bonanova

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.)

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.

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 + + + + + + + + + + + + + +

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?

One approach is to Then it would be helpful to know that the result is

plainglazed suggests an analysis amenable to powers of 2. I'll amend the OP to make it 1024 cars instead of 100 cars.

Actually, two of the boxes will be empty.

@Pickett Interesting result. OP meant for the coins to have equal value, but yours is a great puzzle also, even for finite numbers of coins received. @rocdocmac and @Pickett When midnight has struck an infinite number of events will have transpired and the process will have stopped. Will any of the three be happier than the others?

At an ever increasing pace Al, Bert and Charlie have been receiving into three identical boxes of limitless capacity identical pairs of silver coins engraved with the integers 1, 2, 3, 4, ... etc. These events occur on a precise schedule, each box receiving Coins marked 1 and 2 at 1 minute before midnight Coins marked 3 and 4 at 1/2 minute before midnight Coins marked 5 and 6 at 1/4 minute before midnight Coins marked 7 and 8 at 1/8 minute before midnight etc. But they were instructed at each event to remove a coin from their respective boxes and discard it. After some thought, Al decided each time to discard his lowest-numbered coin; Bert discarded an even-numbered coin; and Charlie thought what the heck and discarded a coin selected at random. Regardless of strategy, at each event the number of coins in each box grew by unity, so that after N events each box held N coins. Needless to say when midnight struck their arms were infinitely tired, but it was a small price to pay for infinite riches. But tell us, now, whether their expectations were met. Describe the contents of each box at midnight.

Problem

Not bad. I like the fact you can throw away the connections you find without fear of removing a potential crossing with a future connection. It’s not the solution I had in mind but I think it works. Nice.

Sizzling. Now finish it.

Just an observation, that may already have been uncovered, but don't know where to go with it:

Final clue Repeating an earlier comment,

Belt "segment" just refers to portions of belt between adjacent bags. Top or bottom is not an issue. Consider bags are randomly placed points on a line if you like. The fraction of length that is not a near-neighbor segment changes with time. To imply a limit, the OP asks "on average" over time.

OK a clue ( and it was plainglazed third answer that's close )

plainglazed is closest so far (but I can’t say which of his guesses). Remember it’s a physics question.

This puzzle has a nice "Aha!" proof.

My freshman physics prof drew these on the board one morning. He asked, if the thing on the left is a centimeter, what is the thing on the right? Your turn (spoilers please.)

At a busy airport a conveyor belt stretches from the runway, where all the planes land, to the baggage claim area inside the terminal. At any given time it may contain hundreds of pieces of luggage, placed there at what we may consider to be random time intervals. Each bag has two neighbors, one of which is nearer to it than the other. Each segment of the belt is bounded by two bags, which may or may not be near neighbors (to each other.) On average, what fraction of the conveyor belt is not bounded by near-neighbors? Example: ----- belt segment bounded by near neighbors ===== belt segment not bounded by near neighbors ... --A-----B==========C---D--------E=============F---G-H---I-- ...

At 5-second intervals, 1024 automobiles enter a straight (and otherwise empty) single-lane highway traveling at initial speeds chosen at random from the interval [50, 70] miles per hour. Cars may not pass nor collide with other cars. When a slower car is encountered, a car must simply reduce its speed, and for the purposes of this puzzle we may consider the cars in such a case become permanently attached, traveling at the slower car's speed. Eventually there will be N clusters of cars. What is the expected value of N? (Equivalently, what is the expected cluster size?)

I'm not sure I recognize your solution 6 from its description of how it's constructed, but your final description of it seems right. Nice solve.