Thue sequences, after Axel Thue, are sequences over an alphabet of three {1, 2, 3} such that no subsequence is immediately repeated. For example, 1213121 is a Thue sequence that cannot be extended 12131211  ×,  12131212  ×,  12131213  ×. However, there are Thue sequences of any length. (There are no solutions longer than 3 for binary sequences, and finding solutions is easier for alphabets larger than three.)

The ''circular program'' below builds a tree of Thue sequences. Note that within function 'build', the tree data-structure 'T' and functions 'toplevel' and 'f' are mutually recursive. The program relies on the fact that a partial soln 'abcde' can be extended with 'f' iff its shadow, 'bcdef', is already in tree at the previous level; this avoids repeating many tests on constraints that have already been treated at a higher level. The subtree for 'abcde' is the subtree of 'bcde' less any 'a' nodes. The tree of all solutions is notionally infinite, but the program prints one branch to a finite depth so only a finite part of the tree is evaluated thanks to lazy evaluation.

See L. Allison. Applications of Recursively Defined Data Structures. Australian Computer Journal, 25(1), pp.14-20, Feb 1993.

let rec
 even = lambda n. (n/2)*2=n,

 repeated = lambda len. lambda s.
   { test:  ? s = s2;s2 ? }
   let rec
     r = lambda half. lambda s2.
       if half=len/2 then
         let rec
           r2 = lambda s1. lambda s2.
             if null s2 then true
             else if hd s1=hd s2 then
               r2  tl s1  tl s2
             else false
         in r2 s  tl s2
       else r (half+1) tl s2
   in r 1 s,

 paths = lambda depth. lambda T.
   {paths from root of T to a given depth}
   let rec
     across = lambda d. lambda Ancestors.
              lambda T. lambda Rest.
       if null T then Rest
       else down d (hd T :: Ancestors) (hd tl T)
                   (across d Ancestors tl tl T Rest),
     down = lambda d. lambda Ancestors.
            lambda T. lambda Rest.
       if d >= depth then Ancestors :: Rest
       else {d < depth} across (d+1) Ancestors T Rest
   in across 1 nil T nil

in let                        { build does the work }
 build = lambda N.
   let rec
     T = toplevel 1,

     toplevel = lambda M.
       if M > N then nil
       else M :: ((f 2 (M::nil) T) ::
                  (toplevel (M+1))),

     f = lambda len. lambda seq. lambda Shadow.
       if null Shadow then nil
       else
       let
         others = f  len seq  tl tl Shadow,
         seq2 = hd Shadow :: seq
       in if even len and repeated len seq2 then others
          else (hd Shadow) ::
               ((f (len+1) seq2  hd tl Shadow)::others)
   in T

in hd( paths 20 ( build 3 ))  {print one}

{\fB Tree of Thue Sequences. \fP}

{see: L. Allison,
      Applications of Recursively Defined Data Structures.
      Australian Computer Journal 25(1) 14-20 Feb 1993
----------------------------------------------------------
tree == list of (elt x subtree)
NB. `T' is a self-referential data structure
    i.e. a circular program.

123  note that a partial soln `abcde'
 32  can be extended with f iff
213  its shadow, `bcdef', is already
 31  in tree at previous level.  The subtree
312  for abcde is the subtree of bcde
 21  less any `a' nodes.
}

let rec even = lambda n. (n/2)*2=n, repeated = lambda len. lambda s. { test: ? s = s2;s2 ? } let rec r = lambda half. lambda s2. if half=len/2 then let rec r2 = lambda s1. lambda s2. if null s2 then true else if hd s1=hd s2 then r2 tl s1 tl s2 else false in r2 s tl s2 else r (half+1) tl s2 in r 1 s, paths = lambda depth. lambda T. {paths from root of T to a given depth} let rec across = lambda d. lambda Ancestors. lambda T. lambda Rest. if null T then Rest else down d (hd T :: Ancestors) (hd tl T) (across d Ancestors tl tl T Rest), down = lambda d. lambda Ancestors. lambda T. lambda Rest. if d >= depth then Ancestors :: Rest else {d < depth} across (d+1) Ancestors T Rest in across 1 nil T nil in let { build does the work } build = lambda N. let rec T = toplevel 1, toplevel = lambda M. if M > N then nil else M :: ((f 2 (M::nil) T) :: (toplevel (M+1))), f = lambda len. lambda seq. lambda Shadow. if null Shadow then nil else let others = f len seq tl tl Shadow, seq2 = hd Shadow :: seq in if even len and repeated len seq2 then others else (hd Shadow) :: ((f (len+1) seq2 hd tl Shadow)::others) in T in hd( paths 20 ( build 3 )) {print one} {\fB Tree of Thue Sequences. \fP} {see: L. Allison, Applications of Recursively Defined Data Structures. Australian Computer Journal 25(1) 14-20 Feb 1993 ---------------------------------------------------------- tree == list of (elt x subtree) NB. `T' is a self-referential data structure i.e. a circular program. 123 note that a partial soln `abcde' 32 can be extended with f iff 213 its shadow, `bcdef', is already 31 in tree at previous level. The subtree 312 for abcde is the subtree of bcde 21 less any `a' nodes. }

output area

Also see:

• L. Allison, Circular programs and self-referential structures, Software Practice and Experience 19(2) pp99-109, Feb 1989.