1 Part 13. Tubular Rulesets

2 Def. For a given state s with rule set R, triple E is tubular if for every triple T that E depends on, T is a tube of E: E  T  E  T Definition of Tubular Edges P P P S C C  E T Example

3 Tubular States, Tubular Rulesets Def. State s is tubular if all its triples are tubular. Def. A ruleset is tubular if all its legal states are tubular. Most example SoP rules given above are tubular. There is no known algorithm to determine if rulesets are tubular.

4 Constructive and Phantomic Tubular Systems Constructive ruleset: a)M  (P M C) b)M  (S) Phantomic ruleset: a)M 1  (P M 2 C) b)M 2  (M 3 ) c)M 3  (M 4 ) d)M 4  (M 2 ) fMgf g d e fPd dMe eCg dSe All states are constructive. Example state: fM 1 g fPd dM 2 e eCg dM 4 e dM 3 e Example tubular state (phantom):

5 Another Example of a Phantom in a Tubular System Ruleset: M 1  (P M 2 C) M 2  (M 3 ) M 3  (L 1 M 4 ) M 4  (M 2 R 1 ) L 1  (L 2 ) L 2  (L 3 L 4 ) L 3  (ID L 4 ) L 4  (L 1 ) R 1  (R 2 ) R 2  (R 2 ) fM 1 g fPd dM 2 e eCg dM 4 e dM 3 e dL 1 d dL 4 d dL 3 d dL 2 d eR 1 e eR 2 e dIDd * Legal sub states, which are phantoms. ** *

6 Part 14. Identic Rulesets and Identic States

7 Six productions of SoP ruleset R: a)t1  (t3) b)t2  (t3) c)t2  ( t4) d)t3  (t4 t5) e)t4  (t5 ID) f)t5  (t3) Ruleset R is identic if it has a sub ruleset r such that v r = v r.  r – {ID} Example. {t3, t4, t5} = {t3, t4, t5, ID} – {ID} Surprise: Being identic does not imply phantoms! But it does imply hidden phantoms. Identic Rulesets: Introduction Notation needs work, as here “ID” is used as triples with ID edges?? t1 t2 t3 t4 t5 ID vrvr V r.  r

8 Def. SoP ruleset R is said to be identic if there is sub ruleset r, r  R, such that v r = v r.  r - {ID} where v r is r’s set of variables. That is, R is identic if the set of variables depended upon according to r by v r variables are exactly v r when ignoring ID constants. Identic Rulesets: Formal Definition Should give algorithm to test to decide if R is identic?? Here, {ID} is a set containing the name “ID” of self-loops. t1 t2 v1 v2 v3 ID vrvr v r.  r Example

9 Suppose state s has these dependencies among triples V1 to V5. a)V1  (V3) b)V2  (V3) c)V2  (V4) d)V3  (V4, V5) e)V4  (V5, ID) f)V5  (V3) State s is identic because there is a non-empty subset of triples t = {V3, V4, V5} with dependencies such that t = t.  r – {ID} where t.  r = {V3, V4, V5, ID} Identic States: Example V1 V2 V3 V4 V5 ID t t.  r Here {ID} is the set of triples that are self loops.

10 Def. Consider SoP ruleset R with sub rle set r, r  R, with legal state s with non-empty sub state t, t  s. If t = t.  r - {ID} we say state s is identic. So, state s is identic if the set of triples depended upon according to r by triples in t are exactly t when ignoring ID constants. Identic States: Formal Definition Here {ID} is the set of triples that are self loops. T1 T2 V1 V2 V3 ID t s Example

11 Identic Rules iff Identic Legal States Theorem: An SoP ruleset is identic iff it has a legal state that is identic. Proof: Ruleset R is identic  Legal identic state s consisting of ID triples of form (a v i a) for each variable in subset rule r. Legal state s is identic  Productions v  (R 1, R 2, …, R n ) for each dependency in s of form (x 0 v x n )  (x 0 R 1 x 1, x 1 R 2 x 2, …, x n-1 R n x n ) v1 v2 v3 v4 v5 Somewhere did I assume that identic states are all legal??

12 Identic Rules Sets Permit Trivial Looping States Theorem. If a ruleset is identic, it has a legal state all of whose edges are ID loops. Example. T  T U  U o T T U a Legal state which is identic Should say: if s is identic phantom then R is identic ruleset??

13 Part 15. Tubular Phantoms are Identic

14 Recap: Tubular Rules Sets Can Have Identic Phantoms Ideally, tubular rules sets should have no phantoms However, with cyclic patterns of dependencies, tubular states and rulesets can permit phantoms. We want to be able detect when these phantoms exist in a tubular ruleset. We will prove what seems obvious, namely that tubular states can be phantoms only if they are identic.

15 Tubular Phantoms are Identic Theorem. Given a tubular state s, s is a phantom  s is identic Proof. … given below … Corollary. Given a tubular state s, s is non-identic  s is constructive Corollary. Tubular non-identic rulesets are constructive. There is an obvious algorithm to test if a ruleset is identic, but no known algorithm to test if a ruleset is tubular.

16 Any phantom state s is recursive, and has a 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 E i and F i 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. Part 1 Proof: Tubular Phantoms Are Identic. Phantoms Have Cycle of Dependency E N-1 T N F N-1 E N T 0 F N E 0 T 1 F 0 …   

17 Since state s is a phantom, there exists a non-empty sequence of triples (T 0 T 1 … T N ) such that T 0  T 1  … T N  T 0 Recall that if tubular triple V depends on triple W (if V  W) then Len(V)  Len(W) so Len(T 0 )  Len(T 1 )  …  Len(T N )  Len(T 0 ) Hence Len(T 0 ) = Len(T 1 ) = … = Len(T N ) In other words, since s is tubular: All triples, T 0 to T N, have the same length. Part 2 Proof: Tubular Phantoms Are Identic. Triples in basic recursion have same lengths T0T0 T1T1 T2T2 V1V1 ID Example

18 We have this pattern of dependency among tuples T i  (E i T i+1 F i ) which is short for T i  (E i,1 E i,2 … T i+1 F i,1 F i,2 … ) Each triple T i depends on a sequence of triples consisting of (1) the triples in E i then (2) triple T i+1 and finally (3) the triples in F i. Recall that if tubular triple V depends on tuple sequence (W 1 W 2... W k ) then Len(V) = Len(W 1 ) + Len(W 2 ) + … + Len(W k ), so Len(T i ) = Len(E i,1 )+Len(E i,2 )+ … Len(T i+1 )+ Len(F i,1 )+ Len(F i,2 )+ … Since we have already determined that Len(T i ) = Len(T i+1 ), it follows that for all i, j: Len(E i,j ) = 0 and Len(F i,j ) = 0 Part 3 Proof: Tubular Phantoms Are Identic. Triples depended on by base recursion are ID’s E N-1 T N F N-1 E N T 0 F N E 0 T 1 F 0 …   

19 If tubular triple V has length zero and V depends on triple W directly or directly, then the length of W must also be zero: Len(V) = 0  V  + W  Len(W) = 0 Since for all i and j, Len(E i,j ) = Len(F i,j ) = 0, it follows that: All triples that E i,j or F i,j depend on transitively have length zero and hence are ID triples. Part 4 Proof: Tubular Phantoms Are Identic. Triples depended on by Ei and Fi are IDs.

20 In the sequence (T 0 T 1 … T N ) each T i depends on T i+1 (where T N+1 is T 0 ): T i  R (Ei T i+1 Fi) with corresponding production from R: t i  R (e i t i+1 f i ) Def. s 0 = {T 0, T 1, … T N } r 0 = Set of productions for T i from R: t i  R (e i t i+1 f i ) Def. s i+1 = s i.  ri (Compute s i ’s targets) r i+1 = union of r i and set of productions corresponding to dependencies for T  R  for each T in s i+1 - s i Part 5 Proof: Tubular Phantoms Are Identic. Construction of Sub-States s i and Sub-Rulesets r i Since T is legal there must exist tuple sequence  in s such that T  R  Example Note: s i  s i+1 T0T0 T1T1 T2T2 V1V1 ID V2V2 V1V1 V1V1 S0S0 S1S1 S2S2 S3S3

21 We previously showed that all triples in E i and F i and all triples transitively depended upon by them are limited to be ID triples. We observe that each T i can depend (according to r 0 ) only on T i+1 or on triples in E i and F i. Therefore: Any constant triple depended upon transitively by any T i is necessarily an ID constant. Part 6 Proof: Tubular Phantoms Are Identic. Triples depended on recursively by T i are IDs. Example T0T0 T1T1 T2T2 V1V1 ID V2V2 V1V1 V1V1 S0S0 S1S1 S2S2 S3S3

22 Part 7 Proof: Tubular Phantoms Are Identic. Induction Hypothesis We define a hypothesis H i, i  0, as follows: H i = def s i  s i.  ri We will show that H i is true for all i  0. We start by proving that H 0 is true. By definition s 0 = {T 0, T 1, …, T N } We conclude that s 0  s 0.  ri because every triple in s 0 is depended upon by a triple in s 0. For example T 1 depends on T 2. Hence H 0 is true. T0T0 T1T1 T2T2 V1V1 ID Example s0s0 s 0.  ri

23 Part 8 Proof: Tubular Phantoms Are Identic. Inductive Proof of Hypothesis H i We will prove that, for all i  0, H i is true, i.e., s i  s i.  ri We have already shown that H 0 is true. We will use induction to prove that H i is true for all i > 0, by assuming H i is true and showing that consequently H i+1 must also be true. Assuming H i is true, then every triple in s i is depended upon according to r i by at least one other triple in s i. Now consider any triple that s i depended upon according to r i by a triple in s i+1, but is not in s i. Any such triple is clearly depended upon by a triple in s i. Since s i+1 consists of such triples along with triples already in s i, it follows that every triple in s i+1 is depended upon according to r i by at least one triple in s i+1. Hence: For all i  0, H i is true.

24 Part 9 Proof: Tubular Phantoms Are Identic. Conclusion of proof Both series s 0, s 1, … and r 0, r 1, … are monotonically increasing in size. Both are limited in size, s i by s and r i by R. It follows there is a limiting value, call it L, after which all states, s L, s i+1, … and all rulesets r i, r i+1, … are identical. Hence, for i  L, there remain no triples outside of s i that are depended upon according to r i by triples in s i, but are not members of s i. Hence, s L is the same s L.  r except for constant triples in s L.  r so: s L = s L.  r - {CONST} We claim that for i  L s i = s i.  r - {ID} This must be true because we have previously established that every constant triple depended upon directly or indirectly by any E i, F i or T i is an ID constant. Since s i is a subset of state s, it follows that phantom tubular state s is identic. QED