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Optimal PRAM algorithms: Efficiency of concurrent writing “Computer science is no more about computers than astronomy is about telescopes.” Edsger Dijkstra.

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Presentation on theme: "Optimal PRAM algorithms: Efficiency of concurrent writing “Computer science is no more about computers than astronomy is about telescopes.” Edsger Dijkstra."— Presentation transcript:

1 Optimal PRAM algorithms: Efficiency of concurrent writing “Computer science is no more about computers than astronomy is about telescopes.” Edsger Dijkstra (11/05/1930-6/9/2002)

2 One more time about PRAM model  N synchronized processors  Shared memory EREW, ERCW, CREW, CRCW  Constant time access to the memory standard multiplication/addition Communication (implemented via access to shared memory)

3 Covered so far… in COMP308  Metrics … efficiency, cost, limits, speed-up  PRAM models and basic algorithms  Simulation of Concurrent Write (CW) with EW  Parallel sorting in logarithmic time TODAY’s topic: how fast we can compute with many processor and how to reduce the number of processors?

4 Two problems for PRAM Problem 1. Min of n numbers Problem 2. Computing a position of the first one in the sequence of 0’s and 1’s.

5 Min of n numbers  Input: Given an array A with n numbers  Output: the minimal number in an array A Sequential algorithm … At least n comparisons should be performed!!! COST = (num. of processors)  (time) Cost = 1  n ? Sequential vs. Parallel Optimal Par.Cost = O(n)

6 Mission: Impossible … computing in a constant time  Archimedes: Give me a lever long enough and a place to stand and I will move the earth  NOWDAYS…. Give me a parallel machine with enough processors and I will find the smallest number in any giant set in a constant time!

7 Parallel solution 1 Min of n numbers  Comparisons between numbers can be done independently  The second part is to find the result using concurrent write mode  For n numbers ----> we have ~ n 2 pairs [a 1,a 2,a 3,a 4 ] (a 1,a 2 ) (a 2, a 3 ) (a 3, a 4 ) (a 2, a 4 ) (a 1, a 3 ) (a 1, a 4 ) 000000000000000000000000000000000000000000000000 10 (a i,a j ) If a i > a j then a i cannot be the minimal number ij 1 n M[1..n]

8 The following program computes MIN of n numbers stored in the array C[1..n] in O(1) time with n 2 processors. Algorithm A1 for each 1  i  n do in parallel M[i]:=0 for each 1  i,j  n do in parallel if i  j C[i]  C[j] then M[j]:=1 for each 1  i  n do in parallel if M[i]=0 then output:=i

9 From n 2 processors to n 1+1/2 Step 1: Partition into disjoint blocks of size Step 2: Apply A1 to each block Step 3: Apply A1 to the results from the step 2 A1

10 From n 1+1/2 processors to n 1+1/4 Step 1: Partition into disjoint blocks of size Step 2: Apply A2 to each block Step 3: Apply A2 to the results from the step 2 A2

11 n 2 -> n 1+1/2 -> n 1+1/4 -> n 1+1/8 -> n 1+1/16 ->… -> n 1+1/k ~ n 1  Assume that we have an algorithm A k working in O(1) time with processors Algorithm A k+1 1.Let  =1/2 2. Partition the input array C of size n into disjoint blocks of size n  each 3. Apply in parallel algorithm A k to each of these blocks 4. Apply algorithm A k to the array C’ consisting of n/ n  minima in the blocks.

12 Complexity  We can compute minimum of n numbers using CRCW PRAM model in O(log log n) with n processors by applying a strategy of partitioning the input ParCost = n  log log n

13 Mission: Impossible (Part 2) Computing a position of the first one in the sequence of 0’s and 1’s in a constant time. 00101000 00000000 00000001 01101000 00010100 00000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000001000 00000000000000010000000000000000000000000010000000000000 00000000000000000000000000000010000000000000000000000000 000000000000000000000000000000100000010000001111111111111 111000000000000000000000001000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000 00000000000000000000100000000000010000000000000000000000 00000000000000000000000000000000000000000000000000000000 0000000000000000111111111111111111111111111111000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000001000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000100000000010000000000000000000000000000000000000000000000000000000100000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 0000000000000000000000000000010000001111111111111111000000000000000000000001000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000010000000000000000001000000000000000000000 000001000000000000000000000000000000000000000000010000000000000000000000000000000000000000000000000000000100000010 0000011111111111111110000000000000000000000010000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000011111111111111111111110000000

14 Problem 2. Computing a position of the first one in the sequence of 0’s and 1’s. FIRST-ONE-POSITION(C)=4 for the input array C=[0,0,0,1,0,0,0,1,1,1,0,0,0,1] Algorithm A (2 parallel steps and n 2 processors) for each 1  i { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/12/3381548/slides/slide_14.jpg", "name": "Problem 2. Computing a position of the first one in the sequence of 0’s and 1’s.", "description": "FIRST-ONE-POSITION(C)=4 for the input array C=[0,0,0,1,0,0,0,1,1,1,0,0,0,1] Algorithm A (2 parallel steps and n 2 processors) for each 1  i

15 Reducing number of processors Algorithm B – it reports if there is any one in the table. There-is-one:=0 for each 1  i  n do in parallel if C[i] =1 then There-is-one:=1 000000000000000000 11 1

16 Now we can merge two algorithms A and B 1.Partition table C into segments of size 2.In each segment apply the algorithm B 3.Find position of the first one in these sequence by applying algorithm A 4.Apply algorithm A to this single segment and compute the final value BBBBBBBBBB A A

17 Complexity  We apply an algorithm A twice and each time to the array of length which need only ( ) 2 = n processors  The time is O(1) and number of processors is n.

18 Homework:  Construct several algorithms A1,A2,A3,A4... for MIN problem reducing number of processors  Define the algorithm Ak for MIN problem in terms of A1 only [not in terms of A(k-1) ]  How to formulate the algorithm Ak in a direct way?


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