Space Complexity

1
Space Complexity
2

• Def
• Let M be a deterministic Turing Machine that
  halts on all inputs.
• Space Complexity of M is the function fN?N,
  where f(n) is the maximum number of tape cells
  that M scans on any input of length n.

Def Let f N?N be a function. SPACE(f(n))L L
is a language decided by an O(f(n)) space
DTM NSPACE(f(n))L L is a language decided by
an O(f(n)) space NTM
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• Eg.
• SAT is NP-complete.
• SAT can be solved in linear(O(m)) space.
• M1On input lt?gt, where ? is a Boolean formula
• For each truth assignment to the variables x1,xm
  of ?.
• Evaluate ? on that truth assignment.
• If ? ever evaluate to 1, ACCEPT if not, REJECT.

x1 x2 x3 xn
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Let ALLNFAltAgt A is a NFA and L(A)?

• Eg.
• ALLNFA can be decided in nondeterministic linear
  space, but not known to be in NP or co-NP.
• NOn input ltMgt where M is an NFA
• Place a marker on the start state of the NFA.
• Repeat 2q times, where q is the number of states
  of M.
• Non-deterministically select an input symbol
• and change the positions of the markers
  on Ms
• states to simulate reading that symbol.
• Accept if Stages 2 and 3 reveal some string that
  M rejects, that is, if at some point none of the
  markers lie on accept states of M. Otherwise,
  reject.

Why 2q ?
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Savitchs Theorem(1970)

• For any function fN?N, where f(n)?n,
• NSPACE(f(n))?SPACE(f2(n)).
• pf
• N an non-deterministic TM deciding a language A
  in space f(n).
• Goal Construct a det TM M deciding A in space
  f2(n).
• Let ? be an input string to N.
• t integer
• c1,c2 configurations of N on ?.

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• CANYIELD(c1,c2,t) accept if starting from c1 N
  has a branch entering configuration c2 in t
  steps o/w rejects.
• CANYIELDOn input c1,c2 and t
• If t1, test whether c1c2 or c1 c2, then
  accept else reject.
• If tgt1, then for each configuration cm of N on ?
  using space f(n).
• Run CANYIELD(c1,cm, )
• Run CANYIELD(cm,c2, )
• If 3 4 both accept, then accept else reject.

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• caccept accept configuration.
• cstart start configuration of N on ?.
• Select a constant d so that N has no more than
  2df(n) configurations using f(n) tape space, n
  ? .
• MOn input ?
• output the result of CANYIELD(cstart, caccept,
  2df(n)).
• CANYIELD is a recursive procedure
• Recursive depthlog22df(n)O(f(n))
• Each recursive call of CANYIELD use O(f(n))
  space.
• Thus, it uses O(f2(n)) space.

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• PSPACE is the class of languages (or problems)
  that are decidable in polynomial space on a det.
  TM.
• PSPACE SPACE(nk).
• Thm
• P?NP ? PSPACENPSPACE ?EXPTIME TIME( )
• Def (PSPACE completeness)
• A language (or problem) B is PSPACE-complete if
  it satisfies 2 conditions
• B is in PSPACE, and
• Every A in PSPACE is poly-time reducible to B.

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universal quantifier,
existential quantifier,
? quantified Boolean formula fully
quantified Boolean formula
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• TQBF given a fully quantified Boolean formula,
  is it true?
• Thm TQBF is PSPACE-complete.
• pf
• TQBF is in PSPACE
• TOn input ? a fully quantified Boolean
  formula
• 1) If ? has no quantifier, then it is an
  expression with only constants, so evaluate ? and
  accept if it is true otherwise reject.

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• If ? ?x?, recursively call T on ?, first with
  x0, and then x1. If either result is accept,
  then accept o/w reject.
• If ? ?x?, recursively call T on ? with x0, and
  x1. If both results are accept, then accept o/w
  reject.
• TQBF is PSPACE-hard
• Let A be a language decided by a TM M in space nk
  for some constant k.
• Show A?pTQBF.

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• M accepts ? iff ? is true.
• c1,c2 configurations, t integer
• Construct a formula ? c1,c2,t
• ? ?Cstart,Caccept,h, h2df(n).
• ? c1,c2,t?m1,?(c3,c4)?(c1,m1),(m1,c2) ?c3,c4,
  
• Configuration size O(f(n))
• Recursion depth lg2df(n)O(f(n))
• Formula size O(f2(n))
• ?c1,c2,t?m, ?c1,m, ? ?m,c2,
• What is the formula size?

f
?
?

• x ?y, z... ?
• x(xy)?(xz)?...

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Winning Strategies for games

• ? ?x1 ?x2 ?x3Qxk?
• Q represents either a ? or an ? quantifier.
• Two players player A player E.
• Take turns selecting the values of x1,, xk.
• Player A selects values for the variables that
  are bound to ?.
• Player E selects values for the variables that
  are bound to ?.
• Player E wins, if ? is TRUE.
• Player A wins, if ? is FALSE.

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Eg.

• Let ?1 ?x1 ?x2 ?x3(x1? x2) ? (x2 ? x3) ? (x2?
  x3)
• Player E has a winning strategy.

• Let ?2 ?x1 ?x2 ?x3(x1? x2) ? (x2 ? x3) ? (x2?
  x3)
• Player A has a winning strategy.

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• FORMULA-GAMElt?gt player E has a winning
  strategy in the formula game associated with ?.
• Thm FORMULA-GAME is PSPACE-complete.
• pf
• ??TQBF exactly when ??FORMULA-GAME.

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Geography Game
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Generalized geography(GG)

• Eg.
• Player 1 moves first.
• Player 2 moves second.
• Player 1 has a winning strategy
• Starts at node 1,
• Player 1 choose 3
• Player 2 choose 5
• Player 1 choose 6
• Player 2 cannot move anywhere
• Thus, player 1 wins.

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• GGltG,bgt player 1 has a winning strategy for
  the generalized geography game played on graph G
  starting at node b.
• Thm GG is PSPACE-complete.
• Pf
• MOn input ltG,bgt, G directed graph b a node
  of G.
• 0. REJECT if b has no outgoing edge.
• 1. Remove node b and all arrows touching it to
  get a new graph G1.
• 2. For each of the nodes b1,b2,,bk that b
  originally pointed at, recursively call M on
  ltG,bigt.
• 3. If all of these accept, player 2 has a winning
  strategy in the original game, so REJECT. O/w
  player 2 doesnt have a winning strategy, so
  player 1 has, thus ACCEPT.

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• The only space required is for storing the
  recursion stack.
• The level of recursion can be at most m, which is
  the number of node in G.
• Each level uses poly. Space, thus in total M uses
  poly. space.

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• Next need to show GG is PSPACE-hard
• FORMULA-GAME?pGG.
• ? ?x1 ?x2 ?x3Qxk?
• ?p
• ltG,bgt
• Player 1 in GG player E in formula game.
• Player 2 in GG player A in formula game.

In conjunctive normal form.
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? ?x1 ?x2 Qxk(x1? x2? x3) ? (x2? x3? ) ??
()
For simplicity, let Q?
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• 1)
• If ? is FALSE, player 2 may win by selecting the
  unsatisfied clause.
• Then any literal that player 1 may pick is FALSE
  and is connected to the side the diamond that
  hasnt been played.
• Thus player 2 may choose the node in the diamond,
  then player 1 cannot move and lose.

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• 2)
• If ? is TRUE, any clause that player 2 picks
  contains a TRUE literal.
• Player 1 choose that TRUE literal.
• ?it is true, ? it is connected to the side of the
  diamond that has been chosen.
• So, player 2 cannot move and lose.

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• The class of L and NL
• LSPACE(log n)
• NLNSPACE(log n)
• Eg.
• PATHltG,s,tgt G is a directed graph that has a
  directed path from s to t.
• PATH Given a directed graph G and two vertices s
  and t, is there a directed path from s to t?

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• Def If M is a TM

read only input tape working tape
A configuration of M on ? is a setting of the
state, work tape, and the position of the tape
heads.
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• If M runs in f(n) space on ?, ?n, then the
  number of configurations of M on ? is
• gf(n)f(n) c n ? 2O(f(n))
• c state number.
• g number of tape symbol

Possible work tape.
Head position of input head.
State position.
Possible state.

f(n)
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• Savitchs theorem is true for f(n)?lgn
• note that the definition of configurations is
  different for small space case. But the proof is
  the same.

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NL-completeness NL?L

• Def (log space reducibility)
• A log space transducer is a TM with (1) a
  readonly input tape (2) a write-only output
  tape (3) a read/write work tape.
• A log space transducer M computes a function
  f???, where f(n) is the string remaining on
  the output tape after M halts when it starts with
  ? on input tape. f is called log space computable
  function.
• A?LB, if A is mapping reducible to B by using a
  log space computable function.

May contain O(logn) symbols.
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• Def A language B is NL-complete if
• B?NL and
• Every A in NL is log space reducible to B.
• Thm If A ?LB and B ?L then A?L.
• Cor If any NL-complete language is in L then
  LNL.

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Thm PATH is NL-complete.

• Pf
• It has been shown in class that PATH?NL.
• Need to show that for any A?NL, A?LPATH.
• Let M be a NTM that decides A in O(log n) space.
  Given an input ?, construct a graph G and 2
  vertices s and t in log space, where G has a
  directed path from s to t iff M accepts ?.
• V(G) configurations of M on ?.
• (c1,c2)?E(G) if c1 c2.
• s start configuration, t accept configuration.

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• Need to show that reduction operates in log
  space. I.e. need to describe Gs vertices and
  edges in log space.
• Each configuration needs c logn space, for some
  constant c.
• Try all c logn bit string and test with M on ? to
  see if it is a legal configuration, if yes, put
  it into vertex set.
• For any c1, c2, test if c1 c2 with M on ?.

Takes O(log n) space.
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• Cor NL? P.
• PATH ?P ? NL?P (?)
• co-NL B B is in NL
• B x x ?B
• Thm NL co-NL.

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Thm NL co-NL.

• Robert Szelepcsenyi, The method of forcing for
  nondeterminitic automata, Bull of the EATCS, 33,
  p.96-100, 1987.
• N. Immerman, Nondeterministic space is closed
  under complementation, SIAM J. on computing 17,
  p.935-938, 1988 .

Gödel award.
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Thm NL co-NL.

• pf
• PATH is NL-complete ? PATH ? co-NL.
• If ltG,s,tgt ?PATH, then there is no directed path
  in G from s to t.
• Every A ?co-NL has A?LPATH.
• The rest shows PATH ?NL.

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• MOn input G, s, t
• c01 / counter
  /
• For i0 to m-1 / mof vertex /
• ci10
• d0
• For each node v in G
• For each node u in G
• Non-det. Either skip or perform the
  following steps
• Non-det. Follow a path of length i
  from s, and if
• non of the nodes met are u,
  reject.
• Increase d.
• If (u,v) ?E(G), then increase ci1
  and goto 6
• with next v.
• 11. If d?ci, then reject.

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• 12. For each node u in G
• 13. Non-det., either skip or perform these steps
• 14. nondeterminitically follow a path of
  length i from s
• and if none of the nodes met are u,
  reject.
• 15. If ut, then reject.
• 16. Increase d.
• 17. If d?cm, then reject, otherwise accept.
• The algorithm needs to store ci, d, i, j and a
  pointer to the head of a path . Thus, it runs in
  log space.

Aiu u is reachable from s within i
steps. CiAi, s Ai?Ai1.
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Nondeterminitic Space is Closed under Complement

• NSPACE co-NSPACE.
• L?NSPACE ? L?co-NSPACE.
• Robert Szelepcsenyi The method of forcing for
  nondeterministic automaya, Bull of the EATCS,
  33, p. 96-100, 1987.
• N. Immerman Nondeterministic space in closed
  under complementation, SIAM J. on computing, 17,
  p. 935-938, 1988.

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• M a nondeterminitic S(n)-bounded-space TM.
• Each configuration of M can be described in
  O(S(n)).
• c0 Initial configuration of M.
• RchM,x(t)
• The number of configurations reachable by
  computation on input x in at most t steps
  starting at c0.

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• M0 Input x, n, t, u
• accept if u is reachable in t1
  steps.
• m0
• for each configuration v of M do
• non-det simulate M on x during at
  most t steps, checking if v is reached.
• if so, then
• mm1
• if M on input x goes from v to u
  then accept.
• if mn then reject
• else abort by halting in a nonfinal state.

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• M1 Input x
• n1 t0
• for t from 1 to t(x) do
• m0
• for each configuration u do
• if u is reachable from c0 in t1 steps
• then mm1
• nm

Call M0(x,n,t,u)
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• n is RchM,x(t) for each t.
• Thm If S(n)?log n and S is space constructible
  then NSPACE(S) is closed under complementation.
• pf
• Call M1 on x to compute RchM,X(tx).
• For each configuration u call M0 on
  ltx,n,t(x),ugt.
• If No accepting configuration accessible, then
  accept.

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Input M, s, x. Compute initial conf. C0.

• NTM accept L(M).

• I10
• for each conf. C do
• if C0 1C then I1I11
• DI1 K1
• While Klt2S(n)
• I20
• for each conf. C2 do
• I10
• for each conf. C1 do
• guess path C0 ?K C1
• if ? a correct path, I1
• if C1 1 C2 then
• if C2 is accept conf., STOP
  REJECT else I2 exit inner loop
• if I1?D, then STOP
• if (DI2) then ACCEPT else K DI2

I1 Rch(k-1) I2 Rch(k)