Kent State University CS 4/ Design and Analysis of Algorithms. Dept. of Math & Computer Science LECT-16. Dynamic Programming

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1 CS 4/560 Desgn and Analyss of Algorthms Kent State Unversty Dept. of Math & Computer Scence LECT-6 Dynamc Programmng 2

2 Dynamc Programmng Dynamc Programmng, lke the dvde-and-conquer method, solves problems by combnng the solutons to sub-problems. Pure dvde-and-conquer: dvdes problems nto ndependent sub-problems, solves the sub-problem recursvely, and then, combnes ther solutons to solve the orgnal sub-problem. Dynamc programmng n contrast s used when the sub-problems are not ndependent, that s subproblems share sub-problems. It s typcally appled to optmzaton problems. LECT-6, S-3 Example: Matrx Chan Multplcaton 4 2

3 Matrx Multplcaton Cost of multplyng A[p][q] x B[q][r] s p.q.r What s the cost of multplyng three matrces A, B, and C of szes 0x00, 00x5, and 5x50? How to fnd the best way of multplyng? LECT-6, S-5 Matrx Chan Multplcaton Gven a chan A, A 2, A 3,.. A n of n matrces, such than A has dmenson p - x p, fnd the sequence of multplcaton that wll result n mnmum number of scalar multplcaton. Recursve Cost Functon Catalan numbers: ( A.( A (( A. A (( A.( A 2 2.( A. A ).( A. A 2. A )). A ))) 4 )) 4 ) p( n) f = n k= 4 P( n + ) = Ω n n = p( k). P( n k), f n 3 2 n > LECT-6, S-6 3

4 Observatons Exstence of Optmal Substructure: In an optmum sequence of decsons, each subsequence must also be optmum. (A A 2 A 3 ). (A 4. A 5. A 6 ) Total cost s C(..3) + C(4..6) + cost of multplyng the two fnal matrces. Recursve Soluton Possble: If m[,] s the optmum cost of multplyng all matrces between th and th matrces.(a A +. A ).. f == then m[,]= 0 otherwse, m[, ] = mn{ m[, k] + m[ k +, ] + p. pk. p } k LECT-6, S-7 Recursve Soluton Runnng tme s exponental O(2 n )! LECT-6, S-8 4

5 Observaton-2 Exstence of Overlappng Sub-problem: the same sub-sequence s part of many super sequences. For a strng of lmted sze, the actual number of subproblems are qute small. O(n 2 ) only! LECT-6, S-9 Dynamc Programmng Soluton: A Bottom up approach Compute the optmum cost for multplyng all matrx chans of sze 2. Store them n a matrx m[,], when - spans two matrces. Use the above values to compute optmum cost for multplyng all matrx chans of sze 3. Then sze 4.. Up to sze n. LECT-6, S-0 5

6 Algorthm No cost for chans of sze 0 Sze of chans ncreases Fnd the best k Best k for optmum dvson of the sequence - LECT-6, S- Example LECT-6, S-2 6

7 Example LECT-6, S-3 Constructng the Optmal Soluton LECT-6, S-4 7

8 Complexty of Algorthm No cost for chans of sze 0 Sze of chans ncreases Fnd the best k Best k for optmum dvson of the sequence - Runnng tme? Space? LECT-6, S-5 Example: Optmal Polygon Trangulaton 6 8

9 Polygon Trangulaton We are gven a convex polygon P=<v 0,v, v n- > and a weght functon w defned on trangles formed by sdes and chords of P. The problem s to fnd a trangulaton that mnmzes the sum of the weghts of the trangles n the trangulaton. v v2 v v2 v7 v7 v3 v3 v6 v4 v6 v4 v5 w ( v v v ) = v v + v v + v k k v5 k v LECT-6, S-7 Observatons Optmum substructure: Overlappng subproblems: v v2 v v2 v7 v7 v3 v3 v6 v4 v6 v4 v5 LECT-6, S-8 9

10 Dynamc programmng Soluton For all degenerated polygon of sze 2, <v -, v > cost = zero. For all polygons of sze 3 the cost s w ( v v v ) = v v + v v + v k For all polygons of sze 4 or more try all dvson pont k and pck the best: t[, ] = mn { t[, k] + t[ k +, ] + w( v vkv k k k v )} LECT-6, S-9 Example: Knap/ Sack Problem 20 0

11 /0 Knapsack Problem You have a shoppng bag (knapsack) wth capacty C lb. There are n tems n a super market. Each tem has a value p and weght w. Whch of the n tems wll you pck to maxmze your proft? LECT-6, S-2 Observatons Optmum substructure: If a soluton s optmum for a large proft P wth weght W tems, each of the smaller subsolutons wth proft P-c and weght W- w are also optmum. Overlappng Subproblems: An optmum soluton wth smaller subset of obects n a bag can be part of a large number of superproblems. LECT-6, S-22

12 Problem Formulaton Obectve s to maxmze n = p x subect to constrants: n = w x c Select the 0/ values for x s. LECT-6, S-23 Soluton Let f(,y) denote the proft value of an optmal soluton wth remanng capacty y and remanng obects,+,.n. When, there s only the last obect (termnatng condton): p f ( n, y) = 0 n f y w 0 y < w n n Otherwse (recurson): max{ f ( +, y), f ( + y, y w ) + p} f (, y) = f ( +, y) 0 y < w f y w LECT-6, S-24 2

13 Recursve Soluton F(,y) { f(=n) return (y<w[n]? 0:p[n]; f(y<w[]) return f(+,y); return max(f(+,y),f(+,yw[])+p[]); } Complexty? LECT-6, S-25 Dynamc Prog. Soluton Matrx f[][y], wll store the best proft value for all remanng capacty y smaller than C, for remanng obects. A bottom up approach, frst computes f(n,*). For all y less that w n t s zero. For all y between w n and C t s W n. (termnatng condton) Now compute f(,*) n the order =n-, n-2.2. (use the recurson condton). Complexty: O(nc). LECT-6, S-26 3

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