saved_chats/#45407

USER: https://codeforces.com/problemset/problem/1942/C1

BOT: Thanks! Now could you provide a link to the editorial.



USER: https://codeforces.com/blog/entry/126942

BOT: Would you like hints to help you solve the problem or an interactive explanation of the editorial (hint/editorial)? 



USER: hint

BOT: What are you current thoughts on the problem? 



USER: if we sorted the vertices that have been chosen then, if any vertex equal the previous vertex + 2 the then we can make a polygon, any for every vertex when there are strictly 2 or more vertices before it then we can also make a triangle 

BOT: Your idea of sorting the vertices and checking for vertices that are exactly two apart aligns with the editorial's approach. This helps in forming additional triangles. 

For a small hint: Focus on forming triangles by connecting non-adjacent vertices and consider how many such connections you can make given the constraints.

Would you like a more detailed hint or further elaboration on this approach?



USER: yes

BOT: To maximize the number of triangles, you should:

1. Sort the chosen vertices.
2. Form a smaller polygon using these vertices.
3. For each pair of adjacent vertices in this smaller polygon, check if there is exactly one vertex between them in the original polygon. If so, this can form an additional triangle.

Would you like further elaboration on how to implement this or any specific part of the solution?



USER: yes

BOT: Here's a more detailed hint:

1. **Sort the Vertices**: Start by sorting the chosen vertices in ascending order.
2. **Form Triangles**: For each pair of adjacent vertices in the sorted list, check if there is exactly one vertex between them in the original polygon. If there is, you can form an additional triangle.
3. **Count Triangles**: The number of triangles formed will be the number of chosen vertices minus two, plus the number of additional triangles you can form by the above method.

Would you like further elaboration on any specific step or part of this approach?



USER: can you proof it

BOT: Sure, let's break down the proof:

1. **Forming a Polygon with Chosen Vertices**:
   - When you have $ x $ chosen vertices, you can form a smaller polygon with these vertices.
   - A polygon with $ x $ vertices can be triangulated into $ x - 2 $ triangles. This is a well-known result in computational geometry.

2. **Additional Triangles**:
   - For each pair of adjacent vertices in the sorted list of chosen vertices, if there is exactly one vertex between them in the original $ n $-sided polygon, you can form an additional triangle.
   - This is because the two chosen vertices and the vertex between them form a triangle.

3. **Counting Triangles**:
   - The base number of triangles from the $ x $ chosen vertices is $ x - 2 $.
   - You then add the number of additional triangles formed by the pairs of adjacent vertices with exactly one vertex between them.

**Example**:
- Suppose you have a polygon with $ n = 6 $ sides and chosen vertices are $ \{1, 3, 5\} $.
- The sorted chosen vertices are $ \{1, 3, 5\} $.
- Forming a polygon with these vertices gives you $ 3 - 2 = 1 $ triangle.
- Checking for additional triangles: 
  - Between vertices 1 and 3, there is vertex 2.
  - Between vertices 3 and 5, there is vertex 4.
  - Both pairs can form additional triangles.
- So, the total number of triangles is $ 1 + 2 = 3 $.

This method ensures you are counting all possible triangles formed by the chosen vertices and any additional triangles formed by vertices in between.

Would you like further elaboration on any specific part of this proof?



USER: why is this true: A polygon with xx vertices can be triangulated into x−2x−2 triangles.
?

BOT: The statement that a polygon with $ x $ vertices can be triangulated into $ x - 2 $ triangles is a well-known result in computational geometry. Here’s a brief proof by induction:

1. **Base Case**: 
   - For $ x = 3 $ (a triangle), the polygon is already a triangle, so no additional diagonals are needed. The number of triangles is $ 3 - 2 = 1 $.

2. **Inductive Step**:
   - Assume that any polygon with $ k $ vertices can be triangulated into $ k - 2 $ triangles.
   - Consider a polygon with $ k + 1 $ vertices. Choose any vertex $ v $ and draw a diagonal to a non-adjacent vertex, splitting the polygon into two smaller polygons.
   - Each of these smaller polygons has fewer than $ k + 1 $ vertices and can be triangulated by the inductive hypothesis.
   - The total number of triangles formed in the $ k + 1 $-vertex polygon is the sum of the triangles in the two smaller polygons plus the one triangle formed by the diagonal, which is $ (k - 2) + 1 = k - 1 $.

Therefore, by induction, a polygon with $ x $ vertices can be triangulated into $ x - 2 $ triangles.

Would you like further elaboration on any specific part of this proof?