How to understand the dynamic programming solution in linear partitioning?

33
votes

I'm struggling to understand the dynamic programming solution to linear partitioning problem. I am reading the The Algorithm Design Manual and the problem is described in section 8.5. I've read the section countless times but I'm just not getting it. I think it's a poor explanation (the what I've read up to now has been much better), but I've not been able to understand the problem well enough to look for an alternative explanation. Links to better explanations welcome!

I've found a page with text similar to the book (maybe from the first edition of the book): The Partition Problem.

First question: In the example in the book the partitions are ordered from smallest to largest. Is this just coincidence? From what I can see the ordering of the elements is not significant to the algorithm.

This is my understanding of the recursion:

Lets use the following sequence and partition it into 4:

{S1...Sn} =  100   150   200   250   300   350   400   450   500
k = 4

Second question: Here's how I think the recursion will begin - have I understood it correctly?

The 1st recursion is:

100   150   200   250   300   350   400   450 | 500 //3 partition to go
100   150   200   250   300   350   400 | 450 | 500 //2 partition to go 
100   150   200   250   300   350 | 400 | 450 | 500 //1 partition to go
100   150   200   250   300 | 350 | 400 | 450 | 500 //done

The 2nd recursion is:

100   150   200   250   300   350   400   450 | 500 //3 partition to go
100   150   200   250   300   350   400 | 450 | 500 //2 partition to go 
100   150   200   250   300   350 | 400 | 450 | 500 //1 partition to go
100   150   200   250 | 300   350 | 400 | 450 | 500 //done

The 3rd recursion is:

100   150   200   250   300   350   400   450 | 500 //3 partition to go
100   150   200   250   300   350   400 | 450 | 500 //2 partition to go 
100   150   200   250   300   350 | 400 | 450 | 500 //1 partition to go
100   150   200 | 250   300   350 | 400 | 450 | 500 //done

The 4th recursion is:

100   150   200   250   300   350   400   450 | 500 //3 partition to go
100   150   200   250   300   350   400 | 450 | 500 //2 partition to go 
100   150   200   250   300   350 | 400 | 450 | 500 //1 partition to go
100   150 | 200   250   300   350 | 400 | 450 | 500 //done

The 5th recursion is:

100   150   200   250   300   350   400   450 | 500 //3 partition to go
100   150   200   250   300   350   400 | 450 | 500 //2 partition to go 
100   150   200   250   300   350 | 400 | 450 | 500 //1 partition to go
100 | 150   200   250   300   350 | 400 | 450 | 500 //done

The 6th recursion is:

100   150   200   250   300   350   400   450 | 500 //3 partition to go
100   150   200   250   300   350   400 | 450 | 500 //2 partition to go 
100   150   200   250   300 | 350   400 | 450 | 500 //1 partition to go
100   150   200   250 | 300 | 350   400 | 450 | 500 //done

The 7th recursion is:

100   150   200   250   300   350   400   450 | 500 //3 partition to go
100   150   200   250   300   350   400 | 450 | 500 //2 partition to go 
100   150   200   250   300 | 350   400 | 450 | 500 //1 partition to go
100   150   200 | 250   300 | 350   400 | 450 | 500 //done

The 8th recursion is:

100   150   200   250   300   350   400   450 | 500 //3 partition to go
100   150   200   250   300   350   400 | 450 | 500 //2 partition to go 
100   150   200   250   300 | 350   400 | 450 | 500 //1 partition to go
100   150 | 200   250   300 | 350   400 | 450 | 500 //done

The 9th recursion is:

100   150   200   250   300   350   400   450 | 500 //3 partition to go
100   150   200   250   300   350   400 | 450 | 500 //2 partition to go 
100   150   200   250   300 | 350   400 | 450 | 500 //1 partition to go
100 | 150   200   250   300 | 350   400 | 450 | 500 //done

etc...

Here's the code as it appears in the book:

partition(int s[], int n, int k)
{
    int m[MAXN+1][MAXK+1];                  /* DP table for values */
    int d[MAXN+1][MAXK+1];                  /* DP table for dividers */ 
    int p[MAXN+1];                          /* prefix sums array */
    int cost;                               /* test split cost */
    int i,j,x;                              /* counters */
    
    p[0] = 0;                               /* construct prefix sums */
    for (i=1; i<=n; i++) p[i]=p[i-1]+s[i];
    
    for (i=1; i<=n; i++) m[i][3] = p[i];    /* initialize boundaries */
    for (j=1; j<=k; j++) m[1][j] = s[1];
    
    
    for (i=2; i<=n; i++)                    /* evaluate main recurrence */
        for (j=2; j<=k; j++) {
            m[i][j] = MAXINT;
            for (x=1; x<=(i-1); x++) {
                cost = max(m[x][j-1], p[i]-p[x]);
                if (m[i][j] > cost) {
                    m[i][j] = cost;
                    d[i][j] = x;
                }
            }
        }

    reconstruct_partition(s,d,n,k);         /* print book partition */
}

Question about the algorithm:

  1. What values are being stored in the m and d?
  2. What does 'cost' mean? Is it simply the total of the elements values within a partition? Or is there some additional more subtle meaning?
3
algorithmpartitioningdynamic-programming
BTW, even if you can't answer my questions I would appreciate comments on the quality of the source material. I'd like some confirmation that it's not just me that finds the explanation poor (It's made me feel fairly stoopid). – Benedict Cohen
I don't think you will find many people here able to answer you question without giving a succinct explanation of the problem you need to solve. There are many variations of partitioning problems and pasting long tables of the algorithm being executed by hand doesn't halo make things clearer. – hugomg

3 Answers

38
votes

Be aware that there's a small mistake in the explanation of the algorithm in the book, look in the errata for the text "(*) Page 297".

About your questions:

  1. No, the items don't need to be sorted, only contiguous (that is, you can't rearrange them)
  2. I believe the easiest way to visualize the algorithm is by tracing by hand the reconstruct_partition procedure, using the rightmost table in figure 8.8 as a guide
  3. In the book it states that m[i][j] is "the minimum possible cost over all partitionings of {s1, s2, ... , si}" into j ranges, where the cost of a partition is the larges sum of elements in one of its parts". In other words, it's the "smallest maximum of sums", if you pardon the abuse of terminology. On the other hand, d[i][j] stores the index position which was used to make a partition for a given pair i,j as defined before
  4. For the meaning of "cost", see the previous answer

Edit:

Here's my implementation of the linear partitioning algorithm. It's based on Skiena's algorithm, but in a pythonic way; and it returns a list of the partitions.

from operator import itemgetter

def linear_partition(seq, k):
    if k <= 0:
        return []
    n = len(seq) - 1
    if k > n:
        return map(lambda x: [x], seq)
    table, solution = linear_partition_table(seq, k)
    k, ans = k-2, []
    while k >= 0:
        ans = [[seq[i] for i in xrange(solution[n-1][k]+1, n+1)]] + ans
        n, k = solution[n-1][k], k-1
    return [[seq[i] for i in xrange(0, n+1)]] + ans

def linear_partition_table(seq, k):
    n = len(seq)
    table = [[0] * k for x in xrange(n)]
    solution = [[0] * (k-1) for x in xrange(n-1)]
    for i in xrange(n):
        table[i][0] = seq[i] + (table[i-1][0] if i else 0)
    for j in xrange(k):
        table[0][j] = seq[0]
    for i in xrange(1, n):
        for j in xrange(1, k):
            table[i][j], solution[i-1][j-1] = min(
                ((max(table[x][j-1], table[i][0]-table[x][0]), x) for x in xrange(i)),
                key=itemgetter(0))
    return (table, solution)
3
votes

I've implemented Óscar López algorithm on PHP. Please feel free to use it whenever you need it.

 /**
 * Example: linear_partition([9,2,6,3,8,5,8,1,7,3,4], 3) => [[9,2,6,3],[8,5,8],[1,7,3,4]]
 * @param array $seq
 * @param int $k
 * @return array
 */
protected function linear_partition(array $seq, $k)
{
    if ($k <= 0) {
        return array();
    }

    $n = count($seq) - 1;
    if ($k > $n) {
        return array_map(function ($x) {
            return array($x);
        }, $seq);
    }

    list($table, $solution) = $this->linear_partition_table($seq, $k);
    $k = $k - 2;
    $ans = array();

    while ($k >= 0) {
        $ans = array_merge(array(array_slice($seq, $solution[$n - 1][$k] + 1, $n - $solution[$n - 1][$k])), $ans);
        $n = $solution[$n - 1][$k];
        $k = $k - 1;
    }

    return array_merge(array(array_slice($seq, 0, $n + 1)), $ans);
}

protected function linear_partition_table($seq, $k)
{
    $n = count($seq);

    $table = array_fill(0, $n, array_fill(0, $k, 0));
    $solution = array_fill(0, $n - 1, array_fill(0, $k - 1, 0));

    for ($i = 0; $i < $n; $i++) {
        $table[$i][0] = $seq[$i] + ($i ? $table[$i - 1][0] : 0);
    }

    for ($j = 0; $j < $k; $j++) {
        $table[0][$j] = $seq[0];
    }

    for ($i = 1; $i < $n; $i++) {
        for ($j = 1; $j < $k; $j++) {
            $current_min = null;
            $minx = PHP_INT_MAX;

            for ($x = 0; $x < $i; $x++) {
                $cost = max($table[$x][$j - 1], $table[$i][0] - $table[$x][0]);
                if ($current_min === null || $cost < $current_min) {
                    $current_min = $cost;
                    $minx = $x;
                }
            }

            $table[$i][$j] = $current_min;
            $solution[$i - 1][$j - 1] = $minx;
        }
    }

    return array($table, $solution);
}
1
votes

The following is a modified implementation of Skienna's Linear partitioning algorithm in python that does not calculate the last k column values except of the answer itself : M[N][K] (a cell calculation only depends on the previous )

A test against the input {1,2,3,4,5,6,7,8,9} (used in Skienna's example in the book ) yields a slightly different matrix M ( given the above modification) but correctly returns the final result (in this example the minimum-cost partitioning of s into k ranges is 17 , and matrix D is used to print the list of dividers positions that lead to this optimum) .

import math   
    
def partition(s, k):
    # compute prefix sums

    n = len(s)
    p = [0 for _ in range(n)]
    m = [[0 for _ in range(k)] for _ in range(n)]
    d = [[0 for _ in range(k)] for _ in range(n)]

    for i in range(n):
        p[i] = p[i-1] + s[i]

    # initialize boundary conditions
    for i in range(n):
        m[i][0] = p[i]

    for i in range(k):
        m[0][i] = s[0]

    # Evaluate main recurrence
    for i in range(1, n):
        """
          omit calculating the last M's column cells 
          except for the sought minimum cost M[N][K]
        """
        if i != n - 1:
            jlen = k - 1
        else:
            jlen = k

        for j in range(1, jlen):

            """
            - computes the minimum-cost partitioning  of the set {S1,S2,.., Si} into j partitions .
            - this part should be investigated more closely .
            
            """
            #
            m[i][j] = math.inf

            # This loop needs to be traced to understand it better
            for x in range(i):
                sup = max(m[x][j-1], p[i] - p[x])
                if m[i][j] > sup:
                    m[i][j] = sup
                    # record which divider position was required to achieve the value s
                    d[i][j] = x+1

    return s, d, n, k


def reconstruct_partition(S, D, N, K):
    if K == 0:
        for i in range(N):
            print(S[i], end="_")
        print(" | ", end="")
    else:
        reconstruct_partition(S, D, D[N-1][K-1], K-1)
        for i in range(D[N-1][K-1], N):
            print(S[i], end="_")
        print(" | ", end="")

# MAIN PROGRAM

S, D, N, K = partition([1, 2, 3, 4, 5, 6, 7, 8, 9], 3)

reconstruct_partition(S, D, N, K)