Modify your algorithm for Problem 417 so that it determines the optimal investment strategy. in .NET Embed Data Matrix ECC200 in .NET Modify your algorithm for Problem 417 so that it determines the optimal investment strategy.

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Modify your algorithm for Problem 417 so that it determines the optimal investment strategy. using none toembed none on web,windows application Microsoft Windows Official Website After graduation none none , you start working for a company that has a hierarchical supervisor structure in the shape of a tree, rooted at the CEO. The personnel o ce has ranked each employee with a conviviality rating between 0 and 10. Your rst assignment is to design a dynamic programming algorithm to construct the guest list for the o ce Christmas party.

The algorithm has input consisting of the hierarchy tree, and conviviality ratings for each employee. The output is to be a guest list that maximizes the total conviviality, and ensures that for every employee invited to the party, his or her immediate supervisor (that is, their parent in the tree) is not invited. 428.

Design such a dynamic programming algorithm, and analyze it.. Sec. 8.7. Hints You are not sure none for none if the CEO should get invited to the party, but you suspect that you might get red if he is not. Can you modify your algorithm to ensure that the CEO does get invited . HINTS 378. Start by dr none for none awing a picture of the array entries on which m[i, j] depends. Make sure that the array is lled in such way that the required entries are there when they are needed.

409. Construct an array B[1..

n] such that B[i] contains the length of the longest ascending subsequence that starts with A[i], for 1 i n. Fill B from back to front. From the information in B, you should be able to nd the length of the longest ascending subsequence.

You should also, with a little thought, be able to output a longest ascending subsequence (there may be more than one of them, obviously) in time O(n). 413. In addition to the table of costs for subproblems, you will need to keep track of the number of blanks in the last line of text in the solution of each subproblem.

The base case is more than just lling in the diagonal entries. 415. An O(nC) time algorithm can be constructed by exploiting the similarity between the coin changing problem and the knapsack problem (see Section 8.

2). It is easy to formulate an inferior algorithm that appears to run in time O(nC 2 ), but with a little thought it can be analyzed using Harmonic numbers to give a running time of O(C 2 log n), which is superior for C = o(n/ log n)..

SOLUTIONS 1 1 2 3 4 5 0 2 none none 48 0 3 352 114 0 4 1020 792 684 0 5 1096 978 936 2394 0. 384. Suppose n = 3, and r0 = 1, r1 = 2, r2 = 32, r3 = 12. The suggested algorithm parenthesizes the product as M1 (M2 M3 ), at a cost of 2 23 12+1 2 12 = 792.

The most e cient way to parenthesize the product is (M1 M2 ) M3 , at a cost of 1 2 32 + 1 32 12 = 448. For a larger example, take n = 7, and dimensions 1, 2, 32, 12, 6, 5, 17, 12. 407.

For each nonterminal symbol S N , keep a Boolean table tS [1..n, 1.

.n]. We.

Chap. 8. Dynamic Programming will ll the tab les so that for 1 i j n, tS [i, j] is true i the string xi xi+1 xj is in the language generated by S. tS [i, j] is true i either i = j and S xi , or there exists k such that i k < j and S T U , where tT [i, k] and tU [k + 1, j] are true. function c (x1 x2 xn ) 1.

for each nonterminal S do 2. for i := 1 to n do 3. tS [i, i] :=true i S xi 4.

for d := 1 to n 1 do 5. for i := 1 to n d do 6. j := i + d 7.

for each nonterminal S do 8. tS [i, j] :=false 9. for each production S T U do 10.

tS [i, j] := tS [i, j] j 1 (tT [i, k] tU [k + 1, j]) k=i 11. return(tR [1, n]) where R is the root Since there are O(1) nonterminals: The for-loop on lines 1 3 costs O(n). Line 10 costs O(n) (a single for-loop).

The for-loop on lines 7 10 costs O(n). The for-loop on lines 5 10 costs O(n2 ). The for-loop on lines 4 10 costs O(n3 ).

. Therefore, the a none for none lgorithm runs in time O(n3 )..
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