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

Lua DSA Implementation

basic.lua 
local function row_string(row)
  local parts = {}
  for i, value in ipairs(row) do parts[i] = tostring(value) end
  return "[" .. table.concat(parts, ", ") .. "]"
end
local function table_string(table_rows)
  local parts = {}
  for i, row in ipairs(table_rows) do parts[i] = row_string(row) end
  return "[" .. table.concat(parts, ", ") .. "]"
end

local weights = {2, 3, 4, 5}
local values = {3, 4, 5, 6}
local capacity = 5
local dp = {}
for item = 1, #weights + 1 do
  dp[item] = {}
  for cap = 1, capacity + 1 do dp[item][cap] = 0 end
end
for item = 1, #weights do
  local weight = weights[item]
  local value = values[item]
  for cap = 0, capacity do
    if weight > cap then
      dp[item + 1][cap + 1] = dp[item][cap + 1]
    else
      local skip = dp[item][cap + 1]
      local take = value + dp[item][cap - weight + 1]
      dp[item + 1][cap + 1] = math.max(skip, take)
    end
  end
end
print(dp[#weights + 1][capacity + 1])
print(table_string(dp))

Implementation notes

  • The source does not build item records; it uses parallel Lua tables: weights = {2, 3, 4, 5} and values = {3, 4, 5, 6}.
  • local capacity = 5, so each DP row has six list slots for logical capacities 0 through 5.
  • Because this Lua code stores rows as 1-based lists, logical capacity cap is stored at column cap + 1.
  • dp is a table of row tables. The setup loop creates #weights + 1 rows: one row for zero items plus one row after each item.
  • Every cell starts at 0; the trace shows the zero-item row as [0, 0, 0, 0, 0, 0].
  • The outer loop is for item = 1, #weights do; each step reads weight = weights[item] and value = values[item].
  • The inner loop scans capacities forward with for cap = 0, capacity do. This is not the 1D reverse-scan version.
  • When weight > cap, the code copies from the previous row with dp[item + 1][cap + 1] = dp[item][cap + 1].
  • Otherwise it compares skip = dp[item][cap + 1] with take = value + dp[item][cap - weight + 1].
  • math.max(skip, take) writes the best value into the next row, so each item can be used at most once.
  • The traced rows are: after item 1 (w=2, v=3), [0, 0, 3, 3, 3, 3]; after item 2 (w=3, v=4), [0, 0, 3, 4, 4, 7].
  • Item 3 (w=4, v=5) updates capacity 4 to 5 but leaves capacity 5 at 7, giving [0, 0, 3, 4, 5, 7].
  • Item 4 (w=5, v=6) does not beat the existing capacity-5 value, so the final row stays [0, 0, 3, 4, 5, 7].
  • The final answer is dp[#weights + 1][capacity + 1], which prints 7.
  • table_string(dp) renders each row with row_string, then joins the rows, matching the printed full table in the replay.

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]]