程序代写代做代考 algorithm Imperial College London – Department of Computing

Imperial College London – Department of Computing

MSc in Computing Science

580: Algorithms

Tutorial: Dynamic Programming

1. A thief can carry k kilograms of loot in his knapsack. He robs a shop containing N items.
Item i is worth bi bitcoin and weighs ki kilos. The thief wants to decide which items to
take to maximise the total value he steals.

(a) How would you decompose this problem into subproblems? Does the problem have
optimal substructure and overlapping subproblems?

(b) Write an algorithm that, given an array B such that B[i] is the value of item i and
an array K such that K[i] is the weight of item i, and a maximum weight k, solves
the problem in O(kN) time.

(c) Since he is greedy, the thief attempts to use the following strategy: the next item
chosen should always be the one with the greatest value per kilogram, from those
remaining. Show that this strategy is not guaranteed to give the optimal solution.

Answer:

(a) First, it is important to note a difference between this problem and the rod
cutting problem. In this problem, each item can only be stolen once. So, item
i can only be used once in a solution, whereas you can make as many rods of
length i as you like, within the total of N . So, this problem (and its subproblems)
has two dimensions: the space available in the knapsack, and the items (still)
available. So, the solution you are looking for is V (N, k), the maximum value
that can be achieved for a knapsack of size k, by choosing from items 1 to N .

A possible trap is to start with the problem of finding V (N, k) and divide this
into one possibility for each available item: bi + V (?, k − ki). This is awkward
to implement because you have to track which items are still available somehow
(hence the ”?”). It also duplicates options by considering the items in all possible
orders. When considering item N , there are just two possibilities. Either you
take item N , giving a possible solution bN + V (N − 1, k − kN ), or you do not
giving a possible solution V (N − 1, k). There is no need to separately consider
picking item 1 or item 2 first, because these items can still be chosen within the
second option V (N − 1, k).

So, V (i, w), the maximum value that can be gained if there is room for w kilos
in the knapsack and items 1 to i are available is:

V (i, w) = max(bi + V (i− 1, w − ki), V (i− 1, w))

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Including base cases and edge cases we have:

• V (i, 0) = 0, for all i

• V (0, w) = 0, for all w

• V (i, w) = V (i− 1, w), if ki > w

• V (i, w) = max(V (i− 1, w), V (i− 1, w − ki) + bi), otherwise

and the overall solution is V (N, k). The optimal solution is defined using optimal
solutions to subproblems, and these subproblems occur multiple times, so the
problem has optimal substructure and overlapping subproblems.

(b)
1: procedure Knapsack(B = [B1, …, BN ], K = [K1, …,KN ], k)
2: V = new array[N + 1][k + 1]
3: for i = 0 to N do
4: V [i][0] = 0
5: end for
6: for i = 0 to k do
7: V [0][i] = 0
8: end for
9: for i = 1 to N do

10: for w = 1 to k do
11: if K[i] > w then
12: V [i][w] = V [i− 1][w]
13: else
14: V [i][w] = Max(B[i] + V [i− 1][w −K[i]],V [i− 1][w])
15: end if
16: end for
17: end for
18: return V [N ][k]
19: end procedure

(c) Since the strategy must always work, it is sufficient to find a single counter
example to show that it is incorrect. So, if the shop contains three items with
weights and values as follows:

i 1 2 3

ki 5 10 12

bi 5 10 13

and the thief’s knapsack can hold 15 kg, then choosing the item with the greatest
value per kg (item 3) will give a total value of 13, whereas choosing the other 2
items together will give a value of 15. So, the strategy is not optimal.

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