Request for Web page critique.

Quick… Fix the link.

How do I request being able to edit? This is the correct link

@johnpowell Oh how I hate when that happens. Email a moderator and ask if they can post the link within the actual question details for you.

Why do you begin with the 9 coin problem under the heading for the 12 coin problem? I know you need to talk about the 9 coin problem first, but perhaps it could have its own heading? Also, for the last weighing, why do you need to shuffle the coins? Wouldn’t it work simply to weigh two coins from the stack that you determined had the counterfeit? I’m not sure what including the extra coins accomplishes. Naturally, I have the same question regarding the 12 coin problem, but I’m sure the answer follows from the answer here.

For the 12 coin problem, it would be good to include an explanation of what you would do if the coin were in one of the three stacks that you kept together instead of being one of the three isolated coins. I’m sure this has something to do with the shuffling I didn’t understand above, but some explanation might be helpful. Even if it follows directly from the argument from the 9 coin problem, including a sentence about it would make the argument more readable and more complete.

After reading it a couple times, I follow your argument for the 39 coin problem, but an illustration would make it more clear. Since you’re dealing with stacks of three, you could make your first pictures smaller by putting 1–3 on one circle to illustrate those three coins (obviously, the same goes for each of the following groups of coins). Once again, I’m not understanding the need to shuffle the coins for the final step. I’m beginning to think there’s a really simple answer to that question, but I’m not seeing it.

I need to get going, but I will look at the general case later.

@bobbinhood, Thanks for your comments.

The coins have to be shuffled on the last weighing in order to guarantee that no coin is in the same position for the first two weighings when the single coins are replaced by stacks of 3 in the next stage. If that was not done then at the next stage there would be a stack of 3 coins that remained in the same pan for the first two weighings in addition to the two additional coins that remain in the left and right pans. There would be no way of distinguishing the one case from the other.

I will consider your suggestion about what is done for the 12 coin case if the counterfeit is in one of the stacks of 3. It is not too difficult to see that if you have 3 coins with one that is heavier/ lighter than the others, that weighing any 2 will determine the counterfeit. If the counterfeit is included among the 2 then that coin will cause the scale to tip. Otherwise, if the two coins balance, then the coin not on the scale is the counterfeit.

Under the three coin problem, you misspelled lighter as lightter. This may seem minor, but you will be surprised how much misspellings can affect the perceived credibility of the site. You should run the entire page through a spellchecker.

Like people above have said, you should probably split the nine coin problem from the twelve coin problem.

Also, since the main question at the top of the page is the 12 coin problem, you should have statement in the introduction saying why you are starting with simpler cases.

Thanks. I wish the NVu Web page editor I used had a spell checker. I will give serious consideration to doing the 9 coin problem separately. I need to emphasize that this is just an intermediate step toward the 12 coin problem using the same number of weighings.