xkcd

redrogue

If you do, however, incorporate the date/time into your algorithm, the plan works better for the short run. The Thurns will eventually break any formulaic code with enough data points. However, if meetings are never held on the same date or at the same time of day, and if they aren't very frequ...

redrogue

Perfect control over finger joints, eh? I don't have perfect control over my last joints, but I thought I could get three states per finger. That failed. To see what I mean, extend your ring finger, then try to put your middle finger all the way down vs. half way down. :) I can get 16,384 with relat...

redrogue

Chess is 'solvable' in the sense that we could map the entire tree and have a game with perfect play. The question is whether or not black or white will win (or if there will be a draw) once we have the technological ability to solve it. Until we solve it, we don't know. However, there is some discu...

redrogue

I was going to say "All of their vowels are in alphabetical order," but then I discovered Arkansas, which isn't on the list, but shares that quality. Possibly Maryland, too, if you're lax on the 'and sometimes y' criteria.

redrogue

If it is a word problem, perhaps it is in Spanish? These are Peruvian second graders, after all. This may be one of those 'pee in the pipe to get the ping pong ball out' type problems that my adult brain can't process. Tonight, on 'Are You Smarter Than A Peruvian Second Grader,' I don't walk away wi...

redrogue

See also: Montmort's problem. A neat party trick/great unfair bet is to shuffle two decks of cards and flip over pairs, one from each deck, stopping if an exact match (rank and suit) is found. Finding an exact match shares the same probability as this puzzle. What are the odds of simply flipping two...

redrogue

actually there are supposed to by hyphens between fish and and and and and chips sorry i couldn't resist. Your post would have made more sense if there had been apostrophes between fish and and and and and and and and and and and and and and and and and and and and and chips, plus one before fish a...

redrogue

Spoiler:

redrogue

Just use negative numbers for the dimensions on the x, y, z box. A 2x2x2 box will totally fit inside an inside-out 3x3x3 box.

redrogue

krucifi, it is solvable with 3 prisoners, but not with 4 (unless I'm very, very mistaken). With 4 prisoners, you'll have one prisoner turn the switch on and never touch the switch again. The next prisoner will turn it off... but the warden can keep sending him back in until he turns it on again. The...

redrogue

krucifi, The warden can defeat this by increasing the number of times each prisoner visits the room in a row. Even more simple, if the prisoner switches the bulb 'off,' keep returning that guy until he turns it on again. Each non-observer will leave the switch in the state they found it. Edit: You c...

redrogue

Think I got it: Don't got it. You must always decrease your count if the light is off (and your count is > 0). Do so by switching the light on. You may only increase your count if the light is on, your count is > 0, and you saw the light was on the last time you entered the room. Do so by s...

redrogue

Indeed I do. :oops: This is a little better, since only one person needs to announce: Everyone starts at one. If the light bulb is on AND their count is 1 or more , flip it off and increment their count by one. If the bulb is off AND their count is 1 or more, decrement their count by 1 (to a min...

redrogue

Unfortunately this won't work with a malicious warden, but here's the general strategy for solving it without a specified counter. Each prisoner has a count that starts at zero. If they enter the room and the light bulb is on, they flip it off and increment their count by one. If they enter and the ...

redrogue

You've just asked a question I don't fully comprehend. If I answer 'yes,' then I concede that the tortoise hasn't run an infinite distance, which is clearly not the case. If I answer 'no,' then I must somehow rationalize how Achilles hasn't outpaced the tortoise. I'd like to answer that yes, there a...

redrogue

Right, but won't there also be an infinite distance between Achilles and the tortoise? Just because I can't name a *finite* distance both haven't covered in an infinite race doesn't mean the tortoise will have caught him.

redrogue

So you maintain that even in the "remove the lowest numbered ball" case, asking for a ball in the jug is trolling? Do I have that right? Yes. I'm probably wrong here, but yes. By trolling, I mean 'asking a question with an indeterminate answer.' My position is that the jug is not empty in...

redrogue

Follow me on this. We add groups of marbles to the jug, 10 at a time, then remove 1 at random. Lets do two groups. Group A is #1-#10, group B is #11-#20. After the first cycle, balls from group A have a 0.9 probability of remaining, of course. During the second cycle, balls from group A have 9/19 ch...

redrogue

Fine, I'm a troll. :) Say you have a balance: You can tell me the blue jug weighs twice as much as the red: lim(x->inf) 2x/x However, you can't tell me how much more the blue one weighs: lim (x->inf) 2x - x Also, with regards to probability: Ball #1 might have a near-zero percent chance of still bei...

redrogue

{sarcasm} We only start with infinite balls, but we put 10 of them into the jar at each of the infinite steps?!!! We'll exhaust the supply early at that rate! {/sarcasm} I actually have no problem with the 'there are infinite balls in the jug' answer for all three, and find the "What is the num...

redrogue

Spoiler:

Edit: That's wrong. Realized A loses money when he loses a bet.

Spoiler:

redrogue

By your logic:

If all men are mortal, and this group of statements is a paradox, then there exists at least one man who is not mortal!

redrogue

Are you familiar of the tale 'The Emperor's New Clothes'? There's a point in the story that's particularly relevant to this discussion: When the boy yells that the emperor is naked and the crowd starts laughing. Before that point, everyone knew the emperor was naked, but the members of the crowd did...

redrogue

I believe it's 2. Assume the grins travel in a loop that goes 1->4->7->8->5->2->1. Treat each row individually. The first row travels from 8 one step to 5. Then two steps to 1... so 1 step then two steps. The second row travels from 1 two steps to 7. Then 3 steps to 2... so 2 steps then three s...

redrogue

I, for one, groaned at your joke. I don't believe Mary needs to be human. For example, my dog could have four puppies. I can point to one of them and explain that Mary's mum has four children, and then finish the joke. Linguistically, we know that the proper word would be 'puppies,' but even my two ...

redrogue

I've been stumped on this one as well. It may have something to do with missions/objects launched. There are a few that have Greek names (Apollo, Gemini, etc.), and Russia generally named theirs after Russian things. Unfortunately, I haven't found any that were named after Greek cities. It isn't spa...

redrogue

...which is why I specified '2D, non-mobius mazes lacking teleporters' in my above solution. :) For your cylinder problem, wouldn't two circles require a larger diameter cylinder? I know that the slight difference in curves between 999 linked underground symbols and 1000 on a cylinder will probably ...

redrogue

This is rotationally symmetric. Sorry. I intended that to refer to a portion of the maze, not the entire maze. Assume other paths and intersections exist. For symmetrical, 2D, non-mobius mazes lacking teleporters: Find the path that travels along the outermost perimeter of the maze, as well as the ...

redrogue

I'd assume infinite time. This would be a great problem for robotics -- contestants will be provided two identical robots, and you must provide a single piece code that will be copied onto each by the event organizers, before they are dropped randomly into a closed maze. You win when the robots meet...

redrogue

Assuming the maze has intersections (and is non-symmetrical): Perform a comprehensive search, noting the location of three-way, four-way, five-way, ... n-way intersections as separate groups. Choose the group with the shortest non-cyclical route that travels through its intersections (one with c...

redrogue

(Taken straight from The Canteburry Puzzles: http://www.gutenberg.org/files/27635/27635-h/27635-h.htm#p32 -- stumbled upon this while looking up Troll Pi discussions... see #28) Bandy-ball, cambuc, or goff (the game so well known to-day by the name of golf), is of great antiquity, and was a special ...

redrogue

Found this puzzle while looking at Richard Wiseman's Blog: http://richardwiseman.wordpress.com/2010/09/24/its-the-friday-puzzle-78/ The idea is four brothers have an orchard, and they want to split it. You must divide this orchard into four parts of the same shape, with the same number of trees in e...

redrogue

It's not a stupid answer. Let's take the N=4 case. Each of the four people will concurrently have this recursive thought pattern: I see three people with blue eyes. I'll assume my eyes are brown, and that three people are having this thought process: I see two people with blue eyes. I'll assume my e...

redrogue

Right. If you switch doors, you'll win it 2/3 of the time. I was trying to illustrate that the odds of the bottom card in a shuffled deck should stay 50/50 even if you reveal more cards. In the Monte Hall problem, if you keep your initial door choice, your odds of winning STAY at 1/3 after one of th...

redrogue

This means that while the bottom one (or any other one) has to be 50/50, there's only 1 red card and 5 blacks. How does this square? Because the deck order stays put once it's been randomized. If you shuffled the remaining cards each time, the odds would change to reflect the number of black and re...

redrogue

<- hangs head in shame. In that case, replace the last bit with "Alright everyone, go into the room take envelope (your number+1) % n. We shuffled the envelopes already, so this will make a random hamiltonian cycle. We'll just forget about the leaking information rule this year. Whoever gets me...

redrogue

From the initial post: Given a set of n > 3 people... Your scenario wouldn't come up. Try it with a base case of 4 people and you'll find you end up with either two loops (Fran<->Jack, Alice<->Zoey) or one loop (Bob->Joe->Jill->Matt->Bob). Besides, secret santa with n = 3 isn't fun. Nobody can pull ...

redrogue

I think this one might work, assuming you can trust your co-workers to follow the rules: Number envelopes 1-N. Blindly pass them out at random so no one knows anyone else's number. Everyone puts their own name on a card and seals it in their envelope. Remember your number. Collect the envelopes ...

redrogue

I don't see a useful way of transmitting information solely using the switches. Since the initial switch order is unknown, my interpretation of the state '01' could be the same as 10 for another prisoner. Furthermore, since the states (say red/green) aren't known prior, my 00 state might be another ...

redrogue

ircmaxell, that's true if we only consider reaching the ending position, but the robot might be avoiding predictable obstacles, shooting targets with a laser (that automatically fires after each movement step), scanning crates, or trying to win a dance competition. I looked up DRoD and came up with ...

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