Peter van Emde Boas: Games and Complexity, Guangzhou 2009 THE PEBBLE GAME Games and Complexity Guangzhou 2009 Peter van Emde Boas References available at: http://staff.science.uva.nl/~peter/teaching/gac09.html © Games Workshop © Games Workshop
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
THE PEBBLE GAME
Games and Complexity
Guangzhou 2009
Peter van Emde Boas
References available at: http://staff.science.uva.nl/~peter/teaching/gac09.html
© Games Workshop© Games Workshop
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Stoning of st Stephen
Acts VIII 57--60
Doré bible
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Origin Pebble Game
7 + (1+x)(5-z) - ((1+x)/(u-t) + 2z)/v
+
+
+
-
-
/
/
-
*
*7
1 x5 z u t
2 z
v
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Origin Pebble Game
+
+
+
-
-
/
/
-
*
*7
1 x 5 z u t2 z
v
r1 = 1r2 = xr3 = r1 + r2r1 = 5r2 = zr4 = r1 - r2r1 = 7r2 = r3 * r4r5 = r1 + r2.... How to evaluate this expression on a
Von Neuman Computer using fewregisters and little time?
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Formalization
nodeinputoutput
Storing value in register =placing a pebble on the node
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Formalization
Directed Acyclic GraphBounded IndegreeMultiple Outputs & Sharing
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Rules of the Pebble Game
• 1: You can always pebble an input
• 2: You can always remove a pebble
• 3: You may pebble a node provided its ancestors are pebbled
• 4: You must pebble all outputs at least once
• Use as few pebbles as possible
• Use as few moves as possible
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Interpretation
• 1: You can always pebble an input• 2: You can always remove a pebble• 3: You may pebble a node provided its
ancestors are pebbled• 4: You must pebble all outputs at least
once• Use as few pebbles as possible• Use as few moves as possible
load
free
compute
# registers
time
results
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Alternative Version
3: You may pebble a node provided its ancestors are pebbled
Ri = Rj § Rk
3a: You may shift a pebble from an ancestor to a node provided its ancestors are pebbled
Ri = Ri § Rk
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Impact Alternative Rule 3a
If the DAG entirely consists of isolated nodesboth versions allow a pebbling with a singlepebble in real time.
In all other cases rule 3a saves exactly onepebble.
Proof by transformation of pebbling strategies
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Impact Alternative Rule 3a
G DAG;
S(G): # pebbles reqired to pebble G in standard game
S’(G): # pebbles requied to pebble G using rule 3a
S(G) = S’(G) = 1 iff G is a collection of isolated nodes
S(G) = S’(G) + 1 otherwise
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
ProofsIsolated Nodes case : trivial
S’(G) ≤ S(G) : don’t use 3a
S(G) ≤ S’(G) + 1 : substitute shift by a place/remove combination of moves,
S’(G) ≤ S(G) - 1 , for non isolated case:follow an optimal standard play andremove the peaks in pebble consumption
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
ProofsMove j requires peak consumption ( S(G) pebbles) so the next move j+1 removes a pebble; what sort of move was it:
a: remove a pebble: not applicable : this can’t provokepeak consumption
b: pebble a node (with ancestors): put x ; remove y, x ≠ y :if y is ancestor of x play a shift; otherwiseinterchange move j and move j+1 : this saves apebble without loss of time
c: pebble an input: put x ; remove x (only useful if x isalso an output) . Shift these moves to start of play.
d: pebble an internal output: put x ; remove x : replacemove by a shift (saving a pebble) , and remove everything and start all over but forget output x.This saves a pebble but causes loss of time.
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Consequences
In the worst case the transformed strategy requires an (almost complete) replay forevery output.
This may square the time of the play
Worst case example is easy to obtain.in the standard game this DAG requires2 pebbles and runs in real time. With rule3a a single pebble suffices in quadratic time.
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Cook’s Pyramid
C(7) THEOREM:
C(k) requiresprecisely k pebbles
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Cook’s Pyramid
input node blocking the last free path
When the last free pathto an input is blockedthere are k-2 disjointside paths, each havingat least one pebble on it.This blocking places thek-1-th pebble
The next move is a placement of the k-thpebble
C(k)k=7
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Saving the last pebble?Saving the last pebble may require recomputationwhich requires possibly substantial timeHow bad can it be?
T(G,k) : time needed for pebbling DAG G in the standard game using S(G) + k pebbles
T’(G,k) : time needed for pebbling DAG G in the standard game using S’(G) + k pebbles
T’(G,0) T(G,0) T’(G,1) T(G,1) ..... T’(G,k) T(G,k) ...
less rules butextra pebble
extra rule butsame numberof pebbles
Eventually enough pebbles available for real time pebbling
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Exponential Gap
G[ k + 1 ]
G[ k ]
u
v
T[ k+1 ]
H[ k+1,k+1 ]
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
RequirementsG[ k ] can be pebbled in real time (shifting rules)with k+1 pebbles; doing it with k pebbles requirestime > k!
T[ k +1] : a component requiring k+1 pebbles to reach u
H[ k+1,k+1 ] : with k+2 pebbles all outputs can be pebbledin any required order in real time
with k+1 pebbles any output can be pebbled, butat this move all other outputs obtain free path to v
Example components: T[k+1] node with k+1 inputs;H[k+1,k+1] complete bipartite graph on k +1 + k +1 nodes.These components are not binary!
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Proof by Induction
Theorem: if G[ k ] satisfies the requirements so does G[ k+1 ]
With k + 2 pebbles: pebble u ; keep a pebble on thepath from u to v while pebbling G[k] with k+1 pebbles in real time. Having reached v pebble the k+1successors. Remove the pebble from v and use iton the k+1 outputs.
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Proof by Induction
Theorem: if G[ k ] satisfies the requirements so does G[ k+1 ]
With k + 1 pebbles: pebble u ; keep a pebble on thepath from u to v while pebbling G[k] with k pebbles.This now requires time > k!. Having reached v pebble the k+1 successors, with a shift for the last one. Nextuse a shift to pebble an output. Now v becomes visible (and so u and its inputs) and we must start to repebble u which requires all our pebbles.Consequently we have to pebble the embedded G[k]k+1 times using k pebbles. Total time > (k+1)k! = (k+1)!
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Binary Components
For T[ k ] : Take Cook’s Pyramid C(k+1) with base k (requires kpebbles in the shifting game)
For H[k,k] : a pyramidal grid of height k*(k+1)
Properness condition: pebblesstay “close” together. Whenviolated free paths are created.
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
A General Upper BoundGiven a DAG G with n nodes and max indegree k;How many pebbles may be required in the worst case?
Sk(G) n (trivial bound)
without bound on the indegree this can’t be improved!
THEOREM: Sk(G) Cn / log(n)where the constant C depends on k
Sk(G) = O( n/ log(n) )
n-1
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
A General Upper BoundEquivalent Formulation:
Rk(m) := least number of edges in a DAG with max indegree k which requires m pebbles.
THEOREM: Rk(m) D m log(m)where the constant D depends on k
Rk(m) = ( m log(m) )
R’k(m) := least number of nodes in a DAG with max indegree k which requires m pebbles.R’k(m) Rk(m) k R’k(m)
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Equivalence?
G n nodes max indegree k requiring m pebbles
m C n / log(n) <==> n 1/C m * log(n)<==> n D m * log(m)
provided log(n) log(m) for these worst case DAGs
Evidently m n , but the Cook Pyramid alreadyprovides us with examples where m nshowing that for the worst case m, log(m) log(n)/2so indeed log(n) log(m)
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Proof by induction
inputs
output
Gcheappart
expensivepart
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Induction
inputs
output
Gcheappart: G1
expensivepart: G2
G requires m pebblescheap part: nodes can be pebbles with m/2 pebblesexpensive part: nodes require > m/2 pebbles.
three classes of edges:
edges within G1 : E1edges within G2 : E2edges from G1 to G2 : Dno edges from G2 to G1
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Induction
inputs
output
Gcheappart: G1
expensivepart: G2
Some node in G1 needsat least m/2 - k pebbles,otherwise an “input” ofG2 becomes “cheap”
So G1 contains at leastRk(m/2 - k) edges.
Some node in G2 (in isolation) requires m/2 - k pebbles,otherwise the entire G can be pebbled with <m pebbles.
So G2 contains at least Rk(m/2 - k) edges.
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Counting cross edges in D
inputs
output
Gcheappart: G1
expensivepart: G2
Two cases:
A: D contains > m/4edges
B: D contains m/4edges
In case B part G2 (in isolation) must require at least 3m/4 pebbles!
A node with indegree j which requires t pebbleshas an ancestor requiring at least t - j pebbles.
j
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Counting Cross Edges in D
expensivepart
Walking down from the mostexpensive node to its mostexpensive ancestor we starta path to a node requiring tpebbles with m/2-k < t m/2.Remove the nodes along this path: this removes at least m/4edges.
So in both cases:
Rk(m) Rk(m/2 - k) + Rk(m/2 - k) + m/4
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Solving the Recurrence
H(m) H(m/2 - k) + H(m/2 - k) + m/4 + C
This solves to H(m) U m log(m) for some U
Let K(m) := H(m-2k) then K satisfies:
K(m) K(m/2) + K(m/2) + m/4 + (C-k/2) , so for m > 20(k/2 - C)
K(m) K(m/2) + K(m/2) + m/5
which solves to K(m) 1/5 m log(m) ( in fact K(m) ≥ 1/5 C’ m log(m) due to initial effect )
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Just a Upper Bound ?• Matching Lowerbound (n/log(n)) for
Sk(G) is known ( Paul, Tarjan & Celoni )
• Heavy Combinatorial and Probabilistic Graph Theory involved (or Algebraic Number Theory if you aim for explicit constructions...)– Superconcentrators– Grates – Communication Networks
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
The Hopcroft Paul Valiant Theorem
THEOREM: A j-tape TM in time T(n) can be simulatedin space S(n) = O( T(n)/log(T(n)) ) ( T(n) > n reasonable )
Consequence: TM’s can do more in space S(n) thanin time S(n) (requires standard diagonalization results)
Proof steps:1: Make computation block preserving2: Introduce Computation Graph of head configurations3: Establish correspondence between pebbling
strategies and simulation strategies4: Correlate Space and Pebble consumption5: Apply Upper Bound Result from Pebbling Theory6: Eliminate Nondeterminism
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
k-tape Turing Machine
Finite Control
input tape
work tapes
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Block Preserving
Finite Control
input tape
work tapes
Compute for k steps; only cross borders at end of block of steps
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Block Preserving
w2 w5w4w3
w3R w6Rw5Rw4R
w1R w4Rw3Rw2R
By having three tracks on every tape information for a block of k steps is always accessible.After block of k steps the tracks must be updated.Must be done in block preserving way.Computation delayed by a constant factor.
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Head ConfigurationsFirst configuration in a block of k steps
Full Block Determined byA: Position input head & State Finite ControlB: Positions (also within block) on WorktapesC: Contents Worktape Blocks
Space required for head configuration:A: O(log(n))B: j. log(n)C: O(j.k)
But what for the next k steps ???!!!
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Configuration Graph
Nodes: Head Configurations: H(0), H(1), ....Edges: Logical Dependencies !
Type 0: Time dependency : H(i) ---> H(i+1)
Type a: Tape dependency : H(i) ---> H(m) (a)if the tape block on tape a visited duringblock m was last visited during block i
Claim: This Configuration Graph is a DAG withindegree bounded by j+1
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Configuration Graph
Number of nodes: T(n) / k
Secret: take k dependent on T(n) : k = (T(n))2/3
Now the Configuration Graph can be described in space O( T(n)1/3 . log(T(n)) )
Assume for a moment that the correctConfiguration Graph is Available, togetherwith the state and head position information.
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Computing and PebblingBlock H(m) can be simulated provided theHead Configurations of its Direct Ancestorsin the Configuration Graph are known.Missing Information: the Tape Block Contents !Solution: Write them down in Memory; [ Costs O(k) space per configuration ]and call it a pebble.
Number of Pebbles Required: O( T(n)1/3 / log(T(n) )Space needed for a pebble: O(k) = O(T(n)2/3)
Total Space Needed: O( T(n) / log(T(n)) )
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
The Wrong Theorem
THEOREM: DTIME(T(n)) NSPACE( T(n) / log(T(n)) )
Proof:a: Guess states and head positions of the T(n)1/3
head configurations; based on this information build the Configuration Graph
b: Play a pebble strategy with O(T(n)1/3/ log(t(n)) )pebbles, using nondeterminism for guessingthe right next move.
c: When the last configuration is pebbled read offthe answer
d: Abort on time-overflow (repeating moves is futile)
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Making this Simulation Deterministic
Step a: Replace “Guess and Verify” by “Try out allpossibilities”
This is possible, since we are economizing on space rather than time
Step b: Trying out all possibilities is not possible; they take too long and there are too many of them...
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Consulting the NPSPACE Oracle
what is the rightnext move ?
I can’t answer thatquestion
is move z the rightnext move ?
YES / NO
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Asking the Right QuestionGIVEN: the DAG G, with a given collection of pebbles
on it, and the number of the move tQUESTION: is z the correct next move in some
pebbling strategy using no more than m’ pebblesand succeeding within 2m - t moves.
This problem is solvable in NSPACE( m.log(m) )
SAVITCH: This problem is solvable in DSPACE( m2.log(m)2 )
Since in our situation m = O(T(n)1/3) this space consumption is acceptable.
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
A Reasonable Simulation?
Re-pebbling a node amounts to re-computing ablock of steps: time overhead!
Retrieving the correct next move in the pebble gamerequires a Savitch-based simulation for every move.
Trying out all possibilities for the configuration graphamounts to performing this awesome simulation anexponential number of times.
WHHAAARGHHH !
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Gilbert Lengauer & Tarjan
THEOREM: Deciding whether a given DAG can be pebbled with a given number of pebbles
is PSPACE-Complete.
Proof: Reduction from QBF
Requires: Intricate ConstructionGood Control over Conceivable pebblingStrategies
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Reference Frame
• Single output DAG’s only
• Pebbling Game with Shifting Rule
• Elimination of inessential moves
• Normalization of Strategies
• Non-standard Encoding of Propositional Valuations
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Essential Moves
The first pebbling of the output node is essential
The last pebbling of a internal node preceeding anessential pebbling of one of its successors isessential
That’s all folks.
Lemma: inessential pebblings can be eliminated.Lemma: inessential pebblings can be eliminated.
(G. L. & T. call these moves necessary)
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Essential MovesLet the placement on node x at time t1 be inessential.This pebble is removed at time t2.Claim: During the interval [t1,t2] all placements on directsuccessors of x are inessential.So during the interval [t1,t2] all placements on allconnected successors of x are inessential.In particular, the first pebbling of the output is outsideinterval [t1,t2] (or unconnected) .A move placing a pebble on an internal node y whichis not followed by a placement on some successorcan be removed. Such a move must occur after t1 !Eliminate it and use same argument again. Eventuallyall placements on connected successors of x are eliminated. Next remove the placement on x. QED.
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Essential Moves
A pebbling strategy without inessential movesis called Frugal.
For Frugal strategies:
After the first pebbling of some internal node somepath connecting it to the output is pebbled.
After the last pebbling of some internal node allpaths connecting it to the output are pebbled.
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Normal Strategies
C(7)
6
=
The DAG’s used in the reduction contain severalembedded Pyramids withthe purpose to force someorder on the pebbling ofthe embedded components.
In a Normal Strategy thesePyramids are pebbled inconsecutive sequences ofmoves, leaving at most a pebble on the Apex.
(7 --> 6 due to shifts)
Apex
Apex
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Normal Strategies
k
g
Assume Frugal Strategy.
First placement on P(k) at t0
Last Removal from P(k) at t1
Maximal pebble consumptionon P(k) at t2 (k pebbles); where t2 [to ,t1]
Reschedule all moves on P(k)to a block of steps at time t2
This is possible since P(k) hasno ancestors of its inputs....
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
The Construction
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
The Construction
input
output5 quantifiers
4 clauses with3 literals each
18 pebbles18 = 5*3 + 3
Pyramids arepebbled from top to bottom.Yet it mustremain possibleto pebble the path along theclauses....
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
The Construction
quantifier blocks
variablesetters
clause blocks
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
The Components
x x
x’ x’
x x
x’ x’
x x
x’ x’
x x
x’ x’
variable x true setting false setting double false setting
A formula can be made true by the illogical double falsesetting if its clauses are made true by other literals ....
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
The Components
clause C[ j ] = ( l1 l2 l3 )
l1’ l2’ l3’
l1 l2 l3pj-1
pj
If current setting makesclause true 3 pebblessuffice.
If current setting failsto make clause true4 pebbles are neededor some other pebble must move
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
The Components
k
k-1
k-2
xx
x’ x’
qi
ai
bi
ci
qi+1
di
fi
gi
i+1-th quantifiervariable: xUniversalk = 3(m-i) + 3
when pebbledk pebblesavailable
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
The Components
k
k-1
k-2
xx
x’ x’
qi
ai
bi
ci
qi+1
di
i+1-th quantifiervariable: xExistentialk = 3(m-i) + 3
when pebbledk pebblesavailable
fi
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Overall Idea
The embedded pyramids enforce that the quantifierblocks are pebbled in order, leaving a variableassignment. (Proper configurations in the block)For the i+1-th quantifier 3(m-i) + 3 pebbles areavailable. In a proper configuration exactly 3 areinside the block.The clauses part can be traversed (pebbled) whenthe formula evaluates to true for the assignment.For the existential quantifier a value can be chosen;for the universal quantifier the entire block must betreated twice, unless the double false assignmentalready makes the formula true.Nesting of universals causes exponential time!
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Variable Settings
k
k-1
k-2
xx
x’ x’
qi
ai
bi
ci
qi+1
di
fi
gi
k
k-1
k-2
xx
x’ x’
qi
ai
bi
ci
qi+1
di
fi
gi
k
k-1
k-2
xx
x’ x’
qi
ai
bi
ci
qi+1
di
fi
gi
true setting false setting
doubly false setting
The proper settings in the i+1-th quantifier block.Universal quantifierCalled Ni in paper
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Variable Settings
kk-1
k-2
x x
x’ x’
qi
ai
bi
ci
qi+1
di
fi
kk-1
k-2
x x
x’ x’
qi
ai
bi
ci
qi+1
di
fi
kk-1
k-2
x x
x’ x’
qi
ai
bi
ci
qi+1
di
fi
true setting false setting
doubly false setting
The proper settings in the i+1-th quantifier block.Existential quantifierCalled Ni in paper
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Induction Hypothesis: H(i)Given formula: Qx1 Qx2 .... Qxm[ F(x1,x2, ..., xm) ]
Partial truth assignment for x1,x2, ..., xi , say wNB! requires good understanding for what it means to assign doubly false! Think of substituting true/false and removing thedoubly false variables (and reduce to a proper logical form afterwards)
Equivalent are:Resulting QBF formula is true <==>Given the initial proper configuration for this wwith 3i pebbles on the i quantifier blocks, node qi
can be pebbled using 3(m-i) + 3 further pebblesleaving the 3i given pebbles fixed.
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Order of ProofH(m): All variables are assigned a truth value(or doubly false). The resulting formulaevaluates either to true or false. The DAG can be traversed up to qm iff the formula evaluates to true, and we have just 3 pebbles available....
H(0): No variable assigned yet and the DAG iscompletely empty. We have 3m + 3 pebbles.This is what we need for proving the Theorem...
Proof order: H(m) => H(m-1) => .... => H(0)
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
H(m): Base of Induction
Forward: If the fully assigned formula is trueevery clause block has at least one node inits base pebbled, so 3 more pebbles suffice.
Backward: If the fully assigned formula isfalse some clause block has an empty baseso at least 4 pebbles are required. Note thatother pebbles can’t be moved....
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
H(i+1) => H(i) , Induction Step
In all cases there are 3(m-i) + 3 pebbles available, 3 of which can remain on the i-th quantifier block.Note that 3(m-i) + 3 is also the order of the largestpyramid connected to this block.
When proper configuration is reached invokeinduction hypothesis.
Cases of Universal and Existential quantifierare treated separately.
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Forward Universal
k
k-1
k-2
x x
x’ x’
qi
ai
bi
ci
qi+1
di
fi
gi
x’ ; di; x’; (k, leave 3)x’ => x ;qi+1 (IH) ; (k-3)qi+1 => ci => bi => ai ;x’ => x ; remove x ;x’; (k-2)qi+1 (IH) ; (k-3)x’ => gi => fi => qi ;
If the doubly false assignment makes the formula truepebbles can remain on x’ and x’ . The at least 2 freepebbles can be used to traverse the block pebbling qi+1
just once.
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Forward Existential
kk-1
k-2
x x
x’ x’
qi
ai
bi
ci
qi+1
di
fi
x’ ; di; fi; fi => x’; (k, leave 3)if x = true then x’ => x ; qi+1 (IH) ; (k-3) x ; (1) qi+1 => ci => bi => ai => qi ;else { x = false } x’ => x ; qi+1 (IH) ; (k-3) qi+1 => ci => bi ; remove x ; fi => x’; (k-2) bi => ai => qi
fi
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Backward Universal
k
k-1
k-2
x x
x’ x’
qi
ai
bi
ci
qi+1
di
fi
gi
Since only k pebbles areavailable we must start withx’ ; di; x’;Before qi+1 is pebbled theonly possible actions toproceed are x’ => x, x’ => x or no move at all, alwaysleaving a proper configuration. By IH the involved partial assignmentmakes the formula true.From qi+1 either ai or fi canbe reached but not both;
which one depends onthe value assigned to xUnless x is assigneddoubly false qi+1 mustbe pebbled again.
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Backward Existential
kk-1
k-2
x x
x’ x’
qi
ai
bi
ci
qi+1
di
fi
Since only k pebbles areavailable we must start withx’ ; di; fi; fi => x’ must precedethe removal from x’.Before qi+1 is pebbled theonly possible actions toproceed are x’ => x, x’ => xor no move at all, alwaysleaving a proper configuration. Move x’ => x’ is a looser!By IH the involved partial assignment makes the formula true.
Once qi+1 is pebbled it issimple to reach qi withthe remaining two freepebbles.
Peter van Emde Boas: Games and Complexity, Guangzhou 2009
Final Remarks
One extra pebble eliminates the need for repebblingthe Universal Quantifier blocks, yielding an exponential speed-up. This gives an independent prooffor the exponential slowdown by saving the last pebble(with in fact a graph of order n3 in stead of n4).
A modification shows that also the question whethera DAG can be pebbled with k pebbles within a restricted amount of time is hard for PSPACE. Note that this problem is in NP if the time allowance ispolynomial. (Sethi has proven NP-hardness in this case.)