Part 13. Tubular Rulesets

1. Part 13.Tubular Rulesets

2. Definition of Tubular Edges 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 d T  E ³ T S C Example P T C P  d P E

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 ruleset: M D (P M C) M D (S) Phantomic ruleset: M1D (P M2 C) M2D (M3) M3D (M4) M4D (M2) Constructive and Phantomic Tubular Systems All states are constructive. Example state: Example tubular state (phantom): dSe dM4e dM3e d e dMe dM2e fPd eCg fPd eCg f fMg g fM1g

5. Another Example of a Phantom in a Tubular System Ruleset: • M1D (P M2 C) • M2D (M3) • M3D (L1 M4) • M4D (M2 R1) • L1D (L2) • L2D (L3 L4) • L3D (ID L4) • L4D (L1) • R1D (R2) • R2D (R2) * * * dIDd dL3d eR2e dL4d dL2d dL1d eR1e dM4e dM3e dM2e fPd eCg fM1g * Legal sub states, which are phantoms.

6. Part 14.Identic Rulesets and Identic States

7. Identic Rulesets: Introduction Vr . dr vr Six productions of SoP ruleset R: • t1 D (t3) • t2 D (t3) • t2 D(t4) • t3 D (t4 t5) • t4 D (t5 ID) • t5 D (t3) Ruleset R is identic if it has a sub ruleset r such that vr=vr . dr – {ID} Example. {t3, t4, t5} = {t3, t4, t5, ID} – {ID} Surprise: Being identic does not imply phantoms! But it does imply hidden phantoms. t5 ID t4 t3 t2 t1 Notation needs work, as here “ID” is used as triples with ID edges??

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

9. Identic States: Example Suppose state s has these dependencies among triples V1 to V5. • V1  (V3) • V2  (V3) • V2  (V4) • V3  (V4, V5) • V4  (V5, ID) • 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 . dr – {ID} where t . dr = {V3, V4, V5, ID} t . dr t V5 ID V4 V3 V2 V1 Here {ID} is the set of triples that are self loops.

10. Identic States: Formal Definition 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 . dr - {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. Example s t V3 ID V2 V1 T2 T1 Here {ID} is the set of triples that are self loops.

11. Identic Rules iff Identic Legal States Theorem: An SoP ruleset is identic iff ithas a legal state that is identic. Proof: Ruleset R isidentic Legal identic state sconsisting of ID triples of form (a vi a) for each variable in subset rule r. Legal state s is identic Productions v D (R1, R2, …, Rn) for each dependency in s of form (x0 v xn)  (x0 R1 x1, x1 R2 x2, …, xn-1 Rn xn ) v5 v4 v3 v2 v1 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 Legal state which is identic T a U 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. EN T0 FN D E0 T1 F0 D D … EN-1 TN FN-1 Part 1 Proof: Tubular Phantoms Are Identic.Phantoms Have Cycle of Dependency Any phantom state s is recursive, and has a cycle of dependencies: T0D (E0 T1 F0) T1D (E1 T2 F1) … TiD (Ei Ti+1 Fi) … TN-1 D (EN-1 TN FN-1) TND (EN T0 FN) 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.

17. Example T2 ID V1 T1 T0 Part 2 Proof: Tubular Phantoms Are Identic.Triples in basic recursion have same lengths Since state s is a phantom, there exists a non-empty sequence of triples (T0 T1 … TN) such that T0d T1d … TNd T0 Recall that if tubular triple V depends on triple W (if V d W) then Len(V) £ Len(W) so Len(T0) £ Len(T1) £ … £ Len(TN) £ Len(T0) Hence Len(T0) = Len(T1) = … = Len(TN) In other words, since s is tubular: All triples, T0 to TN, have the same length.

18. EN T0 FN D E0 T1 F0 D D … EN-1 TN FN-1 Part 3 Proof: Tubular Phantoms Are Identic.Triples depended on by base recursion are ID’s We have this pattern of dependency among tuples TiD (Ei Ti+1 Fi) which is short for TiD (Ei,1 Ei,2 … Ti+1 Fi,1 Fi,2… ) Each triple Ti depends on a sequence of triples consisting of (1) the triples in Ei then (2) triple Ti+1 and finally (3) the triples in Fi. Recall that if tubular triple V depends on tuple sequence (W1 W2 ... Wk) then Len(V) = Len(W1) + Len(W2) + … + Len(Wk), so Len(Ti) = Len(Ei,1)+Len(Ei,2)+… Len(Ti+1)+ Len(Fi,1)+Len(Fi,2)+… Since we have already determined that Len(Ti) = Len(Ti+1), it follows that for all i, j: Len(Ei,j) = 0 and Len(Fi,j) = 0

19. Part 4 Proof: Tubular Phantoms Are Identic.Triples depended on by Ei and Fi are IDs. 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 d+ W  Len(W) = 0 Since for all i and j, Len(Ei,j) = Len(Fi,j) = 0, it follows that: All triples that Ei,j or Fi,j depend on transitively have length zero and hence are ID triples.

20. S3 V1 S2 ID V1 S1 ID V2 V1 S0 T2 T1 T0 Part 5 Proof: Tubular Phantoms Are Identic. Construction of Sub-States si and Sub-Rulesets ri In the sequence (T0 T1 … TN) each Ti depends on Ti+1 (where TN+1 is T0): TiDR (Ei Ti+1 Fi) with corresponding production from R: tiDR (ei ti+1 fi) Def. s0 = {T0, T1, … TN} r0 = Set of productions for Ti from R: tiDR (ei ti+1 fi) Def. si+1 = si . dri (Compute si’s targets) ri+1 = union of ri and set of productions corresponding to dependencies for T DR for each T in si+1 - si Example Note: si si+1 Since T is legal there must exist tuple sequence  in s such that T DR

21. S3 V1 S2 ID V1 S1 ID V2 V1 S0 T2 T1 T0 Part 6 Proof: Tubular Phantoms Are Identic.Triples depended on recursively by Ti are IDs. Example We previously showed that all triples in Ei and Fi and all triples transitively depended upon by them are limited to be ID triples. We observe that each Ti can depend (according to r0) only on Ti+1 or on triples in Ei and Fi. Therefore: Any constant triple depended upon transitively by any Ti is necessarily an ID constant.

22. Example s0 . dri s0 T2 V1 ID T1 T0 Part 7 Proof: Tubular Phantoms Are Identic.Induction Hypothesis We define a hypothesis Hi, i ³ 0, as follows: Hi =def si  si . dri We will show that Hi is true for all i ³ 0. We start by proving that H0 is true. By definition s0 = {T0, T1, … , TN} We conclude that s0 s0 . dri because every triple in s0 is depended upon by a triple in s0 . For example T1 depends on T2. Hence H0is true.

23. Part 8 Proof: Tubular Phantoms Are Identic.Inductive Proof of Hypothesis Hi We will prove that, for all i ³ 0, Hi is true, i.e., si  si . dri We have already shown that H0 is true. We will use induction to prove that Hi is true for all i > 0, by assuming Hi is true and showing that consequently Hi+1 must also be true. Assuming Hi is true, then every triple in si is depended upon according to ri by at least one other triple in si. Now consider any triple that si depended upon according to ri by a triple in si+1, but is not in si. Any such triple is clearly depended upon by a triple in si. Since si+1 consists of such triples along with triples already in si, it follows that every triple in si+1 is depended upon according to riby at least one triple in si+1. Hence: For all i ³ 0, Hi is true.

24. Part 9 Proof: Tubular Phantoms Are Identic.Conclusion of proof Both series s0, s1, … and r0, r1, … are monotonically increasing in size. Both are limited in size, si by s and ri by R. It follows there is a limiting value, call it L, after which all states, sL, si+1, … and all rulesets ri, ri+1 , … are identical. Hence, for i ³ L, there remain no triples outside of si that are depended upon according to ri by triples in si, but are not members of si. Hence, sL is the same sL. dr except for constant triples in sL. dr so: sL = sL. dr - {CONST} We claim that for i ³ L si = si. dr - {ID} This must be true because we have previously established that every constant triple depended upon directly or indirectly by any Ei,Fi or Ti is an ID constant. Since si is a subset of state s, it follows that phantom tubular state s is identic. QED