1
Language Theory and Infinite Graphs
Colin Stirling
School of Informatics
University of Edinburgh
cps
marktoberdorf
today
2
Example
a
X
a
A
b
ε
b
C
b
YX
b
b
a
ε
a
AA
b
XX
YXX
CA
b
b
b
b
AAA
XXX
....
.
.
.
....
.
.
.
·
·
A=X
·
²=²
C =YX
UNF
·
AA = XX
UNF
·
·
A=X
CA = Y XX
BAL(L)
·
CX = Y XX
CUT
·
C =YX
cps
marktoberdorf
UNF
today
3
Decidability
Sketched decidability proof of language equivalence for simple grammars using
tableau proof system
Propositions
• Every tableau is finite
·
• α ∼ β iff the tableau with root α = β is successful
·
Main point, given α = β, then every goal in the tableau has a bounded size (in
terms of |α|, |β| and the size of the PDG)
cps
marktoberdorf
today
4
Comments
What are key features that underpin decidability?
• Bisimulation equivalence is a congruence with respect to stack prefixing
if L(α) = L(β) then L(δα) = L(δβ)
Congruence allows us to tear apart a configuration α0α and replace its tail
with a potentially equivalent configuration β, α0β
• Bisimulation equivalence supports cancellation of postfixed stacks
if L(αδ) = L(βδ) then L(α) = L(β)
Cancellation, as used in CUT, has the consequence that goals become small
cps
marktoberdorf
today
5
Back to DPDA
• How can we tear apart DPDA (stable) configurations pα and replace parts
with parts? What is a part of a configuration?
A configuration has state as well as stack
• L(pα) = L(qβ) does not, in general, imply L(pδα) = L(qδβ)
• L(pαδ) = L(qβδ) does not, in general, imply L(pα) = L(qβ)
cps
marktoberdorf
today
6
a
Even worse
a
a
a
a, c
pX
b
b
c
pYX
b
pYYX
c
pYYYX
c
...
b
....
pε
• A main attack on the decision problem in the 1960/70s examined differences
between stack lengths and potentially equivalent configurations that eventually
resulted in a proof of decidability for real-time DPDAs
• However, L(pY nX) = L(pY mX) for every m and n
cps
marktoberdorf
today
7
However
Many of these problems disappear with pushdown grammars.
Despite
nondeterminism, stacking is a congruence with respect to pre and postfixing
for language and bisimulation equivalence and both equivalences support pre and
postfixed cancellation
1. L(α) = L(β) iff L(αδ) = L(βδ)
2. L(α) = L(β) iff L(δα) = L(δβ)
3. α ∼ β iff αδ ∼ βδ
4. α ∼ β iff αδ ∼ βδ
cps
marktoberdorf
today
8
Therefore ...
• This suggests that we should try to remain with PDGs
• Deterministic PDGs are too restricted
• Arbitrary PDGs are too rich (as they generate all the non-empty context-free
languages)
•
cps
What is needed is a mechanism for constraining nondeterminism in a PDG
marktoberdorf
today
9
Textbook transformation of PDAs into PDGs
• From PDA to language equivalent PDGs (with ²-transitions)
w
• Introduce stack symbols [pXq] so that L([pXq]) = {w : pX −→ q²}
• Convert basic transitions into sets of transitions:
a
a
pX −→ q² is {[pXq] −→ ²}
a
a
pX −→ rY is {[pXq] −→ [rY q]}
a
a
pX −→ rY Z is {[pXq] −→ [rY p0][p0Zq] : p0 ∈ P}
• w ∈ L(pX1 . . . Xn) iff w ∈ L([pX1q1][q1X2q2] . . . [qn−1Xnqn]) for some qi
cps
marktoberdorf
today
10
Example
q²
↑b
pX
↓b
r²
b
←−
qX
↑b
a
−→ pXX
↓b
²
←− rX
b
←−
qXX
↑b
a
−→ pXXX
↓b
²
←− rXX
b
←− . . .
a
−→ . . .
²
←− . . .
a
• [pXq] −→ [pXp][pXq] . . .
a
• [pXq] −→ [pXq][qXq] . . .
• Introducing extra nondeterminism
cps
marktoberdorf
today
11
For DPDA
• Transform into normal form PDG without ²-transitions
• We only convert transitions for a ∈ A
²
• A stack symbol [pXq] is an ε-symbol, if pX −→ q² ∈ T. All ε-symbols are
erased from the right hand side of any transition in the SDG
• Next, the SDG is normalised by removing all unnormed stack symbols
cps
marktoberdorf
today
12
Example
a
a
a
a
a, c
pX
c
pYX
b
b
pYYX
c
pYYYX
c
...
b
b
....
pε
a
[pXr] −→ [pXr]
c
[pXr] −→ [pXr]
a
[pY r] −→ ²
c
[pY r] −→ [pY p][pY r]
[pXp] −→ [pXp]
c
[pXp] −→ [pXp]
[pY p] −→ ²
c
[pY p] −→ [pY r][rY p]
a
[pXp] −→ ²
b
b
[pY p] −→ [pY p][pY p]
c
[pY r] −→ [pY r][rY r]
c
There are two ²-stack symbols, [rXp] and [rY r]
cps
marktoberdorf
today
13
Example
a
a
a
a
a, c
pX
c
pYX
b
pYYX
b
c
pYYYX
c
...
b
b
....
pε
a
[pXp] −→ ²
a
[pY r] −→ ²
c
[pY r] −→ [pY r]
[pXp] −→ [pXp]
[pY p] −→ ²
c
[pY r] −→ [pY p][pY r]
b
[pXp] −→ [pXp]
c
b
[pY p] −→ [pY p][pY p]
c
Extra nondeterminism: consider G([pY r])
cps
marktoberdorf
today
14
Main problem
• Transformation does not preserve bisimulation equivalence
• How to overcome this ?
• Need to preserve structure
cps
marktoberdorf
today
15
Sum configurations
• Convert transitions into transitions over SUMS:
a
a
pX −→ q² is [pXq] −→ ²
a
a
pX −→ rY is [pXq] −→ [rY q]
a
a P
pX −→ rY Z is [pXq] −→ {[rY p0][p0Zq] : p0 ∈ P}
• Convert pX1 . . . Xn to
P
{[pX1q1][q1X2q2] . . . [qn−1Xnqn] : qi ∈ P}
• Language of a sum is union of languages of summands
• For DPDA determinize transitions for sum configurations: preserve structure
cps
marktoberdorf
today
16
Example
a
a
a
a
a, c
pX
c
pYX
b
pYYX
b
c
pYYYX
c
...
b
b
....
pε
a
[pXp] −→ ²
a
[pY r] −→ ²
c
[pY r] −→ [pY r]
[pXp] −→ [pXp]
[pY p] −→ ²
c
[pY r] −→ [pY p][pY r]
b
[pXp] −→ [pXp]
c
b
[pY p] −→ [pY p][pY p]
c
• Let A = [pXp], B = [pY p], C = [pY r]. So, pY X is BA + C
cps
marktoberdorf
today
17
Determinized transitions
a
a
a
a
a, c
pX
c
pYX
b
b
b
pYYX
c
c
pYYYX
...
b
....
pε
a
A −→ ²
b
A −→ A
a
C −→ ² B −→ BB
b
c
A −→ A
B −→ ²
c
C −→ BC + C
c
• Let A = [pXp], B = [pY p], C = [pY r]. So, pY X is BA + C
cps
marktoberdorf
today
18
Strict deterministic grammars
Due to Harrison and Havel in 1970s
Add an extra component to a PDG
• An equivalence relation ≡ on S (stack symbols)
≡ partitions the stack symbols S into disjoint subsets S1, . . . , Sk :
for each i, and pair of stack symbols X, Y ∈ Si, X ≡ Y
INTUITION FROM DPDA: X ≡ Y iff X = [pZq] and Y = [pZr]
cps
marktoberdorf
today
19
Sequences
Extend ≡ on S to a relation on S∗
α ≡ β iff
• α = β, or
• there is a δ ∈ S∗, α = δXα0, β = δY β 0, X ≡ Y and X 6= Y
For instance, if Y ≡ Z then XY α ≡ XZ for any α
≡ on stacks is not an equivalence relation
cps
marktoberdorf
today
20
Properties
• αβ ≡ α iff β = ²
• α ≡ β iff δα ≡ δβ
• If α ≡ β and γ ≡ δ then αγ ≡ βδ
• If α ≡ β and α 6= β then αγ ≡ βδ
• If αγ ≡ βδ and |α| = |β| then α ≡ β
cps
marktoberdorf
today
21
Strict deterministic
≡ on S is strict when
a
1. if
X −→ α ∈ T
|||
a
Y −→ β ∈ T
2. if
X −→ α ∈ T
|||
a
Y −→ α ∈ T
then
α
|||
β
then
X
||
Y
a
A PDG with ≡ is strict determinisitic, an SDG, if its relation ≡ is strict
cps
marktoberdorf
today
22
Example
X −→ X
a
X −→ ²
b
X −→ X
c
Z −→ ²
b
Z −→ Z
Y −→ Y Y
c
Y −→ ²
c
Z −→ Y Z
a
a,c
c
X
b
ε
c
Z
a
c
a
YZ
c
a
YYZ
c
...
b
ε
Strict: partition {{X}, {Y, Z}}
c
c
c
Y −→ Y Y , Z −→ Z, Z −→ Y Z and Y Y ≡ Z, Y Y ≡ Y Z and Z ≡ Y Z
cps
marktoberdorf
today
23
Constrained nondeterminism
• If each set in the partition is a singleton then the SDG is deterministic,
a simple grammar, and α ≡ β iff α = β
• In general, constrained nondeterminism
a
a
X −→ ² ∈ T, X ≡ Y and X 6= Y implies no a-transition Y −→ β ∈ T
a
Suppose Y −→ β. By condition 1 of being strict ² ≡ β, so β = ².
By condition 2 of being strict X = Y , which is a contradiction.
cps
marktoberdorf
today
24
Key property
Strictness conditions generalise to words and stacks
w
1. if
α −→ α0
|||
w
β −→ β 0
2. if
α −→ α0
|||
a
β −→ α0
then
α0
|||
β0
then
α
||
β
w
Proof in the notes page 25
cps
marktoberdorf
today
25
Corollaries
• If α ≡ β and w ∈ L(α), then for all words v, and a ∈ A, wav 6∈ L(β)
• If α ≡ β and α 6= β then L(α) ∩ L(β) = ∅
DPDA intuition: properties hold for [pXq], [pXr]
cps
marktoberdorf
today
26
Sum configurations
• Extend configuration to sets of sequences of stack symbols, {α1, . . . , αn}
• Write it as a sum α1 + . . . + αn without repeated elements
• Degenerate case is the empty sum, ∅
• L(α1 + . . . + αn) =
• L(∅) = ∅
cps
S
{L(αi) : 1 ≤ i ≤ n}
marktoberdorf
today
27
Admissible sums
• β1 + . . . + βn is admissible if βi ≡ βj for each i and j
• Admissible configurations are preserved by transitions
If α1 + . . . + αn admissible then for all w the sum configuration
w
w
{βi1 : α1 −→ βi1} ∪ . . . ∪ {βin : αn −→ βin}
is admissible
• DPDA intuition: {[pX1q1] . . . [qn−1Xnqn] : qi ∈ P} is admissible
cps
marktoberdorf
today
28
Example
a
a,c
c
X
b
ε
c
Z
a
YZ
c
a
YYZ
c
...
b
ε
Strict: partition {{X}, {Y, Z}}
Admissible: XX, ZZZ + ZZY , Y X + Z, Z + Y Z, Z + Y Z + Y Y Z
Not admissible X + Z, Y + ²
cps
marktoberdorf
today
29
Determinization
• T is changed to Td. For each stack symbol X and a ∈ A, the family of
a
a
transitions X −→ α1, . . . , X −→ αn ∈ T is determinized
a
X −→ α1 + . . . + αn ∈ Td
a
• For each X, a a unique transition X −→
cps
P
marktoberdorf
αi ∈ Td as sum could be ∅
today
30
Determinization cont.
• Prefix rule for generating transitions is generalised
a
If Xi −→ α1i + . . . + αni i for each i : 1 ≤ i ≤ m then
a
X1β1 + . . . + Xmβm −→ α11β1 + . . . + αn1 1 β1 +
...
...
+
α1mβm + . . . + αnmm βm
• Determinized graph Gd(α1 + . . . + αn) using Td and the extended prefix rule
cps
marktoberdorf
today
31
Example
S = {X, Y, Z}, A = {a, b, c} and the partition of S = {{X}, {Y, Z}}. And Td is
a
a
a
b
b
X −→ X Y −→ ² Z −→ ∅ X −→ ² Y −→ ∅
b
c
c
c
Z −→ ² X −→ X Y −→ Y Y Z −→ Y Z + Z
Gd(Y X + Z) is
a
a,c
X
b
a
YX + Z
b
c
a
YYX + YZ + Z
b
c
...
...
ε
cps
marktoberdorf
today
32
Which is ∼
a
a
a
a
a, c
pX
b
b
c
pYX
b
pYYX
c
pYYYX
c
...
b
....
pε
cps
marktoberdorf
today
33
Characterising DPDA
• A DPDA can be converted into a determinized SDG, and configuration pα is
converted into sum configuration sum(pα) of the SDG where
Gc(pα) ∼ Gd(sum(pα))
• (Conversely, a determinized SDG can be transformed into a DPDA)
• DPDA equivalence problem = to deciding whether two admissible
configurations of a determinized SDG are bisimulation equivalent
cps
marktoberdorf
today
34
Notation
• E, F , G, . . . range over admissible configurations
• + can be extended: if E and F are admissible, E ∪ F is admissible, E and F
are disjoint, E ∩ F = ∅, then E + F is the admissible configuration E ∪ F
• Also use sequential composition, if E and F are admissible, then EF is the
admissible configuration {βγ : β ∈ E and γ ∈ F }
• E = E1G1 + . . . + EnGn is a head/tail form , if the head E1 + . . . + En is
admissible and at least one Ei 6= ∅, and each tail Gi 6= ∅
If we write E = E1G1 + . . . + EnGn then E is a head/tail form
cps
marktoberdorf
today
35
Congruence
• Decision question E ∼ F ?
• Assume E = E1G1 + . . . + EnGn
– If each Hi 6= ∅ and each Ei 6= ε and for each j such that Ej 6= ∅, Hj ∼m Gj ,
then E ∼m+1 E1H1 + . . . + EnHn
– If each Hi 6= ∅ and for each j such that Ej 6= ∅, Hj ∼ Gj , then
E ∼ E 1 H1 + . . . + E n Hn
• We can tear apart configurations and replace parts with equivalent parts
cps
marktoberdorf
today
36
Notation
v
• For X ∈ S, w(X) is the unique shortest word v ∈ A+ such that X −→ ²
• w(E) is the unique shortest word for configuration E
• M is the maximum norm of the SDG
u
• “E following u”, written E · u, is the unique F such that E −→ F
cps
marktoberdorf
today
37
Decision procedure
• Given by a deterministic tableau proof system where goals are reduced to
subgoals
·
• Goals and subgoals have the form E = F , is E ∼ F
• Rules are (generalisations of) unfold, UNF, and balance rules,
BAL(L) and BAL(R). (CUT free)
cps
marktoberdorf
today