Puzzles
A set of selected puzzles compiled by my dear friend and puzzle buddy, Mayank Mohta. This is just a refresher course of more or less cliched yet challenging puzzles. Enjoy!
Shall upload the corrected question statements and a few more interesting puzzles in a short time.
Unexpected Hanging Paradox : Heard about it first on a Numb3rs show. Pretty anti-intuitive but fun when figured out.
A judge tells a condemned prisoner that he will be hanged at noon on one weekday in the following week but that the execution will be a surprise to the prisoner. He will not know the day of the hanging until the executioner knocks on his cell door at noon that day. Having reflected on his sentence, the prisoner draws the conclusion that he will escape from the hanging. His reasoning is in several parts. He begins by concluding that the "surprise hanging" can't be on a Friday, as if he hasn't been hanged by Thursday, there is only one day left - and so it won't be a surprise if he's hanged on a Friday. Since the judge's sentence stipulated that the hanging would be a surprise to him, he concludes it cannot occur on Friday. He then reasons that the surprise hanging cannot be on Thursday either, because Friday has already been eliminated and if he hasn't been hanged by Wednesday night, the hanging must occur on Thursday, making a Thursday hanging not a surprise either. By similar reasoning he concludes that the hanging can also not occur on Wednesday, Tuesday or Monday. Joyfully he retires to his cell confident that the hanging will not occur at all. The next week, the executioner knocks on the prisoner's door at noon on Wednesday — which, despite all the above, will still be an utter surprise to him. Everything the judge said has come true.
ADDITIONAL PUZZLES
Once upon a time, there was a girl named Pandora, who wanted a bright groom so she made up a few logic problems for the wannabe. This is one of them.
Based upon the inscriptions on the boxes (none or just one of them is true), choose one box where the wedding ring is hidden.
Golden box
The ring is in this box.
Silver box
The ring is not in this box.
Lead box
The ring is not in the golden box.
And here is the second test. At least one inscription is true and at least one is false. Which means the ring is in the...
Golden box
The ring is not in the silver box.
Silver box
The ring is not in this box.
Lead box
The ring is in this box.
Two Masters of Logic sit down for a friendly or not so friendly game of Iterated Prisoners' Dilemma. The rules of the game are quite simple:
Each player has 2 cards, marked "COOPERATE" or "DEFECT".
In each round of the game, each player chooses one card and plays it face down, then both cards are revealed.
If both players played "COOPERATE", they are awarded a point each.
If they both played "DEFECT", they get nothing.
If one player played "DEFECT" and the other "COOPERATE", the defector gets 2 points, and the cooperator loses 1 point for being a sucker.
The objective of the game is to amass as many points as possible*, as these will be converted into beer tokens after the game, and paid to the players by the Grand Master who is hosting the game.
It doesn't matter whether the players score more or less than each other, their sole objective is to maximise their own score.
The players do not know each other and may not confer or agree on a combined strategy, but they are both Masters of Logic, so will both play the very best strategy possible for their own gain. Each player knows that the other is also a Master of Logic.
The length of the game is not decided at the beginning, but is announced after the tenth round. On this occasion it happens to be 50 rounds.
How many points will each player get in total?
As a prince in a far-off kingdom, you decide to travel to an adjacent kingdom to find a princess to marry. The king there has three daughters: a pathological liar who cannot possibly tell the truth, a brutally honest girl who never lies, and a whimsical girl who is known to answer just about anything to any question. The girls are all beautiful and kind and would make wonderful brides, but you are certain that you don't want to marry the whimsical girl: you couldn't ever ask her anything and get a meaningful answer. One of the other two would be OK, since even if she lied all the time, at least you would then know the truth.
The three girls line up in front of you and you may ask exactly one yes or no question to one girl to find who you will marry. Can you successfully avoid the random answerer and find a suitable bride?
Three boxes are all labeled incorrectly, and you must get the labels right.
The labels on the boxes read as follows:
[box 1] nails
[box 2] screws
[box 3] nails and screws
To gain the information you need to move the labels to the correct boxes,
you may remove a single item from one of the boxes. You may not look
into the boxes, nor pick them up and shake them, etc.
Can this be done? If so, how? If not, why not?
A 6 inch hole is drilled through a sphere. What is the volume of the remaining portion of the sphere? Calculus/ Logic solution.
Three Masters of Logic wanted to find out who was the wisest amongst them. So they turned to their Grand Master, asking to resolve their dispute. “Easy,” the old sage said. "I will blindfold you and paint either red, or blue dot on each man’s forehead. When I take your blindfolds off, if you see at least one red dot, raise your hand. The one, who guesses the color of the dot on his forehead first, wins." And so it was said, and so it was done. The Grand Master blindfolded the three contestants and painted red dots on every one. When he took their blindfolds off, all three men raised their hands as the rules required, and sat in silence pondering. Finally, one of them said: "I have a red dot on my forehead."
How did he guess?
After losing the “Spot on the Forehead” contest, the two defeated Puzzle Masters complained that the winner had made a slight pause before raising his hand, thus derailing their deductive reasoning train of thought. And so the Grand Master vowed to set up a truly fair test to reveal the best logician amongst them. He showed the three men 5 hats – two white and three black. Then he turned off the lights in the room and put a hat on each Puzzle Master’s head. After that the old sage hid the remaining two hats, but before he could turn the lights on, one of the Masters, as chance would have it, the winner of the previous contest, announced the color of his hat. And he was right once again.
What color was his hat? What could have been his reasoning?
The ruler of the island of Honestants and Swindlecants wishes to establish diplomatic relations with your country (the population of the island are all either Honestants, who speak only the truth, or Swindlecants, who always lie). He sends four envoys: an Admiral, a Bishop, a Consul, and a Duke, whose names are Edwards, Fitzroy, Gerrard, and Harris, though perhaps not in that order. The envoys introduce themselves to you:
Envoy 1: "My name is Edwards. Fitzroy is an Admiral."
Envoy 2: "My name is Fitzroy. There are at least two Swindlecants in this delegation."
Envoy 3: "My name is Gerrard. If Harris is a Bishop, then Edwards is a Swindlecant."
Envoy 4: "I am the Consul. My name is not Harris."
Well, that's enough chit-chat. You have a job to do, which is to introduce the envoys one at a time to your king, using their correct names and official titles. Any mistakes or omissions would be politically embarrassing, so naturally you would be beheaded for your incompetence. Ah! Here comes the king now...
A river has six islands in the middle, connected by 13 bridges as shown:
Enemy tanks are approaching on the other side of the river, so you've set up explosives to blow up the bridges. Unfortunately your detonators came from a dodgy batch, so each bridge has only a 50% chance of blowing up (independently of the others). What is the probability that the enemy tanks will still be able to find a route across the river?
One of twelve tennis balls is a bit lighter or heavier (you do not know which) than the others. How would you identify this odd ball if you could use an old two-pan balance scale only 3 times?
You can only balance one set of balls against another, so no reference weights and no weight measurements.
In a collection of 2n+1 persons, for any group of n persons there exists a person, not in the group, who knows each person of the group. Prove that there exists a person, who knows everybody.
I'm thinking of two integer numbers, each if them is more than 1 and their sum is less than 100.
I tell my friend Serge the sum of these two numbers, and another friend, Polly, product of these two numbers.
Then such a dialog took place:
P: I can't determine what are these numbers.
S: Ah, i knew you wouldn't be able to do this.
P: Oh, then i know what they are!
S: Oh, then i know them too!
Can you determine the numbers?
More Puzzles :) and some more