91.304 Foundations of
(Theoretical) Computer Science
Chapter 1 Lecture Notes (more on Section 1.4)
David Martin
[email protected]
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1
Simplifying RPP proofs
 I find it easier to forget about
contradiction proofs and instead
prove directly that a language is not
in RPP
 So we need a direct, formal version of
of the statement that L 2 RPP
2
Unraveling RPP (repeat)
Rephrasing L is a member of RPP if
 There exists p¸0 such that
 For every s2L satisfying |s|¸ p,
 There exist x,y,z 2 * such that
1.
2.
3.
4.
s=xyz
|y|>0
|xy|· p
For all i ¸ 0,
x yi z 2 L
(9 p) (8 s) (9 x,y,z) (8 i) !!!
Pretty complicated
3
Unraveling non-RPP
Rephrasing L is not in RPP if
 For every p¸0
 There exists some s2L satisfying |s| ¸ p
such that
 For every x,y,z 2 * satisfying 1-3:
1.
2.
3.

s=xyz,
(8 p) (9 s) (8 x,y,z) (9 i)
Still complicated — but you don't have
|y|>0, and
to use contradiction now
|xy|· p
There exists some i ¸ 0 for which
x yi z 2 L
4
A direct proof of nonregularity
2
 Let D={an | n¸0 } = {,a1,a4,a9, } ('a' is
just some character). Then D is not regular.
Proof idea The pumping lemma says there's a
fixed-size loop in any DFA that accepts long
strings. You can repeat the characters in that
loop as many times as you want to get longer
strings that the machine accepts. Each time
you add a repetition you grow the pumped
string by a constant length.
But the spacing between strings in D above
keeps changing; it's never constant. So D
doesn't have the pumping property.
5
A direct proof of nonregularity
2
 Let D={an | n¸0 } = {,a1,a4,a9, }.
Then D is not in RPP and thus not regular.
2
(p+1)
Proof Let p¸0 and set s=a
. Then s2D
and |s|>p (so such an s certainly exists).
Now let x,y,z2* be any strings satisfying
1. xyz = s
= a(p+1)2
2. |y|>0, and
3. |xy|· p
Our goal is to produce some i such that
xyiz 2 D
6
Direct proof continued
(We'll actually show that xy0z 2 D)
Observe that y=aj for some 1· j · p, so
2
|xy0z| = |a(p+1) -j| < (p+1)2
Since j · p we know that -j ¸ -p and thus
|xy0z| = (p+1)2 - j
¸ (p+1)2 - p
= p2 + p + 1
> p2
2
In other 2words, xy0z has > p characters and
< (p+1) characters. So |xy0z| is not a
perfect square and thus xy0z 2 D. QED
7
Direct or contradiction proof?
 Both work fine... it's your choice
 But you must clearly state what you
are doing
 If proof by contradiction, say so
 If direct proof, say so
8
Game theory formulation
 The direct proof technique can be
formulated as a two-player game
 You are the player who wants to
establish that L is not pumpable
 Your opponent wants to make it difficult
for you to succeed
 Both of you have to play by the rules
9
Game theory continued
The game has just four steps.
1. Your opponent picks p¸0
2. You pick s2L such that |s|¸ p
3. Your opponent chooses x,y,z 2 *
such that s=xyz, |y|>0, and |xy|· p
4. You produce some i ¸ 0 such that
xyiz 2 L
10
Game theory continued
 If you are able to succeed through step 4,
then you have won only one round of the
game
 Like winning one round of Tic-tac-toe
 Do example for a member of D
 To show that a language is not in RPP you
must show that you can always win,
regardless of your opponent's legal moves
 Realize that the opponent is free to choose the
most inconvenient or difficult p and x,y,z
imaginable that are consistent with the rules
11
Game theory continued
 So you have to present a strategy for
always winning — and convincingly
argue that it will always win
 So your choices in steps 2 & 4 have to
depend on the opponent's choices in
steps 1 & 3
 And you don't know what the opponent
will choose
 So your choices need to be framed in
terms of the variables p, x, y, z
12
Game theory continued
 Ultimately it is not very different from
the direct proof
 But it states clearly what choices you
may make and what you may not — a
common cause of errors in proofs
 Repeat previous proof in this framework
13
A direct proof of nonregularity
2
Step 1,
opponent's
choice
Step 2,
your choice
and
reasoning
4 9
 Let D={an | n¸0 } = {,a1,a ,a , }.
Then D is not in RPP and thus not regular.
2
(p+1)
Proof Let p¸0 and set s=a
. Then s2D
and |s|>p (so such an s certainly exists).
Now let x,y,z2* be any strings satisfying
1. xyz = s
= a(p+1)2
Step 3,
opponent's
2. |y|>0, and
choice
3. |xy|· p
Our goal is to produce some i such that
xyiz 2 D
14
Direct proof continued
(We'll actually show that xy0z 2 D)
Observe that y=aj for some 1· j · p, so
2
Step
2 4,
|xy0z| = |a(p+1) -j| < (p+1)
your
Since j · p we know that -j ¸ -pchoice
and thus
|xy0z| = (p+1)2 - j
¸ (p+1)2 - p
= p2 + p + 1
Step 4,
2
your
>p
2
reasoning
In other 2words, xy0z has > p characters
and
< (p+1) characters. So |xy0z| is not a
perfect square and thus xy0z 2 D. QED
15
Unraveling RPP (repeat)
Rephrasing L is a member of RPP if
 There exists p¸0 such that
 For every s2L satisfying |s|¸ p,
 There exist x,y,z 2 * such that
1.
2.
3.
4.
s=xyz
|y|>0
|xy|· p
For all i ¸ 0,
x yi z 2 L
Theorem REG µ RPP
16
Structural facts about RPP
1. If L 2 RPP(p) (meaning "strings in L
with length ¸ p are pumpable") and
q>p then L 2 RPP(q)
2. If L 2 RPP(q) and q>p then L
(contrapositive of 1)
2
RPP(p)
 Thus if you have a proof that establishes
L 2 RPP(q) only when q¸5, that's good
enough: it follows that L is not regular
 Relevant for “C is not regular” problem
17
Structural facts about RPP
3. If L 2 FIN and the longest string in L has
length n, then
a) L 2 RPP(n+1)
b) L 2 RPP(q) for all q < n+1
Note RPP is a class of languages that's only
interesting because of its relation to REG.
It is not a reasonable proposal for a
computation model!
18
Unraveling non-RPP (repeat)
L is not in RPP if
 For every p¸0
(opponent choice)
 There exists some s2L satisfying |s| ¸ p
such that
(your choice)
 For every x,y,z 2 * satisfying 1-3:
1.
2.
3.

s=xyz,
(opponent's)
|y|>0, and
|xy|· p
There exists some i ¸ 0 for which
(yours)
x yi z 2 L
19
Another example
 Let C = {0m 1n | m  n }. Is C regular? Try
to prove it isn't:
 Set s=0p 12p. If opponent chooses
x=, y=0p, z=12p, then we can set i=2 and
win because xy2z=02p 12p 2 C.
 What if opponent chooses a shorter y?
 Looks like it's relatively easy to be a member of
C and hard to not be a member of C
 Can force opponent to choose y 2 0+
 So try to arrange it so that no matter what |y|
is, some number of repetitions of it will match
the target number of '1's
20
Direct proof?
 Hmmm
21
Using closure properties
 Can simplify argument a great deal
Fact If L is not regular then Lc is not regular
either.
Proof If L is not regular but Lc were regular,
c
c
then (L ) would also be regular because
c
REG is closed under complement. But (Lc)
=L
QED
 Recall the languages
B = { 0m 1n | m = n }
C = { 0m 1n | m  n }
C is similar to Bc...
22
Using closure properties
 Start over:
B = { 0m 1n | m = n }
C = { 0m 1n | m  n }
 Certainly B µ Cc
(known nonreg)
(suspected nonreg)
 If m=n then it's true that (not m  n)
 But B  Cc
 Find example x 2 Cc - B...
 On the other hand, B = 0*1* Å Cc
23
Using closure properties
Fact If L1ÅL2 2 REG and L1 2 REG, then
L2 2 REG
Proof Suppose
(a) L1Å L2 2 REG and L12 REG and
(b) L22REG. Since REG is closed
under Å we know that L1ÅL2 2 REG,
but that contradicts assumption (a).
Thus (a) and (b) can't both be true.
QED
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