BrainDen.com - Brain Teasers

# bonanova

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1. ## Pegity - white to win

OK so I was a kid once, back in the 40s and 50s, and we had this game called Pegity. It had a board with a 16x16 array of holes, and in turn each of 2-4 players inserted a peg of his own color into a hole. The object was to get 5 pegs of your color in a row: vertically, horizontally or diagonally. Not every game was won: similar to tic-tac-toe, you could run out of holes. But with only two players, there was usually a win. For simplicity let's shrink to a 9x9 board, mark the holes like in chess (A1, E7, etc.) and say that after O and X have each made three moves we have this position, with O to move: 1 2 3 4 5 6 7 8 9 A + + + + + + + + + B + + + + + + + + + C + + + + + + + + + D + + + + + + + + + E + + + + O + + + + F + + + O + O + + + G + + + + X + + + + H + + + X + X + + + I + + + + + + + + + With best play by both players, how soon can O win? Use chess notation (naming the holes, in two columns) to list the moves.
2. ## Next card is Red

@CaptainEd shuffles a standard deck of playing cards and begins dealing the cards face up on the table. At any time @plainglazed can say Stop and bet \$1 that the next card will be red. If he does not interrupt, the bet will automatically be placed on the last card. What is the best strategy? How much better than 50% can @plainglazed do?
3. ## Green and Yellow hats again, but harder

That is amazing. And you've shown that 1 fewer than a power of 2 is the optimal value for n. Can you reduce it to only q memorized strings, and the success rate to (q-1)/q, where q is the highest power of 2 less than or equal to n? (Purposely leaving this un-spoilered, cuz this is a tough problem.)
4. ## Baggage on a conveyor belt

Yup. And a bit of possibly helpful thinking: Edit: The math gets a little demanding here. Integrals and stuff.
5. ## A big, big, big hotel

Yeah, I hate annoyances, too...
6. ## Modified checkerboard destruction

Can you think of a reason?
7. ## A big, big, big hotel

OK I think I just found your first blue man.
8. ## Worth the Weight

@aiemdao - Nice solve. @plainglazed - Nice puzzle.
9. ## Modified checkerboard destruction

@CaptainEd brilliantly answered a puzzle that I mis-worded into a much tougher one. Nice job. Here's the puzzle I had intended to post: You have just lost your 143rd straight game of checkers and have vowed never to play another game. To confirm your vow you decide to saw your wooden checkerboard into pieces that contain no more than a single (red or black) square. With each use of the saw you may pick up a piece of the board and make one straight cut, along boundaries of individual squares, completely through to the other side. You wish to inflict as much damage as possible with each cut, so you first calculate the minimum number of saw cuts needed to finish the job. And that number is ... (spoilers appreciated.)
10. ## When midnight strikes

Isn't this fun? The { real numbers } between 0 and 1 comprise two infinite groups: { rationals } and { irrationals }. Rationals (expressible as p/q where p and q are integers) are countably infinite. We can order them, by placing them in an infinite square of p-q space and drawing a serpentine diagonal path. But the irrationals are not countable. The cardinality (notion of size used for infinite sets) of the rationals is called Aleph0 and for the irrationals (and reals) it's called Aleph1 or C (for continuum.) So first off, what we can do with the { numbers } between 0 and 1 depends on { which numbers } we want to deal with. Next, there's a problem that points are not objects that can be moved from place to place, as coins can. Points are more descriptions of space than of autonomous objects. If 0 and 1 are on a number line, the location midway between them is the point denoted by 0.5. It can't be removed. (We could erase the line, I guess, but that would not produce an interval [0 1] devoid of numbers, either.) But we can finesse this matter by "painting" a point. Kind of like what we did when we inscribed a number on each coin. Painting does a little less - it puts a point into a particular group or class but it does not distinguish among them. We can tell coin 2 from coin 3 and also distinguish both from a coin with no inscription. Two blue points, however, are distinguishable from unpainted points, but not from each other. But for our purpose here, painting is actually enough. Since the { rationals } are countable, we can paint them, in sequence, and if we do that at times of 1, 1/2, 1/4 .... minutes before midnight, all the rationals between 0 and 1 will be painted blue by midnight. (I don't know of a similar scheme for completing uncountable task sets, so the irrationals alas must remain unpainted.) Next, instead of asking whether all the points in [0 1] were removed, we can ask, with pretty much the same meaning, whether at this point the entire interval [0 1] has been painted blue. It should be, right? After all, between any two rationals lie countably infinite other rationals. So the paint must have covered the entire interval, right?. Well ... actually ... the answer is counter-intuitively No. And this is because even though the { rationals } are "dense" meaning there are no "gaps" between them, as noted above, the { irrationals } are even "more dense." That is, between any two irrational numbers there are uncountably many other irrationals. That means, for every blue point we created in the interval [0 1] there are uncountably many unpainted points. We can't magnify the line greatly enough to "see" individual points, but "if we could," if we were lucky enough to come across a single blue point we'd have to pan our camera over uncountably many unpainted points before we saw another blue one. In measure theory, the measure of the rationals over any interval is zero. The measure of the irrationals over [0 1] is unity. So what of our quest of emptying [0 1] of points? It's the same as our quest to paint [0 1] blue by painting all the rational numbers in that interval. (Reminding ourselves that there were too many irrationals to paint one at a time.) Turning again to measure theory, the measure of the "blueness" of the interval would be zero. That means we would not see even the hint of a faint blue haze. Back to our "empty the interval by removing the rationals" quest, Nope. [0 1] would not be empty: uncountably many points (numbers) would remain.

12. ## Worth the Weight

This is a nice puzzle.
13. ## When midnight strikes

That is what infinity does to our brains. Al retains coin 2 at step 1. But Al discards coin 2 at step 2. What is true for coin 2 is true for every coin. Every coin has a scheduled pre-midnight discard date. So "what happened to the N coins not discarded?" If N is finite, then it's not midnight yet, and the box does in fact contain coins. We have to be patient. The process has to run its course. And, specifically, {coin n+1} will be discarded at step n+1, at time t n+1, prior to midnight. So after midnight, it will be gone. Along with all the others. Every coin, identified by the number engraved on it, has a well-defined pre-midnight discard date. But then if we're Bert, who never schedules the discard of an odd coin, we'll have a ton of coins.
14. ## When midnight strikes

I would dispute your first point. I can't immediately think of a scenario that permits a discard which will not eventually happen given an infinite number of opportunities. Can you provide one? We agree on the second point. My previous post answers precisely that question.

16. ## Destroying a checkerboard

@CaptainEd Wow. Not the answer I had in mind -- a much better one! Nice. Bonus (slightly modified) version. Same question, but this time no cut may end short of the opposite edge. It must go entirely through the piece being cut.

19. ## Destroying a checkerboard

That would not be permitted. The OP says with each use of the saw you may pick up one piece of the board and make a straight cut.

22. ## Destroying a checkerboard

You have just lost your 143rd straight game of checkers and have vowed never to play another game. To confirm your vow you decide to saw your wooden checkerboard into pieces that contain no more than a single (red or black) square. With each use of the saw you may pick up a piece of the board and make one straight cut, along boundaries of individual squares. You wish to inflict as much damage as possible with each cut, so you first calculate the minimum number of saw cuts needed to finish the job. And that number is ... (spoilers appreciated.)
23. ## Baggage on a conveyor belt

Gah! Mine was not a useful clue. The denominator (greater than 6 but not huge) is too large for a simulation to tell you the fraction. Also I think your simulation value is high. But you're thinking in the right direction. Here may be more useful clues.
24. ## Increase the square

Four pegs begin at the corners of a unit square on a grid having integer coordinates. At any time one peg may jump a second peg along any straight line and land an equal distance on its other side. The jumped peg remains in place. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + O + + + + + + O O + ==> + + + O O + + + + O O + + + + O + + + + + + + + + + + + + + Is it possible to maneuver the pegs to the corners of a larger square? + + + + + + + O + O + + + + + + + + + O + O + + + + + + + + + + + + + +
25. ## Pairing the points

Nice. A construction is certainly a proof of existence. A pairing without intersects exists and ... here it is! Now I wonder if for any groups of n blue and n red points there is only one pairing without crossings?
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