Dear Marcel,
                             maybe that last was too terse.
     The point is, verifying the rnd function returns all non-zero numbers,
is of linear order ( or less).  I take it you checked the period of the
(rnd)->(rnd)' function and found it to be, 2^32-1, for a non-zero value.
This and 0->0 are enough to prove that (rnd)->(rnd)' is 1-to-1.
 (rnd)->(rnd)' is of the form:
          (rnd)' = G( F( (rnd) ) )
.  The form of the take-off function, rnd, is:
          rnd = F( (rnd) )
.  As long as F() is 1-to-1, all values of (rnd) will also eventually be
 returned by  rnd.  So, you need only check that the period of F()-iterated
is 2^32-1 for a nonzero argument.
                       James Gere

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