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Questions With Answers (cf. the "Universe Threads")

Fjones said:
You are overlooking something. If you drop ball 1 from floor 50 and it gets marked, you are screwed. You now have only 1 ball left with 50 floors to rule out.
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It's certainly no elegant solution, but I'mma find that goddamn radiofloor if it's the end of me.

Use ball 1 from floor 1, use ball 1 from floor 2, etc, but since I have two balls I can rule out 1/2 the floors or more until my binary search yields a positive.

There's probably a much better way to do it, so my only other solution is to drop both balls at the same time such that ball 1 lands on top of ball 2, smashing it:p
 
RedLeader said:
Consider two boys, "John" and "Michael" and their teacher. Each of the two boys writes down a positive integer on a piece of paper and hands it to the teacher, without being able to see the other's number. The teacher then writes down two distinct positive integers on the blackboard and states that one of the two numbers is the sum of the boys' integers. The teacher then turns to John and says "John, can you for sure tell me which one is the sum?" If John, says "no," then Michael is asked the same question. In turn, if Michael says "no," then the question is repeated for John.

Prove that if the boys are suitably intelligent, eventually one will answer with the correct sum (without guessing).
Click to expand...

I suspect that providing a solution to this is largely dependant on ones expository writing skills. But I do sorta have an algorithmicish way, given John and Michael's numbers and the numbers on the board, to determine which boy will arrive at the answer first.
 
^ It has nothing to do with writing skills. Present your algorithm way. The problem is solved completely and without confusion using algebra and a bit of number theory.

The solution to John and Michael (when anyone's ready):

NSFW:


Ahh, typos in soltution:

  • Any subscript i should be a subscript k.
  • the final k in the problem (seocond-to-last line) should be a k'


johnmichael.jpg
 
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RedLeader said:
Present your algorithm way
Click to expand...

Certainly, my good man.

It is by no means formal, but it gives you a ***good idea*** about which round the game will end.

Example: Suppose John's number is 5, Michael's number is 7, and the blackboard numbers are 12 and 15.

We construct a tree like this: The top two nodes are what John thinks Michael's number could be. In this case, 7 or 10. And beneath each node, we write the two numbers that a person with the node's number would think the other person would have. So, below the seven, we put 5 and 8, and below the 10, we put 2 and 5. And we do this until we come to a row that has numbers less than one, or greater than or equal to 12 (the smaller blackboard number). And, what ever row that happens to be (were calling 7, 10 the first row), the game should end within 1 round of that number. In this case, we get 13 in the 3rd row, so we expect the game to end between the 2nd and 4th rounds.

***by "good idea", I tend to say accurate within one round.
 
Okay that's kind of getting on the right track. A first hint to proving it technically would be to think about the difference between the two sums, say d= S_2 - S_1 for S_2 > S_1. You want to prove this by contradiction, so assume an infinate sequence of "no"s. Now Upon hearing John say "no," Michael learns that 0 < John's # < S_1, for if it were greater than S_1 John would have for sure claimedS_2 as the sum. Now what does John, in turn, learn upon hearing Michael's first no? First, he'll know that Michael's # is also less than S_1 by the same logic as above. But can he establish a stronger inequality involving d? (yes) Then you can get a nice finite procedure, which hones in on a contradiction.

Think about it a bit more, but the solution's posted above!
 
^Yup, I read your solution. This is the kind of problem that, if I had seen it on something like the Putnam, would have tripped me up really badly in the "write up." In other words, it's a problem where I would have instantly known the conclusion was true, but I would have struggled to provide a clear proof.


BTW*, "write ups" used to be an interest of mine. I read a lot of books by guys like Halmos, and always thought of mathematical exposition as an art in and of itself.

BTW**, I notice you post a lot of logic puzzles. Is logic a specialty of yours?
 
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That problem, well actually a generalized verson of it involving n kids/integers, was on some European equivalent of the Putnam -- that's where I got it from. I guess since I graduated, I've gotten a lot more into logic (moreso like proper logic - symbolic, predicate, modal, etc. as you said you have studied than these "logical puzzles" though) for recreational reading. I didn't really do any logic at all in college or grad school other than what was implicit - my research was in stochastic calculus mostly. So I've done the analysis and measure theory and some advanced calculus, but I'm kind of too burnt out to get any actual fun out of grinding out such problems now that I am out of school. The puzzler ones, or the ones with more of a Putman flaire, kind of keep the fun in it, IMO.
 
Here is a puzzler that may or may not be a Binge Artist original.

You're on some kind of fucked up prison camp, in some kind of fucked up country. And you're about to be executed. But the warden is an armchair logician, and decides to set you free if you can solve this puzzle.

On your prison uniform is a number. It means either one, two, three, four, five, six, seven, or eight. But you don't know which one it is, because you don't speak the native tongue.

Anyway, here's the game. The warden will pick three gaurds at random. Now, each gaurd either always tells the truth or always lies. But you don't know which gaurds are honest or dishonest. You are allowed to ask each of the three a single yes or no question.

Your job is to figure out your uniform number. What questions should you ask? (they all understand English).
 
Here's another liar/truth-teller problem. Be warned, though...it's not easy (not a trick either...just technical).

Question: You find three people in front of you. One of them always tells the truth. One of them always lies. And the third person aswers haphazardly, essentially guessing "yes" or "no." You get three questions - one per person. Determine who is who.
 
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^sol'n

NSFW:
Call the guys Larry, Curly, and Moe.

Ask this question to Larry:

Is at least one of the following statements true?
1. You are the truth-teller and Moe is the guesser.
2. You are the liar and Moe is not the guesser.

THEOREM: A "yes" answer will imply Curly is not the guesser, and a "no" answer will imply Moe is not the guesser.

PROOF: Clearly this is true if Larry is the guesser. Now suppose Larry is the truth-teller. A "yes" answer means either 1 or 2 is true, but obviously 2 is false. Hence 1 is true, hence Moe is the guesser, hence Curly is not the guesser. A "no" answer means both 1 and 2 are false, hence 1 is false, hence Moe is not the guesser. Now suppose Larry is the liar. A "yes" answer means both 1 & 2 are false. Hence, 2 is false. Hence, Moe is the guesser, hence curly is not the guesser. A "no" answer means either 1 or 2 is true. Since 1 is false, 2 is true. Hence, Moe is not the guesser.

After getting the answer to the first question, we know who is NOT the guesser. From this point on, it's smooth sailing. We can find out if the non-guesser is the liar or truth teller by asking something like "Is the sky blue?"

We then ask the final question to the same guy, "Is Larry the guesser?", and from there, it's easy to see we can now deduce who is who.
 
^ I think that's good. I may have to compare it a bit to the solution I know (which is different), but I think you got it.

Seems like we're going to have to kick this up a notch!


Question: Four cars -- A, B, C and D are traveling on the same road at constant speeds. A passes B at 8:00am, A passes C at 9:00am, and meets D at 10:00am. D passed B at 12:00am and passed C at 2:00PM. What time did B pass C?
 
RedLeader said:
Question: Four cars -- A, B, C and D are traveling on the same road at constant speeds. A passes B at 8:00am, A passes C at 9:00am, and meets D at 10:00am. D passed B at 12:00am and passed C at 2:00PM. What time did B pass C?
Click to expand...

Answer

NSFW:
10:40

Based on the times presented for the intersections, the most salient interpretation is that A, B, & C are all traveling one direction, and D is going the opposite. If they are all going the same direction, then evidently, the times posted would all be on different days, which would make the problem...nightmarishly hard.

Let a and c be the speeds of A and C. On the number line, @8:00, we say that A&B start at point 0, and C starts at point a-c. Then, @9:00, A meets C at point a. Then, @ 10:00, A meets D at point 2a, and C is at point a + c. Since C & D meet 4 hours after 10:00, we know they meet at point a + 5c. So, in that time, D has traveled a distance of a-5c. So his hourly rate is (a-5c)/4. So, 2 hrs after 10:00, D meets B at point 2a - (a-5c)/2. So, in 4 hrs, B goes from 0 to 2a-(a-5c)/2. So his hourly rate is a/2 - (a-5c)/8. Since he's chasing C from an initial distance of a-c, he catches C at [a-c] / [a/2 - (a-5c)/8 - c] hours after 8:00. And that expression simplifies to 8/3. Thus, B meets C at 2 hrs 40 min after 8:00, or 10:40. #
 
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And now for a new puzzler.

Two players play a game involving Tetris pieces and a chessboard.

They take turns putting a tetris piece on the chessboard (each of the four squares of each tetris piece must cover exactly one square of the chessboard). No piece is allowed to hang off the board. The game is over when a player no longer has any room to place a piece (overlapping of pieces is not allowed). The player unable to place a piece loses.

Assuming both players are super mega geniuses, which one wins?

BTW, this thread has a pretty shitty title. Needs to be something more along the lines of "Mathematical Puzzlers"
 
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RedLeader, I think I may have figured out your analysis problem.

NSFW:
Notation: Let I denote integration (and we assume the product "dx" without writing it, for convenience).

Define F(x) = I[f(x)] over [0, x].

Integrating by parts, we have xF(x) = I{xf(x) + F(x)} over [0,x].

Taking x = 1, we get F(1) = I[xf(x)] over [0,1] + I[F(x)] over [0,1].

Since F(1) = 0 and I[xf(x)] over [0,1] = 1, we have I[F(x)] over [0,1] = -1.

Hence, I[|F(x)|] over [0,1] > (or =) 1.

Since we are doing a proof by contradiction, we assume that |f(x)| = |F'(x)| < 4 for all x in [0,1]. So, by the mean value theorem (or whatever theorem it is), we must assume that both |F(x)|/x < 4 and |F(x)|/(1-x) < 4 for all x in (0,1). Equivalently, |F(x)| < 4x if x is in (0, 1/2] and |F(x)| < 4-4x if x is in [1/2, 1).

[That is, suppose there were some point (x, F(x)) such as (.5, 2.0001). Then there would be some point y, such that F'(y) = 2.0001/.5 > 4, I THINK that's the mean value theorem, but really, it's been too long. Visually, we're saying that the positive (non zero) values of |F(x)| must be in the interior of the triangle (0,0), (1/2, 2), (1, 0), because for any point (x, F(x)) on the exterior of the triangle, we would have either |(F(x) - 0)/(x-0)| > 4 or |(F(x)-0)/(x-1)| > 4, which would mean that for some point y in (0,x) or (x,1), respectively, f(y) = F'(y) > 4.]

Thus, I[|F(x)|] over [0,1] < I[4x] over [0,1/2] + I[4 - 4x] over [1/2, 1] = 1.

Contradiction (via the red statements)!
 
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Oh, I completely missed the Tetris problem. I'll have to get on that!

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Question: Seventeen painters live in 17 houses, each of which is either white or black. The houses are built around one circular road (entrance irrelevant). Each month, a single (novel) painter, starting from his own house, goes counterclockwise along the road and repaints houses according to the following rule:

1) Determine if the house is white or black.
1.a) If white, paint it black, move on to next house, and begin again at 1).
1.b) If black, paint it white, stop, and return home.

Prove that if at least one of the houses is originally white, then in 17 months time each house will have its initial color (note that in 17 months, each painter goes once).
 
Tetris Solution (re: #56):

NSFW:

The first player will win. By the following strategy:

Player 1 places the box piece in the center four squares of the board.
Player 2 makes a placement.
Player 1 responds by placing the same piece on the board in a radially-symmetric fasion (For example, if 2 placed another square piece at the extreme edge of Q1, then 1 will place a square piece at the extreme edge of Q3.).

Player 1 effectively will "drive" Player 2 off of the board, as eventually 2 will run out of space.
 
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