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Should have given me the second one two years ago when I could actually do it. And as for the first, can't think of one. 4 becomes 9 but -1 isn't a perfect square so I have no idea.
EDIT: Actually as for the knights and knaves. There was a wierd one that had some really wacked out logic that actually works. You are at a fork in a road. You know that there is rest house down one of the forks but you only have enough energy to travel down one. There is a single man at the crossroads but you don't know if he is a knight or a knave. You can ask the man one question before you must choose which fork to travel down.
@ILoveLiza: Doesn't the question you must ask go somewhat like this?
"If a knight was standing here and I asked him what you would say if I asked you whether the left road is the path to the rest house, what would he say?"
Something along those lines I think. It invovles the truth tables for sure(duh).
Substitute knave for knight, and/or right for left, and/or path not to rest house for path ro rest house as you wish, though the logic of how to interpret the answer will change.
EDIT: There's a simpler one: "If I asked you whether the left road led to the rest house, what would you have said?"
Destiny Warrior; there are many variants of puzzles along those lines. The idea is that you ask a question about what answer they would give, which means that instead of the answer revealing anything about the speaker, it reveals things about the truth of the matter instead.
For example: "If I were to ask you if the left road is the correct one, would you answer yes?"
If the man is a knight, and the left road is correct, then he would answer yes, and so he will answer yes this time.
If the man is a knave, and the left road is correct, then he would answer no, so he will answer yes this time (lying about what he would say).
If the man is a knight and the left road is wrong, he would say no, and so he will truthfully tell you no this time.
If the man is a knave and the left road is wrong, he would say yes, so he'll lie this time and say no.
Consequently, if the answer to the question you ask is 'yes', then the left road is correct. If the answer to the question you ask is 'no', the left road is wrong. You cannot tell whether or not the speaker is a knight or a knave, though.
Here's a more complicated version of the same, taken from a book of mine.
There are five wandering minstrels; a bear, a drummer, a piper, a juggler, and a fool. Among them, are three normals, who speak both truth and lies interchangably, but one of them is a knight and one of them is a knave. Noone knows which is which. You meet them trying to find a particular statue, and they know the way. This is the dialogue they give you:
Drummer: When asked how to find the statue, I say: "You must take the road to the town of Tabor."
Bear: You say no such thing.
Piper: You must take the road to the city of Mandolin.
Fool: Indeed, you must take the road to Mandolin.
Drummer: At the crossroads, you must go to Castle Gargoylia.
Fool: You must go to the Castle of Arc.
Bear: You must not go to Castle Gargoylia.
Juggler: You must go to the Castle Gargoylia.
Piper: You must head to Tabor or Mandolin.
Drummer: I tell a mixture of truth and lies.
Juggler: That is not true.
Fool: If the bear is the knight, the juggler is the knave.
Bear: That is false.
Drummer: At the castle, you must find the sage.
Piper: The drummer is the knight.
Fool: The piper is the knave.
Juggler: You must find the page boy.
Bear: You must find the cook.
Who is the Knight, and who the Knave? What is the road you must take, what is the castle you must go to, and who is the man you must seek?
Here are some other classes of puzzles that I am fond of. I can present more in the same vein as either if requested.
Alice and the Forest of Forgetfulness:
Alice, when she visited the Forest of Forgetfulness in Wonderland, forgot many things, but not everything. She remembered that the two inhabitants of the forest, The Lion and the Unicorn, tell the truth on some days and lies on others. She remembers the Lion tells the truth on all days of the week except Monday, Tuesday, and Wednesday. The Unicorn lies on Thursdays, Fridays, and Saturdays, but tells the truth on other days.
One day, Alice met the pair.
Lion: Yesterday was one of my lying days.
Unicorn: Yesterday was one of my lying days too.
What day is it?
Another occasion, Alice met only the Lion.
Lion: I lied yesterday.
Lion: I will lie again two days after tomorrow.
What day was it?
On what day of the week will the lion make the statements:
Lion: I lied yesterday.
Lion: I will lie again tomorrow.
On what day of the week will the lion make the statement:
Lion: I lied yesterday and I will lie again tomorrow.
Portia's Caskets
Portia, of the Merchant of Venice, had three caskets made. Inside one of them is her portrait, and any suitor seeking her hand must use the inscriptions on the caskets to work out where it is to win her.
Gold Casket: The portrait is in here.
Silver Casket: The portrait is not in here.
Lead Casket: The portrait is not in the gold casket.
If Portia tells her suitor that at most one of the statements is true, which casket should he pick to be able to marry her?
Inspector Craig's Case Files:
Three people, Arthur, Benjamin, and Carl, are brought in for questioning for a robbery where the thieves escaped in a getaway car. The following facts are known:
1) Noone other than these three were involved in the robbery.
2) Carl never pulls a job without using Arthur as an accomplice.
3) Benjamin doesn't know how to drive.
Is Arthur innocent or guilty?
Another case, Damien, Erica, and Fillip are arrested. The following facts are known:
1) If Damien is guilty and Erica is innocent, Fillip is guilty.
2) Fillip never works alone.
3) Damien never works with Fillip.
4) Noone other than these three are involved, and at least one of them is guilty.
When you start doing more advanced maths, it gets pretty fucking awesome.
Bayesian Statistics is great (Remember the Monty Hall problem? I have a lot of good Bayesian puzzles), but my favourite maths course was in third year: Lagrangian and Hamiltonian Dynamics.
It is MAGIC. You start with two equations, the Lagrange Equation and L = T - V. From this, you can deduce EVERYTHING about the system. Equations of motion, laws like conservation of momentum, etc.
The first is trivial, but the second, while most obviously being monday, could be any lying day provided he didnt speak the day previously and has no intention of speaking the day afterward. Which is the only way I can justify the second to last one.
consider the case where there are only two people and they flip coins instead of rolling dice, can you find an answer and generalize it?
A man dies and finds himself at the gates of heaven. He sees three gods. A messenger of the gods appears before him and introduces him to the three gods. They are TRUE FALSE and RANDOM, Whenever asked a yes/no question TRUE will answer truthfully, FALSE will lie, and RANDOM will pick a random answer. The man is allowed to ask the gods 3 yes/no questions each question directed to one of the gods. After these three questions he must say which god is which. If the man is unable to do so, or if at any point he attempts to ask a question which is not yes/no he will be refused entry into heaven. To make the game more challenging the gods have decided to answer in code. the will say either Ja or Da, one of these means yes and the other means no and the man does not know which. What questions can the man use to ensure entry into heaven?
There are 10 boxes of bouncy balls each box has 1,000,000 balls in it. 9 of the boxes contain balls with a mass of 10g each while the other has balls with a mass of 11g each. What is the minimum number of weighings on a digital scale necessary to find the box that has the heavier balls
there are 20 bags, what is the minimum number of balls needed so that we can place a different number of balls in each bag?
there are 12 weights, they all weigh the same except for one of them which weighs slightly more or slightly less. using a balance scale 3 times find the different weight and deduce whether it weighs more or less.
Find functions f and g from the reals to the reals such that f(g(x)) and g(f(x) are nonconstant linear functions and f(g(x))+g(f(x))=0 for all real x
There are 10 boxes of bouncy balls each box has 1,000,000 balls in it. 9 of the boxes contain balls with a mass of 10g each while the other has balls with a mass of 11g each. What is the minimum number of weighings on a digital scale necessary to find the box that has the heavier balls
190. The first bag is empty, so you have nineteen bags with balls in them. Then you need a total of 1+2+3+...+19, which is the same as 10+ 9 lots of 20, which is 190
there are 12 weights, they all weigh the same except for one of them which weighs slightly more or slightly less. using a balance scale 3 times find the different weight and deduce whether it weighs more or less.
This is the same pattern as the above scale question.
@Hip: The second last Alice one is more an illustration of the syntax for the last one than an actual puzzle in itself.
Most of them are simple, because they're only the introductory problems of each different kind. I can post more complicated ones if you like a particular class of them, though.
Also, the one with the minstrels isn't that trivial.
Here's another interesting one:
I have a bag in which there are 9 red balls and 7 green balls. You close your eyes, reach into the bag, pull out a ball and put it in your pocket without looking. What is the probability the ball in your pocket is red?
Without putting the first ball back, you draw out another ball with your eyes open. This ball is red. What is the probability the ball in your pocket is red?
19. First bag is empty, second bag has 1 ball and is put inside the second bag which has one additional ball etc.
1. You get another box, that you know the weight of, and put in one ball from the first box, two from the second, etc. Then you look at how much heavier than anticipated your result is.
I have a bag in which there are 9 red balls and 7 green balls. You close your eyes, reach into the bag, pull out a ball and put it in your pocket without looking. What is the probability the ball in your pocket is red?
Without putting the first ball back, you draw out another ball with your eyes open. This ball is red. What is the probability the ball in your pocket is red?
Okay then, how about this: "Find a perfect square which remains a perfect square when 5 is added or subtracted". *Give the solution in rational numbers. This problem actually has a nice history to it and is harder than it seems, I wonder if you can get it...
Well, I tried to devise a formula, didnt go as planned.
Im thinking it is 4, because 9 is a perfect square, and according to imaginary numbers, -1 is also. However it does say rational.
It cannot go beyond 4, because higher than that the gap gets bigger between perfect squares.
I really would like to know the answer, and feel free to prove anything wrong
If you know what the sum of all 6 numbers is mod 6 then given the other 5 numbers there is only one possibility for what your number is. each person is assigned a different number mod 6 then all residues are assigned once. each person then guess their number by assuming that the sum of all the numbers is congruent to their residue mod 6, one person has to be right about the sum mod 6 and he will then guess his number right.
Well, I tried to devise a formula, didnt go as planned.
Im thinking it is 4, because 9 is a perfect square, and according to imaginary numbers, -1 is also. However it does say rational.
It cannot go beyond 4, because higher than that the gap gets bigger between perfect squares.
I really would like to know the answer, and feel free to prove anything wrong
The question was originally set by Emperor Frederick II, during a visit to Pisa, as a challenge to the great mathematician Leonardo (later Fibonacci). After hearing of his reputation, the emperor had set up a mathematical tournament in which Leonardo went up against the likes of John of Palermo and Theodore. Not only was Leonardo correct, he managed to find the general solution; as explained in his Book of Squares.
So basically, if we replace 5 with an arbitrary constant, say d, we can solve:
y^2 - d = x^2 and y^2 + d = z^2
This is similar to the familiar equation a^2 + b^2 = c^2 whose area is 1/2 * a * b. If we let y = c/2, then it can be shown that y^2 - ab/2 and y^2 + ab/2 are both perfect squares (you might like to try this). To get d=5, we have to start with the 9-40-41 triangle with area 180 = 5 * 36. Then divide by 6 (6^2=36) to get the triangle with sides 3/2, 20/3 and 41/6, whose area is 5. Now:
If you know what the sum of all 6 numbers is mod 6 then given the other 5 numbers there is only one possibility for what your number is. each person is assigned a different number mod 6 then all residues are assigned once. each person then guess their number by assuming that the sum of all the numbers is congruent to their residue mod 6, one person has to be right about the sum mod 6 and he will then guess his number right.
Whatever the numbers rolled, they will add up to a number that when divided by 6 will give a remainder of 0-5.
Person A assumes this number will be 0. He then guesses the number that his dice would have to be showing for that to be the case. So for example lets say everyone has rolled a 1. A can see 5 1s, he then guesses 1 (in this case he is right).
Person B, then considers what he should guess. He knows that if the remainder is going to be 0, then he has rolled a 1, but also that A has correctly guessed that his own dice will be a 1. So instead he assumes the remainder is 1, and guesses 2.
Person C then knows that if the remainder is going to be 0, then he should guess 1, but in that case a has already won. If the remainder is going to be 1, then he must guess 2. But in that case, b would have correctly guessed 1. So he assumes the remainder will be 2 and guesses 3.
The first three are mathematically based problems. Problem four is more interesting though,,
Problem one
I have a large supply of counters, colours red, blue and white. I place one counter on each square of an 8x8 regular chess board. A particular pattern of coloured counters is called an arrangement. Determine whether there are more arrangements with an odd number of red counters or more arrangements with an even number of red counters. (0 is neither even nor odd)
Problem two
I have 6 separate lengths of string. One end is picked at random and is tied to one of the 11 other ends. This process is then repeated until no end of string is left untied. I then count the number of circles of string at the end. What is the average number of circles?
Problem three
The nonnegative real coefficents a1... a n-1 of the polynomial f(x) = x^n + an-1x^n-1 + ... + a1x+1 are such that all n roots are real. Prove f(3) >= 4^n (for all n>=1)
Problem four
Two players are playing a number subtraction game using two four-digit numbers. The first player calls out a digit between 0 and 9 (these may be repeated), and the second player places it in any free space.
i.e. abcd-efgh, where each is a digit from 0 to 9, digits can be repeated (e.g. a,d and f can all be 3), and the first player calls out the number in order, but the second player allocates where each number is placed in the subtraction.
The first player tries to make the difference as large as possible, while the second tries to make the difference as small as possible. Assuming both players play to win, find the maximum possible difference.
I can post solutions later, although by the looks of how quickly other problems have been solved, it probably won't be necessary.
Also, "why is a raven like a writing desk" > when the "never" answer was submitted by the author, proofreaders corrected the intended "nevar" (raven backwards) to "never", which just confused the situation even more. There was no intended answer but some solutions are quite witty nonetheless.
Assuming optimal play, it shouldn't matter. But for the sake of clarity, yes. Also, I think I should re-emphasize the fact that player 1 can call out the same number in succession, but they're both trying to win.
hint:
The fact that there are 8 digits is irrelavent. Try singular digit numbers if it helps, then apply the logic. Also, consider the digits 0,1,2,3,4,5,6,7,8,9, and their respective subtraction values, then think what player 2's tactic would be in advance. How would player 1 combat this?
If player one is trying to make the difference between the two numbers in the initial equation as great as possible, I would say all 0s until there is a 0 in the leftmost digit of one of the numbers then say all 9s. I am not sure this is what you are asking though. It doesnt fit with your hint. But if you say any number other than 0 or 9 it could be placed in one of the leftmost places, in which case your maximum difference would be less than what is guaranteed with this system.
I can see your logic, but the difference must be positive. That answer is actually correct for a negative difference.
Unfortunately, this is quite ambiguous. Most games need to be played to be understood, maybe just having a quick real practice will help you understand it. It's like trying to explain monopoly to someone who has never played, without giving away too many clues.
I'll give a better overview.
There are 8 spaces, once filled with a digit, that digit remains in that place.
abcd
-efgh
> player 1 calls any digit
> player 2 puts that digit anywhere
> player 1 calls any digit
> player 2 puts that digit anywhere but where the first digit was placed.
> repeat until all 8 digits have been placed.
Using abcd-efgh, the first digit can be put at a,e,b,f, or anywhere.
Player 2 wants the outcome to be as small as possible, but does not control what digits are shouted out. Player 1 wants the largest difference possible.
hint:
There are two ways to obtain the largest possible difference.
hint:
The key principle is for player1 to understand how player2 would optimise their play. From this, we see player1 is really the one in full control, as they decide the numbers. Player 1 uses the limitations of player 2 (player 2 can't select the numbers) to get the largest possible difference
Big hint:
The first number called out is by far the most important, and the one that decides the end result.
Basically the answer:
The 2nd, 3rd, 4th, 5th, 6th, 7th and 8th numbers called will all be the same when player1 goes for the largest possible difference. If you read this i want both solutions and a method :P
