December 8, 2010 § Leave a comment
Anyone remember this puzzle from school? For me, it was from middle school. It is one of those puzzles that you just have to work out step by step, each step often requiring a moment of great insight. The lore behind this puzzle is that Einstein was stumped by this test, and could not solve it. Supposedly only 2% of the world population can get it, or something like that. Seems kinda bogus to me, but it is pretty hard.
There are 5 houses in 5 different colors. In each house lives a person with a different nationality. The 5 owners drink a certain type of beverage, smoke a certain brand of cigar, and keep a certain pet. No owners have the same pet, smoke the same brand of cigar, or drink the same beverage.
Somebody owns a fish. The question is: who?
The Brit lives in the red house.
The Swede keeps dogs as pets.
The Dane drinks tea.
The green house is on the left and next to the white house.
The green homeowner drinks coffee.
The person who smokes Pall Mall rears birds.
The owner of the yellow house smokes Dunhill.
The man living in the center house drinks milk.
The Norwegian lives in the first house.
The man who smokes Blends lives next to the one who keeps cats.
The man who keeps the horse lives next to the man who smokes Dunhill.
The owner who smokes Bluemaster drinks beer.
The German smokes Prince.
The Norwegian lives next to the blue house.
The man who smokes Blends has a neighbor who drinks water.
It helps to make a chart of the possibilities. Actually, it can be thought of kinda similar to sudoku for the solving strategy.
The answer is the German who lives in the green house number 4, drinks coffee and smokes Prince, owns the Fish! We have solved the Einstein riddle.
December 1, 2010 § 1 Comment
|The warden of the Arizona State Penitentiary decides to have some fun one day… he decides he will give three prisoners a chance to go free… but only if they can prove themselves worthy of being released!
The warden brings out five hats, three of them red and two of them blue. The three prisoners are then blindfolded and one of the hats, at random, is placed on each of their heads.
He lines up each prisoner in a row, facing a brick wall. The blindfolds are then removed. Prisoner C can see both Prisoners A and B and the color hats that each of them is wearing. Prisoner B can only see Prisoner A and his hat. Prisoner A, being in the front, cannot see any hats. And no prisoner can, of course, see the color of his own hat.
|The problem he puts to the prisoners is simple:“If you can determine the color of the hat on your own head, I will release you from this prison. If you answer incorrectly, I will add another 20 years to your sentence! However, if you do not know and choose not to answer at all, nothing will come of it.”
He first asks Prisoner C, “What is the color of the hat on your head?” Prisoner C thinks for a short while and then replies, “I’m sorry, I do not know.”
He then asks Prisoner B, “What is the color of the hat on your head?” Prisoner B thinks for a short while and then also replies, “I’m sorry, I do not know.”
He finally asks Prisoner A, “What is the color of the hat on your head?” Prisoner A thinks for a short while and then replies, “I know! The color of my hat is….”
What color hat is Prisoner A wearing and how did he know?
If you need a hint, think of it in terms of informational steps. When Prisoner C says he doesn’t know what hat he has on, what do Prisoners A and B know about Prisoner C? When Prisoner B says he doesn’t know what hat he has on, what do Prisoner A about Prisoner B and C now?
Here is a handy chart of the possibilities to look at:
Remember, the answer is written in white letters. Highlight the text with your mouse to see it, or click here to go to the website answer.
The Answer is that Prisoner A had a red hat. See the link for a detailed description of why he knew, but it essentially boiled down to him knowing that since the others didn’t speak up the only possible scenarios were 1-4. In all of those scenarios Prisoner A has a red hat, so he is safe saying he is wearing a red hat. If you got it, congrats!
November 28, 2010 § 1 Comment
Here is a logic puzzle for you to solve. I’ll try to put up a new one each week. Try to work it out yourself and post your answer in the comments section before you check your answer.
You have 2 pieces of string of different, unspecified length, and some matches. Each piece of string takes exactly an hour to burn, but the burn rate is not constant. This means that it could take 59 minutes to burn the first 1/4, and 1 minute for the rest. The strings have different burn rates, and of course you don’t know the rates anyway.
Using only the matches and the strings, measure 45 minutes.
Warning, hint below.
If you are stuck, try measuring half an hour using one piece of string.
Answer below in white letters. To see it drag select the text so it is highlighted.
The answer is to light one end of string A and both ends of string B. Since an entire string takes an hour to burn, lighting both ends would burn all of string B in 30 minutes. At this point you know 30 minutes worth of string A has been burned, so you light the other end of what is left of string A. The remaining 30 minutes is burned up in half the time, and so burning the rest of string A takes 15 minutes. 45 minutes have then passed since you first lit the strings.