Convex hull trick

The convex hull trick is a technique (perhaps best classified as a data structure) used to determine efficiently, after preprocessing, which member of a set of linear functions in one variable attains an extremal value for a given value of the independent variable. It has little to do with convex hull algorithms.

History

The convex hull trick is perhaps best known in algorithm competitions from being required to obtain full marks in several USACO problems, such as MAR08 "acquire", which began to be featured in national olympiads after its debut in the IOI '02 task Batch Scheduling, which itself credits the technique to a 1995 paper (see references). Very few online sources mention it, and almost none describe it. It has been suggested (van den Hooff, 2010) that this is because the technique is "obvious" to anybody who has learned the sweep line algorithm for the line segment intersection problem.

The problem

Suppose that a large set of linear functions in the form $y = m_i x + b_i$ is given along with a large number of queries. Each query consists of a value of $x$ and asks for the minimum value of $y$ that can be obtained if we select one of the linear functions and evaluate it at $x$. For example, suppose our functions are $y=4$, $y=4/3+2/3\,x$, $y=12-3x$, and $y=3-1/2\,x$ and we receive the query $x=1$. We have to identify which of these functions assumes the lowest $y$-value for $x=1$, or what that value is. (It is the function $y=4/3+2/3\,x$, assuming a value of 2.)

The variation in which we seek the maximal line, not the minimal one, is a trivial modification and we will focus our attention on finding the minimal line.

Graphical representation of the lines in this example

When the lines are graphed, this is easy to see: we want to determine, at the $x$-coordinate 1 (shown by the red vertical line), which line is "lowest" ( has the lowest $y$-coordinate). Here it is the heavy dashed line, $y=4/3+2/3\,x$.

Naive algorithm

For each of the $Q$ queries, of course, we may simply evaluate every one of the linear functions, and determine which one has the least value for the given $x$-value. If $M$ lines are given along with $Q$ queries, the complexity of this solution is $\mathcal{O}(MQ)$. The "trick" enables us to speed up the time for this computation to $\mathcal{O}((Q+M)\lg M)$, a significant improvement.

The technique

Consider the diagram above. Notice that the line $y=4$ will never be the lowest one, regardless of the $x$-value. Of the remaining three lines, each one is the minimum in a single contiguous interval (possibly having plus or minus infinity as one bound). That is, the heavy dotted line is the best line at all $x$-values left of its intersection with the heavy solid line; the heavy solid line is the best line between that intersection and its intersection with the light solid line; and the light solid line is the best line at all $x$-values greater than that. Notice also that, as $x$ increases, the slope of the minimal line decreases: 2/3, -1/2, -3. Indeed, it is not difficult to see that this is always true.

Thus, if we remove "irrelevant" lines such as $y=4$ in this example (the lines which will never give the minimum $y$-coordinate, regardless of the query value) and sort the remaining lines by slope, we obtain a collection of $N$ intervals (where $N$ is the number of lines remaining), in each of which one of the lines is the minimal one. If we can determine the endpoints of these intervals, it becomes a simple matter to use binary search to answer each query.

Etymology

The term convex hull is sometimes misused to mean upper/lower envelope. If we consider the "optimal" segment of each of the three lines (ignoring $y=4$), what we see is the lower envelope of the lines: nothing more or less than the set of points obtained by choosing the lowest point for every possible $x$-coordinate. (The lower envelope is highlighted in green in the diagram above.) Notice that the set bounded above by the lower envelope is convex. We could imagine the lower envelope being the upper convex hull of some set of points, and thus the name convex hull trick arises. (Notice that the problem we are trying to solve can again be reformulated as finding the intersection of a given vertical line with the lower envelope.)

Adding a line

As we have seen, if the set of relevant lines has already been determined and sorted, it becomes trivial to answer any query in $\mathcal{O}(\lg N)$ time via binary search. Thus, if we can add lines one at a time to our data structure, recalculating this information quickly with each addition, we have a workable algorithm: start with no lines at all (or one, or two, depending on implementation details) and add lines one by one until all lines have been added and our data structure is complete.

Suppose that we are able to process all of the lines before needing to answer any of the queries. Then, we can sort them in descending order by slope beforehand, and merely add them one by one. When adding a new line, some lines may have to be removed because they are no longer relevant. If we imagine the lines to lie on a stack, in which the most recently added line is at the top, as we add each new line, we consider if the line on the top of the stack is relevant anymore; if it still is, we push our new line and proceed. If it is not, we pop it off and repeat this procedure until either the top line is not worthy of being discarded or there is only one line left (the one on the bottom, which can never be removed).

How, then, can we determine if the line should be popped from the stack? Suppose $l_1$, $l_2$, and $l_3$ are the second line from the top, the line at the top, and the line to be added, respectively. Then, $l_2$ becomes irrelevant if and only if the intersection point of $l_1$ and $l_3$ is to the left of the intersection of $l_1$ and $l_2$. (This makes sense because it means that the interval in which $l_3$ is minimal subsumes that in which $l_2$ was previously.) We have assumed for the sake of simplicity that no three lines are concurrent.

Analysis

Clearly, the space required is $\mathcal{O}(M)$: we need only store the sorted list of lines, each of which is defined by two real numbers.

The time required to sort all of the lines by slope is $\mathcal{O}(M \lg M)$. When iterating through them, adding them to the envelope one by one, we notice that every line is pushed onto our "stack" exactly once and that each line can be popped at most once. For this reason, the time required overall is $\mathcal{O}(M)$ for this step; although each individual line, when being added, can theoretically take up to linear time if almost all of the already-added lines must now be removed, the total time is limited by the total number of times a line can be removed. The cost of sorting dominates, and the construction time is $\mathcal{O}(M \lg M).$

Another example: APIO 2010 Commando

You are given:

• A sequence of $N$ positive integers ($1 \le N \le 10^6$).
• The integer coefficients of a quadratic function $f(x) = ax^2 + bx + c$, where $a < 0$.

The objective is to partition the sequence into contiguous subsequences such that the sum of $f$ taken over all subsequences is maximized, where the value of a subsequence is the sum of its elements.

An $O(N^2)$ dynamic programming approach is not hard to see. We define:

1. sum(i,j) = x[i] + x[i+1] + ... + x[j]
2. adjust(i,j) = a*sum(i,j)^2 + b*sum(i,j) + c

Then let

$\operatorname{dp}(n) = \max_{0 \le k < n} [\operatorname{dp}(k) + \operatorname{adjust}(k+1, n)]$.

Pseudocode:

dp[0] ← 0
for n ∈ [1..N]
    for k ∈ [0..n-1]
        dp[n] ← max(dp[n], dp[k] + adjust(k + 1 , n))

Now let's play around with the function "adjust".

Denote $\operatorname{sum}(1,x)$ by $\sigma(x)$. Then, for some value of $k$, we can write

$\begin{array}{rl} \operatorname{dp}(n) &= \operatorname{dp}(k) + a(\sigma(n)-\sigma(k))^2 + b(\sigma(n)-\sigma(k)) + c \\ &= \operatorname{dp}(k)+ a(\sigma(n)^2 - 2\sigma(n)\sigma(k) + \sigma(k)^2) + b(\sigma(n) - \sigma(k)) + c \\ &= (a\sigma(n)^2 + b\sigma(n) + c) + \operatorname{dp}(k) - 2a\sigma(n)\sigma(k) + a\sigma(k)^2 - b\sigma(k) \end{array}$

Now, let

1. $z = \sigma(n)$
2. $m = -2a\sigma(k)$
3. $p = \operatorname{dp}(k) + a\sigma(k)^2 - b\sigma(k)$

Then, we see that $mz + p$ is the quantity we aim to maximize by our choice of $k$. Adding $a\sigma(n)^2 + b\sigma(n) + c$ (which is independent of $k$) to the maximum gives the correct value of $\operatorname{dp}(n)$. Also, $z$ is independent of $k$, whereas $m$ and $p$ are independent of $n$, as required.

Unlike in task "acquire", we are interested in building the "upper envelope". We notice that the slope of the "maximal" line increases as $x$ increases. Since the problem statement indicates $a < 0$, the slope of each line is positive.

Suppose $k_2 > k_1$. For $k_1$ we have slope $m_1 = -2a \sigma(k_1)$. For $k_2$ we have slope $m_2 = -2a \sigma(k_2)$. But $\sigma(k_2) > \sigma(k_1)$, and since the given sequence is positive, so $m_2 > m_1$. We conclude that lines are added to our data structure in increasing order of slope.

Since $\sigma(n) > \sigma(n-1)$, query values are given in increasing order and a pointer walk suffices (it is not necessary to use binary search.)

Fully dynamic variant

The convex hull trick is easy to implement when all insertions are given before all queries (offline version) or when each new line inserted has a lower slope than any line currently in the envelope. Indeed, by using a deque, we can easily allow insertion of lines with higher slope than any other line as well. However, in some applications, we might have no guarantee of either condition holding. That is, each new line to be added may have any slope whatsoever, and the insertions may be interspersed with queries, so that sorting the lines by slope ahead of time is impossible, and scanning through an array to find the lines to be removed could take linear time per insertion

It turns out, however, that it is possible to support arbitrary insertions in amortized logarithmic time. To do this, we store the lines in an ordered dynamic set (such as C++'s std::set). Each line possesses the attributes of slope and y-intercept, the former being the key, as well as an extra $left$ field, the minimum $x$-coordinate at which this line is the lowest in the set. To insert a new line, we merely insert it into its correct position in the set, and then all the lines to be removed, if any, are contiguous with it. The procedure is then largely the same as for the case in which we always inserted lines of minimal slope: if the line to be added is $l_3$, the line to the left is $l_2$, and the line to the left of that is $l_1$, then we check if the $l_1$-$l_3$ intersection is to the left of the $l_1$-$l_2$ intersection; if so, $l_2$ is discarded and we repeat; similarly, if lines $l_4$ and $l_5$ are on the right, then $l_4$ can be removed if the $l_3$-$l_5$ intersection is to the left of the $l_3$-$l_4$ intersection, and this too is performed repeatedly until no more lines are to be discarded. We compute the new $left$ values (for $l_3$, it is the $l_2$-$l_3$ intersection, and for $l_4$, it is the $l_3$-$l_4$ intersection).

To handle queries, we keep another set, storing the same data but this time ordered by the $left$ value. When we insert or remove lines from that set (or update, in the case of $l_4$ above), we use the $left$ value from the element in that set to associate it with an element in this one. Whenever a query is made, therefore, all we have to do is to find the greatest $left$ value in this set that is less than the query value; the corresponding line is the optimal one .

References

• Brucker, P. (1995). Efficient algorithms for some path partitioning problems. Discrete Applied Mathematics, 62(1-3), 77-85.
• Christiano, Paul. (2007). Land acquisition. Retrieved from an archived copy of the competition problem set at http://tjsct.wikidot.com/usaco-mar08-gold
• Peng, Richard. (2008). USACO MAR08 problem 'acquire' analysis. Retrieved from http://ace.delos.com/TESTDATA/MAR08.acquire.htm
• van den Hooff, Jelle. (2010). Personal communication.
• Wang, Hanson. (2010). Personal communication.

Code

• C++ solution for "acquire"
• C++ solution for "commando"

External links

• APIO 2010