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v 1
v 2
u 1
u 2
p 2
Figure 11.8 A two-input butterfly graph with pebbles p 1 and p 2 resident on inputs.
p 1
obtain an upper bound on the number of pebbled vertices if we assume that both of them
are pebbled. In this proof we let
denote the S pebbles available to pebble
G . We assign an integer cost num ( p i ) (initialized to zero) to the i th pebble p i in order to
derive an upper bound to the total number of pebble placements made on G .
Consider a matching pair of output vertices v 1 and v 2 of a two-input butterfly graph
and their common predecessors u 1 and u 2 , as suggested in Fig. 11.8 . Suppose that on the
next step we can place a pebble on v 1 . Then pebbles (call them p 1 and p 2 )mustresideon
u 1 and u 2 .Advance p 1 and p 2 to both v 1 and v 2 . (Although the rules stipulate that an
additional pebble is needed to advance the two pebbles, violating this restriction by allowing
their movement to v 1 and v 2 can only increase the number of possible moves, a useful effect
since we are deriving an upper bound on the number of pebble placements.)
After advancing p 1 and p 2 ,if num ( p 1 )= num ( p 2 ) , augment both by 1; otherwise,
augment the smaller by 1. Since the predecessors of two vertices in an FFT graph are in
disjoint trees, there is no loss in assuming that all S pebbles remain on the graph in a
pebbling that maximizes the number of pebbled vertices. Because two pebble placements
are possible each time num ( p i ) increases by 1 for some i , ρ ( S , G )
{
p i |
1
i
S
}
2 1 ≤i≤S num ( p i ) .
We now show that the number of vertices that contained pebbles initially and are con-
nected via paths to the vertex covered by p i is at least 2 num ( p i ) .Thatis,2 num ( p i )
S
or num ( p i )
log 2 S , from which the upper bound on ρ ( S , G ) follows. Our proof is by
induction. For the base case of num ( p i )= 1, two pebbles must reside on the two immedi-
ate predecessors of a vertex containing the pebble p i . Assume that the hypothesis holds for
num ( p i )
1. We show that it holds for num ( p i )= e . Consider the first point in
time that num ( p i )= e . At this time p i and a second pebble p j reside on a matching pair
of vertices, v 1 and v 2 . Before these pebbles are advanced to these two vertices from u 1 and
u 2 , the immediate predecessors of v 1 and v 2 , the smaller of num ( p i ) and num ( p j ) has a
value of e
e
1. This must be p i because its value has increased. Thus, each of u 1 and u 2
has at least 2 e− 1 predecessors that contained pebbles initially. Because the predecessors of u 1
and u 2 are disjoint, each of v 1 and v 2 has at least 2 e
= 2 num ( p i ) predecessors that carried
pebbles initially.
This upper bound on the S -spaniscombinedwithTheorem 11.4.1 to derive a lower
bound on the I/O time at level l to pebble the FFT graph. We derive upper bounds that match
to within a multiplicative constant when the FFT graph is pebbled in the standard MHG. We
develop bounds for the red-blue pebble game and then generalize them to the MHG.
 
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