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1 Parallel Parentheses Matching Plus Some Applications

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2 Parentheses Matching Problem Definition: Given a well-formed sequence of parentheses stored in an array, determine the index of the mate of each parentheses stored in the array

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3 Example 12345678 ((())()) 85432761

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4 Sequential Solution Traditional solution uses a stack Push left parentheses, pop for a right parenthesis; these are a pair Can this method be implemented in parallel? Why or why not?

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5 Parallel Solution: Divide & Conquer Lemma 1: The mate of a parenthesis at an odd position in a balanced string lies in an even position (and vice versa). Lemma 2: If a balanced string has no left parenthesis at an even location (or, equivalently, a right at an odd location), then the mate of each left parenthesis in the string lies immediately to its right.

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6 Lemma 2 Any string that satisfies Lemma 2 is of form ( ) ( ) ( ) ( )….( ) and is referred to as form F.

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7 Algorithm Overview Each left position at an odd position and each right position are marked with a 0. All others are marked with a 1. These 2 disjoint sets are copied (packed) into a new array. Repeat for the 2 sets. Stop when each new substring is of form F.

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8 Algorithm Match For i = 1 to log n – 1 do if “(“ & index is odd then mark 0 else mark 1 Use segmented prefix sums to compute new index for each parenthesis Move parentheses to new location Determine if string is now in form F; if not, terminate – unbalanced string Match parentheses and store in original array

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9 Example 12345678 ((())()) 01001110 (())()() 0110 ()()()()

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10 Example – Keep Index 12345678 1 (2 (3 (4 )5 )6 (7 )8 ) 01001110 1 (3 (4 )8 )2 (5 )6 (7 ) 0110 1 (8 )3 (4 )2 (5 )6 (7 )

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11 Segmented Prefix Sum Problem Definition Given an array containing elements, some marked 0 and some marked 1. Compute the prefix sum of each subset. (For this application the sums will be on values of 1, to number the items.)

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12 Segmented Prefix Sum - Example 12345678 ((())()) 01001110 11232344 How can this be accomplished with one prefix sums operation?

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13 Parentheses Matching on Hypercube Use the Divide & Conquer strategy Consider 2 processors Each PC – assign 0/1 P0 send 1’s to P1; P1 send 0’s to P0 Each solve the sub-problem Does the problem split evenly? Consider Large problem – P0 & P2 take 0 items, P1 & P3 take 1 items

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14 PPM - Hypercube Overview of Algorithm 2-Cube: Special case of 2 pc hypercube 4-Cube: Used to partition large sub-problems consisting of 4 pc cubes Match: The Driving Algorithm

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15 Data Distribution INPUT array is divided into P equal partitions of size n/p First n/p items are given to P0, next n/p items to P1, etc. Final MATCH information for each item is stored in the original PC

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16 Data Distribution Array consists of elements INPUT which holds the parentheses & MATCH which will hold the final matching information Local Match: if the match is determined by the pc in which the match information is to be stored Non-local: otherwise

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17 Algorithm 2-Cube Mark left and right parentheses with 0/1 as previously discussed P0 & P1 exchange entries – P0 contains 0 and P1 contains 1 Each PC use stack to sequentially match parentheses Send non-local match operation to appropriate processor

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18 Algorithm 2-Cube – Step 1 12345678 (()(())) 01110010 Mark 0/1 P0 P1

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19 Algorithm 2-Cube – Step 2 & 3 15682347 (())()() 00001111 86513274 Exchange & Match P0 P1

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20 Algorithm 2-Cube – Step 4 12345678 (()(())) 01110010 83276541 Send non-local match information P0 P1

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21 Algorithm 4-Cube Basis of Algorithm Match Insures near-equal data distribution Overview Phase 1: local matches determined sequentially & communicated Phase 2: Unmatched parentheses marked, redistributed; P0 & P1 have 0’s, P2 & P3 have 1’s (half each)

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22 Algorithm 4- Cube Phase 1: Sequential Processing 1. Each PC use stack to match 2. Send non-local Match to other PC 3. Count unmatched parentheses; prefix sum to reindex

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23 Algorithm 4-Cube Phase 2 1. Mark parentheses with 1/0 2. P0 & P2 exchange: P0=0 & P2=1 Likewise, P1=0 and P3=1 3. P0 & P1 exchange number information; likewise for P2 & P3 4. P0 & P1 exchange entries; P0 obtains 1 st half: likewise for P2 & P3

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24 Algorithm 4-Cube Distribution of Data – 64 entries P0 Init.1-16 P1 Init.17-32 1-21-16,33-48 (0)1-217-32, 49-64(0) 3-41-32 (0)3-433-64 (0) P2 Init.33-48 P3 Init.49-64 1-21-16,33-48 (1)1-217-32, 49-64(1) 3-41-32 (1)3-433-64 (1)

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25 Algorithm Hypercube Match 1. Each subcube of size 4 executes 4-Cube // Logically P/2 subcubes with independent subproblems 2. Each subcube (p/2) prefix sums to determine new index 3. Each subcube from Step 1 recursively repeat 1, 2, 3 until each subcube is of size 2 4. Execute 2-Cube to complete the solution

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26 Algorithm Hypercube Match Complexity Analysis O (log 2 p + n/p log p) For p=n/log n simplifies to O(log 2 n) Is the Algorithm Optimal?

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