Triangular (and tetrahedral) lattice

An equilateral triangle $ABC$ with side $n$ a positive integer is divided into triangular lattice consisting of unit triangles.

Each point on the lattice is colored red, white, or blue such that:
1. No point on $AB$ is colored red
2. No point on $BC$ is colored blue
3. No point on $AC$ is colored white

Prove that there exists a unit triangle whose vertices are colored with 3 different colors.

Challenge problem: Same problem but for tetrahedron and four colors.

Solution

We first state the one-dimensional version of this problem:

Given a sequence of points on a line colored red or blue, and the first point must be colored red, and the last point must be colored blue, then we have an ODD number of segments that has two differently-colored end points.

This one-dimensional version is trivial to prove.

Now we turn to the 2-dimensional version as stated in the problem.
For each side that has both red and blue endpoints, we "mark" that side.
Then for each unit triangle, we give it a score based on how many marked sides it has. Now, if we consider the "outside" area as another "surrogate unit triangle", then we can also give it a score based on how many marked sides are facing the outside.

First, note that the total score of all areas must be even, since any marked side contribute one score to two areas.

Now, applying the 1-dimensional version of the problem, we know that side AC has an odd number of marked sides. Side AB and BC does not have any marked sides. So the outside area has an odd score. Thus, there is an odd number of unit triangles that has an odd score. In order for a unit triangle to have an odd score, it must have three differently-colored vertices.

Number of black cells

Each cell on an $n \times n$ checkerboard is to be painted black or white.

How many ways are there to paint the board such that each column and each row contains an even number of black cells?

How many ways are there to paint it such that each column and row contains an odd number of black cells?

Circles and Coloring

Several circles are drawn on a piece of paper to form several regions on the paper. We wish to color these regions such that no two regions that share a common boundary would have the same color.

(It is okay if two regions that share a common point to have the same color, but not boundary)

How many colors minimum are needed?

Room and Lights Game

This problem is identical to this one but reworded for better clarity.

Alice is playing a game against Bob and Charlie. In a room, there are 27 lights with individual switches, some of them could be turned on or off. Bob enters the room while Charlie waits outside. Alice will tell Bob a number from 1 to 27. Bob then is allowed to flip at most 3 switches if he wishes. Then Bob exits and Charlie enters the room. He has to, upon examining the lights, guess the number that Alice told Bob. Neither Bob nor Charlie knows the configuration of the lights before Bob entered the room.

How can Bob and Charlie agree on a strategy to win this game?

Harder version: 15 lights, but Alice tells Bob a number from 1 to 16, and Bob can only flip at most one switch.

Prisoners and hats

2047 prisoners are kept in a dungeon. One day, the King decided to do an experiment. They will each be given either a blue hat or a red hat and then collected in a room. Each prisoner can see the color of everyone else's hats but not his own. No form of communication is allowed. After a few minutes, the warden will come and ask each of them individually what he thinks the color of his hat is. They can answer "red" or "blue." Each prisoner must answer by whispering to the warden's ear, so that the other prisoners can't hear his answer. Those who can answer correctly will be freed, but those who answer incorrectly will be executed on the spot. The prisoners found out about this experiment the night before, so they had some time to devise a scheme to ensure they can save as many people as possible.

Question 1: What is the scheme that allows everyone to live in this case?

Suppose now the King decided to alter the experiment a little bit. The warden will come to the room immediately after everyone get their hats. Each prisoner is allowed to answer "red", "blue", or "I don't know." If everyone answers "I don't know", they will all be executed. If someone answers "red" or "blue" and he is wrong, they will all be executed. Otherwise, everyone is freed at the same time. In other words, in order for them to live, at least one person must answer "red" or "blue" and he must be right. As before, the prisoners can't hear everyone's answers, and the release or execution is delayed until everyone had given their answers.

Question 2: What is the scheme that allows everyone to live with the highest probability? What is that probability?