bonanova

phaze, I added a little guidance in the spoiler at the bottom. If may help you think about what to figure out first. Or, better, what can you figure out first. Order matters, and it's not the usual order for this type of puzzle. Ask yourself, what are all the things I eventually need to know. Then, which of these can I figure out from the clues given. Make any charts or schedules or grids that need to be filled in, and go as far as you can ... Good Luck.

That does it. BrooksPuzzel, are you going to make more like this? This type is a favorite of mine, and there aren't many like it in the forum.

Your question makes sense, and it's an important point. When a person acts as a Knave, that person looks back to his OWN previous statement. In your example, What the Knave said on Monday doesn't matter. What matters is whether the Knight that is acting as a Knave has just told the truth, or lied. Since the Knight is acting like a Knave, he does what Knave would do - look back to his OWN previous statement and alternate the truth value for his next statement. Again, quoting the OP: L[N] is a Liar ACTING like a Knave. If the LIAR's last previous statement was a lie, then L[N] will tell the truth. etc. Also, if L[N] is making his [the Liar's] first statement, he [the Liar] can choose T or F, just as a real Knave could do when making his first statement. So L[N] means the character is a Liar, but he ACTs exactly the way he would if he were a Knave. Hope that helps.

To get the intent of this, you may change "available" to "present" Quoting from the OP: For example, if K[L] and L[C] spoke on the same day, L[C] preceded, i.e. spoke before, K[L], because [C] comes before [L]. Since the Liar ACTED as a Chameleon, he speaks before the Knight. That's because the Knight ACTED as a Liar, and C comes before L in the alphabet. The order is alphabetical according to the ACTED character.

This reminds me of a puzzle question that shows five diagrams. Four of the five have a distinguishing characteristic. Something that the others do not have. The fifth diagram has no distinguishing characteristic. The puzzle asks: Which of these five is the most different from the others? The answer is the one that is not different from the others in any distinguishable manner. Except that it is the ONLY one that is not in some manner distinguishable. That fact makes it different from the others! That is, the others are unique. Just like all the others. Except for this one. That makes this one ... unique.

Not true. By the OP, it's tied to your choice. It can't be "a previously set amount" unless your choice has previously been set. If your choice has previously been set, there's no question to answer.

The man is correct that if he gets lost he can always return to the beginning. But it is not correct to believe the second path is only half as long. Suppose mile markers were placed on the two paths. The first would have markers at 0, +/-1, +/-2, +/3, ... +/-N, .... The second would have markers at 0, 1, 2, 3, ..., N, .... For any finite distance N from his starting point, the first path would be twice as long: 2N miles as compared with N miles. But as N increases without bound, limN->inf [N] = limN->inf [2N] = inf. Similarly, there are the same "number" of even integers as integers. With infinite sets, the algebra of cardinality that holds for finite sets breaks down.

Let J be the number of days John reads and R be the number of days Robert reads. For any finite number of days, [1] J = R + 1. Now the question is worded, "they both read for an infinite amount of time" [then] who will have read for longer? The verb tense is future perfect. That is, it refers to a time in the future [future] when the deed mentioned [read for an infinite amount of time] is completed [perfect]. That is, the reference point of the question is after an infinite amount of time has passed. Lay aside the impossibility of being able to look back after an infinite period, we would have to do it from a finite period and then grow the period without bound. We thus write [2] limJ->inf [J] = limJ=>inf [R] = inf. Equations [1] and [2] do not contradict. inf is not a number. So inf+1 = inf is not a contradiction. The paradox comes when we write [3] J - R = 1 [4] limJ->inf [J - R] = limJ->inf [1] = 1 = limJ->inf [J] - limJ->inf [R] = 0. [if we were to imagine that inf-inf=0.] The answer to this paradox is: inf - inf is not a proper construction. inf is not a number or the symbol for a number. So inf - inf cannot be said to be 0; it can be anything. For example, the number I of positive integers 1, 2, 3, 4, 5.... is inf. The number O of positive odd integers 1, 3, 5, ... is inf. The number E of positive even integers 2, 4, 6, 8, ... is inf. Now it's tempting to say I = O + E. And that's true for any finite even cutoff point, say 1000: I = 1000, E = 500 and O = 500. But including all integers, then I = U = E = inf. So here you see I - E = O gives: inf - inf = inf! Moral to the story. Don't do arithmetic with inf. The resolution of this is also that, for the infinite set of integers, I, E, and O are not numbers, either. They are the cardinality of the lowest order of infinity, which is the cardinality of the natural numbers. For that reason it's called a countable infinity.There are larger infinities.

So maybe to Nimrod meters become light years or inches. So what? Why would anything change for another observer? Suppose Nimrod had never been born. Would anything change then? I think Nimrod might be alone in his confusion.

Thanks for asking! First, zero is the identity element for addition. But there's

The statement in the OP each citizen made a series of statements, one statement on each of three consecutive days. Should be interpreted: each citizen speaks on three consecutive days and makes one statement each of those days. Three days applies to each citizen, not the entire group. As we see later on, that makes 12 statements in all. You're correct as far as you go, just don't stop after three days. I'll edit the OP to make that point clearer. Thanks! - bn

Hi BrooksPuzzel and welcome to the Den! Is there another clue? Seems like we need more information.

. Think of sacrifice... This "solution" is amazingly seductive! It continues to be posted by people who don't even check to see whether it works.

They are fair coins.

The Mother of all Truth-teller puzzles [Edits to clarify a point shown in red] Citizens of the Land of KNLC are of four character types: Knights [K], Knaves [N], Liars [L], and Chameleons [C]. Citizens normally behave like this: . [K] will tell the truth. . [N] statements alternate in truth value; the first statement that a Knave makes is freely chosen [TLT... or LTL...] If another citizen is available [i.e. is present and available to speak], a Knave will permit him or her to be the initial speaker. . [L] will lie. . [C] will mimic the truth value of the most recent previous speaker. If a lie has just been told, [C] will lie. etc. If no one has yet spoken, either today or the preceding day, then [C] will sit silently in the corner. Sometimes citizens ACT as if they were of a different character. Let's introduce the notation Character[ACTING_LIKE]. Then, for example, . K[L] is a Knight ACTING like a Liar: K[L] will lie. . C[K] is a Chameleon ACTING like a Knight: C[K] will tell the truth. . L[N] is a Liar ACTING like a Knave. If the LIAR's last previous statement was a lie, then L[N] will tell the truth. etc. . L[C] is a Liar ACTING like a Chameleon. If the LAST PREVIOUS SPEAKER spoke truthfully, L[C] will also tell the truth. etc. This deviant behavior is confusing, but it's predictable. Here's the ACTING schedule in the land of KNLC: . Mon: K[K], N[N], L[L] and C[C] - Every citizen ACTS according to his own character. Tue: K[N], N[L], L[C] and C[K] - Knaves ACT like Liars, etc. Wed: K[L], N[C], L[K] and C[N] - Liars ACT like Knights, etc. Thu: K[C], N[K], L[N] and C[L] - Chameleons lie, etc. Fri: K[K], N[N], L[L] and C[C] - Everyone's back to original character. Sat: K[N], N[L], L[C] and C[K] - Back to Tuesday behavior. Sun: Citizens are not allowed to speak on Sunday. [/code] [color=#afeeee].[/color] [b]Please meet four KNLC Citizens: Andy, Dave, Iago and Lisa. [/b] Here's a little anecdote to help you know them better. Yesterday they were given a single written Y/N question to be answered verbally. [color=#afeeee] .[/color] [/font] [font=Verdana]Two of them, now senile and without any memory to speak of, independently blurted out contradictory answers, as they always do. [color=#afeeee].[/color] [/font][font=Verdana]Another was deaf, but answered easily after reading the question, only because some memory still remained. [color=#afeeee].[/color] [/font][font=Verdana]The last citizen was deaf, memory intact, but needed a hearing aid to answer. [/font][/list][font=Verdana][color=#afeeee].[/color] [b]Now for the puzzle:[/b] Last week, each citizen made a series of statements, one statement on each of three consecutive days. [color=#8b0000][[b]Edit to clarify:[/b]] In all, twelve statements were made, over a period of six days.[/color] [color=#afeeee].[/color] [/font][list][font=Verdana]No two citizens made their first statement on the same day. [color=#afeeee].[/color] [/font][font=Verdana]On days when more than one statement was made, the statements that day were alphabetically ordered according to the character that was being ACTED: [color=#afeeee].[/color] [/font][list][font=Verdana]In any given day, therefore, the order was *[C], *[K], *[L], *[N]. Where * is a wild card notation, denoting any citizen who may have spoken that day. [color=#afeeee].[/color] [/font][font=Verdana]This doesn't imply there were ever four speakers in a day; it just gives the order of any who did speak. For example, if K[L] and L[C] spoke on the same day, L[C] preceded K[L], because [C] comes before [L]. [/font][font=Verdana][color=#afeeee].[/color] [/font][*][font=Verdana]The characters that were being ACTED in the final two statements are in strict alphabetic order. [No ties.][/font] [font=Verdana][color=#afeeee].[/color] [b]Here are the twelve statements, in the order they were made:[/b] Note: Only declarative statements have truth value. Questions, exclamations and imperatives [commands] can safely be ignored. [color=#afeeee].[/color] [/font][list=1][*][font=Verdana]Snazzlefartz! [color=#8b0000][Note from OP: the meaning of this term is uncertain.][/color][/font][*][font=Verdana]Don't ask me what Snazzlefartz means. I never use that word.[/font][*][font=Verdana]Looking for Iago? Look somewhere else. He's not I.[/font][*][font=Verdana]Hi! I'm Lisa. Are you enjoying your visit to KNLC?[/font][*][font=Verdana]None of our initials agree with the initial letters of our real characters.[/font][*][font=Verdana]If the Knave's final statement is True, then my name is Lisa.[/font][*][font=Verdana]If the majority of these statements are True, then most of them are False.[/font][*][font=Verdana]If the Chameleon's final statement is false, then the Knight's first statement is true.[/font][*][font=Verdana]I hate having to appear in these puzzles. You'd think they'd at least pay us something.[/font][*][font=Verdana]I'm not Lisa. Can I leave now?[/font][*][font=Verdana]I'm not Lisa, either. Can I leave, too?[/font][*][font=Verdana]If all 4 of us have made at least two statements that appear consecutively in this list, then I am not Dave. [color=#afeeee].[/color] [/font] [font=Verdana] Now you should know the character types of the four Citizens and the order in which they spoke. [color=#8b0000][color=#0000ff]Which was Andy's final statement?[/color] [b][color=#000000]Enjoy![/color][/b] [/color][/font][font=Verdana][spoiler=A very weak hint and a hint about hints]A precise sequence of 8 steps leads to the solution. i.e., there are eight puzzles, and solving the first permits solving the second, which permits solving the third, etc. Thankfully, they get easier as you go along. One hint would be to describe the sequence: e.g., first determine ...... then find out whether .... I'll wait a while before giving these hints.[/spoiler][/font]

It was a joke. I thought branching to the counter reset on failure was ... almost subtle. This bit of silliness reminds me of an NCAA basketball tournament office pool a few years ago. The $5 entry fee was due end of business day to an office in another building. I couldn't get there in time, so I faxed him a $5 bill. I amuse myself, at least, sometimes ...

SP, I'm kind of with you on this puzzle. But I don't think you can use just any combination of numbers [or letters] at the corners. [Perhaps only] the 117 'pairings' that solved the puzzle. But take that only as a guess.

Not that many that satisfy the OP: Here are two threesomes: ABC and ABD. Both pair A with B. " ... and no one is paired with the same person more than once."

This is a variant of prisoners with hats where they are freed if one and only one guesses right. It's a beautiful solution.

Here's the first dent in this hideous problem. Re-reading the OP,

Maybe too late to help. a. 10.1118.., so 10.1 looks good. b. Use only the northward component of the boat's velocity, not [a], to compute this time = 32.5 sec. c. Use 32.5 sec and the river's speed to get downstream distance. d. displacement = ? 325m N + [c] E? e. Angle east of North would be arc tan (river speed/boat speed) = tan-1 (.15) = 8.53o f. Don't understand. It's constant at [a], so why ask for average? Good luck. - bn