Recursion and Dynamic Programming

0/1 Knapsack (Small)

Fill a small 0/1 knapsack table where each row decides whether one more item is available.

Algorithm

Steps

  1. Create a table with one extra row for using zero items.
  2. Process the four items in fixed order.
  3. For each capacity, inherit when the item is too heavy.
  4. Otherwise compare skip and take from the previous row.
  5. Print the best value and the full deterministic table.
state transition `dp[i][w]` compares skipping item `i` with taking it and reading the remaining capacity from the previous row.

Complexity

  • Time: O(item_count * capacity)
  • Space: O(item_count * capacity)

Visual walkthrough

The pinned items have weights [2, 3, 4, 5], values [3, 4, 5, 6], and capacity 5. The diagrams show each item row, the overweight inherit branch, and the skip-versus-take comparison that produces 7.

Step 1 - Zero items baseline

With no items available, every capacity has value 0.

Row 0: the empty knapsack is worth zero at every capacity.rowcap 0cap 1cap 2cap 3cap 4cap 50 items000000

Step 2 - First item fills a row

Item 1 has weight 2 and value 3, so capacities below 2 inherit 0 and capacities 2 through 5 can take it.

Row 1 after processing item 1 (weight 2, value 3).rowcap 0cap 1cap 2cap 3cap 4cap 5item 1003333choiceinheritinherittaketaketaketake

Step 3 - Overweight capacity inherits

For item 2 (weight 3, value 4) at capacity 2, weight 3 is too large, so the cell inherits skip = 3 from the previous row.

The overweight branch does not read a negative-capacity cell.decisionitemcapskiptakeresultinherit2 (w=3, v=4)23too heavy3row 2 cap 2w > cap2row 1—inherit

Step 4 - Skip or take at capacity 5

At capacity 5 for item 2, skip = 3 while take = 4 + dp[1][2] = 7, so max(skip, take) is 7.

Item 2 improves capacity 5 by taking the item and using the previous row at capacity 2.cellitemcurrent capskip: prev row cap 5take: prev row cap - weight 2resultcompare2 (w=3, v=4)534 + 3 = 77readprevious rowcap 5dp[1][5]dp[1][2] (5 - 3)max

Step 5 - Final processed row reaches value 7

After processing item 4, the final DP row is [0, 0, 3, 4, 5, 7]; item 4 is only the last processed row, not a reconstructed selected-item set.

Complete 0/1 knapsack table for weights [2, 3, 4, 5] and values [3, 4, 5, 6]; the final row means all four items have been processed.rowcap 0cap 1cap 2cap 3cap 4cap 50 items000000item 1003333item 2003447item 3003457item 4 (final processed row)003457

Fortran DSA Implementation

basic.f90 
program main
  implicit none
  integer :: weights(4), values(4), dp(0:4,0:5)
  integer :: item, cap, weight, value, skip, take
  weights = [2, 3, 4, 5]
  values = [3, 4, 5, 6]
  dp = 0
  do item = 1, 4
    weight = weights(item)
    value = values(item)
    do cap = 0, 5
      if (weight > cap) then
        dp(item, cap) = dp(item - 1, cap)
      else
        skip = dp(item - 1, cap)
        take = value + dp(item - 1, cap - weight)
        dp(item, cap) = max(skip, take)
      end if
    end do
  end do
  print '(I0)', dp(4,5)
  call print_table(dp)
contains
  subroutine print_table(table)
    integer, intent(in) :: table(0:4,0:5)
    integer :: i, w
    write(*, '(A)', advance='no') '['
    do i = 0, 4
      if (i > 0) write(*, '(A)', advance='no') ', '
      write(*, '(A)', advance='no') '['
      do w = 0, 5
        if (w > 0) write(*, '(A)', advance='no') ', '
        write(*, '(I0)', advance='no') table(i,w)
      end do
      write(*, '(A)', advance='no') ']'
    end do
    print '(A)', ']'
  end subroutine print_table
end program main

Output

7
[[0, 0, 0, 0, 0, 0], [0, 0, 3, 3, 3, 3], [0, 0, 3, 4, 4, 7], [0, 0, 3, 4, 5, 7], [0, 0, 3, 4, 5, 7]]