## 25 July 2007

### Propp's self-referential aptitude test

Check out the Self-referential aptitude test by Jim Propp. The test consists of twenty questions, each with five possible answers; the answer to each question depends in various ways on the other questions! There are various routes to deriving the answers. I'd say to see if you can solve it, but that takes quite some time!

I tried to come up with the answers by first assuming that the answers were a random string of letters from A to E and then changing the answers to various questions in order to agree with each other; unfortunately this didn't work, because the way in which the answers depend on each other is too complicated. I expected this process to converge to a solution fairly quickly, but it didn't.

There is a probabilistic solution, though, via genetic algorithms; Mark Van Dine writes about it. The idea is that we can tell how many of the questions are correct, and then "mutate" the answers a bit hoping that we gradually find the right answer. This took thousands of generations, though, which is why I couldn't do it by hand.

I give a (deterministic) solution below. First, I want to give a couple thoughts on it. Presenting a solution like this is tricky; one wants to minimize the number of "intermediate results" where we know, say, that a question can have one of two or three answers but we don't know which it is. These sorts of results are hard to hold in one's head. (My original path to the solution had more such intermediate results; I don't claim that this way of writing the solution is optimal.) Another thing one wants to reduce is contradictions that take a long time to derive; there are a couple points in the solution where I know that a question has two possible answers, so I assume one of them and seek a contradiction. If it takes a long time to find the contradiction, then one forgets which assumptions need to be "undone" when the contradiction is finally reached. These two principles that I've given here are probably good principles for the writing of mathematical proofs, as well. It's a good idea to minimize the number of things your reader has to remember at any given time, because remembering things is hard.

It looks to me like there's no trick to finding a solution, from looking at the other solutions given at Propp's web site; the other solutions given there seem about as complicated as mine.

Anyway, here we go.

First, the answer to #12 has to be (A): the question is "the number of questions whose answer is a consonant is: even, odd, a perfect square, prime, divisible by 5". So I assume on that there is only one "right" answer to this question; therfore we need a number between 0 and 20 which is exactly one of those things. Clearly any number is even or odd, but not both, so we need a number which isn't a square, prime, or divisible by 5. Thus the number of questions whose answer is a consonant is 6, 8, 12, 14, or 18, and the answer is (A), for "even".

I also let knowledge from the outside world enter; the answer to #20 is (E). #12 and #15 have the same answer, so #15 is (A). This means that #13 must be (D). And the answer to #5 must be (E), by the principle that questions have only one correct answer and (E) is always a correct answer to that question.

Next, #6 and #17 only refer to each other, so we can hope to work out the answers to them without reference to the other questions. First, neither one can have the answer (E), since "all of the above" doesn't make sense in this context. Thus neither can have answer (C), because if either 6 or 17 has answer (C) then the other must have answer (E). Reasoning similarly, neither can have answer (A). So both #6 and #17 have either (B) or (D) as the answer; they must be one of each but we can't determine the order.

Similarly, #10 and #16 only refer to each other; they're (A) and (D), respectively.

So far, then, we have the answer string

where the two *s are one B and one D. Furthermore, we know that 6, 8, 12, 14 or 18 answers are consonants. So we move on to problem 8, which asks how many answers are vowels; this must be 2, 6, 8, 12, or 14. The only ones of these which are among the choices are (C) 6 or (E) 8. Looking at #7, then, we must have DC, DE, or EE for the answers to #7 and #8.

Next, we consider #3, #4, and #8. The numerical answers to the first two must sum to the numerical answer to the third. Furthermore, we've already fixed two answers of (E), so the answer to #3 is (C), (D), or (E). From #4 we know there are at least four A's. The answer to #8 is C or E, as we've already shown. Since there are at least two E's and at most eight vowels, there are at most six A's, so #4 is A, B, or C. The possible combinations of answers for these three questions are: CAC, CCE, DBE, EAE. If we take EAE here, then there are four E answers (#3, #5, #8, #20) already and no other E answers are possible; in particular #7 is D. Thus this are only six possibilities for the answers to #3, #4, #7, #8: CADC, CCDE, CCEE, DBDE, DBEE, EADE.

#9 can't be C (because #12 isn't C). And it can't be (E), because #13 is (D) and so the "next question with the same answer as this one" can't be #14.

#1 asks: "The first question whose answer is B, is question: (A) 1 (B) 2 (C) 3 (D) 4 (E) 5". It can't be (A) (because then the question would contradict itself) or (E) (because the answer to #5 isn't (B).) It can't be (C), because we just showed the answer to #3 isn't (B). So it must be either (B) or (D). Now, it seems that we have to make an arbitrary choice. We say #1 is (B). Thus #2 is also (B). So #7 and #8 must be the same; we see that they both have to be (E). Thus #5, #7, #8, and #20 are all (E); this is four (E)s, so #3 must be (E). But now there are five (E)'s, which isn't possible by examining the choices to #3. So our assumption that #1 had answer (B) is wrong; #1 must be (D).

Thus #2 isn't (B), meaning that #7 and #8 aren't the same. And #4 is (B). Thus the answers to #3, #4, #7, and #8 must be (D), (B), (D), (E) respectively, by looking at our previous list of possible answers to those four questions. The answers so far are

and we can at least see the solution taking shape.

Next, we look at #2 again. The answer to #10 is (A). Now, #11 can't be (A), because one of the questions preceding it has answer (B). Thus the answer to #2 isn't (E). Next, #2 can't be (C), because then #8 and #9 would be the same; but #8 is (E) and #9 can't be (E). So #2 is (A) or (D). But #2 can't be (D), because #2 refers to "the only two consecutive questions with identical answers" and #1 is already (D). So #2 is (A). Ths means #6 and #7 have the same answer, (D). Thus #17 is (B). The answer string so far is

There are five answers (A); we know this from #4. But there are less than five answers (C), since there are only five blanks left and #9 is not (C). So #18 isn't (B). Similarly, there are more than five answers (D) from the choices to #14, and exactly three answers (E) from #3. So #18 isn't (C) or (D) either; thus #18 is (A) or (E). But we've already accounted for the three answers (E) -- they're #5, #8, and #20. So #18 is (A). We now know there are the same number of answers (A) and (B), namely five of each. We've identified the five (A) answers, and two of the answers (B). The answer string so far is

and there are only two answers (B) and four blanks left. Three of the blanks have the answer (B).

Now, #11 must be (B) or (C), since there are 1 or 2 (B) among the first ten answers. Say it's (C); this means that there are two (B)s among the first ten answers, so #9 must be (B). But this means that #9 and #11 have the same answer, and we just said they didn't. So #11 is (B). This means that there is one (B) among the first ten answers, so #9 is not (B). (A) and (E) have already been ruled out for #9 (as the answers to the adjacent questions), and (C) was previously ruled out. So the answer to #9 is (D). The answer string is now

with three (B)s and two blanks; there are five answers (B), so the blanks must be (B), and we have

Max said...

#1. The first question whose answer is B, is question: (A) 1 (B) 2 (C) 3 (D) 4 (E) 5"

Since this says the first question that is #B, not the next question, it can't be A (due to contradicting itself) or B (since #1 is the first B, not #2). So you didn't have to go down that particular tree :)

frank said...

A very fun puzzle, thanks for pointing it out. I cannot imagine writing down the thoughts in my mind as I solved it and making it remotely logical to a reader.

I did cheat a little and use a piece of paper to cross off answers that could not be correct, and in doing so found the answers in a different order. For example, if you start by solving 10 & 16, then you know from 2 which order 6 & 17 fall in (can't have 2 D in a row), which is how one of the solutions on the website starts, incidentally.

The website also had a link to another similar puzzle, but I'll need to save that for another day.

Isabel said...

Max,

thanks for pointing that out. I think I knew that when I actually solved it, but the solution I give is actually slightly different from what was going through my mind and I suspect I got confused at some point in the editing.

John Armstrong said...

Your solution seems less than maximally-elegant, since it starts by assuming that only one of the conditions listed in #12 is true, and then justifies this by the asserted uniqueness of the overall solution. Thus you've proved the existence of that solution, but you haven't really proved its uniqueness.

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