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Here is a (slightly abridged) problem from Kleinberg and Tardos:


Consider a complete balanced binary tree with $n$ leaves where $n$ is a power of two. Each edge $e$ of the tree has an associated length $\ell_e$, which is a positive number. The distance from the root to a given leaf is the sum of the lengths of all the edges on the path from the root to the leaf.

Now, if all leaves do not have the same distance from the root, then the signals we send starting from the root will not reach the leaves at the same time, and this is a big problem. We want the leaves to be completely synchronized, and all to receive the signal at the same time. To make this happen, we will have to increase the lengths of certain edges, so that all root-to-leaf paths have the same length (we’re not able to shrink edge lengths). If we achieve this, then we say the tree has zero skew.

Give an algorithm that increases the lengths of certain edges so that the resulting tree has zero skew and the total edge length is as small as possible.


This problem was in the "Greedy Algorithms" chapter.

I know the solution to this problem is as follows:

Let the subtrees below the root be $L$ and $R$. If the maximum sum of the edge lengths on the path to a leaf in $L$ starting from the root is greater than the maximum sum of the edge lengths on a path to a leaf in $R$, then increase the edge length from the root to the right subtree by the positive difference between the two.

Conversely, if the maximum sum of the edge lengths on a path to a leaf in $R$ starting from the root is greater than the maximum sum of the edge lengths on a path to a leaf in $L$, then increase the edge length from the root to the left subtree by the positive difference between the two.

I am having trouble proving the following:

  • $1)$ The resulting tree upon termination of this algorithm is a zero-skew tree (the length from the root to any leaf is the same).

  • $2)$ The sum of the edge lengths is minimized with this algorithm.

I have tried to prove these facts in many ways, such as induction and direct proof, but I am having a lot of difficulty doing so. The algorithm makes sense to me intuitively, but I'm having trouble explaining why it works formally.

I would greatly appreciate some help in solving this problem.

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  • $\begingroup$ Are formulas correct? Doesn't it yield a wrong answer for the simplest tree with 2 leaves, if edges have different weight? It seems you won't change anything as both subtrees have the same (zero) height. $\endgroup$ – Vladislav Mar 14 at 16:19
  • $\begingroup$ Sorry I did not make it clear. The height of the tree is the sum of the edge lengths on the path from the parent to the leaf. In your example for a tree with two leaves, if we let $\ell_{L}$ denote the edge length to the left child and $\ell_{R}$ denote the edge length to the right child, then we increase the smaller one of $\ell_{L}$ and $\ell_{R}$ by $|\ell_{L} - \ell_{R}|$. Hopefully that makes more sense. $\endgroup$ – user100598 Mar 14 at 16:27
  • $\begingroup$ @John L. Yes, you are correct. I can't figure out how to edit my previous comment. I will update my post. $\endgroup$ – user100598 Mar 14 at 16:34
  • $\begingroup$ @hom this definition makes the height of the root undefined. Wouldn't it be better to define heights the usual way and correct formulas? $\endgroup$ – Vladislav Mar 14 at 16:35
  • $\begingroup$ I've gotten rid of the term "height" entirely. Let me know if there is any confusion. $\endgroup$ – user100598 Mar 14 at 16:39
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Notation. Given a binary tree $T$ and vertex $v$ of $T$, let $\ell_v$ and $r_v$ be the left and right children of vertex $v$ in $T$, respectively. Let $T_v$ be subtree of $T$ rooted at $v$. Let $d_T(v)$ be the weighted distance between the root of $T$ and $v$ and $h(T) = \max_{v \in T} d_T(v)$. If $(u,v)$ is an edge of $T$, let $w_T(u,v)$ be its weight. With a slight abuse of notation, define $w(T) = \sum_{(u,v) \in T} w(u,v)$. Finally, we define the skew of $v$ as $s_T(v) = w_T(v, \ell_v) + h(T_{\ell_v}) - w_T(v, r_v) - h(T_{r_v})$ if $v$ is an interval vertex of $T$ and $s_T(v)=0$ if $v$ is a leaf of $T$.

1) The resulting tree upon termination of this algorithm is a zero-skew tree (the length from the root to any leaf is the same).

We will show by induction on $i=0,1,\dots$ that, for all trees $T$ with $2^{i+1} - 1$ vertices, the algorithm returns a tree in which all vertices have skew $0$.

The base case $i=0$ is trivially true since $T$ must have exactly one vertex and the algorithm does nothing.

Suppose now that $i>0$ and let $v$ be the root of $T$. The trees $T_{\ell_v}$ and $T_{r_v}$ have exactly $\frac{(2^{i+1} - 1) - 1}{2} = 2^i - 1$ vertices and, by inductive hypothesis, the recursive calls on $T_{\ell_v}$ and $T_{r_v}$ ensure that all vertices, except possibly $v$, have skew $0$. The reweighing of $(v, \ell_v)$ and $(v, r_v)$ only affects the skew of $v$ and ensures that $s(v)=0$.

The sum of the edge lengths is minimized with this algorithm.

Let $T'$ be the tree returned by the algorithm. Suppose that an optimal zero-skew tree $T^*$ obtained from the input tree $T$ by lengthening edges is such that $w(T^*) < w(T')$.

Let $(u,v)$ be an edge of $T$ (where $u$ is the parent of $v$) such that $w_{T^*}(u,v) < w_{T'}(u,v)$ and the depth of $u$ is maximized.

Since $s_{T^*}(u)=0$ and since $h(T_v^*) \ge h(T_v) = h(T'_v)$, the sum of the weights in $T^*$ on each path from $v$ to a leaf in $T_v$ must have increased by at least $w_{T'}(u,v) - w_{T^*}(u,v)$. Consider the subgraph of $T^*_v$ induced by the edges whose weight has not been increased w.r.t. $T$ and let $C$ be the connected component containing $v$. Let $E$ be the set of edges with one endpoint in $C$ and the other in $T^*_v \setminus C$. Let $\Delta = \min_{(x,y) \in E} \{ w_{T^*}(x,y) - w_T(x,y) \}$.

Consider a new tree $T''$ that is obtained from $T'$ by reducing the weight of all edges in $E$ by $\Delta$ and increasing the weight of $(u,v)$ by $\Delta$. The tree $T''$ is a feasible solution to your problem and weighs $w(T^*) - |E| \Delta + \Delta < w(T^*)$ (since $|E|\ge 2$). This is a contradiction.

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  • $\begingroup$ Thank you so much. I've been reading this over, and I'm wondering how we know $h(T_v) = h(T'_v)$ holds in the second-to-last paragraph. I know why $h(T^{*}_v) \geq h(T_v)$ holds, but I don't see why $h(T_v) = h(T'_v)$ holds. Could you please clarify? $\endgroup$ – user100598 Mar 14 at 21:18
  • $\begingroup$ @hom, A generic recursive call of the algorithm on vertex $v$ only modifies the weight of one edge incident to $v$. Therefore the heights $h(T_u)$ of all subtrees $T_u$ where $u$ is not an ancestor of $v$ remain unchanged ($v$ is an ancestor of itself). Moreover, the new weight is computed in a way that does not change $h(T_v)$ either. Therefore the heights $h(T_u)$ where $u$ in an ancestors of $v$ also remain unchanged. In conclusion, no invocation of the algorithm changes the height of any subtree, i.e., all subtrees of $T'$ have exactly the same height of the corresponding subtree of $T$. $\endgroup$ – Steven Mar 14 at 22:23
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The proof of the first part is quite straightforward if you strengthen the induction proposal.

Statement:

The algorithm applied to the tree with root $v$ (we denote it as $f(v)$) yields a zero-skew tree which has the same height as $v$.

Proof:

By structural induction: if $v$ has no children, it is trivially true. Otherwise, it has children $v_l, v_r$ and edges of length $w_l, w_r$ respectivetely. Without loss of generality, we assume $h(v_l) + w_l \geq h(v_r) + w_r$. Therefore, $v$ has height $h(v_l) + w_l$. By induction proposal, $f(v_l)$ and $f(v_r)$ are zero-skew with height $h(v_l)$ and $h(v_r)$ respectivetely. Thus, every path from $f(v)$ to its leaf from left subtree has length $$w_l + h(v_l)$$ and every path from $f(v)$ to its leaf from the right subtree has length $$w_r + ((h(v_l) + w_l) - (h(v_r) + w_r)) + h(v_r) = w_l + h(v_l)$$ i.e. $f(v)$ is zero-skew with height $h(v_l) + w_l$.

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