bonanova

In the land of Truth-tellers, Liars and Alternators, every citizen either always answers truthfully T, always answers falsely L, or alternates true and false answers A. You encounter, and find yourself extremely attracted to, a citizen. Before exploring a possible relationship, however, you wonder whether you'll be dealing with a T, L or A. You may ask a number of yes/no questions to satisfy your curiosity. How many questions will it take?

Oh so close ... ! But guessing binary digits seems doomed. Grouping is the right idea, need better groups. Bingo.

Hi kevink2, welcome to the Den. You're on the right track. Any yes/no question is permitted. The question can have multiple parts; the answer can have only one part. It can depend on previous answers.

You can guess a number between 1 and 1,000,000 with 20 yes/no questions. This owes to the fact that any positive integer not greater than one million can be represented using 20 binary digits, and you can determine them, one per question. In general, N questions will determine any positive integer not larger than 2N. Better results can be had if it is known that some numbers are more likely than others to be the target to be guessed. If nines and higher are removed from a deck of cards [counting Aces low] 3 guesses would determine the value [1-8] of a randomly chosen card. [but not the suit.] If the nines were added, a 4th guess would be needed. But suppose the numbers had different likelihoods of being chosen. This would be the case if the deck [ignoring suits] comprised 1 Ace, 2 deuces, 3 treys ... 8 eights and 9 nines. Can you determine a guessing strategy [a series of yes/no questions] that on average would determine the value of a card chosen at random from such a deck, in three guesses?

Very close, and heroic attempt at the summation. Good job all!

Very close, and your analysis is on the right track. But you can still get a little better result.

Nope, sorry. You noted a necessary part of the puzzle. Did you get SP?'s answer below? Bingo.

Johnny Dangerously spotted a sale on small pets for his store that he couldn't resist. Giddy with visions of profit, he bought a certain number of hamsters and half that many pairs of parakeets. He paid $2 each for the hamsters and $1 for each parakeet. On every pet he [carefully] stuck a retail price that was a 10 per cent mark-up over what he paid for it. After all but seven of the creatures had been sold, Johnny found that he had taken in for them an amount of money exactly equal to what he had originally paid for all of them. Wow. How cool was that. His potential profit was represented by the combined retail value of the seven remaining animals. What was this value?

It was ladies' night at Morty's, and Alex came prepared to dazzle the girls by using his newly claimed powers of clairvoyance. As usual, his compatriots Ian and Jaimie were present. Davie was home, unfortunately, with a cold. I didn't ask for it, Alex began, in a voice loud enough to be heard over the slightly annoying music Morty was playing to encourage couples out onto the dance floor, it just came on me one night last week. And now I can see, dimly like, it seems, right through objects. Not all the time, mind you, but often enough and clearly enough that I can beat the odds in most simple games of chance. By now three pretty ladies were at their table, and Alex proposed the following demonstration of his powers. He tore a sheet of paper in half four times and asked the other five to write numbers on the resulting 16 pieces. Think of any number you like - they don't have to be whole numbers. They can be the probability of winning the lottery, the number of beers you expect to drink tonight, or the number of molecules in the universe. They just have to be positive numbers, and they all have to be different. My job then is to find the largest number. I will pull them out of a hat, one by one, and stop when I think that the last number I've pulled from the hat is the highest of the bunch. I get to read the numbers as I pull them, of course. Now if I used no strategy at all, or did it blindfolded, I'd have a 1/16 chance of getting it right. But even one of you fellows could do better than that! But ... given my [here Alex cleared his throat, softly, out of an abundance of feigned humility] substantial intellect, and my newly acquired powers of clairvoyance, I'm prepared to do it four times better. I'm willing to do this all night, my quid against your four, and I expect you'll be buying my drinks. What do you say? Alex's friends exuded a silence born of past experience with his other schemes, even tho a 1/4 chance of his picking the highest of 16 truly unpredictable numbers seemed to make an attractive bet. And, oh, not for a moment did they buy the clairvoyance claim. Finally Jaimie spoke up. Make the odds three to one and you're on! Ouch! Alex responded, That's asking a lot. But, remembering his goal of impressing the ladies, Alex agreed. Did Alex make a good bet? Or did Jaimie finally out-fox his buddy? Specifically, what is Alex's optimal strategy, and what are his odds of winning?

It is, unless you can show a dissection with fewer.

Bravo.

I'm starting a genre of puzzles that describes an effect for a magician to present, then asks how would you create the effect. The best answers are the simplest ones. Hidden cameras, tiny mirrors, wireless microphones, one hundred accomplices, quantum transport, infinitely short yellow lines :-) and the like, would not be considered simple methods. Here's the first puzzle, one described more than 50 years ago by Martin Gardner, which I'm calling the Intelligent Match Sticks. The effect: You place three fair dice on a table, in full view of the audience. With your back turned to the table you have a volunteer roll the dice and stack them, three high, in any order and any orientation. You briefly turn and hand him a coin, and have him place it on top of the stack, so that now all the horizontal die faces are obscured. With your back again turned, the volunteer picks up the dice and adds the numbers from the two touching faces of the top and middle dice. He adds to that sum the numbers from the two touching faces of the middle and bottom dice. Finally he increases the sum by the number on the bottom face of the bottom die. He writes that final sum on a sheet of paper which he folds and puts into his pocket. You then ask him to roll the dice and tell you the total of the three top faces. All this is done out of your sight. You then turn to the volunteer and hand him some match sticks from your pocket. The number of match sticks equals the number written on the paper. How is the magician able to hand the volunteer the correct number of match sticks?

I wonder whether the OP envisioned lateral thinking answers to the question. Karchimex [welcome to the Den, btw] certainly found a few! Reminds me of the ways to measure the height of a building, using a barometer.

Traffic to and from each city is separated by a yellow line. The Board was willing to concede that a rose by any other name smelled just as sweet. But when I forwarded your proposal to paint a yellow line of a non-yellow color, they just kind of got glassy eyed and looked confused.

The second question was aimed at who would win if played competitively. Is your "minimum" solution a statement that B wins if the players compete?

That's it, and very nice explanation.

Total PT dates with 2-digit years.

Start the ball rolling.

Back in the day, swapping x <---> + in computations made a difference. p.c. Answer was just using time while I worked on the real answer .

OK I think it can be done, but it will cost you

To stretch the point, if you convert to polar coordinates (a mere 4 squares, 2 additions, 2 square roots and 2 arc tangents, it'll take only 1 multiplication and 1 addition. Well, to recover a Cartesian result, add on 1 sine, 1 cosines and 2 more multiplications. Still only 3 multiplications instead of 4.