1. Undecidable problems for CFGs
The Chinese University of Hong Kong
Fall 2008
CSC 3130: Automata theory and formal languages
Undecidable problems for CFGs
Andrej Bogdanov

2. Decidable vs. undecidable
“DFA M accepts w”
“TM M accepts w”
“PDA M accepts w”
“TM M halts on w”
“DFA M accepts all inputs”
“TM M accepts some input”
“TM M and M’ accept same inputs”
“CFG G generates all inputs”
“CFG G is ambiguous”
other kinds of problems? ?

3. Representing computations
L1 = {w$w: w ∈{a, b}*}
a/aR
b/bR
x/xR
$/$R
a/aL
abbaa$abbaa q0 q1 q2
b/bL
a/aL
a/xL
b/bL
xbbaa$abbaa q1
x/xR
x/xL
a/xR
xbbaa$abbaa
... q1
$/$R
☐/☐R
$/$R q0 q7 qa q5 q6
b/xR
b/xL
xbbaa$abbaa q2 q3 q4
xbbaa$xbbaa q5
$/$R
...
xxxxx$xxxxx qa
a/aR
x/xR
b/bR
x/xR

4. Turing Machine configuration
time 0
abbaa$abbaa
#q0abbaa$abbaa☐ q0
time 1 q1
xbbaa$abbaa
#xq1bbaa$abbaa☐
...
time 6
xbbaa$abbaa q2
#xbbaa$q2abbaa☐
A configuration of TM M at time t is a string
over alphabet G∪Q∪{#} that represents the state, tape head, and “touched” part of tape of M after t steps of work (possibly with ☐s)

5. Computation histories
a/aR
b/bR
x/xR
#0q0a0b0$0a0b0☐
#0x0q1b0$0a0b0☐
#0x0b6q1$0a0b0☐
#0x0b6$0q2a0b0☐
#0x0b6q5$0x1b0☐
#0x0q6b0$0x0b0☐
$/$R
a/aL q1 q2
b/bL
a/aL
a/xL
b/bL
x/xR
x/xL
a/xR
$/$R
☐/☐R
$/$R q0 q7 qa q5 q6
b/xR
b/xL
... q3 q4
$/$R
#0x0x0$0x0x0q7☐
#0x0x0$0x0x0☐aqa
a/aR
x/xR
b/bR
x/xR
computation history

6. Computation histories as strings
If M halts on w, We can represent the computation history by a string hist over alphabet G∪Q∪{#}
#0q0a0b0$0a0b0☐
#0x0q1b0$0a0b0☐
...
#q0ab$ab☐#xq1b$ab☐#...#xx$xx☐qa#
#0x0x0$0x0x0q7☐
#0x0x0$0x0x0☐aqa
M accepts w
qa occurs in string hist
M rejects w
qa does not occur in hist

7. Undecidable problems for PDAs
ALLPDA = {〈P〉: P is a PDA that accepts all inputs}
Theorem
Proof: We will show that
The language ALLPDA is undecidable.
If ALLPDA can be decided, so can ATM.

8. Undecidable problems for PDAs
accept if P accepts all inputs
reject if not
accept
if M rej/loops w A
〈P〉
〈M〉, w
reject
if M accepts w
P accepts all inputs if M rejects or loops on w
P does not accept some input if M accepts w

9. Undecidability via computation histories
P accepts all inputs if M rejects or loops on w
P does not accept some input if M accepts w
candidate computation
history of M on w P
reject accepting histories
#q0ab$ab☐#xq1b$ab☐#...#xx$xx☐qa#
reject
every other string
accept
M accepts w
P rejects hist
M rej/loops on w
no accepting histories
P accepts everything

10. Undecidability via computation histories
Task: Design a PDA P such that
candidate computation
history hist of M on w P
reject accepting histories
Input: hist of the form w1#w2#...#wk#
#0q0a0b0$0a0b0☐
#0x0q1b0$0a0b0☐
If w1 ≠ #q0w , accept hist.
If hist does not contain qa, accept hist.
...
#0x0x0$0x0x0q7☐
#0x0x0$0x0x0☐aqa
If two consecutive blocks wi#wi+1
do not correspond to a proper transition of M, accept hist.

11. Computation is local #0q0a0b0$0a0b0☐ #0x0q1b0$0a0b0☐ #0x0b6q1$0a0b0☐
a/aR
b/bR
x/xR
#0q0a0b0$0a0b0☐
#0x0q1b0$0a0b0☐
#0x0b6q1$0a0b0☐
#0x0b6$0q2a0b0☐
#0x0b6q5$0x1b0☐
#0x0q6b0$0x0b0☐
$/$R
a/aL q1 q2
b/bL
a/aL
a/xL
b/bL
x/xR
x/xL
a/xR
$/$R
☐/☐R
$/$R q0 q7 qa q5 q6
b/xR
b/xL
... q3 q4
$/$R
#0x0x0$0x0x0q7☐
#0x0x0$0x0x0☐aqa
a/aR
x/xR
b/bR
x/xR
The changes between configurations occur in a 2x3 window

12. Implementing P valid transition
On input hist:
Nondeterministically make one of the following choices
Look in the first block w1 of hist
If w1 ≠ #q0w, accept hist.
Look for the appearance of qa
If t does not contain qa, accept hist.
#0x0b6$0q2a0b0☐
#0x0b6q5$0x1b0☐
Look for the beginning of the ith block of hist
If two consecutive blocks wi#wi+1 do not represent a valid transition of M, accept hist.
valid transition
wi#wi+1 represents a valid transition if all 3x2 windows correspond to possible transitions of M

13. Valid and invalid windows
… 6a3b0x0 …
… 0a6b0x0 …0
… 6q2a0b0 …
… 0a6b0q2 …0
… 6a3q2a0 …
… 0q5a6x0 …0
… 6q2q2a0 …
… 0q2q2x3 …0 q2
a/xL
… 6a3b0a0 …
… 0a6b0q5 …0
… 6a3q2a0 …
… 0q5a6b0 …0 q5
… 6a3a0☐0 …
… 0x6a0☐0 …0
… 6a3q2a0 …
… 0a6q5x0 …0

14. Implementing P How to check wi#wi+1 represents a valid transition:
Nondeterministically look for a 3x2 window
#0x0b6$0q2a0b0☐
#0x0b6q5$0x1b0☐
Remember window offset on stack
valid transition
Remember first row of window in state
Locate the second row of window
Go past next #, then keep reading/popping
until stack is empty
offset
Remember second row of window in state
If window is not valid, accept;
Otherwise reject.

15. The Post Correspondence Problem
Input: A set of tiles like this
Given an infinite supply of such tiles, can you match top and bottom?
bab cc c ab a ab
baa a a
baba
bab e a ab
baa a
bab e c ab c ab
bab cc a
baba

16. Undecidability of PCP
PCP = {〈T〉: T is a collection of tiles that contains a top-bottom match}
The language PCP is undecidable.
Proof: We will show that
If PCP can be decided, so can ATM.

17. Undecidability of PCP 〈M, w〉 T
(collection of tiles)
If M accepts w, then T can be matched
If M rej/loops on w, then T cannot be matched
Idea: Matches represent accepting histories
#q0ab$ab#xq1b$ab#...#xx$xx☐qa#
#q0ab$ab#xq1b$ab#...#xx$xx☐qa#
e #q0ab$ab
#q0a
#xq1 b a $ ☐ #
xq1$
x$q2 …

18. Some technicalities We will assume that
TM M never attempts to move left of first cell
One of the PCP tiles is marked as a starting tile
These assumptions can be made without loss of generality (we will see why later) s
bab cc c ab a ab
baa a a
baba

19. for each valid window with state qi in top middle
Undecidability of PCP
〈M, w〉 T
(collection of tiles)
If M accepts w, then T can be matched
If M rej/loops on w, then T cannot be matched
To decide ATM, we construct these tiles for PCP s e
#q0w
x1qix2
x3x4x5
☐#qix1
☐#x2x3 ☐# xx
xqa qa
qax qa
qa## #
for each valid window with state qi in top middle
for all x in G∪{#}

20. Undecidability of PCP tile type purpose s e #q0w
represent initial configuration
represent transitions between configurations
x1qix2
x3x4x5 xx
add blank spaces if necessary
☐#qix1
☐#x2x3 ☐#
complete match if computation accepts
xqa qa
qax qa
qa## #

21. Undecidability of PCP #q0a$ab#xq1$ab#...#xx$xxq7☐#xx$xx☐qa#
accepting computation history
#q0a b $
ab#
xq1b$ab#...
#xx$x
xq7☐ #
#q0ab$ab
#xq1 b $
ab#
...#xx$xxq7☐
#xx$x
x☐qa # s e
#q0w
x1qix2
x3x4x5
☐#qix1
☐#x2x3 ☐# xx

22. Undecidability of PCP
Once the accepting state symbol occurs, the last two tiles can “eat up” the rest of the symbols
#xx$xx
☐qa
#xx$x
xqa
#xx$xqa#... #
qa##
#xx$xx☐qa
#xx$xx qa
#xx$x qa
#...#qa # # xx
xqa qa
qax qa
qa## #

23. for each valid window of this form
Undecidability of PCP
If M rejects on input w, then qreject appears on bottom at some point, but it cannot be matched on top
If M loops on w, then matching keeps going forever s e
#q0w
a1qia3
b1b2b3
☐#qia2
☐#b1b2 ☐# xx
xqa qa
qax qa
qa## #
for each valid window of this form
for all x in G∪{#}

24. “starting tile” begins with *
A technicality
We assumed that one tile marked as starting tile
We can remove assumption by changing tiles a bit s a
baba
bab cc c ab
*a*
*b*a*b*a
b*a*b*
*c*c c*
*a*b  *
“starting tile” begins with *
“middle tiles”
“ending tile”
matches last *

25. Ambiguity of CFGs AMB = {G: G is an ambiguous CFG} Theorem
Proof: We will show that
The language AMB is undecidable.
If AMB can be decided, so can PCP.

26. Ambiguity of CFGs T G Proof: Step 1: Number the tiles
(collection of tiles)
(CFG)
If T can be matched, then G is ambiguous
If T cannot be matched, then G is unambiguous
Proof: Step 1: Number the tiles 1 2 3
bab cc c ab a ab

27. Ambiguity of CFGs T G Terminals: a,b,c,1,2,3 Variables: S, T, B
(collection of tiles)
(CFG)
Terminals:
a,b,c,1,2,3
Variables:
S, T, B
Productions:
T → babT1
T → cT2
S → aT3
B → ccS1
B → abB2
B → abB3 1 2 3
bab cc c ab a ab
S → T | B

28. Ambiguity of CFGs T G Terminals: a,b,c,1,2,3 Variables: S, T, B 1 2 3
(collection of tiles)
(CFG)
Terminals:
a,b,c,1,2,3
Variables:
S, T, B 1 2 3
Productions:
bab cc c ab a ab
S → T | B
T → babT1
T → cT2
T → aT3
T → bab1
T → c2
T → a3
B → ccS1
B → abB2
B → abB3
B → cc1
B → ab2
B → ab3

29. Ambiguity of CFGs Each sequence of tiles gives two derivations
If the tiles match, these two derive the same string 1 2 2
bab cc c ab c ab
S ⇒ T ⇒ babT1 ⇒ babcT21⇒ babcc221
S ⇒ B ⇒ ccB1 ⇒ ccabB21⇒ ccabab221

30. Ambiguity of CFGs T G ✓ ✓ Argue by contradiction:
(collection of tiles)
(CFG) ✓
If T can be matched, then G is ambiguous ✓
If T cannot be matched, then G is unambiguous
Argue by contradiction:
If G is ambiguous then ambiguity must look like this S S
Then n1...ni = m1…mj T B
So there is a match a1 T n1 b1 m1 B n1 n2 ni … a2 n2 b2 … m2 a1 b1 a2 b2 ai bi T B … ai ni bj mj