Vol. 07: Puzzle This
MAKE's favorite puzzles.
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Answers
Jelly Beans
Mountain Man
River Crossing
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Showing messages 1 through 4 of 4.

Harder version of Jellybeans
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These were all pretty easy  but there is a neat variation on the Jellybeans one that seems impossible:
This time, the three jars each contain an infinite number of jellybeans and have blank labels. As before, one contains all red, another all blue and the third contains a mixture. What is the least number of beans you have to take out in order that you can correctly write 'Red', 'Blue' or 'Mixed' on the three labels?
WARNING  STOP READING NOW BECAUSE THE ANSWER FOLLOWS:
Four  or three at a stretch...
OK  you read ahead didn't you?
WARNING  STOP READING NOW BECAUSE THE ACTUAL ANSWER FOLLOWS:
Yes  it really is four...or maybe three...First take one bean from each jar. You now have two beans of one colour (call that colour 'A') and one bean of the other colour (call it 'B'). One of the two jars that you got an 'A' from is all 'A' and the other is a mixture of 'A' and 'B'  but we don't know which is which yet. The jar you got the 'B' from must be all 'B' so you can write that on its label right now. But we can never guarantee which of the other two jars is all 'A' and which is the mixture without removing beans from each of the two jars until we find a 'B'. Since there is an infinite number of beans in each jar, we can never empty them and therefore we may never know whether there is a Bcoloured bean lurking down at the bottom underneath an infinite sea of A's. So is the problem insoluable? Not at all! Take a second 'B' from the jar of knownB's. You now have two A's and two B's in your hand  pop one of each into each of the two unknown jars  and now you can label both of them as "Mixed" with a clear conscience! You could do it with only three beans by cutting the 'B'coloured bean in half and putting half into each of the two nonB jars  but cutting up jelly beans without eating them is hard.
Posted by 3D_geek on August 19, 2006 at 06:43:04 Pacific Time

River Crossing Alternate Solution
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The first two steps of my solution, contrary to the official answer, were:
1. C and C cross
W = { A, A, A, C }
E = { C, C }
2. C returns
W = { A, A, A, C, C }
E = { C }
Essentially the same (it doesn't matter who brings the first cannibal across the river), but I figured if you're an anthropologist, and "the cannibals will follow all directions given to them by the anthropologists," why not minimize your workload? Call it the "lazy antropologist corollary."
Walaszek, I like the concreteness of the graph illustration. I did think, however, that the ghost was an excellent way of illustrating the answer  in fact, I was surprised that the illustration accompanying the puzzle in the magazine gave away the solution so readily. When the hiker passes his ghost, they are by definition at the same point at the same time. And since they travel opposite routes with identical stop and start times, they must pass each other.Posted by TPIRman on August 15, 2006 at 00:42:51 Pacific Time

Mountain man  alternate explanation
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A graphical way of thinking about this is to draw two graphs, both with time (6 am to 6 pm) on the xaxis and place on the mountain on the yaxis.
A steadily upward sloping line on the first graph represents the ascent. A steadily downward sloping line on the second graph represents the descent. Horizontal parts of either line are rests.
Now, superimposing the two graphs shows that the two lines must intersect. The intersection demonstrates that the two paths cross in the same place (ycoordinate) at the same time (xcoordinate).
I don't think the ghost analogy explains why the intersection must occur at the same time of day.Posted by walaszek on August 12, 2006 at 16:58:52 Pacific Time

Mountain man  alternate explanation (with math)
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This is just another example of what, I guess, is a large class of similar examples.
The essence of the problem is that the path is continuous (it has no breaks or instantaneous jumps). Once that is known you can apply the fact that the sum of two continous functions is continuous. Let up(t) be the position of the climber on the way up at time t where t=0 at 0600 and t=12 at 1800. And similarly for down(t) but superimpose both days. Take the difference d(t)=down(t)up(t). Note that at t=0 (at 0600 on both days) down(0) is the height of the mountain while up(t) is at the base of the mountain so d(0) is positive. At t=12 the situation is reversed and d(12) is negative. So it was above zero and then it went below zero (negative) and it is continuous (no breaks or jumps) so it must have crossed zero somewhere at a certain time t.
There is a better example than a trip up a mountain: there exist two points on the Earth's Equator directly opposite each other (antipodal points) that have the same temperature at any instant in time.
A metereological theorem! An oxymoron ... almost.Posted by arouse on August 17, 2006 at 20:15:24 Pacific Time
Showing messages 1 through 4 of 4. 
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