1. Lecture 3: Parallel Algorithm Design

2. Techniques of Parallel Algorithm Design
Balanced binary tree
Pointer jumping
Accelerated cascading
Divide and conquer
Pipelining
Multi-level divide and conquer 　

3. Balanced binary tree
Processing on binary tree: Let the leaves correspond to input and internal nodes to processors.
Example
Find the sum of n integers (x1, x2, ... , xn).

4. Problem of finding Prefix Sum
Balanced binary tree
Problem of finding Prefix Sum
Definition of Prefix Sum
Input：　n integers put in array A[1..n] on the shared memory
Output： array B[1..n], where for each B[i] (1≦i≦n)
B[i] = A[1] + A[2] A[i]
Example　Input: A[1..5] = (5, 8, -7, -10, 3)，
　　 Output: B[1..5] = (5, 13, 6, -4, -1)
Sequential algorithm for Prefix Sum
main (){
B[1] = A[1];
for (i = 2; i≦n; i++) {
B[i] = B[i-1] + A[i]; }

5. Solving Prefix Sum problem on balanced binary tree (1)
Outline of the parallel algorithm for prefix sum
To simplify the problem, let n = 2k (k is an integer)
Calculate the sub-sum from the leaves to the root in bottom up style.
Using the sub-sum obtained in (1) , calculate the prefix sum from the root to the leaves in up down style.

6. Solving Prefix Sum problem on balanced binary tree (2)
First read the input at the leaves. Then,
calculate the sub-sum from the leaves
to the root in bottom up style.
(2) From the root to the leaves, do
the following: send the right son
its sub-sum obtained in (1), and
send the left son the value of
(its sub-sum) – the right son’s sub-sum).
12-(-2) P 4 3 2 1 6 7
=14 12
12-5 =7
-(-3)
=10
-(-4) =6 -5
=-3 -2 =4
12-10 =2

7. Solving Prefix Sum problem on balanced binary tree (3)
Correctness of the algorithm
　When step (1) finished, the sub-sum in each internal node is the sum of its subtree.

8. Solving Prefix Sum problem on balanced binary tree (4)
Correctness of the algorithm - Continue
In step (2), at each internal node
The sub-sum sent to the right son is the summation of its subtree.
(b) The sub-sum sent to the left son is the sum of its subtree subtracted by the sum of its right son’s subtree. P 4 3 2 1
12-10 =2 7 5
(b)
(a) 12

9. Solving Prefix Sum problem on balanced binary tree (5)
Algorithm Parallel-PrefixSum (EREW PRAM algorithm)
main (){
if (number of processor == i) B[0, i] = A[i];
for (h=1; h≦log n; h++) {
if (number of processor j ≦ n/2h) {
B[h, j] = B[h-1, 2j-1] + B[h-1, 2j]; }
C[log n, 1] = B[log n, 1]
for (h = (log n) - 1; h≧0; h--) {
if (j is even) C[h, j] = C[h+1, j/2];
if (j is odd) C[h, j] = C[h+1, (j+1)/2] - B[h, j+1];

10. Solving Prefix Sum problem on balanced binary tree (6)
(First step)
B[3,1]
B[0,4]
B[0,3]
B[0,2]
B[0,1]
A[1]
A[2]
A[3]
A[4]
B[1,2]
B[1,1]
B[2,1]
B[0,8]
B[0,7]
B[0,5]
B[0,6]
A[5]
A[6]
A[7]
A[8]
B[1,4]
B[1,3]
B[2,2]

11. Solving Prefix Sum problem on balanced binary tree (7)
(Second step)
Ｃ[3,1] C
[0,4]
[0,3]
[0,2]
[0,1]
[1,2]
[1,1]
[2,1]
[0,8]
[0,7]
[0,5]
[0,6]
[1,4]
[1,3]
[2,2]

12. Solving Prefix Sum problem on balanced binary tree (8)
Analysis of the algorithm
　Computing time: for loop repeated log n times and each loop can be executed in O(1) time →　O(log n) time
　Number of processors: Not larger than n →　n processors
　Speed up: O(n/log n)
　Cost: O(n log n)
It is not cost optimal since the running
time of the optimal Θ(n).

13. Accelerated cascading
Balanced binary tree
Accelerated cascading
To reduce the cost, solve the problem sequentially when the size of the problem is small.
Accelerated cascading is used, usually, with balanced binary tree
and divide and conquer techniques.

14. Balanced binary tree Accelerated cascading for Prefix Sum problem
Policy for improving the algorithm
To make the algorithm cost optimal, we decrease the number of processors from n to n/logn.
(Note: Computing time of the algorithm is O(logn).)
Steps:
Instead of processing n elements in parallel, divide n elements into n/logn groups with logn elements each.
To each group assign one processor and solve the problem for the group sequentially.

15. Balanced binary tree Accelerated cascading for Prefix Sum problem
Improved algorithm Parallel-PrefixSum
Divide n elements in A[1..n] in to n/log n groups with log n elements each.
　（O(1) time，O(n/log n) processors）
(2)　Assign each group one processor and find the prefix sum for each group.
（O(log n) time，O(n/log n) processors）
A[1..n]
log n elements

16. Balanced binary tree Accelerated cascading for Prefix Sum problem (3)
Improved algorithm Parallel-PrefixSum - continue
(3)　Let S be the set of the last element in each group (it is the sum of the group). Use algorithm Parallel-PrefixSum to find the prefix sum of S.
( O(log (n/log n) ) = O(log n) time，O(n/log n) processors)
Algorithm
Parallel-PrefixSum
Last element
in each group

17. Balanced binary tree Accelerated cascading for Prefix Sum problem (4)
Improved algorithm Parallel-PrefixSum - continue
(4)　Use the prefix sum of S to find the prefix sum of the input
A[1..n].
（O(log n) time，O(n/log n) processor）
Result of (3)

18. Balanced binary tree Accelerated cascading for Prefix Sum problem (5)
Analysis of the improved algorithm
Computing time and the number of processors：　
Each step: O(log n) time, O(n/log n)
→ Totally, O(log n) time, O(n/log n) processors
　Speed up = O(n/log n)
　Cost：　O(log n × n/log n) = O(n)
　It is cost optimal.
　It is also time optimal
( Don’t show the proof here)
　 It is optimal algorithm.

19. Divide and Conquer Divide and conquer (1) 2 divide and conquer
(2) n divide and conquer（ε＜１） ε

20. Divide and Conquer Divide and conquer technique
　Well known technique in algorithm design
　Solving problems recursively
　Used very often in both sequential and parallel
algorithms
How to divide and conquer
Dividing step: dividing the problem into a number of subproblems.
Conquering step: solving each subproblem recursively.
(3) Merging step: merging the solutions of subproblems to the solution of the original problem.

21. Divide and Conquer Convex hull problem
Input: a set of n points in the plane.
Output: the smallest convex polygon which contains all points of the
input. (The convex polygon is represented by the list of its
vertices in order of clockwise.)
　Basic problems in computational geometry.
　A lot of applications.
　Solved in O(nlogn) time sequentially.
In the following we only consider
the upper convex hull.
（Upper convex hull: ( P9, P8, P1, P0 ) ） P 1 2 3 9 4 5 6 7 8
Output: ( P ,P ,P ,P ,P ,P )
Lower convex hull
Upper convex hull

22. Merging of two upper convex hulls
Divide and Conquer
Merging of two upper convex hulls
Finding the upper common tangent p 1 2 3 9 4 5 6 7 8 10
Common upper tangent = (p ,p )
It is known that common tangents can be found in O(log n) time sequentially.

23. Divide and Conquer 2 divide and conquer (1)
Outline of the algorithm Parallel-UpperConvexHull
Preprocessing　Sort all the points according to their x coordinates, and let the result is the sequence (p1, p2, p3, ... , pn).
If the size of sequence is 2, return the sequence.
Divide (p1, p2, p3, ... , pn) to the left half part and the right half part, and find the upper convex hull of each recursively.
(3) Find the upper common tangent of two upper convex hulls obtained in (2), and output the solution of the problem. 　

24. Divide and Conquer 2 divide and conquer (2)
How 2 divide and conquer works
Find the upper common
tangent for two upper convex
hulls of two vertices each.
Find the upper common
tangent for two upper convex
hulls of four vertices each.
Find the upper common
tangent for two upper convex
hulls of eight vertices each.

25. Divide and Conquer 2 divide and conquer（３） Recursive execution
When the problem is divide once, the size of the subproblem becomes half.
Suppose the size of the subproblems becomes 2 when the
problem is divided k times.
n/2k= 2　⇒　 k = log2 n - 1 n
n/2
n/4 2
Height=
log

26. Divide and Conquer ２divide and conquer (4) Complexity of the algorithm
Preprocessing：　O(log n) time，n processors　
Steps (1)〜(3)：　each step runs O(log n) time，use n/2 processors
　 T(n) = T(n/2) + O(log n) Therefore, T(n) = O(log n)
∴　The algorithm runs in O(log n) time using O(n) processors.
Computational model: There is no concurrent access　⇒　EREW PRAM
Proprocessing　Sort the sequence of the points according to their x coordinates.
If the size of the sequence is 2, return the sequence.
Divide the sequence into the left half part and the right half part,
and find the upper convex hull of each recursively.
Find the upper common tangent of two upper convex hulls obtained in (2),
and output the upper convex hull of the sequence. 2 2

27. Divide and Conquer 2divide and conquer (5)
Finding the complexity of the algorithm from recursive tree
Computing time
Number of processors
At the level of the leaves, n/2 processors are used at the same time. ⇒　n/2 processors n
n/2
n/4 2
Height
log c
Time
Totally
c log n
c log n/2
c log n/4
O(log n)
Processors １ 4
T(n)×P(n)=O(nlog n)　　 It is not cost optimal 2

28. Divide and Conquer n divide and conquer Outline of the algorithm
1/2
Outline of the algorithm
Preprocessing
Sort the sequence of the input points according to their x coordinates, and let the result be sequence (p1, p2, p3, ... , pn).
(1)　If the size of the sequence is 2, return the sequence.
(2)　Divide (p1, p2, p3, ... , pn) to equally-sized subsequence, and find the upper convex hull of each recursively.
(3)　Merge upper convex hulls into the upper convex hull of the sequence． n
1/2 n
1/2

29. Divide and Conquer n Merging upper convex hull
1/2
Merging upper convex hull n
1/2
Assign each upper convex hull processors to find the upper common tangents in 　　O(log n) time, and then determine the edges which belong to the solution．　　　
Case 1
Case 2

30. Recursive tree of n divide and conquer
1/2
When the problem is divided once, the size of the subproblems becomes
Suppose that the size of the subproblems becomes 2 when the problem is divided k times. 　
= 2　⇒　 k = log log n n
1/2 n
1/(2 ) k
Height=
loglog n n
1/2
1/4 2

31. Analysis of the algorithm
Divide and Conquer
Analysis of the algorithm
Preprocessing：　O(log n) time，n processors.　
Steps (1)〜(3)：　 each step O(log n) time，n processors.
　 T(n) = T(n　 ) + O(log n), therefore, T(n) = O(log n)
∴　Totally, the algorithm runs in O(log n) time using O(n) processors.
Computational model
Concurrent reading happens in the procedure of finding the upper common tangents
⇒　CREW PRAM
Preprocessing　Sort the sequence of the points in their x coordinates.
(1)　If the size of the sequence is 2，return the sequence.
(2)　Divide the sequence into　 equally-sized subsequences,
and find the upper convex hull of each recursively.
(3)　Find the upper common tangents of the upper convex hulls
obtained in (2), and determine the solution. n
1/2
1/2
T(n)×P(n)=O(nlog n)　　Optimal !!!

32. Exercise
1. Suppose nxn matrix A and matrix B are saved in two dimension arrays. Design a PRAM algorithm for A×B using n and nxn processors, respectively. Answer the following questions:
What PRAM models that you use in your algorithms?
What are the runings time?
Are you algorithms cost optimal?
Are your algorithms time optimal?
2. Design a PRAM algorithm for A×B using k (k <= nxn processors). Answer the same questions.