Recursion and Dynamic Programming

Fibonacci with Memoization

Compute fib(n) recursively. Cache each fib(k) in a memo map so each subproblem is solved at most once.

Algorithm

Canonical input n = 6 produces fib(6) = 8. Replay highlights every memo write and every cache hit.

memoization A `HashMap<Int, Int>` keyed by `n` stores each completed subproblem. Before recursing, check `memo.containsKey(n)`: a hit returns immediately, a miss descends.
explicit memo state The memo is threaded through the recursion as `memo: HashMap<Int, Int>` so the lesson stays about caching, not global state.

Basic Implementation

basic.kt
Replay: real traced execution (multi-file project)
fun fib(n: Int, memo: HashMap<Int, Int>): Int {
	if (memo.containsKey(n)) {
		return memo[n]!!
	}
	if (n < 2) {
		memo[n] = n
		return n
	}
	val value = fib(n - 1, memo) + fib(n - 2, memo)
	memo[n] = value
	return value
}

fun main() {
	val memo = HashMap<Int, Int>()
	val result = fib(6, memo)
	println(result)
}

  1. memo ← {}, action ← miss -> descend fib(5)

    8}9val value = fib(n - 1, memo) + fib(n - 2, memo)10memo[n] = value
    values this step{}memomiss -> descend fib(5)action6n

  2. memo ← {}, action ← miss -> descend fib(4)

    8}9val value = fib(n - 1, memo) + fib(n - 2, memo)10memo[n] = value
    values this step{}memomiss -> descend fib(4)action5n

  3. memo ← {}, action ← miss -> descend fib(3)

    8}9val value = fib(n - 1, memo) + fib(n - 2, memo)10memo[n] = value
    values this step{}memomiss -> descend fib(3)action4n

  4. memo ← {}, action ← miss -> descend fib(2)

    8}9val value = fib(n - 1, memo) + fib(n - 2, memo)10memo[n] = value
    values this step{}memomiss -> descend fib(2)action3n

  5. memo ← {}, action ← miss -> descend fib(1)

    8}9val value = fib(n - 1, memo) + fib(n - 2, memo)10memo[n] = value
    values this step{}memomiss -> descend fib(1)action2n

  6. memo ← {1: 1}, action ← base 1; memo[1] = 1; return

    8}9val value = fib(n - 1, memo) + fib(n - 2, memo)10memo[n] = value
    values this step{1: 1}memobase 1; memo[1] = 1; returnaction1n

  7. memo ← {0: 0, 1: 1}, action ← base 0; memo[0] = 0; fib(2)=1; memo[2] = 1

    8}9val value = fib(n - 1, memo) + fib(n - 2, memo)10memo[n] = value
    values this step{0: 0, 1: 1}memobase 0; memo[0] = 0; fib(2)=1; memo[2] = 1action0n

  8. memo ← {0: 0, 1: 1, 2: 1, 3: 2}, action ← hit 1; fib(3)=2; memo[3] = 2

    8}9val value = fib(n - 1, memo) + fib(n - 2, memo)10memo[n] = value
    values this step{0: 0, 1: 1, 2: 1, 3: 2}memohit 1; fib(3)=2; memo[3] = 2action1n

  9. memo ← {0: 0, 1: 1, 2: 1, 3: 2, 4: 3}, action ← hit 1; fib(4)=3; memo[4] = 3

    8}9val value = fib(n - 1, memo) + fib(n - 2, memo)10memo[n] = value
    values this step{0: 0, 1: 1, 2: 1, 3: 2, 4: 3}memohit 1; fib(4)=3; memo[4] = 3action2n

  10. memo ← {0: 0, 1: 1, 2: 1, 3: 2, 4: 3, 5: 5}, action ← hit 2; fib(5)=5; memo[5] = 5

    8}9val value = fib(n - 1, memo) + fib(n - 2, memo)10memo[n] = value
    values this step{0: 0, 1: 1, 2: 1, 3: 2, 4: 3, 5: 5}memohit 2; fib(5)=5; memo[5] = 5action3n

  11. memo ← {0: 0, 1: 1, 2: 1, 3: 2, 4: 3, 5: 5, 6: 8}, action ← hit 3; fib(6)=8; memo[6] = 8

    8}9val value = fib(n - 1, memo) + fib(n - 2, memo)10memo[n] = value
    values this step{0: 0, 1: 1, 2: 1, 3: 2, 4: 3, 5: 5, 6: 8}memohit 3; fib(6)=8; memo[6] = 8action4n

  12. stdout ← 8

    16	val result = fib(6, memo)17	println(result)18}
    values this step8stdout8result

Complexity

  • Time: O(n) with memoization (vs. O(2^n) without)
  • Space: O(n) memo + O(n) call stack

Implementation notes

  • Kotlin: the recursion takes the memo as a HashMap<Int, Int> argument rather than a companion object field, which keeps state explicit without hiding the lesson behind a shared global. The containsKey + !! indexer pair stays parallel to the lesson spec instead of leaning on getOrPut.
  • The replay shows the call stack on one side and the memo map on the other so memo writes and cache hits are visually distinct.