bonanova

Hi interested, Thanks for the kind words. Hope you're enjoying Brain Den as much as I do.

Hi tearz, Feel free to post your answers, even if you're not first the the finish line. Everyone approaches the solution in his/her own way, so if you outline how you did it, no need to worry about being thought of as a copycat. We're on the honor system here, anyway. No glory in copying...

Yeah, you're missing this paragraph: A computer, one that actually does have a random-number generator, has placed n blue points and n red points at random on a sheet of paper so that no three points are collinear. You are to connect them, pairwise, red to blue, with straight line segments. If I get to place the points myself, I'll place them at {0, y} and {1, y} for infinitely many y values. You specified n. The computer threw 2n dots at random onto the x-y plane, half red and half blue. You connect the pairs with non-intersecting straight line segments, no two of which intersect. How big a value of n can you specify, and still guarantee that you will succeed? btw - i got a date for tomorrow def, and maybe next tuesday

leamboy and foolonthehill have the correct method. Bonus question: Ralph's method [see OP] was wrong, but was it wrong high? or wrong low?

Overheard at Morty's this afternoon: Did you hear? Davey took Alex to the cleaners last night!

drydung and chuck rampart independently have it. Honorable Mention to cryptic chic's daughter!

Kudos to the effort put into this proof. I think what's been said shows that the paintable area lies within the smallest rectangle that contains the initially painted squares. OK - that's not a spoiler, since I think everyone who's posted has said it, or at least realized it. Now. How does a consequence of that fact get to the proof?

Being a puzzle freak I have no social life to speak of, so ... last night I made some blue dots on a sheet of paper. Be still my heart. But wait, there's more. Then I made an equal number of red dots on the paper, being careful not to have any three points, regardless of color, in a straight line. I imagined they were boys [blue dots] and girls [red dots] who didn't solve puzzles, and therefore were capable of a social life. So I endeavored to pair them up, using straight line segments, each line segment connecting a boy [blue dot] with a girl [red dot]. This way, I reasoned, they could conveniently meet for a date. But I found a problem: some of the lines crossed over previously drawn lines. That might be confusing, I thought. A boy might take a wrong turn at a crossing point, and end up meeting a girl who had already met her date. [i assumed the dots were heterosexual.] That suggested a puzzle. Heaven forbid it should suggest a date. Here's the puzzle: A computer, one that actually does have a random-number generator, has placed n blue points and n red points at random on a sheet of paper so that no three points are collinear. You are to connect them, pairwise, red to blue, with straight line segments. How large can n be so that, no matter where the dots are placed on the paper, a set of connecting lines can be found such that no two of the lines intersect? Assume the paper is large enough and the pencil is sharp enough.

Hi taliesin, It's close, maybe it's complete. Consider: You've given procedures for painting i2 squares from i initially painted squares. I think it's clear the procedures work for arrays of any size. So you proved i initial squares are sufficient. And you've shown - by looking at all the cases - that 7 initial squares can't paint an 8x8 array. You just need so show that holds for arrays of any size. Maybe you did ... Here are two other approaches: What happens [or doesn't happen] when a square is automatically painted? There is a particular answer to this question that will lead to a proof for arrays of any size. In particular for arrays that are so large we can't explore all the possibilities. Observe that it requires at least 2 initially painted squares to paint a 2x2 array. Then prove that if it's true for an nxn array it must also be true for an [n+1]x[n+1] array. [Proof by induction.]

It's been far too long, said Alex to no one in particular as he tossed his hat skillfully onto a coak hook at Morty's last night. Ian and Davey knew that was their cue, and they joined Alex at his favorite table. Not surprisingly, he pulled out a deck of cards. I'm feeling generous tonight laddies, he began. I'll play a game with you that guarantees that you can double your money. So to make it fair, I'll collect your initial stake from you, and then you can keep what ever you win. That way you can assure no worse than to break even. Jamie had been listening, from the bar, and what he heard drew him over to join the others. OK tell us, he said, how does the game work? I'll shuffle this deck here, and then deal them face up one at a time. You start with $1. You can bet any fraction of your current worth, before I deal the next card, on the color of the next card. Even odds on each bet, no matter what. So how does the guarantee work then? asked Ian. Drink yer ale, matey! Alex replied, If yer paying attention at all, you'll know the color of the last card. You can just wait, and then bet everything on the last card! I'll take yer stake, you get your winnings, and you break even. Now, who wants to play? Jamie and Ian figured that if they were really lucky they could win every bet and parlay their stake into about $50 quadrillion. But that would be offset by the fact that if they lost any of the 52 bets, they'd win nothing, and Alex would get their stake. Seems like a bunch of coin tosses to me, said Jamie. Not worth my time just to break even, said Ian, And the two of them left. Davey, trance-like for a moment as he stroked his beard, finally said, I'll play. Deal the cards. Did Davey have any expectation of winning money? If so, how much?

Your post is very nearly a proof - it asserts all the right points. All you need to do is include an observation about what happens whenever a square is automatically painted, and you have a proof. Anybody see what that statement is?

I have invented magic paint that works on square arrays of squares, like checkerboards. Here's how it works: If a square has two or more painted neighbors, it becomes painted automatically. Neighbor squares are those who share a side: each square has at most 4 neighbors. If I paint the diagonal squares, then, since all the squares that border a diagonal square also border another diagonal square, the bordering squares get automatically painted and that creates a new, fatter diagonal. As this process continues, the entire square eventually becomes painted. Can you get an entire 8x8 checkerboard automatically painted, using my magic paint, by initially painting fewer than 8 squares? If so, specify the initial 7 [or fewer] painted squares. If not, can you prove it's impossible?

Read my post again. You have only asserted it was not random. Your task is to prove is was not at random.

Yet another assertion -- no wonder these discussions go on forever. Can you prove that infinity is not even?

Thank you for admitting it is possible to extract a finite subset at random from an infinite set. You have thus contradicted your contention that such a process is impossible. Selecting .314 at random from the reals in [0, 1] is the selection at random of a finite subset of an infinite set. I chose the number of decimal places at random from the infinite set of positive integers. Then I chose .314 at random from the 1000 possibilities that provided. If you say that the choice of 3 decimal places was not a random choice, that's one thing, but if you want others to share that belief, it requires proof. I think your options here are three in number [and since 3 is finite, you should be able to make the choice] Prove that 3 can not have been a random selection from among the positive integers, admit that you're simply wandering around some previously unexplored territory with BrainDen as an audience, or agree that you're just pulling our collective leg.

I stand in awe [i think it's awe ...] of your thinking. I'll close my thoughts by relaying a dream I had last night. I know it was a dream, because in real life I don't have a butler. Enjoy. One night I had a group of friends to dinner. Because they had not all previously met, I provided them with name tags. At one point, hoping to determine whether they had all arrived, I sent my butler to the drawing room to count them. Forty-two guests are present, he informed me, we are waiting on the final six to arrive. After the string quartet had finished their last minuet, I asked the butler to count them once again. When he returned he said: I'm sorry sir, but I'm afraid the guests have removed their name tags, so I have no idea how many are here. I fired the butler.

Well that's fewer than 60, but ... still too many.

On a good set of scales, the pans have equal mass, and the arms have equal length. Down at Ralph's Pretty Good Grocery Store, the scales weren't all that good. The pans had equal weight: each weighed m pounds. But the arms had different lengths - the left arm was a cm long and the right arm was b cm long. When Ralph was asked to weigh out 2 pounds of grapes, he had an idea. First, he balanced some grapes in the left pan against a 1-pound weight in the right pan. Then, he balanced some other grapes in the right pan against a 1-pound weight in the left pan. Together, Ralph reasoned, the grapes would weigh 2 pounds. Was Ralph correct? Is there another way?

The eccentric old man wanted 12 rows of apple trees in his orchard, with 5 trees in each row. But I don't want a four-sided orchard, he told his gardener. How many trees did the gardener have to plant?

Every station on the commuter line sells tickets to every other station on the line. Last week some new stations were added, and 46 additional sets of tickets had to be printed. How many stations were there before last week? How many stations were added? No bickering now ... at no time does the number of stations go to infinity.

There's no difference between removing half of the balls by taking every other ball or taking the first half of the balls. The number of balls that remain is the same. But since you're still wriggling around in your quicksand, try this: Wipe all the numbers off the balls. Move two balls from A to B. Move one ball from B to C. Repeat. Note this is equivalent toMove one ball from A to B. Move one ball from A to C. Repeat. I'll either enjoy - or be embarrassed for you - if you try to get bin B empty this time. Have at it.

Hi S_W, From my other posts, you may be aware that I'm interested in proofs that you can't choose a number at random from an infinite set. You seem to offer such a proof in your post, and I'd like to explore it. If I understand your post, you prove your main assertion by making a secondary assertion. Namely that, among the set of positive integers, elements have differing likelihood of being chosen. Let me choose one at random: 314. If I understand your post, you would object to 314 as a random choice, because you have established that the probability of choosing 314 differs from the probability of choosing say 1024885926414553033465570345667158541759501107168465168464013834058687365. Let's explore that thought. Define p1 as the probability of choosing 314 from among the positive integers and p2 as the probability of choosing 1024885926414553033465570345667158541759501107168465168464013834058687365 from among the positive integers. You have established that these probabilities differ. So, can you tell us by how much? That is, what is the value of p1 - p2?