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I have been working on this for a few days, brushing up on graph theory and disjoint sets, but I have not yet found an algorithm that is guaranteed to produce the optimal result and runs in better than $O(n^2)$ time.

Problem Statement

The problem is to divide a tree graph, which is a connected undirected graph with no cycles, into 3 equal parts. More formally, given a tree graph $G$ with $n$ vertices, find the most even partition of $G$ into $3$ components $A, B, C$ with $a, b, c$ vertices, respectively. For this problem, "most even" is defined as having the largest value of $abc$ possible for the given graph $G$.

What I am most interested in learning is if there are properties of tree graphs I can leverage into proving that a trisection is the best (most even) possible without (in pathological cases) having to try all possible trisections. For example, if the most even bisection is so lopsided that the smaller "half" has $s$ vertices such that $s < n/3$, that means that the vertex of the larger half that just had its edge cut only has remaining cuts of size $r \le s$. So now we know $s < n/3$, $r < n/3$ and neither can get any bigger in a 3-cut.

Discussion

I believe it is correct to extend this reasoning to say that for the trisection of $G$ into $A$, $B$, and $C$, where $B$ is the component in the middle (i.e. that could be connected to either $A$ or $C$ by restoring a single cut edge), the trisection is optimum iff $(A, B)$ is the optimum bisection of $G - C$ and $(B, C)$ is the optimum bisection of $G - A$. Unfortunately, I have not thought enough about it yet to even be sure the optimum trisection is the only $(A, B, C)$ for which that condition holds.

For the optimum trisection of $G$ into $(A, B, C)$ with sizes $a$, $b$, and $c$, we can arbitrarily label them according to size, such that $a \ge b \ge c$. Then we know that $c\le n/3$ and $c \le {(b + c)}/2$.

We also know that the optimum trisection has $c$ as large as possible. Similarly we know the optimum bisection of $B+C$ or $A+C$ has the largest $c$ possible, even though there is no guarantee that both $B+C$ and $A+C$ are fully connected in the first place.

Put them together and they prove that given any cut of $G$ into to components $X$ and $Y$, the best trisection of G given that cut is either $X$ plus the best bisection of $Y$ or it is $Y$ plus the best bisection of $X$. I'm going to call vertices "nodes" from here on because it is a shorter word.

It is easy to cut the graph optimally in half in $O(n)$ time. Every edge has $x$ nodes on one side and $(n-x)$ on the other. Because there is only one path from any node to any other, you can compute these values for every edge in $O(n)$ time, and while you are calculating that, you can just keep track of which edge has the closest to $n \over 2$ nodes on one side of it. Because you have checked every edge, you know this is the best you can do, even if $x \ll \frac n2$. It might not be the only edge that produces those size partitions, but that is not a concern.

So far, however, I have not found a way to prove that a given pair of edges are the best choice to cut to partition non-trivial $G$ unless either (a) the result is perfect, or (b) I compare it to essentially every other combination of 2 edges. I do not really have to compare every combination all the time, often I can cut the search space down, but I have not figured out how to prove that I have found the best possible partition in less that $O(n^2)$ time.

I can precompute a lot, but when I pick a candidate for the first of 2 edges to trisect the tree, I have not found anything I can precompute in $O(n \log n)$ time that results in an $O(1)$ way to tell which of the remaining edges are in which of the 2 resulting components and keep the edges from one component out of the binary search for the best matching edge in the other component.

Everything I have come up with so far fails for various kinds of edge cases. The simplest case is a star tree, $n$ nodes all connected to one central node of degree $n$. If all I am doing is naïvely comparing edges and cuts, I have to try all possible combinations.

I can rank the edges, like I did with bisecting the graph, except this time rank them on how close the are to having $n \over 3$ nodes on one side, but what if I have a tree like this:

One-sided tree

$H$ is connected to 5 leaves and there are a total of 12 nodes in $G$. The edge closest to having $\frac n3 = 4$ nodes on one side is the edge between $4$ and $5$, $(4,5)$, which splits $G$ into sets of 4 and 8 nodes. If I cut that edge, the next best edges to cut (if I am only looking at edges of $G$, and going by the ranking in the previous paragraph) are $(3,4)$ or $(5,6)$, but would result in partitions of $\{8,1,3\}$ and $\{7,1,4\}$ respectively. It will be a long time before I find the optimal solution of $\{6,3,3\}$.

Looking at this, I considered finding the best $\frac 23$ cut and then cutting the larger set in half. Except for this graph, the best $\frac 23$ cut is the same as the best $\frac 13$ cut, so we would cut edges $(4,5)$ and $(H,6)$ giving us a $\{6,2,4\}$ partition.

Since bisecting the graph is easy, I looked at recursively bisecting the graph. I could build up a balanced binary tree of subgraphs, but I don't know what that would get me for a star graph or various kinds of lopsided graphs I can come up with. It might work to find an approximate answer, but I want to be sure I have the exact best answer.

The real graphs I am dealing with can be huge, up to $1,000,000$ vertices, so the $O(n^2)$ solution is not practical. The only real constraint I can rely on is that the diameter of the graph will not get out of control huge like that.

I am looking for either help finding the best edges to cut or help verifying that the partition I have found is the best possible for the given graph (or, of course, both.) Thanks.

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  • $\begingroup$ Not relevant to the question as asked, but: often in situations like this there can be algorithms that find an approximately correct solution that are faster than anything guaranteed to find the correct solution. Are you interested in such solutions or do you really need optimality? $\endgroup$ – Daniel McLaury Dec 26 '18 at 3:26
  • $\begingroup$ @DanielMcLaury I already know how to make a good guess, a.k.a. find an approximately correct solution, so for the purposes of this question I'm only interested in something that guarantees optimality. $\endgroup$ – Old Pro Dec 26 '18 at 4:10
  • $\begingroup$ It looks that it is not very easy to find a required algorithm. So if you will not obtain an answer here I recommend you to ask this question at CS Theory.SE. $\endgroup$ – Alex Ravsky Dec 26 '18 at 8:50
  • $\begingroup$ I am confused. About looking for a bisection: Are you looking for the minimum number of edges to cut to bisect the tree into two forests $F_1$ and $F_2$ such that $F_1$ and $F_2$ have the same number of vertices (within 1). OR are you looking for the one edge to cut to get as close to a bisection as possible, cutting only one edge. Your title and intro says you are looking for the former but the rest of your text says you are looking for the latter. $\endgroup$ – Mike Dec 26 '18 at 17:46
  • $\begingroup$ @Mike For a tree, any edge you cut will bisect the tree into a forest of 2 trees, and any 2 edges cut will trisect the tree. I changed the title for you anyway. $\endgroup$ – Old Pro Dec 26 '18 at 23:53
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The Solution

The missing piece of the puzzle

The property of tree graphs that I did not know or intuit and that I needed to know so I could find the $O(n \log n)$ solution is that for every tree of $n$ nodes, you can choose a root such that no subtree has more than $n/2$ nodes, or, as @Dap put it, every tree graph $G$ has at least $1$ vertex $r$ such that the components of $G-r$ all have order at most $n/2$. Once you know that, everything else falls into place.

Intro

Answers from Dap and SmileyCraft contained helpful nuggets, but were overcomplicated and I could not prove their worst case complexity as being better than $O(n^2)$, though they probably are better than that. My solution is pretty close to Dap's, but simpler. For me, though, it was the simplification I was more interested in than the actual solution.

Because tree graphs are so special, they are talked about in graph theory, in set theory, and in computer science, with sometimes different names and symbols, as you may have noticed if you read the other answers to this question. They also have a lot of specially named properties and operations. So much so that I made a Glossary and included it at the bottom of this answer. Please refer to it if you are confused by any terms I use or the way I use them.

When I talk about a tree (rather than a Tree Graph), I am referring to a hierarchical tree as is commonly used in computer science, with a root, branches, children, etc. When I refer to "a cut on subtree $b$" of a tree, I mean either cutting $b$ from the root (and therefore the rest) of the tree or cutting an edge belonging to subtree $b$. In graph theory, the weight of a cut is the number of edges cut, but with trees, that number is also 1, so the weight of a cut is not that interesting, and I sometimes use "weight of cut" as shorthand to actually mean the size of the subtree created by the cut.

Whenever I talk about weight or heaviness, I am referring to the number of nodes of some subtree or the number of nodes remaining in the other tree after removing one or more subtrees. This corresponds to the order of the component of the Tree Graph represented by those nodes of the tree.

I am not going to try to prove everything. I started to but it was getting long and tedious. Leave me a comment if you think I asserted something that is untrue and maybe I will add a proof later.

  • The goal is to find the optimum trisection of a Tree Graph in $O(n \log n)$ time.
  • "Optimum" is defined as $\max(abc)$, where $a$, $b$, and $c$ are the sizes of the three components remaining after the trisection.

Part 1: Converting the Tree Graph into the right kind of Tree

  • Find vertex $r$ such that the components of $G-r$ all have order at most $n/2$. There is guaranteed to be at least $1$ such vertex.
  • Convert to a hierarchical tree by rooting the Tree Graph at $r$.

The resulting tree has some great properties. In particular it has a subtree I will be calling the "heavy subtree" that contains as many or more nodes as any other subtree, but at the same time is guaranteed to have weight $h \le n/2$.

Part 2: Operating on the Tree

Finding best bisection in $O(n)$

We can create the tree described above, including calculating the weight of every node and finding the heavy subtree, in $O(n)$.

The optimum bisection of the Tree Graph is obtained by cutting the edge between the root of the tree and the heavy subtree.

Show that this is true

Keep in mind that for integers, $x > y \iff (x-1)(y+1) > xy$.

By definition, the weight of the heavy subtree $h \le n/2$, and therefore the number of nodes not in it are $r = n - h$. Also, $h \ge$ any other subtree's weight $s$. Given that any subtree of a tree is smaller than the full tree, $h > s \iff h >$ the weight of any subtree of $s$. So $r \ge h \ge s \ge$ any other subtree. So $rh$ is the product of the $2$ largest and closest numbers we can make and is therefore the best we can do.

Finding the best trisection in $O(n \log n)$

The other missing piece

The best trisection requires at least $1$ cut on the heavy subtree. (Actually, because of symmetry, that is a slight overstatement. Formally, at least $1$ element of the set of equivalently optimal trisections can be made by a cut on the heavy subtree.)

The same reasoning that proves cutting the heavy subtree $h$ makes the best bisection proves that any trisection made with 2 cuts not on $h$ would not be made worse by cutting $h$ instead of either of the other 2 cuts.

Let $h$ be the weight of the heavy subtree, $x$ and $y$ be the weight of the 2 subtrees cut to make a trisection and leaving $r = n - (h + x + y)$ nodes not in the heavy subtree or in $x$ or $y$. $$h(r+x)y \ge (h+r)xy$$ $$h(r+x) \ge (h+r)x$$ $$hr + hx >= hx + rx$$ $$hr >= rx$$ $$h >= x $$

Since we specifically chose $h \ge$ all possible values of $x$, we can count on being able to find an optimal trisection while limiting ourselves to trisections that include at least $1$ cut on the heavy subtree. (Notice that $y$ completely drops out of the equation, so it does not matter if $x$ is greater or less than $y$, which means we do not have to worry that $y$ might be a subtree of $x$. If $y$ was a subtree of $x$, preventing us from keeping $y$ while replacing $x$ with $h$, we could replace $y$ with $h$ instead.)

By reasoning similar to the above bisection and trisection arguments, we can also be sure that if the optimal trisection requires 2 cuts on the heavy subtree, then it will be the optimal bisection of the heavy subtree cut from the rest of the tree. If you make $2$ cuts on the heavy subtree, you are committing to $r \ge n/2$ being $a$, the largest of the 3 components of the trisection. With $r > n/3$ you want to make it smaller, so you want $b+c$ to be as large as possible, which means you want a cut between the heavy subtree and the root. With that done and the requirement that the second cut be in the heavy subtree, the best second cut is by definition the optimal bisection of the heavy subtree.

Grinding it out

With that established, we only need some simple data structures. We have already found all possible subtree sizes for the whole tree. We can find all possible cuts on the heavy subtree and the best bisection of the heavy subtree in $O(n)$.

We can then put the sizes of all of the subtrees not on the heavy subtree into a binary search tree in $O(n \log n)$. Then, given a cut on the heavy subtree, we can find the best possible cut not on the heavy subtree in $O(\log n)$ time.

We then try all $O(n)$ cuts of the heavy subtree against all cuts not on the heavy subtree for $O(n \log n)$ total running time.

We have then calculated all the possible trisections we have not ruled out, and if we were paying attention, we remembered which one was $max(abc)$. We are done.

Optimization to $O(n)$ total running time

If we really care about the time complexity of this, we can get rid of the sorts and binary searches at a cost of increasing the number of $O(n)$ operations and making the solution even more of a grind.

We can, in one $O(n)$ pass through the data structure, find any one number that constitutes the "best" choice according to some arbitrary criteria, so long as either there is always a "best" choice or we do not care which of several equally good choices we pick as "best", which is our case here.

Let us divide the full tree $T$ of size $n$ into the heavy subtree $H$ and the rest of the tree $R$. Using the analysis that got us to the $O(n \log n)$ solution, we only need to find 10 numbers. I will use the shorthand "pair closest to x" to mean the pair of numbers made up of the smallest number $\ge x$ and the largest number $\le x$, even though of course the 2 numbers closest to $x$ might both be greater or less than $x$.

  • $h = |H|$, the weight of $H$
  • $h_{half}$: the weight of the subtree of $H$ closest to $h \over 2$
  • 2 weights of subtrees of $H$: the pair closest to $n \over 3$.
  • 2 weights of subtrees of $R$: the pair closest to $n \over 3$
  • 2 weights of subtrees of $H$: For each of weights $w$ of the pair of subtrees in $R$ closest to $n \over 3$: the weight of the subtree closest to $(n-w)\over 2$
  • 2 weights of subtrees of $R$: For each of weights $w$ of the pair of subtrees in $H$ closest to $n \over 3$: the weight of the subtree closest to $(n-w)\over 2$

The best trisection is either going to be $h_{half}(h - h_{half})(n-h)$ or some combination of 1 of the 4 subtrees of $H$ with 1 of the 4 subtrees of $R$ found above. Since some of the subtrees could be the same (and we would know right away), we we do not necessarily need to find all 10 numbers. We could get away with just finding $h$ and 2 subtrees of weight exactly $n\over 3$ and not have to look further. So we find 3-10 numbers to make from 1 to 1+4+4 = 9 possible partitions to try and pick the best, all $O(n)$ or $O(1)$ operations, and we have and answer in $O(n)$ time.

Glossary

  • A hierarchical tree of size or weight $n$ = a tree graph $G = (V,E)$ of order $n$
  • is a collection of $n$ nodes = $|G|$ = $|V|$ vertices (singular: vertex) connected by
  • $n-1$ connections = $|V|-1$ edges = $|E|$ edges,
  • such that every node is reachable = connected to every other node by a single unique path, which means there are no
  • nodes with more than one parent = cycles.
  • The $order$ of a component or a graph is the number of vertices it contains.
  • The $degree$ of a node/vertex is the number of connections to the node = edges having the vertex as an endpoint.
  • The $weight$ of a node in a hierarchical tree is the number of nodes in the subtree rooted at that node, which equals the number of the node's descendants plus $1$ for the node itself.
  • Something is heavier than something else if it has greater weight. Likewise, something with less weight is said to be lighter.
  • A component is a maximal connected graph = a set of vertices such that there is a path from any vertex to any other vertex. A tree graph is a special kind of graph that is a single component with no cycles or parallel edges. It has the minimal number of edges possible to make the vertices a single component (completely connected).
  • A hierarchical tree is a specialization of a tree graph. It has a single node designated as the root. All nodes except the root have exactly $1$ parent. (The root does not have a parent). All nodes connected to a node except for its parent are said to be children of that node and have that node as their parent. Thus for all nodes, if $a$ is $b$'s parent, $b$ is $a$'s child, and all nodes connected to the root are children of the root. Children are said to be below their parents and parents are said to be above their children. All the children of a given node are called siblings of each other. A node's children's children are called its grandchildren, and names consistent with genealogy are likewise used to describe the rest of the relationships, except that a node with no children is called a leaf. A node's descendants are its children and all of their descendants.
  • A subtree of a tree is a tree itself, consisting of a node and all of its descendants. The subtree corresponding to the root node is the entire tree.
  • A tree graph may be rooted at a vertex. Any node/vertex may be chosen as the root.
  • To root a tree graph at a vertex $r$ means to map it into a hierarchical tree with $r$ as the root, all the neighbors of $r$ as the root's children and all their parents the root, and then recursively for every vertex $v$ starting with the children of the root, designating neighbors of $v$ except for its parent as its children and designating its children's parent as $v$, recursing on $v$'s children until all the vertices are mapped. At the end, every vertex in the tree graph corresponds to exactly 1 node in the tree, and every edge in the tree graph corresponds to exactly 1 parent-child connection, and so they can be considered interchangeable.
  • When you disconnect a node from its parent, it is said to become the root of a disjoint subtree, and can be referred to as the subtree rooted at that node.
  • Similarly, when you cut an edge of a tree graph, you split (or $bisect$) it into $2$ tree graphs, one containing those vertices that remain connected to the vertex on one side of the edge (and all the edges that connect them) and those that remain connected to the vertex on the other side of the edge (and all the edges that connect them). This is called a $bisection$. Remove $2$ edges and you $trisect$ the tree (a $trisection$).
  • A node with no children has no descendants and is called a leaf. A vertex of degree $1$ is called a leaf. It is possible to root a tree graph at a vertex of degree $1$, but in a hierarchical tree the root is not considered a leaf unless it is the only node in the tree.
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Find a vertex $r$ such that the components of $G-r$ all have order at most $n/2.$ Denote these components by $T_i$ for $1\leq i\leq k,$ with $|T_1|\geq |T_2|\geq |T_3|\geq\cdots.$ By recursion on these trees we can efficiently compute the set $S_i$ of possible cardinalities $|T'|$ of sets $T'\subseteq T_i$ such that $T'$ and $G\setminus T'$ are connected.

The trisection will have $r\in A.$ I'll consider several cases for $B$ and $C,$ which are exhaustive up to permuting $A,B,C.$ In each case the algorithm should find an optimum trisection subject to the constraint for that case.

  1. $B$ and $C$ both lie in some $T_i.$

    It's always best to make $A\cap T_i$ as small as possible because $(a-1)(b+1)c\geq abc$ for $a\geq n/2>b,c.$ So we only need to check the case $A\cap T_i=\emptyset,$ or in other words when $B,C$ are an optimum bisection of $T_i.$ This can be found in time $O(|T_i|).$

  2. $B$ and $C$ lie in different $T_i$'s, and $b,c\leq n/3$

    In this case we should pick $b$ and $c$ as large as possible without exceeding $n/3,$ because $c\leq n/3$ and $b<b'\leq n/3$ imply $(n-b-c)bc \leq (n-b'-c)b'c.$ Among the $k$ values $\max(S_i\cap \{1,\dots,\lfloor n/3\rfloor\}),$ taking the top two gives values for $b$ and $c$; look at the corresponding $T_i$'s to see what trisection this came from.

  3. $B$ and $C$ lie in different $T_i$'s, and $b,c\geq n/3$

    We just need to check the case $B\subseteq T_1$ and $C\subseteq T_2$ because the other $T_i$ are too small. Pick $b$ and $c$ as small as possible, so $b=S_1\cap\{\lceil n/3\rceil,\dots\}$ and similarly for $c,$ and work out what trisection this corresponds to.

  4. $B$ and $C$ lie in different $T_i$'s, and $b\geq n/3$ and $c\leq n/3$

    Split into cases for $1\leq i\leq 2.$ For each case we want to find the maximum of $(n-b-c)bc$ over $b\in S_i$ and $c\in S_i'$ where $S_i'=\bigcup_{j\neq i}S_j.$ This can be done efficiently, for example by sorting $S'_i$ and then for each $b$ find the $c\in S'_i$ as close to $(n-b)/2$ as possible using a binary search. (It should be possible to optimize this to linear time: as $b$ increases, the optimum $c$ decreases or stays the same.)

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  • $\begingroup$ Thanks, but that is way more complicated than I suspect it needs to be. $\endgroup$ – Old Pro Jan 4 at 9:47
  • $\begingroup$ The problem with the way you have structured Case 2 is that the upper bound on $i$ is $n-1$, and a large $i$ is consistent with $b,c\leq n/3$, so checking each $\max S_i$ against the others could be $O(n^2)$. Unfortunately, you cannot just put them all in one big sorted set and pick the best 2, because once you combine the sets, you lose the guarantee that one is not a subtree of the other. This and the other cases all get a lot simpler once you realize that $B$ can always be either all or part of $T_1$. See my answer for an explanation of why. $\endgroup$ – Old Pro Jan 5 at 20:45
  • $\begingroup$ @OldPro: I agree your case analysis is simpler. But I'm not checking each $S_i$ against the others in case 2 - I'm just looking at a list of $k$ values, and taking the top two. This takes $O(k)$ time. Computing the max values is not difficult, for example it can be done in $O(n\log n)$ total by storing the sets $S_i$ as sorted lists and using bisection, or could be done in $O(n)$ total by iterating over all the subtrees. $\endgroup$ – Dap Jan 6 at 7:07
  • $\begingroup$ @OldPro: there are $k$ values, let's call them $m_i:=\max(S_i\cap \{1,\dots,\lfloor n/3\rfloor\})$ with $1\leq i\leq k.$ It takes $O(n)$ to find the top two elements of a list (breaking ties arbitrarily), giving some $i\neq j$ with $m_i$ and $m_j$ being the top two values. These correspond to subtrees of different trees $T_i$ and $T_j$ so they cannot intersect. $\endgroup$ – Dap Jan 6 at 7:25
  • $\begingroup$ Yes, of course you are right. Sorry, and thank you for explaining. $\endgroup$ – Old Pro Jan 6 at 10:17
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I believe you can use Heavy Light Decomposition to get an $\mathcal{O}(n\log^2n)$ time algorithm. This is quite complicated, so I will not bother with too many details. The idea is as follows.

Fix some root and determine the HLD. Then for every node we consider the subtree rooted at that node. We will call such subtrees rooted subtrees. We can easily do a linear time DFS to determine for every rooted subtree how many nodes it contains. Less obvious is that we can then do an $\mathcal{O}(n\log^2n)$ time DFS on all heavy chains to determine for each rooted subtree the optimal $2$-cut of that subgraph. With all this information processed, we can find the optimal $3$-cut of the original graph in $\mathcal{O}(n\log^2n)$ time.

The DFS on the heavy chains works as follows. For each chain, we work from bottom to top. We can keep track of all the sizes of rooted subtrees below the current node in a BST. So for each node $N$ in the heavy chain, from bottom to top, we add all sizes of rooted subtrees of its descendants not already in the BST (these can easily be found in linear time with a DFS), and then do a query to find the size closest to $s/2$ where $s$ is the size of the rooted subtree of $N$. Since each path from a root to a node intersects $\mathcal{O}(\log n)$ heavy chains, every node is only considered $\mathcal{O}(\log n)$ times. There are $n$ such nodes and BST operations cost $\mathcal{O}(\log n)$ time, so this step takes $\mathcal{O}(n\log^2n)$ time in total.

Finally, finding the optimal $3$-cut of the original graph in $\mathcal{O}(n\log^2n)$ time is still quite tedious. Any $3$-cut is itself a tree graph, when we connect two components if they are connected in the original graph. Notice there is only one possible tree graph with $3$ nodes. The root of our original graph will be in one component. We can consider two cases. The root component is connected to both other components, or only one.

If the root component is connected to only one other component, we can simply consider every possible rooted subtree and their optimal $2$-cut. Then you will find the optimal $3$-cut where the root component is connected to only one other component. This takes linear time.

If the root component is connected to both other components, things get a lot more complicated, and we will need to use our HLD again. We need a global BST for this step. It initially contains all sizes of rooted subtrees. For each heavy chain, we work from bottom to top. We want to consider for every node in the chain the optimal $3$-cut where one of the components is its rooted subtree. This means the other component that does not contain the root must be a disjoint rooted subtree. So we want to keep track of all such rooted subtrees in our BST.

If we traverse the heavy chains in a DFS, we can ensure that every rooted subtree containing the current heavy chain is already removed from the BST. Then, working from bottom to top, we can ensure that every rooted subtree below the current node is also removed from the BST. We can then do an $\mathcal{O}(\log n)$ BST query to find the optimal $3$-cut per node. Finally, we add all the rooted subtrees back to the BST. By similar analysis to before, this step will take $\mathcal{O}(n\log^2n)$ time in total.

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  • $\begingroup$ Thank you for thinking hard about this. I'm not completely following your answer yet, but so far it seems to me this still might be problematic in pathological cases. For example, there is no guarantee about how many heavy chains you end up with in HLD: in a star, all but 1 vertex has degree 1, so every edge is a heavy chain (or light edge, depending on your flavor of HLD). And if finding the optimal 2-cut is O(n), doing it n times makes it O(n^2) as you know. Plus I'm still not sure how you keep track in O(1) time and O(n) space of which vertices are in which component after cutting 2 edges. $\endgroup$ – Old Pro Dec 29 '18 at 3:35
  • $\begingroup$ @Old Pro Very good observations indeed. I have thought about these problems, and come up with answers / solutions, however this gets really technical, so I did not feel like elaborating too much on this in the answer. For the first question, finding the optimal $2$-cut for each node in a heavy chain requires $\mathcal{O}(n\log n)$ time where $n$ is the amount of nodes descendant from the heavy chain. Summing over all heavy chains gives $\mathcal{O}(n\log^2 n)$ time where $n$ is the total amount of nodes. I don't understand your second question to be honest. $\endgroup$ – SmileyCraft Dec 29 '18 at 3:42
  • $\begingroup$ I'm worried about your analysis of the HLD for pathological cases. Consider a starlike graph with 4 arms rooted at the center, a linear graph rooted at a leaf, and another rooted in the center, etc. My "second question" that you referred to is about keeping track of which vertices are in which component after the 3-cut. In your bottom up search for the optimal 3-cut, it is not sufficient that you removed the subtree from the BST. If the root component is connected to both other components, the BST will often find a match for best value in the edge above you. That is invalid. $\endgroup$ – Old Pro Dec 29 '18 at 5:03
  • $\begingroup$ I do not understand your "pathological" case. But the principle of a HLD is that every path from the root to a node crosses $\mathcal{O}(\log n)$ heavy chains. There are no exceptions to this, and this is all I used in the analysis. For the "second question" the importance is the order in which we traverse the heavy chains. "If we traverse the heavy chains in a DFS, we can ensure that every rooted subtree containing the current heavy chain is already removed from the BST." So all edges above you should have already been removed from the BST. If you want, I might write some pseudocode tomorrow. $\endgroup$ – SmileyCraft Dec 29 '18 at 12:16
  • $\begingroup$ HLD only guarantees that every path from the root to the node crosses at most $\log_2(n)$ heavy chains. One pathological case is all the nodes connected in a line and rooted at one of the 2 leaves, yielding 1 heavy chain of length n-1 containing all the nodes. Another is all the nodes connected to the same central node, yielding n-1 heavy chains of length 1. These show that some of your assumptions about being able to find the second edge of the 3-cut are wrong. In particular, the optimal pair of cuts may be on the same heavy chain. $\endgroup$ – Old Pro Dec 30 '18 at 5:08

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