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[Math Puzzle] Day4 - Page 2

Blogs > evanthebouncy!
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Nytefish
Profile Blog Joined December 2007
United Kingdom4282 Posts
April 30 2009 13:01 GMT
#21
On April 30 2009 20:35 freelander wrote:
Show nested quote +
On April 30 2009 20:23 raiame wrote:
Why am I still up...
+ Show Spoiler +
n colors, assign each color a different number from 1 to n. also, assign each person a number from 1 to n. have each person add up the values of the colors of each hat he sees, and we'll call this number s(k), where k is the number of the prisoner. Have prisoner k say the color corresponding to k - s(k) (mod n). This works because for each person, s(k) plus the color of their own hat is the same total sum, so subtracting s(k) from this total sum gives your own hat. You make sure that each possible total sum (mod n) is accounted for by assigning each prisoner a different number to subtract from.


but they can't communicate at all


I think when he said "say" he meant "write".
No I'm never serious.
raiame
Profile Joined December 2007
United States421 Posts
April 30 2009 14:46 GMT
#22
On April 30 2009 21:13 stenole wrote:
Show nested quote +
On April 30 2009 20:23 raiame wrote:
Why am I still up...
+ Show Spoiler +
n colors, assign each color a different number from 1 to n. also, assign each person a number from 1 to n. have each person add up the values of the colors of each hat he sees, and we'll call this number s(k), where k is the number of the prisoner. Have prisoner k say the color corresponding to k - s(k) (mod n). This works because for each person, s(k) plus the color of their own hat is the same total sum, so subtracting s(k) from this total sum gives your own hat. You make sure that each possible total sum (mod n) is accounted for by assigning each prisoner a different number to subtract from.


I bet the warden didn't see that one coming. Sounds like you've been in this situation before prisoner raiame.

I have! Except we didn't have the wonderful hats last time.
Also, the last time I was the one who correctly guessed my own. I hope to do so again.
Bebop Berserker
Profile Joined April 2009
United States246 Posts
Last Edited: 2009-04-30 17:13:51
April 30 2009 17:11 GMT
#23
+ Show Spoiler +

On April 30 2009 20:23 raiame wrote:
Why am I still up...
+ Show Spoiler +
n colors, assign each color a different number from 1 to n. also, assign each person a number from 1 to n. have each person add up the values of the colors of each hat he sees, and we'll call this number s(k), where k is the number of the prisoner. Have prisoner k say the color corresponding to k - s(k) (mod n). This works because for each person, s(k) plus the color of their own hat is the same total sum, so subtracting s(k) from this total sum gives your own hat. You make sure that each possible total sum (mod n) is accounted for by assigning each prisoner a different number to subtract from.



+ Show Spoiler +

Maybe I don't understand but that's isn't right at all. Example: there are three hats red blue and white(1 2 and 3 respectively.) There are three prisoners(1 2 and 3.) 1 is wearing white 2 is wearing white and 3 is wearing blue. Now your theory states I should take 3 -5 % n. so, prisoner 1 is blue prisoner 2 is red and prisoner 3 is white. Therefore they all die.




+ Show Spoiler +

Rhaegar99 is who i agree with. Everyone chooses a different color beforehand and everyone memorizes everyone's color. Then they pick an interval(10 or 15 seconds is more realistic) Then we warden says go they wait 10 seconds everyone who doesn't see their color writes down their color. Everyone who does see their color waits 10x seconds were x is the the number of hats their color. Now out of the people who are left they have to watch to make sure everyone remaining doesn't write down at the same time(i.e. the warden doesn't put one of every color hat on the persons head who doesn't have that color.(there are more examples.) If he sees everyone writing down at the same time he can see tell what color hat he has on and he writes that down.(for example there are 4 prisoners and two have blue and two have red. the blue and red would wait thirty seconds and then write down both their colors. since the other two wrote down their colrs at ten seconds they know their are no other colors. they then know there are 2 red and 2 blue. so they write down the color they see only one of.)
Whatever happens, happens.
Nytefish
Profile Blog Joined December 2007
United Kingdom4282 Posts
Last Edited: 2009-04-30 17:45:47
April 30 2009 17:42 GMT
#24
Doesn't waiting to see when people start writing count as communicating though?

And to above poster, + Show Spoiler +
It's n - 5 (mod 3) so I think person 2 writes the correct colour.
No I'm never serious.
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