Logic Challenge #20 - The IPA Game

[lrthompson]

E
D
C
A
E
C
E

[lrthompson]

Well, since no one else is going for it...
1. Which one of the following could represent the order in which the beers are tasted, from first to last?

(A) no, N and Q separated by more than one
(B) no, M before N, AND P before R
(C) no, neither M before N, nor P before R
(D) no, O not in third spot.
(E) yes!

2. Which one of the following must be true?

(A) no, not a must
(B) no, not a must
(C) no, not a must
(D)yes, N and Q are either in 2 and 4 or 4 and 6
(E) no, not a must.

3. If Roundabout is tasted first, which one of the following must be true?

(A) No, not a must.
(B)No, not a must.
(C) Yes, M must be second to be before N, since R is already before P.
(D) No, not a must
(E) No, not a must

4. Each of the following could be true EXCEPT:

(A) Yes, this could not be true, sine it leaves only R and M as variable positions. Neither can go in the first position since if R is first, neither P is before R nor is M before M (R P O N L Q M) and if M is first, both P is before R AND M is before N (M P O N L Q R)
(B) no, could be true
(C) no, could be true
(D) no, could be true
(E) no, could be true

5. If Newtonbury is tasted at some point after Pumpkin Pale, which one of the following could be true?
(A) No, this puts either Neither M->N nor P->R, or it puts both P->R, M->N L, M/R, O, Q, P, N, R/M
(B) No, same problem as in (a)
(C) no, same problem as in (a) - when P must be 5th, and M must be last, neither M->N nor P->R
(D) no, same problem as (a)
(E) Yes, could happen . M R O Q P N L

6. If Pumpkin Pale is tasted first, the seven beers could be tasted in how many different orders?
P _ O _ L _ _
P, (N/Q), O, (N/Q), L, (M/R), (M/R) there are 2 x 2 variations here (4)
P, R. O, (N/Q), L, (N/Q), M there are 2 variations here.

the total is (C) Six

7. Which one of the following, if substituted for the condition that either Pumpkin Pale is tasted before Roundabout or Moon Rune is tasted before Newtonbury, but not both, would have the same effect on the order in which the beers are tasted?

(A) No, not as restrictive
(B) No, not as restrictive
(C) No, not as restrictive
(D) No, just as restrictive, but results in different order
(E) Yes, results in the same order since it is the same rule as before (just written as opposites)

[jonathanleslie73]

This is the explanation that I wrote on my bus ride from Baltimore to Philly tonight. I hope it is helpful (I had to add in _ or - into the diagrams order for it to post correctly):



-?----?----O--N/Q-L/P--?--?-
___ ___ ___ ___ ___ ___ ___
1___ 2__3___4___5__6___7
XN____________________XN
XQ____________________XQ

rule three: N/Q_Q/N
rule four: P < R and N< M OR R < P AND M < N.

The inference we make is that, because P/Q must (by process of elimination) be on two of the three slots, 2, 4, in 6, in order to accord with ruler three, one of the two beers MUST be drunk fourth in accordance with rule three. Rule Four in essence dictates two possible alternatives, which we write down.

1. Which one of the following could represent the order in which the beers are tasted, from first to last?

(A) Long Lake, Newtonbury, Olenguard, Pumpkin Pale, Quest, Roundabout, Moon Rune
(B) Long Lake, Moon Rune, Olenguard, Newtonbury, Pumpkin Pale, Quest, Roundabout
(C) Long Lake, Quest, Olenguard, Newtonbury, Roundabout, Pumpkin Pale, Moon Rune
(D) Long Lake, Newtonbury, Moon Rune, Quest, Pumpkin Pale, Olenguard, Roundabout
(E) Long Lake, Quest, Olenguard, Newtonbury, Pumpkin Pale, Roundabout, Moon Rune

This question is easy because we can cancel out each option except the last, with the first three rather simple rules. Cancel out (D) because O is not on the fifth spot and (A) because neither L or P is on the fifth. Then, looking the fourth rule, we can see that (B) cannot be true as it has BOTH P before R AND M before N;(C) cannot be true because it has NEITHER P before R (N)OR M before N. Thus, (E) must be correct.

2. Which one of the following must be true?

(A) Long Lake must be the first beer tasted.
(B) Moon Rune must be the last beer tasted.
(C) Pumpkin Pale must be the fifth beer tasted.
(D) Either Newtonbury or Quest must be tasted fourth.
(E) Either Long Lake or Pumpkin Pale must be tasted first.

The best way is to look back onto our initial representation in which we made the exact inference that is suggested by (D).

3. If Roundabout is tasted first, which one of the following must be true?

(A) Pumpkin Pale is tasted before Moon Rune.
(B) Pumpkin Pale is tasted before Long Lake.
(C) Moon Rune is the second beer tasted.
(D) Newtonbury is the sixth beer tasted.
(E) Quest is the sixth beer tasted.

Here, we should make a side-diagram:

R -------- O--N/Q-L/P
___ ___ ___ ___ ___ ___ ___
1___ 2__3___4___5__6___7

We can also make the inference, looking at rule four, that M MUST be before N, as R is clearly before P.
So:
M<N

With this inference, we can quickly see that M can only be in one spot, which is 2. This is because N can only be, as we know, in spots 2, 4, 6, according to the general rules of the puzzle. All other spots for M that are before spot 6 (the highest spot that N can have), including 1, 3 4, and 5, are, as it were, reserved according to other rules by other beers. Thus, we can scribble M in spot 2:

R-----M---O---N/Q-L/P
___ ___ ___ ___ ___ ___ ___
1___ 2__3___4___5__6___7

So we know that we have the answer in (C), as soon as we look at it.

4. Each of the following could be true EXCEPT:

(A) Pumpkin Pale is tasted second.
(B) Moon Rune is the first beer tasted.
(C) Moon Rune is the seventh beer tasted.
(D) Roundabout is tasted second.
(E) Roundabout is tasted seventh.

The best way to approach this question, which asks what must be universally excluded from any answer, in my view, is by a reverse process of elimination, viz. starting with the likeliest answer and working our way down. We should therefore view our initial diagram of the puzzle:

-?----?----O--N/Q-L/P--?--?-
___ ___ ___ ___ ___ ___ ___
1___ 2__3___4___5__6___7
XN--------------------------- XN
XQ--------------------------- XQ


We know, in general, that the most "open" spots, viz. those that the least rules governing what what must be drunk on them, are 6 and 2 (note how there is nothing above or below them.) Therefore, it is likely that the answer will involve a statement about these spots, as this would logically lead to the greatest amount of conclusions, by which there is therefore more likely to be a contradiction (in what is otherwise a fairly open puzzle) and therefore an answer. We note, therefore, that only (A) and (D) deal with these spots, so we start with them. With either of these, we can make the inference that spot 6 must be taken up by either N/Q, by matter of elimination. We can also note that it might be smarter to start with (A) because here we can make an extra immediate inference, which is that L is on spot 5. As there are only two spots left, for two beers (M and R), we should indicate that as well. Hence:

M/R -P -O N/Q L- N/Q M/R
___ ___ ___ ___ ___ ___ ___
1___ 2__3___4___5__6___7
XN----------------------------XN
XQ----------------------------XQ

Then, we review rule four, and see that we are trapped in a contradiction. If M is on space 1 and R is on space 7 then M is before N AND P is before R. However, if R is on space one and M is on space 7 then P is NOT before R AND M is NOT before N. Both of these are violations of rule four and the only two logical possibilities, so we have an answer, which is (A.)

5. If Newtonbury is tasted at some point after Pumpkin Pale, which one of the following could be true?
(A) Long Lake is tasted before Pumpkin Pale.
(B) Neither Newtonbury nor Quest is tasted before Long Lake.
(C) Neither Moon Rune nor Pumpkin Pale is tasted before Olenguard.
(D) Olenguard is tasted after Long Lake.
(E) Moon Rune is tasted before Roundabout.

This one is a bit harder, as it states a relationship ( P < N) rather than a fact. It also requires that we only eliminate something when we know it to be possible. On the face of it, the most improbable statements are (B) and (C) because they involve two rather than one statements, basically, so we can begin with the others. Also, (D) appears on its face improbable because there are only possible places for L. So we begin with either (A) or (E). We can try (A). Looking at the general diagram and rules, we can add in the new rule and scribble in a few new conditions:

------------O--N/Q-L/P
___ ___ ___ ___ ___ ___ ___
1___ 2__3___4___5__6___7
XN---------------------- XP XN
XQ-----------------------XP XQ
-------------------------------XP


rule three: N/Q_Q/N
rule four: P<R U M<N
new rule: P< N

Because P < N, we know that P CANNOT be in spot 6 or 7, as it would therefore by definition be beyond N.

Assuming, according to (a) that L is before P, P MUST BE on spot 5, as if it were not then L would be on spot 5, in violation of the new rule. N would have to be on spot 6 and Q on spot 4; L on either 1 or 2 and, by definition, M/R would have to be on 7. We can write these inferences, for (a), softly over our diagram if we want:

L/ --/L- O -- Q -- P --- N - M/R
___ ___ ___ ___ ___ ___ ___
1___ 2__3___4___5__6___7
XN---------------------- XP XN
XQ-----------------------XP XQ
-------------------------------XP


Then, we can look back to rule 4 and recognize that, if M is on 7, then R is on 1 or 2, and neither P is before R NOR is M before N; likewise, the reverse, both are TRUE (if we are smart, we can recognize that this is the same trap of contradiction that we saw in question (4) and that therefore (a) cannot be the answer.

So, moving along to (e), in which M<R, we erase our previous inferences and make new ones. We see that we have a cluster of rules about relationships involving the same letters (M<R, P<N, rule four) and want to combine them for purposes of simplicity. The obviously best (or it should be anyway) way to do this is to think about the two different alternatives required by rule four: (either P < R and N< M OR R < P AND M < N. with an exclusive OR here.) In the first, in this case, we know, assuming M < R, that N < (M, P) < R. However, this cannot be the case, as the questions asks us to assume that P < N. So, only the second possibility is left, which in the end requires the following: M <R < P < N.

The question then is if we can plug this into our diagram, with the other rules (that hopefully we've memorized by this point.) The answer is that we can can!, so (E) is the answer. Here is what you get:

M R O Q P N L
___ ___ ___ ___ ___ ___ ___
1 2 3 4 5 6 7

6. If Pumpkin Pale is tasted first, the seven beers could be tasted in how many different orders?

(A) Four
(B) Five
(C) Six
(D) Seven
(E) Eight

First, we diagram what we know and indicate all the possibilities:

P----------O---N/Q--L
___ ___ ___ ___ ___ ___ ___
1___ 2__3___4___5__6___7
XN---XM--------------------- XN
XM--------------------------- XQ


We go back to rule four and can see that, as P is before R, therefore N < M. This means that M must be in either spot six or seven. The obvious way to proceed is then to count each of the alternatives that stem from making either assumption (and then add them.) If M is in spot 6, then there is only two possible alternatives, in which N and Q switch places (all the other spots are fixed.) If M is in spot 7, then R can be in spot 2 or 6, and in each of these spots N/Q can switch places, so there are four possible alternatives. Four plus two is six, thus the answer is (C.)

7. Which one of the following, if substituted for the condition that either Pumpkin Pale is tasted before Roundabout or Moon Rune is tasted before Newtonbury, but not both, would have the same effect on the order in which the beers are tasted?

(A) Either Roundabout is tasted after Pumpkin Pale or before Moon Rune, but not both.
(B) Either Long Lake or Roundabout is the first beer tasted.
(C) Either Newtonbury or Quest is tasted sixth.
(D) Either Pumpkin Pale is tasted before Moon Rune, or Roundabout is tasted before Newtonbury, but not both.
(E) Either Roundabout is tasted before Pumpkin Pale, or Newtonbury is tasted before Moon Rune, but not both.

We begin with the most plausible-sounding answer, which would disqualify B, C, and probably A (if we could make such inferences, then we probably already would have done so.)

Remember that we wrote rule four, which this question quotes, as: (P < R and N< M) OR (R < P AND M < N.) We transcribe both (D) and (E.) D becomes: (P< M and N < R) OR (M<P and R<N); E is (R<P AND M<N) OR (P<R AND N<M.) This is exactly rule four, so we have our answer, which is (e). Voila.