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Potential for Parallel
Computation
Chapter 2
Fundamentals of Parallel
Processing
G. Alaghband
Fundamentals of Parallel
Processing
1, Potential for Parallel
Computation
Main Topics
• Prefix Algorithms
• Speedup and Efficiency
• Amdahl's Law
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Examples of Parallel Programming Design
• Sequential/Parallel Add
• Sum Prefix Algorithm
• Parameters of Parallel Algorithms
• Generalized Prefix Algorithm
• Divide and Conquer
• Upper/Lower Algorithm
• Size and Depth of Upper/Lower
Algorithm
• Odd/Even Algorithm
• Size and Depth of Odd/Even
Algorithm
• A Parallel Prefix Algorithm with Small Size
and Depth
• Size and Depth Analysis
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A Simple Algorithm :
Adding numbers:
Assume a vector of numbers in V[1:N]
Sequential add:
S:= V[1];
for i := 2 step 1 until N
S := S + V[i];
Data dependence graph for sequential summation
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Data Dependence Graph for Parallel Summation
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A Slightly More Complicated Algorithm
Sum Prefix:
For i := 2 step 1 until N
V[i] := V[i-1] + V[i];
Dependence Graph for Sequential Prefix
Each term is the sum of all numbers in V[1:i], i  N
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Questions to think about and try to answer in this
section are:
• Do sequential computations involve different amount
of parallelism?
• What can be done in parallel in an arbitrary algorithm?
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PARAMETERS OF PARALLEL ALGORITHMS
SIZE:
Number of operations
DEPTH:
Number of operations in the longest chain from any input to
any output.
EXAMPLES
Sequential sum of N inputs:
SIZE = N - 1
DEPTH = N - 1
Parallel sum of N inputs (pair-wise summation):
SIZE = N - 1
DEPTH = Log2 N
Sequential Sum Prefix of N inputs:
SIZE = N - 1
DEPTH = N - 1
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A simply stated problem having several different algorithms is the
Generalized Prefix Problem:
Given an associative operator O, and N variables V1, V2, ..., VN, form the N
results:
V1, V1OV2, V1OV2OV3,
…,
V1OV2OV3O...OVN .
There are several different algorithms to solve this problem, each with
different characteristics.
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Divide and Conquer
A general technique for constructing non-trivial parallel algorithms
is the divide and conquer technique.
The idea is to split a problem into 2 smaller problems whose
solution can be simply combined to solve the larger problem.
The splitting is continued recursively until problems are so small
that they are easy to solve.
In this case we split the prefix problem on V1, V2, ..., VN into 2
problems:
Prefix on V1, V2, ..., VN/2 ,
and
Prefix on VN/2+1 , VN/2+2, ..., VN
That is, we split inputs to the prefix computation into a lower half
and an upper half, and solve the problem separately on each half.
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The Upper/Lower Construction
The solution to the 2 half size problems are combined by the construction
below:
Recall that the ceiling of X, X is the least integer  X and the floor of X,
X, is the greatest integer  X.
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Recursively applying the Upper/Lower construction will eventually result in
prefix computations on no more than 2 inputs, which is trivial.
For example:
For 4 inputs we obtain:
N=4
Size = 4
Depth = 2
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A larger example of the parallel prefix resulting from recursive
Upper/Lower construction Pul(8):
N=8
Size = 12
Depth = 3
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Finally Pul(16)
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N = 16
Size = 32
Depth = 4
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Having developed a way to produce a prefix algorithm which
allows parallel operations, we should now characterize it in terms
of its size and depth.
The depth of the algorithm is trivial to analyze.
The construction must be repeated log2 N  times to reduce
everything to two inputs.
For each application of the construction, the path from the
rightmost input to the rightmost output passes through one more
operation.
Therefore
Depth = log2 N 
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Analysis of Size of Upper/Lower
Assume N a power of 2 ( easiest to analyze).
Theorem: Let s(N) = Size(Pul(N)), then for N a power of 2:
s(2k) = k 2k-1 =
N/2 log2N
where N = 2k.
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Proof: s(2k) = k 2k-1 = N/2 log2N
by induction on k.
The initial condition for k = 1 is : s(21) = Size(Pul(2)) = 1 = 21-1
Assume the result is true for k = i and prove it for k = i + 1.
We assume
s(2i) = i2i-1
The size of s(2i+1) is related to s(2i) by counting operations in the recursive
constructions:
s(2i+1) = 2 s(2i)
2i
+
= 2(i2i-1) + 2i
= (i+1)2i+1-1 =
k.2k-1
Thus if results holds for k = i, it holds for k = i + 1.
The result follows for arbitrary integer k by finite induction.
k.2k-1 = N/2log2N, ( N = 2k , K = log2N)
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If we have unlimited processors (arithmetic units)
available then the minimum depth algorithm finishes
soonest.
The Upper/Lower construction gives an algorithm
with minimum depth.
If number of processors are limited then we have to
keep the size small:
ODD/EVEN Algorithm
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Parallel prefix with larger depth but smaller size:
Divide the inputs into sets with odd and even index values.
Combine odds with the next higher evens at input.
Do the parallel prefix on the new combined evens and combine evens with
next higher odds at output.
Recursive application
of odd/even
construction
continues until a
prefix of 2 inputs is
reached. Poe(N)
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Examples:
Poe(4): The odd/even construction for 4 inputs is presented first:
N=4
Size = 4
Depth = 2
Note : Poe(4) is a special case. It is equivalent to Pul(4) . Notice that
the longest path is only one more than the 2-input case.
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Odd/Even construction for 8 inputs shows the recursive construction:
N=8
Size = 11
Depth = 4
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Size and Depth
The size and depth analysis of Odd/Even algorithm is simple for N a power of 2.
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Thus size of Odd/Even algorithm is less than the size of Upper/Lower
but its depth is greater (~ twice)
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The sequential algorithm is very deep,
Odd/Even
about
twice as
deep asis
Upper/Lower
The sizeisof
sequential
algorithm
smallest,
both are much
shallower
sequential
The sizebut
of Upper/Lower
grows
fasterthan
with the
N than
the sizecase.
of Odd/Even.
The size of Odd/Even is less than twice the size of sequential algorithm.
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The sequential algorithm is very deep,
Odd/Even is about twice as deep as Upper/Lower
but both are much shallower than the sequential case.
The size of sequential algorithm is smallest,
The size of Upper/Lower grows faster with N than the size of
Odd/Even.
The size of Odd/Even is less than twice the size of sequential
algorithm.
It is possible to find a parallel prefix algorithm with
minimum depth which also has a size proportional to
N instead of Nlog2N.
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A Parallel Algorithm with Small Depth and Size
Reference:
Ladner, R. E. and Fisher, M. J., “Parallel Prefix Computation,” JACM,
vol. 27, no. 4, pp. 831-838, Oct. 1980.
By combining the 2 methods:
(Upper/Lower and Odd/Even),
we can define a set of prefix algorithms Pj(N).
For j  1, Pj(N) is defined by Odd/Even construction
using Pj-1(N/2). (until we get to P0)
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P0(N) is defined differently, using Upper/Lower construction with P1
and P0 of fewer inputs.
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To get P0(N), only P0 and P1 are needed.
For example:
To get P0(16) we need
P1(8) and P0(8)
To get P1(8) we need
P0(4)
To get P0(8) we need
P1(4) and P0(4).
Note that P1(4) happens to be the same as P0(4).
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Depth of P0 can be seen by realizing that while the
depth of P1(N) is log2N + 1, the longest path from
any input to the highest numbered output is only
log2N.
Depth grows by one for each doubling of N.
Remember that P1 is applied to N/2 in P0(N).
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Size of P0
Size of P1
Upper/Lower
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Theorem: If N = 2k, then
S0(N) = 4N -F(2+k) -2F(3+k) + 1 and
S1(N) = 3N - F(1+k) -2F(2+k)
where F(m) is the m-th Fibonacci number.
Recall :
F(0) = 0, F(1) = 1,
F(m) = F(m-1) + F(m-2), m  2
0, 1, 1, 2, 3, 5, 8, ...
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S0(N) = 4N -F(2+k) -2F(3+k) + 1
S1(N) = 3N - F(1+k) -2F(2+k)
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S0(N) = 4N -F(2+k) -2F(3+k) + 1
S1(N) = 3N - F(1+k) -2F(2+k)
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Two problems remain in understanding Ladner and
Fisher’s P0.
1. What happens if N is not a power of 2?
In this case there is an upper bound:
for N  1:
Sj(N) < 4N
so P0(N) is no more than a few times larger than the
sequential algorithm.
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2. Some insight into the behavior of Fibbonacci numbers would help as most
people have little intuition about how Fibbonacci numbers behave.
We can do this by using an asymptotic formula for large m:
Which shows size of P0(N)
Is less than 4N
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Speedup and Efficiency
• Speed up and Efficiency of Parallel
Algorithms
• Arithmetic Expression Evaluations
• Vector and Matrix Algorithms
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Speedup and Efficiency of Algorithms
For any given computation (algorithm):
Let TP be the time to perform a computation with P
processors. (arithmetic units, or PEs)
We assume that any P independent operations can be
done simultaneously.
Note: The depth of an algorithm T , the minimum
execution time.
The speedup with P processors is
,
and efficiency is
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These numbers,
SP and EP,
refer to an algorithm and not to a machine.
Similar numbers can be defined for specific hardware.
The time T1 can be chosen in different ways:
To evaluate how good an algorithm is,
it should be the time for the
“BEST” sequential algorithm.
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The Minimum Number of Processors
Giving the Maximum Speedup:
Let P be the minimum number of P of processors such that
TP = T
i.e.
P = min { P | TP = T }
Then
TP, SP, EP
are the best known
time, speedup, and efficiency
respectively.
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If we have one processor, we should use the smaller sequential algorithm
This makes the efficiency,
Thus the efficiency of Upper/Lower really decreases as
the problem size grows.
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Evaluation of Arithmetic Expressions
Most problems are not so simple that best sequential algorithm is known,
to say nothing of the best parallel algorithm.
Arithmetic expression evaluation is a case in which general results are
known:
Definition:
An atom is a constant or variable appearing in an expression.
Let E<N> be an expression in +, -, , /, (, ) having N atoms.
The minimum time for sequential evaluation of any such
algorithm is N-1 steps.
Evaluation of E<N> must take a minimum amount of time as
shown by the next lemma.
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Lemma 1:
For any number P of processors, the time to evaluate an
expression in N atoms satisfies:
The proof is based on the fact that all atoms are combined into one result.
Proof:
Since there is a single result and all operations are dyadic (they
take 2 operands), then there is only one result at the last step and no more
than 2 intermediate results at the nest to the last step.
In general, there are at most 2l intermediate values at the last-l step.
Since there must be N atoms at stage l, then N  2 no. of stages and
QED
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Expressions can be transferred by mathematical operations into more
parallel forms.
Using associative, commutative, and distributive laws we can reduce the
height of an expression tree.
Consider the expression:
E<8> = A + B(CDE + F + G) + H
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E<8> = A + B(CDE + F + G) + H
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Including the distributive law, we can get an even smaller depth.
But the number of operations will increase.
E<8> = A + B(CDE + F + G) + H
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Using associativity and commutativity, evaluation of an expression is
bounded.
In addition to lower bound on the evaluation time for arithmetic expression,
we can also get an upper bound when associativity and commutativity are
used to put the expression into the most parallel form possible:
Theorem1:
If E<N,d> is an arithmetic expression in N atoms with depth d of
parenthesis nesting, then using commutativity and associativity only,
E<N,d> can be transformed so that
Reference:
J.L. Baer and D.P. Bovet, “Compilation of Arithmetic Expressions for Parallel
Computation,” Proc. IFIP Congress 1968, North Holland, Amsterdam, pp.
340-346.
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If distributivity is also used, the upper bound is independent of parenthesis
nesting.
But the size of the computation may increase, so the bound on P also will
increase.
Theorem:
An expression E<N> in N atoms can be transformed by
associativity, commutativity, and distributivity so that
Note: The time bound only applies to transformed expressions. The
transformation itself takes an order of Nlog2N steps.
For computations larger than single expressions,
we must look at specific cases.
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Amdahl’s Law
Let T(P) be the execution time with hardware parallelism P.
Let S be the time doing the sequential part of the work ,
Time to do the parallel part of the work sequentially is Q,
i.e., S and Q are the sequential and parallel amounts of work
measured by time on one processor,
The total time with P processors is
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Amdahl’s Law
Expressing this in terms of the fraction
of serial work
Amdahl’s law states that
Speedup
Efficiency
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There are several consequences of this simple
performance model:
In order to achieve at least 50% efficiency
on the program with hardware parallelism P,
f can be no larger than 1/(P-1).
This becomes harder and harder to achieve
as P becomes large.
Amdahl used this result to argue that
sequential processing was best,
But
it has several useful interpretations in different
parallel environments:
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• A very small amount of unparallelized code can
have a very large effect on efficiency if the amount of
parallelism is large;
• A fast vector processor must also have a fast
scalar processor in order to achieve a sizeable
fraction of its peak performance;
• Effort in parallelizing a small fraction of code that
is currently executed sequentially may pay off in
large performance gains;
• Hardware that allows even a small fraction of new
things to be done in parallel may be considerably
more efficient.
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Although Amdahl’s law is a simple performance
model, it need not be taken simplistically.
The behavior of the sequential fraction, f, for example, can be quite
important.
System sizes, especially the number, P, of processors are often
increased for the purpose of running larger problems.
Increasing the problem size often does not increase the absolute
amount of sequential work significantly.
In this case, f is a decreasing function of problem size,
and if problem size is increased with P, the somewhat pessimistic
implications of equations look much more favorable.
see Problem 2.16 for a specific example.
The behavior of performance as both problem and
system size increase is called scalability.
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