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.

Short Answer

Expert verified
  1. There are 8possible patterns which can be legally occur according to the given constraint.
  2. For each pattern, there is set of compatible patterns. Split the problem into subproblems and find the optimal value by pebbling column i with patten j. Create separate arrays for each pattern type.

Step by step solution

01

Consider the given information

The checkboard has 4 rows and ncolumns. 2npebbles need to be placed on the check board such that they are legal patterns. The condition for the pattern to be legal is that no two pebbles can be placed on vertically or horizontally adjacent squares.

02

Determining the legal patterns

The following table shows are possible patterns with the given constraint:

Pattern

Description

_ _ _ _

No Pebble in any Row

X _ _ _

Pebble at 1st Row

_ X _ _

Pebble at 2nd Row

_ _ X _

Pebble at 3rd Row

_ _ _ X

Pebble at 4th Row

X _ _ X

Pebble at 1st and 4th Row

_ X _ X

Pebble at 2nd and 4th Row

X _ X _

Pebble at 1st and 3rd Row

Here, ‘X’ denotes Pebble position and ‘_’ denotes empty place.

Thus, the possible legal patterns are 8.

03

Determine the dynamic programming to compute optimal placement

Consider the initial checkboard be M[i,j]. The method compatibility (i) gives the patterns compatible with i. In the array A[i,j], i denotes pattern and j denotes column. Sum (i, j) method returns the sum of numbers on column i when pattern j is applied.

Ai,0=0foralliforxfrom1ton:  foryfrom0to7    max=0    m=Sumx,y    forpincompatibilityy      n=Sumx1,p      ifm+n>max         thenmax=m+n      Ax,y=maxreturnMax(Ai,Nforall i

The time complexity of the algorithm is O(n).

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Most popular questions from this chapter

A subsequence is palindromic if it is the same whether read left to right or right to left. For instance, the sequence

A,C,G,T,G,T,C,A,A,A,A,T,C,G

has many palindromic subsequences, including A,C,G,C,Aand A,A,A,A(on the other hand, the subsequence A,C,Tis not palindromic). Devise an algorithm that takes a sequence X[1...n]and returns the (length of the) longest palindromic subsequence. Its running time should be0(n2).

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?

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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.

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  1. Show that if we just want to compute the value of the edit distance (rather than the optimal sequence of edits), then only O(n) space is needed, because only a small portion of the table needs to be maintained at any given time.
  2. Now suppose that we also want the optimal sequence of edits. As we saw earlier, this problem can be recast in terms of a corresponding grid-shaped dag, in which the goal is to find the optimal path from node (0,0) to node (n,m). It will be convenient to work with this formulation, and while we’re talking about convenience, we might as well also assume that is a power of 2.
    Let’s start with a small addition to the edit distance algorithm that will turn out to be very useful. The optimal path in the dag must pass through an intermediate node (k,m2) for some k; show how the algorithm can be modified to also return this value k.
  3. Now consider a recursive scheme:
    Procedure find-path((0,0)(n,m))
    Compute the value kabove
    find-path ((0,0)k,m2)
    find-path k,m2n,m
    concatenate these two paths, with kin the middle.
    Show that this scheme can be made to run inO (mn) time and O(n) space.

You are given a convex polygon P on n vertices in the plane (specified by their x and y coordinates). A triangulation of P is a collection of n-3diagonals of such that no two diagonals intersect (except possibly at their endpoints). Notice that a triangulation splits the polygon’s interior into n-2 disjoint triangles. The cost of a triangulation is the sum of the lengths of the diagonals in it. Give an efficient algorithm for finding a triangulation of minimum cost. (Hint: Label the vertices of P by 1,....,n, starting from an arbitrary vertex and walking clockwise. For 1i<jn, let the subproblem A(i,j)denote the minimum cost triangulation of the polygon spanned by vertices i,i+1,...,j.).

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