Suppose there are b black, w white and r red tiles on a table. We are allowed to replace two tiles of different colors with the third color. Note one thing here that this third colored tile is from our pool containing infinite black, white and red tiles.

If we start replacing tiles, can we reach at a state where there is only one tile left on the table?
If YES, then how? and what will be the color of that last tile?
If NO, then why not?
You have to solve this problem with the help of pre-conditions, post-conditions and invariant.
[Example: Let’s say, there are three tiles (1 black, 1 white, 1 red) on the table.
If we pick two tiles (say 1 black and 1 white), we have to replace them with 1 red tile. Now there will be 2 red tiles on the table with no white or black tiles. ]

TELL ME THE NSWER

15

Pre-conditions:

- We have b black tiles, w white tiles, and r red tiles on the table.
- Each tile is considered identical to others of the same color.

Post-conditions:
- We will have only 1 tile left on the table.

Invariant:
- At each step, the sum of black, white, and red tiles on the table remains the same.

To solve this problem, we can start by examining the parity of the number of colors.
- If all three colors are present (i.e., b > 0, w > 0, r > 0), we can choose any two colors and replace them with the third color. This operation will reduce the count of original colors by 1, so if we repeat it until only one color remains, we will eventually have only 1 tile left on the table.
- If only two colors are present, we cannot reach a state with only 1 tile left. This is because each replacement operation either maintains the count of two colors or reduces it by 2, resulting in an even number of tiles.
- If only one color is present, there is no need for further replacement as only 1 tile is already left on the table.

The color of the last tile will depend on the original configuration of tiles. If black is the last remaining color after the replacement operations, the last tile will be black. The same applies to white and red tiles.

To solve this problem, we can analyze the initial state and apply logical reasoning to determine if we can reach a state with only one tile remaining on the table. Let's start by defining the pre-conditions and post-conditions:

Pre-conditions:
1. There are b black, w white, and r red tiles on the table.
2. We are allowed to replace two tiles of different colors with the third color.

Post-conditions:
1. There is only one tile left on the table.

To analyze whether we can reach a state with only one tile left, we can consider the number of tiles of each color. Let's define an invariant:

Invariant:
The parity (odd or even) of the sum of black, white, and red tiles always remains the same.

Based on the given problem, we start with b black, w white, and r red tiles. The initial parity of the sum of these tiles is even (b + w + r is even).

Now, every time we replace two tiles of different colors with the third color, the parity of the sum of the tiles remains the same. This means that if the sum of the tiles was initially even, it will always stay even after each replacement step. The same applies to an odd sum.

Therefore, it is not possible to reach a state with only one tile left on the table. This is because no matter how many replacement steps we perform, the parity of the sum of the tiles will remain the same.

So the answer to the question is NO, we cannot reach a state where there is only one tile left on the table.

Note: This reasoning is based on the premise that we are only replacing two tiles of different colors with the third color. If additional operations or conditions are introduced, the answer may change.