# Divide and Conquer Optimization in Dynamic Programming

#### Algorithms Dynamic Programming (DP)

Divide and conquer optimization is used to optimize the run-time of a subset of Dynamic Programming problems from $O(N^2)$ to $O(N logN)$.

For a two-variable function $f(x, y)$ :

• Convex quandrangle inequality : $\forall_{a \leq b \leq c \leq d}(f(a, c) + f(b, d) \leq f(a, d) + f(b, c))$

• Concave quadrangle inequality : $\forall_{a \leq b \leq c \leq d}(f(a, c) + f(b, d) \geq f(a, d) + f(b, c))$

## Optimization criteria

A dynamic programming problem of the form:
$$dp(i, j) = min_{k \leq j}(f(i, j, k))$$
where,
$$f(i, j, k) = dp(i - 1, k) + cost(k, j)$$
Can be optimized using Divide and Conquer optimization if the function $cost(x, y)$ satisfies the convex-quadrangle inequality (sufficient but not necessary).

It can be noticed that the above recurrence, takes $O(n^3)$ time to be solved.

Note: Concave quadrangle inequality should be satisfied in case of maximization problem.

## Optimization

If $cost(x, y)$ obey's the optimization criteria, it results in a useful property:
Note: In some cases even if the criteria is not satisfied, the following property can be observed and the optimization is still possible.
$$h(i, j) \leq h(i, j + 1)$$
where,
$$h(i, j) = argmin_{k \lt j} (f(i, j, k))$$
The above property can be generalized as,
$$h(i, j^{\prime}) \leq h(i, j) \text{ , } j^{\prime} \lt j$$
$h(x, y)$ is the smallest position where $dp(x, y)$ is optimal. This clearly tells us that the solution for $dp(x, y^{\prime})$ will always occur before the solution for $dp(x, y)$, where $y^{\prime} \lt y$ (monotonic).

Let,
$$rec(x, yl, yr, kl, kr)$$
be a function which recursively computes $dp(x, yl..yr)$ for a fixed $x$, given that the solution lies between $kl$ and $kr$.
This can be done by first computing $dp(x, mid)$ by iterating $k$ from $kl$ to $kr$, and then recursively calling
$$rec(x, yl, mid - 1, kl, h(x, mid))$$
and
$$rec(x, mid + 1, yr, h(x, mid), kr)$$
where, $mid = (yl + yr) / 2$.
The initial call will be $rec(x, 1, n, 1, n)$.
This works because, the solution for $dp(x, yl..mid - 1)$ will lie before the solution for $dp(x, mid)$ and the solution for $dp(x, mid + 1..yr)$ will lie after the solution for $dp(x, mid)$ because of the monotonicity of $h(x, y)$.

### Complexity

For a fixed $x$, the number of operations of the above function $rec$ will be of the form:
$$T(n) = 2T(n/2) + O(n)$$
Because of two recursive calls, each with half of the original problem size and another $O(n)$ for finding the solution for the $mid$. By master's theorem, the function will have a complexity of $O(nlogn)$.
Overall compexity will be $O(knlogn)$, because $x$ can take values from 0 to $k-1$.

## Example to demonstrate Divide and Conquer Optimization

Problem

There are $p$ people at an amusement park who are in a queue for a ride. Each pair of people has a measured level of unfamiliarity. The people will be divided into $g$ non-empty contiguous groups. Each division has a total unfamiliarity value which is the sum of the levels of unfamiliarity between any pair of people for each group. Determine the minimal possible total unfamiliarity value.
Given a $pxp$ matrix $unfamiliarity$, where $unfamiliarity[x][y]$ is the measure of unfamiliarity between person $x$ and $y$, find the minimum possible total unfamiliarity after dividing the $p$ people into $g$ non-empty contiguous groups. ($unfamiliarity[x][y] = unfamiliarity[y][x]$)
Example: If there are 3 ($p$) people and they have to be divided into 2 non-empty contiguous groups ($g$) where unfamiliarity between person 0 and 1 is 2 ($unfamiliarity = unfamiliarity = 2$), between person 1 and 2 is 3 ($unfamiliarity = unfamiliarity = 3$) and between person 0 and 2 is 0 ($unfamiliarity = unfamiliarity = 0$).
Minimal unfamiliarity of value 2 is obtained when 0 and 1 is present in one group and 2 is present in the otherjj.
The only other division into 2 non-empty contiguous groups is {{0}, {1, 2}} and that has a total unfamiliarity of 3.

Solution

Let $cost(l, r)$ be the unfamiliarity of a contiguous group from $l$ to $r$ (that is the value if all the people from $l$ to $r$ are grouped together). $cost(l, r)$ can be found in constant time by finding the two-dimensional prefix sum matrix associated with the $unfamiliarity$ matrix.

State: Let $dp[x][y]$ represent the minimum unfamiliarity when the people $1..y$ are split into $x$ groups.
Transition: To compute $dp[x][y]$, the position where the $x$-th contiguous group should start is required. This can be found by iterating $k$ from $0..y$ and computing the unfamiliarity by cutting at each $k$:
$$dp[x][y] = min_{0 \leq k \lt y}(dp[x - 1][k] + cost(k + 1, y))$$

Unoptimized implementation:

const int P = MAX_PEOPLE;
const int G = MAX_GROUPS;

int pre[P][P];
int dp[G][P];

int cost(int l, int r) {
--l;
return pre[r][r] - pre[l][r] - pre[r][l] + pre[l][l];
}

void prefixSum(vector<vector<int>>& unfamiliarity) {
for(int i = 1; i < unfamiliarity.size(); ++i) {
for(int j = 1; j < unfamiliarity.size(); ++j) {
pre[i][j] = unfamiliarity[i][j] + pre[i - 1][j] + pre[i][j - 1] - pre[i - 1][j - 1];
}
}
}

int rec(int x, int y) {
if(x == 0 && y == 0) { return 0; }
if(x == 0) { return (1 << 30); }
int& xy = dp[x][y];
if(xy != -1) { return dp[x][y]; }
xy = (1 << 30);
for(int k = 0; k < y; ++k) {
xy = min(xy, rec(x - 1, k) + cost(k + 1, y));
}
return xy;
}

int minimumUnfamiliarity(int p, int g, vector<vector<int>> unfamiliarity) {
prefixSum(unfamiliarity);
memset(dp, -1, sizeof dp);
return rec(g, p) / 2;
}


The above solution uses prefixSum function to find the prefix sum of unfamiliarity and the rec function to find the minimal unfamiliarity after dividing into g contiguous sequences.

Notice that the cost function satisfies the convex-quadrangle inequality (because it's based on prefix sums). So the above implementation can be optimized using divide and conquer.

const int P = MAX_PEOPLE;
const int G = MAX_GROUPS;

int pre[P][P];
int dp[G][P];

int cost(int l, int r) {
--l;
return pre[r][r] - pre[l][r] - pre[r][l] + pre[l][l];
}

void prefixSum(vector<vector<int>>& unfamiliarity) {
for(int i = 1; i < unfamiliarity.size(); ++i) {
for(int j = 1; j < unfamiliarity.size(); ++j) {
pre[i][j] = unfamiliarity[i][j] + pre[i - 1][j] + pre[i][j - 1] - pre[i - 1][j - 1];
}
}
}

int rec(int x, int yl, int yr, int kl, int kr) {
if(yl > yr) { return 0; }
int mid = yl + (yr - yl) / 2;
int pos = 0;
dp[x][mid] = 1 << 30;
for(int k = kl; k <= min(mid - 1, kr); ++k) {
int cur = dp[x - 1][k] + cost(k + 1, mid);
if(cur < dp[x][mid]) {
dp[x][mid] = cur;
pos = k;
}
}
rec(x, yl, mid - 1, kl, pos);
rec(x, mid + 1, yr, pos, kr);
}

int minimumUnfamiliarity(int p, int g, vector<vector<int>> unfamiliarity) {
prefixSum(unfamiliarity);
for(int i = 0; i < P; ++i) { dp[i] = 1 << 30;  }
dp = 0;
for(int x = 1; x <= g; ++x) {
rec(x, 1, p, 0, p - 1);
}
return dp[g][p] / 2;
}


This implementation, divides the problem into two-equal halves as described before. The function rec computes $dp(x, yl..yr)$ for a fixed x by recursively computing for the left and right halves of $yl$ to $yr$ after finding $dp(x, mid)$ - dp[x][mid] and $h(x, mid)$ - pos (position where $dp(x, mid)$ is minimum). The function minimumUnfamiliarity makes a call to rec for every value of x.
Reducing the complexity from $O(kn^2)$ to $O(knlogn)$.

## Question

#### What is the space complexity of minimumUnfamiliarity?

O(1)
O(kn^2)
O(kn)
O(nlogn)
The algorithm uses the dp table which is of O(kn) size.

With this article at OpenGenus, you must have the complete idea of Divide and Conquer Optimization in Dynamic Programming. Enjoy.

#### Anand Saminathan

Intern at OpenGenus