I wonder what do you mean by “sandra failing the Turing test”?

Some blogs are configured so that the first comment by an individual commenter goes into moderation. Once manually approved, subsequent comments are approved automatically.

Faced with that configuration, spammers sometimes post innocuous-looking comments without a spammy link in the hope that they will be approved, allowing the real spammy payload to be sent later.

My spam catcher allowed her to pass.

No spam catcher is perfect, and this kind is more difficult to recognize automatically than the usual. What marks them out to human eyes is that they are completely generic. The comment could just as well have been posted to any other blog. There is nothing in it specific to your blog that would indicate a real person had read it and wanted to comment on it. The other big giveaway is that Sandra’s link doesn’t point to anything.

Sandra could prove me wrong by replying to this comment.

]]>Thank you for pointing out my arithmetic mistake. And sorry that it confused you. I corrected it. I wanted the revealing coefficient to compare what you are supposed to prove with the info you actually reveal.

Also, I wonder what do you mean by “sandra failing the Turing test”? My spam catcher allowed her to pass.

]]>the number of equally probable possibilities was 100 choose 2, which is 5050

I was initially puzzled by this, as 100 choose 2 is 4950. I then realized that what you meant was 100 choose 1 or 2, which is 5050.

For the elementary version of the puzzle, two facts are apparent.

1. Every weighing must set aside an odd number of coins, not one.

2. Every weighing must balance (otherwise the heavy side would be revealed to contain only genuine coins).

The task then is to exhibit a sequence of weighings, all of which balance, but which, if there had been only a single fake coin, some would not have balanced.

Setting three coins aside, weigh 48 coins against 48 to show that they balance.

Swap the three coins for two from one of the pans and one from the other, and show that they balance.

Before you weighed them all, one or two coins out of 99 could have been fake, so the number of equally probable possibilities was 99 choose 1 or 2, which is 4950.

After weighing, one fake could be among the 46 coins never swapped in the first pan, and the other in the 47 never swapped in the other, for a total of 2162 possibilities. Or they could be among the three swapped out, or the three swapped in, for an additional 4 possiblities. The revealing coefficient is then (4950 – 2166) / 4950 = 0.562.

This is the best we can do. If, for example, we instead set aside 5 coins, and swap 3 and 2 to the two pans respectively then after weighing the number of possibilities would be 45 * 46 + 3 * 2 + 3 * 2 = 2082

Turning to the original Shapovalov problem, I note that fact 1 above is replaced by a requirement that it is always an even number of coins set aside (including none). Also the requirement that the pans always balance no longer applies. Mary’s solution, which is analogous to the one given above, is probably the best with all weighings balancing. It is not clear to me that solutions with non-redundant non-balancing weighings can’t exist, but despite considerable brain-wracking, I can’t find one.

]]>Yes, I invented the revealing coefficient for this problem.

Also, your solution with 4 piles will not work as the coins in the piles where all of the coins are good would be proven to be all good. Thus, the information about a particular coin is revealed.

]]>Mary, I don’t agree with your computation of the revealing coefficient. If I understand you correctly, the process we follow is this: we have initially three piles of size 32 and a pile of size 4. Let’s name the coins in the pile of size 4 {a, a, b, c}. Then we switch {a, a} for two coins {d, d} from the first of the three big piles, switch {b} for a coin {e} from the second big pile and switch {c} for a coin {f} from the third big pile. Then the following sets of coins could be the counterfeit coins:

* {a, b, c} for one of the a-s (2 ways)

* {d, e, f} for one of the d-s (2 ways)

* a triple of coins, one from each of the big piles, that includes none of a, b, c, d, e, f (30*31*31 = 28830 ways).

This leaves a slightly larger revealing coefficient than you suggested (82.168…%).

An easier-to-explain but less-good solution is to divide the original pile into four piles of size 25 so that there is one counterfeit in three of the piles. Weighing each of those piles against the one “all honest” pile shows that there must be at least 3 counterfeit coins. This gives a revealing coefficient of 1 – 25^3/(100 C 3) = 90.3…%.

]]>My initial impression was that your were swapping all 4 coins together to the same pile. Now I agree with you.

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