0/1 Knapsack (Small)

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

Algorithm

@steps

Create a table with one extra row for using zero items.
Process the four items in fixed order.
For each capacity, inherit when the item is too heavy.
Otherwise compare skip and take from the previous row.
Print the best value and the full deterministic table. @end @complexity

Time: O(item_count * capacity)
Space: O(item_count * capacity) @end

state transition `dp[i][w]` compares skipping item `i` with taking it and reading the remaining capacity from the previous row.

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))

@end @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]] @end