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Slope Trick

Author: Benjamin Qi

Slope trick refers to a way to manipulate piecewise linear convex functions. Includes a simple solution to USACO Landscaping.

Tutorials

Resources
CF3 problems using this trick
CFclarifying the above and another example problem

From the latter link (modified):

Slope trick is a way to represent a function that satisfies the following conditions:

  • It can be divided into multiple sections, where each section is a linear function (usually) with an integer slope.
  • It is a convex/concave function. In other words, the slope of each section is non-decreasing or non-increasing when scanning the function from left to right.

It's generally applicable as a DP optimization.

Pro Tip

Usually you can come up with a slower (usually O(N2)O(N^2)) DP first and then optimize it to O(NlogN)O(N\log N) with slope trick.

The rest of this module assumes that you know the basic idea of this trick. In particular, you should be at least somewhat familiar with the O(NlogN)O(N\log N) time solution to the first problem in zscoder's tutorial:

StatusSourceProblem NameDifficultyTagsSolution
CFEasy
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Slope Trick

Check CF

It's ok if you found the explanations confusing; the example below should help clarify.

Buy Low Sell High

StatusSourceProblem NameDifficultyTagsSolution
CFEasy
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Slope Trick

Slow Solution

Let dp[i][j]dp[i][j] denote the maximum amount of money you can have on day ii if you have exactly jj shares of stock on that day. The final answer will be dp[N][0]dp[N][0]. This solution runs in O(N2)O(N^2) time.

Slow Code

If we run this on the first sample case, then we get the following table:

Input:

9
10 5 4 7 9 12 6 2 10

Output:

dp[0] = {  0}
dp[1] = {  0, -10}
dp[2] = {  0,  -5, -15}
dp[3] = {  0,  -4,  -9, -19}
dp[4] = {  3,  -2,  -9, -16, -26}
dp[5] = {  7,   0,  -7, -16, -25, -35}
dp[6] = { 12,   5,  -4, -13, -23, -35, -47}
dp[7] = { 12,   6,  -1, -10, -19, -29, -41, -53}
dp[8] = { 12,  10,   4,  -3, -12, -21, -31, -43, -55}
dp[9] = { 20,  14,   7,  -2, -11, -21, -31, -41, -53, -65}

However, the DP values look quite special! Specifically, let

dif[i][j]=dp[i][j]dp[i][j+1]0.dif[i][j]=dp[i][j]-dp[i][j+1]\ge 0.

Then dif[i][j]dif[i][j+1]dif[i][j]\le dif[i][j+1] for all j0j\ge 0. In other words, dp[i][j]dp[i][j] as a function of jj is concave down.

Full Solution

Explanation

My Code

Extension

Stock Trading (USACO Camp): What if your amount of shares can go negative, but you can never have more than LL shares or less than L-L?

Potatoes & Fertilizers

StatusSourceProblem NameDifficultyTagsSolution
LMiONormal
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Slope Trick

Simplifying the Problem

Instead of saying that moving fertilizer from segment ii to segment jj costs ij|i-j|, we'll say that it costs 11 to move fertilizer from a segment to an adjacent segment.

Let the values of a1,a2,,aNa_1,a_2,\ldots,a_N after all the transfers be a1,a2,,aNa_1',a_2',\ldots,a_N'. If we know this final sequence, how much did the transfers cost (in the best case scenario)? It turns out that this is just

C=i=1N1j=1i(ajaj).C=\sum_{i=1}^{N-1}\left|\sum_{j=1}^i(a_j-a_j')\right|.

We can show that this is a lower bound and that it's attainable. The term D=j=1i(ajaj)D=\sum_{j=1}^i(a_j-a_j') denotes the number of units of fertilizer that move from segment ii to segment i+1i+1. Namely, if DD is positive then DD units of fertilizer moved from segment ii to segment i+1i+1; otherwise, D-D units of fertilizer moved in the opposite direction. Note that it is never optimal to have fertilizer moving in both directions.

Let difi=aibidif_i=a_i-b_i and define dj=i=1jdifid_j=\sum_{i=1}^jdif_i for each 0jN0\le j\le N. Similarly, define difi=aibidif_i'=a_i'-b_i and dj=i=1jdifid_j'=\sum_{i=1}^jdif_i'. Since we want difi0dif_i'\ge 0 for all ii, we should have d0=d0d1dN=dN.d_0=d_0'\le d_1'\le \cdots\le d_N'=d_N. Conversely, every sequence (d0,d1,,dN)(d_0',d_1',\ldots,d_N') that satisfies this property corresponds to a valid way to assign values of (a1,a2,,aN)(a_1',a_2',\ldots,a_N').

Now you can verify that C=i=1N1didiC=\sum_{i=1}^{N-1}|d_i-d_i'|. This makes sense since moving one unit of fertilizer one position is equivalent to changing one of the did_i by one (although d0,dNd_0,d_N always remain the same).

Slow Solution

For each 0iN0\le i\le N and 0jdN0\le j\le d_N, let dp[i][j]dp[i][j] be the minimum cost to determine d0,d1,,did_0',d_1',\ldots,d_i' such that dijd_i'\le j. Note that by definition, dp[i][j]dp[i][j+1]dp[i][j]\ge dp[i][j+1]. We can easily calculate these values in O(NdN)O(N\cdot d_N) time.

Full Solution

Explanation

My Code

USACO Landscaping

StatusSourceProblem NameDifficultyTagsSolution
PlatHard
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Slope Trick

External Sol

This looks similar to the previous task (we're moving dirt instead of fertilizer), so it's not too hard to guess that slope trick is applicable.

Slow Solution

Let dp[i][j]dp[i][j] equal the number of ways to move dirt around the first ii flowerbeds such that the first i1i-1 flowerbeds all have the correct amount of dirt while the ii-th flowerbed has jj extra units of dirt (or lacks j-j units of dirt if jj is negative). The answer will be dp[N][0]dp[N][0].

Full Solution

Explanation

My Solution

Extension

We can solve this problem when Ai+Bi\sum A_i+\sum B_i is not so small by maintaining a map from jj to dif[j+1]dif[j]dif[j+1]-dif[j] for all jj such that the latter quantity is nonzero. Then the operation "add ZjZ\cdot |j| to DP[j]DP[j] for all jj" corresponds to a point update in the map (advance() in the code below).

Code (from Alex Wei)

Problems

Although we haven't provided any examples of this, some of the problems below will require you to merge two slope containers (usually priority queues).

StatusSourceProblem NameDifficultyTagsSolution
CFNormal
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Slope Trick

Check CF
CCNormal
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Slope Trick

Check CC
CEOINormal
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Slope Trick

View Solution
NOI.sgHard
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Slope Trick

View Solution
CFHard
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Slope Trick

Check CF
CFHard
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Slope Trick

Check CF
APIOHard
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Slope Trick, Small to Large

View Solution
CFVery Hard
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Slope Trick

Check CF
ICPC WFVery Hard
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Slope Trick, Small to Large

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