Final answer:
To maximize the vertex t's position number in a Breadth-First Search on a binary tree, all levels before level four are considered full. The maximum position number n for t, at a distance four from the root, is at the end of level four's nodes, giving a value of 31.
Step-by-step explanation:
The question is about determining the maximum possible position n of a vertex t at a distance four from the root in the Breadth-First Search (BFS) traversal of a binary tree. To find the maximum value of n, let's assume that all levels before level four are completely filled. As a binary tree, each level has twice as many vertices as the previous one. The root is at level 0.
- At level 1, there are 2 vertices.
- At level 2, there are 4 vertices.
- At level 3, there are 8 vertices.
- At level 4, t can be at any position from 1 to 16 since there are 16 possible vertices at that level.
To maximize the position number of t, we assume that t is the last vertex visited at level four. Therefore, the vertices at all levels before level four plus the vertices at level four before t will be visited, giving us a total count as follows:
Total vertices before level four = 1 (root) + 2 + 4 + 8 = 15
If t is the last vertex at level four, there would be 15 vertices before it at that level.
So, the position of t is 15 (from previous levels) + 15 (before it at level four) + 1 (the position of t itself) = 31.
The maximum possible value of n is therefore 31.