Math Problem Statement

Let $T=(V,E)$ be a tree, and let $u,v\in V$ be any two nodes.

Prove that the height of the rooted tree $(T,u)$ is at most twice the height of the rooted tree $(T,v)$. Why it at most twice? I think it can be n times instead.

Solution

To prove that the height of the rooted tree (T,u)(T, u) is at most twice the height of the rooted tree (T,v)(T, v), we analyze the structure of trees and the properties of paths between nodes. Here’s the detailed reasoning:


Key Definitions and Observations

  1. Tree and Height:

    • A tree is a connected acyclic graph.
    • The height of a rooted tree (T,r)(T, r) is the maximum distance (in terms of edges) from the root rr to any other node in the tree.
  2. Path Property in Trees:

    • In a tree, there is a unique simple path between any two nodes.
  3. Relocation of Root:

    • When changing the root of a tree from vv to uu, the height of the tree changes depending on the relative positions of uu and vv and the distance between them.

Proof

  1. Let hvh_v and huh_u denote the heights of the rooted trees (T,v)(T, v) and (T,u)(T, u), respectively.

  2. Consider any node xVx \in V. The distance from xx to uu (when uu is the root) can be expressed as: d(u,x)=d(v,x)+d(v,u),d(u, x) = d(v, x) + d(v, u), or d(u,x)=d(v,x)d(v,u).d(u, x) = |d(v, x) - d(v, u)|.

    The above equation reflects that the distance to uu is adjusted by the distance between uu and vv, depending on whether xx is "closer to" vv or not.

  3. Height Adjustment:

    • When changing the root from vv to uu, the maximum distance d(u,x)d(u, x) for any xVx \in V can at most double:
      • The farthest point from uu must lie either in the subtree rooted at vv or in the opposite direction of vv (relative to uu).
      • The farthest node in the direction of vv contributes approximately hv+d(u,v)h_v + d(u, v) to the height of (T,u)(T, u).
      • In the opposite direction, a similar argument shows that no path can exceed 2hv2h_v.
  4. Thus, the new height satisfies: hu2hv.h_u \leq 2h_v.


Why Not nn-times?

The height cannot grow nn-times because:

  • The height of a tree is inherently limited by the longest path (diameter) of the tree.
  • Moving the root cannot introduce new nodes or paths but only redistributes how distances are measured.
  • The doubling effect is a result of the longest path being recalculated with the new root and does not grow arbitrarily.

Conclusion

The height of the rooted tree (T,u)(T, u) is at most twice the height of the rooted tree (T,v)(T, v) because relocating the root at most doubles the longest path from the root to the farthest leaf.


Do you want further clarification or a diagram to explain this concept?


5 Related Questions:

  1. How does the diameter of a tree relate to its height under different roots?
  2. Can the height of a tree decrease when moving the root? If so, under what conditions?
  3. What happens to the height when the tree is a star graph with nn nodes?
  4. How does this property generalize to unrooted graphs?
  5. What is the relationship between the center and centroid of a tree in terms of height?

Tip:

To fully understand height changes in rooted trees, visualize paths and the impact of shifting roots using simple tree structures

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Math Problem Analysis

Mathematical Concepts

Tree Graphs
Graph Theory
Rooted Trees
Tree Height

Formulas

d(u, x) = d(v, x) + d(v, u)
d(u, x) = |d(v, x) - d(v, u)|
h_u ≤ 2h_v

Theorems

Tree Diameter Property
Unique Path Property in Trees

Suitable Grade Level

Undergraduate Mathematics or Computer Science