than the alphabet.

   * routine stolen from gimpel
   * algorithm from peck and schrack 

    define('p(s)t,n,c,k','p_init') :(p_end)
   p_init n = size(s)
    r = array('2:' n, 0)
    &alphabet len(n) . y
    x = array('2:' n, y)
    k = n + 1
   p_0 k = k - 1
    x[k] len(1) . s1 tab(k) . s2 = s2 s1  :s(p_0) 
    define('p(s)i,k') 
    p = s :(return)
   p k = size(s)
   p_1 s = replace(x[k],y,s) :f(p_2)
    r[k] = r[k] + 1
    k = eq(remdr(r[k], k),0) k - 1 :s(p_1)
    p = s :(return)
   p_2 define('p(s)t,n,s1,s2','p_init') :(freturn)
   p_end

   * example: all permutations of string abcdefgh
    s = 'abcdefgh'
   abc output = p(s) :s(abc)f(end)
   end

Here is another solution that only returns the unique permutations. The items of the set must first be sorted or grouped, e.g, a string like "cabcd" could be given as "ccabd", "adbcc", "abccd", etc. Duplicate items must be adjacent.

    define('r(s,ors)c,f,s1,a,d,os') 
    :(r_end)
   r ors rtab(1) len(1) . c :f(freturn)
    s (span(c) | null) . f =
    s arb . s1 len(1) . d c = :f(r_1)
    r = s1 f c d s :(return)
   r_1 ors break(c) . os
    r = r(s,os) f :s(return)f(freturn)
   r_end

    s = 'abcdefgh' 
    output = s
   x01 output = r(output,s) :s(x01)
   end