Question

My homework task is: Let a linear, binary, array A and two positives real numbers a and b, with a < b. Give a dynamic programming algorithm using bottom-up approach, that splits table A to the minimum numbers of continuous sub-arrays, so that the percentage of units (1s) <= a, OR percentage of units (1s) >= b.. Analyze the run time of the algorithm. For example, with input A = [1; 0; 1; 1; 1; 0; 0; 1; 0; 0; 0; 1; 1], a = 23, and b = 64, the Array b can break into the three subsets [1; 0; 1; 1; 1], [0; 0; 1; 0; 0], and [0; 1; 1] with unit rates 80%, 20%, and 66.6%, respectively

My thought: Instead of counting the units (1s) of the sub-array, we can calculate the sum of its elements and divide it with the total numbers of its elements (0s and 1s).I have studied about DP and I know that I have to define a function, a recurrence and a table for storing data, but I find difficulties putting them all together writing the algorithm

Answer 1

IDEA:

There is a famous dynamic programming problem called matrix multiplication.There we will make a 2 dimensional table and fill the values in that table in bottom up approach to finally achieve the required answer.Even in the given problem,we need to make a 2 dimensional matrix(M) such that M(i,j) = answer for the sub array A[ i to j ].

Eg: 1.M(0,0)= A[0] which is equal to 1 since only one element is left

2.M(1,4)={0,1,1,1} and it is equal to 1 since on the whole it satisfies the required condition.

3.M(0,6)={1,0,1,1,1,0,0} and it is equal to 2 as it should be divided into {1,0,1,1,1,0} and {0}.

ALGORITHM:

In this way, we can use the below code in dynamic programming approach to find the value of M(i,j):

if( satisfy(A[ i to j ]) )M(i,j)=1;

else{

M[i][j] = INT_MAX;

for (k=i; k<=j-1; k++)

{

q = M[i][k] + M[k+1][j];

if (q < M[i][j])

M[i][j] = q;

}

}

So,in the bottom up approach by filling the table we can find the required value.The algorithm of the entire code is :

int Array_Division(int A[], int n,int a, int b)

{

    int M[n][n];

    int i, j, k, L, q;

    for (i=1; i<n; i++)

M[i][i] = 1; //Since only 1 element is there its value is 1

    for (L=2; L<n; L++) //Size of sub array

    {

        for (i=1; i<n-L+1; i++)

        {

            j = i+L-1;

if( satisfy(A[ i to j ]) )M(i,j)=1;

else{

M[i][j] = INT_MAX;

for (k=i; k<=j-1; k++)

{

q = M[i][k] + M[k+1][j];

if (q < M[i][j])

M[i][j] = q;

}

}

        }

    }

    return M[0][n-1];

}