Dynamic Programming. Sequence Of Decisions. 0/1 Knapsack Problem. Sequence Of Decisions

Transcription

1 Dyamic Programmig Sequece Of Decisios Sequece of decisios. Problem state. Priciple of optimality. Dyamic Programmig Recurrece Equatios. Solutio of recurrece equatios. As i the greedy method, the solutio to a problem is viewed as the result of a sequece of decisios. Ulike the greedy method, decisios are ot made i a greedy ad bidig maer. 0/1 Kapsack Problem Let x i = 1 whe item i is selected ad let x i = 0 whe item i is ot selected. maximize p i x i i = 1 subject to w i x i <= c i = 1 ad x i = 0 or 1 for all i All profits ad weights are positive. Sequece Of Decisios Decide the x i values i the order x 1, x 2, x 3,, x. Decide the x i values i the order x, x -1, x -2,, x 1. Decide the x i values i the order x 1, x, x 2, x -1, Or ay other order.

2 Problem State The state of the 0/1 kapsack problem is give by ƒ the weights ad profits of the available items ƒ the capacity of the kapsack Whe a decisio o oe of the x i values is made, the problem state chages. ƒ item i is o loger available ƒ the remaiig kapsack capacity may be less Problem State Suppose that decisios are made i the order x 1, x 2, x 3,, x. The iitial state of the problem is described by the pair (1, c). ƒ Items 1 through are available (the weights, profits ad are implicit). ƒ The available kapsack capacity is c. Followig the first decisio the state becomes oe of the followig: ƒ (2, c) whe the decisio is to set x 1 = 0. ƒ (2, c-w 1 ) whe the decisio is to set x 1 = 1. Problem State Suppose that decisios are made i the order x, x -1, x -2,, x 1. The iitial state of the problem is described by the pair (, c). ƒ Items 1 through are available (the weights, profits ad first item idex are implicit). ƒ The available kapsack capacity is c. Followig the first decisio the state becomes oe of the followig: ƒ (-1, c) whe the decisio is to set x = 0. ƒ (-1, c-w ) whe the decisio is to set x = 1. Priciple Of Optimality A optimal solutio satisfies the followig property: ƒ No matter what the first decisio, the remaiig decisios are optimal with respect to the state that results from this decisio. Dyamic programmig may be used oly whe the priciple of optimality holds.

3 0/1 Kapsack Problem Suppose that decisios are made i the order x 1, x 2, x 3,, x. Let x 1 = a 1, x 2 = a 2, x 3 = a 3,, x = a be a optimal solutio. If a 1 = 0, the followig the first decisio the state is (2, c). a 2, a 3,, a must be a optimal solutio to the kapsack istace give by the state (2,c). x 1 = a 1 = 0 maximize p i x i subject to w i x i <= c ad x i = 0 or 1 for all i If ot, this istace has a better solutio b 2, b 3,, b. p i b i > p i a i x 1 = a 1 = 0 x 1 = a 1 = 1 x 1 = a 1, x 2 = b 2, x 3 = b 3,, x = b is a better solutio to the origial istace tha is x 1 = a 1, x 2 = a 2, x 3 = a 3,, x = a. So x 1 = a 1, x 2 = a 2, x 3 = a 3,, x = a caot be a optimal solutio a cotradictio with the assumptio that it is optimal. Next, cosider the case a 1 = 1. Followig the first decisio the state is (2, c-w 1 ). a 2, a 3,, a must be a optimal solutio to the kapsack istace give by the state (2,c -w 1 ).

4 x 1 = a 1 = 1 maximize p i x i subject to w i x i <= c- w 1 ad x i = 0 or 1 for all i x 1 = a 1 = 1 x 1 = a 1, x 2 = b 2, x 3 = b 3,, x = b is a better solutio to the origial istace tha is x 1 = a 1, x 2 = a 2, x 3 = a 3,, x = a. So x 1 = a 1, x 2 = a 2, x 3 = a 3,, x = a caot be a optimal solutio a cotradictio with the assumptio that it is optimal. If ot, this istace has a better solutio b 2, b 3,, b. p i b i > p i a i 0/1 Kapsack Problem Therefore, o matter what the first decisio, the remaiig decisios are optimal with respect to the state that results from this decisio. The priciple of optimality holds ad dyamic programmig may be applied. Dyamic Programmig Recurrece Let f(i,y) be the profit value of the optimal solutio to the kapsack istace defied by the state (i,y). ƒ Items i through are available. ƒ Available capacity is y. For the time beig assume that we wish to determie oly the value of the best solutio. ƒ Later we will worry about determiig the x i s that yield this maximum value. Uder this assumptio, our task is to determie f(1,c).

5 Dyamic Programmig Recurrece f(,y) is the value of the optimal solutio to the kapsack istace defied by the state (,y). ƒ Oly item is available. ƒ Available capacity is y. If w <= y, f(,y) = p. If w > y, f(,y) = 0. Dyamic Programmig Recurrece Suppose that i <. f(i,y) is the value of the optimal solutio to the kapsack istace defied by the state (i,y). ƒ Items i through are available. ƒ Available capacity is y. Suppose that i the optimal solutio for the state (i,y), the first decisio is to set x i = 0. From the priciple of optimality (we have show that this priciple holds for the kapsack problem), it follows that f(i,y) = f(i+1,y). Dyamic Programmig Recurrece The oly other possibility for the first decisio is x i = 1. The case x i = 1 ca arise oly whe y >= w i. From the priciple of optimality, it follows that f(i,y) = f(i+1,y-w i ) + p i. Combiig the two cases, we get ƒ f(i,y) = f(i+1,y) wheever y < w i. ƒ f(i,y) = max{f(i+1,y), f(i+1,y-w i ) + p i }, y >= w i. Recursive Code f(i,y) */ private static it f(it i, it y) { if (i == ) retur (y < w[])? 0 : p[]; if (y < w[i]) retur f(i + 1, y); retur Math.max(f(i + 1, y), f(i + 1, y - w[i]) + p[i]); }

6 Recursio Tree f(1,c) f(2,c) f(2,c-w 1 ) f(3,c) f(3,c-w 2 ) f(3,c-w 1 ) f(3,c-w 1 w 2 ) f(4,c) f(4,c-w 3 ) f(4,c-w 2 ) f(4,c-w 1 w 3 ) Time Complexity Let t() be the time required whe items are available. t(0) = t(1) = a, where a is a costat. Whe t > 1, t() <= 2t(-1) + b, where b is a costat. t() = O(2 ). f(5,c) f(5,c-w 1 w 3 w 4 ) Solvig dyamic programmig recurreces recursively ca be hazardous to ru time. Reducig Ru Time f(1,c) f(2,c) f(2,c-w 1 ) f(3,c) f(3,c-w 2 ) f(3,c-w 1 ) f(3,c-w 1 w 2 ) f(4,c) f(4,c-w 3 ) f(4,c-w 2 ) f(4,c-w 1 w 3 ) f(5,c) f(5,c-w 1 w 3 w 4 ) Time Complexity Level i of the recursio tree has up to 2 i-1 odes. At each such ode a f(i,y) is computed. Several odes may compute the same f(i,y). We ca save time by ot recomputig already computed f(i,y)s. Save computed f(i,y)s i a dictioary. ƒ Key is (i, y) value. ƒ f(i, y) is computed recursively oly whe (i,y) is ot i the dictioary. ƒ Otherwise, the dictioary value is used.

7 Iteger Weights Iteger Weights Dictioary Assume that each weight is a iteger. The kapsack capacity c may also be assumed to be a iteger. Oly f(i,y)s with 1 <= i <= ad 0 <= y <= c are of iterest. Eve though level i of the recursio tree has up to 2 i-1 odes, at most c+1 represet differet f(i,y)s. Use a array farray[][] as the dictioary. farray[1:][0:c] farray[i][y] = -1 iff f(i,y) ot yet computed. This iitializatio is doe before the recursive method is ivoked. The iitializatio takes O(c) time. No Recomputatio Code private static it f(it i, it y) { if (farray[i][y] >= 0) retur farray[i][y]; if (i == ) {farray[i][y] = (y < w[])? 0 : p[]; retur farray[i][y];} if (y < w[i]) farray[i][y] = f(i + 1, y); else farray[i][y] = Math.max(f(i + 1, y), f(i + 1, y - w[i]) + p[i]); retur farray[i][y]; } Time Complexity t() = O(c). Aalysis doe i text. Good whe c is small relative to 2. = 3, c = w = [100102, , 6327] p = [102, 505, 5] 2 = 8 c =