First of all, Hi! I'm new to RHC and am very excited to have found such a great group of people. I look forward to meeting as many of you as I can in here or on the other side of a board.
The four hats puzzle really needs to be drawn on a piece of paper to understand it, but I will try. I haven't read all the threads in this forum so I hope this has not already been mentioned.
There are four people, a wall, 2 red hats, and two blue hats.
Three people are placed on one side of the wall in a line perpendicular to and facing the wall. They are not allowed to turn around. One person is placed on the other side of the wall. Noone can see through the wall. So it looks like this:
Illustration:
Man1 Man2 Man3 Wall Man4
Man1 can see Man2 and Man3, but not Man4(on the other side of the wall)
Man2 can see Man3, but not Man1 (because he's facing the wall) or Man4
Man3 cannot see anyone.(only the wall)
Man4 cannot see anyone.(only the wall)
You place the red hats on Man1 and Man3. You place the blue hats on Man2 and Man4. Nobody can see their own hat color, only the hats of the people in front of them. Again they are all facing the wall, not allowed to move and not allowed to turn around. So now it looks like this:
Illustration:
Man1(RedHat) Man2(BlueHat) Man3(RedHat) Wall Man4(BlueHat)
You say On you're heads I put a hat. There are two red hats and two blue hats. If you know what color your hat is, speak up? After 15 seconds, one person says I know. He did.
Who was it and how did he know?
I did this with my students once. It was fun.
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Originally posted by IronPawnXmen 3 and 4 have the same information - so they would speak at once & only 1 spoke up.
First of all, Hi! I'm new to RHC and am very excited to have found such a great group of people. I look forward to meeting as many of you as I can in here or on the other side of a board.
The four hats puzzle really needs to be dr ...[text shortened]... did he know?
I did this with my students once. It was fun.
Man 1 has the most information - so would work it out in less than in 15 seconds if he could.
Therefore it must be man 2. 🙂
Originally posted by IronPawnX<Shameless plug> If you want to see some (in fact most) of the puzzles that have been posted on this forum, go to:
First of all, Hi! I'm new to RHC and am very excited to have found such a great group of people. I look forward to meeting as many of you as I can in here or on the other side of a board.
The four hats puzzle really needs to be dr ...[text shortened]... did he know?
I did this with my students once. It was fun.
http://www.srcf.ucam.org/~cdr29/rhppuzzles.html
</plug>
Originally posted by IronPawnXhere is how it breaks down.
Great. But if you think it's Man2, how does he know?
man 1 sees a blue hat on the person standing in front of him and a red one on the head ahead of that one.
he therefore cannot determine what color hat he has.
since man 1 is silent, man 2, who sees the red hat, {i]knows[/i] that his hat must be blue since, if it were red, man 1 would have said he was wearing a blue hat.
problem solved!
a similar puzzle:
there are three men in a room. one is blind and the others are blindfolded. there is a box in the room containing five metal headbands, three gold and two silver. the bands feel identical and weigh the same.
each of the men removes a band from the box and places it on his head. naturally, no man knows what his band contains.
the first blindfolded man removes his blindfold, looks t the other two, and states, "i do not know what metal my band contains." he then replaces his blindfold.
the other blindfolded man takes his eye covering off and states, "i do not know what my band is made of either." he re-covers his eyes.
the blind man speaks up.
"from what i have heard, i know what metal my band contains."
what is it, and how does he know?
Originally posted by BarefootChessPlayerthe second man and the blind man can not both have silver. likewise, the first man and the blind man can not both have silver, but when the second man looked, he knew that he and the blind man could not have silver. thus, if the blind man had silver, then he would have said gold. thus the blind man knows that he has gold.😀
here is how it breaks down.
man 1 sees a blue hat on the person standing in front of him and a red one on the head ahead of that one.
he therefore cannot determine what color hat he has.
since man 1 is silent, man 2, who sees the red hat, {i]knows[/i] that [b]his hat must be blue since, if it were red, man 1 would have said he was w ...[text shortened]... d, i know what metal my band contains."
what is it, and how does he know?[/b]
Somewhere in the realm of fantasy lives a large leprechaun population. One day their king got bored and summoned all leprechauns to his throneroom.
He adresses his loyal subjects;
"In this room are now all leprechauns. Each and every one of you wears a hat. One of our elven friends has cast a spell wich makes it either blue or red. I tell you now that there is at least one red hat. Each day all of you will gather here before the middle hour of the day. Exactly on the middle of the day, exactly those will step forward who know what colour their own hat is. If no-one comes forward, we shall repeat the process the next day.
No-one is allowed to mention anything related to this assignment to others, nor will one take of his hat or by any other means find the colour of the hat on their head, other then looking at others. This under penalty of death."
The leprechauns dispersed again, and the process began.
Supposing there are N hats, the king wears a white hat, the king is always honest, the leprechauns fear death more then anything, and the elves know what they are doing when casting spells; How many days after this speech, will how many leprechauns step forward? Explain yourselves, for the king will not be pleased with mere guesses.
Ton
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Originally posted by TheMaster37Consider if there is one red hat. The leprechaun with the red hat will see all other leprechauns with blue hats, therefore he will know his own hat is red. Therefore on day 1 he will step forward.
Somewhere in the realm of fantasy lives a large leprechaun population. One day their king got bored and summoned all leprechauns to his throneroom.
He adresses his loyal subjects;
"In this room are now all leprechauns. Each and every one of you wears a hat. One of our elven friends has cast a spell wich makes it either blue or red. I tell you now th ...[text shortened]... step forward? Explain yourselves, for the king will not be pleased with mere guesses.
Ton
If there are two red hats then each leprechaun with a red hat will see one other red hat. No leprechauns will step forward on day 1, but on day two each red-hatted leprechaun will know that the 1 red hat case (above) didn't happen - so will know there must be two red hats and on day 2, two leprechauns will step forward.
If there are 3 red hats, it takes until day 3 for each red hatted leprechaun to see that cases 1 and 2 above didn't happen and so the 3 red hatted leprechauns will all step forward on day 3.
Likewise, with N red hats, it takes N-1 days for the red hatted leprechauns to eliminate the possibility of there being N-1 red hats. By day N they will know that there are N red hats and they will step forward.
Originally posted by iamatigerBravo!
Consider if there is one red hat. The leprechaun with the red hat will see all other leprechauns with blue hats, therefore he will know his own hat is red. Therefore on day 1 he will step forward.
If there are two red hats then each leprechaun with a red hat will see one other red hat. No leprechauns will step forward on day 1, but on day two each red-h ...[text shortened]... N-1 red hats. By day N they will know that there are N red hats and they will step forward.
Originally posted by fearlessleaderwell reasoned! hats off to you!
the second man and the blind man can not both have silver. likewise, the first man and the blind man can not both have silver, but when the second man looked, he knew that he and the blind man could not have silver. thus, if the blind man had silver, then he would have said gold. thus the blind man knows that he has gold.😀
i think we have thoroughly covered this concept (including the leprechauns) so we might be able to cap it here.
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Originally posted by BarefootChessPlayerI think I knew the answer to this one once, but I've forgotten it. Maybe I will remember soon!
well reasoned! hats off to you!
i think we have thoroughly covered this concept (including the leprechauns) so we might be able to cap it here.
Assume there are 1000 leprechauns and 50 different hat colours. Just after noon, as the last puzzle finishes, the king sets another puzzle:
"Tomorrow at noon I will sit you in a line so each one can see all the leprechauns in front, but none of the leprechauns behind. I will then get my chief elf to cast a spell which randomises the color of your hat. Starting with the leprechaun at the back, who can see the hats of all the leprechauns except his own. I will ask each leprechaun what his/her hat color is. On pain of death, each leprechaun must say one of the 50 possible hat colours as an answer. Each leprechaun who states the color of the hat on his own head will go free."
The leprechauns can spend the next 24 hours discussing strategy - they decide to cooperate so that as many of them go free as possible.
On noon tomorrow what is the best strategy the leprechauns can follow?
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Originally posted by iamatigerRemembered the answer now - solve away folks!
I think I knew the answer to this one once, but I've forgotten it. Maybe I will remember soon!
Assume there are 1000 leprechauns and 50 different hat colours. Just after noon, as the last puzzle finishes, the king sets another puzzle:
"Tomorrow at noon I will sit you in a line so each one can see all the leprechauns in front, but none of the leprec ...[text shortened]... hem go free as possible.
On noon tomorrow what is the best strategy the leprechauns can follow?
I like this one!
Okay, this might not be the best way, but they can get at least a half of the Leprechauns out if every other one calls out the colour of the hat in front of him/her (are there female Leprechauns?), and the next person simply repeats that colour to go free. I search for something better though...
- Mike