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Everything posted by bonanova

  1. Three numbers that add to 15.
  2. I've gotten hooked on Ken Ken. It's like Sudoku with extra constraints. Numbers 1-9 must occur in every row and column. Clusters of numbers must also obey rules. Within a cluster, 15+ means they must sum to 15. e.g. 6 and 9. 5- means the (two) numbers must differ by 5. e.g. 3 and 8. 72x means their product must be 72, e.g. 2 4 9. 4/ means the (two) numbers must have 4 as a quotient. e.g. 2 and 8. Clusters for + and x can have any number of members. For - and / there obviously can be only two members in a cluster. I find the "expert" 9x9 puzzles can provide 30-60 minutes of fun. Here's one for you:
  3. This is one version of Martin Gardner's "Unexpected Hanging" paradox. The resolution turns on the truthfulness of the person saying the test (hanging) will happen at a time that cannot be anticipated.
  4. Nothing. The + does not move at all.
  5. Illustrating gavinksong's description of the progression of points: The graphic shows the progression of smaller and smaller triangles that fill the outer boundary. A1: 1 red triangle, 1/4 of the size of the outer boundary A2: 3 green triangles, each 1/4 of the size of the red triangle A3: 9 blue triangles, each 1/4 of the size of the green triangles. A4: 27 red triangles, each 1/4 of the size of the blue triangles A5: 81 green triangles, each 1/4 of the size of the red triangles. .... The trajectories of two dots placed in A1 is shown. A black dot starts in the upper right corner, and a red dot starts in the center. With each step, each dot progresses through red, green, blue, red green ... triangles, maintaining its relative position within each triangle. This underscores how sparse the dots are. Only a single dot is placed in each of the areas A1, A2, A3, ... That is, at each level, there is a collection of triangles. One and only one of them receives a single point. The other triangles at that level are totally white. This accounts for the appearance of white triangular spaces in the simulation. Since a single dot will never color an entire triangle, no matter how small the triangle is, no triangle can be said to be black. Every triangle is white. The entire triangle is therefore white.
  6. Is this easier than just computing the average directly?
  7. [spoiler='The clues explained'] Rotten, Denmark The setting It either begins in 'and' or else __ __ in 'or' Elsinore The castle moe: one final question, sir? what would you say is the difference between involvement and commitment? joe: certainly. Let me just say this about that: the Capitalized Words to make Ham and eggs, the chicken is involved, while the pig is committed. upon a few bits 0010 1011 of text was cast a spell hex. => 2B The quote is English, actually, and, although quite well known, the answer might require some calculation OR. FF = 1111 1111 = (0010 1011) v (1101 0100) = 2B v ~2B And that (FF) is the question! FF? tragically, There are more things in the clues than are dreamt of in the answers.
  8. Yes. This is exactly what I found. bona_gold_star.gif Three fourths of the triangle are self-similar copies and the rest is white. The fractional whiteness of the four parts must be equal. That's the simplest and most elegant of the three proofs. If you like, try dissecting the trapezoid and do the same proof. It has twelve triangles of three varieties and takes two equations to solve. Cool. I did white space, used 12 triangles of three types and had to throw a 1 in the mix. Your way is simpler.
  9. I love this. I didn't do the series to quantify things, so, being lazy I'll just ask: Do all the questions in the OP have answers? e.g. if x = 1.0000001, is it impossible? Or just take much longer?
  10. We traveled a kind of intuitive path in doing this analysis. So to be clear of the result, we have to make statements that are a little more precise. In fact, the statement of the puzzle uses the intuitive term "white space." This leads to the idea that the triangle is partitioned somehow into disjoint regions that are intrinsically white or black. Or, as you say "the 'white space' that the points could never occupy." -- regions where points are not allowed to land. Such regions do not exist. Not even a single point is off limits: it could be the initial point. This beautiful puzzle forces us to view the triangle not as two sets of tiny triangles, some white and some black, (more precisely some that have No Trespassing signs, others that do not) but as two sets of points: one set that is rationally related to the arbitrarily selected first point, and all the others. Consider a one-dimensional analog. Color the interval [0, 1] of the x-axis white. Now paint the point x=.5 black. Now move half-way to 0 or 1, randomly chosen, and paint that point black. Repeat ad infinitum. The rational numbers will be painted black The non-rational numbers like 1/pi and 1/e remain white. What fraction of the interval is white? What you will find is there is no interval of infinitesimally small length that is free of black points, yet we can say that the "measure" of the interval that is white is 1. Or, define f(x) to be 1 if x is rational and 0 otherwise. What is the integral of f(x) on the interval? Our normal measuring device for lengths and areas is the ordinary Riemann integral. But it only works for functions that are reasonably continuous. When you get down to point-size regions, Mr. Riemann throws up his hands and says, don't bother me, I have no idea. Go talk with Mr. Lebesgue over there. And Mr. Lebesgue says, don't talk to me about length and area, I don't know what they are. I do know about measure, however. The measure of the rationals on [0, 1] is zero. The measure of the reals on [0,1] is 1. The measure of non-rational reals on [0, 1] is 1 - 0 = 1. On the triangle, the measure of the points with rational coordinates is 0. The rest of the points have measure equal to the area of the triangle - 0. In the Lebesgue sense, (and no other "sense" applies to this question,) the "area" of the triangle that is white is the area of the triangle minus something that has the value zero. What I really liked about your analysis was talking about the decomposition into triangles of ever decreasing size. Call the sequence of areas A1, A2, A3, ..., each value 1/4 of the preceding. That decomposition does not depend on the initial point. What you said, and I think you are correct in this, that wherever the initial point is placed, say it's in a triangle of area A13, none of the following points will land in triangles with area A1, A2, ... A12, or the other regions with area A13. The following points will be in triangular regions respectively with area A14, A15 ..., and so on. That analysis shows that once the initial point is placed, there ARE point-free regions. There are regions that will never (for that initial point placement) have a point in them. But those regions (A1, A2, ... A12) are not intrinsically "white areas". The initial point could land in a triangle of area A7. Then the A1, A2, ... A6 regions, and the other A7 regions will be point-free. And the initial point (my musings...) could land in A1. The probability of that happening is 0.25, but it is not 0. One final observation. my one-dimensional case in blue above behaves differently from the triangle, and makes the triangle at once more beautiful and intriguing to analyze. In one dimension, the point can return to any region arbitrarily close to the initial point. If we simulated this case, the whole interval would eventually look black because we use dots of non-zero size. Yet the interval remains white in the Lebesgue sense. In the triangle that is not possible; the point cannot back-track. It never moves directly away from a vertex, so it can never move back in the same direction. That is why the A1 region looks white every time the simulation is run. And the A2 regions do, also, and the A3 regions. Each of all the triangles with a given area contain at most one black dot. Perhaps this is why the square shape showed no structure.
  11. Yes. This is exactly what I found. Three fourths of the triangle are self-similar copies and the rest is white. The fractional whiteness of the four parts must be equal. That's the simplest and most elegant of the three proofs. If you like, try dissecting the trapezoid and do the same proof. It has twelve triangles of three varieties and takes two equations to solve.
  12. This touches on a crucial idea required by the mathematicians' strategy. It must be understood before being used. It's the idea of equivalence classes. We'll first explore what they are by finding a way to construct them. Then we'll see how the math guys used them. So, let's take a minute to gather all the infinite sequences of real numbers of infinite length that exist, and lay them out here on my dining room table. Pick a sequence at random; call it sequence A. Quickly glance at all the other sequences, ignoring the initial finite number of terms. When a sequence, (call it sequence X) is found for which the remaining infinite terms agree with the corresponding terms in A, then we group X with A. Put that process into an infinite set of infinite DO loops [my FORTRAN is showing] and finally we get an infinite number of classes of infinite sequences. These are the equivalence classes needed to implement the math guys' strategy. Mull over that process so that you can define these classes in your own words. The classes by themselves are too clumsy to use. We have to simplify matters. Enter the Axiom of Choice. Without it, the math guys remain locked in their rooms. AC is just this: There is a particular sequence, chosen from each class, that can be used to represent that class. So, for every infinite sequence of real numbers, it belongs to a unique equivalence class, and there exists a single, unique infinite sequence that can be used to represent it. Those are the vital ingredients: classes and their uniquely chosen representatives. AC has importance only for infinite sets. Unique representatives of finite sets are not troublesome. But for infinite sets unique representatives are problematical. It cannot be proven. It must be inserted into the process as an axiom. And by the way, asserting a unique representative is not the same as finding it. So our math guys are only theoretical winners by this strategy. We don't get picky about that, even though the numbers in the boxes are real, the math guys aren't. This is only a puzzle. Without AC, the math guys can't translate an infinite sequence into a unique representative sequence. So they can't predict any term of any real sequence they encounter in their rooms. But with AC, they can. They just run a quick check of the final infinite number of terms. When they find congruence, they know its equivalence class. Using AC they find [but see the above disclaimer as to how they do this] the representative of that class. That identifies all the remaining real numbers that follow a finite number of differing ones. They just have to get sufficiently far out into the sequence to know that have skipped the differing numbers. The rest of the strategy is easy to understand. One of them might have the wrong cutoff point, where the differing numbers end, and give the wrong number contained in a selected box. But there is at most one such guy. And 99 out of 100 is a win.
  13. DeGe has an insightfully creative solution. Without loss of generality the triangle can be scaled by a rational number times the inverse of the coordinates of the vertices and first random point. Or we could say without loss of generality the vertices and initial point can be assigned rational coordinates. Either way, all the computed points will have rational coordinates, and as such, although infinite in number, have the cardinality of the rational numbers, Aleph Null. Their Lebesgue measure is zero. Another way to say it, as DeGe points out, they are points. But so is the interior of the triangle. The distinction is that the rationals are not dense - they are isolated points. The entire set of points inside the triangle has the cardinality of the continuum. Those points have real coordinates and are dense; thus, their Lebesgue measure equals the area of the triangle. The computed points can be removed without changing that measure. Therefore the area of the white space is the area of the triangle. One Gold star remains to be achieved. A geometrical proof that does not use the word infinity or its implications.
  14. Well, for a triangle, if we add up the white space, it is just the infinite sum of a geometric series with a factor of 3/4 (the big empty triangle in the middle is one-fourth of the total area, the three smaller triangles are one-fourth of the remaining area, and so forth). Strangely, it converges at the total area of the triangle. It we calculate the black space, it is three-fourths raised to the infinite power, which is zero. So apparently, the entire triangle is supposed to be prohibited space. I think here bonanova's musings becomes relevant: Instead of looking backwards, as we did to solve the problem initially, we should look forwards from the initial point. This is all intuition, but you will notice, that if the initial point is in the big empty triangle in the middle, then in the next step, it will always move to one of the three smaller white triangles. And from there, it will always move into one of its three smaller "shadows" (I think that is an appropriate term, and hopefully you understand what I mean), and so on. The important point here is that we always move into smaller and smaller "shadows". When the initial point is thrown into the triangle at random, the probability that it lies into a white triangle is 100%. However, as the simulation progresses, the points are eventually forced into smaller and smaller white triangles. In other words, the amount of "prohibited space" is not actually an absolute value, but simply increases with step number (in a sense). As in, there is no prohibited space when you throw in the first point, but the second point cannot be in the large white triangle, the third point cannot be in the large white triangle or any of its shadows, the fourth cannot be in the large white triangle or its shadows or any of its shadows' shadows, and so on. So to us, it seems as though there is some sort of absolute law because we see the larger empty white triangles, but in reality, it is only because we do not see the infinitesimally small white triangles with the dots in them. The entire process is just endless error-correction, moving toward an impossible figure, trailing behind an interesting asymptotic pattern in the process. gavinksong has the answer. His method was the infinite sum of triangular areas, and he gives an intuitive description of the point generation process and its implications. Kudos for explaining my wonderings about placement of the initial point. It's an amazing mental image to visualize an initial point at the center try to search for a "black area" to land on, but instead finding at each step it simply lands on a smaller white area. Every point travels a path of white areas; or putting it another way, there simply are no black areas! [As an aside, therefore, my puzzle is a bit of a ruse. The figures clearly show black areas. A moment's reflection tells us why. I constructed the figures using dots that have non-zero area. Eventually they touch and overlap each other. It's OK that the construction used black dots: it's just that to avoid being misleading they should have had zero area. If I had done that, it is perfectly clear that the entire triangle would remain white forever, or until the computer reboots, whichever comes first.] My first crack at calculating the white space used a geometrical approach, the answer to which surprised me enough to do the math. When that agreed, a third argument suggested itself to explain what seems an impossible result. Two more Gold Stars are available for proofs that don't require infinite sums. As for the trapezoid, I think we can safely modify whatever is proved for the triangle, without considering it a different problem.
  15. BMAD's gave an algorithm for recursively placing points, ad infinitum, within the interior of a triangle. The beautifully surprising result was the appearance of well-defined areas, within which points could not be placed. Call these areas white space. The puzzle inspired Pickett to page-2#entry334664, and the page-3#entry334674 that generates them, of several other shapes. The most interesting seemed to be BMAD's original triangle, and the page-2#entry334665, reproduced below. My puzzle is a simple calculation: What fraction of the triangle is white space? What fraction of the trapezoid is white space? Edit: Creativity will be rewarded: Got a different approach? Get another star.
  16. And now, for the coveted bonanova Gold Star, let's have ... da da da daaaa, da daaaaaaaa ... the quotation.
  17. The total area that is white.
  18. has all the additional conditions necessary to solve. (Including the Edit in my previous post.)
  19. Solved clues are edited in red. The main clue is bold. Still to be decoded is the speaker and the text of the quotation: FF?. ----------------- Do you sense a Scandinavian odor? [Something is rotten in Denmark] It either begins in 'and' or else __ __ moe: one final question, sir? what would you say is the difference between involvement and commitment? joe: certainly. Let me just say this about that: to make Ham and eggs, the chicken is involved, while the pig is committed. upon a few bits of text was cast a spell. The quote is English, actually, and, although quite well known, the answer might require some calculation. tragically, There are more things in the clues than are dreamt of in your answers. [Mimics a line spoken to Horatio in a famous tragedy].
  20. What fraction of the area is white (prohibited) For the triangle? (straightforward) For the trapezoid? (more thought-provoking)
  21. Shall we assume that mates are always of the same generation? That is, would a man be prohibited from mating with his mother-in-law? Does the OP mean to say that once a person becomes a grandparent, then s/he dies? Wondering why this is necessary, unless cross-generational marriages could happen. When children come of age, may their parents mate (again), with a different opposite-sex person from their own generation, and bear more children? Or does each person parent two and only two offspring? Shall we assume that the blood relationship prohibition for marriage is absolute? Often, by civil law, second cousins, but not first cousins, may marry.
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