bonanova

DejMar has it. The puzzle has no solution for the clues as given, which contain an error in clue 3. My bad; there is one even prime number. Nice solve.

Is this what you mean? Any evidently false statement (A standard deck of cards has five aces.) Is not true Therefore cannot be proved to be true. Which proves that it cannot be proved to be true Can be disproved by counting the aces. You disprove a statement by negating a necessary condition. You prove a statement by establishing a sufficient condition. So you can or you can't prove or disprove (something), depending on the existence and truth or falsity of necessary or sufficient conditions. In the OP I don't know of sufficient conditions that can be shown to exist. k-man discovered a successful tiling, but I don't see how to prove a priori that one exists. In the case of tiling with dominoes a chessboard with opposite corners removed, there is a necessary condition (equal numbers of white and black squares) that does not exist. You know a priori that you can't do it.

In fact, I wonder: Might a close study of the colored triangle disclose an algorithm for moving methodically from one color to the next until red is reached? Or do the triangles for each N differ qualitatively enough that each N would need its own algorithm?

I think it could only be disproven. As in removing opposite corners from a chessboard and tiling the result with dominos.

From any starting point a peg can be emptied. I know of two proofs. One that seeks to increase the number of rings on one of the pegs until another peg is empty, another that systematically decreases the rings on one of the pegs, after securing a mild starting position (where exactly one peg has an odd number.) Agree about plasmid's use of trilinear coordinates, and I think it's a solution, so long as all the elements of arbitrarily large triangles can be given colors. I won't mark it solved until that is clarified.

Nice progress. We now have two conditions that lead to a win.

Someone is willing to bet, at even odds, that if he throws six fair dice, exactly four distinct numbers will show. For example 3 3 4 4 5 6 or 3 3 3 4 5 6. Do you take the bet?

Nope. I hit the Post button before proofreading the text. See amended OP.

The diameter of an arbitrary closed region is the greatest distance between any two of its points. Can every region in the plane of diameter 1 fit inside a circle of diameter 2? Edit: It can be shown that among all closed regions of diameter 1 in the plane, the circular disk has the greatest area. Does it follow that every region of diameter 1 will fit inside a circle of diameter 1?

I decided to post a separate thread to show how Russell's paradox relates to the powerset proof.

The paradox of self reference Statements have attributes, among them is their truth value. And the act of making a statement is the act of asserting its truth. The sky may be blue. Making that statement is tantamount to saying I assert that 'the sky is blue' is a true statement. Thus, this statement is false is immediately a paradox, for it asserts both truth and falsity: I assert that the truth value of 'false' is 'true.' Or, what I assert to be true is false. Self-referential paradoxes lead logicians to modify the rules, to accommodate (or avoid) them. It does not help to deny the premise that by making a statement you have asserted that something is true. If you deny that you are a truth asserter, by saying I am a liar (as if that would remove the problem of your saying This statement is false) all you have done is to revoke your permission to speak. Thus you cannot be the person who informs others that you lie. It's perfectly alright for a person to say bonanova is a liar, but bonanova can't logically be that person. Impossible statements of this type provide us not only with paradoxes, but with proofs. Specifically, those of the reductio ad absurdum variety. I found it interesting after posting to find one such connection: Russell's paradox lays the foundation for a simple proof that the cardinality of a set is less than that of its powerset. Do you see the connection between the two sets R in what follows? Russell's paradox According to naive set theory, any definable collection is a set. Let R be the set of all sets that are not members of themselves. Does R contain itself? This question is paradoxical, for if R is not a member of itself, then its definition dictates that it must contain itself; but if it contains itself, then it contradicts its own definition as the set of all sets that are not members of themselves. Cardiality of powersets Suppose a set and its powerset have the same cardinality. Then there is a bijection (pairing) between a set's subsets and its elements. Color an element blue if it is a member of its paired subset and red if it is not. Consider the subset R that comprises all the red elements. It cannot be paired with a blue element (it contains no blue elements) nor with a red element (which cannot be a member of its paired subset.) R is thus a subset that is not paired with an element. This violates our assumption, and it shows by example that sets have more subsets than elements.

Post 12 clears up the suggested injection from a powerset to its set. Which leaves us just with the need for a simple word proof. The one I have in mind refers closely to Russell's paradox. I used the wrong term (superset) for powerset in my above posts. Brain fart.

@karthickgururaj, S-B theorem seems to be exactly what we're looking for. The part that seems difficult, but I can't yet say it's impossible, is the injection from the superset into the original set. In your solution, can you identify (or give examples of) sets A and B? That would complete the proof. Or, apply it to say A = {1, 2, 3}.

Captain Kirk has landed the Enterprise in your front yard and invites you onto the holodeck, where you meet a King with a beautiful daughter (aren't they all?) The King wishes to give her hand in marriage only to the wisest of her suitors. So he has devised the following test, and you're first in line. (For puzzle solvers of the female persuasion, the King has a handsome prince ... etc. Actually, you are free to write the prologue in the manner that provides the greatest motivation.) On the Princess' vanity table sits a ring holder comprising three vertical pegs, each holding a random positive number of royal rings. (Reminiscent of the old Tower of Hanoi puzzle, but here the objective is different.) You must transfer rings from one peg to another, in discrete steps, with the objective of emptying one of the pegs of all its rings. A step consists of doubling the number of rings on some peg using rings from another peg that initially has at least as many rings as the first. Say the rings on two pegs initially number a and b, respectively, where a is not less than b. After the move, the pegs will have respectively (a - b) and 2b rings. Clearly the winning position occurs when you have two pegs with equal numbers of rings (a=b.) You may make as many moves as you like. Will you have a pleasurable afternoon on the holodeck? If so, how will you empty one of the pegs? By the way, when you go home, the Princess stays on the deck.

See if there's a similar case (of oppositely directed injections) that applies to finite sets.

karthikgururaj proposes two injections (both of which are far from being surjections also) rather than a single bijection. So the question is, are two (one each direction) injections equivalent to a bijection?

The OP requires Bob to have uniform speed, so he needs an abrupt change from 0 at the start. Alice does not have that restriction, only that she runs the same speed profile each lap.

Can we infer from this, that you tried it?