This Blog is Just the List of Problems for Dynamic Programming Optimizations.Before start read This blog. I solved Problem 1. Seriously? Stay Updated and Keep Learning! The … General Idea for Solving Chess based problems, Number of subarrays with sum less than K, using Fenwick tree, AtCoder Regular Contest #111 Livesolve [A-D], Codeforces Round #318 [RussianCodeCup Thanks-Round] Editorial, Why rating losses don't matter much (alternate timelines part II), Educational Codeforces Round 99 Editorial. So dp[i] = min(f(k, i)) for k < i, where f(k, i) = dp[k] + (suffix[k] - suffix[i + 1] >= 0). Define $$dp_{i,j}$$$to be the minimum cost if we split the first $$i$$$ plants into $$j$$$groups. Seriously? 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). Programming competitions and contests, programming community. 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. I know it's necroposting, but still. We'll define bad(a, b) = dp[a] != dp[b] ? I think it is meaningful, but when discussed on reddit many seemed to disagree. 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. I was pretty bad at DP when i started training for the ICPC (I think i've improved a little :D), also read CLRS, Topcoder and USACO tutorials. Normally Iterative-DP runs faster than Recursive-DP. We then use CHT to get values for all $$j$$$ on $$i$$$-th row in $$O(n)$$$ time. Optimizations for Competitive Programming By vatsal , history , 5 years ago , Hello Readers, I have seen that many people don't use scanf or cin or cout or printf but their own input output functions . The same is also true if dp[a] == dp[b] && suffix[a] >= suffix[b]. 2). Cp-algorithms has added official translation of Divide and Conquer Optimization, thought it might be useful to add. 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. Always choose the largest coin you can and you'll always choose … dp[a] > dp[b] : suffix[a] >= suffix[b] which is true if a will never be the optimal value for k in the future. It's not hard to see that $$b$$$ is a decreasing function and $$a$$$is an increasing function while $$F$$$ is computable in $$O(1)$$$. After some simplification, we have: where $$sum(k) = \sum\limits_1^k w(i)$$$ and $$wsum(k) = \sum\limits_1^k i \cdot w(i)$$$which we can precompute. 2) post-contest discussion Programming competitions and contests, programming community. I slightly disagree with the second part, $$F[j]$$$ is just computed from $$j$$$and not from $$dp[j]$$$ but that's just being nitpicky with notations. 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. CodeChef - A Platform for Aspiring Programmers. Define $$dp_{i,j}$$$to be the minimum cost if we split the first $$i$$$ plants into $$j$$$groups. Please update link of Convex Hull Optimize1 that link is now dead. So dp[i] = min(f(k, i)) for k < i, where f(k, i) = dp[k] + (suffix[k] - suffix[i + 1] >= 0). But wasn't able to formally prove it. 0.00/5 (No votes) See more: algorithms. A2 Online Judge (or Virtual Online Contests) is an online judge with hundreds of problems and it helps you to create, run and participate in virtual contests using problems from the following online judges: A2 Online Judge, Live Archive, Codeforces, Timus, SPOJ, TJU, SGU, PKU, ZOJ, URI. Dynamic programming overkill. We can generalize a bit in the following way: It is claimed (in the references) that the recurrence. The following table summarizes methods known to me. Dynamic Programming Optimizations. 1 1 1 We can have much fewer cases to take care for. 2. 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. 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 $$a[j] \leq a[j + 1]$$$? 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. Yeah I Agree, I was also trying to solve it. 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). 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 $$a[j] \leq a[j + 1]$$$? C++14. (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). where $$cost_{k,j}$$$ is the cost of grouping plants from $$k$$$to $$j$$$. The same is also true if dp[a] == dp[b] && suffix[a] >= suffix[b]. 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. Are there any other optimization techniques? can someone help me plz <3. This makes the solution O(n). Programming competitions and contests, programming community. Say the state is as follows: I am finding it hard to reduce it to any less than O(n * m) for finding dp[n][m], I guess you should ask it after the Long Challenge. Alien's optimization? Programming competitions and contests, programming community. In all other cases, f(a, i) <= f(b, i) for all i > b. orz indy256 meta-san. Can anyone explain how to bring the DP of NKLEAVES to the form of Convex Hull Optimization 2 from above? Can anyone give a link to tutorial or something(in English) for the trick used in IOI16 Aliens problem? Sorry if I misunderstanding your questions, but here are some DP-optimization techniques, By 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 $$wordsize = 32$$$), In 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). https://cp-algorithms.com/dynamic_programming/divide-and-conquer-dp.html. We can generalize a bit in the following way: It is claimed (in the references) that the recurrence. You said "at most n" what brings 0 bits of information since k<=n. We can have much fewer cases to take care for. 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 Assume that for certain $$j$$$ the optimal $$k$$$for the formula above is $$\hat k$$$ i.e. Here is a list I gathered a few weeks ago: Arabic (Youtube Videos and Playlists): i dont know how to use knuth optimizaation?? Take any $$k<\hat k$$$and prove: which is true because adding plant $$j+1$$$ to a bigger group of plants from $$k+1$$$to $$j$$$ certainly costs no more than adding plant $$j+1$$$to a smaller group of plants from $$\hat k+1$$$ to $$j$$$. ... Codeforces Global Round 1 - coding contest stream 1/2 - … Sorry if I misunderstanding your questions, but here are some DP-optimization techniques, By 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 $$wordsize = 32$$$), In 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). Read This article before solving Knuth optimization problems. This series of videos are focused on explaining dynamic programming by illustrating the application of digit DP through the use of selected problems from platforms like Codeforces, Codechef, SPOJ, CSES and … Can humans use it? 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. 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 dp[b], then f(a, i) >= f(b, i) for any i > b. The link given for convex hull optimization is dead now please update. In combinatorics, C(n.m) = C(n-1,m) + C(n-1,m-1). algorithms solutions competitive-programming data-structures codeforces dynamic-programming greedy-algorithms codeforces-solutions competitive-coding Updated Sep 2, 2020 Java We define dp[i] to be the minimum number of Guernsey majority or tied districts with the first i pastures. Please update link of Convex Hull Optimize1 that link is now dead. [leetcode 32][dynamic programming] Dynamic query with dynamic records in SQL. Atcoder: ARC 111 post-contest discussion Normally Iterative-DP runs faster than Recursive-DP. Alien's optimization? I found a good Tutorial on 2d-dp Matrix Exponentiation. (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). Also exists Alien's optimization (link1, link2). dp[a] > dp[b] : suffix[a] >= suffix[b] which is true if a will never be the optimal value for k in the future. 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]. First, for each suffix, we compute the number of Guernseys minus the number of Holsteins and store it in the suffix array. Programming competitions and contests, programming community. Can anyone give a link to tutorial or something(in English) for the trick used in IOI16 Aliens problem? Say the state is as follows: I am finding it hard to reduce it to any less than O(n * m) for finding dp[n][m], I guess you should ask it after the Long Challenge. Dynamic or not dynamic. 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]. use this archive https://web.archive.org/web/20181030143808/http://wcipeg.com/wiki/Convex_hull_trick. He asked if N or K will be removed from the complexity, I answered. But in some specific case as$$$Knapsack_{\ 0/1}$$problem, where we can have some good observations with optimizing algorithm by remove as many not-good cases as possibles. For ex. DYNAMIC PROGRAMMING PROBLEM - CONSECUTIVE SUBSEQUENCE (Codeforces) Algopedia. For two indices a < b, if dp[a] > dp[b], then f(a, i) >= f(b, i) for any i > b. Dynamic Programming versus Memoization is a blog post by Shriram Krishnamurth that covers the subtle distinction between the two techniques. This shows that when we advance$$$j$$to$$$j+1$$we can ignore any$$$k<\hat k$$which allows us to apply divide and conquer. Both the loops popping from the front and back are amortized O(1) since there are a total of N indices, each of which can only be popped off at most once. First, for each suffix, we compute the number of Guernseys minus the number of Holsteins and store it in the suffix array. Assume that for certain$$$j$$the optimal$$$k$$for the formula above is$$$\hat k$$i.e. This problem is based on Dynamic Programming on Trees. orz indy256 meta-san. i dont know how to use knuth optimizaation?? where$$$cost_{k,j}$$is the cost of grouping plants from$$$k$$to$$$j$$. Problem 1 Problem 2 Problem 3 ( C) Problem 4 Problem 5 Problem 6. What is wrong here? He asked if N or K will be removed from the complexity, I answered. 2). Can anyone explain how to bring the DP of NKLEAVES to the form of Convex Hull Optimization 2 from above? So solution by dynamic programming should be properly framed to remove this ill-effect. After some simplification, we have: where$$$sum(k) = \sum\limits_1^k w(i)$$and$$$wsum(k) = \sum\limits_1^k i \cdot w(i)$$which we can precompute. *has extra registration I think dynamic programming is one of those techniques that is hard to grasp at first, even with examples. What is wrong here? At the end, there is a simple, nicely commented code. What is the meaning of table , how to apply them , suppose row 1 in table 1 , what to do ? The actual video starts at 30 mins. Here's a similar problem with a nice description of the algorithm used. We can maintain a deque q such that !bad(u, v) for u < v. Before we insert a new index i, we pop off indices from the end of the queue while bad(q.back(), i) since those indices will never be better than i. k). In all other cases, f(a, i) <= f(b, i) for all i > b. Please Sign up or sign in to vote. This shows that when we advance$$$j$$to$$$j+1$$we can ignore any$$$k<\hat k$$which allows us to apply divide and conquer. optimizing the optimization technique itself? algorithms solutions competitive-programming data-structures codeforces dynamic-programming greedy-algorithms codeforces-solutions competitive-coding Updated Dec 30, 2020 Java But wasn't able to formally prove it. But it still baffles me how to bring my formula to the neat one in the post. The link given for convex hull optimization is dead now please update. Codeforces. 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