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Given an unlimited supply of coins of denominations, we wish to make change for a value ; that is, we wish to find a set of coins whose total value is . This might not be possible: for instance, if the denominations are and 10 then we can make change for 15 but not for 12. Give an dynamic-programming algorithm for the following problem.Input:,; .Question: Is it possible to make change for using coins of denominations ?

Short Answer

Expert verified

The algorithm is as follows:

Create an array of size

D0=Truefori=1 to V   Di=Falseforv=1 to V   forj=1 to V

      ifxjv         Dv=DvORDvxi      else         Dv=Falsereturn DV

Step by step solution

01

Define dynamic programming 

Dynamic programming is a paradigm used for writing algorithm which helps to solve some particular type of problems more efficiently by saving the solution of subproblems and using them to get the final solution. Rather than performing the same calculation again and again, the optimal solution to subproblems are calculated and stored.

02

Determine the subproblem

ConsiderDu as sub-problem such that u=1,.,v.

Thus, according to question,Du is true if it is possible to make change for v using coins denomination.x1,x2,xn

Du=TRUE;ifitispossibletomakechangeforvFALSE;otherwise

Now, takingDu as sub-problem, if it is possible to make change forv using denomination,x1,x2,xnthen it is also possible to make change for by using same denomination with the coin.

The desired recursion is:

Here, ‘’ is the value for which we finding denomination.

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Most popular questions from this chapter

Pebbling a checkerboard. We are given a checkerboard which has 4 rows and ncolumns, and has an integer written in each square. We are also given a set of 2n pebbles, and we want to place some or all of these on the checkerboard (each pebble can be placed on exactly one square) so as to maximize the sum of the integers in the squares that are covered by pebbles. There is one constraint: for a placement of pebbles to be legal, no two of them can be on horizontally or vertically adjacent squares (diagonal adjacency is fine).

  1. Determine the number of legal patterns that can occur in any column (in isolation, ignoring the pebbles in adjacent columns) and describe these patterns.

Call two patterns compatible if they can be placed on adjacent columns to form a legal placement. Let us consider subproblems consisting of the first columns 1kn. Each subproblem can be assigned a type, which is the pattern occurring in the last column.

  1. Using the notions of compatibility and type, give an O(n)-time dynamic programming algorithm for computing an optimal placement.

Let us define a multiplication operation on three symbols a,b,caccording to the following table; thus ab=b,ba=c, and so on. Notice that the multiplication operation defined by the table is neither associative nor commutative.

Find an efficient algorithm that examines a string of these symbols, say bbbbac, and decides whether or not it is possible to parenthesize the string in such a way that the value of the resulting expression is . For example, on input bbbbacyour algorithm should return yes because((b(bb))(ba))c=a.

Reconstructing evolutionary trees by maximum parsimony. Suppose we manage to sequence a particular gene across a whole bunch of different species. For concreteness, say there are n species, and the sequences are strings of length k over alphabet={A,C,G,T}. How can we use this information to reconstruct the evolutionary history of these species?

Evolutionary history is commonly represented by a tree whose leaves are the different species, whose root is their common ancestor, and whose internal branches represent speciation events (that is, moments when a new species broke off from an existing one). Thus we need to find the following:

• An evolutionary tree with the given species at the leaves.

• For each internal node, a string of length K: the gene sequence for that particular ancestor.

For each possible tree T annotated with sequencess(u)kat each of its nodes , we can assign a score based on the principle of parsimony: fewer mutations are more likely.

localid="1659249441524" score(T)=(u.v)E(T)(numberofpositionsonwhichs(u)ands(v)disagree)

Finding the highest-score tree is a difficult problem. Here we will consider just a small part of it: suppose we know the structure of the tree, and we want to fill in the sequences s(u) of the internal nodes u. Here’s an example with k=4 and n=5:


(a) In this particular example, there are several maximum parsimony reconstructions of the internal node sequences. Find one of them.

(b) Give an efficient (in terms of n and k ) algorithm for this task. (Hint: Even though the sequences might be long, you can do just one position at a time.)

Here is yet another variation on the change-making problem (Exercise 6.17). Given an unlimited supply of coins of denominations x1,x2,x3.........xnwe wish to make change for a value v using at most k coins; that is, we wish to find a set ofkcoins whose total value is v. This might not be possible: for instance, if the denominations are 5 and 10 and k=6, then we can make change for 55 but not for 65. Give an efficient dynamic-programming algorithm for the following problem. Input: ; x1,x2,x3.........xn;k;v.Question: Is it possible to make change for v using at most k coins, of denominations x1,x2,x3.........xn?

Alignment with gap penalties. The alignment algorithm of Exercise 6.26 helps to identify DNA sequences that are close to one another. The discrepancies between these closely matched sequences are often caused by errors in DNA replication. However, a closer look at the biological replication process reveals that the scoring function we considered earlier has a qualitative problem: nature often inserts or removes entire substrings of nucleotides (creating long gaps), rather than editing just one position at a time. Therefore, the penalty for a gap of length 10 should not be 10 times the penalty for a gap of length 1, but something significantly smaller.

Repeat Exercise 6.26, but this time use a modified scoring function in which the penalty for a gap of length k is c0 + c1k, where c0 and c1 are given constants (and c0 is larger than c1).

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