### Блог пользователя 300iq

Автор 300iq, история, 13 месяцев назад, перевод, ,

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Разбор задач Codeforces Round #453 (Div. 1)
Разбор задач Codeforces Round #453 (Div. 2)

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 » 13 месяцев назад, # | ← Rev. 8 →   -25 Can anyone explain in detail problem c div2?
•  » » 13 месяцев назад, # ^ |   -11 Wow you got downvoted hard.
 » 13 месяцев назад, # | ← Rev. 2 →   -12 Can Some Explain Div2 B with more detail and Example.
•  » » 13 месяцев назад, # ^ | ← Rev. 4 →   0 Div2 B: You must first build the tree using the normal process and then do a bfs from the root. We're doing a bfs for 2 reasons: 1. because we will colour a subtree only if its parent has a different colour and a bfs is the optimal way to decide that. 2.We need to minimize the number of colourings so its best if we go from top to bottom one layer at a time than any other order.
•  » » » 13 месяцев назад, # ^ | ← Rev. 2 →   0 Is is possible to use strategy like this ? Do tree traversal root-left-right. If child's color different from parent's then increase answer.
•  » » » » 13 месяцев назад, # ^ | ← Rev. 2 →   0 Yes that's exactly the same as doing a bfs(breadth first search) :). But we cant be sure that a parent node will only have only two nodes, left and right. it can have any number of children.
•  » » » » » 13 месяцев назад, # ^ |   0 Thank you !
•  » » » » » 13 месяцев назад, # ^ |   0 SIMPLE APPROACH:without tree building it can be done just maintaining the two arrays at a time one for checking the parent and other for which colour it has earlier..and if the colour assigned to it due to its parent is not the required colour then increment the steps and set the colour value to the above assigned colour for the remaining subtrees...so the crux is just check for every possible parent and iterate through its all possible childrens and do the above criteria :)
•  » » » » » » 13 месяцев назад, # ^ |   0 Nice Logic!! But i figured this approach out only after the contest. I guess contest pressure pushed me into implementing what came to my mind first and hence the tree implementation. :P
•  » » » 13 месяцев назад, # ^ |   +1 Thanks For Help :)
•  » » » 13 месяцев назад, # ^ |   +1 Much simpler code... without using bfs or dfs :) 33419770
•  » » » » 13 месяцев назад, # ^ |   0 Nicely Done!!! Easier the problem the more the ways of solving it i guess!
•  » » » » 13 месяцев назад, # ^ |   0 Maybe I am wrong, but I think you are doing a dfs in your code, you have built the edges and go throw them. You just did not create a separate function and marked vertex already visited.
•  » » » » » 13 месяцев назад, # ^ |   0 No I am not using dfs , I am just checking for any given node if there is any first-level descendant which has a different color than its parent,if it is then I increment the answer. In dfs you traverse from one node to its descendant until no further descendants are left to traverse. Hope this helps :)
•  » » 13 месяцев назад, # ^ | ← Rev. 2 →   0 in the div2 if you want color the tree,you can color the tree from Leaf to root， if anyone node's color don't same as its father,you should color the father first, then color this node,so if c[i] != c[fa[i]] the ans need ++ansfinally, the root need color once,so you need ans + 1;that's allhere is my code http://codeforces.com/contest/902/submission/33418162
 » 13 месяцев назад, # |   +3 Could anyone explain this sentence "if remainders sequence has k steps while you consider numbers by some modulo it will have at least k steps in rational numbers" in problem 901B - GCD of Polynomials?
•  » » 13 месяцев назад, # ^ |   +21 You can perform Euclid's algorithm using field of remainders modulo some prime number instead of rationals. If at some point you will get B = 0 in rationals, you will also have B = 0 at this step using remainders field. It is possible though that you obtain 0 even earlier but we generate sequence in such way that you don't get B = 0 for n steps modulo prime number 2 thus you won't get B = 0 in rationals during these n steps also.
•  » » » 13 месяцев назад, # ^ |   0 Hey, what's wrong with this comment? I'm just explaining things :(
•  » » » » 13 месяцев назад, # ^ |   0 I don't know, it's entirely correct... this is how I solved the problem.
•  » » 13 месяцев назад, # ^ |   0 Let's say you do one step of the Euclidean algorithm, for polynomials p and q: p = aq+r, where a and r are polynomials with deg r
•  » » » 13 месяцев назад, # ^ |   0 Do you mean that we will not receive even denominators for any sequence of polynomials or for the sequence of polynomials from the solution of the problem?Apparently, the first variant is incorrect: . And when applying the Euclidean algorithm to the pairs like (X3 + X + 1,  - 2X - 1) in and it doesn't seem like steps of the algorithm correspond to each other.And I still can't find the obvious reason why it is true for the sequence from the solution.
•  » » » » 13 месяцев назад, # ^ |   0 You're right. I'll have to think of it some more. Hopefully it's not a fatal error...
•  » » » » 13 месяцев назад, # ^ |   +10 OK I think it still works.The reason is that in the defined sequence, the degrees decrease by 1. So, you can get extra terms with even coefficients, but the leading term will always have an odd coefficient.
•  » » » » » 13 месяцев назад, # ^ |   0 Oh, that makes sense. Thanks!
 » 13 месяцев назад, # | ← Rev. 2 →   0 In Div 2 C, is there any solid prove to this: if there is such a height k that a[k] > 1 and a[k + 1] > 1, then we can create 2 non-isomorphic trees from the current sequence? I mean how can we know if 2 created trees from that sequence are non-isomorphic or not?
•  » » 13 месяцев назад, # ^ |   0 You have one longest path from root to farthest leaf. In first tree for each vertex either it is on the path or its parent is at such path. In second tree however it is possible that parent of some vertex still not on longest path.
•  » » 13 месяцев назад, # ^ |   0 Take all the nodes in height k+1 and set their parents to only one node of depth k (call it v). This is a rooted tree that has a[k+1]-1 leaves with height k+1.Now, do the same thing, but take one node in height k+1 and set its parent to another node different from v. This is a rooted tree that has a[k+1]-2 leaves with height k+1. These two trees cannot be isomorphic because the number of leaves with height k+1 are different.That's why there is the condition of the two consecutives a[i] such that a[i] > 1. If this doesn't happen in the sequence, we can guarantee that there isn't any non-isomorphic pair of trees because: For every height k such that a[k] > 1 and a[k+1] = 1 (because there aren't consecutive terms greater than 1), picking a parent for this node with height k+1 will not change the isomorphism. For every height k such that a[k] = 1 and a[k+1] > 1, there is only one possible parent for these nodes with height k+1.
•  » » » 13 месяцев назад, # ^ | ← Rev. 2 →   0 This is a rooted tree that has a[k+1]-2 leaves with height k+1. These two trees cannot be isomorphic because the number of leaves with height k+1 are different.. Sorry if i'm wrong, but i think they want to find 2 non-isomorphic trees with same sequence of vertices amount at heights, and i still don't understand why two trees cannot be isomorphic there is the condition of the two consecutives a[i] such that a[i] > 1.
•  » » » » 13 месяцев назад, # ^ |   +1 Yes, they want you to give 2 trees (non-isomorphic) with the same sequence of vertices.The thing is: If there are two consecutives a[i] such that a[i] > 1 then there exists two non-isomorphic trees with the given sequence. In fact, there are some trees with this sequence. Some of them are isomorphic among each other, some aren't. So you just need to pick a pair of non-isomorphic trees.If there aren't these consecutives a[i], all the possible trees are isomorphic (by the argument above).To check if two trees are isomorphic, every node in the first tree must have one corresponding node in the second tree. In the example above, there is a node in the first tree with height k and a[k+1] children, and there isn't any corresponding node in the second tree satisfying this (looking at the nodes with height k, there are some with 0 children, one with 1 child and one with a[k+1]-1 children).
•  » » » » » 13 месяцев назад, # ^ | ← Rev. 4 →   0 But what if in the second tree, there is a node with a[k + 1] children tbat isn't at height k?
•  » » » » » » 13 месяцев назад, # ^ |   0 The two trees still won't be isomorphic since these two nodes (the node from the first tree and the node you mentioned from the second tree) can't be matched (because they don't have the same height).
•  » » » » » » » 13 месяцев назад, # ^ |   0 The height of a node depends on the node which you chose to be the root doesn't it?
 » 13 месяцев назад, # | ← Rev. 3 →   +14 Can we solve 1B through a random algorithm? As we all know, there are hundreds of thousands of possible solutions, and I have no idea on construction of such a sequence, so I used a random algorithm and passed system test.Here is my code. Hack is welcomed.
•  » » 13 месяцев назад, # ^ |   +9 Kinda hard to hack if there are 150 possible inputs and all of them can be tested in no time xD
•  » » 13 месяцев назад, # ^ |   +1 Or even better, you only really need to solve this problem for n = 150. Then you can just work your way down from there.
•  » » » 13 месяцев назад, # ^ |   0 Not really, intermediate polynomials don't have to have coefficients in {-1, 0, 1} which is kinda crucial here (I think).
•  » » » » 13 месяцев назад, # ^ | ← Rev. 2 →   -10 I think not if you do it modulo 2Edit: I agree, but when you figure out that you can do the everything modulo 2 you've practically solved the whole thing
 » 13 месяцев назад, # |   +13 How does the checker work for problem B div 1? It would seem that it might require large (how large?) bignums. I guess it can compute the GCD sequence with a random prime modulus. If it works then great. If not then try another modulus. Once the product of the primes you try is larger than the largest number you'd ever need, I guess you can declare it "good".
•  » » 13 месяцев назад, # ^ |   0 It computes pseudoremainders using big integers. Generally size of coefficients grows exponentially but if you divide each term by gcd of coefficients then size of coefficients will grow only linearly of n. I wanted to use prime numbers at first but KAN decided it's not the best idea. However I suppose one needs much smaller number of primes than maximum number to have very high probability of correct verdict...
 » 13 месяцев назад, # |   0 can someone please explain the idea behind DIV2-D. How fibonacci polynomial leads to the desired answer?
•  » » 13 месяцев назад, # ^ |   0 Need it too QWQ
 » 13 месяцев назад, # |   +3 I was upsolving problem C in Div2, using PyPy2 and got a TLE on test case 14. The complexity is O(sum(a)) which is fine, but still got a TLE. However it was accepted when the same code was submitted in Python2 and Python3. I had always known PyPy to be faster than native Python. This has been true in Codechef platform. Can anyone reason as to why it TLE'd in PyPy2?
•  » » 13 месяцев назад, # ^ |   +5 This link might help you. Your solution seems to have a lot of string manipulation, and pypy might not be optimizing them as much as cpython, leading to TLE.
 » 13 месяцев назад, # |   0 For Div1B:Why does xp_n +/- p_{n-1} always have coefficients in {-1,0,1} for some choice of plus/minus? Is there a reason, or does it just happen to work for the given bounds?
•  » » 13 месяцев назад, # ^ | ← Rev. 2 →   +11 That's tough one to prove strictly. You can check this mathoverflow question.Spoiler: The answer is by Terry Tao.
•  » » » 13 месяцев назад, # ^ |   +3 Talk about bringing in the big guns...Is this something you noticed from playing around with polynomials, or did it come up in research or something?
•  » » » » 13 месяцев назад, # ^ |   0 I was thinking on what's the worst case for Euclid's algorithm and apparently found out that this sequence seems to be infinite and has some peculiar properties (like you can choose  +  or  -  only at powers of 2, others are determined by that). Surely such approach shouldn't be model solution since it's hard to prove theoretically but there's modular solution which is just fine for this purpose.I'm not doing any serious research now :)
•  » » » » » 13 месяцев назад, # ^ |   +5 I was just looking at this year's Putnam problems. http://kskedlaya.org/putnam-archive/2017.pdfLook at A2... curious isn't it? This sequence appears everywhere it seems.
•  » » » » » » 13 месяцев назад, # ^ |   0 How is it same?
•  » » » » » » » 13 месяцев назад, # ^ |   +5 I guess you have to solve the problem to see that. :D
•  » » » 13 месяцев назад, # ^ |   0 So was it supposed to just believe in this fact for the model solution? It was hard even to understand, don't think much people could prove this during the contest.
•  » » » » 13 месяцев назад, # ^ | ← Rev. 2 →   0 It was supposed to either guess the fact or to use sum modulo 2 which is easy to prove. I doubt that's hard to understand. You want exactly n steps so on each step you want to have . Obviously, you would like to construct such sequences and that they're answers for n and first step lead you from to . Thus But this is the same as . And it's not about believe because it can be checked in no time.
•  » » » » » 13 месяцев назад, # ^ |   +5 or to use sum modulo 2 which is easy to prove Sorry, I somehow haven't noticed second solution in the editorial. Yes, I agree that it's easy to guess and to prove (ability to use  - 1 misleads a bit though)
 » 13 месяцев назад, # |   +5 Could you tell me which knowledges are required to understand div2 D's solution?
 » 13 месяцев назад, # | ← Rev. 2 →   +15 To come up with the solution like in Div2D / Div1B, I would rather thinking like this:We already know that it takes two steps to find (x2 + 1, x).Let's say we need to find the pair that takes three steps. If we call it (A, B), such that the polynomial division A / B will have the divisor equal to x2 + 1 and the remainder equal to x. By that way, we will find (A, B) by continuing finding (x2 + 1, x), which takes two steps. Therefore, it takes three step to find (A, B). Now, as I stated, the divisor equal to x2 + 1, then B = x2 + 1. The remaining task is to find A which has the form of A = (x2 + 1)Q + x.We can freely choose Q to obtain A. However, to satisfy the restriction that coefficents must equal to  - 1, 0, or 1, Q should be x or  - x. This solution is pretty close to what is stated in the tutorial. (Mine is pn + 1 =  ± xpn + pn - 1. You can easily figure out that if the tutorial's solution satisfies the restriction then so does mine) Proof by Terry TaoAfter we have A and B, we continue applying this algorithm with four steps, five steps, ..., and finally n steps.Using the characteristic of two consecutive Fibonacci numbers to propose a solution of two consecutive Fibonacci polynomials is, though correct, a little vague (!)
•  » » 13 месяцев назад, # ^ |   0 How can you decide when to choose -x and when to choose x?
•  » » » 13 месяцев назад, # ^ |   0 Just try to assign Q = x and Q =  - x. Select the case that brings up a polynomial A that satisfies the coefficents contraint. If both cases satisfy, choose either one of them.
•  » » » » 13 месяцев назад, # ^ |   0 Is it easy to figure out why always either to assign Q=x,Q=-x will lead to a polynomial that satisfies the coefficients constant?
•  » » » » » 13 месяцев назад, # ^ |   +2 Not really. It’s pretty difficult. Did you check the proof by Terence Tao?
•  » » » » » » 13 месяцев назад, # ^ | ← Rev. 2 →   0 yep, I agree is too difficult. I finally decided to stick with the version of the problem with mod 2. I have heard that it is easier to prove yet I couldn't. Do you know how why the algorithm using mod 2 works?
•  » » 13 месяцев назад, # ^ |   +3 Thanks a lot, this comment made it seem very simple! (Though, I would be scared to use it because I can't really understand the proof of why it always works.)
•  » » 12 месяцев назад, # ^ | ← Rev. 3 →   0 but does your method (pn + 1 =  ± xpn + pn - 1) guarantee that the leading coefficient will always be 1 (not -1) ??
 » 13 месяцев назад, # |   -19 Unly nubz nid tutorial becoz they hev alwayz been spoonfed by their fat mommas
 » 13 месяцев назад, # |   +3 In div2B why dfs gives the minimum answer.when to apply dfs or bfs i am weak at it plzz help
•  » » 13 месяцев назад, # ^ | ← Rev. 2 →   +3 Okay, I will try to explain. You need to understand that if there is a vertex A and a vertex B, and A is parent of B, you ALWAYS need to paint A first. Why? Because if we paint B, then A, the color of B becomes the same as the color of A (don't forget that A is parent B). So you will need to paint it again, which is not optimal. That's why we need to start painting from the root of tree.
 » 13 месяцев назад, # | ← Rev. 5 →   +1 Simple solution for 2B: int ans=1; for(int i=0;i
 » 13 месяцев назад, # |   0 HASHING TREES:why the output of 2 sample case is ambiguous 0 1 1 3 3 0 1 1 3 2 
 » 13 месяцев назад, # |   +8 Div1B can be solved with backtracking — http://codeforces.com/contest/901/submission/33452328
•  » » 13 месяцев назад, # ^ |   -12 nicely done but in two loops everyhthing can be done easily bro can u plzz explain me how to solve C i am not getting the editorial there can be easy way to solve it ?
 » 13 месяцев назад, # |   +7 Can someone elaborate on the idea behind div2/E? I still don't understand after reading the editorial.
•  » » 13 месяцев назад, # ^ |   +3 I too had some trouble understanding it. Convince yourself that there could be no intersecting cycles. You can prove this by contradiction and by considering the parity of length of intersection of the cycle. Now, for every cycle, find the maximum node index in it and the minimum node index in it. Then a segment is good only if it does not contain both the minimum and maximum indices of any cycle in it. So, basically, you have some intervals and now in every query, you are asked to find the number of segments which do not completely cover any of these given intervals. For reference, I used this code: http://codeforces.com/contest/901/submission/33440062
•  » » 13 месяцев назад, # ^ |   +17 My solution: First of all, let's try to reformulate the problem statement using this observation: A bipartite graph is a graph with no odd cycles (Proof). Since the given graph has no even cycles by definition, the only cycles it (and its subgraphs) will have must be odd. Therefore, a subsegment of this graph is bipartite iff it is acyclic. A subsegment of the graph will only be acyclic if it is either a tree, or a forest. Therefore, the problem can be reduced to this: "Find the number of subsegments [x, y] (l ≤ x ≤ y ≤ r) that are forests or trees.".The second observation is this: If we only take vertex x from the graph and incrementally add vertices x + 1, x + 2, x + 3 and so on, at some point the graph will become cyclic, and from that point on it can never be acyclic again. Let's call that point mx (that is, after the addition of vertex mx, the subsegment [x, mx] becomes cyclic). Note that m1 ≤ m2 ≤ m3 ≤ ... ≤ mn. Therefore, we can use a two pointer approach to calculate m (I used the solution to this problem to remove vertices, see my code: 33436708). The rest of the solution is identical to the one described in the editorial.
•  » » » 13 месяцев назад, # ^ |   +1 Thanks a lot for your answer. Can you please elaborate a bit more on how you are computing the mx? I referred to your code but could not quite get it. A link to some resource your code follows would also be fine.
•  » » » 13 месяцев назад, # ^ |   +1 All editorials should be written with this level of precision.
•  » » » » 13 месяцев назад, # ^ |   0 Can you explain your implementation?
•  » » » » » 13 месяцев назад, # ^ |   +1 yep! wait some hours I will make you an explanation
•  » » » » » » 13 месяцев назад, # ^ |   0 That would be really kind. Thank you so much.
•  » » » » » 13 месяцев назад, # ^ | ← Rev. 7 →   0 1] We read the tree, CONN vector array stores the connections2] We run a dfs to search for cycles. The trick is to maintain updated in array vis the deep level of each node that is part of the current track the dfs algorithm analyzes. When we collide with a node member of the track we interpret there is a cycle and with vis[] array we re-build the cycle and detect the min-max values on it (that is what we will need later). We give each cycle a code We add to start[] and finish[] indexed by the min-max value the code of the cycle. (I will explain in 3 why we need them)3] We start a two-pointers processing to compute mi (that is defined in the editorial as the lower index such that [i, mi] is cyclic) Note that in this case, like f2lk6wf90d says, it is true that a sub-graph is cyclic<=>not bipartiteAlso, as f2lk6wf90d says, it can be proved that mi <  = mi + 1 1 <  = i <  = N - 1 (for example by contradiction)This is the reason now we can use or two-pointers technique or iterate N binary searches. I decided to use two pointers method in the following way:Use pointers i, j maintaining this invariant [i, j] is cyclic. Iterate over i and update j pointer in each iteration by incrementing it until the sub-graph with nodes of codes [i, j] becomes cyclic thus by definition mi = j. To be able to check if the subgraph is cyclic on each moment we need to make use of start[] vector array, finish[] vector array and min-max values of cycles stored, that gives us the information needed to update if the subgraph [i, j] is or is not a cycle (the interval [i, j] is member of cycle k if and only if i <  = mink <  = maxk <  = j). Each time when we increment j we check if a cycle k ends in that j value and have already started (mink >  = i) (using start,finish vector arrays and min-max cycles data). When this happens we know the graph is cycle.Also when we increment i we make sure we can check which open cycles stop from existing, using start[] vector array. So I explained almost all details, I assure you that by this process you compute mi4] Now we have mi computed 1 <  = i <  = N and this is the hardest part. We precompute prei to make prefix sums. pre0 = 0prei = prei - 1 + mi 1 <  = i <  = Nnow 1 <  = i <  = j <  = NWith this information, there is a way to answer the queries in O(log(n)) each one. with tricks obviously.The query is an interval [l, r] we need to check for every x, y, l <  = x <  = y <  = r if [x, y] is not cyclic (thus bipartite). We iterate over x and get the amounts of y for each x such that [x, y] is not cyclicTo develop the formula we need to analyze two cases:if for an x mx <  = r then there are (mx - 1) - x + 1 = mx - x values of y such that [x, y] is not cyclic.if for an x mx > r then we've got r - x + 1 values of y.So for each x we need to compute index u such that u is the lower value such that mu > r We know m is increasing the binary search makes this possible. (We could also use two pointers) And now the answer to the query should be  +  This should be computed on O(R - L). But by working algebraically we can show it is equivalent to — (r(r + 1) - l(l - 1)) / 2  +  (r + 1)(r - u + 1)Using pre-array we can compute the sum so this is O(log(n)) per query that is more than enough to solve the problem.
•  » » » » » » 13 месяцев назад, # ^ | ← Rev. 2 →   0 I have written lot of details. Tell me if I made some mistake. It was long and this could be a possibility.
•  » » » » » » 12 месяцев назад, # ^ |   +1 Thank you very much! Your explanation and your submission were really helpful
•  » » » 13 месяцев назад, # ^ |   0 In problem Div 1C / Div 2E, can someone please explain, how are we finding the value of mx, in detail. Details in the editorial are not clear to me.
•  » » » 13 месяцев назад, # ^ |   0 Awesome dude! That's what we need!
 » 13 месяцев назад, # |   +13 In Editorial of Div 1.C, this sentenceTo answer the query, we need to take the sum over mxi - i + 1 for those who have mxi ≥ r and the sum over r - i + 1 for those who have mxi ≥ rI think it should beTo answer the query, we need to take the sum over mxi - i + 1 for those who have mxi < r and the sum over r - i + 1 for those who have mxi ≥ r
•  » » 13 месяцев назад, # ^ |   0 then the vertex segment is good - if there is no loopWhat is 'loop'?
•  » » » 13 месяцев назад, # ^ |   0 then the vertex range (l<=r) is good — if there is no cycle
•  » » 13 месяцев назад, # ^ |   0 Can you explain your implementation?
 » 13 месяцев назад, # |   +3 In DIV 1D's description，what does it mean "It is guaranteed that the given graph is connected and does not contain loops and multiple edges." especially 'loop' here, isn't it crashed with the sample?
•  » » 13 месяцев назад, # ^ |   0 'loop' which you mentioned is an edge that connect a vertex to itself. But 'loop' in the editorial is like circle.
•  » » » 13 месяцев назад, # ^ |   0 When they say loop here, they actually mean cycle.
 » 13 месяцев назад, # | ← Rev. 2 →   +1 Div 1 B. Can anyone prove that mod 2 of the coefficients of series p(n+1) = x*p(n) + p(n-1) will give the correct answer? The given explanation isn't very clear. :/
•  » » 13 месяцев назад, # ^ |   0 I am interested in seeing answer to this question.
 » 13 месяцев назад, # |   0 problem A(div 2) says It is guaranteed that ai ≥ ai - 1 for every i (2 ≤ i ≤ n). but my submission failed at test 17 which was: 1 10 0 10 so is 0 greater than 1 ? link :http://codeforces.com/submissions/tibialAcorn98
•  » » 13 месяцев назад, # ^ |   0 1 10 0 10.1 — the number of teleports;10 — the point we want to reach;0 — a[1]; 10 — b[1].
•  » » » 13 месяцев назад, # ^ |   0 thnx i got my mistake .
 » 13 месяцев назад, # |   0 I can not understand the isomorphic statement . i got WA on test 2 .How these two trees are isomorphic ?
•  » » 13 месяцев назад, # ^ |   0 This was a confusion I had during contest.If u read the question carefully,it is written that if u REENUMERATE the vertices in some way, here REENUMERATE means that assigning the vertices in some order to make the trees equal.Now if u see ur output, if in the 2nd tree, if I interchange vertices 2 and 3, then it becomes exactly equal to the 1st tree, which shows that they are both isomorphic.In question U have to output two non-isomorphic trees instead.Hope this helps :)
•  » » » 13 месяцев назад, # ^ | ← Rev. 2 →   0 if I interchange vertices 2 and 3, then how it becomes exactly equal ? In my output 1st tree -> node 4, 5 connected with node 3 and In 2nd tree node 4,5 connected with node 2 then how it becomes equal ?
•  » » » » 13 месяцев назад, # ^ |   0 By interchanging or reenumerating the vertices I meant that rename vertex 2 as vertex 3 AND vertex 3 as vertex 2.See image
•  » » » » » 13 месяцев назад, # ^ |   +3 THANKS :)
•  » » » » » » 13 месяцев назад, # ^ |   0 But if we have 1child for root2 and root3 for first tree,i build second my tree as picture shows .Here,how two trees are equal?
 » 13 месяцев назад, # |   -10 How to Solve This Problems help me plz.
 » 13 месяцев назад, # |   0 Can someone please some implementation details about Div 2E? It would be really helpful.
 » 12 месяцев назад, # | ← Rev. 3 →   0 My thinking of why the mod approach works for div2 D:consider always making A = B*x + C where degree of C is less than degree of B, the problem we will face is that sometimes the result will have coefficients greater than 1, but the leading term coefficient will never be greater than 1, because degree of B*x is higher than degree of C, so the leading term will always be existing once in the expression B*x+C (and so will not be affected by the mod).now think of A%2, this is = (B*x + C)%2, and similarly B%2 = (C*x + D)%2, and C%2 = (D*x + E)%2, and so on till the rightmost term (E in the last expression) is 0. before we reach this point, all expressions A, B, C, ...etc will always have the leading element coefficient = 1 after and before the mod. so with the mod, we took the same number of steps we would take without the mod.
 » 10 месяцев назад, # | ← Rev. 2 →   0 Div 1.EDo we need a root of unity of order 2n because when we evaluate polynomial at points {zi} with Bluestein's algorithm we work with elements which are in cyclic subgroup of order 2n? (z has order n) Just making sure I got everything right.