Pickover Problems 20 through 22

SPOILER

This greenness question is interestingly, a question about the statement of weak duality of nonlinear programming. For the rectangular array: Let us index the rows as x (row_k is x_k) belonging to set X Let us index the columns, as y (column_i is y_i, say) belonging to set Y, and Let F (x, y) be the function denoting 'greenness'. In the rectangular array, every (i,k) pair (where x_k belongs to X, where y_i belongs to Y), will have a value given by F(x_k, y_i) that denotes the 'greenness' of the alien-entry at the (i, k)th position in the array. Cliff's question, in other words, is: Is max(over Y) (min (over X) F(x, y)) <= min(over X) (max (over Y) F(x,y)), Or Otherwise ? Weak Duality's implication is: max-over-Y (min-over-X (F(x,y)) <= min-over-X (max-over-Y (F(x, y)). Provable by an argument that requires very little mathematical depth, as follows:

We know that, for any general x' belonging to X, and y' belonging to Y: a) min-over-X ( F(x, y')<= F(x', y' <

Problem 21: It's not easy being green

The palest of the greenest aliens in each row will be greener (well, perhaps not, but certainly no paler) than the greenest of the palest aliens in each column.

For simplicity, call the palest of the green POG, and the greenest of the pale GOP. (Curiously, these do in fact happen to be common alien names.) Now look at the alien where POG's row crosses GOP's column. Clearly this alien is at least as green as GOP but no greener than POG. QED (which, disappointingly, is not an alien name at all).

Al Zimmermann White Plains, NY

Hi!

SPOILER for problem 20 follows the quotation of the original puzzle . . . .

For part 1, suppose someone videotapes all three shots and reorders the tape clips so that the shots are shown in order, from best to worst. Then the first and second shot can be in (I) first and second clip, (II) first and third clip, or (III) second and third clip. Only in one case is the third shot better than the first; two in three hence is the probability that the last shot was further from the center than the first shot.

For part 2, suppose someone videotapes all 1000 shots and reorders the tape clips so that the shots are shown in order, from best to worst. Then the first and last shots are among the 1000 clips in either of two orders (I) the first shot before the last shot, or (II) the last shot before the first shot. Both cases being equally probable, one in two hence is the probability that the last shot was further from the center than the first shot.

With proper odds, I'd place a wager. My answer does not depend on the shape of the target. The solution would be different if I knew that another player alternates turns with Mr. Eck.

Cheers!

- Risto -

The videotape editing is a nice thought experiment construct. It doesn't help, though, if Dr. Eck is manipulating his results. I'm not ready to place a wager at any odds. I have no reason to trust Dr. Eck, and I would suspect that he has placed his first two (or 999) shots far enough apart that he can put the last one inside or outside them, whichever is needed to win his confederate's

Hi!

Well...

Cheers!

- Risto -

Yes, but a skillful person need not be aiming at the same point for each shot.

The point being that f(n) = 10^(n+1)/81 - (9n+10)/81 so sqrt(f(n)) is very close to 10^((n+1)/2)/9.

Well, you want something that will be very close to the square of a rational number with a denominator of 9, 99 or 999, or a factor thereof. e.g, let's try f(n) = 100 f(n-1) + 36 n - 128, f(0)=1, which results in f(n) = 9/121*100^n + (112 - 44 n)/121, and say sqrt(f(30)) = 272727272727272727272727272727.272727272727272727272727 2727 0896969696969696969696969696969696969696969696969696969 6969 0828013468013468013468013468013468013468013468013468013 4676...

Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2

Hi!

That's true. The quoted sentence from Cliff's article lead me to believe that Dr. Eck is merely an observable experiment and that the actual betting was to take place against a third party. I was silly not to have thought of the interpretation that you suggest (that the other party to the bet could be Dr. Eck himself).

Thanks for pointing out! My revised answer is that only with proper odds, I'd place a wager.

Cheers!

- Risto -

This incorrectly assumes that (I), (II) and (III) are equally likely.

If Dr. Eck fires two shots at a target, each taken with equal accuracy, then it is equally likely for the first or the second to be more accurate than the other. Doesn't matter how many shots he or anyone else takes in between, nor the shape of the target.

Hi!

Then please provide the actual probabilities. Or, failing that, order the three events from most to least likely. Please also explain reasoning.

True, if I understood correctly (I think you mean any two arbitrary shots, one of which you call 'first' and the other one 'second' in the above paragraph).

Note, however, that the problem statement explicitly says that the very first shot is better than the very next, if that is what you mean by 'first' and 'second'.

False. Every shot that is worse than the first one gives information about how hard it is to beat - and so does every shot that is better than the first one.

So if he shoots 2001 times and 285 shots are better than the first one, and 1715 shots are worse than the first, would you still say that the 2002nd shot has a 1/2 chance of being better than the first shot?

(ObPuzzle: What is the actual probability of the 2002nd shot being better than the very first one in the situation described above?)

Cheers!

- Risto -

Whoops, good catch. But the original problem's 1000-shots case is a 50-50 chance, because in that case we aren't told whether any of the intermediate shots are better or worse than the first.

Trivia: Is there a word meaning 'something that intentionally appears to be a red herring, but actually is not'?