1. 1 Part 11. Extending Concept of Dependency, as Defined by Permission

2. 2 Review. In a ruleset R, if variable (type) t has a product  1 o  2 … o  n as in t  …   1 o  2 … o  n  … then we say type t depends on sequence according to R We write this as a production t  R ( 1  2 …  n ) or as t  R  We also say t depends on each symbol of the sequence t  R  i Review: A type (variable) depends on a sequence of types    11 22 ii nn t …… The R “subscript” on  and is omitted when obvious from context

3. 3 Suppose SoP ruleset R has production t  R (  1  2 …  n ) Suppose state s contains triple T = (x 0 t x n ) as well as this path (sequence of triples)  = ( 1  2 …  n ) = (x 0  1 x 1, x 1  2 x 2, …, x n-1  n x n ) then we say T depends on  written as T  R  We also say triple T depends on each triple  i written as T  R  i Extension: A triple (tuple) depends on a sequence of triples    11 22 ii nn T ……

4. 4 Suppose SoP ruleset R has this production R i  R (V 1 V 2 … V m ) Consider these two type sequences:  1 = (R 1 R 2 … R i … R n )  2 = (R 1 R 2 … V 1 V 2 … V m … R n ) In  1, R i is replaced by (V1 V2 … Vm) to form  2 We say sequence  1 depends on sequence  2, written as:  1  R  2 We can apply transitive closure to :  1  R +  2 means there exist a 1, a 2, …, a n such that  1  R a 1  R a 2 …  R a n  R  2 Extend type dependency so: a type sequence depends on a type sequence R2R2 V1V1 V2V2 VmVm RnRn … R1R1 … … R2R2 RiRi RnRn … R1R1 … 

5. 5 Suppose SoP ruleset has production R i  R (V 1 V 2 … V m ) Suppose state s contains these triple sequences (paths):  1 = (x 0 R 1 x 1 R 2 x 2 … x i-1 R i X i … x n-1 R n x n )  2 = (x 0 R 1 x 1 R 2 x 2 … x i-1 V 1 y 1 V 2 y 1 … y m-1 V m X i … x n-1 R n x n ) then we say  1 depends on  2, written as:  1  R  2 We can apply transitive closure to :  1  R +  2 means there exist A 1, A 2, …, A n such that  1  R A 1  R A 2 …  R A n  R  2 Extend tuple dependency so: a tuple sequence depends on a tuple sequence Note analogy to CFG productions and sentences

6. 6 Tuple Length Equals Length of Its Lifted Path of Tubes Theorem. In a tubular state, if triple T depends on a sequence of triples ( 1  2 …  n ), that is, if T  ( 1  2 …  n ) (Note that each  i is a tube of T) then Len(T) = Len( 1 ) + Len( 2 ) + … + Len( n ) … T 11 22 nn Len(x t y) = def min number of P and C edges from x to y

7. 7 Useless Productions Def. Suppose SoP ruleset has production R  (V 1 V 2 … V n ) If for every triple T = (x 0 R x n ) in every legal state there is no triple sequence  = (x 0 V 1 x 1, x 1 V 2 x 2, …, x n-1 V n x n ) such that T   then we say the production is useless. Omit this definition. Instead “useless” applies to constructive states??

8. 8 Legality as Dependency Theorem. State s is legal iff for every triple T in s there exists triple sequence  such that T   Proof. To prove s is legal, we need to show L f (s) is true. We will re-state this successively: s  f(s) for every T in s, there is W in f(s) such that V=W for every T in s, there is  such that W is the composition of the triples in  for every T in s, there is  such that T  . QED Recall similar result: s is legal iff every triple of s is legal

9. 9 Def. For a given SoP ruleset, type (variable) t is type recursive, or simply, recursive, if t transitively depends on itself, that is if t   t Example. t  t  v o P v  t Types t and v are both recursive Recursive Types

10. 10 Def. For a given SoP ruleset, type (variable) t is (type) multi-recursive, if t transitively depends on a sequence that includes more than one instance of t: t   ( … t … t …) If type t is recursive, but not multi-recursive, we say it is linearly recursive. Example. t  (t P v P) v  (t P) So, types t and v are both recursive, e.g., t   (t P t P P) Multi-Recursive Types

11. 11 Def. For a given SoP ruleset and a given state s, triple T is tuple recursive, or triple recursive, or simply recursive, if T   T Example. t  t v  v o t In state s, triple T = (a v a) is recursive because (a v a)   a v a) Triple V = (a v a) is also recursive. Recursive Triples t v a Legal state s a v a a t a   

12. 12 Part 12. Phantoms and Recursive Dependency

13. 13 Phantoms Have Recursive Dependency Theorem. Every phantom state contains a recursive triple T: T  + T Proof. Assume the contrary (assume a phantom p has no -recursion). Then p can be de-constructed (and then re-constructed) one triple at a time. So p is constructive, which is a contradiction.

14. 14 Properties of Tuple Recursion Where there is tuple recursion, there exists a cycle of N triples: T 0 = (a 0 t 0 b 0 ), T 1 = (a 1 t 1 b 1 ), … T N = (a N t N b N ), that depend successively on the next triple in the cycle T 0  T 1  …  T N  T 0

15. 15 Properties of Tuple Recursion When state s is tuple recursive, there exists a triple T 0 = (a 0 t 0 b 0 ) (node a 0 is connected by type t 0 to node b 0 ) and path of triples  such that T 0 transitively depends on path  T 0  +  where T 0 is an element of , that is, T 0  + ( … T 0 … )

16. 16 Properties of Tuple Recursion When state is recursive we have this cycle of dependencies: T 0  E 0 T 1 F 0  T 1  E 1 T 2 F 1  … T i  E i T i+1 F i  … T N-1  E N-1 T N F N-1  T N  E N T 0 F N  where we are using the convention that Ei and Fi are sequences (really, paths) of triples. Note that if (a V b) and (c W) d are successive triples in a path of triples, then necessarily nodes b and c are identical: b = c. E N-1 T N F N-1 E N T 0 F N E 0 T 1 F 0 …   