bonanova
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Posts posted by bonanova
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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.)
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On 2/19/2018 at 12:06 AM, CaptainEd said:
Ok I tried using random numbers, got a fraction...
.8237
Bonanova, Since you said rational, I’ll guess 5/6. I’ll bet you have a simple but deep proof. I’m eager to see it! I can see that it has to be more than 2/3, because a nearest neighbor gap is smaller than either adjacent gap, hence is less than 1/3.
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.
SpoilerThe intervals can be classified small if both neighboring intervals are larger, large if both neighboring intervals are smaller, and medium otherwise.
SpoilerEvery interval has an equal probability of being large, medium or small. i.e. for every interval P(small) = P(medium) = P(large) = 1/3.
SpoilerThe (Poisson) probability that any interval is smaller than say a distance x can be taken, with appropriate scaling, to be 1 - e-x.
SpoilerPart of what is needed is the expected size <large> of a large interval.
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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 + +
+ + + + + +
+ + + + + +
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Nice. A construction is certainly a proof of existence. A pairing without intersects exists and ... here it is!
SpoilerI like this puzzle because the result is at first dubious. What if there were billions of points? But (think diagonals of a square) removing intersects always reduces connection length. So the pairing with the shortest connection length (which must exist) has no intersects.
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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One approach is to
Spoilersimulate the bags with a random number generator.
Then it would be helpful to know that the result is
Spoilera rational number.
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plainglazed suggests an analysis amenable to powers of 2. I'll amend the OP to make it 1024 cars instead of 100 cars.
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Actually, two of the boxes will be empty.
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@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?
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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.
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Spoiler
Another answer is the matching that minimizes the total length of the connecting line segments, which has a simple proof
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Problem
Spoiler12 balls, one is heavy or light (24 cases.) 24 <= 33 so we're good.
SpoilerReduce the cases to <= 32 in number.
Number the balls and compare { 1 2 3 4 } with { 5 6 7 8 }.- If they balance, one of { 9 10 11 12 } is heavy or light (8 cases.)
- If they don't balance one of the lighter group is light or one of the heavier group is heavy (8 cases.)
Balance
SpoilerReduce the cases to <= 3 in number.
Compare { 10 11 12 } with { 1 2 3 }.- If they balance, { 9 } is heavy or light (2 cases.) Compare { 9 } with any other ball to decide.
- Otherwise one of { 10 11 12 } is known heavy or light (3 cases.) Compare { 10 } with { 11 } to decide.
Don't balance
SpoilerReduce the cases to <= 3 in number.
Set aside { 4 7 8 } and compare { 1 2 5 } with { 3 6 12 }.If { 1 2 3 4 } was heavier, then
Spoiler- If { 1 2 5 } is heavier, 1 or 2 is heavy or 6 is light. Compare 1 with 2 to decide.
- If they balance, 4 is heavy or 7 or 8 is light. Compare 7 with 8 to decide.
- If ( 1 2 5 ) is lighter, 3 is heavy or 5 is light Compare 3 with 12 to decide
If { 5 6 7 8 } was heavier, then
Spoiler- if { 1 2 5 } is heavier, 5 is heavy or 3 is light. Compare 3 with 12 to decide.
- If they balance, 4 is light or 7 or 8 is heavy. Compare 7 with 8 to decide.
- If { 1 2 5 } is lighter, 1 or 2 is light or 6 is heavy. Compare 1 with 2 to decide.
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14 minutes ago, plasmid said:
Draw a convex perimeter around the set of points (the shape of a taut rubber band with all the points inside). Unless every point on the perimeter is the same color, you will be able to find a segment on the perimeter with alternating colors at its vertices. Draw a line connecting those points, and you can remove them from the puzzle and be left with a smaller set of points where the line you just drew is outside of its convex perimeter and therefore won’t prevent you from drawing any more lines, so you can just repeat until you’ve joined all the points with lines.
But what if you reach a state where every point on the convex perimeter is the same color? Find a line that can divide the set of points into two separate sets of points that each have the same number of red and blue points, and then apply that algorithm to each of those two subsets. If the subsets also have monochromatic perimeters, divide those subsets into subsets and keep repeating the process as much as necessary until you can start connecting opposite color points.
Need a proof that a line must exist that can divide a set of points into two separate sets of points with equal numbers of red and blue? Draw any old line and it will divide the points into two regions; if they have equal numbers of red and blue points then you’re done, and if not then call the region with more blue than red points region B and call the region with more red than blue points R. If you gradually rotate the line 180 degrees, then regions B and R will be swapped so now B has more red than blue points and R has more blue than red. At some point during that rotation, if you’re allowed to nudge the line around a bit to make sure that only one point crosses the line at a time (I know I’m doing some hand waving but I think it’s still reasonable), the B and R regions must have reached a point where the number of blue and red vertices were equal.
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.
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Just an observation, that may already have been uncovered, but don't know where to go with it:
Spoiler14 has two prime factors, one odd, one even.
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Final clue
SpoilerAnswer has three letters
Repeating an earlier comment,
Spoilerplainglazed's third guess is like a phonetic clue.
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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.
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OK a clue ( and it was plainglazed third answer that's close )
Spoilerit has to do with death or, more precisely, near death.
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plainglazed is closest so far (but I can’t say which of his guesses). Remember it’s a physics question.
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This puzzle has a nice "Aha!" proof.
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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.)
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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-- ...
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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?)
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On 2/8/2018 at 1:48 AM, Donald Cartmill said:
I have it ...the 6th solution
1) a square with 4 sides and two diagonals =6 lines
2) would be a diamond shape created by two equilateral triangles; 5 sides equal and one long diagonal
3) An equilateral triangle ,3 equal sides ; The 4th point being in the center equidistant from the other 3 points providing 3 other lines of equal length but obviously shorter
4) an equilateral triangle is 3 lines connecting points "A" ,"B" and "C", Assume "A" is the vertex and "B" and "C" are the base points. A 4th point "X"gives us a line the same length as a sides of the triangle ,and directly off of vertex point "A" away from the triangle , but perpendicular to the opposing side "B"C"; This arrangement gives us 4 lines of equal length . Now the lines connecting the 2 base points "B"and "C", and point "X" ,are of equal length and longer than the other 4 lines
5) an equilateral triangle is 3 lines connecting points "A" ,"B" and "C", Assume "A" is the vertex and "B" and "C" are the base points. A 4th point "X"gives us a line the same length as a sides of the triangle ,and directly off of vertex point "A",but downward bisecting line BC = 4 lines of equal length; 2 short lines BX and CX
6) 4 points A,B,C,D, lay on two intersecting circles. point A is the center of one circle C and D lay on this circle. D is the center of the 2nd circle with A and B laying on the circle. The 4 points create a trapezoid in which the distance BC , is the same length as chords AB and CD (that is 3 short lengths ) The diagonals AC ,AD and BD are equal all being equal to the radius.
That should be all of them ????
6) 4 points A,B,C,D, lay on two intersecting circles having the same radius. point A is the center of one circle C and D lay on this circle. D is the center of the 2nd circle with A and B laying on the circle. The 4 points create a trapezoid in which the distance BC , is the same length as chords AB and CD (that is 3 short lengths ) The diagonals AC ,AD and BD are equal all being equal to the radius.
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.
SpoilerAnother way to describe it is to take a 5-pointed star (which fulfills the 2- length criterion) and remove one of the points.
When midnight strikes
in New Logic/Math Puzzles
Posted
Al:
On step 1 Al discards coin 1. On step 2 A discards coin 2. ... On step N Al discards coin N. As N -> inf, every coin is discarded.
If there were a coin in Al's box at midnight it would have a number. What number would that be?
Bert:
On step 1 coin 1 enters Bert's box and is never discarded.
On step 2 coin 3 enters Bert's box and is never discarded. etc.
At midnight every odd coin remains in Bert's box.
Charlie:
After N turns, what are the probabilities for coins to remain in Charlie's box?
...
At midnight, the probabilities for coins to remain in Charlie's box have vanished.
One might note that after N turns, PN(coin N) = N/N = 1. That is, at each turn the probability that the last coin to appear in the box is in the box with probability of unity. That may appear to be a counter argument, but it is not. In particular we can't meaningfully take the limit of that ratio as N -> inf to prove there must be at least one coin in Charlie's box at midnight, for two reasons.
First, the ratio has heterogeneous "units." The numerator is the number of a coin whose condition (in the box or not) we wish to know, while the denominator is the index number of the successive events that lead up to midnight. So the numerator and denominator should be represented by distinct variables.
Second, to say the last coin to enter the box is in the box, and taking this to infinity would provide relvant information only if there were a last integer. Which there is not. Nor is there a last coin or a last event.
What we can say is that for every coin k, PN(coin k) = k/N ->0.