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Part 13.Tubular Rulesets

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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

PS C

C

E

T

Example

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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.

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Constructive and Phantomic Tubular SystemsConstructive ruleset:a) M (P M C)b) M (S)

Phantomic ruleset:a) M1 (P M2 C)

b) M2 (M3)

c) M3 (M4)

d) M4 (M2)

fMgf g

d e

fPd

dMe

eCg

dSe

All states are constructive.Example state:

fM1g

fPd

dM2e

eCg

dM4edM3e

Example tubular state (phantom):

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Another Example of a Phantom in a Tubular SystemRuleset:

• M1 (P M2 C)

• M2 (M3)

• M3 (L1 M4)

• M4 (M2 R1)

• L1 (L2)

• L2 (L3 L4)

• L3 (ID L4)

• L4 (L1)

• R1 (R2)

• R2 (R2) fM1g

fPd

dM2e

eCg

dM4edM3e

dL1d

dL4d

dL3d

dL2d

eR1e

eR2e

dIDd

* Legal sub states, which are phantoms.

* **

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Part 14.Identic Rulesets and Identic States

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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

vr = vr . 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

t5ID

vrVr . r

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Def. SoP ruleset R is said to be identic if there is sub ruleset r, r R, such that

vr = vr . r - {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.

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

v3ID

vr

vr . 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 thatt = 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.

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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

V3ID

t sExample

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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 vi a) for each variable in subset rule r. Legal state s is identic Productions v (R1, R2, …, Rn) for each dependency in s of form(x0 v xn)

(x0 R1 x1, x1 R2 x2, …, xn-1 Rn xn )

v1 v2

v3 v4

v5

Somewhere did I assume that identic states are all legal??

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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 TU U o T T

U

a

Legal state which is identic

Should say: if s is identic phantom then R is identic ruleset??

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Part 15.Tubular Phantoms are Identic

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Recap: Tubular Rules Sets Can Have Identic Phantoms

Ideally, tubular rules sets should have no phantomsHowever, 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.

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Tubular Phantoms are IdenticTheorem. Given a tubular state s,

s is a phantom s is identicProof. … given below …Corollary. Given a tubular state s,

s is non-identic s is constructiveCorollary. 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.

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Any phantom state s is recursive, and has a cycle of dependencies:

T0 E0 T1 F0T1 E1 T2 F1…

Ti Ei Ti+1 Fi…

TN-1 EN-1 TN FN-1TN 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.

Part 1 Proof: Tubular Phantoms Are Identic.Phantoms Have Cycle of Dependency

EN-1 TN FN-1

EN T0 FN

E0 T1 F0

…

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Since state s is a phantom, there exists a non-empty sequence of triples (T0 T1 … TN) such that

T0 T1 … TN T0

Recall that if tubular triple V depends on triple W (if V W) then Len(V) Len(W) so

Len(T0) Len(T1) … Len(TN) Len(T0)

HenceLen(T0) = Len(T1) = … = Len(TN)

In other words, since s is tubular: All triples, T0 to TN,

have the same length.

Part 2 Proof: Tubular Phantoms Are Identic.Triples in basic recursion have same lengths

T0T1

T2V1ID

Example

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We have this pattern of dependency among tuples

Ti (Ei Ti+1 Fi)which is short for

Ti (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

Part 3 Proof: Tubular Phantoms Are Identic.Triples depended on by base recursion are ID’s

EN-1 TN FN-1

EN T0 FN

E0 T1 F0

…

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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) = 0Since 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.

Part 4 Proof: Tubular Phantoms Are Identic.Triples depended on by Ei and Fi are IDs.

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In the sequence (T0 T1 … TN) each Ti depends on Ti+1 (where TN+1 is T0):

Ti R (Ei Ti+1 Fi)

with corresponding production from R: ti R (ei ti+1 fi)

Def. s0 = {T0, T1, … TN}

r0 = Set of productions for Ti from R:

ti R (ei ti+1 fi)

Def. si+1 = si . ri (Compute si’s targets)

ri+1 = union of ri and set of productions

corresponding to dependencies for T R

for each T in si+1 - si

Part 5 Proof: Tubular Phantoms Are Identic. Construction of Sub-States si and Sub-Rulesets ri

Since T is legal there must exist tuple sequence in s such that T R

Example

Note: si si+1

T0T1

T2

V1

IDV2V1

V1

ID

S0

S1

S2

S3

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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.

Part 6 Proof: Tubular Phantoms Are Identic.Triples depended on recursively by Ti are IDs.

Example

T0T1

T2

V1

IDV2V1

V1

ID

S0

S1

S2

S3

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Part 7 Proof: Tubular Phantoms Are Identic.Induction Hypothesis

We define a hypothesis Hi, i 0, as follows:

Hi =def si si . ri

We will show that Hi is true for all i 0. We start by proving that H0 is true.

By definition s0 = {T0, T1, … , TN}

We conclude that s0 s0 . ri

because every triple in s0 is depended upon by a triple in s0 . For example T1 depends on T2. Hence H0 is true.

T0T1

T2V1ID

Example

s0s0 . ri

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Part 8 Proof: Tubular Phantoms Are Identic.Inductive Proof of Hypothesis Hi

We will prove that, for all i 0, Hi is true, i.e.,

si si . ri

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 ri by at least one triple in si+1. Hence:

For all i 0, Hi is true.

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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 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. r except for constant triples in sL. r so:

sL = sL. r - {CONST}

We claim that for i L si = si. r - {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