### Блог пользователя indy256

Автор indy256, 5 лет назад, ,

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
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. WJMZBMR 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

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 » 5 лет назад, # | ← 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).
•  » » 5 лет назад, # ^ |   +3 could you please give some example problems?
•  » » 6 месяцев назад, # ^ |   0 Please give a sample problem or describe by a problem.
 » 5 лет назад, # |   +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.
•  » » 5 лет назад, # ^ |   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 >= & <=?
•  » » » 18 месяцев назад, # ^ |   0 There is both concave & convex quadrangle inequalities. concave is for minimization problems, while convex is for maximization problems. refer to Yao'82.
•  » » 11 месяцев назад, # ^ |   0 How do you prove that if this condition is met, then A[i][j]<= A[i][j+1]?
 » 5 лет назад, # |   +18 There is one more optimization of dimanic progamming: 101E - Candies and Stones (editoral)
•  » » 20 месяцев назад, # ^ |   +3 More Problem Collection.
 » 5 лет назад, # |   +13 you have put problem "B. Cats Transport" in "Convex Hull Optimization1", actually it belongs to "Convex Hull Optimization2"
•  » » 5 лет назад, # ^ |   +5 fixed
 » 5 лет назад, # | ← Rev. 2 →   +55 For this moment it's the most useful topic of this year. Exactly in the middle: June 30th, 2013.
 » 5 лет назад, # |   +8 this one seemed a nice dp with optimization to me:https://www.hackerrank.com/contests/monthly/challenges/alien-languages
 » 5 лет назад, # | ← 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)
 » 5 лет назад, # |   +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
•  » » 2 года назад, # ^ | ← Rev. 2 →   0 GOODG can be solved with Type 1EDIT: I explain that below.
•  » » » 2 года назад, # ^ |   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.
•  » » » » 2 года назад, # ^ | ← 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 :/ .
•  » » » 11 месяцев назад, # ^ | ← Rev. 2 →   +13 Yeah, I'm sorry about saying this and not explaining. Actually I should give credit because ItsYanBitches 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, samier_aldroubi and synxazox. And sorry in advance for any mistake, the ideia for the solution is there.
•  » » » » 11 месяцев назад, # ^ |   +3 Why the downvotes? Is it wrong?
•  » » 2 года назад, # ^ |   +3 New link for Commando:http://www.spoj.com/problems/APIO10A/
 » 5 лет назад, # |   0 For some reason I cannot open the links with firefox because they go over the Top Rated table.
•  » » 5 лет назад, # ^ |   +4 Try to zoom out, pressing Ctrl + -
 » 5 лет назад, # | ← 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?
 » 4 года назад, # |   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.
•  » » 4 года назад, # ^ |   +1 Check this one : Guardians of the Lunatics
•  » » » 4 года назад, # ^ |   0 Learnt Divide and Conquer Optimization just from there. :P That is why I'm asking for more problems to practice. :D
•  » » » 3 года назад, # ^ |   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?
•  » » » » 3 года назад, # ^ |   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.
 » 4 года назад, # |   +3 can anyone provide me good editorial for dp with bitmask .
 » 4 года назад, # |   0 Has matrix-exponent optimizations been included here?
•  » » 4 месяца назад, # ^ |   0 wtf do you mean by matrix-expo optimization?
 » 4 года назад, # |   +2 Can matrix chain multiplication problem b also optimized by knuth optimization? If not, dn why?
•  » » 4 года назад, # ^ |   +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.
•  » » » 3 года назад, # ^ |   0 Link to the Hu and Shing algorithm?
•  » » » » 23 месяца назад, # ^ |   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.
 » 3 года назад, # |   +1 What are some recent USACO questions that use this technique or variations of it?
 » 3 года назад, # | ← 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 
•  » » 3 года назад, # ^ |   0 I don't think so, but I guess it can be solved by Divide And Conquer optimization.
•  » » » 6 месяцев назад, # ^ | ← 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].
 » 3 года назад, # |   0 Could you elaborate a little me more in the "Convex Hull Optimization2" and other sections for the clearer notations. For example, You have "k" — a constant in O(kn^2). So the first dimension is of the length K and the second dimension is of the length N?I think it would be clearer if you can write dp[n], dp[k][n] ... instead of dp[i], dp[i][j] .Best regards,
 » 2 года назад, # |   0 I don't get it why there is a O(logN) depth of recursion in Divide and conquer optimization ? Can someone explain it ?
•  » » 2 года назад, # ^ | ← Rev. 2 →   +3 Because each time range is decreased twice.
•  » » » 2 года назад, # ^ | ← 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
 » 2 года назад, # |   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 :)
 » 23 месяца назад, # |   +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?
•  » » 23 месяца назад, # ^ |   0 I think you're right
 » 22 месяца назад, # |   +15 ZOJ is currently dead. For the problem "Breaking String" (Knuth opt.), please find at here
•  » » 22 месяца назад, # ^ |   +13 fixed
 » 22 месяца назад, # |   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
•  » » 21 месяц назад, # ^ |   +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.
 » 18 месяцев назад, # |   +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?
 » 10 месяцев назад, # | ← Rev. 2 →   +3 It is also worthwhile to mention the DP Optimization given here http://codeforces.com/blog/entry/49691 in this post.
 » 7 месяцев назад, # |   0 Can we have the same dp optimizations with dp[i][j] = max (dp....)?
•  » » 7 месяцев назад, # ^ |   0 Yes.
•  » » » 7 месяцев назад, # ^ |   0 In that case (dp[i][j] = max (dp...) ) the condition still unchanged : A[i][j] ≤ A[i][j + 1]. Is that true? Thanks!
•  » » » » 7 месяцев назад, # ^ |   0 Yes.
 » 5 месяцев назад, # |   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.
 » 4 месяца назад, # |   0 You can try LARMY on spoj for Divide and Conquer Optimization.
 » 2 месяца назад, # |   +5
 » 2 месяца назад, # |   0 Fresh problem. Can be solved with CHT.
 » 4 недели назад, # |   +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.
•  » » 4 недели назад, # ^ |   +13 I would like to mention Zlobober, rng_58, Errichto, Swistakk here politely. It would be really helpful if you kindly answered my query.
•  » » 4 недели назад, # ^ |   +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).
•  » » » 4 недели назад, # ^ |   +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?
•  » » » » 4 недели назад, # ^ |   +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).
•  » » » 4 недели назад, # ^ |   -10 How does it help in any way xd?
•  » » » » 4 недели назад, # ^ |   +3 He asked if N or K will be removed from the complexity, I answered. What is wrong here?
•  » » » » » 4 недели назад, # ^ |   -10 You said "at most n" what brings 0 bits of information since k<=n.
•  » » 4 недели назад, # ^ |   +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).
 » 2 недели назад, # | ← 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)