Can you find the unique solution to Jim Propp's twenty-question Self-Referential Aptitude Test?

[ Editor: the following solutions were mailed to Chris Cole by
Ucalegon@aol.com in June and July 1996 ("Here are the SRAT solutions in
all their NPL-specific glory"), as compiled for an NPL publication.  The
solvers use their National Puzzlers' League names.  Notation: = means
"is", ~ means "is not".  Comments in square brackets inside the
solutions are the compiler's. ]

     Solution by Ostrich to the Propp self-ref quiz on 13.4.21b: DADBE
DDEDA BADBA DBABE. = means is, ~ means is not. With 10 & 16, only case
is 10=A 16=D. By 2, 17~16D; so with 6 & 17, only case is 6=D 17=B.
With 7 & 8, cases are AE, BE, CA, CE, DC, DE, EE; so with 3 & 4 & 8,
cases are AAA, ACC, BBC, CAC, AEE, BDE, CCE, DBE, EAE. If 1A or 1B,
contradiction; if 1E, then 5B, then 2B, contradiction; so 1C or 1D; so
at least one of 3B, 4B; also 2~B. By 2, 3~4, so with 3 & 4 & 8, cases
become BDE, DBE, so 8=E; by 8, 12=A, so 15=A, so 13=D. If 3B 4D, then
1C; by 2~B, 5D or 5E [don't follow that]; by 2, 5~6D; by 3, 2 E's is
contradiction; so 3=D 4=B, so 1=D; by 2~B, 5=E. By 2, 2~3D; if 2E,
then 11A, but by 2, 11~12A, contradiction; if 2C, then 9E, then 14E,
but by 3, 4 E's is contradiction; so 2=A; so 7=D. 4 A's, 2 B's, 6 D's,
2 E's. Of 6 ?'s, by 3, one is E, by 4, one is A. With 11, cases are B,
C; by 2, 9~10A; if 11C, 9B; if 11B, then if 9E, then 14E, but < 10 D's
is contradiction, so 9D; so with 9 & 11, cases become DB, BC; so of 6
?'s, one is B; also < 10 D's; so 14~E. By 2, 14~13A, 14~15D; so 14B or
14C. By 5 total A's, by < 4 C's, 18~B; by 6 D's, 18~C; by 3 total E's,
18~D; so 18A or 18E. If 18A, then with 6 ?'s cases are (unordered)
ABBBCE, ABBBDE; so 14B; so ABBBDE; so 9D 11B; so with 6 ?'s cases
become [1] DBBABE, [2] DBBAEB. If 18E, then if 14C, with 6 ?'s only
case is (unordered) ABCDDE; so 9D 11B; by 13, 19~A, so 19D 20A; then
with 6 ?'s only case is [3] DBCEDA. If 18E 14B, by 5 A's 20A; by 18,
<> 5 B's, so with 6 ?'s only case is (unordered) ABBCDE; if 9D 11B,
then 19C; if 9B 11C, then 19D; so with 6 ?'s cases become [4] BCBEDA,
[5] DBBECA. Thus 5 cases satisfy 1-19. Switch to other half of brain
for 20. ((A few minutes before this point, I read 20 for the fifth
time and thought about it for the first time: a barometer measures air
pressure! It doesn't match any of these! But then I realized ':' means
not 'roughly measures', but 'does not actually measure'. Ha ha. So any
answer works somewhat. Then, when I found there were five choices I
recognized that by the strictness of phrasing E works best. Switch
back.)) So 20=E; so only case is [1]; so 9=D, 11=B, 14=B, 18=A, 19=B,
QED, DQ.
     Jim Propp (a former NPLer) and I both placed in the top eight in
the USA Math Olympiad in our respective high school years, making us
alumni of a select group (about 100), but contacting him has been
unsuccessful. After a challenge like this one, I feel like measuring
the height of a building with a barometer. /B-)

     Qaqaq: In the detailed solution below, a number in brackets
represents a qestion that should be referred to in order to follow
that particular step's logic. (As you'll see, qestion 2 proves to be
hugely important.) If the logic seems self-evident or follows from the
preceding step, I don't provide a number. This is generally in the
order I did it, although I have condensed and reordered a bit for
clarity, and to reduce the embarrassment factor for those steps where
I noticed something blindingly obvious far later than I should have.
However, my path can probably still be shortened. [I'll condense the
solution with Ostrich's notation.]
     5=E. 1-4~E. 3~A. 1~A or B. 2~B. 10=A, 16=D. 11~A [2]. 15,17~D
[2]. 7,19~A [13]. 13~A,C. 9~A. 2~D. 6~E [2]. 17~C. 6~A. 17~E. 6~C.
17~A, so 17=B. 6=D. 18~B [2]. 12=A/B. 15=A/B. 9~C. 13~B/E so 13=D.
15=A. 12=A. 14~A/D [2]. If 9=B, 11=B, but 3/4=B [1], so 11~B, so 9~B.
11~D/E; the maximum is two--3/4 and 7/8 [2]. 4~D; if it were, then 2,
8, 18, and 20 all =A, but if 2=A, 7=D, and 8 cannot then be A. 14~E;
because of [2], the maximum is nine--the known 6, 13, and 16, plus
possibilities 1, 3, either 7 or 8, 9, 18, and 20. 9~E, so 9=D. 8~B/D
[12]. 2~C, so 2=A. 7=D. 8~A. 4~A, as that would be the fifth A. 4~C;
if it were, 18=A and 20=A, but if 18=A, there must be six Bs, and only
five are then possible--the known 17 plus either 3 or 4 [2], 11, 14,
and 19. So 4=B. 3~B [2]. 1=D. 11=B. 18~C; there are already six Ds and
a then-maximum of five As--the four known plus 20. 18~D; there are
already four As and a then-maximum three Es--the known 5 plus 8 and
either 19 or 20 [2]. There are five known vowels, and either 18 or 20
must be A [4], and there is at least one more E [3]. Therefore 8=E.
The only remaining qestions whose answers could be vowels are 18, 19,
and 20; since two of these are vowels, 19 is not E and 20 is not B, C,
or D [2]. [I don't follow that.] 18 and 20 are A and E in some order,
so there are three Es, and 3 is D.
     At this point, 14 could be B or C; 18 could be A or E; 19 could
be B, C, or D; and 20 could be A or E. If you treat 20 the same as
19--ignoring the meaning of the qestions--then there are three
combinations of answers that make every answer work correctly:
14/18/19/20 = C/E/D/A, B/A/B/E, or B/E/C/A.
     So we must consider qestion 20. A seems very unlikely, since a
barometer does not measure 'temperature (only)'; and that leads to a
non-uniqe solution to the puzzle. 20 must be E to give B/A/B/E, the
unique answer; can it be justified? I think so. I suspect Mr. Propp's
whole point is that a standardized test does not measure intelligence
any more than a barometer measures any of the given choices in qestion
20. And I would agree with him.
     [Addendum the next day:] Oh, duh. As further proof that E is
right, read the answer letters in order. I can't believe I just
noticed this!

[ Editor: "qestion" and "non-uniqe" are so spelled in the source. ]

-- Qaqaq

     I'll just sketch my solution; the above gives all details needed.
Some steps may be a bit out of order; hard to linearize my 2-D mess of
notes. Good fun all the way. 5=E by uniqueness. 12=A/B, so C-E must be
false, so of the possible 12-16 (from 8) only 12/14 work and 8=C/E,
12=A. 10=A, 16=D. 7 and 19 can be anything. 20=E. (This was the only
non self-referential question so I figured it had to be a key.)
6/17=B/D or D/B. 15=A so 13=D. 3/4/8 imply counts of A/E = 4/2, 4/4,
5/3, or 6/2. For 1, C/E have been ruled out and A/B are self-
contradictory, so 1=D, so 4=B, so 3=D, so 8=E. We have all the Es now.
2/9=A/D or D/A. 18~C (6>5), D (5 3), E (used up), or B (need 5 Cs, but
only 7/11/14/19 can be), so 18=A and, with 2+9=A+D, all As are
accounted for. With 5A 5B 3E we have, from 14, 1C 6D or 0C 7D; 14=B
because we're out of As. Now we have to allocate A 3B 3D to the
remaining 2 6 7 9 11 17 19. If 2=D, 9=A and 6 7 so 11=C; can't be, so
2=A, 9=D, If 6=B then 7=B, 11=D, and 17=D; or if D, then D, B, B. 19=B
either way. But 16=D so, by 2, 17=B. Done.

-- Treesong

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[ Editor: second mail, 29 Jul 1996, "More softcopy S.R.A.T. solution,
to appear in July GotS".  "You" is Treesong. ]

     Ostrich: Just completed my wrap-up of the three SRAT solutions
(some people play Minesweeper, some resolve logic problems with rigor
mortis).  Conclusion is: we all blew it in places.
     1. You caught me in a loose step; for the 'by 2~B' you questioned,
please read 'by 1~A & 2~B & 3~C'. Actually, though, by pointing this out
you show a larger weakness in my solution, that I would be more
efficient solving for 5 first and then proceeding to 3, 4, 1.
     2. Qaqaq shows that I waste a lot of time not immediately
eliminating the impossible case [4] and nailing down 9 and 11; I
could've just used his sentence ending '9~B' (ll. 5-6). Instead I fail
to see that 9=B 11=C is a direct contradiction in 9 (even if it
validates 11). I was also overwordy elsewhere.
     3. Qaqaq's step you questioned does indeed seem a misjudgment;
perhaps a temporary myopia that 18A 19E would violate 2. Not as bad as
my myopia and embarrassment factor in the previous paragraph. Using
this apparently incorrect step, he eliminates my case [2] which appears
valid, and allows only cases [3], [1], [5]. So he said three cases and
I said five when there are in fact four.
     4. You state about halfway down '2/9=A/D or D/A'; you would be a
little better off by explicitly including also A/B and then ruling it
out via 11. I suspect this was in your notes somewhere anyway though.
     5. I don't object to your bringing 20=E in early as a timesaver,
but feel a complete solution should list all four cases.
     6. Qaqaq and you both include the steps 5=E and 12=A/B (and you
add 8=C/E, 12=A) by a 'uniq(u)eness' argument. I think, though, that
these are presumptive. The argument you apply seems to be 'if 5 were
something other than E, then E would also work; but the quiz has a
unique solution; so 5 cannot have two possible answers; so it must be
E.' But this neglects the fact that the other questions limit the range
of answers to 5 (as they do to 19). I might say, e.g., 19=A because it
works and the answer must be 'unique', by this same argument, even
though 13 explicitly contradicts me. Similarly, I might falsely
conclude that if 9=D and 13=D (which are true), by uniqueness 10~A and
11~B (which are false), since if 10=A then 9 could also be A and not be
a 'unique' answer. In the actual event, though, both of you got lucky,
because 5E, 8E, 12A were all right; but you would not have been if
you'd misapplied this argument elsewhere. [I thought that the
uniqueness argument was valid within a single question, but 19 shows
it's not.]
     So here is a semicanonical, total-symbolic solution that improves
on all of ours. Thanks to Qaqaq for use of [] for 'by'.
     SYMBOLS ('is' is implied by context; ,;.])}s are self-evident):
= established to be, ~ not, > implies, / or ('exclusive or'), + or
more, - or fewer, < among questions lower than, [ by, ( hypothetical,
{ case, * contradiction
     10=A, 16=D [10, 16]. 17~16D [2]; 6=D, 17=B [6, 17]. 1~AB, 2~B [1];
5~B; 1~E. 3~4 [2]; 3B / 4B [1]. 8~BD [7]; 8=E [3, 4, 8]; 12=A; 15=A;
13=D; 19~A. 3BD [8]. 5~6D [2]; 5=E [1, 2, 3]; 2+ Es; 3~B; 3=D; 4=B;
1=D; 3 Es, 5 As; 18~D. 9~E [9E > 14E > 4+ Es *]; 2~C [8, 9]; 2~3D,
11~12A [2]; 2~E [10, 11]; 2=A; 7=D. 9~B [9B > 11B > 1 Bs<11 *]; 11=B.
9~C [12]; 9~10A [2]; 9=D. 2- Cs; 18~B. 7+ Ds; 18~C, 14~A. 9- Ds; 14~E.
14~13D [2]. (19=B {1}. 14=B [14C > 8 Ds *]; 5 Bs; 18=A; 20=E [3].)
(19=C {2}. 14=B [14C > 8Ds *]; 4 Bs; 18=E; 20=A [4].) (19=D {3}. 8+
Ds; 14=C; 3 Bs; 18=E; 20=A [4].) (19=E {4}. 18~19E [2]; 18=A; 5 Bs;
14=B, 20=B.) (* Right brain finally interrupts: you dummy, that
barometer doesn't measure any of those four options, just like a
standardized test DOESN'T measure intelligence! *) 20=E; {1}; 14=B,
18=A, 19=B.

-- Ostrich

[ Editor: the answer, from Ostrich's first solution, is
DADBE DDEDA BADBA DBABE. ]