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Advanced Algorithms Piyush Kumar (Lecture 12: Parallel Algorithms) Welcome to COT5405 Courtesy Baker 05.

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Presentation on theme: "Advanced Algorithms Piyush Kumar (Lecture 12: Parallel Algorithms) Welcome to COT5405 Courtesy Baker 05."— Presentation transcript:

1 Advanced Algorithms Piyush Kumar (Lecture 12: Parallel Algorithms) Welcome to COT5405 Courtesy Baker 05.

2 Parallel Models An abstract description of a real world parallel machine. Attempts to capture essential features (and suppress details?) What other models have we seen so far? RAM? External Memory Model?

3 RAM Random Access Machine Model –Memory is a sequence of bits/words. –Each memory access takes O(1) time. –Basic operations take O(1) time: Add/Mul/Xor/Sub/AND/not… –Instructions can not be modified. –No consideration of memory hierarchies. –Has been very successful in modelling real world machines.

4 Parallel RAM aka PRAM Generalization of RAM P processors with their own programs (and unique id) MIMD processors : At each point in time the processors might be executing different instructions on different data. Shared Memory Instructions are synchronized among the processors

5 PRAM Shared Memory EREW/ERCW/CREW/CRCW EREW: A program isnt allowed to access the same memory location at the same time.

6 Variants of CRCW Common CRCW: CW iff processors write same value. Arbitrary CRCW Priority CRCW Combining CRCW

7 Why PRAM? Lot of literature available on algorithms for PRAM. One of the most “clean” models. Focuses on what communication is needed ( and ignores the cost/means to do it)

8 PRAM Algorithm design. Problem 1: Produce the sum of an array of n numbers. RAM = ? PRAM = ?

9 Problem 2: Prefix Computation Let X = {s 0, s 1, …, s n-1 } be in a set S Let  be a binary, associative, closed operator with respect to S (usually  (1) time – MIN, MAX, AND, +,...) The result of s 0  s 1  …  s k is called the k-th prefix Computing all such n prefixes is the parallel prefix computation s 0 s 0  s 1 s 0  s 1  s 2... s 0  s 1 ...  s n-1 1 st prefix 2 nd prefix 3 rd prefix... (n-1)th prefix

10 Prefix computation Suffix computation is a similar problem. Assumes Binary op takes O(1) In RAM = ?

11 Prefix Computation (Akl)

12 EREW PRAM Prefix computation Assume PRAM has n processors and n is a power of 2. Input: s i for i = 0,1,..., n-1. Algorithm Steps: for j = 0 to (lg n) -1, do for i = 2 j to n-1 do h = i - 2 j s i = s h  s i endfor Total time in EREW PRAM?

13 Problem 3: Array packing Assume that we have –an array of n elements, X = {x 1, x 2,..., x n } –Some array elements are marked (or distinguished). The requirements of this problem are to –pack the marked elements in the front part of the array. –place the remaining elements in the back of the array. While not a requirement, it is also desirable to –maintain the original order between the marked elements –maintain the original order between the unmarked elements

14 In RAM? How would you do this? Inplace? Running time? Any ideas on how to do this in PRAM?

15 EREW PRAM Algorithm 1.Set s i in P i to 1 if x i is marked and set s i = 0 otherwise. 2. Perform a prefix sum on S =(s 1, s 2,..., s n ) to obtain destination d i = s i for each marked x i. 3. All PEs set m = s n, the total nr of marked elements. 4. P i sets s i to 0 if x i is marked and otherwise sets s i = Perform a prefix sum on S and set d i = s i + m for each unmarked x i. 6. Each P i copies array element x i into address d i in X.

16 Array Packing Assume n processors are used above. Optimal prefix sums requires O(lg n) time. The EREW broadcast of s n needed in Step 3 takes O(lg n) time using a binary tree in memory All and other steps require constant time. Runs in O(lg n) time and is cost optimal. Maintains original order in unmarked group as well Notes: Algorithm illustrates usefulness of Prefix Sums There many applications for Array Packing algorithm

17 Problem 4: PRAM MergeSort RAM Merge Sort Recursion? PRAM Merge Sort recursion? Can we speed up the merging? –Merging n elements with n processors can be done in O(log n) time. –Assume all elements are distinct –Rank(a, A) = number of elements in A smaller than a. For example rank(8, {1,3,5,7,9}) = 4

18 PRAM Merging A = 2,3,10,15,16B = 1,8,12,14,19 Rank(2)=1 Rank(3)=1 Rank(10)=2 Rank(15)=4 Rank(16)=4 Rank(1)=0 Rank(8)=2 Rank(12)=3 Rank(14)=3 Rank(19)=

19 PRAM Merge Sort T(n) = T(n/2) + O(log n) Using the idea of pipelined d&c PRAM Mergesort can be done in O(log n). D&C is one of the most powerful techniques to solve problems in parallel.

20 Problem 5: Closest Pair RAM Version ? L  = min(12, 21)

21 Closest Pair: RAM Version Closest-Pair(p 1, …, p n ) { Compute separation line L such that half the points are on one side and half on the other side.  1 = Closest-Pair(left half)  2 = Closest-Pair(right half)  = min(  1,  2 ) Delete all points further than  from separation line L Sort remaining points by y-coordinate. Scan points in y-order and compare distance between each point and next 11 neighbors. If any of these distances is less than , update . return . } O(n log n) 2T(n / 2) O(n) O(n log n) O(n)

22 Closest Pair: PRAM Version? Closest-Pair(p 1, …, p n ) { Compute separation line L such that half the points are on one side and half on the other side.  1 = Closest-Pair(left half)  2 = Closest-Pair(right half)  = min(  1,  2 ) Delete all points further than  from separation line L Sort remaining points by y-coordinate. Scan points in y-order and compare distance between each point and next 11 neighbors. Find min of all these distances, update . return . } O(1) T(n / 2) O(log n) O(1) O(log n) In parallel Use sorted lists Use presorting and prefix computation. Again use prefix computation. Recurrence : T(n) = T(n/2) + O(log n)

23 Problem 6: Planar Convex hulls MergeHull (P) HL = MergeHull( Left of median) HR = MergeHull( Right of median) Return JoinHulls(HL,HR) Time complexity in RAM? Time complexity in PRAM?

24 Join_Hulls

25 Towards a better Planar Convex hull Let Q = {q 1, q 2,..., q n } be a set of points in the Euclidean plane (i.e., E 2 - space). The convex hull of Q is denoted by CH(Q) and is the smallest convex polygon containing Q. –It is specified by listing its corner points (which are from Q) in order (e.g., clockwise order). Usual Computational Geometry Assumptions: – No three points lie on the same straight line. – No two points have the same x or y coordinate. – There are at least 4 points, as CH(Q) = Q for n  3.

26 PRAM CONVEX HULL(n,Q, CH(Q)) 1.Sort the points of Q by x-coordinate. 2.Partition Q into k =  n subsets Q 1,Q 2,...,Q k of k points each such that a vertical line can separate Q i from Q j –Also, if i < j, then Q i is left of Q j. 3.For i = 1 to k, compute the convex hulls of Q i in parallel, as follows: – if |Q i |  3, then CH(Qi) = Q i – else (using k=  n PEs) call PRAM CONVEX HULL(k, Qi, CH(Q i )) 4.Merge the convex hulls in {CH(Q1),CH(Q2),...,CH(Q k )} together.

27 Basic Idea

28 Last Step The upper hull is found first. Then, the lower hull is found next using the same method. – Only finding the upper hull is described here –Upper & lower convex hull points merged into ordered set Each CH(Q i ) has  n PEs assigned to it. The PEs assigned to CH(Q i ) (in parallel) compute the upper tangent from CH(Q i ) to another CH(Q j ). –A total of n-1 tangents are computed for each CH(Q i ) –Details for computing the upper tangents will be separately

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30 Last Step Among the tangent lines to CH(Q i ), and polygons to the left of CH(Q i ), let L i be the one with the smallest slope. Among the tangent lines to CH(Q i ) and polygons to the right, let R i be the one with the largest slope. If the angle between L i and R i is less than 180 degrees, no point of CH(Q i ) is in CH(Q). –See Figure 5.13 on next slide (from Akl’s Online text) –Otherwise, all points in CH(Q) between where L i touches CH(Q i ) and where R i touches CH(Q i ) are in CH(Q). Array Packing is used to combine all convex hull points of CH(Q) after they are identified.

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32 Complexity Step 1: The sort takes O(lg n) time. Step 2: Partition of Q into subsets takes O(1) time. Step 3: The recursive calculations of CH(Q i ) for 1  i  n in parallel takes t(n) time (using n PEs for each Qi). Step 4: The big steps here require O(lgn) and are – Finding the upper tangent from CH(Q i ) to CH(Q j ) for each i, j pair. – Array packing used to form the ordered sequence of upper convex hull points for Q. Above steps find the upper convex hull. The lower convex hull is found similarly. –Upper & lower hulls merged in O(1) time to ordered set

33 Complexity Cost for Step 3: Solving the recurrance relation t(n) = t(  n) +  lg n yields t(n) = O(lg n) Running time for PRAM Convex Hull is O(lg n) since this is maximum cost for each step. Then the cost for PRAM Convex Hull is C(n) = O(n lg n).


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