57 replies to this topic

[wolfgang]

An unknown number of prisoners( more than 14)were told a day before their execution, that they are going to be blind folded, and a colored paper(either.. yellow,green,red or blue) will be  glued on forehead of each one of them( no one can see his own paper), then they should inter a big hall one by one to make 4 raws in this order( Yellow,Green,Red,Blue)i.e. a raw of Yallows, a raw of Greens  etc.,each one will be unfolded when he inters the hall.No one of them is allowed to arrenge the others,each one should himslf choose where to stay, once he did,he can not change his place( only the first 2 prisoners can change their minds only once).Any kind of comunication between them is not allowed. If any one of them stands in a wrong raw, all of them will die!

All colores were used.

They should make a plane which saves all of them, can you help them?

[Rainman]

Spoiler for solution

They assign values to each color: yellow = 0, green = 1, red = 2, blue = 3. The first prisoner adds the values of all the other prisoners' hats (modulo 4), then finds the corresponding color to that number and stands in that row. By doing this, everyone else will know the color of their own hat. Now the second prisoner does the same thing as the first one did. This tells the first prisoner the color of his/her hat. Then the first two prisoners change their minds if necessary, moving to the right row. And everyone else can move to the right row as well.

[plasmid]

Spoiler for solution

They assign values to each color: yellow = 0, green = 1, red = 2, blue = 3. The first prisoner adds the values of all the other prisoners' hats (modulo 4), then finds the corresponding color to that number and stands in that row. By doing this, everyone else will know the color of their own hat. Now the second prisoner does the same thing as the first one did. This tells the first prisoner the color of his/her hat. Then the first two prisoners change their minds if necessary, moving to the right row. And everyone else can move to the right row as well.

I think the problem's set up in such a way that you can't use that strategy. The people are blindfolded until they enter the room. So the first person won't see anyone's color, the second person will only see the first person's color, etc.

Just for clarification, are there set positions where each row should form? In other words, should we treat this as if there are four rows of seats and all of the yellows must sit in the first row and all the greens in the second row etc.? Or instead can they form rows wherever they want, so if the first person to enter is yellow he can he stand anywhere in the room as long as all the other colors stand somewhere behind him?

[Pankaj Varma]

The suggested solution assumes that their are equal number of each color slip and the each person, when he enters can see every other forehead and count the color papers glued there. We need a more innovative solution, I guess.

[Prime]

I don't see the OP requiring each prisoner to assume his position before the next one enters the room. So why should we be adding such a restriction?

On the other hand, the OP never specified that the place for each row has been marked with like colored chairs, tile, or in any other way. Thus we must not assume that it has.

Given the above, Rainman's method needs just a slight modification.

Spoiler for how about

How about, after entering the hall, they just hang out there and see what the first two guys do. Then the first prisoner just stands where he is, while the second stands on his left if the first one has a yellow piece of paper, in front for green, on the right for red, and behind for blue.
After that, the first inmate calculates modulus just like Rainman suggested and changes his position to be on the left, front, right, or rear of the the second condemned man for 0, 1, 2, and 3 respectively. Thereafter each man knows his color. Those with the same color as the first convict join him to form a row. The other criminals form their rows as appropriate in relation to that row. When all settled, the second contestant can change his position.

As for the blindfold, it is kept on until each captive enters the hall to prevent them from seeing the color when it is placed upon their heads and thereafter to hinder them from checking the same in the mirror of the dressing room.

 

 

[TimeSpaceLightForce]

Spoiler for

Color Rows
Y _ _ _ _ _ _ _
G _ _ _ _ _ _ _
R _ _ _ _ _ _ _
B _ _ _ _ _ _ _

position  
   2
   1
   3


door
4
5
6
7
8
9
10
11
12
13
14
.
.
L

Plan: using 4-way path
Y   G     R    B
|o   o|    o/    \o   
|o   o|    /o    o\

set-up stage:
1 enter and face the door
2 enter, walk behind 1 and face the door
3 enter, pass 1& 2 ( 1 of 4 ways) according to color of 1,
  then stands behind 2 and face the door ..so 1 knew his own color
1 pass  2&3 (1 of 4 ways) according to color of 2,
 then  stands behind 3 and face the door.. so 2 knew his own color
2 pass 3&1 (1 of 4 ways) according to color of 3,
 then stands behind 1 and face the door ..so 3 knew his own color


sitting down stage:
4 enter , time for 3 to sit down in color row ,
pass 1 & 2 (1 of 4 ways) according to color of  4
Now 4 knew his own color, stand in front of 1 facing door and waits for 5..

5 enter , time for 4 to sit down in color row ,
pass 1 & 2 (1 of 4 ways)  according to color of  5
Now 5 knew his own color, stand in front of 1 facing door and waits for 6..

..goes on until "Last" prisoner has nothing to wait on, but also now knew his color,
just sit down.   Finally,  1 & 2 sit down in color row .

[wolfgang]

 

Spoiler for solution

They assign values to each color: yellow = 0, green = 1, red = 2, blue = 3. The first prisoner adds the values of all the other prisoners' hats (modulo 4), then finds the corresponding color to that number and stands in that row. By doing this, everyone else will know the color of their own hat. Now the second prisoner does the same thing as the first one did. This tells the first prisoner the color of his/her hat. Then the first two prisoners change their minds if necessary, moving to the right row. And everyone else can move to the right row as well.

I think the problem's set up in such a way that you can't use that strategy. The people are blindfolded until they enter the room. So the first person won't see anyone's color, the second person will only see the first person's color, etc.

Just for clarification, are there set positions where each row should form? In other words, should we treat this as if there are four rows of seats and all of the yellows must sit in the first row and all the greens in the second row etc.? Or instead can they form rows wherever they want, so if the first person to enter is yellow he can he stand anywhere in the room as long as all the other colors stand somewhere behind him?

 

They can form raws whenever they want

[wolfgang]

The suggested solution assumes that their are equal number of each color slip and the each person, when he enters can see every other forehead and count the color papers glued there. We need a more innovative solution, I guess.

The colors are not equal

[wolfgang]

Spoiler for

Color Rows
Y _ _ _ _ _ _ _
G _ _ _ _ _ _ _
R _ _ _ _ _ _ _
B _ _ _ _ _ _ _

position  
   2
   1
   3


door
4
5
6
7
8
9
10
11
12
13
14
.
.
L

Plan: using 4-way path
Y   G     R    B
|o   o|    o/    \o   
|o   o|    /o    o\

set-up stage:
1 enter and face the door
2 enter, walk behind 1 and face the door
3 enter, pass 1& 2 ( 1 of 4 ways) according to color of 1,
  then stands behind 2 and face the door ..so 1 knew his own color
1 pass  2&3 (1 of 4 ways) according to color of 2,
 then  stands behind 3 and face the door.. so 2 knew his own color
2 pass 3&1 (1 of 4 ways) according to color of 3,
 then stands behind 1 and face the door ..so 3 knew his own color


sitting down stage:
4 enter , time for 3 to sit down in color row ,
pass 1 & 2 (1 of 4 ways) according to color of  4
Now 4 knew his own color, stand in front of 1 facing door and waits for 5..

5 enter , time for 4 to sit down in color row ,
pass 1 & 2 (1 of 4 ways)  according to color of  5
Now 5 knew his own color, stand in front of 1 facing door and waits for 6..

..goes on until "Last" prisoner has nothing to wait on, but also now knew his color,
just sit down.   Finally,  1 & 2 sit down in color row .

I said in the OP",each one should himslf choose where to stay, once he did,he can not change his place".and with this method each one stays infront  of 1 making a raw,so can not change his place.

[bonanova]

This is a challenge.
A two-color version requires and permits movement after taking one's position.
So I don't think it provides useful guidance for this case

Spoiler for here it is anyway

The first two stand next to each other.
The third and subsequent persons inserts himself between differing colors, if there are any.
This requires movement of previous persons.
if all previous persons have the same colors, he takes an end position.

However, this results in a single line - just with the colors segregated.

In every way, the present puzzle is more demanding.

Great puzzle.
Eager to see the solution.