Parallel Algorithms

Algorithms
Parallel Algorithms
1

An overview
• A simple parallel algorithm for computing
parallel prefix.
• A parallel merging algorithm

• We are given an ordered set A of n
elements
and a binary associative operator .
• We have to compute the ordered set
0 1 2 1, , ,..., nA a a a a
0 0 1 0 1 1, ,..., ... na a a a a a
Definition of prefix computation

• For example, if is + and the input is the
ordered set
{5, 3, -6, 2, 7, 10, -2, 8}
then the output is
{5, 8, 2, 4, 11, 21, 19, 27}
• Prefix sum can be computed in O (n) time
sequentially.
An example of prefix computation

First Pass
• For every internal node of the tree, compute
the sum of all the leaves in its subtree in a
bottom-up fashion.
sum[v] := sum[L[v]] + sum[R[v]]
Using a binary tree

for d = 0 to log n – 1 do
for i = 0 to n – 1 by 2d+1 do in parallel
a[i + 2d+1 - 1] := a[i + 2d - 1] + a[i + 2d+1 - 1]
• In our example, n = 8, hence the outer loop
iterates 3 times, d = 0, 1, 2.
Parallel prefix computation

• d = 0: In this case, the increments of 2d+1 will be
in terms of 2 elements.
• for i = 0,
a[0 + 20+1 - 1] := a[0 + 20 - 1] + a[0 + 20+1 - 1]
or, a[1] := a[0] + a[1]
When d= 0

• d = 1: In this case, the increments of 2d+1 will be
in terms of 4 elements.
• for i = 0,
a[0 + 21+1 - 1] := a[0 + 21 - 1] + a[0 + 21+1 - 1]
or, a[3] := a[1] + a[3]
• for i = 4,
a[4 + 21+1 - 1] := a[4 + 21 - 1] + a[4 + 21+1 - 1]
or, a[7] := a[5] + a[7]
When d = 1

• blue: no change from last iteration.
• magenta: changed in the current iteration.
The First Pass

Second Pass
• The idea in the second pass is to do a
topdown computation to generate all the prefix
sums.
• We use the notation pre[v] to denote the prefix
sum at every node.
The Second Pass

• pre[root] := 0, the identity element for the
operation, since we are considering the
operation.
• If the operation is max, the identity element will
be .
Computation in the second phase

pre[L[v]] := pre[v]
pre[R[v]] := sum[L[v]] + pre[v]
Second phase (continued)

Example of second phase
pre[L[v]] := pre[v]
pre[R[v]] := sum[L[v]] + pre[v]

for d = (log n – 1) downto 0 do
for i = 0 to n – 1 by 2d+1 do in parallel
temp := a[i + 2d - 1]
a[i + 2d - 1] := a[i + 2d+1 - 1] (left child)
a[i + 2d+1 - 1] := temp + a[i + 2d+1 - 1] (right
child)
a[7] is set to 0

• We consider the case d = 2 and i = 0
temp := a[0 + 22 - 1] := a[3]
a[0 + 22 - 1] := a[0 + 22+1 - 1] or, a[3] := a[7]
a[0 + 22+1 - 1] := temp + a[0 + 22+1 - 1] or,
a[7] := a[3] + a[7]

• blue: no change from last iteration.
• magenta: left child.
• brown: right child.

• All the prefix sums except the last one are now
in the leaves of the tree from left to right.
• The prefix sums have to be shifted one position
to the left. Also, the last prefix sum (the sum of
all the elements) should be inserted at the last
leaf.
• The complexity is O (log n) time and O (n)
processors.
Exercise: Reduce the processor complexity to
O (n / log n).

Parallel merging through
partitioning
The partitioning strategy consists of:
• Breaking up the given problem into many
independent subproblems of equal size
• Solving the subproblems in parallel
This is similar to the divide-and-conquer
strategy in sequential computing.

Partitioning and Merging
Given a set S with a relation , S is linearly
ordered, if for every pair a,b S.
• either a b or b a.
The merging problem is the following:

Input: Two sorted arrays A = (a1, a2,..., am) and
B = (b1, b2,..., bn) whose elements are drawn
from a linearly ordered set.
Output: A merged sorted sequence
C = (c1, c2,..., cm+n).

Merging
For example, if A = (2,8,11,13,17,20) and B =
(3,6,10,15,16,73), the merged sequence
C = (2,3,6,8,10,11,13,15,16,17,20,73).

Merging
A sequential algorithm
• Simultaneously move two pointers along the
two arrays
• Write the items in sorted order in another
array

• The complexity of the sequential algorithm is
O(m + n).
• We will use the partitioning strategy for
solving this problem in parallel.

Definitions:
rank(ai : A) is the number of elements in A less
than or equal to ai A.
rank(bi : A) is the number of elements in A less
than or equal to bi B.

Merging
For example, consider the arrays:
A = (2,8,11,13,17,20)
B = (3,6,10,15,16,73)
rank(11 : A) = 3 and rank(11 : B) = 3.

Merging
• The position of an element ai A in the sorted
array C is:
rank(ai : A) + rank(ai : B).
For example, the position of 11 in the sorted
array C is:
rank(11 : A) + rank(11 : B) = 3 + 3 = 6.

Parallel Merging
• The idea is to decompose the overall merging
problem into many smaller merging
problems.
• When the problem size is sufficiently small,
we will use the sequential algorithm.

Merging
• The main task is to generate smaller merging
problems such that:
• Each sequence in such a smaller problem has
O(log m) or O(log n) elements.
• Then we can use the sequential algorithm since
the time complexity will be O(log m + log n).

Parallel Merging
Step 1. Divide the array B into blocks such that each
block has log m elements. Hence there are m/log m
blocks.
For each block, the last elements are
i log m, 1 i m/log m

Parallel Merging
Step 2. We allocate one processor for each last
element in B.
•For a last element i log m, this processor does
a binary search in the array A to determine two
elements ak, ak+1 such that ak i log m ak+1.
•All the m/log m binary searches are done in
parallel and take O(log m) time each.

Parallel Merging
• After the binary searches are over, the array
A is divided into m/log m blocks.
• There is a one-to-one correspondence
between the blocks in A and B. We call a pair
of such blocks as matching blocks.

Parallel Merging
• Each block in A is determined in the following
way.
• Consider the two elements i log m and(i + 1)
log m. These are the elements in the (i + 1)-th
block of B.
• The two elements that determine rank(i log m
: A) and rank((i + 1) log m : A) define the
matching block in A

Parallel Merging
• These two matching blocks determine a smaller
merging problem.
• Every element inside a matching block has to be
ranked inside the other matching block.
• Hence, the problem of merging a pair of matching
blocks is an independent subproblem which does
not affect any other block.

Parallel Merging
• If the size of each block in A is O(log m), we can
directly run the sequential algorithm on every pair of
matching blocks from A and B.
• Some blocks in A may be larger than O(log m) and
hence we have to do some more work to break
them into smaller blocks.

Parallel Merging
If a block in Ai is larger than O(log m) and the
matching block of Ai is Bj, we do the following
•We divide Ai into blocks of size O(log m).
•Then we apply the same algorithm to rank the
boundary elements of each block in Ai in Bj.
•Now each block in A is of size O(log m)
•This takes O(log log m) time.

Parallel Merging
Step 3.
• We now take every pair of matching blocks from A
and B and run the sequential merging algorithm.
• One processor is allocated for every matching pair
and this processor merges the pair in O(log m)
time.
We have to analyse the time and processor
complexities of each of the steps to get the overall
complexities.

Parallel Merging
Complexity of Step 1
• The task in Step 1 is to partition B into
blocks of size log m.
• We allocate m/log m processors.
• Since B is an array, processor Pi, 1 i m/log
m can find the element i log m in O(1) time.

Parallel Merging
• In Step 2, m/log m processors do binary
search in array A in O(log n) time each.
• Hence the time complexity is O(log n) and
the work done is
(m log n)/ log m (m log(m + n)) / log m (m + n)
for n,m 4. Hence the total work is O(m + n).

Parallel Merging
• In Step 3, we use m/log m processors
• Each processor merges a pair Ai, Bi in O(log m)
time.Hence the total work done is m.
Theorem
Let A and B be two sorted sequences each of
length n. A and B can be merged in O(log n) time
using O(n) operations in the CREW PRAM.

Parallel Algorithms

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