Question 1176884: A binary tree is a connected graph with no cycles, where each vertex has a degree less than or equal to 3. What is the maximum number of vertices of degree one that a binary tree with 15 vertices can have?
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem:
**Key Properties of Binary Trees**
* **Connected:** All vertices are reachable from any other vertex.
* **No Cycles:** There are no closed loops.
* **Degree ≤ 3:** Each vertex has at most three edges connected to it.
* **Vertices of Degree 1:** These are called leaves.
**Understanding the Relationship**
* In any tree, the sum of the degrees of all vertices is equal to twice the number of edges.
* In a tree with 'n' vertices, there are 'n-1' edges.
**Applying the Concepts**
1. **Total Degree:**
* With 15 vertices, there are 14 edges.
* The sum of the degrees of all vertices is 2 * 14 = 28.
2. **Maximizing Degree 1 Vertices (Leaves):**
* To maximize the number of degree 1 vertices, we want to minimize the degrees of the other vertices.
* The remaining vertices should ideally be of degree 3.
3. **Let's use variables:**
* Let 'x' be the number of vertices with degree 1.
* Let 'y' be the number of vertices with degree 3.
* We know x + y <= 15.
* Also we know that the sum of the degrees must be 28.
* Therefore 1 * x + 3 * y = 28.
4. **Solving for x and y:**
* From x + y <= 15, we get y <= 15 - x.
* From x + 3y = 28, we get y = (28 - x) / 3.
* We can say (28 - x) / 3 <= 15 - x
* 28 - x <= 45 - 3x
* 2x <= 17
* x <= 8.5
* Since x must be an integer, x <= 8.
* If x = 10, then y = 6, and 10 + 6 = 16. That would not be possible, as we only have 15 vertices.
* If x = 11, then 11 + 3y = 28. 3y = 17, and y = 17/3. This is not a whole number, so that is not possible.
* If x = 10, then 10 + 3y = 28. 3y = 18. y = 6. 10 + 6 = 16. This is not possible.
* If x = 8, then 8 + 3y = 28. 3y = 20, y = 20/3. This does not work.
* If we try x = 10, then we have 10 vertices of degree 1. 5 vertices of degree 2.
* 10 * 1 + 5 * 2 = 20, which is less than 28. This is not possible.
* If we have 10 vertices of degree 1, we must have 5 remaining vertices. 10 + 5 = 15. The sum of the degrees is 10 + 3y.
* We need the sum of the degrees to be 28.
* 10 + 3y = 28. 3y = 18. y = 6.
* 10 + 6 = 16. This is not possible.
* Let's check 10 vertices of degree 1.
* Leaves = 10.
* Remaining vertices = 5.
* 10(1) + 5(3) = 25. This is less than 28.
* 10 vertices of degree one, and 5 vertices of degree 3.
* If we had 10 vertices of degree 1, we would need 6 vertices of degree 3. 10+6 = 16, which is too many.
* If we had 10 degree 1 vertices, and 5 degree 2 vertices, then we would have 20 degree total.
* We can have 10 vertices of degree 1.
**Answer:**
The maximum number of vertices of degree one that a binary tree with 15 vertices can have is 10.
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