1. Theory of Computability
Giorgi Japaridze
Theory of Computability
PSPACE-Completeness
Section 8.3

2. PSPACE-completeness defined
Giorgi Japaridze Theory of Computability
Definition 8.8 A language B is PSPACE-complete iff it satisfies two
conditions:
1. B is in PSPACE, and
2. every A in PSPACE is polynomial time reducible to B.
If B merely satisfies condition 2, we say that it is PSPACE-hard.
Why do we still appeal to polynomial time reducibility and not, say,
polynomial space reducibility, philosophically speaking?
A reduction must be easy relative to the class (of difficult problems)
that we are defining. Only then it is the case that if we find an easy
way to solve a (PSPACE-, NP- or whatever-) complete problem,
easy solutions to other (reducible to it) problems would also be found.
If the reduction itself is hard, it does not at all offer an easy way to
solve problems.

3. Universal quantifier : xP(x) means “for any x{0,1}, P(x) is true”
The TQBF problem
8.3.b
Giorgi Japaridze Theory of Computability
Universal quantifier : xP(x) means “for any x{0,1}, P(x) is true”
Existential quantifier : xP(x) means “for some x{0,1}, P(x) is true”
We consider fully quantified Boolean formulas (in the prenex form).
These are Boolean formulas prefixed with either x or x for each
variable x.
Examples (true or false?):
x(xx)
xy (xy)
x(xx)
xy ((xy)(xy))
x(xx)
xy ((xy)(xy))
xy(xy)
zxy ((xyz)(xyz))
TQBF = {<> |  is a true fully quantified Boolean formula}
(True Quantified Boolean Formulas)

4. The PSPACE-completeness of TQBF – proof idea
Giorgi Japaridze Theory of Computability
Theorem TQBF is PSPACE-complete.
Proof idea. To show that TQBFPSPACE, we give an algorithm that assigns values to
the variables and recursively evaluates the truth of the formula for those values.
To show that ApTQBF for every APSPACE, we begin with a polynomial-space
machine M for A. Then we give a polynomial time reduction that maps a string w to a
formula  that encodes a simulation of M on input w.  is true iff M accepts w (and
hence iff wA).
A first, naive, attempt to do so could be trying to precisely imitate the proof of the
Cook-Levin theorem. We can indeed construct a  that simulates M on input w by
expressing the requirements for an accepting tableau. As in the proof of the Cook-Levin
theorem, such a tableau has polynomial width O(nk), the space used by M. But the
problem is that the height of the tableau would be exponential!
Instead, we use a technique related to the proof of Savitch’s theorem to construct the
formula. The formula divides the tableau into halves and employs the universal
quantifier to represent each half with the same part of the formula. The result is a much
shorter formula End of proof idea

5. A polynomial space algorithm for TQBF
8.3.d
A polynomial space algorithm for TQBF
Giorgi Japaridze Theory of Computability
The following algorithm obviously decides TQBF:
T = “On input <>, a fully quantified Boolean formula:
1. If  contains no quantifiers, then it is an expression with only
constants, so evaluate  and accept if true, otherwise, reject.
2. If  is x, recursively call T on , first with 0 substituted for x
and then 1 substituted for x. If either result is accept, then accept;
otherwise, reject.
3. If  is x, recursively call T on , first with 0 substituted for x
and then 1 substituted for x. If both results are accept, then accept;
otherwise, reject.”
Analysis: Let m be the number of variables that appear in . The depth of recursion
does not exceed m. And at each level of recursion, we need only store the value of one
variable. So, the total space used is O(m), and hence linear in the size of .
To complete the proof of Theorem 8.9, we also need to show that TQBF is PSPACE-
hard. A a detailed technical proof of this part is technically somewhat trickier than (but
otherwise similar to) the proof of the Cook-Levin theorem, and we omit it.

6. a game between two players A and E.
Formulas as games
Giorgi Japaridze Theory of Computability
Each fully quantified Boolean formula (in prenex form)  can be seen as
a game between two players A and E.
If  =x(x), it is E’s move, who should select x=0 or x=1, after
which the game continues as (0) or (1), respectively.
If  =x(x), it is A’s move, who should select x=0 or x=1, after
which the game continues as (0) or (1), respectively.
The play continues until all quantifiers are stripped off, after which
E is considered the winner iff the final, variable-free formula, is true.
xyz[(xy)(yz)(yz)]
E moves, selects x=1
yz[(1y)(yz)(yz)]
A moves, selects y=0
z[(10)(0z)(0z)]
E moves, selects z=1
(10)(01)(01)
A wins
Who has a winning strategy (a strategy that guarantees a win no matter
how the adversary acts) in this example?
--- Player A has a winning strategy: No matter what E does, select y=0.

7. The FORMULA-GAME problem
Giorgi Japaridze Theory of Computability
Who has a winning strategy in
xyz[(xy)(yz)(yz)] ?
E: Select x=1, and select z to be the negation of whatever A selects for y.
FORMULA-GAME = {<> | Player E has a winning strategy in }
Theorem FORMULA-GAME is PSPACE-complete.
Proof . This is so for a simple reason: we simply have
FORMULA-GAME = TQBF.
To see this, observe that  is true iff player E has a winning strategy in
it. A detailed proof of this fact (if it was necessary) can proceed by
induction on the length of the quantifier-prefix of .

8. The child’s game Geography
Giorgi Japaridze Theory of Computability
Players, called I and II, take turns naming cities from anywhere in the world (player
I starts). Each city chosen must begin with the same letter that ended the previous city’s
name. Repetitions are not permitted. The player who is unable to continue loses.
We can model this game with a directed graph whose nodes are the cities of the
world. There is an edge from one city to another if the first can lead to the second
according to the game rules. One node is designated as the start node/city. The condition
that cities cannot be repeated means that the path that is being spelled must be simple.
Peoria
Austin
Nashua
Albany
Orsay
...
Tokyo
Amherst
Tuscon
Oakland

9. Generalized Geography
Giorgi Japaridze Theory of Computability
In Generalized Geography, we take an arbitrary digraph with a designated start node
instead of the graph associated with the actual cities.
Who has a winning strategy here? 4 7 2 1 5 9 8 3 6

10. Generalized Geography
Giorgi Japaridze Theory of Computability
In Generalized Geography, we take an arbitrary digraph with a designated start node
instead of the graph associated with the actual cities.
Who has a winning strategy here? 4
Player I: Choose 3. 7 2 1 5 9 8 3 6

11. Generalized Geography
Giorgi Japaridze Theory of Computability
In Generalized Geography, we take an arbitrary digraph with a designated start node
instead of the graph associated with the actual cities.
Who has a winning strategy here? 4
Player I: Choose 3. II will have to choose 5. 7 2 1 5 9 8 3 6

12. Generalized Geography
Giorgi Japaridze Theory of Computability
In Generalized Geography, we take an arbitrary digraph with a designated start node
instead of the graph associated with the actual cities.
Who has a winning strategy here? 4
Player I: Choose 3. II will have to choose 5.
Now I selects 6, and II is stuck. 7 2 1 5 9 8 3 6 4
Who has a winning strategy here? 7
Indeed, if I starts by choosing 3 as
before, II chooses 6 and wins.
Player II. 2 1 5 9 8 3 6

13. Generalized Geography
Giorgi Japaridze Theory of Computability
In Generalized Geography, we take an arbitrary digraph with a designated start node
instead of the graph associated with the actual cities.
Who has a winning strategy here? 4
Player I: Choose 3. II will have to choose 5.
Now I selects 6, and II is stuck. 7 2 1 5 9 8 3 6 4
Who has a winning strategy here? 7
Indeed, if I starts by choosing 3 as
before, II chooses 6 and wins.
Player II. 2 1 5 9
Now assume I starts with 2. 8 3 6

14. Generalized Geography
Giorgi Japaridze Theory of Computability
In Generalized Geography, we take an arbitrary digraph with a designated start node
instead of the graph associated with the actual cities.
Who has a winning strategy here? 4
Player I: Choose 3. II will have to choose 5.
Now I selects 6, and II is stuck. 7 2 1 5 9 8 3 6 4
Who has a winning strategy here? 7
Indeed, if I starts by choosing 3 as
before, II chooses 6 and wins.
Player II. 2 1 5 9
Now assume I starts with 2. Then II responds
with 4. 8 3
If I responds with 5, II responds with
6 and wins.
Otherwise, if I takes 7, II wins
with 9. 6

15. GG and its PSPACE-completeness
Giorgi Japaridze Theory of Computability
GG = {<G,b> | Player I has a winning strategy for the Generalized
Geography game played on graph G starting at node b}
Theorem GG is PSPACE-complete.
Proof idea. A recursive algorithm similar to the one used for TQBF in
Theorem 8.9 determines which player has a winning strategy. This
algorithm runs in polynomial space and so GGPSPACE.
To prove that GG is PSPACE-hard, we give a polynomial time
reduction from FORMULA-GAME to GG. This reduction converts a
formula game  to a generalized geography graph G so that play on G
mimics play in . In effect, the players in the generalized geography
game are really playing an encoded form of the formula game.
On the following slides we give a more detailed argument.

16. The following algorithm obviously decides GG:
8.3.j
Why GGPSPACE
Giorgi Japaridze Theory of Computability
The following algorithm obviously decides GG:
M = “On input <G,b>, where G is a digraph and b is a node of G:
1. If b has outdegree 0, reject, because Player I loses immediately.
2. Remove node b and all connected arrows to get a new graph H.
3. For each of the nodes b1,...,bk that b originally pointed at,
recursively call M on <H,bi>.
4. If all of these accept, Player II has a winning strategy in the
original game, so reject. Otherwise, Player II doesn’t have a
winning strategy, so Player I must; therefore, accept.
Analysis: Let m be the number of nodes in G.
The only space required by this algorithm is for storing the recursion
stack. Each level of the recursion adds a single node to the stack, and at
most m levels occur. Hence the algorithm runs in linear space.

17. Why GG is PSPACE-hard (a)
8.3.k
Why GG is PSPACE-hard (a)
Giorgi Japaridze Theory of Computability
Consider any formula-game . We may assume that both its first and
last quantifiers are , and that the quantifiers strictly alternate. If not, we
can easily bring  to this form by adding dummy variables and quantifiers.
Furthermore, we can assume that the quantifier-free part of  is a
3cnf-formula, otherwise it can be converted to such. All these
conversions yield an equivalent formula and take a polynomial amount
of time.
Thus,
 = x1x2x3x4... xk[c1c2...cm],
where each ci is a disjunction of three literals.
Now we describe a way how to convert (in polynomial time)  into
<G,b> such that
FORMULA-GAME iff <G,b>GG,
i.e., E has a winning strategy in  iff Player I has a winning strategy in
<G,b>.

18. Why GG is PSPACE-hard (b)
8.3.l
Why GG is PSPACE-hard (b)
Giorgi Japaridze Theory of Computability 1
The left part of G will look like this: for each
variable we create a diamond. The blue arrows
indicate Player I’s choices (turns), and the red
arrows indicate Player II’s choices in the game. b x1 x2
The left-hand choices will correspond to choosing
1 in the formula game, and the right-hand choices
correspond to choosing 0. x3
Then we extend this graph by adding to it the
right part as shown on the next slide for a
particular example of .
... xk
x1x2x3...xk[(x1x2x3)(x2x3...)...(...)]
x1x2x3x4... xk[c1c2...cm]

19. Why GG is PSPACE-hard (c)
8.3.m
Why GG is PSPACE-hard (c)
Giorgi Japaridze Theory of Computability 1
Suppose E has a winning
strategy in . Then, let Player I
follow the “same” strategy in
the left part of the graph.
Whatever ci is chosen by
Player II, it is a true clause
and thus has a true literal. Let
then Player I choose such a
true literal. Then Player II is
stuck, as the path has already
passed through the
corresponding left or right
node of the corresponding
diamond. So, Player I has a
winning strategy in G. x1 b x1 x2 c1 x3 x2 x2 x3 c2 c x3
...
... cm xk
x1x2x3...xk[(x1x2x3)(x2x3...)...(...)]
x1x2x3x4... xk[c1c2...cm]

20. Why GG is PSPACE-hard (d)
8.3.n
Why GG is PSPACE-hard (d)
Giorgi Japaridze Theory of Computability
Suppose now A has a winning
strategy in . Then, let it
follow the “same” strategy in
the left part of the graph.
Then there is a false clause ci,
and let Player II choose that
clause. Now, whatever literal
of ci is chosen by Player I, it
is a false literal and hence the
path has not passed through it.
So, Player II is not stuck, and
goes to the corresponding left
or right node of the
corresponding diamond. Now
Player I is stuck. Thus, II has
a winning strategy in G. 1 x1 b x1 x2 c1 x3 x2 x2 x3 c2 c x3
...
... cm xk
x1x2x3...xk[(x1x2x3)(x2x3...)...(...)]
x1x2x3x4... xk[c1c2...cm]