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. @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.

Scala DSA Implementation

basic.scala 
object Main extends App {
  def rowString(row: Array[Int]): String = row.mkString("[", ", ", "]")
  def tableString(table: Array[Array[Int]]): String = table.map(rowString).mkString("[", ", ", "]")
  val weights = Array(2, 3, 4, 5)
  val values = Array(3, 4, 5, 6)
  val capacity = 5
  val dp = Array.fill(weights.length + 1, capacity + 1)(0)
  for (item <- 1 to weights.length) {
    val weight = weights(item - 1)
    val value = values(item - 1)
    for (cap <- 0 to capacity) {
      if (weight > cap) dp(item)(cap) = dp(item - 1)(cap)
      else {
        val skip = dp(item - 1)(cap)
        val take = value + dp(item - 1)(cap - weight)
        dp(item)(cap) = math.max(skip, take)
      }
    }
  }
  println(dp(weights.length)(capacity))
  println(tableString(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