Knapsack - save time and memory - c

Question

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]);

Answer 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]);

Answer 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;