Logic Puzzles

[chaos_defrost]

Weigh one black ball and one grey ball on one side, and the other black ball and one white ball on the other.

If it balances, We have BL = WR = GO, and GL = BR = WO. (L is left balance, R is right, O is off the scale). Weigh any pair to determine which is the heavy trio and which is the light ones.

If it fails to balance, then we KNOW the heavy side has the heavy black ball, and the light side has the light black ball: There's no way for the side with the light black ball to be any "lower" on the scale than balancing. We also know that the other ball on the heavy side is at least as heavy as the other ball on the light side, as if it weren't the scale would balance instead. So we know the weights of the black balls, and that one of the grey/white balls is at least as heavy as one of the other color. I will call this ball that we know is at least as heavy as one other ball "heavyish."

Take the two white/grey balls you weighed and place them on one side, and the two you didn't and place them on the other. If the side with the "heavyish" ball is lighter, then you weighed the two light grey/white balls in weighing one. If it's heavier, then you have the heavy white/grey balls on that side. If they balance, you have one of each, but remember that the "heavyish" ball is on a known side, and MUST be paired with the lighter ball of the other color. No matter what, you know which grey and which white ball is light, and which is heavy.

Good puzzle. I'd not seen it and it took me almost an hour to actually get to the answer.

[chaos_defrost]

And a follow-up to my "guess my number" riddle. I thought of it a while back but I am not actually sure of the answer.

I am thinking of an integer between 1 and n inclusive (assume you know n when you are asking). You can ask me any three yes or no questions this time, but the following things are changed:

1) Instead of answering yes or no, I will instead respond by raising my left or right hand. One means yes and one means no, but you don't know which is which to start.
2) If you ask a question I cannot answer truthfully, instead of remaining silent, I will either answer the yes or no question with either the choice with the larger probability of being true, or that with the smaller probability. I will answer all ambiguous questions consistently also; that is, I will ALWAYS give either the larger or smaller probability answer, but you again do not know which to start. If you ask a question with a 50/50 chance of being yes or no, I'll answer randomly.

What is the largest n for which there exists a definite system to always guess my number correctly? n = 4 is pretty trivial (ask factual question, do binary search), but is it possible for n>4? I get this feeling it isn't, but I mean

[Proxima]

I was all set to give my solution to Simon's puzzle, then I noticed there was a new page of the topic http://www.lemmingsforums.com/Smileys/lemmings/tongue.gif" alt=":P" title="Tongue" class="smiley" /> Here it is anyway. It's very similar to Steve's.

Weigh one black, one grey against one black, one white.

If they balance, the heavy black is on the same pan as a light ball and vice versa, so simply weighing the blacks against each other will give the full result.

If one side goes down, it contains the heavy black. In addition, the grey and white you used are either both heavy or both light, or the one on the side that went down is heavy. These possibilities can be distinguished by weighing the same grey and white together against both blacks.

Insane Steve's number puzzle:

"Is it the case that your number is greater than 4 if and only if raising your right hand means yes?"

Regardless of which hand means yes, you will raise the right hand if the number is greater than 4, the left if it isn't. It's similar to the classic truth-teller / liar puzzle: by asking "If I were to ask you which road leads to the village, would you say yes?" his truth-teller / liar status is made irrelevant by being called on twice, so that if he lies, his lies cancel each other out. Here, calling the detail of which hand means yes twice makes it cancel out, so it becomes equivalent to the situation where the right hand always means yes. Therefore, by binary search we can distinguish the numbers 1 to n where n is at most 8.

[chaos_defrost]

Ha, for some reason I didn't think of it like this, but that certainly looks good, and it shouldn't be hard to show that's as good as you can do, also. Nice job  http://www.lemmingsforums.com/Smileys/lemmings/thumbsup.gif" alt=":thumbsup:" title="Thumbs Up" class="smiley" />

[chaos_defrost]

Also double post for something else I found:

http://puzzle.cisra.com.au/" class="bbc_link" target="_blank">http://puzzle.cisra.com.au/ -- here's a site that runs a puzzle contest every year, and this year's contest looks like it's going to start any day now: though most of the questions look like they are going to be presented the week of the 1st of October.

I'd definitely be up for making a team if there's three other people here interested in these kinds of puzzles. Given that we aren't Australian students we wouldn't be eligible for prizes but it might still be fun.

Also, this contest is now open for registration. So far geoo and I have expressed interest and we can add two more people to the team, so if anyone else wants to do this, let us know.

[Proxima]

I'll join in, if you like. I've had a look through some of the previous puzzles, and they look interesting, but very difficult!

[Simon]

I shall offer myself as the 4th person. I cannot tell how much time I'll have available, and I'm probably also not the brightest at solving without instructions or making words from random things. If someone would like to join, he can certainly do so instead of me.

The puzzles seem to be shower puzzles (you might get an idea while showering just as well as while pondering about them full power), so time shouldn't be an issue.

Steve's and Proxima's solutions to the 6-ball weighing puzzle were correct!

-- Simon

[chaos_defrost]

This contest is entirely for fun, so if you don't have a ton of time it's all good.

We do have four interested now, but we'd need a team name before we can register. It also asks for each person's name and an e-mail for each team member but since we wouldn't be going for prizes I don't see a problem using pseudonyms if you aren't wanting to give your real name.

[chaos_defrost]

New puzzle that I found on a website with an IQ test designed for people on the very extreme end of the IQ spectrum. It's called the "Hoeflin Power Test" if you want to try it. I've been messing around with the problems on it, probably not going to have it scored because it costs $50 lol but there's some very interesting challenges.

May not be the best idea to post an answer publicly since it is from a currently-being-scored test and there's really very few actually decent tests like this, but I have an answer that I'm pretty sure is right (EDIT: On second glance I've made a mistake and I've not got a neat way to do it yet. One moment please).

Suppose you build a tetrahedral pyramid out of spheres, such that the first row has one sphere, the second 3 in a triangle, the third 6, and so forth. Without using a calculator or any sort of program, how many spheres are there in a one million (American million, so one thousand thousand) row high tetrahedral pyramid?

[ccexplore]

Hmm, now I think I see why you made a mistake and have not found a neat way to do it.  At first I too thought this is just a 3D version of what Gauss allegedly did as a child with 1+2+...+n, but the shift to 3D does add some complications from the 2D version, and I haven't gotten the details straighten out myself. http://www.lemmingsforums.com/Smileys/lemmings/XD.gif" alt=":XD:" title="XD" class="smiley" />

I've included the correct answer as calculated by computer program below (which means I too don't have a way to do it yet that satisfies the puzzle's rule of no rote machine computation).  The idea is so that when you think you have the answer, you can check it against the number below to see if you have it right or not.

166667166667000000

[Simon]

Hmm, I had

166,668,666,667,000,000

I sent the method to Steve via PM, the multiplications are broken into hand-computed steps, so maybe he'll find an error later.

Edit: Yeah, found the mutliplication error myself, so cc's number is verified. As I already wrote in the PM: The deem careful bookkeeping a major part of intelligence as they don't allow computers, but require everything computed by hand.

-- Simon

[chaos_defrost]

Just as well, I'm not offhand sure I get the notation you're using. Which I'm fine with, since I want to try and get it without help  http://www.lemmingsforums.com/Smileys/lemmings/laugh.gif" alt=":D" title="Laugh" class="smiley" />

My other favorites, as a probability type person:

An ant is placed on every vertex of each of the following solids. At the same time, and at the same speed, they randomly pick an adjacent vertex with equal probability and try to move there along the edge of the solid connecting them. Calculate the probability that, after the move, all ants end up on different vertices and no ants ever cross paths if they are placed on:

I) A regular tetrahedron
II) A cube
III) A regular octahedron
IV) A regular dodecahedron
V) A regular icosahedron

I've managed I-III (barring math errors) because I'm really bizarrely good at probability maps, but I get the feeling I'm going to need to find some kind of relation to get IV and especially V. Seriously, V is apparently so lol it's not even on the Power test, it was a later addition to another test and is supposed to be one of the most difficult problems.

[Simon]

The notation from the PM was merely binomial coefficients, with (n k) as shorthand for binom(n, k) = n! / ( k! * (n-k)! ). I still used brackets for grouping all over the place, so a space between two numbers was actually important. The usual notation in plaintext seems to be binom(n, k), which I will use in the future for clarity.

-- Simon

[Proxima]

Solution to Insane Steve's tetrahedron puzzle. (By the way, a million is the same everywhere, it's only numbers above that that vary.)

The triangular numbers (I assume I'm allowed to take this as known) are governed by the quadratic formula n^2/2 + n/2. Therefore the tetrahedral numbers, being their sums, must be governed by a cubic formula.

The first five tetrahedral numbers are 1, 4, 10, 20, 35. Take the difference of each pair of terms: 3, 6, 10, 15. (Yes, we knew this already, but it's part of the method.) Do the same again: 3, 4, 5. And again: 1, 1.

If we start with the cubes (1, 8, 27, 64, 125) and take differences three times, we get a sequence of 6s. Therefore the tetrahedral sequence has a cubic term of n^3/6.

Multiply the tetrahedral numbers by 6 and subtract n^3 from each: 5, 16, 33, 56, 85. Take differences of this sequence and we get 11, 17, 23, 29 and then 6, 6, 6. The squares have second differences of 2, so this is a quadratic sequence with quadratic term 3n^2.

Subtract 3n^2 from each term and we get 2, 4, 6, 8, 10 which is obviously 2n.

This gives the full formula for the tetrahedral numbers: (n^3 + 3n^2 + 2n) / 6.

Thus T(1000000) = 10^18 / 6 + 10^12 / 2 + 10^6 / 3 = 16666... + 50000... + 33333 = 16666716666700000.

[Proxima]

An ant is placed on every vertex of each of the following solids. At the same time, and at the same speed, they randomly pick an adjacent vertex with equal probability and try to move there along the edge of the solid connecting them. Calculate the probability that, after the move, all ants end up on different vertices and no ants ever cross paths if they are placed on:

I) A regular tetrahedron
II) A cube
III) A regular octahedron
IV) A regular dodecahedron
V) A regular icosahedron

My answers to I-III:

I. 2/27
II. 8 / 3^7
III. 15 / 2^10