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### indy256's blog

By indy256, 10 years ago,

Several recent problems on Codeforces concerned dynamic programming optimization techniques.

The following table summarizes methods known to me.

Name Original Recurrence Sufficient Condition of Applicability Original Complexity Optimized Complexity Links
Convex Hull Optimization1 b[j] ≥ b[j + 1]
optionally a[i] ≤ a[i + 1]
O(n2) O(n) 1 2 3
p1
Convex Hull Optimization2 dp[i][j] = mink < j{dp[i - 1][k] + b[k] * a[j]} b[k] ≥ b[k + 1]
optionally a[j] ≤ a[j + 1]
O(kn2) O(kn) 1
p1 p2
Divide and Conquer Optimization dp[i][j] = mink < j{dp[i - 1][k] + C[k][j]} A[i][j] ≤ A[i][j + 1] O(kn2) O(knlogn) 1
p1
Knuth Optimization dp[i][j] = mini < k < j{dp[i][k] + dp[k][j]} + C[i][j] A[i, j - 1] ≤ A[i, j] ≤ A[i + 1, j] O(n3) O(n2) 1 2
p1

Notes:

• A[i][j] — the smallest k that gives optimal answer, for example in dp[i][j] = dp[i - 1][k] + C[k][j]
• C[i][j] — some given cost function
• We can generalize a bit in the following way: dp[i] = minj < i{F[j] + b[j] * a[i]}, where F[j] is computed from dp[j] in constant time.
• It looks like Convex Hull Optimization2 is a special case of Divide and Conquer Optimization.
• It is claimed (in the references) that Knuth Optimization is applicable if C[i][j] satisfies the following 2 conditions:
•      monotonicity:
• It is claimed (in the references) that the recurrence dp[j] = mini < j{dp[i] + C[i][j]} can be solved in O(nlogn) (and even O(n)) if C[i][j] satisfies quadrangle inequality. YuukaKazami described how to solve some case of this problem.

Open questions:

1. Are there any other optimization techniques?
2. What is the sufficient condition of applying Divide and Conquer Optimization in terms of function C[i][j]? Answered

References:

• "Efficient dynamic programming using quadrangle inequalities" by F. Frances Yao. find
• "Speed-Up in Dynamic Programming" by F. Frances Yao. find
• "The Least Weight Subsequence Problem" by D. S. Hirschberg, L. L. Larmore. find
• "Dynamic programming with convexity, concavity and sparsity" by Zvi Galil, Kunsoo Park. find
• "A Linear-Time Algorithm for Concave One-Dimensional Dynamic Programming" by Zvi Galil, Kunsoo Park. find

• +388

| Write comment?
 » 10 years ago, # | ← Rev. 4 →   +27 Here is another way to optimize some 1D1D dynamic programming problem that I know.Suppose that the old choice will only be worse compare to the new choice(it is quite common in such kind of problems).Then suppose at current time we are deal with dpi, and we have some choice a0 < a1 < a2, ..., ak - 1 < ak. then we know at current time ai should be better than ai + 1. Otherwise it will never be better than ai + 1,so it is useless.we can use a deque to store all the ai.And Also Let us denote D(a, b) as the smallest i such that choice b will be better than a.If D(ai, ai + 1) > D(ai + 1, ai + 2),we can find ai + 1 is also useless because when it overpass ai,it is already overpass by ai + 2.So we also let D(ai, ai + 1) < D(ai + 1, ai + 2). then we can find the overpass will only happen at the front of the deque.So we can maintain this deque quickly, and if we can solve D(a, b) in O(1),it can run in O(n).
•  » » 10 years ago, # ^ |   +3 could you please give some example problems?
•  » » 3 years ago, # ^ |   0 Hi. I'm trying to solve https://codeforces.com/contest/321/problem/E with this approach (e.g. O(n)) I know there is a solution with D&C, but I would like to try this approach.We can easily prove quadrangle inequality: $C[i][j] = \sum\limits_{k = i}^j\sum\limits_{l = i}^j a[k][l]$$C[a][d] + C[b][c] - C[a][c] - C[b][d] = \sum\limits_{k = a}^{b-1}\sum\limits_{l = c}^d a[k][l] * 2 >= 0.$ Now let's maintain deque $Q$ with potential candidates(opt). As we know that $opt[i] <= opt[i+1]$, so $Q[0]$ will hold the best candidate for the current time.Let $Solve(i,j) = dp[i-1][0] + \sum\limits_{i}^j\sum\limits_{i}^j a[k][l]$Mine approach:1) Try to add $i$ to the candidates list. We can compare the last candidate and current: while (Solve(Q.back(), i) > Solve(i,i)) Q.pop_back(); 2) Add i to the candidate list: $Q.pushback(i)$3) Check if Q[0] is still the best answer: while (Solve(Q[0], i) > Solve(Q[1],i)) Q.pop_back(); 4) update dp: \$dp[i][1] = Solve(Q[0], i);This approach gives me WA10. My code. Any ideas why this approach is wrong?
 » 10 years ago, # |   +5 For question 2: The sufficient condition is: C[a][d] + C[b][c] ≥ C[a][c] + C[b][d] where a < b < c < d.
•  » » 10 years ago, # ^ |   0 Is it quadrangle inequalities? ∀i≤ j,w[i, j]+w[i+1, j+1]≤w[i+1, j]+w[i, j+1], and are these two inequalities equivalent except the >= & <=?
•  » » 6 years ago, # ^ |   0 How do you prove that if this condition is met, then A[i][j]<= A[i][j+1]?
 » 10 years ago, # |   +18 There is one more optimization of dimanic progamming: 101E - Конфеты и Камни (editoral)
•  » » 7 years ago, # ^ |   +3 More Problem Collection.
 » 10 years ago, # |   +13 you have put problem "B. Cats Transport" in "Convex Hull Optimization1", actually it belongs to "Convex Hull Optimization2"
•  » » 10 years ago, # ^ |   +5 fixed
 » 10 years ago, # | ← Rev. 2 →   +55 For this moment it's the most useful topic of this year. Exactly in the middle: June 30th, 2013.
 » 10 years ago, # |   +8 this one seemed a nice dp with optimization to me:https://www.hackerrank.com/contests/monthly/challenges/alien-languages
 » 10 years ago, # | ← Rev. 4 →   +29 The problem mentioned in the article (Breaking Strings) is "Optimal Binary Search Tree Problem" , traditional one.It can be solved by simple DP in O(N^3), by using Knuth's optimization , in O(N^2) . But it still can be solved in O(NlogN) — http://poj.org/problem?id=1738 (same problem but bigger testcases) (I don't know how to solve it. I hear the algorithm uses meld-able heap)
•  » » 3 years ago, # ^ |   0 Are you sure that there is an O(N log n) solution to finding the Optimal Binary Search Tree Problem?
•  » » » 3 years ago, # ^ |   +5
 » 10 years ago, # |   +20 Convex Hull Optimization 1 Problems: APIO 2010 task Commando TRAKA ACQUIRE SkyScrapers (+Data Structures) Convex Hull Optimization 2 Problems: Convex Hull Optimization 3 Problems (No conditions for a[] array and b[] array) : GOODG BOI 2012 Day 2 Balls Solution-Video
•  » » 8 years ago, # ^ | ← Rev. 2 →   0 GOODG can be solved with Type 1EDIT: I explain that below.
•  » » » 8 years ago, # ^ |   0 How? I noticed that, in this problem, b[j] follows no order and a[i] can be either decreasing or increasing, depending on how the equation is modeled. I was able to solve it using the fully dynamic variant, but I can't see how to apply the "type 1" optimization.
•  » » » » 8 years ago, # ^ | ← Rev. 2 →   0 Can you add a link to your code I tried to implement the dynamic variant few weeks ago but there were so many bugs in my code :( .Maybe yours can help :/ .
•  » » » 6 years ago, # ^ | ← Rev. 2 →   +13 Yeah, I'm sorry about saying this and not explaining. Actually I should give credit because yancouto first realized the fully dynamic approach was not necessary. Here's my code.Maybe the most natural approach for this problem is to try to solve the following recurrence (or something similar) where f(0) = 0 and d0 = 0: f(i) = maxj < i(f(j) - dj * (i - j)) + aiWell, this recurrence really requires a fully dynamic approach. We'll find one that doesn't. Instead of trying to solve the problem for each prefix, let's try to solve it for each suffix. We'll set g(n + 1) = 0, a0 = d0 = 0 and compute g(i) = maxj > i(g(j) - di * (j - i) + aj)which can be written as g(i) = maxj > i( - di * j + aj + g(j)) + di * i)now we notice that the function inside the max is actually a line with angular coefficient j and constant term aj + g(j) (which are constant on i) evaluated at  - di. Apply convex trick there (the standart one) and we're done.Notifying possibly interested people after a long delay (sorry about that again): fofao_funk, Sgauwjwjj and -synx-. And sorry in advance for any mistake, the ideia for the solution is there.
•  » » » » 6 years ago, # ^ |   +3 Why the downvotes? Is it wrong?
•  » » 8 years ago, # ^ |   +3 New link for Commando:http://www.spoj.com/problems/APIO10A/
 » 10 years ago, # |   0 For some reason I cannot open the links with firefox because they go over the Top Rated table.
•  » » 10 years ago, # ^ |   +4 Try to zoom out, pressing Ctrl + -
 » 10 years ago, # | ← Rev. 2 →   +8 One more problem where Knuth Optimization is used: Andrew Stankevich Contest 10, Problem C. BTW, does anybody know how to insert a direct link to a problem from gyms?
 » 9 years ago, # |   0 I need some problems to solve on Divide and Conquer Optimization. Where can I find them? An online judge / testdata available would be helpful.
•  » » 9 years ago, # ^ |   +1 Check this one : Guardians of the Lunatics
•  » » » 9 years ago, # ^ |   0 Learnt Divide and Conquer Optimization just from there. :P That is why I'm asking for more problems to practice. :D
•  » » » 8 years ago, # ^ |   0 Is this the best complexity for this problem? Can't we do any better? Can't we somehow turn the logL needed into a constant?
•  » » » » 8 years ago, # ^ |   0 We can, using that opt[i-1][j] <= opt[i][j] <= opt[i][j+1].Key thing is to see that opt function is monotone for both arguments. With that observation, we don't need to use binary search.Check out my submission.
 » 9 years ago, # |   +3 can anyone provide me good editorial for dp with bitmask .
 » 9 years ago, # |   0 Has matrix-exponent optimizations been included here?
 » 9 years ago, # |   +2 Can matrix chain multiplication problem b also optimized by knuth optimization? If not, dn why?
•  » » 9 years ago, # ^ |   +3 Quote from the first of the references above: The monotonicity property for the division points does not hold for the matrix multiplication chain problem...Consider the matrices M1,M2,M3,M4 with dimensions 2x3, 3x2, 2x10, and 10x1, respectively. As can be easily verified, the proper order to compute M1M2M3 is to parenthesize it as (M1M2)M3, while the optimal computation of M1M2M3M4 corresponds to M1(M2(M3M4)). The second reference gives O(n2) dynamic programming solution, based on some properties of the matrix chain multiplication problem.There is also an algorithm by Hu and Shing.
•  » » » 9 years ago, # ^ |   0 Link to the Hu and Shing algorithm?
•  » » » » 7 years ago, # ^ |   0 Here is a link to a 1981 version of the thesis. The original was published in two parts in 1982 and 1984.http://i.stanford.edu/pub/cstr/reports/cs/tr/81/875/CS-TR-81-875.pdfHowever, I doubt that this will be used in competitive programming.
 » 9 years ago, # |   +1 What are some recent USACO questions that use this technique or variations of it?
 » 8 years ago, # | ← Rev. 6 →   0 Can this problem be solved using convex hull optimization?You are given a sequence A of N positive integers. Let’s define “value of a splitting” the sequence to K blocks as a sum of maximums in each of K blocks. For given K find the minimal possible value of splittings.N <= 105K <= 100 Input: Output: 5 2 6 1 2 3 4 5 
•  » » 8 years ago, # ^ |   0 I don't think so, but I guess it can be solved by Divide And Conquer optimization.
•  » » » 6 years ago, # ^ | ← Rev. 3 →   0 Divide and Conquer optimization doesn't work here since the monotonicity of the argmin doesn't hold, consider e.g. 2 3 1 5. The optimal partition is [2] [3 1 5] but when you remove the 5 it becomes [2 3] [1].
 » 7 years ago, # |   0 I don't get it why there is a O(logN) depth of recursion in Divide and conquer optimization ? Can someone explain it ?
•  » » 7 years ago, # ^ | ← Rev. 2 →   +3 Because each time range is decreased twice.
•  » » » 7 years ago, # ^ | ← Rev. 2 →   0 Oh, that was very trivial.I get it now, we spend total O(N) for computing the cost at each depth 2N to be specific at the last level of recursion tree. And therefore O(N * logN) is the cost of whole computation in dividing conquer scheme for relaxation. Thanks
 » 7 years ago, # |   0 Hello , I have a doubt can anyone help?In the divide and conquer optimization ,can we always say that it is possible to use in a system where we have to minimize the sum of cost of k continuous segments( such that their union is the whole array and their intersection is null set) such that the cost of segment increases with increase in length of the segment?I feel so we can and we can prove it using contradiction Thanks :)
 » 7 years ago, # |   +5 For convex hull optimizations, with only b[j] ≥ b[j + 1] but WITHOUT a[i] ≤ a[i + 1],I don't think the complexity can be improved to O(n), but only O(n log n) Is there any example that can show I am wrong?
•  » » 7 years ago, # ^ |   0 I think you're right
 » 7 years ago, # |   +15 ZOJ is currently dead. For the problem "Breaking String" (Knuth opt.), please find at here
•  » » 7 years ago, # ^ |   +13 fixed
 » 7 years ago, # |   0 please someone tell me why in convex hull optimization should be b[j] ≥ b[j + 1] and a[i] ≤ a[i + 1] in APIO'10 Commando the DP equation is Dp[i] = -2 * a * pre_sum[j] * pre_sum[i] + pre_sum[j]^2 + Dp[j] -b * pre_sum[j] + a * pre_sum[i]^2 + b * pre_sum[i] + c we can use convex hull trick so the line is y = A * X + B A = -2 * a * pre_sum[j] X = pre_sum[i] B = pre_sum[j]^2 + Dp[j] -b * pre_sum[j] Z = a * pre_sum[i]^2 + b * pre_sum[i] + c and then we can add to Dp[i] += Z , because z has no relation with j the question is , since a is always negative (according to the problem statement) and pre_sum[i],pre_sum[j] is always increasing we conclude that b[j] ≤  b[j + 1] and a[i] ≤ a[i + 1] I've coded it with convex hull trick and got AC , and the official solution is using convex hull trick someone please explain to me why I'm wrong or why that is happening thanks in advance
•  » » 7 years ago, # ^ |   +8 if b[j] >= b[j + 1], then the technique is going to calculate the minimum value of the lines, if b[j] <= b[j + 1], then it's going to calculate the maximum value of the lines, as this problem requires.
 » 7 years ago, # |   +10 Is it necessary for the recurrence relation to be of the specific form in the table for Knuth's optimization to be applicable? For example, take this problem. The editorial mentions Knuth Optimization as a solution but the recurrence is not of the form in the table. Rather, it is similar to the Divide-and-Conquer one, i.e. dp[i][j] = mink < j{dp[i - 1][k] + C[k][j]}. Does anyone know how/why Knuth's optimization is applicable here?
 » 6 years ago, # | ← Rev. 2 →   +3 It is also worthwhile to mention the DP Optimization given here http://codeforces.com/blog/entry/49691 in this post.
 » 6 years ago, # |   0 Can we have the same dp optimizations with dp[i][j] = max (dp....)?
•  » » 6 years ago, # ^ |   0 Yes.
•  » » » 6 years ago, # ^ |   0 In that case (dp[i][j] = max (dp...) ) the condition still unchanged : A[i][j] ≤ A[i][j + 1]. Is that true? Thanks!
•  » » » » 6 years ago, # ^ |   0 Yes.
 » 6 years ago, # |   0 can someone please give me intuition on proof of A[i, j - 1] ≤ A[i, j] ≤ A[i + 1, j] given for knuth optimization of optimal binary search tree.
 » 6 years ago, # |   +5
 » 6 years ago, # |   0 Fresh problem. Can be solved with CHT.
 » 6 years ago, # |   +8 What will be the actual complexity of Knuth optimization on a O(n^2 * k) DP solution ? Will it be O(n^2) or O(n*k)? Here n = number of elements and k = number of partitions.
•  » » 6 years ago, # ^ |   +13 I would like to mention Zlobober, rng_58, Errichto, Swistakk here politely. It would be really helpful if you kindly answered my query.
•  » » 6 years ago, # ^ |   +3 It can only remove n from the complexity because it (more or less) gets rid of iterating over all elements of the interval (up to n elements).
•  » » » 6 years ago, # ^ |   +8 Thanks a lot for replying to me. So what you are saying is — a O(n^2 * k) solution will turn into a O(n * k) solution? I got TLE in problem SPOJ NKLEAVES (n=10^5, k=10) using Knuth optimization, but passed comfortably using divide & conquer optimization in O(n k log n). That's why I am curious to know whether knuth optimization reduces a n factor or a k factor from the original O(n^2 * k) solution, since a O(n*k) solution should have definitely passed. Would you please have a look?
•  » » » » 6 years ago, # ^ |   +3 Can you show the code? I'm not sure but I think that Knuth optimization can be used here to speed up the O(n3·k) solution with states dp[left][right][k]. The improved complexity would be O(n2·k).
•  » » » 6 years ago, # ^ |   -10 How does it help in any way xd?
•  » » » » 6 years ago, # ^ |   +3 He asked if N or K will be removed from the complexity, I answered. What is wrong here?
•  » » » » » 6 years ago, # ^ |   -10 You said "at most n" what brings 0 bits of information since k<=n.
•  » » 6 years ago, # ^ |   +13 It is O(n2) as the complexity upper bound is proven by summating the running time of DP value calculation over the diagonals of the n × k DP matrix. There are n + k - 1 diagonals, on each of them running time is at most O(n) due to the given inequalities, so the total running time is O((n + k)n) = O(n2).
 » 5 years ago, # | ← Rev. 2 →   0 Someone know where i can find an article about Lagrange optimization? (i know that this can be used for reduce one state of the dp performing a binary search on a constant and add this on every transition until the realized number of transition for reach the final state become the desired)
 » 5 years ago, # | ← Rev. 2 →   +6 I know this can be treated as off-topic by many of us, but what about solving DP states which are direct arithmetic (specifically sum/difference) of previous states:Say the DP state is 1D, like in Fibonacci series (dp[i] = dp[i-1] + dp[i-2]), we already know of a solution in O(k^3 * log n) using matrix binary exponentiation, where k represents that dp[i] depends on dp[i-1], dp[i-2], .. dp[i-k].Is there an optimisation for 2D states like the same? Say the state is as follows: dp[i][j] = dp[i][j-1] + dp[i-1][j-1] + dp[i-1][j-2] dp[i][j] = 0, i < j I am finding it hard to reduce it to any less than O(n * m) for finding dp[n][m]
•  » » 5 years ago, # ^ |   +8 I guess you should ask it after the Long Challenge
•  » » » 5 years ago, # ^ | ← Rev. 2 →   -30 .
•  » » » » 5 years ago, # ^ | ← Rev. 2 →   0 Yeah I Agree, I was also trying to solve it.I found a good Tutorial on 2d-dp Matrix Exponentiation.
 » 5 years ago, # |   0 https://www.youtube.com/watch?v=OrH2ah4ylv4 a video by algorithms live for convex hull trick. The link given for convex hull optimization is dead now please update.
•  » » 5 years ago, # ^ |   0
 » 5 years ago, # |   0 I know it's necroposting, but still. Can anyone give a link to tutorial or something(in English) for the trick used in IOI16 Aliens problem?
•  » » 5 years ago, # ^ |   0 Here's a similar problem with a nice description of the algorithm used
 » 5 years ago, # |   0 Can someone help me prove that the sufficient conditions for Divide and Conquer Optimization are satisfied in this problem.I was able to solve it using the same technique with some intuition. But wasn't able to formally prove it.
•  » » 3 years ago, # ^ |   0 Define $dp_{i,j}$ to be the minimum cost if we split the first $i$ plants into $j$ groups. $dp_{i,j} = \min\limits_{k  » 5 years ago, # | +9 Please update link of Convex Hull Optimize1 that link is now dead.https://web.archive.org/web/20181030143808/http://wcipeg.com/wiki/Convex_hull_trick  » 4 years ago, # | +5 I solved Problem 1. Redistricting from the Platinum division of the USACO 2019 January contest in O(n) using the deque optimization described in this comment while the official solution is O(n log n). SolutionFirst, for each suffix, we compute the number of Guernseys minus the number of Holsteins and store it in the suffix array. We define dp[i] to be the minimum number of Guernsey majority or tied districts with the first i pastures. So dp[i] = min(f(k, i)) for k < i, where f(k, i) = dp[k] + (suffix[k] - suffix[i + 1] >= 0). For two indices a < b, if dp[a] > dp[b], then f(a, i) >= f(b, i) for any i > b. The same is also true if dp[a] == dp[b] && suffix[a] >= suffix[b]. In all other cases, f(a, i) <= f(b, i) for all i > b. We'll define bad(a, b) = dp[a] != dp[b] ? dp[a] > dp[b] : suffix[a] >= suffix[b] which is true if a will never be the optimal value for k in the future.We can maintain a deque q such that !bad(u, v) for u < v. Before we insert a new index i, we pop off indices from the end of the queue while bad(q.back(), i) since those indices will never be better than i. After popping off indices from the front of the deque which are more than K away from i, the optimal k value is always at the front. Both the loops popping from the front and back are amortized O(1) since there are a total of N indices, each of which can only be popped off at most once. This makes the solution O(n). Code#include #include #include #include using namespace std; constexpr int maxn = 300005; int n, k; bool guernsey[maxn]; //true if a pasture has a guernsey int dp[maxn]; //min number of guernsey majority or tied districts with first i pastures, 1 indexed int suffix[maxn]; //number of guernseys minus number of holsteins for pastures starting from i, 0 indexed bool bad(int a, int b) //returns true is b is always at least as good as a, 1 indexed { return dp[a] != dp[b] ? dp[a] > dp[b] : suffix[a] >= suffix[b]; } int main() { ifstream fin("redistricting.in"); ofstream fout("redistricting.out"); fin >> n >> k; for (int i = 0; i < n; i++) //input { char t; fin >> t; guernsey[i] = t == 'G'; } for (int i = n - 1; i >= 0; i--) //compute suffixes { suffix[i] = suffix[i + 1] + (guernsey[i] ? 1 : -1); } deque q; //stores the indices of potential minimum values q.push_back(0); for (int i = 1; i <= n; i++) { assert(!q.empty()); if (i - q.front() > k) //too far away { q.pop_front(); } assert(!q.empty()); int best = q.front(); dp[i] = dp[best] + (suffix[best] - suffix[i] >= 0); while (!q.empty() && bad(q.back(), i)) { q.pop_back(); } q.push_back(i); } fout << dp[n] << '\n'; }   » 4 years ago, # | -10 orz indy256 meta-san. optimizing the optimization technique itself?  » 4 years ago, # | +5 Cp-algorithms has added official translation of Divide and Conquer Optimization, thought it might be useful to add. https://cp-algorithms.com/dynamic_programming/divide-and-conquer-dp.html  » 4 years ago, # | 0 Also exists Alien's optimization (link1, link2).  » 3 years ago, # | +10  » 3 years ago, # | ← Rev. 6 → +6 Can anyone explain how to bring the DP of NKLEAVES to the form of Convex Hull Optimization 2 from above? More specifically, how to I define$b$and$a$from$dp[i][j] = \min\limits_{k < j}(dp[i - 1][k] + b[k] \cdot a[j])$so that$b[k] \geq b[k + 1]$and$a[j] \leq a[j + 1]$?My DP so far is$dp[i][j] = \min\limits_{k < j}(dp[i - 1][k] + cost(k, j))$where$cost(k, j)$is the cost to move leaves from$(k+1)$-th to$j$-th into a pile at location$j$. After some simplification, we have:$ dp[i][j] = \min\limits_{k < j}(dp[i - 1][k] + j \cdot (sum[j] - sum[k]) - (wsum[j] - wsum[k])  dp[i][j] = \min\limits_{k < j}(-sum(k) \cdot j + wsum(k) + dp[i - 1][k] + sum[j] \cdot j - wsum[j]) $where$sum(k) = \sum\limits_1^k w(i)$and$wsum(k) = \sum\limits_1^k i \cdot w(i)$which we can precompute.If we fix$j$then in the formula above,$sum[j] \cdot j - wsum[j]$is a constant that we can ignore and focus on$-sum(k) \cdot j + wsum(k) + dp[i - 1][k]$and for each$k$define a line$hx + m$where$h = -sum(k)$and$m = wsum(k) + dp[i - 1][k]$. We then use CHT to get values for all$j$on$i$-th row in$O(n)$time.But it still baffles me how to bring my formula to the neat one in the post. •  » » 3 years ago, # ^ | -8 Actually, right under the table in the post the author wrote: We can generalize a bit in the following way:$dp[i] = \min\limits_{j
 » 3 years ago, # | ← Rev. 5 →   +5 Are there any other optimization techniques? Sorry if I misunderstanding your questions, but here are some DP-optimization techniques1. DP-bitmaskBy storing all states by bit numbers, in most problems, you can have some bitwise tricks that reduce both constant time (calculating by bitwise cost less) and complexity ($O(f(x))$ -> $O(\frac{f(x)}{w})$, $w$ is normally use as $wordsize = 32$)2. Space-optimizationIn some problems where we only care about $f[n][..]$ and $f[n][..]$ depends on $f[n - 1][..]$ then we can reduce to $2$ DP-array $f[0][..]$ & $f[1][..]$ (reduced $O(n)$ times of space)Some other linear reccurence $f[n][..]$ depends on $f[n][..], f[n - 1][..], ... f[n - k][..]$ and when $k$ is much smaller than $n$ we can use this trick with $k$ DP-array (reduced $O(\frac{n}{k})$ times of space)3. Recursive-DP with branch-and-boundNormally Iterative-DP runs faster than Recursive-DP. But in some specific case as $Knapsack_{\ 0/1}$ problem, where we can have some good observations with optimizing algorithm by remove as many not-good cases as possibles. We can have much fewer cases to take care for.Ofcourse in some cases it will break the $space-optimization$ tricks, and we have to store data by another ways to reduce the time (like using global variables, passing reference data)4. DP-swaplabelSometimes by changing DP-states will results in another way of computing the results, thus will change the complexity** 5. Alien trick**
 » 3 years ago, # |   0 Can you add ioi 2016 alien's trick to the tables with the conditions required to be satisfied for the trick to be used ?