Math Puzzle - 7 Hats

i don't think they get to pick which hat they take. the hats are placed on the prisoners' heads

Before sitting down on the table and getting the hats, the prisoners agree to write down the color of the hat of the person sitting clockwise to them once they sit down and get the hats. Afterwards, they all pass their papers clockwise by one person and turn in the answers.

First guy writes down a color he sees. Everyone copies it.

Prisoners meet and introduce themselves before getting the hats and sitting at the table. One person introduces himself as "Bob," and all the prisoners agree to write down the color of Bob's hat, whatever it happens to be, once they all sit down and get hats.

All the prisoners before hand pick one guy. The six other players agree that if the designated prisoner is wearing a certain colored hat, one of them will begin write that hat color down on their paper (so one of the players will begin to write if the designated one is wearing a red hat, another will begin to write if he is wearing an orange hat, etc). The designated prisoner, seeing the person who is writing, will then know his hat color. If none of the prisoners are writing down anything, then his hat is the seventh color. He writes down his hat color and everyone goes free

If Guy1 sees 6 different colors, he picks the one not there, then the other 6 guys know to ALL pick one of the 6 colors.

If Guy1 sees 5 different colors, he picks one not there, and if Guy2 sees 5 different colors, he knows to pick the other not there, then everyone else knows to ALL pick one of the 5 colors.

And so on... until the last scenario where Guy1 sees only one color, so he'll pick one not there. Guy2 will see one color, he'll pick the second one not there, and so on until Guy7 sees 1 color, but everyone already guessed the other 6 colors so he knows no one else sees those 6 colors, so he picks the 1 color he sees.

Two hat colors, black and white.

Guy X's strategy:
If he sees black, he guesses black.
If he sees white, he guesses white.

Guy Y's strategy:
If he sees black, he guesses white.
If he sees white, he guesses black.

Hat combinations:
BB -> X gets it right
BW -> Y gets it right
WB -> Y gets it right
WW -> X gets it right

Assign a number to each color (0..6)
Assign a number to each prisoner (0..6)
Each prisonner adds the colors he sees on all other guys + his number
Sum%7 gives the color the prisoner has to announce
In any possible case one (exactly one) prisoner will be right.

Let's say three colors, color 1, color 2, and color 3. The prisoners arrange themselves as prisoner 1, 2 and 3. Each prisoner has a different strategy, which combined will let them live.

Case 1: both hats you see are same color (11, 22, or 33)
prisoner 1 writes down the same color
prisoner 2 writes down color of one greater
prisoner 3 writes down color of two greater

for example, if you see that the other hats are both color 1, (if you see 11):
--prisoner 1 writes down hat 1 (same)
--prisoner 2 writes down hat 2 (one greater)
--prisoner 3 writes down hat 3 (two greater)

If you see 22
--prisoner 1 writes down 2 (same)
--prisoner 2 writes down 3 (one greater)
--prisoner 3 writes down 1 (two greater)

If you see 33
--prisoner 1 writes down 3
--2 writes down 1
--3 writes down 2


Case 2: if the hats you see are different color (12, 23, or 31)
prisoner 1 writes down the remaining color
prisoner 2 writes down the the left color
prisoner 3 writes down the right color

*note i don't mean traditional left and write.

for example if you see 12
--1 writes down 3
--2 writes down 1
--3 writes down 2

if you see 23
--1 writes down 1
--2 writes down 2
--3 writes down 3

if you see 31
--1 writes down 2
--2 writes down 3
--3 writes down 1


This strategy will cover all 3^3 = 27 possibilities.

for cases where all numbers are same, prisoner 1 guesses right
111, 222, 333

for cases where all numbers are different, prisoner 1 guesses right
123, 231, 312
132, 213, 321 (symmetrical to first)

for cases where two numbers are same, then either prisoner 2 or prisoner 3 will guess right
(prisoner 2 first case, prisoner 3 second case)
121, 113 (prisoner 2 sees 11, so he guesses 2. prisoner 3 sees 11, so he guesses 3)
232, 221
313, 332

112, 131 (prisoner 2 sees 12, so he guesses 1. prisoner 3 sees 31, so he guesses 1)
223, 212
331, 323

211, 311 (symmetrical to case above, since person 2 still sees 12, person 3 still sees 31)
322, 122
133, 233

And those are all the cases.

before sitting down, everyone agrees to only consider one person's hat, person A. Then, they will each assign each other a color based on the 7 colors available. If person A is wearing red, person B will be the first to give his paper to the king; it doesn't matter what person B guesses. Then person A knows his hat is red. If person A's hat is orange, person C will be the first to give his paper to the king, and person A will know his hat is orange, and so on. If person A is wearing the 7th color that is assigned to him, nobody will turn in the paper to the king and person A will know his hat is the 7th color.

modulo 2

We assign each color a number from 0-6

Now we can say that the prisoners are essentially getting a number, c0,c1,c2,c3,c4,c5,c6.

S = (c0 + c1 + c2 + c3 + c4 + c5 + c6) modulo 7

S has seven possible values, 0-6

At this point, what each prisoner is doing is trying to guess what S is, and solving for their own number to match S.

Prisoner 0 assumes S = 0, and solves the equation c0 = S - (c1+c2+c3+c4+c5+c6)
Prisoner 1 assumes S = 1, and solves the equation c1 = S - (c0+c2+c3+c4+c5+c6)
and so on..

At least one of the 7 prisoners will assume the correct S value. That prisoner will be able to guess the correct number for their own hat.

Assign each color a value 0-6. Each person takes sum of all the colors he sees and says the result in mod 7 (says the corresponding color to the sum). If you let a_1 be the first person's color, a_2 second guy's, etc, and S the sum of all, the first person will say S-a_1 (mod 7), second S-a_2, etc. Now it is possible to express all a_2, a_3, ... a_7 in terms of a_1. Add all values (expressed in terms of a_1) and you get 7a_1 + X, for some X. X = S (mod 7). Now the only thing the last person has to do is subtract and get his color.

This doesn't break any rules in the OP. Two prisoners agree beforehand to write each other's names on their pieces of paper. They also write the color of the hat that the other person is wearing. Bam, foolproof answer.

We start with the fact that the summation of the values of the hats modulo 7 is a number between 0 and 6, inclusive, and it doesn't change once the hats have been assigned.

That is, [hat1 + hat2 + hat3 + hat4 + hat5 + hat6 + hat7] % 7 = C, for some number C that can't change.

For each prisoner, there is one unknown hat (his own). There is also the unknown value C. If he guesses a value for C, then he can solve for his own hat value.

Since there are 7 possible values for C, each prisoner agrees to choose a different value to guess for C.

Then, exactly 1 prisoner will choose the correct value (the other 6 will choose the wrong values). He will then proceed to guess his own hat value correctly.

Note that this won't work if you have fewer prisoners than hat colors.