Divide and Conquer DP

Divide and Conquer is a dynamic programming optimization.

Preconditions

Some dynamic programming problems have a recurrence of this form: $$dp(i, j) = \min_{0 \leq k \leq j} \{ dp(i - 1, k - 1) + C(k, j) \}$$ Where $C(k, j)$ is a cost function and $dp(i, j) = 0$ when $j \lt 0$.

Say $0 \leq i \lt m$ and $0 \leq j \lt n$, and evaluating $C$ takes $O(1)$ time. Then the straightforward evaluation of the above recurrence is $O(m n^2)$. There are $m \times n$ states, and $n$ transitions for each state.

Let $opt(i, j)$ be the value of $k$ that minimizes the above expression. If $opt(i, j) \leq opt(i, j + 1)$ for all $i, j$, then we can apply divide-and-conquer DP. This is known as the monotonicity condition. The optimal "splitting point" for a fixed $i$ increases as $j$ increases.

This lets us solve for all states more efficiently. Say we compute $opt(i, j)$ for some fixed $i$ and $j$. Then for any $j' < j$ we know that $opt(i, j') \leq opt(i, j)$. This means when computing $opt(i, j')$, we don't have to consider as many splitting points!

To minimize the runtime, we apply the idea behind divide and conquer. First, compute $opt(i, n / 2)$. Then, compute $opt(i, n / 4)$, knowing that it is less than or equal to $opt(i, n / 2)$ and $opt(i, 3 n / 4)$ knowing that it is greater than or equal to $opt(i, n / 2)$. By recursively keeping track of the lower and upper bounds on $opt$, we reach a $O(m n \log n)$ runtime. Each possible value of $opt(i, j)$ only appears in $\log n$ different nodes.

Note that it doesn't matter how "balanced" $opt(i, j)$ is. Across a fixed level, each value of $k$ is used at most twice, and there are at most $\log n$ levels.

Generic implementation

Even though implementation varies based on problem, here's a fairly generic template. The function compute computes one row $i$ of states dp_cur, given the previous row $i-1$ of states dp_before. It has to be called with compute(0, n-1, 0, n-1). The function solve computes m rows and returns the result.

int m, n;
vector<long long> dp_before(n), dp_cur(n);

long long C(int i, int j);

// compute dp_cur[l], ... dp_cur[r] (inclusive)
void compute(int l, int r, int optl, int optr) {
    if (l > r)
        return;

    int mid = (l + r) >> 1;
    pair<long long, int> best = {LLONG_MAX, -1};

    for (int k = optl; k <= min(mid, optr); k++) {
        best = min(best, {(k ? dp_before[k - 1] : 0) + C(k, mid), k});
    }

    dp_cur[mid] = best.first;
    int opt = best.second;

    compute(l, mid - 1, optl, opt);
    compute(mid + 1, r, opt, optr);
}

int solve() {
    for (int i = 0; i < n; i++)
        dp_before[i] = C(0, i);

    for (int i = 1; i < m; i++) {
        compute(0, n - 1, 0, n - 1);
        dp_before = dp_cur;
    }

    return dp_before[n - 1];
}

Things to look out for

The greatest difficulty with Divide and Conquer DP problems is proving the monotonicity of $opt$. Many Divide and Conquer DP problems can also be solved with the Convex Hull trick or vice-versa. It is useful to know and understand both!

Practice Problems

References