Part 13.
Tubular Rulesets
1
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  T
Example
P
P
P
S
T
C
C
d 
E
2
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.
3
Constructive and Phantomic
Tubular Systems
Constructive ruleset:
Phantomic ruleset:
a) M D (P M C)
b) M D (S)
a)
b)
c)
d)
All states are constructive.
Example state:
dMe
fPd
f
(P M2 C)
(M3)
(M4)
(M2)
dM4e
dM3e
e
eCg
fMg
D
D
D
D
Example tubular state (phantom):
dSe
d
M1
M2
M3
M4
g
dM2e
fPd
eCg
fM1g
4
Another Example of a Phantom
in a Tubular System
Ruleset:
• M1 D (P M2 C)
• M2 D (M3)
• M3 D (L1 M4)
• M4 D (M2 R1)
• L1 D (L2)
• L2 D (L3 L4)
• L3 D (ID L4)
• L4 D (L1)
• R1 D (R2)
• R2 D (R2)
*
*
*
dIDd
dL3d
eR2e
dL4d
dL2d
dL1d
eR1e
dM4e
dM3e
dM2e
fPd
eCg
fM1g
* Legal sub states, which are phantoms.
5
Part 14.
Identic Rulesets and Identic
States
6
Identic Rulesets: Introduction
Six
a)
b)
c)
d)
e)
f)
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)
vr Vr . dr
t5
ID
t3
t4
t1
t2
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.
Notation needs work, as here “ID” is used as triples with ID edges??
7
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
v3
vr
vr . dr
ID
v1
v2
t1
t2
Should give algorithm to test to decide if R is identic??
Here, {ID} is a set containing the name “ID” of self-loops.
8
Identic States: Example
Suppose state s has these dependencies among triples V1
to V5.
t . dr
a) V1 D (V3)
t
V5
b) V2 D (V3)
ID
c) V2 D (V4)
d) V3 D (V4, V5)
V4
V3
e) V4 D (V5, ID)
f) V5 D (V3)
V1
V2
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}
Here {ID} is the set of triples that are self loops.
9
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
V3
ID
V1
V2
T1
T2
Here {ID} is the set of triples that are self loops.
s
t
10
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 D (R1, R2, …, Rn) for
each dependency in s of form
(x0 v xn) D
(x0 R1 x1, x1 R2 x2, …, xn-1 Rn xn
)
v5
v3
v4
v1
v2
Somewhere did I assume that identic states are all legal??
11
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UoT
Legal state which
is identic
T
a
U
Should say: if s is identic phantom then R is identic ruleset??
12
Part 15.
Tubular Phantoms are Identic
13
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.
14
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.
15
Part 1 Proof: Tubular Phantoms Are Identic.
Phantoms Have Cycle of Dependency
Any phantom state s is recursive, and has a cycle of
dependencies:
T0 D (E0 T1 F0)
E T F
D
T1 D (E1 T2 F1)
E T F
D
…
D
…
Ti D (Ei Ti+1 Fi)
E
T F
…
TN-1 D (EN-1 TN FN-1)
TN D (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.
N
0
N
0
1
0
N-1
N
N-1
16
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
T0 d T1 d … TN d 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)
Example
In other words, since s is tubular:
All triples, T0 to TN,
have the same length.
T2
V1
T0
ID
T1
17
Part 3 Proof: Tubular Phantoms Are Identic.
Triples depended on by base recursion are ID’s
We have this pattern of dependency among tuples
Ti D (Ei Ti+1 Fi)
which is short for
Ti D (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
EN T0 FN
D
E0 T1 F0
D
…
D
EN-1 TN FN-1
18
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.
19
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):
S3
Ti DR (Ei Ti+1 Fi)
with corresponding production from R:
S2
ti DR (ei ti+1 fi)
Def. s0 = {T0, T1, … TN}
S1
r0 = Set of productions for Ti from R:
ti DR (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
V1
ID
V1
V1
S0
V2
ID
T2
T0
T1
Note: si  si+1
Since T is legal there must exist tuple sequence  in s such that T DR 
20
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.
S3
V1
S2
S1
ID
V1
V1
S0
T0
V2
ID
T2
T1
21
Part 7 Proof: Tubular Phantoms Are Identic.
Induction Hypothesis
We define a hypothesis Hi, i  0, as follows:
Hi =def si  si . dri
We will show that Hi is true for all i  0. We start by proving
that H0 is true.
Example
T
s0
s0 . dri
2
V1
By definition
ID
s0 = {T0, T1, … , TN}
T1
T0
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 H0 is true.
22
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 . 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 ri by at least one triple in si+1.
Hence:
For all i  0, Hi is true.
23
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. dr except for constant triples in sL. dr
so:
sL = sL. dr - {CONST}
We claim that for i  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
24