Copying books using dynamic programming

Question

I am implementing the dynamic programming solution for copying books problem. The idea for the solution is taken from here and here.

Problem statement:

Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so called scribers. The scriber had been given a book and after several months he finished its copy. One of the most famous scribers lived in the 15th century and his name was Xaverius Endricus Remius Ontius Xendrianus (Xerox). Anyway, the work was very annoying and boring. And the only way to speed it up was to hire more scribers.

Once upon a time, there was a theater ensemble that wanted to play famous Antique Tragedies. The scripts of these plays were divided into many books and actors needed more copies of them, of course. So they hired many scribers to make copies of these books. Imagine you have m books (numbered 1, 2, ...., m) that may have different number of pages ( p_1, p_2, ..., p_m) and you want to make one copy of each of them. Your task is to divide these books among k scribes, k <= m. Each book can be assigned to a single scriber only, and every scriber must get a continuous sequence of books. That means, there exists an increasing succession of numbers 0 = b_0 < b_1 < b_2, ... < b_{k-1} <= b_k = m$ such that i-th scriber gets a sequence of books with numbers between bi-1+1 and bi. The time needed to make a copy of all the books is determined by the scriber who was assigned the most work. Therefore, our goal is to minimize the maximum number of pages assigned to a single scriber. Your task is to find the optimal assignment.

I am able to obtain the optimal solution for the problem described iteratively, but unable to use that to find the required solution for the problem, that is:

Sample input:
2
9 3
100 200 300 400 500 600 700 800 900
5 4
100 100 100 100 100

Sample Output
100 200 300 400 500 / 600 700 / 800 900
100 / 100 / 100 / 100 100

Where 2 is the number of datasets, 9 is the number of books and 3 is the number of scribes to assign the books to.

Here is my output, for the respective inputs:

100 100 100 
300 300 300 
600 600 600 
1000 700 700 
1500 900 900 
2100 1100 1100 
2800 1300 1300 
3600 1500 1500 
4500 1700 1700 

100 100 100 100 
200 200 200 200 
300 300 300 300 
400 300 300 300 
500 300 300 300 

For the first solution set, I can use 1700 as the optimal number of page assignments to each user and keep on assigning the book pages until, Current scribe page sum >= 1700. However, the second solution does not have any pattern to it whatsoever?

Here is my code to generate the solution:

private void processScribes(){
        int[][] bookScribe = new int[numOfBooks][numOfScribes];
        //set first row to b1 page number
        for (int j = 0; j < numOfScribes; ++j)
            bookScribe[0][j] = bookPages[0];

        //set first column to sum of book page numbers
        for (int row = 1; row < numOfBooks; ++row)
            bookScribe[row][0] = bookScribe[row - 1][0] + bookPages[row]; 

        //calculate the kth scribe using dp
        for (int i = 1; i < numOfBooks; ++i){
            for (int j = 1; j < numOfScribes; ++j){
                //calculate minimum of maximum page numbers
                //from k = l + 1 to i
                //calculate sum 
                int minValue = 1000000;
                for (int k = 0; k < i - 1; ++k){
                    int prevValue = bookScribe[i - k][j - 1];
                    int max = 0;
                    int sumOflVals = 0;
                    for (int l = k + 1; l <= i; ++l){
                        sumOflVals = sumOflVals + bookPages[l];
                    }
                    if (prevValue > sumOflVals){
                        max = prevValue;
                    }
                    else
                        max = sumOflVals;
                    if (max < minValue )
                        minValue = max;
                }
                if (minValue == 1000000)
                    minValue = bookScribe[i][0];
                //store minvalue at [i][j]
                bookScribe[i][j] = minValue;
            }
        }

        //print bookScribes
        for (int i = 0; i < numOfBooks; ++i){
            for (int j = 0; j < numOfScribes; ++j)
                System.out.print(bookScribe[i][j] + " ");
            System.out.println();
        }
        System.out.println();
    }

Any pointers here? Is it the interpretation of solution or something is wrong with how I am translating the recurrence in my code?

Answer 1

Not sure of your solution but here is an intuitive recursive approach with memoization. Let there be n books with ith book having pages[i] pages. Also let there be m subscribers. Also let dp[i][j] be the answer to problem if we were given only books i,i+1.....n and there are only j subscribers to do the job. Following is a recursive pseudo code with memoization

    //dp[][] is memset to -1 from main
    // Assuming books are numbered 1 to n
    // change value of MAX based on your constraints
    int MAX = 1000000000;
    int rec(int position , int sub )
    {
          // These two are the base cases
          if(position > n)
          {
             if(sub == 0)return 0;
             return MAX;
          }
          if(sub == 0)
          {
              if(position > n)return 0;
              return MAX;
          }

          // If answer is already computed for this state return it
          if(dp[position][sub] != -1)return dp[position][sub];

          int ans = MAX,i,sum = 0;
          for(i = position; i <= n;i++)
          {
             sum += pages[i];
             // taking the best of all possible solutions
             ans = min(ans,max(sum,rec(i+1,sub-1)));
          }
          dp[position][sub]=ans;
          return ans; 
    }

    //from main call rec(1,m) which is your answer

You can convert it to an iterative solution by dynamic programming it will be same complexity in time and space .Space is O(n.m) and time is O(n^2.m).

EDIT
Here have a look at running version of the code on your testcases Book Copying Code . It not only finds optimal answer but also prints the optimal assignment with it ( which I have not included in the pseudo code above). ( click on the top right corner fork and it would run on your testcases, input format is same as yours ). Output will optimal answer followed by optimal assignment. Do comment if you have doubts regarding the code.