part 13. tubular rulesets
Post on 19-Jan-2016
42 Views
Preview:
DESCRIPTION
TRANSCRIPT
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
PS 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 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):
5
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.
* **
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
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
8
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.
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
V3ID
t sExample
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 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??
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 TU 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 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.
15
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.
16
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
…
17
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
18
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
…
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) = 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.
20
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
21
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
22
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
23
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.
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 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
top related