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LECTURE 2-9
PSPACE-Complete
SPACE BOUNDED HALTING PROBLEM (SBHP)
Given an DTM M , an input x and a string 0 s ,
can M accept x using space at most s?
THEOREM
SBHP is  mp -complete for PSPACE.
PROOF
Construct an universal DTM M u to simulate M
on input x within space s. Then SBHP  L( M u )  PSPACE.
A  PSPACE  A  L( M ) for an DTM M with
polynomial space bound p.
For input x of M , construct M , x, p(| x | ) .
QUANTIFIED BOOLEAN FORMULA
(x1 )(y1 )(x2 )(y2 )( x1  y1  x2 )( x1  y1  y2 )
QBF
Given a quantified Boolean formula F ,
det ermine whether F is true.
THEOREM
QBF is  mp -complete for PSPACE.
PROOF
A  PSPACE  there exist B  P and polynomial q
such that for any x   * with | x | n,
x  A  (Q1 y1 :| y1 | q (n))(Q2 y2 :| y2 | q (n)) 
(Qk yk :| yk | q (n)) x, y1 ,..., yk  B
where k  q (n).
To show A  mp QBF,
for each x  *, we construct F such that
x  A  (Q1 1 : X 1  {0,1})(Q2 2 : X 2  {0,1})
(Qk k : X k  {0,1}) F |  1 ,  2 ,...,  k  1
Let B  L( M ) where M has polynomial - time bound p.
x  A iff (y1 :| y1 | q (n))(y2 :| y2 | q(n)) 
(yk :| yk | q (n))[ there is a computatio n path of M ,
( 0 , 1 ,...,  m ), m  p (| x, y1 , , yk |),
|  i | p (| x, y1 , , yk |)
such that
(a)  0 is the initial configurat ion;
(b)  m contains the final state;
(c)  i   i 1 ;
Let x, y1 ,, yk  x # y1 # # yk .
Then p( x, y1 ,, yk )  n  k (q(n)  1)  r (n).
zi , j , a
 1 if cell (i, j ) holds symbol a

0 otherwise
Construct Boolean formulas f1 , f 2 , f 3 , f 4 such that
f1  1  Each cell holds exactly one symbol.
f 2  1  (a) holds, i.e.,  0 is the initial configurat ion;
f 3  1  (b) holds, i.e.,  m contains the final state;
f 4  1  (c) holds, i.e.,  i   i 1 ;
 0  [ x1 # s]x2  xn # y1 ## yk B  B
f 2  z11, x1 # s z12, x2  z1n , xn z1,n 1,# z1,n  2, y11  z1,n 1 q ( n ), y1,q ( n ) z1,n  2 q ( n ),#
 z1,r ( n ), B
Obtain f '2 by replacing each z1, j , yi ,h with
( xi ,h  z1, j ,1 )( xi ,h  z1, j , 0 ).
Set F  f1  f '2  f 3  f 4 . Then
x  A  (Q1 1 : X 1  {0,1})(Q2 2 : X 2  {0,1})
(Qk k : X k  {0,1})( k 1 : {zija}  {0,1}) F |  1 ,  2 ,...,  k  k 1  1
3-QBF
Given 3 - CNF quantified Boolean formula F , determine
whether F is true .
THEOREM
3 - QBF is  mp -complete for PSPACE.
Geography Game
This is a 2 - player game on a directed graph G with a marked
node s. Initially, player 0 marks on a node u such that edge
( s, u ) exists. Then player 1 has to mark on a node v such that
edge (u , v) exists. Next, player 0 has to mark a node w such
that edge (u , w) exists until a player cannot find a node to
mark. In such a case, another player win s.
Example 1
BLUE ALWAYS WINS
Example 2
RED WINS
Example 2
BLUE WINS
IF RED DOESN’T MAKE A MISTAKE, THEN
BLUE CANNOT WIN.
Example 3
BLUE HAS A WINNING STRATEGY!!!
Geography Problem
Given a directed graph G with a start node u, determine
whether player 0 has a winning strategy.
WIINNIG STRATEGY
(1st move of player 0)( 1st move of player 1)
( 2nd move of player 0) 
( kth move of player 0)[player 1 cannot make futher move]
THEOREM
p
GEOGRAPHY is  m -complete for PSPACE.
PROOF
3 - QBF  mp GEOGRAPHY
Construction
Boolean formula F ( X , Y )  G, s 
Such that
( 1 : x1  {0,1})(1 : y1  {0,1}) ( k : xk  {0,1})
( k : yk  {0,1})[ F | ,  1]
 [player 0 has a winning strategy on  G, s ]
( 1 : x1  {0,1})(1 : y1  {0,1})( 2 : x2  {0,1})
( 2 : y2  {0,1})[( x1  y1  x2 )( y1  x2  y2 )( x1  x2  y2 ) |  ,   1]
 BLUE HAS A WINNING STRATEGY!!!
c1
x1
x1
y1
x2
y1
x2
c2
y2
y2
c3
(x1 )(y1 )(x2 )(y2 )( x1  y1  x2 )( y1  x2  y2 )( x1  x2  y2 )
Questions
• Is go game on 19x19 grid PSPACEcomplete?
• What do we mean that the go game is
PSPACE-complete?