Knapsack - save time and memory

问题

According to Wikipedia and other sources I had went through, you need matrix m[n][W]; n - number of items and W - total capacity of knapsack. This matrix get really big, sometimes too big to handle it in C program. I know that dynamic programming is based on saving time for memory but still, is there any solution where can you save time and memory?

Pseudo-code for Knapsack problem:

// Input:
// Values (stored in array v)
// Weights (stored in array w)
// Number of distinct items (n)
// Knapsack capacity (W)
for j from 0 to W do
  m[0, j] := 0
end for 
for i from 1 to n do
  for j from 0 to W do
    if w[i] <= j then
      m[i, j] := max(m[i-1, j], m[i-1, j-w[i]] + v[i])
    else
      m[i, j] := m[i-1, j]
    end if
  end for
end for

Lets say, that W = 123456789 and n = 100. In this case we get really big matrix m[100][123456789]. I was thinking how to implement this, but best I have in my mind is just to save which items was selected with one bit (0/1). Is this possible? Or is there any other approach for this problem?

int32 -> 32 * 123456789 * 100 bits
one_bit -> 1 * 123456789 * 100 bits

I hope this is not stupid question and thanks for your effort.

EDIT - working C code:

    long int i, j;
    long int *m[2];
    m[0] = (long int *) malloc(sizeof(long int)*(W+1));
    m[1] = (long int *) malloc(sizeof(long int)*(W+1));
    for(i = 0; i <= W; i++){
        m[0][i] = 0;
    }

    int read = 0;
    int write = 1;
    int tmp;

    long int percent = (W+1)*(n)/100;
    long int counter = 0;

    for(i = 1; i <= n; i++){
        for(j = 0; j <= W; j++){
            if(w[i-1] <= j){
                m[write][j] = max(m[read][j],(v[i-1]) + m[read][j-(w[i-1])]);
            }else{
                m[write][j] = m[read][j];
            }
            counter++;
            if(counter == percent){
                printf(".");    //printing dot (.) for each percent
                fflush(stdout);
                counter = 0;
            }
        }
        tmp = read;
        read = write;
        write = tmp;
    }

    printf("\n%ld\n", m[read][W]);

    free(m[0]);
    free(m[1]);

回答1:

Knapsack problem can be solved using O(W) space.
At each step of the iteration you need only 2 rows - current state of the array m[i] and m[i + 1].

current = 1
int m[2][W]
set NONE for all elements of m # that means we are not able to process this state
m[0][0] = 0 # this is our start point, initially empty knapsack

FOR i in [1..n] do
    next = 3 - current; /// just use 1 or 2 based on the current index
    for j in [0...W] do
       m[next][j] = m[current][j]
    FOR j in [w[i]..W] do
       if m[current][j - w[i]] is not NONE then  # process only reachable positions
           m[next][j] = max(m[next][j], m[current][j - w[i]] + v[i]);
    current = next; /// swap current state and the produced one

Also it is possible to use only 1 array. Here is the pseudocode

FOR i in [1..n] do
    FOR j in [w[i]..W] do
       m[j] = max(m[j], m[j - w[i]] + v[i]);

回答2:

You can decrease the space use from m[100][123456789] into m[2][123456789] by this observation:

Look at this part of the code, at any time, you only need to refer to two rows of the matrix i and i - 1

if w[i] <= j then
  m[i, j] := max(m[i-1, j], m[i-1, j-w[i]] + v[i])
else
  m[i, j] := m[i-1, j]
end if

You can use this trick:

int current = 1;


//.........
if w[i] <= j then
  m[current, j] := max(m[1 - current, j], m[1 - current, j-w[i]] + v[i])
else
  m[i, j] := m[1 - current, j]
end if
current = 1 - current;

来源：https://stackoverflow.com/questions/23777255/knapsack-save-time-and-memory