Logic Puzzles
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BeardyTom
Jamie
Mattel
SamVS
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Re: Logic Puzzles
Is there an assumption it's a 5050 split of hats like in the eye problem?
Mattel Red Meeple
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Join date : 20150316
Re: Logic Puzzles
I'm also having trouble with this one. I figure we have to come up with some rule for every wise man to follow to improve their survival rate, probably based on what his closest neighbours are wearing in front or what they said from behind. Something like "If the next three hats are the same color, say that color. If not, say whatever colour was in the majority from what the last three people said.", which I think would swing the odds at least somewhat in their favour in most situations, but not at all where there are no groups of three or more of the same colour together eg. If all the hats were in the order red/blue/red/blue. Also that would never give an exact number of survivors, which I think is what the puzzle wants.
Last edited by Sam on Fri 29 May 2015, 15:41; edited 1 time in total
SamVS Count of Carcassonne
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Re: Logic Puzzles
Jamie wrote:I think we don't know how many hats of each colour there are.
If not then the key seems to be agreeing what rules they follow at the start. Thus giving them a basis to work out what they are wearing.
If they don't have information on the number of hats each each type used then the only information they have is:
1  the shouts from behind
2  the things discussed prior to the event
So they must agree a set of rules that allows them to work out what they are wearing.
Mattel Red Meeple
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Location : Crookes
Re: Logic Puzzles
Mattel wrote:Jamie wrote:I think we don't know how many hats of each colour there are.
If not then the key seems to be agreeing what rules they follow at the start. Thus giving them a basis to work out what they are wearing.
If they don't have information on the number of hats each each type used then the only information they have is:
1  the shouts from behind
2  the things discussed prior to the event
So they must agree a set of rules that allows them to work out what they are wearing.
Edited by way of a new post :
Actually they have info number 3 also
3  The number and colour of the hats in front of them
So we need to make a set of rules that take what shouts they hear AND what hats they can see in front of them to work out their own colour.
R  R  B  R  B  B
Person 1 can see 2R3B hats but doesn't know his own
Person 2 can see 1R3B but will hear what person 1 says but not know if he is right
Person 3 can see 1R2B but will hear what person 1, and 2 says but will not know if either are right
So given that you don't know if the people behind you are correct (which doesn't help as we don't know total number of each colour) then how do you setup some rules to give info?
Can wen use the change of view?  2R and 3B goes to 1R and 3B  subtraction would tell Person 2 they have a Red hat but ONLY if they knew that the previous person saw 2R hats.
Bah, that doesn't work if they can only say Red or Blue.
Mattel Red Meeple
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Location : Crookes
Re: Logic Puzzles
I think this works. Does this work?
 Solution?:
 99 of the 100 can be guaranteed to survive. The unlucky 1st wise man counts the numbers of red and blue hats in front of him. One of those numbers will be odd and the other will be even and he says the colour which has an even number. He has a 50/50 chance of surviving but from then on each wise man will know for sure what their hat colour is. If the 2nd guy hears the 1st one say "red", but he sees an odd number of reds in front of him then he knows he must be wearing red. For each subsequent wise man they know that there were an even number of reds from the second guy to the 100th and they also know how many of those reds have been accounted by listening to the correct answers of everyone behind them, so they can apply the same logic  e.g "there were an even number of reds, 5 have gone so there should now be an odd number. I can see an even number so mine must be red."
BeardyTom Dominant Species
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Re: Logic Puzzles
BeardyTom wrote:I think this works. Does this work?
 Solution?:
99 of the 100 can be guaranteed to survive. The unlucky 1st wise man counts the numbers of red and blue hats in front of him. One of those numbers will be odd and the other will be even and he says the colour which has an even number. He has a 50/50 chance of surviving but from then on each wise man will know for sure what their hat colour is. If the 2nd guy hears the 1st one say "red", but he sees an odd number of reds in front of him then he knows he must be wearing red. For each subsequent wise man they know that there were an even number of reds from the second guy to the 100th and they also know how many of those reds have been accounted by listening to the correct answers of everyone behind them, so they can apply the same logic  e.g "there were an even number of reds, 5 have gone so there should now be an odd number. I can see an even number so mine must be red."
I THINK that's it!
The information passed from person 12 is that COLOUR has an even number.
If person 2 adds in what they know they can see one of those colour not accounted for the are COLOUR hat if not they are the other.
I think I'd got stuck on working out rules to pass information that allowed person 1 to live.
Glad Tom is more callous than me :)
Mattel Red Meeple
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Re: Logic Puzzles
Didn't even get close to that one! The nearest I got was 1 signalling to 2 and 3 whether or not they wore the same colour hat, 2 and 3 live, repeat the pattern, 66 guaranteed to live. That was a good puzzle.
SamVS Count of Carcassonne
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Re: Logic Puzzles
The Greek Philosophers
One day three Greek philosophers settled under the shade of an olive tree, opened a bottle of Retsina, and began a lengthy discussion of the Fundamental Ontological Question: Why does anything exist?
After a while, they began to ramble. Then, one by one, they fell asleep.
While the men slept, three owls, one above each philosopher, completed their digestive process, dropped a present on each philosopher's forehead, the flew off with a noisy "hoot."
Perhaps the hoot awakened the philosophers. As soon as they looked at each other, all three began, simultaneously, to laugh. Then, one of them abruptly stopped laughing. Why?
I don't even...
SamVS Count of Carcassonne
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Re: Logic Puzzles
I'm guessing this isn't a logic problem? It's more abstract?
Or maybe, one had realised, through observation and logical deduction, that he too must have bird poo on his head? (assuming the other two are only laughing because of the misfortune of the other philosophers, while not realising that they too, share the misfortune).
Or maybe, one had realised, through observation and logical deduction, that he too must have bird poo on his head? (assuming the other two are only laughing because of the misfortune of the other philosophers, while not realising that they too, share the misfortune).
Re: Logic Puzzles
I would have thought so too, but it is under logic puzzles on the site, not lateral puzzles.
SamVS Count of Carcassonne
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Join date : 20130717
Re: Logic Puzzles
Think I have it...
Call them A,B and C.
A, assumes he has no poo on his head. A puts himsef in the shoes of B. B is laughing at C, however C is laughing back at B. So that would tell B he also has poo on his head, and he would stop laughing. However, he (B) doesn't stop laughing, because he can see that A also has poo on his head (which naturally, C finds amusing). So A twiggs, that he (A) must also have poo on the head.
This is why A stops laughing.
Call them A,B and C.
A, assumes he has no poo on his head. A puts himsef in the shoes of B. B is laughing at C, however C is laughing back at B. So that would tell B he also has poo on his head, and he would stop laughing. However, he (B) doesn't stop laughing, because he can see that A also has poo on his head (which naturally, C finds amusing). So A twiggs, that he (A) must also have poo on the head.
This is why A stops laughing.
Re: Logic Puzzles
I've been looking at the very difficult ones on that site Sam linked to. This one seems tantalisingly doable but I cant quite get there. I have managed to resist looking at the hint or solution... so far.
Peter has 4 cards so there are 48 possibilities for the other card while there are only 24 ways of ordering 4 cards. This means that it matters which card of the 5 is chosen.
The Card Trick
I ask Alex to pick any 5 cards out of a deck with no Jokers.
He can inspect then shuffle the deck before picking any five cards. He picks out 5 cards then hands them to me (Peter can't see any of this). I look at the cards and I pick 1 card out and give it back to Alex. I then arrange the other four cards in a special way, and give those 4 cards all face down, and in a neat pile, to Peter.
Peter looks at the 4 cards i gave him, and says out loud which card Alex is holding (suit and number). How?
The solution uses pure logic, not sleight of hand. All Peter needs to know is the order of the cards and what is on their face, nothing more.
Peter has 4 cards so there are 48 possibilities for the other card while there are only 24 ways of ordering 4 cards. This means that it matters which card of the 5 is chosen.
BeardyTom Dominant Species
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Re: Logic Puzzles
Trying to reduce the problem down to it's basics...
You have 4 cards, arranged in such a way, as to indicate one of the other 48 other cards in the pack?
Did I understand that correctly, or am I missing something?
You have 4 cards, arranged in such a way, as to indicate one of the other 48 other cards in the pack?
Did I understand that correctly, or am I missing something?
Re: Logic Puzzles
Jamie wrote:Trying to reduce the problem down to it's basics...
You have 4 cards, arranged in such a way, as to indicate one of the other 48 other cards in the pack?
Did I understand that correctly, or am I missing something?
Given the initial 5 random cards, you've got to be able to choose one which you are able to indicate using the other 4 in some way. There simply aren't enough ways to order four cards to be able to indicate any other card chosen at random so that initial choice of object card must be constrained in some way.
BeardyTom Dominant Species
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Re: Logic Puzzles
You could use a predefined formula. Peter (and 'I') would need to have knowledge of the formula though.
Re: Logic Puzzles
I see what you mean Tom. 24 ways to arrange 4 cards. 48 possible other cards. Though, there are 5 options for the other card....
Re: Logic Puzzles
Convert card values and suits to give a range 152.
Pick 2 of 5 cards that are as close to each other in the range described above, as possible.
There are 20 combinations for this first pick. So the maximum distance of range is approx 52/20 (about 3 or 4), lets just call it 6. *
The lower range value of these 2 cards, I give to Alex.
The other is my first of the 4 cards, that I will later show Peter. The remaining 3 cards, form a value 1 to 6 (there are 6 combinations of 3 cards), which is to be added to the range value (152) of the first card, to reveal the card that Alex has.
* I have a sneaky feeling I may have miscalculated or overlooked something on this part...
Pick 2 of 5 cards that are as close to each other in the range described above, as possible.
There are 20 combinations for this first pick. So the maximum distance of range is approx 52/20 (about 3 or 4), lets just call it 6. *
The lower range value of these 2 cards, I give to Alex.
The other is my first of the 4 cards, that I will later show Peter. The remaining 3 cards, form a value 1 to 6 (there are 6 combinations of 3 cards), which is to be added to the range value (152) of the first card, to reveal the card that Alex has.
* I have a sneaky feeling I may have miscalculated or overlooked something on this part...
Re: Logic Puzzles
Yep, it is only 10 combinations of 2 cards from 5, giving a max range of 52/10, call it 6 still? Also, the range 152 can wrap around. Also, it can be 1 to 6 missing cards in the range between the 2 chosen cards, in step 1 of my solution.
Re: Logic Puzzles
Now think my solution does not work. You can get 5 cards that are spaced too far apart, for example...
1
9
17
25
33
You could not pick 2 of those cards, where there is only a gap of 6 or less between them.
1
9
17
25
33
You could not pick 2 of those cards, where there is only a gap of 6 or less between them.
Re: Logic Puzzles
You could do something along those lnes though, but use an offset of 24 given by all 4 cards...
Re: Logic Puzzles
Jamie wrote:You could do something along those lnes though, but use an offset of 24 given by all 4 cards...
But then how would you indicate which card you're offsetting from?
BeardyTom Dominant Species
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Re: Logic Puzzles
I have officially retired from logic puzzles after my attempts to explain the eye colour problem seemingly confused everyone and gave me high blood pressure.
However...
Since "Alex" can choose which card of the five is the hidden card he can express the suit by making the top card he hands me the same suit as the one I need to guess (Four suits, five cards, there will always be a duplicate).
That leaves three cards with which to express the value. The cards can be ranked from low to high by their value and a preagreed hierarchy of the suits). This means there are six combinations available. So the trick becomes how do we express a number between 1 and 13 with only 6 different values.
If we think of the range of numbers as a continuous cycle: (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A, 2, 3, 4...), since there are 13 distinct values, every value is within 6 of another by either counting forwards or backwards. For example, counting from 2 forwards to 10 is 8, counting from 2 backwards to 10 is 5. So now the only problem is how do we express whether we should count forwards or backwards. The simple answer is, we don't.
Remember, Alex can choose which card to hide and which to use to represent the suit, he can therefore ensure we only have to count forwards. Using our example of 2 and 10, counting from 2 to 10 is 8, however, counting from 10 forwards to 2 is just 5 (J, Q, K, A, 2). So by returning the 10 on top of the cards, and using the remaining three cards to represent the number 5, he can tell us exactly which card he has hidden from us.
Right, back into retirement.
However...
Since "Alex" can choose which card of the five is the hidden card he can express the suit by making the top card he hands me the same suit as the one I need to guess (Four suits, five cards, there will always be a duplicate).
That leaves three cards with which to express the value. The cards can be ranked from low to high by their value and a preagreed hierarchy of the suits). This means there are six combinations available. So the trick becomes how do we express a number between 1 and 13 with only 6 different values.
If we think of the range of numbers as a continuous cycle: (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A, 2, 3, 4...), since there are 13 distinct values, every value is within 6 of another by either counting forwards or backwards. For example, counting from 2 forwards to 10 is 8, counting from 2 backwards to 10 is 5. So now the only problem is how do we express whether we should count forwards or backwards. The simple answer is, we don't.
Remember, Alex can choose which card to hide and which to use to represent the suit, he can therefore ensure we only have to count forwards. Using our example of 2 and 10, counting from 2 to 10 is 8, however, counting from 10 forwards to 2 is just 5 (J, Q, K, A, 2). So by returning the 10 on top of the cards, and using the remaining three cards to represent the number 5, he can tell us exactly which card he has hidden from us.
Right, back into retirement.
PaulC Starting player token
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