Post/Author/DateTime	Post
zzo38 05-27-07, 12:24 AM	I think that D&D is too easy I want to make it harder by adding complicated math to it. Can you tell me how to do so?
Soren_of_Chains 05-27-07, 12:28 AM	If D&D is too easy, then you need to face things with higher CRs and come up with imaginary ways to slaughter them. That's what makes the game challenging and fun.
Maekrix_Waere 05-27-07, 12:36 AM	If D&D is too easy, then you need to face things with higher CRs and come up with imaginary ways to slaughter them. That's what makes the game challenging and fun. I concur.
Forgotten_Warrior 05-27-07, 12:36 AM	What if the characters needed to see a powerful wizard, but he only sees intelligent people so he requires them to solve afew Calculus problems.
Enigmous 05-27-07, 12:43 AM	Don't play D&D, go to class.
the_Conqueror_Worm 05-27-07, 02:09 AM	Think outside the box! What you need is a good whack in the head with a shovel! One or two of those, and all the math with be complicated. Of course, so will things like fine motor coordination and controlling your bowels but, hey, no solution is perfect.
zzo38 05-27-07, 02:33 AM	What if the characters needed to see a powerful wizard, but he only sees intelligent people so he requires them to solve afew Calculus problems. I am not trying to turn D&D into mathematics school!
Comus 05-27-07, 02:41 AM	I am not trying to turn D&D into mathematics school! So, what, you want to add more math to every ingame equasion? Why? Don't make something complicated unless it absolutely has to be complicated. Simpler is nearly always better. Elegant - simple and useful - is the ideal. Really, if you want more complication, try 2nd edition. Or RIFTS. Or, if you're obsessive-compulsive, FATAL.
Bob Loblaw 05-27-07, 02:50 AM	I think that D&D is too easy I want to make it harder by adding complicated math to it. Can you tell me how to do so? Have you considered playing Hero System with their fantasy rules? That should give your calculator a work out.
zzo38 05-27-07, 02:56 AM	Nearly everyone misses the point (I probably wasn't very clear at first)... What I am interested in is if some encounters are difficult and complicated math is useful for figuring out the best way to do the encounter. (PS: I probably also put this in the wrong section it might be better in D&D General)
Optimized_Commoner 05-27-07, 03:20 AM	Statistics.
OleOneEye 05-27-07, 03:25 AM	I am a bit confused about what you are looking for. Do you want: 1. Encounter like riddles and traps whereby the characters must solve a mathematical formula to succeed, or 2. Changing the rules of the system whereby attack values and the like are calculated using Bayesian probability assessments?
nytemare3701 05-27-07, 03:42 AM	I thought he was speaking about probability equations and statistics. Something along the lines of cat vs commoner, or asmodeus vs demogorgon
Roivas 05-27-07, 04:50 AM	Okay from now on everyone's AC is d/3 + BA - LA x initiative=20 solve for d. Remember that BA is base attack and LA is level adjustment. Also just change all the numbers so that they all are prime factors. Now you have some converting to do. Enjoy!
queenfange 05-27-07, 09:23 AM	I think that D&D is too easy I want to make it harder by adding complicated math to it. Can you tell me how to do so? You're kidding, right? Well, I suppose you could use a bell curve instead of d20 for attacks, skills, etc. since that's more realistic. But seriously, take calculus or something. More complicated math will give most of your players a headache.
Ack 05-27-07, 09:39 AM	If you want more complicated mathematics, check out d20 Lethal (the link is in my sig). Funny, most people see the math in it as the drawback... :P
Cincinnati 05-27-07, 10:24 AM	:uh-huh:
Soluphobe 05-27-07, 11:54 AM	Just go to the TO boards and read some stuff by Tlielaxu_Ghola (sp?).
albinomonkeyking 05-27-07, 12:01 PM	You want encounters more mathematical? Uh, how about a monster/villian who is squared everytime he is struck? I guess normal math for encounters is easy. "4 of us, 10 of them...hmmm"
Comus 05-27-07, 12:37 PM	Maybe he means he wants an encounter calculator, to determine the difficulty of an encounter with multiple creatures of differing CR.
Ack 05-27-07, 12:42 PM	zzo38, I'm a little confused as to what you actually want here. Could you explain what you intend to get out of this?
zzo38 05-27-07, 01:55 PM	You're kidding, right? Well, I suppose you could use a bell curve instead of d20 for attacks, skills, etc. since that's more realistic. But seriously, take calculus or something. More complicated math will give most of your players a headache. I am one of the players. I am just interested in many different things. zzo38, I'm a little confused as to what you actually want here. Could you explain what you intend to get out of this? I'm confused as well. But, someone did once answer "statistics", which is a OK answer. I am a bit confused about what you are looking for. Do you want: 1. Encounter like riddles and traps whereby the characters must solve a mathematical formula to succeed, or 2. Changing the rules of the system whereby attack values and the like are calculated using Bayesian probability assessments? Probably something closer to the first one, but it would be much more interesting if it wasn't obvious that math will help at all. Although I can suggest things posted here to the DM, I'm not sure the DM will be able to figure it out. What can be done is if any good suggestions are posted in here then maybe the DM will look at it and make up their own
Darthos the Librarian 05-27-07, 05:46 PM	I THINK I understand where you're going with this. Are we talking about lifting some plots from episodes of Numb3rs, except without butchering the math so as not to confuse the "other half?"
zzo38 05-27-07, 07:18 PM	I THINK I understand where you're going with this. Are we talking about lifting some plots from episodes of Numb3rs, except without butchering the math so as not to confuse the "other half?" I have never seen the show Numb3rs. I have heard about Numb3rs but that's about it. I know nothing about it other than that. Edit: I did look at it on Wikipedia just now. But still some people in the talk page debate the truth of the article. More edit: I also found a numb3rs blog, it is interesting (half the stuff is stuff I have already seen, but some of it does explain how it is related to the show NUMB3RS).
Candin 05-27-07, 07:51 PM	Give movement rates in terms of acceleration and as a function of x. So a barbarian can move at a rate of e^sinx / ln(root[x-lnx]). Now you can play D&D and use "complicated math". Or play a world more hyberbolic than spherical. Determine their position relative to the equator, and each time they take a move (after determining how far they can move using the above mentioned idea) make them calculate exactly how far they are from their target. Remember, they aren't traveling in a line, they're traveling along a hyperbola, so this needs to be accounted for. I guess you could do this with a sphere too. Or if you want even more math fun, remember that the earth isn't a sphere, it's a geoid. As for plot devices, you're going to find math is pretty shallow in the roleplaying sector. The closest we ever got was a riddle with golems. Each were numbered: 5, 7, 9, 11, 13, and we had one minute to pick which one didn't belong. I figured they had some deformity that we had to identify. Turns out it was 9, because 9 is the only number in the group that was not prime. I guess you could do something similar to the Seven Bridges of Königsberg and Euler. I think Wikipedia has an interesting math problem about competing interests building bridges for certain ends. I've never heard anyone want D&D to be math intensive...
Pangeon 05-27-07, 07:58 PM	Well, here's (http://boards1.wizards.com/showthread.php?t=821842) a start, I suppose.
Optimized_Commoner 05-27-07, 08:06 PM	Each were numbered: 5, 7, 9, 11, 13, and we had one minute to pick which one didn't belong. I figured they had some deformity that we had to identify. Turns out it was 9, because 9 is the only number in the group that was not prime. God those puzzles are stupid. It could equally be 13 because it's the only one with two different numbers in it. Reminds me of Corwin from Amber ranting at the sphynx "No, I gave an answer that satisfied all the criteria of the riddle. It's hardly my fault that you had a different answer in mind."
Tempest Stormwind 05-27-07, 08:18 PM	Just go to the TO boards and read some stuff by Tlielaxu_Ghola (sp?). Tleilaxu Ghola. (Read up on your Dune; that's where the name comes from.) Although it pertains to character creation rather than Numb3rs-style solutions, the Ghola Character project (http://boards1.wizards.com/showthread.php?t=777565) is the most advanced math used in D&D. It sprung out of Pun-Pun, which can do the same sort of approach but far cheesier (the Ghola Character was an attempt to figure out how to do it without using the cheese). As for applications of math through problem solving, that's strictly in the realm of adventure design. Get a mathematician for a DM -- especially one who enjoys discrete mathematics -- and you'll have your chance for using them. For instance, a very simple problem would be scrying on your target and finding out that he's in an underground room (hewn stone) with three doors. You know he teleported directly there, and thanks to a trace, you've gotten it down to a few square miles that he could be in. However, the locals know that there's not one, but two underground labyrinths nearby: one has two entrances, the other has three. The villain likely won't stay there long enough for you to sweep out both labyrinths; figure out which one he's in. That can be solved with very, very simple graph theory (it's a parity problem). It still leaves the actual dungeon crawl to do, but math will tell you which dungeon to tackle. One of my favorites was a puzzle lock with four sliders -- three of them had the Sylvan symbols for the digits of 0-9, and the one on the left of those three had two symbols (for 0 and 1), thus being able to represent any number between 0 and 1999. An inscription above them said that it was a combination lock that only allowed ONE guess -- if you pulled the lever and the number was wrong, it would vanish and the gate it locked would be sealed for another decade. Reading the inscription would trigger an illusory voice effect, informing you that it was put there by some more capricious fey who wanted to give the mortals a staggering chance, but only if they were smart. It would allow you to ask it fifteen yes or no questions relating to the correct combination... but, unlike most puzzles of this sort, it would NOT respond until the 15th question had been asked, and it MIGHT lie on one of its answers (you don't know ahead of time if it's going to lie, or if it does, which one of its answers was wrong). After it answers the fifteen questions, it bids you luck and disappears. In this case, it's simple code theory (in particular, a Hamming code can get the right answer every time). The penalty for getting it wrong is that you don't go through the door. If you get it right, well, the reward's up to the DM. If your DM doesn't like sticking math puzzles in his games, then... well, there's not much we can help with. (Cookie for anyone who figures out the second problem (that is, what 15 questions to ask); the first is trivial.)
Johnny_Angel 05-27-07, 08:49 PM	A man rode into town on Friday. He stayed in town for 3 days. The man left on Friday. :P
Tempest Stormwind 05-27-07, 08:58 PM	A man rode into town on Friday. He stayed in town for 3 days. The man left on Friday. :P Lateral thinking, not math. Randall Munroe (http://www.xkcd.com) put it best:I always hated those ‘lateral thinking’ puzzles — the ones where they said “oh HO, I never told you that the doctor’s mother was a midget and this was all happening a space station! Don’t you feel dumb now?‘ Because once I figured out that language was really, really flexible and imprecise, it seemed that the key to communication was just figuring out what they probably meant. And figuring out what they could possibly mean if you use all the wrong definitions and stuff is interesting, but I don’t think it teaches all that much more than bad communication. These should be kept very separate from actual logic puzzles, which are really neat.The comic accompanying that statement (http://www.xkcd.com/c169.html) sums it up nicely. The closest you'll get to lateral thinking puzzles in math are logic puzzles, not wordplay. Speaking of Mr. Munroe, his own Blue Eyes problem (http://xkcd.com/blue_eyes.html) is a classic example of such a logic problem.
Johnny_Angel 05-27-07, 09:19 PM	I know it's a lateral thinking puzzle, but a lot of people try to solve it with math... hence the 3. You're right though.
OleOneEye 05-27-07, 10:28 PM	Any bookstore will have mathematical puzzle books. Buy one, give it to your DM. Ask hims to place some in the adventures. Odds are, if he enjoys mathematical puzzles he will agree. If he doesn't like them, he won't.
Optimized_Commoner 05-28-07, 12:33 AM	(Cookie for anyone who figures out the second problem (that is, what 15 questions to ask); the first is trivial.) I can't quite get it, I think I may be doing the hamming code wrong (I've never used it before). If I had 2 more questions then I could crack it. So, I'll show my working and see if anyone else can take it that last step or let me know where I've made the mistake. 1. Is the number of yes answers to questions 1, 3, 5, 7, 9, 11, 13 and 15 even? 2. Is the number of yes answers to questions 2, 3, 6, 7, 10, 11, 14 and 15 even? 3. Is the first digit odd? 4. Is the number of yes answers to questions 4, 5, 6, 7, 12, 13, 14 and 15 even? 5 Is the second digit odd? 6 Is the third digit odd? 7 Is the last digit odd? 8. Is the number of yes answers to questions 8, 9, 10, 11, 12, 13, 14 and 15 even? 9 Is the second digit greater than 4? 10 Is the third digit greater than 4? 11 Is the last digit greater than 4? 12 Is the second digit either a 1, 2, 5 or 6? 13 Is the third digit either a 1, 2, 5 or 6? 14 Is the last digit either a 1, 2, 5 or 6? And the last three questions, if I could've squeezed them in would have been whether the respective digits were a 0 or a 9, since should theoretically cover all available options and the parity bits can detect the single lie if there is one. So, what am I doing wrong here? Edit: If my brain were working properly I'd change the wording in those hamming code phrases to make it less ambigous about whether it's their answers or the correct answers. Obviously we want the correct answers but I can't seem to word that properly. Second edit: Just using the questions above and adding in 15 as "is the last digit a 0 or 9", I should be able to guess with a 25% chance of a correct answer which isn't too bad since we started with 1,999 permutations.
Optimized_Commoner 05-28-07, 12:40 AM	A man rode into town on Friday. He stayed in town for 3 days. The man left on Friday. :P Good for him. I've been to towns like that. You leave the same day as you got there and still feel like you've been stuck there for days. Hay is like that. You stop off there on a road-trip to stretch your legs and it feels like you've been stuck there forever by the time you leave.
Tempest Stormwind 05-28-07, 12:55 AM	I can't quite get it, I think I may be doing the hamming code wrong (I've never used it before). If I had 2 more questions then I could crack it. ... So, what am I doing wrong here? You're actually doing pretty well (although I *did* tell you what method to use). To improve your accuracy, note that a Hamming code with N error-detecting/correcting bits can correctly encode any problem with (2^N)-1 total bits (data bits plus error detecting/correcting bits). Note that with 15 questions (bits: Y=1, N=0), for it to work as a Hamming code, N must equal 4 (2^4 -1 = 15). Thus, 4 of those bits have to be error-detecting/correcting, leaving you with 11 bits to encode the actual data itself. Do note as well that with 11 digits, you can represent any natural number up to 2048 in a format that lends itself well to yes/no questions. Last I checked, 1999 was less than 2048. With that, you've got more than enough information to solve the problem; I'm convinced you can handle it without me freely handing you the last key word.
Optimized_Commoner 05-28-07, 01:19 AM	(although I *did* tell you what method to use). I'll freely admit that I couldn't have figured it out without that hint. Then again I did actually have to learn Hamming code first which I think is at least worth something. Note that with 15 questions (bits: Y=1, N=0), for it to work as a Hamming code, N must equal 4 (2^4 -1 = 15). Thus, 4 of those bits have to be error-detecting/correcting, leaving you with 11 bits to encode the actual data itself. I thought that's what I did. Do note as well that with 11 digits, you can represent any natural number up to 2048 in a format that lends itself well to yes/no questions. Last I checked, 1999 was less than 2048. "So binary, my old adversary, we meet again. Your math is weak old man, you should not have come. Yeah... well I did." I understand the nature of the problem and somewhat see the solution, (basically composing the 11 questions in such a way as to mimic the functioning of 11 digits of binary), but I don't think I'm able to do it just at the moment. If no one else has solved it by then, I'll give it another try when I wake up. I really need to get some rest. Thanks for posing a really interesting puzzle. (Oh, I have no idea whatsoever about the blue eyes puzzle)
Tempest Stormwind 05-28-07, 02:20 AM	"So binary, my old adversary, we meet again. Your math is weak old man, you should not have come. Yeah... well I did."Yeah, it does have the uncanny ability to show up as the most convienient way of represenging a lot of things, now, doesn't it? Of course, I say this as one who can handle base conversions in his head (not arbitrary bases yet, but I'm working on it) and is actually quicker in base 8 than base 10... so I'm probably not the best person to speak on that. ;)Thanks for posing a really interesting puzzle.Thanks for rising to the challenge; I didn't think anyone would have tried it. Which I find odd, since discrete mathematics is just so useful if you bother to study elements of it, even at a cursory level. (Code theory, graph theory, modular arithmetic, and math/logic relations come to mind as some of the more common applications.) Here's a similar one that I'm actually adapting to one of my current games. I play a gnome illusionist who has an almost pathological need to flex his intellectual might over those around him, and may or may not have some grand plan in place that he's working on even as he adventures. He's from a totalitarian magocracy that he takes great issue with (something about them putting undue pressure on social and intellectual liberties; telling people how to think really gets on his nerves), although he's currently outside its borders and faking his death. Soon, he'll need to send a message to friendly contacts behind the magocracy's borders without it being intercepted by the magocracy's intelligence division. He doesn't have unlimited resources, but he does have the brilliance needed to construct a code that cannot be broken unless the entire message he sends is read together. This means that if he breaks his message up into parts, he has to find a way to get every single part to his contact without letting enough components fall into the magocracy's hands -- which would not only fill them in on the nature of his plans but also alert them that he's still alive. The magocracy isn't stupid: They ARE going to intercept some of the messengers. He has to account for that. For the sake of keeping the variables simple, let's say that he can afford to hire a small number of capable bards and rogues, each one likely to make it past the border undetected. My wizard knows safe enough routes that he's certain no more than 2 of them will be captured en route (naturally, he won't be informing them of this; they'd demand hazard pay). However, due to the particular nature of this message, he can break his message up into 4 parts, which we'll call A, B, C, and D (which can each be copied any number of times). How do you divide the copies of the message parts up amongst the couriers to be certain that his contact will get all four parts and the magocracy will not have enough information to decipher the full message? Try to do it hiring as few bards and rogues as possible; the smaller the operation is, the more likely it is to remain secret. (To get you in the right mindset, know that it works with nine bards/rogues, and might work with less. In the 9 solution, no two of them hold all four parts, but any seven DO hold all four. The actual code I'm having my wizard use doesn't have a limit on four parts, though -- essentially, I have a coded message that can be split into K parts, using a code that cannot be broken by using only K-1 parts. I know for a fact that I can do it with five bards/rogues, assuming I'm right on the "no more than two are captured" bit. It should be obvious by now that if all the components don't reach the contact, he won't be able to crack the code either.) ...Yes, I'm a geek. So sue me: the DM tends to make the villains a little *too* smart at times, so I naturally try to blind them with SCIENCE! (...well, math, here, technically, but you get the idea.) Math. It's not just for DMs anymore.(Oh, I have no idea whatsoever about the blue eyes puzzle)It's far simpler than you think. I got it in about ten minutes, and that was a general-case solution. I've never seen a physicsist or mathematician take less than forty minutes to solve it. (EDIT: Then again, none of them ever saw it so close to bedtime...) Mind you, I've also seen philosophers, logicians, and the like spend weeks on it, often taking hours even after I provide the key term needed to solve it...
Optimized_Commoner 05-28-07, 09:15 AM	Yeah, it does have the uncanny ability to show up as the most convienient way of represenging a lot of things, now, doesn't it? Unfortunately for me, yes. Of course, I say this as one who can handle base conversions in his head (not arbitrary bases yet, but I'm working on it) and is actually quicker in base 8 than base 10... so I'm probably not the best person to speak on that. All your base are belong to us. (and you thought you were geeky, I'm hitting the math puns). I have trouble with different bases, honestly. My brother can crank out hexadecimal or binary with fluent ease but I've never had the knack for it. ;)Thanks for rising to the challenge; I didn't think anyone would have tried it. Which I find odd, since discrete mathematics is just so useful if you bother to study elements of it, even at a cursory level. (Code theory, graph theory, modular arithmetic, and math/logic relations come to mind as some of the more common applications.) Yeah, unfortunately high school crushed my love of maths under tons of equations and pointless work (I swear, I was better at Maths before I started the advanced maths course than when I left it). I've only just been getting back into it and my brain's still a little rusty at even basic maths now, but it's enjoyable doing it on my own terms for sheer enjoyment. Same thing largely happened with physics. I went from receiving a solid 100% on every physics test or assignment to failing in my last year when they removed the conceptual element and turned it into drudge work. Lately I've been reading a lot of Game Theory, which is enjoyable and gives me a chuckle every time I see Min/Max theory. It was good seeing a fun, and practical, maths problem there. Its helping me get back my interest in the subject. For the sake of keeping the variables simple, let's say that he can afford to hire a small number of capable bards and rogues, each one likely to make it past the border undetected. My wizard knows safe enough routes that he's certain no more than 2 of them will be captured en route (naturally, he won't be informing them of this; they'd demand hazard pay). However, due to the particular nature of this message, he can break his message up into 4 parts, which we'll call A, B, C, and D (which can each be copied any number of times). How do you divide the copies of the message parts up amongst the couriers to be certain that his contact will get all four parts and the magocracy will not have enough information to decipher the full message? Try to do it hiring as few bards and rogues as possible; the smaller the operation is, the more likely it is to remain secret. 1. AB 2. AC 3. AD 4. B 5. B 6. C 7. C 8. D 9. D Edit: Blerg, never mind about what I deleted here. Misread what you said.
Optimized_Commoner 05-28-07, 09:19 AM	One my grandad used to get me and my brother to try was a puzzle where (in 2 dimensions) you draw 3 houses anywhere on the page, 3 supply sources (1 each for gas, water and electricity) and need to connect the houses to the supply sources with lines that don't cross each other or pass through a house. We struggled with that one for ages before we finally realised it's impossible.
Optimized_Commoner 05-28-07, 09:37 AM	It's far simpler than you think. I got it in about ten minutes, and that was a general-case solution. I've never seen a physicsist or mathematician take less than forty minutes to solve it. (EDIT: Then again, none of them ever saw it so close to bedtime...) Mind you, I've also seen philosophers, logicians, and the like spend weeks on it, often taking hours even after I provide the key term needed to solve it... Well, not so much bedtime as the fact that I'm stuck in bed with a crazy fluu that makes me need to sleep every few hours and fills my head with cotton wool, so my brain isn't exactly functioning at it's fullest level at the moment. Partly I don't think I understand the problem, particularly: "The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following: "I can see someone who has blue eyes." I don't get what it's saying here. So far as I can see the problem it's 201 people. 100 with blue eyes, 100 with brown, 1 with green. One of them needs to guess their own eye colour and the only data they possess is that they can see everyone else's eye colour and know that out of the 200 people who aren't the guru one has blue eyes (information they should already possess). There simply doesn't seem to be enough data available to solve the puzzle. The guru's declaration doesn't help the guru, because she's simply announcing something she already sees herself. The guru's declaration doesn't help a person with blue eyes, because they already know there are people with blue eyes (they can see them themselves). The guru's declaration doesn't help a person with brown eyes, because they already know there are people with blue eyes (they can see them themselves). Thus the guru's declaration helps no one. And their only other source of data specified in the puzzle is that they know the eye colour of everyone except themselves. In short everyone on the island possesses two pieces of data, one of which is guaranteed to be insufficient by one piece of information and the other of which is guaranteed to contain no information the other one doesn't. So yeah, a logic tree says the problem is unsolvable, so I must be missing something or misunderstanding something. I think it's the guru's announcement thing. I'm not too clear on what that's supposed to mean. Edit: Okay, I've thrown logic trees, Venn diagrams and simulataneous equations at the problem, none of which solve it. (though simultaneous equations were fun). Blue Eyed dataset 1. 99A + 100B + 1C + 1D = F 2. 1A + 200E = F Solve for D. (A, B and C are known. D, E and F aren't.) Brown Eyed dataset 1. 100A + 99B + 1C + 1D = F 2. 1A + 200E = F Solve for D. (A, B and C are known. D, E and F aren't.) Geen Eyed dataset. 1. 100A + 100B + 1C = E 2. 1A + 200D = E Solve for C. (A and B are known. C, D and E aren't.) ? What am I missing ?
Tempest Stormwind 05-28-07, 10:45 AM	The Guru's declaration gives them a second perspective.The guru's declaration doesn't help a person with blue eyes, because they already know there are people with blue eyes (they can see them themselves). The guru's declaration doesn't help a person with brown eyes, because they already know there are people with blue eyes (they can see them themselves).The mistake lies here. Any more than that would be giving it away. It *is* true that no one would leave if the guru said nothing.
Optimized_Commoner 05-28-07, 11:07 AM	I know that you know the answer, and I trust you enough to believe that you're right. This creates something of a Dilemna, since I am equally certain that the problem is not solveable. An analysis of this suggests that the most likely causes of the dilemna are that A) My interpretation of the problem text is inaccurate in some way or B) I am ignorant of some principle of logic which can be used to solve this problem. If on a single occasion, the guru simply states "One of the 200 people on this island who isn't me, has blue eyes" as the text appears to imply. And that people are capable of seeing all other eye colours except their own. And that no further data is available to the islanders. Then the problem can not be solved. Repeating a single data point which is already in everyones dataset is not capable of solving the problem. Either they already have sufficient data or they will not have sufficient data even after the guru speaks.
Optimized_Commoner 05-28-07, 11:26 AM	Oh. Okay. Because no one leaves the island at midnight, they can base their reactions on that. yeah... Edit: Honestly, that was more annoying than satisfying.
Tempest Stormwind 05-28-07, 11:26 AM	The problem can be solved. There is another source of information within the problem itself that is absolutely key to the solution. Let's try a similar problem to Blue Eyes. At the Secret Convention of Logicians, the Master Logician placed a band on each attendee's head, such that everyone else could see it but the person themselves could not. There were many, many different colours of band. The Logicians all sat in a circle, and the Master instructed them that a bell was to be rung in the forest at regular intervals: at the moment when a Logician knew the colour on his own forehead, he was to leave at the next bell. Anyone who left at the wrong bell was clearly not a true Logician but an evil infiltrator and would be thrown out of the Convention post haste; but the Master reassures the group by stating that the puzzle would not be impossible for anybody present. How did they do it?The solution to this one is very similar to the Blue Eyes solution, except that it requires a bit of creative thinking to realize you have enough information. (Simply knowing you have enough info won't help you with the solution, though.)
Optimized_Commoner 05-28-07, 11:43 AM	The problem can be solved. There is another source of information within the problem itself that is absolutely key to the solution. Yeah but really, strip away the additional maths required and it's just one of those stupid lateral thinking word puzzles. It's in no way different from the "What's the third word in the english language" comic strip, except for the fact that once you catch the hidden factor provided in the original sentence you have to do a bunch of maths. Plus they made it arbitrarily difficult by increasing the number to 201. I'm supposed to anticipate, what, 100 or so days of nothing happening as the solution. Sorry, didn't like that puzzle. Normally when I don't get a puzzle for ages and finally get the solution it triggers a "Oh god, how stupid I was not to get that earlier" reaction... this time my reaction is "what a stupid puzzle". Ironically the realization of self-stupidity is more enjoyable for some weird and wacky reason. Edit: I apologize since I appreciate you providing an interesting and extremely difficult puzzle. Just not my cup of tea.
Tempest Stormwind 05-28-07, 11:55 AM	No worries; it's just an induction puzzle. Induction, by its definition, involves starting with a simple base case and building up from there. In this case, the base case was 1 person of each color (Blue sees Brown, knows his own eyes must be blue, and leaves on Night 1. Brown sees Blue, doesn't leave since the guru could have been talking about him. When Blue leaves, Brown knows his eyes aren't blue but doesn't know if they're brown, green, hazel, or whatever, so he stays. Net result: 1 blue left on night 1). Solving it for two people of each color produces a similar result. You spot a pattern very quickly and then realize that the number of people who do NOT have blue eyes doesn't matter to this problem at all. Then you amp it up to 100 blue-eyed people and instantly get the solution since the pattern holds. Induction puzzles use the same technique as mathematics, it's just that they don't involve symbolic manipulation unless you really want them to. That's what makes them seem like lateral-thinking puzzles. You may occasionally need to make a leap of logic, but all the information needed to do it is always included. (For instance, in the Logician's Convention puzzle, the leap of logic is to understand that there's at least two of any band around the circle. If a band only appeared once, whoever wore it couldn't possibly solve the problem, and the Master said that it could be solved by everyone. This SEEMS like something you'd find in lateral thinking puzzles, until you look at it like a mathematical formulation (that is, if it were unsolvable for one person, the information we were given is false).)
Optimized_Commoner 05-28-07, 12:05 PM	Yeah, guess I'm just unused to inductive reasoning. Probably just a xenophobic reaction to strange "foreign logic"... that's not the deductive reasoning I've grown comfortable with. Funny how most things we're exposed to focus on the deductive side of reasoning. The puzzle is starting to grow on me a little. Actually, I think after your explanation I do like the puzzle. Not as stupid as I thought. But definitely as annoying as I thought :P I think I need to learn a bit more about induction. Clearly it's a gap in my skill set.
Man in the Funny Hat 05-28-07, 12:32 PM	I think that D&D is too easy I want to make it harder by adding complicated math to it. Can you tell me how to do so? Play 1st Edition AD&D.
StoneStokes 05-29-07, 04:32 AM	Tempest, have you seen this one: A mailman meets Mr. Jones on the street. The mailman says, "Mr. Jones, how many sons do you have now?" Mr. Jones says, "Three." The mailman asks, "And what are their ages?" Mr. Jones replies, "Well, the product of their ages is your age." The mailman says, "That doesn’t help me." Mr. Jones says, "Well, the sum of their ages is the number of windows in that building over there." The mailman says, "That still doesn’t help me." Mr. Jones says, "If it helps, my youngest son has red hair." The mailman says, "Oh, now I know their ages." How old are Mr. Jones’s sons?
Soluphobe 05-29-07, 08:43 AM	Tempus: Ah! Time to rerereread the Dune trilogy again... As to the mailman problem, Jones has three sons. When multiplied together, they give a reasonable number. When added together, they give a reasonable number. And there is a youngest son, meaning that the lowest number is not repeated. 3, 4, 4 works, as does 2, 5, 5. I'm afraid this is a lateral puzzle (I detest those). Heck, 1, 2, 19 works. If the mailman is Mr. Jones's son I shall..I dunno, take a couple deep breaths? Oh, and Tempus: I think you just made me waste a significant portion of my time on the Blue-Eyes Problem. Of course, I'll have to work on it for longer before I'm sure...
northern_bear 05-29-07, 08:49 AM	Reminds me of Corwin from Amber ranting at the sphynx "No, I gave an answer that satisfied all the criteria of the riddle. It's hardly my fault that you had a different answer in mind." Or like the time on Cheers when Cliff was on Jeopardy. The "answer" was "Somebody, somebody, and somebody, have this in common." Cliff responded "who are three people who have never been in my kitchen?" Which of couse is a 100% correct sollution. I hate the whole "we provide the answer, you provide the question" format of Jeopardy. But anyway, if the OP wants ridles in his game, then say riddles. Don't say "math" unless you want the PCs to actually have to solve math problems. If you want to use more complicated mathimatical formulas, as part of the game mechanics, that can be done. In some situations it might provide more game realism. But it would slow things way way down.
maharai23 05-29-07, 09:23 AM	God those puzzles are stupid. It could equally be 13 because it's the only one with two different numbers in it. Reminds me of Corwin from Amber ranting at the sphynx "No, I gave an answer that satisfied all the criteria of the riddle. It's hardly my fault that you had a different answer in mind." Or 11, since it's the only one with two identical digits.
northern_bear 05-29-07, 09:33 AM	Is this the sollution to the blue eyed guru problem? Supose there was only one person on the island with blue eyes. When the guru says "I see someone with blue eyes", then that person would know it had to be him, since he can clearly see that no one else has blue eyes. So he would leave the island imediately. The fact that no one leaves the island on the first night, proves that there must be more than one person with blue eyes. Everyone realises this. Now, supose there are two people on the island with blue eyes. Everybody knows that there is more than one person on the island with blue eyes. Both of the two people with blue eyes, should be saying to themselves, that I must have blue eyes. If I didn't have blue eyes, then that other guy would have realised yesterday that he was the only one with blue eyes, and would have left. He didn't leave yesterday, so he must se at least one other person with blue eyes. Since I don't see anybody else besides him who has blue eyes, it must be me who also has blue eyes. So both of these blue eyed people comes to the same conclusion, and leaves the island the second day. This doesn't happen, so there must be at least 3 people with blue eyes. On the third day, if there were a total of 3 people with blue eyes, then these three blue eyed people would say to themselves, I know there are at least 3 people here with blue eyes, and I only see two people with blue eyes, so I must also have blue eyes. And they the three of them leave that night. If this doesn't happen, it proves that there are a minimum of four people with blue eyes. . . . . On the 99th day, nobody leaves. This proves that there must be at least 100(or more) people with blue eyes. Looking around them and seing only 99 other people with blue eyes, all 100 of the blue eyed people will realise that they have blue eyes, and they will all leave on day 100. On day 101, everyone who is left behind will then known that obviously they do not have blue eyes. However, they don't know if they have brown, green, or red eyes. So none of them can ever leave. I'm very confident about the first part of it. I'm unsure about the second part though.
Optimized_Commoner 05-29-07, 09:43 AM	Is this the sollution to the blue eyed guru problem? Yes. That's the right answer. Incidentally Tempest, that puzzle has been growing on me all night. I'm starting to really like it. Guess it just took a while to get used to it. I realized that it really is math/logic rather than lateral thinking. And the problem with most lateral thinking puzzles is that they want you to think in a unique and varied way, to correspond with the exact answer they want... anyone else notice the hypocrisy?
Tempest Stormwind 05-29-07, 11:02 AM	Yes. That's the right answer. Incidentally Tempest, that puzzle has been growing on me all night. I'm starting to really like it. Guess it just took a while to get used to it. I realized that it really is math/logic rather than lateral thinking.Interestingly, I originally resisted it myself. I'd been told I had to try it out, but I was convinced that it, like other "logic puzzles" I'd heard, was nothing more than communicating badly and acting smug when you're misunderstood. Turns out I'd just been using "logic puzzle" incorrectly. Or rather, the people feeding me garbage had been calling it logic. Induction puzzles are now among my favorite mind games, especially when training people to think rationally, if unconventionally (which is really the core of inductive reasoning: Pure rationality in the face of the almost-instinctive deductive reasoning in which we're trained).And the problem with most lateral thinking puzzles is that they want you to think in a unique and varied way, to correspond with the exact answer they want... anyone else notice the hypocrisy?Everybody! Repeat after me! I! I! Am! Am! Unique! Unique! ...No, no hippocracy at all. :rolleyes: For the record, StoneStrokes, Soluphobe: Please read my name. "Tempest" means "storm" (and also happens to be one of my favorite Shakespeare plays, but that's not why I chose the name; I've had the name for nearly fifteen years now and I first read The Tempest twelve years ago). "Tempus" means "time" (and is also the name of another player in the Soap Opera of the Gods: Everyone Loves Mystra -- er, sorry, the Forgotten Realms). StoneStrokes: I repeat this from before. (http://www.xkcd.com/c169.html) That is not a logic puzzle, but a lateral thinking puzzle. All you can deduce from it is that he has a youngest son (meaning that two of his sons are at least 1; we have no way of knowing if the older two are repeated ages or not), and that their ages are all small (small enough that the product of the three is not above, oh, let's say 60, as no mailman in their right mind works at that age). There's numerous solutions that could work, all being equally likely, and the only way I can deduce which is correct is by figuring out what I know of YOU and basing my answer on that. Busting out the quote from where I linked that earlier link:I always hated those ‘lateral thinking’ puzzles — the ones where they said “oh HO, I never told you that the doctor’s mother was a midget and this was all happening a space station! Don’t you feel dumb now?‘ Because once I figured out that language was really, really flexible and imprecise, it seemed that the key to communication was just figuring out what they probably meant. And figuring out what they could possibly mean if you use all the wrong definitions and stuff is interesting, but I don’t think it teaches all that much more than bad communication. These should be kept very separate from actual logic puzzles, which are really neat.Let's stick to the math. (By the way, speaking of the math: Yep, Optimized Commoner, that's one way of solving the courier problem from earlier. The one I'm working with is fundamentally the same though I do use more variables. For the record, using an arbitrary number of pieces, I can do it with five bards/rogues.) If you guys want a math problem that involves a touch of lateral thinking, take a classic cake-cutting problem. Everyone knows how to cut a classic (cylindrical) cake into eight equal peices with three straight cuts, correct? Well, what if I said I wanted you to cut it into sixteen equal pieces with four cuts, but make sure that each piece has the same amount of frosting on it? For those of us interested in purely mathematical problems, here's another classic. I feel kind of bad playing this one after mentining Hamming codes earlier (since it is also a code theory problem), but at least this time I won't drop a hint.The "Switch Bug", as it came to be known, was a problem that emerged in the phone system at a VERY large university as it expanded during the early 70s. Due to a mechanical defect in the internal telephone's automated switch system, it would periodically switch two consecutive numbers out of the last five that had been dialed. For instance, if I had dialed 421-2345, the last five digits were 12345, but the phone might connect me to someone who had the last five digits of 21345, 13245, 12435, or 12354. This was especially frustrating, since every single person with a university phone number had the same first two digits (42, in this case), meaning I could only be certain of reaching campus, but have no guarantee I would speak to the person I wanted to ! Naturally the administration was up in arms about what to do with this. Engineering solutions would be too expensive since the switch with the defect was everywhere in the system. The university was too large for them to revert to manual operators (over 80,000 phone numbers!), and this was before the dawn of the internet, which gives you a good idea of the state of computer controls at the time. The administration predicted they could implement a new system from a new company in a few years, but by then the university would probably have hit 100,000 phone numbers. A bored information theorist sitting in the math department was equally frustrated, but devised a solution to the problem in an hour. As a result, all the university needed to do was add a "contact code" -- a single extra digit to every phone number. After it did, it had no problems whatsoever: every single call reached the person it was intended to, until the phones were completely replaced with a new system. Replicate the mathematician's result. I've got more that aren't code-theory based, but I figured after priming people on it earlier, well...
northern_bear 05-29-07, 11:17 AM	If you guys want a math problem that involves a touch of lateral thinking, take a classic cake-cutting problem. Everyone knows how to cut a classic (cylindrical) cake into eight equal peices with three straight cuts, correct? Well, what if I said I wanted you to cut it into sixteen equal pieces with four cuts, but make sure that each piece has the same amount of frosting on it? Given your apparent disdain for "lateral thinking" type problems, I guess this is some sort problem where we can assume things you didn't mention. Like: Slice it from above three times, making 8 equal pieces, and then slice it once more horizontally. It works fine if you do any of the following: Assume that there is no icing, so all the pieces have the same amount (zero) Assume that there is no icing on the top. Assume that it is a double decker cake, with a layer of icing in the midle. Also assume that the top and bottom are of exactly equal sizes. Then slice it in such as way that you perfectly seperate the top from the bottom, leaving all the icing from the midle on the bottom layers. [edit] I just re read your problem. Slice a cylindrical cake into 8 pieces using only three cuts. I thought you meant four. If you realy meant three, then I don't know how to do that. How? Without getting into weird horizontal cuts.
Tempest Stormwind 05-29-07, 11:25 AM	Slice it from above three times, making 8 equal pieces, and then slice it once more horizontally.Three cuts produces six pieces, not eight. For the record, this is an everyday cake, like the kind you probably got for your birthday when you were little. Cylindrical in shape, icing on the top and sides. There's nothing particularly special about the cake in the solution.I just re read your problem. Slice a cylindrical cake into 8 pieces using only three cuts. I thought you meant four. If you realy meant three, then I don't know how to do that. How? Without getting into weird horizontal cuts.The classic problem is the cake into eight equal pieces with three cuts, ignoring frosting. Since I assumed everyone'd seen that before, I didn't elaborate on the old-school solution: a pair of "standard" cuts (quartering the cake) and then a horizontal cut (top and bottom pieces like an Oreo) produces eight pieces by three cuts. However, this won't help you get sixteen pieces with equal frosting with only four cuts.
Downtym 05-29-07, 12:29 PM	What if the characters needed to see a powerful wizard, but he only sees intelligent people so he requires them to solve afew Calculus problems. Then you, as a character, don't go see the powerful wizard and just let the evil bad guys take over the world because if the powerful wizard doesn't care enough to not be a jerk, then you shouldn't care enough to try to save his skin.
Downtym 05-29-07, 12:35 PM	The problem with riddles is best illustrated by Mirrormask: What's green, hangs on a wall, and whistles?
Optimized_Commoner 05-29-07, 12:58 PM	I! I! Am! Am! Unique! Unique! Or my favourite monty python quote ever. Brian: "You are all individuals" Crowd: "We are all individuals" One person in crowd: "I'm not!" (By the way, speaking of the math: Yep, Optimized Commoner, that's one way of solving the courier problem from earlier. The one I'm working with is fundamentally the same though I do use more variables. For the record, using an arbitrary number of pieces, I can do it with five bards/rogues.) It's vastly easier with an arbitrary number of pieces. Of course, the ideal mechanism for something like this (barring extremely complex modern mathematics), is to have the decoding system hidden inside each of the pieces. This way unless you have all the pieces, they're completely meaningless. Of course, magic could probably solve this quite easily, "Contact other plane: Oh god of cryptography, what is the cypher for this code?", etc. With 4 pieces you can't get any less than 9 bards/rogues and keep the solution ideal. Any less and you risk your allies not getting the entire message or the magocracy getting the whole thing. Not quite mathematics, but on the topic of cryptography I loved this one "puzzle" faced by the characters in a TV Show, where the secret information was written in three books to keep it safe with the sentence jumping from book to book at random intervals marked by changes in grammar. I thought that was an awesome detail to add to the story. The "Switch Bug", as it came to be known, was a problem that emerged in the phone system at a VERY large university as it expanded during the early 70s. Amusing note (yet largely irrelevant), my grandfather was a telephonic engineer in the 70s.
LCD2YOU 05-29-07, 01:04 PM	Use the equations of wind resistance for terminal velocity so you can really get the "feel" of falling damage. As for "smart players/dumb characters" and riddles/etc.: Make the riddle answered with a "Knowledge, Enginneering" and have a DC of 25 or so. That wakes people up. The other thing, and I've done this, have players with "smart characters" have easier things to solve. One guy who was absolutely brilliant so always nerfed the Int on his character because he could, found that I gave him PDEs to solve while others got basic word puzzles. He complained, but I reminded him that is how his character sees things, not him. :)
StoneStokes 05-29-07, 01:38 PM	StoneStrokes: I repeat this from before. (http://www.xkcd.com/c169.html) That is not a logic puzzle, but a lateral thinking puzzle. All you can deduce from it is that he has a youngest son (meaning that two of his sons are at least 1; we have no way of knowing if the older two are repeated ages or not), and that their ages are all small (small enough that the product of the three is not above, oh, let's say 60, as no mailman in their right mind works at that age). There's numerous solutions that could work, all being equally likely, and the only way I can deduce which is correct is by figuring out what I know of YOU and basing my answer on that. Tempest (my apologies on the incorrect name), the puzzle I presented is a logic puzzle. It is not a lateral-thinking puzzle. It is reasonable to assume that the mailman is under 65 (manditory retirement and all). However, for his age under 65 there is only ONE solution that works! It has already been correctly pointed out that Mr. Jones must have a "youngest son." But there is much more to the puzzle than that. Have fun, StoneStokes. Edit: For the record, my username is StoneStokes (as in the mathematician), not Strokes. What's good for the goose is good for the gander. :P
StoneStokes 05-29-07, 01:51 PM	Tempus: Ah! Time to rerereread the Dune trilogy again... As to the mailman problem, Jones has three sons. When multiplied together, they give a reasonable number. When added together, they give a reasonable number. And there is a youngest son, meaning that the lowest number is not repeated. 3, 4, 4 works, as does 2, 5, 5. I'm afraid this is a lateral puzzle (I detest those). Heck, 1, 2, 19 works. If the mailman is Mr. Jones's son I shall..I dunno, take a couple deep breaths? It is NOT, I repeat, NOT a lateral-thinking puzzle. Moreover, I'm afraid that none of the three solutions you have presented are correct.
northern_bear 05-29-07, 02:03 PM	It is not a lateral-thinking puzzle. . . for his age under 65 there is only ONE solution that works! No. There is not any unique sollution. At least not with the information you've provided. Here's what you provided: The product of their ages is your age. [Assumed to be between 18 and 65] The sum of their ages is the number of windows in that building over there. [Could reasonably be anywhere between 0 and 500. But let's assume the "building" is small, so that it has between 5 and 30 windows.] If it helps, my youngest son has red hair. [So presumably, one son is at least a year younger than the others. Actually there could be two that are the same age, measured in years, but with one youger. We'll assume though that this isn't the case. Let the boy's ages be integers: X, Y, and Z There is a youngest son (X) so that X<Y and X<Z. The product of their ages is such that 18 < XYZ < 65. The sum of their ages are such that 5 < X+Y+Z < 30. There are plenty of different values for X, Y, and Z that satisfy these criteria. 1,2,19 or 2,5,5, or 3,4,4 or 2,4,5 or, or, or. . . . .
northern_bear 05-29-07, 02:04 PM	Moreover, I'm afraid that none of the three solutions you have presented are correct. I should have known it. You are posting some problem that you read or heard somewhere else, but you've forgotten to write it down correctly. If you're so sure that there is a unique answer, then you must not have presented the problem correctly.
Tempest Stormwind 05-29-07, 02:19 PM	Tempest (my apologies on the incorrect name), the puzzle I presented is a logic puzzle. It is not a lateral-thinking puzzle. It is reasonable to assume that the mailman is under 65 (manditory retirement and all). However, for his age under 65 there is only ONE solution that works! It has already been correctly pointed out that Mr. Jones must have a "youngest son." But there is much more to the puzzle than that. Have fun, StoneStokes.Not really. 2, 5, 5 works. So does 4, 4, and 4 (in the case of sons who were born in the same year months apart, or perhaps twins/triplets born sequentially). There's others as well. *edit* Northern Bear provided a sounder example than I did.Edit: For the record, my username is StoneStokes (as in the mathematician), not Strokes. What's good for the goose is good for the gander. :PAck, sorry about that. (The ironic part is the reason I suck at pure math courses is that I honestly can never remember the theorem names even as I master their procedures. For instance, my intro calculus final, one of the problems was "Use Green's Theorem to prove the following". My actual response was "I have no idea what the :censored: Green's Theorem is, but here's the proof. I assume, somewhere in there, that I used Green's Theorem, but don't make me point to it." I did get the question correct, if I recall correctly (it's been about five years now).) Optimized Commoner: I'd have to work it out again, but I believe you can do the courier problem with 4 parts, 2 caputures, but only 8 couriers. I'm positive you can't do it with 7. A related problem, by the way: The camper's dilemma. This can be adapted to D&D fairly easily by changing some names, or even used verbatim (perhaps as a metaphor). A camp counselor is leading a his troop of seven (that is, eight people total counting himself) to a campsite. They're not entirely certain of the route, as the trail signs are poor. They come to a fork in the road, with four possible paths. The sun's setting, and there's only an hour of daylight left -- and it's dangerous to travel in these woods at night. Judging from the signs at the trailhead, they know it's only about 20 minutes away, but they don't know which path it's down. Naturally, the easiest way to figure out which is the right path is to split up, sending groups 20 minutes down paths, 20 minutes back, deciding which route is the right one, and then 20 minutes to the campsite. (We'll assume these are super-campers and are able to set up camp very quickly, or that the campsite is very secure (perhaps a lodge), or something similar.) The twist? Two of the campers might lie - if they found the lodge, they'd say that they didn't, and vice versa. Do note that they only MIGHT lie -- they might also tell the truth. The counselor doesn't know which two are the potential liars, and has no way of telling if they lied anyway. How should you split the group up to figure out which is the right trail every time? Two notes: 1) Variants on this problem (# of trails, campers, and liars) are common. I will admit I DO NOT have my notes on me and may have forgotten the original telling; I'm pretty sure 4/7/2 is the basic one, but I could be wrong. I DO have a far harder one in store 2) The counselor never lies. (This is actually critical to the solution, despite seeming trivial.)
Ralp 05-29-07, 02:20 PM	I should have known it. You are posting some problem that you read or heard somewhere else, but you've forgotten to write it down correctly. If you're so sure that there is a unique answer, then you must not have presented the problem correctly. I didn't solve this problem, but I think I know how to solve it. I'm pretty sure the hint that you are missing is... ...is the fact that with the information available, the mailman is able to find a unique solution. In other words, the fact that there is a unique solution is the key to finding a unique solution. Put another way: you're not solving the same problem that the mailman solved. The mailman knows how old he is, and we don't. The mailman knows how many windows are in the building, but we don't. Fortunately, we know two things that the mailman didn't know: 1. There's a unique solution. We know that since the mailman figured it out, but above and beyond that, the puzzle-giver came right out and told us so. But moreover... 2. There's only a unique solution after all three hints. Given some values of the mailman's age, and the windows on the building, the solution still wasn't clear (i.e. there's more than one possible solution after Jones's first 2 hints). But knowing that there's a youngest kid eliminates all those potential solutions except one.
northern_bear 05-29-07, 03:05 PM	I think I figured out what you meant by the postman problem. Thanks to Ralp for helping to clarify. But I still don't think there is a unique sollution. Supose the mail man is 40 years old, and the building accross the street had 14 windows. The three sons could be aged 2,2, and 10. Or they could be aged 1,5, and 8. Both multiply out to be 40, and add up to be 14. So the mail man doesn't know which is correct. When he is told that one of them is the "youngest" that eliminates the possibility that they are 2, 2, and 10. So they must be 1, 5, and 8. The problem is, there is still more than one such sollution. If the mail man is 36, and the number of windows is 13, then the values (2,2,9) and (1,6,6) both solve the base conditions. They both add to 13, and multiply out to 36. Saying that there is a youngest reduces the sollution to (1,6,6). So from the informaiton given, we don't know if the boys are are 1,5, and 8. Or if they are 1, 6, and 6.
Ralp 05-29-07, 03:21 PM	I'm not confident enough in my players' abstract critical reasoning skills to drop puzzles on them like those in this thread. But I do have a pretty neat mathematical adventure planned for when they get tired of slaying zombies: a casino. I've designed several games for this casino, some of which are based on some famous math paradoxes. These aren't paradoxes in the strict sense of the word, but rather mathematical properties that are contrary to what one would expect. (Which makes them perfect sucker bets.) Here's one: The Shell Game is run by a half-orc, Dumont "Monty" Lorgan. He runs his game on a small collapsible cloth table in a shady corner of the establishment. Players will probably be immediately suspicious of a game described like this, but instead of the usual three-card con game, here the roles are reversed: the PCs play the house, and Monty plays the customer. The house (that is, the PCs) sets the stakes, by placing at least 1 gold, silver, and copper, under three shells. (As much or as little as they like, but the gold must be more in total value than the silver, and the silver must be more than the copper.) "Since the house always has the advantage", Monty says, he graciously matches 150% of the gold (if he can afford it), and places it under the same shell with the rest of the gold. The shells have permanent silence and nondetection effects (which is evident to players who chose to rent a detect magic eyepatch provided by this pirate-themed casino, but this is irrelevant to the casino game!) to ensure that Monty doesn't cheat. The players rearrange the shells however they like while Monty looks away; then he picks one shell (but does not look under it). Then the house reveals any one shell other than the one Monty has picked. Then Monty picks any of the three shells and wins what's underneath. The house wins the rest. In actuality, Monty is not the bumbling incompetent con-artist that he appears to be; he is much wiser and smarter than he lets on. This is actually a sanctioned game run by the casino (the magical effects cast on the shells should be a clue to this), because despite the 150% house "advantage" that Monty offers, the PCs will lose money at this game in the long run. Players will soon discover that Monty, without any magic or trickery, can pick the shell with the gold 67% of the time (or more if the player is foolish enough to ever reveal the gold). Clever players can try to skew the odds in their favor by employing sleight of hand, or perhaps subtle magic, although anyone caught using magic in the casino is politely "invited" to leave. (The half-orc's nickname "Monty" is not a reference to "three card monte" as one might assume, but rather the "Monty Hall problem" upon which this game is based.)
northern_bear 05-29-07, 03:41 PM	. . . (monty hall game). . . Realy good idea. It probably won't work though if your players have heard of this before. And if they have taken some statistics courses, there's a good chance they have.
Tempest Stormwind 05-29-07, 03:56 PM	The shell game's a good one if you're acting it out (I really like how it's constructed, actually, and may borrow it in one of my upcoming games!), but since it's probabilistic instead of deterministic, it's vulnerable to cheat (a fun, seldom-remembered spell; one might claim it's truth in advertising) if a roll is involved. (That said, the whole point of the game is to act it out...) I'd love to see your other games. I tend to suck at gambling problems in general; I can't visualize them in the same way I can with physical problems.
Optimized_Commoner 05-29-07, 04:18 PM	Optimized Commoner: I'd have to work it out again, but I believe you can do the courier problem with 4 parts, 2 caputures, but only 8 couriers. I'm positive you can't do it with 7. AB AC BC BC A D D D Of course. Sorry, should have figured that one out sooner. My obsession with symetry kept doing me in. Edit: The limitations for this are basically: Each part must appear at least 3 times, in case that part is captured with both of the captures. No messenger can carry 3 or more parts, or the magocracy can figure out the message with the right 2 captures. Doubles must either all include the same letter, or none of them can include a particular letter.
Optimized_Commoner 05-29-07, 04:25 PM	Path 1: Camp Counsellor Path 2: Child 1, Child 2, Child 3 Path 3: Child 4, Child 5, Child 6 Path 4: Child 7 The obvious solution doesn't appear to work. Edit: I assume these are 4 possible paths to the campsite, one of the paths isn't the way they came from, correct?
Einvaldurinn_mikli 05-29-07, 04:27 PM	How should you split the group up to figure out which is the right trail every time? Would splitting the group in four parts, two groups of three and two singles, with the counselor going alone work?
Optimized_Commoner 05-29-07, 04:58 PM	Would splitting the group in four parts, two groups of three and two singles, with the counselor going alone work? That adds up to 9...
Tempest Stormwind 05-29-07, 05:17 PM	Yes, it's four potential paths not including the one they came from. Just as the archetypal Y fork would be the same as coming along two paths. Do note that some of the solutions involve leaving paths BLANK (not followed). If you can conclusively rule out the other three, the fourth ("the road not taken", if you'll excuse the pun) is the correct path. I'm not entirely certain if you need to do that with this particular version of the problem, though. I'll post more conclusively when I get my notes at home.
Optimized_Commoner 05-29-07, 05:53 PM	Do note that some of the solutions involve leaving paths BLANK (not followed). If you can conclusively rule out the other three, the fourth ("the road not taken", if you'll excuse the pun) is the correct path. Oh, of course. I'm such an idiot. Path 1: Camp counsellor, child 7 Path 2: Child 1, child 2, Child 3 Path 4: Child 4, child 5, child 6 Still doesn't seem right though. You can get an inconclusive result, as depicted below Assuming the path is on camp 3 Path 1: No camp, No camp. Path 2: Camp, Camp, No camp <- could be either only 1 person lying or 2. Path 4: No camp, No camp, No camp Result: Inconclusive. You require 5 children in a party not including the councillor to guarantee accuracy. Perhaps 6 children, 2 camp counsellors and 4 paths would be a better number for the puzzle? Still difficult (you have to deduce the 5 children safe number, camp counsillors as safe options and leaving a path blank). Or 3 paths rather than 4. This puzzle would work well for an adventure. I'd dress it up a little naturally, but with the counsellors representing PCs and the children representing NPCs of an untrustworthy nature. Yeah, this could work well.
Einvaldurinn_mikli 05-29-07, 06:07 PM	That adds up to 9... The councilor is one of the singles.
StoneStokes 05-29-07, 06:23 PM	The puzzle I presented can be solved with all of the information I have given. I have written the puzzle correctly, I am 100% certain. There is one unique answer (up to age 60 for the postman). No one has come up with that answer yet; Ralp is very close (see his spoilers for a very big hint). It is not a lateral-thinking puzzle (at least no more so than the Blue Eyes puzzle). I know the answer, and I solved it without hints or help. I know other people who have solved it without hints or help. I will tell the solution to anyone who asks exactly how to solve the problem. I will give a hint to anyone who asks that will lead them toward the solution without giving it away. Tempest, Green's Theorem is a consequence of Stokes' Theorem; as are Gauss's Divergence Theorem, the Kelvin-Stokes' Theorem, and The Fundamental Theorem of Calculus. Now you know that you only need to remember the name of one theorem! :)
StoneStokes 05-29-07, 07:31 PM	Here is another puzzle that may be interesting for smart players: The PCs enter a room with two doors, both locked magically. A sphinx appears to them and speaks, "Greetings adventurers. Beyond one of these two doors is a great treasure, but beyond the other is a Balor. Before you are twelve gold coins and a merchant's scale. But alas, one of the coins is a fake. You may use the scale three times. I will open the door to the treasure if you correctly identify the fake coin, and tell me whether it is too light or too heavy. On the other hand, if you choose a genuine coin or cannot tell me whether the fake is too light or too heavy, then I must release the Balor. Any other manipulation of the coins will force me to release the Balor, as will a fourth weighing on the scale. You have one hour." This too, is a logic puzzle, and the solution is not based on word-play or anything silly like I have been (wrongly) accused of regarding the last puzzle. I know the solution to this puzzle as well.
Tempest Stormwind 05-29-07, 08:27 PM	Here is another puzzle that may be interesting for smart players: The PCs enter a room with two doors, both locked magically. A sphinx appears to them and speaks, "Greetings adventurers. Beyond one of these two doors is a great treasure, but beyond the other is a Balor. Before you are twelve gold coins and a merchant's scale. But alas, one of the coins is a fake. You may use the scale three times. I will open the door to the treasure if you correctly identify the fake coin, and tell me whether it is too light or too heavy. On the other hand, if you choose a genuine coin or cannot tell me whether the fake is too light or too heavy, then I must release the Balor. Any other manipulation of the coins will force me to release the Balor, as will a fourth weighing on the scale. You have one hour." This too, is a logic puzzle, and the solution is not based on word-play or anything silly like I have been (wrongly) accused of regarding the last puzzle. I know the solution to this puzzle as well. A nice and classic problem (made more difficult than the usual ones since you don't know whether the coin is heavier or not) and I'd probably include it in my games. I've already used a similar one: you're presented with two sets of 8 coins, and two balance pans. The coins in each set are numbered, and one coin in each set is SLIGHTLY heavier than the others (the two heavy coins have the same number). The scales ans are enchanted: they will not tip until both scales are loaded (that is, you can't weigh one set of coins, make a decision, and then weigh the other), and coins from the other set fall through the wrong balance (that is, you can't put coins from group A on balance B, and vice versa). You have ONE weighing. Find the heavy coin. (Mathematically, this is identical to devising a nonadaptive solution to the "8-coin, 1's heavier" problem with two weighings. It's easier than yours, though -- I saw the solution to mine instantly when I was first presented with it, but don't immediately see the solution to yours (though I suspect I see how to solve it; I'll work on it right now.).) IMPORTANT NOTICE regarding the CAMPERS' DILEMMA problem: I'm looking at my notes right now and I made a mistake when I presented it. The classic incarnation is 4 paths, 2 possible liars, but eight total campers (other than the counselor). Seven will not be enough in all cases. In recompense, I'm looking at the later versions of this puzzle now, which I'll post up straightaway. 1) 4 paths, but five liars. What's the bare minimum number of campers you'd need to solve this? 2) 4 paths, 2 liars, but only four campers. However, unlike the last time, you have 100 minutes until sunset (enough for two round trips plus the final walk to the site). Do note that this time, since you have multiple trips, the liars can lie whenever they feel like -- they could tell the truth on the first run and lie on the second, lie on both, tell the truth on both, or lie on the first and tell the truth on the second. As before, though, you don't know who these two liars are, although they're always the same campers. (This is harder and solved differently from the other two versions. You will need one more piece of information, although you already knew it: the campsite is down only one path. Good luck.) I also have one more problem for you that works well as a trapped room which seals behind the party. In the room are seven unlabelled caskets, and suspended above them is an image of Death. He informs you that six of the caskets are trapped with deadly spells and gases that will obliterate everyone in the room if even one is opened. The seventh has the key to leave the room. None of the caskets are locked, but all of them register absolutely identically to all forms of magical detection (that is, you can't detect magic or traps on any of them; you have to solve this numerically). The Death image indicates that the key is in the casket numbered 54321. He then demonstrates how to count them: He touches one, then the one to its right, and so on, until he hits the seventh casket, when he reverses direction. Graphically, you might represent it as follows: A B C D E F G 1 2 3 4 5 6 7 13 12 11 10 9 8 14 15 16 17 18 19 25 24 23 22 21 20 (The letters at the top are the individual caskets, the columns correspond to what casket Death is on when he says the relevant number. It's easier if you set it up physically for yourself and see.) However, Death decides to make things interesting. He announces that you're timed -- the six deadly caskets will open "when your time is up". He then inverts his hourglass and fades away, leaving an image of an hourglass in the middle of the room, counting down to your demise. Note that this isn't enough time to bruteforce-count your way to the fifty-thousands. Mathematically, this is trivial to solve, but it's a real ***** if you pop it on your unsuspecting players.
Comus 05-29-07, 08:31 PM	I know the solution to this puzzle as well. So do I... I waste the sphinx with my crossbow! :P
PwnageLlama 05-29-07, 10:44 PM	:uh-huh: Uh-Huh?
StoneStokes 05-30-07, 12:35 AM	Tempest, I like the death puzzle. Here is the answer: A = 1 (mod 12) B = 2 (mod 12) etc. 54321 = 9 (mod 12), which works out to casket E. A very fast (no divisions) way to see that 54321 = 9 (mod 12) is to notice that: 54321 = 1 (mod 2), because the last digit is odd (this is where I started, but it is actually irrelevant); 54321 = 0 (mod 3), because the sum of the digits is 15 which is divisible by 3; 54321 = 1 (mod 4), because the last two digits are 21, which is 1 (mod 4); Then, the only number between 1 and 12 that is concurrently divisible by 3 and has remainder 1 mod 4 is 9...hence E. To make it more challenging, but have the exact same answer, give the correct casket the number 987,654,321 The only real difference is that division takes longer, but the same modular arithmatic does not.
RadicalTaoist 05-30-07, 12:36 AM	I have a suspicion about the last hint Mr. Jones gives the mailman. What I saw was this: if the mailman needed to know hair color (theoretically, eye color or any other distinguishing feature) to catch the youngest, that means that he couldn't distinguish the youngest apart from the middle son (and maybe the oldest, if they're close enough) by age-obvious features (like height, signs of puberty, etc.). That means that the age range between the youngest and the middle has to be so low the mailman couldn't figure it out without distinguishing the youngest some way. Am I getting warmer, Stones?
pmurray@bigpond.com 05-30-07, 01:51 AM	I think that D&D is too easy I want to make it harder by adding complicated math to it. Can you tell me how to do so? Treat anything that requires a d20 roll as a complex number with a modulus of 1, and opposed checks of any sort (to hit vs armour class, DC vs saves) as an interference effect. The whole game becomes deterministic, but insanely complicated.
StoneStokes 05-30-07, 01:57 AM	Okay, I spoke too hastily when I said I was 100% correct about the wording of the Mailman problem. Before I made that claim, I had double-checked my source, and it matched what I wrote for the problem completely. However, I now think that indeed there is an error that I inherited from said source. I believe it should be corrected to say that Mr Jones's oldest son has red hair. In this case, Northern_Bear's solution is correct (and was the impetus for me double-double checking the source). My humblest apologies. :embarrass I will make amends with another D&D-worthy puzzle shortly.
pmurray@bigpond.com 05-30-07, 03:22 AM	One of my favorites was a puzzle lock with four sliders -- three of them had the Sylvan symbols for the digits of 0-9, and the one on the left of those three had two symbols (for 0 and 1), thus being able to represent any number between 0 and 1999. ... It would allow you to ask it fifteen yes or no questions relating to the correct combination... but, unlike most puzzles of this sort, it would NOT respond until the 15th question had been asked, and it MIGHT lie on one of its answers I can do it in 15, but only for the range 0-1023, not 0-1999 1: is the remainder mod 2 >= 1 ? 2: is the remainder mod 4 >= 2 ? 3: is the remainder mod 8 >= 4 ? 4: is the remainder mod 16 >= 8 ? 5: is the remainder mod 32 >= 16 ? 6: is the remainder mod 64 >= 32 ? 7: is the remainder mod 128 >= 64 ? 8: is the remainder mod 256 >= 128 ? 9: is the remainder mod 512 >= 256 ? 10: is it >= 512 ? 11: was the answer to question 2,4,6,8, or 10 a lie? 12: was the answer to qusetion 3,4,7,8 or 11 a lie? 13: was the answer to question 5,6,7, or 8 a lie? 14: was the answer to question 9 or 10 a lie? 15: was the answer to question 11, 12, 13 or 14 a lie?
northern_bear 05-30-07, 07:09 AM	I believe it should be corrected to say that Mr Jones's oldest son has red hair. In this case, Northern_Bear's solution is correct (and was the impetus for me double-double checking the source). My humblest apologies. :embarrass If you wanted to frame this same problem in purely mechanical terms (thus removing any confusion about semantics) then you could phrase it as follows: Let A and B be constant fixed integers, where 18 < A < 65 and 5 < B < 30 Find integers A, B, X, Y, and Z, such that XYZ = A and X+Y+Z = B has multiple sollutions, but XYZ = A, X+Y+Z = B, X < Z, and Y < Z has a unique sollution.
High Octane 05-30-07, 07:11 AM	I think that D&D is too easy I want to make it harder by adding complicated math to it. Can you tell me how to do so? Play First edition.
Drausktanchache 05-30-07, 10:43 AM	You want math to find out HOW to go about beating an encounter here it is: You - fighter 6 Jimmy - 20 expert/20 aristocrat/20 commoner If Jimmy (20th lvl expert) has 200 gold, and a 3 magic Items, and you want ONE HUNDRED GOLD and TWO magic items, then you take out your sword, disembowl Jimmy, and decide to work from there... Complicated? I think not...
Soluphobe 05-30-07, 11:59 AM	Tempest: Ugh. Sorry about that. Probably the same reason I misspelled Tleilaxu Ghola. :embarrass To make up for it, here's a logic puzzle I found. Its sort of similar to the man at the crossroads style, but I found it entertaining: Three gods A, B, and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which. It could be that some god gets asked more than one question (and hence that some god is not asked any question at all). What the second question is, and to which god it is put, may depend on the answer to the first question. (And of course similarly for the third question.) Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely. Random will answer 'da' or 'ja' when asked any yes-no question
Zemyla 05-30-07, 12:50 PM	Here is another puzzle that may be interesting for smart players: The PCs enter a room with two doors, both locked magically. A sphinx appears to them and speaks, "Greetings adventurers. Beyond one of these two doors is a great treasure, but beyond the other is a Balor. Before you are twelve gold coins and a merchant's scale. But alas, one of the coins is a fake. You may use the scale three times. I will open the door to the treasure if you correctly identify the fake coin, and tell me whether it is too light or too heavy. On the other hand, if you choose a genuine coin or cannot tell me whether the fake is too light or too heavy, then I must release the Balor. Any other manipulation of the coins will force me to release the Balor, as will a fourth weighing on the scale. You have one hour." This too, is a logic puzzle, and the solution is not based on word-play or anything silly like I have been (wrongly) accused of regarding the last puzzle. I know the solution to this puzzle as well. I know the solution to this puzzle! Start buffing. You're going to fight a balor and you've got an hour to prepare.
Downtym 05-30-07, 01:27 PM	Every problem or puzzle thus far suggested share one common flaw: They are all incredibly boring to a player. I don't think I'm alone when I say that if you're going to include puzzles in the game they should be more kinetic. Which is to say that the solution should involve the PC's doing something and not just sitting around with a pad of paper and a pencil scratching out the proper solution.
Nathreet 05-30-07, 01:37 PM	I think that D&D is too easy I want to make it harder by adding complicated math to it. Can you tell me how to do so? 1. Drink heavily. Now DnD isn't too easy anymore and you've lost your desire to add complicated math to it. 2. Don't play on a battle grid. Install push-pins in the bottom of miniatures, and stick them on a cork board. Measure all distances with a ruler (1" = 5'). Use pythagoreans theorom when necessary. It's especially useful in 3D. 3. Ditch iniative. Assign a time to every single action in DnD (see "combat" chapter). Certain actions may be combined with others, and others might be quicker if a player has a higher stat. Track time by the second. A player takes his turn when his previous action is complete. 4. Do both #2 & #3. Admit defeat, and go back to regular DnD. Despite the temptation, avoid plummeting into #1, as it is no longer necessary and might cause things to go too far in the other direction.
Nathreet 05-30-07, 02:36 PM	Tleilaxu Ghola. (Read up on your Dune; that's where the name comes from.) Although it pertains to character creation rather than Numb3rs-style solutions, the Ghola Character project (http://boards1.wizards.com/showthread.php?t=777565) is the most advanced math used in D&D. It sprung out of Pun-Pun, which can do the same sort of approach but far cheesier (the Ghola Character was an attempt to figure out how to do it without using the cheese). As for applications of math through problem solving, that's strictly in the realm of adventure design. Get a mathematician for a DM -- especially one who enjoys discrete mathematics -- and you'll have your chance for using them. For instance, a very simple problem would be scrying on your target and finding out that he's in an underground room (hewn stone) with three doors. You know he teleported directly there, and thanks to a trace, you've gotten it down to a few square miles that he could be in. However, the locals know that there's not one, but two underground labyrinths nearby: one has two entrances, the other has three. The villain likely won't stay there long enough for you to sweep out both labyrinths; figure out which one he's in. That can be solved with very, very simple graph theory (it's a parity problem). It still leaves the actual dungeon crawl to do, but math will tell you which dungeon to tackle. One of my favorites was a puzzle lock with four sliders -- three of them had the Sylvan symbols for the digits of 0-9, and the one on the left of those three had two symbols (for 0 and 1), thus being able to represent any number between 0 and 1999. An inscription above them said that it was a combination lock that only allowed ONE guess -- if you pulled the lever and the number was wrong, it would vanish and the gate it locked would be sealed for another decade. Reading the inscription would trigger an illusory voice effect, informing you that it was put there by some more capricious fey who wanted to give the mortals a staggering chance, but only if they were smart. It would allow you to ask it fifteen yes or no questions relating to the correct combination... but, unlike most puzzles of this sort, it would NOT respond until the 15th question had been asked, and it MIGHT lie on one of its answers (you don't know ahead of time if it's going to lie, or if it does, which one of its answers was wrong). After it answers the fifteen questions, it bids you luck and disappears. In this case, it's simple code theory (in particular, a Hamming code can get the right answer every time). The penalty for getting it wrong is that you don't go through the door. If you get it right, well, the reward's up to the DM. If your DM doesn't like sticking math puzzles in his games, then... well, there's not much we can help with. (Cookie for anyone who figures out the second problem (that is, what 15 questions to ask); the first is trivial.) I can do it in 15, but only for the range 0-1023, not 0-1999 1: is the remainder mod 2 >= 1 ? 2: is the remainder mod 4 >= 2 ? 3: is the remainder mod 8 >= 4 ? 4: is the remainder mod 16 >= 8 ? 5: is the remainder mod 32 >= 16 ? 6: is the remainder mod 64 >= 32 ? 7: is the remainder mod 128 >= 64 ? 8: is the remainder mod 256 >= 128 ? 9: is the remainder mod 512 >= 256 ? 10: is it >= 512 ? 11: was the answer to question 2,4,6,8, or 10 a lie? 12: was the answer to qusetion 3,4,7,8 or 11 a lie? 13: was the answer to question 5,6,7, or 8 a lie? 14: was the answer to question 9 or 10 a lie? 15: was the answer to question 11, 12, 13 or 14 a lie? Doesn't work if the sphinx lies on any of questions 11-15, I don't think. Okay, time for my attempt. This question is easier to answer in binary. You can convert the answer to decimal, but I am going to use binary to make things easier on myself. You need 11 questions to find out the 11 digits of the answer, in binary. That part is simple. In a similar fashion, you use the other 4 questions to give you a number between 0 and 15. 1 to 15 means that question is a lie. 0 means none of the questions is a lie. Use questions 1, 9, 13 and 15 to determine which question is the lie (if any). If he lies on any of these: 1111 becomes 1101 (15 becomes 13 b/c question 15 asks about the 3rd digit) 1101 becomes 1001 (13 becomes 9 b/c question 13 asks about the 2nd didgit) 1001 becomes 0001 (9 becomes 1 b/c question 9 asks about the 1st digit) 0001 becomes 0000 (1 becomes 0 b/c question 1 asks about the 4th digit) If the sphinx claims to lie on question 13, he actually lied on question 15, etc. Regardless, if he lies on any of these 4 questions, or doesn't lie at all, or claims that he didn't lie at all (lies on question 1), you still know for certain that he didn't lie on the other 11 questions. Give me my cookie.
Optimized_Commoner 05-30-07, 02:53 PM	In a similar fashion, you use the last 4 questions to give you a number between 0 and 15. 1 to 15 means that question is a lie. 0 means none of the questions is a lie. EDIT: If you ask which question is the lie using other questions, not questions 12 through 15, it will work. Still haven't figured out which questions to use to detect the lie, though. Check my post, the lie-correction code was right. (I'm 99% certain). My only problem was I wasn't good enough with binary to write the other 11 questions. Combine my 4 hamming codes (the questions about whether the number of yes answers were even) with 11 questions that can determine the answer through binary and you should (assuming I was right) have the solution.
Nathreet 05-30-07, 03:07 PM	Edited my post. It has the answer now. I checked the Hamming code. Turns out it's slightly different from my method. Optimized_Commoner, it seems like you tried to use the exact Hamming code. Unfortunately I cannot check your work without taking a closer look at the Hamming Code.
elondir 05-30-07, 03:46 PM	You want complex math in D&D? I have a task you might be interested in: create a realistic economy for D&D that still feels like D&D.
Optimized_Commoner 05-30-07, 03:49 PM	You want complex math in D&D? I have a task you might be interested in: create a realistic economy for D&D that still feels like D&D. I'm working on that project at the moment... <points to patches of pulled out hair>
Tempest Stormwind 05-30-07, 03:52 PM	Express the number in base 2 (binary). If it is less than 11 digits long, prepend 0s to it until the final result is 11 digits long. (This is mathematically identical to the original number.) Regarding this particular way of expressing the combination: 1) Is the 1st digit a 1? 2) Is the 2nd digit a 1? 3) Is the 3rd digit a 1? 4) Is the 4th digit a 1? 5) Is the 5th digit a 1? 6) Is the 6th digit a 1? 7) Is the 7th digit a 1? 8) Is the 8th digit a 1? 9) Is the 9th digit a 1? 10) Is the 10th digit a 1? 11) Is the 11th digit a 1? 12) Is there an odd number of ones on the subset of digits 1, 2, 3, 4, 5, 6, 8? 13) Is there an odd number of ones on the subset of digits 1, 2, 3, 5, 6, 9, 10? 14) Is there an odd number of ones on the subset of digits 1, 2, 4, 5, 7, 9, 11? 15) Is there an odd number of ones on the subset of digits 1, 3, 4, 5, 8, 10, 11? There are other question sets that are mathematically identical to this one, but I like this set because it's easy to represent graphically for people who don't understand it (I use it as my example when I teach this). For those who need a graphical representation to see how it works, here's an example:A A A A A A A | A B B B B B B B | B C C C C C C C | C D D D D D D D | D 1 0 1 1 0 1 1 1 0 0 1 | 1 1 1 0 Each column is a question, with 1 representing "yes" and 0 representing "no". The letters are a unique combination of letters for each column that are used to detect and correct errors with the four checkbits (that is, the four questions at the end). What you do now is look at each column with an A in it and count the number of 1s under them (including the four at the end). The result should be even -- here, though, it is not. This means that there was a lie told somewhere, and it's in a column with an A in it. Repeat this for each of the other letters. You'll find there are an odd number of 1s under A, C and D, but an even number under the Bs. This means the error happened in a column marked ACD - and since the only possible numbers are 0 and 1, you change the 1 to a 0. This gives us 10 100 111 001 in the first 11 digits (those which represent the number itself), which if we convert from binary to decimal we get 1,337. (The actual answer was not "leet", but I'm using that here as an in-joke.) If the error appeared only under column A, for instance, we know he lied on one of the four error-detecting bits.Does that help, if you still hadn't solved it?
Nathreet 05-30-07, 03:57 PM	Check my answer. I did something similar, then realized there's a problem if the Sphinx lies on any of questions 12-15. I then changed it to the correct answer.
_Jayne_Cobb_ 05-30-07, 04:34 PM	A man rode into town on Friday. He stayed in town for 3 days. The man left on Friday. :P Clearly, "Friday" is the name of his horse. I know that for me a declaration by the DM that he wants to introduce complex math to the game would be greeted with only slightly more enthusiasm than an announcement that the players will all be forcibly sodomized with a curling iron before each session.
Optimized_Commoner 05-30-07, 04:59 PM	Does that help, if you still hadn't solved it? Well, I was trying to come up with a way of using binary without using binary. If that makes any sense (I don't imagine that fae are familiar with it). So I was trying to come up with 11 questions that would effectively give the answer in the same fashion as binary, without using binary itself.
northern_bear 05-31-07, 06:25 AM	Clearly, "Friday" is the name of his horse. Good answer. That's not the answer I was thinking of. I would have said, that he simply left town 7 days later. The question says that he stayed for 3 days. It does not say that he ONLY stayed for three days. He could have stayed for 3 days, and then stayed for another four days. Then left. And it would still be gramatically correct (although misleading) to say that he stayed in town for 3 days. I like your answer better though. It's neater.
Drausktanchache 05-31-07, 10:52 AM	Make all player/DM notes in binary, that will make it harder, and use math... or speak only in binary...
zzo38 05-31-07, 02:21 PM	About the problem with the 15 questions and up to one lie, I haven't figured it out yet but I have been thinking about it and realized that you can have 2 bits and 5 questions up to 1 lie: - 1. Is first bit set? - 2. Is first bit set? - 3. Is second bit set? - 4. Is second bit set? - 5. Is first and second bits different? But I also thought of something else, if you can't figure out the answer then attempt to break the machine (tell your familiar or whatever to ask the questions so you won't get cought in the explosion): Half the questions will be "Is the correct number 0000?" and the other half "Is 'no' the correct answer to all of these questions?"