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Knuth Division

  • Title: Knuth Division
  • Judge / source: CSES
  • Original URL: https://cses.fi/problemset/task/2088/
  • Secondary topics: Interval DP, Prefix sums, Opt monotonicity
  • Difficulty: hard
  • Subtype: Split-point interval DP with additive interval cost and Knuth opt window
  • Status: solved
  • Solution file: knuthdivision.cpp

Why This Note Matters

This is the cleanest in-repo first anchor for the exact Knuth lane.

The recurrence is the standard merge-style interval DP:

\[ dp[l][r] = \min_{l \le k < r}(dp[l][k] + dp[k+1][r] + sum(l, r)). \]

Nothing here is disguised.

That makes it the right flagship for:

  • seeing the plain interval recurrence first
  • then shrinking the split search with Knuth windows

Recognition Cue

Reach for Knuth optimization when:

  • the state is a contiguous interval [l, r]
  • choosing one split k produces two smaller solved intervals
  • the extra price is the total interval sum, independent of k
  • the intended optimization is stronger than generic Interval DP, but not a prefix-row optimization like Divide and Conquer DP

Problem-Specific Transformation

Let:

\[ dp[l][r] = \text{minimum cost to fully split subarray } [l, r] \text{ into singletons}. \]

If the first split happens between k and k + 1, then:

  • left subarray [l, k] is solved optimally
  • right subarray [k + 1, r] is solved optimally
  • we pay the current segment sum once

So:

\[ dp[l][r] = \min_{l \le k < r}\left(dp[l][k] + dp[k+1][r] + sum(l, r)\right). \]

This is exactly the Knuth form because the added term sum(l, r) does not depend on k.

Precomputation

Use prefix sums:

\[ sum(l, r) = pref[r + 1] - pref[l]. \]

Then every candidate split costs:

left answer + right answer + one O(1) interval sum

Exact Knuth Route

Keep:

  • dp[l][r]
  • opt[l][r]

Base:

  • dp[i][i] = 0
  • opt[i][i] = i

Transition window:

start = opt[l][r - 1]
end   = opt[l + 1][r]

and only test k inside that range.

Why The Optimization Fits

This note is not about proving the full theorem from scratch.

It is about recognizing the exact standard lane:

  • merge-cost interval DP
  • additive interval sum
  • monotone split windows imported from the Knuth optimization family

That is enough to route safely to the exact starter.

Implementation Plan

  1. read n and the array
  2. build prefix sums
  3. initialize dp[i][i] = 0, opt[i][i] = i
  4. iterate interval length from 2 to n
  5. for each interval [l, r], scan only k in [opt[l][r-1], opt[l+1][r]]
  6. output dp[0][n-1]

Complexity

  • prefix sums: O(n)
  • DP states: O(n^2)
  • total transition work with Knuth windows: O(n^2)

Main Traps

  • mixing inclusive [l, r] intervals with prefix sums incorrectly
  • forgetting opt[i][i] = i
  • using this route when the extra cost still depends on k
  • confusing it with Divide and Conquer DP, which is a different state shape

Reopen Path

Solution