Metapuzzles

A metapuzzle is a kind of puzzle in which you are not given complete information but can solve the puzzle only on the basis of knowing that someone who had more information was or was not able to solve it. These puzzles tend to be pretty easy and interesting to solve, though they may look difficult at first. (excerpts from 'The Riddle of Scheherazade'-Raymond Smullyan)

(1) Ages of Children

A: What are the ages in years of your three children?
B: Let's see, the product of their ages is thirty-six.
A: That doesn't tell me their ages.
B: Also, the sum of their ages is your own age.
A: I still don't have enough information.
B: Fine. My son is more than a year older than both his sisters.
A: Okay, now I know their ages.
B: And they are...?

Discussion - The first clue starts us off with eight distinct set of possible solutions. The metapuzzle lies in the second clue and subsequent reply. The fact that A was still unable to figure out the answer means that there must be more that one set of possible solutions whereby the sum of their ages equal A's age. Thus summing the ages for all the sets, we find only one case involving (1,6,6) and (2,2,9), which both total 13. Note: We have hence managed to deduce that A is 13 years old, which was the additional information not told. The last clue eliminates (1,6,6), telling us that B has a son aged nine and twin daughters aged two.

(2) Prize Box

There are three boxes on the table coloured yellow, red and blue. B hides a prize in one of them and A has to find it by asking B any number of yes or no questions. However, B informs A that he can choose either to answer truthfully to all the questions or lie to all the questions. After asking the following two questions(and getting the answers), A was able to correctly deduce where the prize was. So where was the prize?
Q1:Is the prize in the yellow box?
Q2:Is the prize in the red box?

Discussion - Consider B's answers(which we were not told). He could have given similiar answers (Y,Y)/(N,N) or dissimiliar answers (Y,N)/(N,Y). If he gave dissimiliar answers, it would not be possible for A to deduce where the prize was without knowing whether B told the truth or lied. Hence, B gave similiar answers. If he answered (Y,Y), he must have lied; conversely if he answered (N,N), he must be telling the truth. In both cases, the prize must be in the blue box.

(3) Chests and Jewals

Four men(A,B,C,D) were given a test. In front of each man was a chest of three drawers. Each drawer contained either an emerald(E) or a ruby(R). One chest contained three Es, one contained two Es and one R, one contained one E and two Rs and the last contained three Rs. Four labels (E,E,E),(E,E,R),(E,R,R) and (R,R,R) had been put on the four chests and the four men were each told that his label was wrong. No man could see any label but his own. Each was to open two drawers of his chest and try to determine what jewal lay in the third drawer. They started the test and sounded off in the order:
A: I've found two Es, and I know what my remaining jewal is.
B: I've found one E and one R, and I know what my remaining jewal is.
C: I've found two Rs.
D: And do you know what is your remaining jewal?
After C replied to D, he was able to deduce the contents and labels of each of the four chests without even opening any of his drawers! So what were they?

Discussion - Since A knew his last jewal, he either got label (E,E,E) or (E,E,R). Since B knew his last jewal, he either got label (E,R,R) or (E,E,R). For C, case 1 is if he knows his last jewal, case 2 if he doesn't. Hence for case 1, he either got label (E,R,R) or (R,R,R). After eliminating duplicate arrangements and labels, this actually gives us only two possible sets of solutions -
A:'E,E,R':(E,E,E), B:'E,R,R':(E,E,R), C:'R,R,R':(E,R,R), D:'E,E,E':(R,R,R)
A:'E,E,E':(E,E,R), B:'E,E,R':(E,R,R), C:'E,R,R':(R,R,R), D:'R,R,R':(E,E,E)

Since there are two possible solutions, D would not have been able to deduce the correct arrangements. Hence we eliminate case 1 and conclude that C does not know what his last jewal is. So C got either of the other two labels - (E,E,E) or (E,E,R). True enough, this time round, after eliminating duplicate arrangements and labels, we have only one final solution:
A:'E,E,E':(E,E,R), B:'E,E,R':(E,R,R), C:'R,R,R':(E,E,E), D:'E,R,R':(R,R,R)

Hosted by www.Geocities.ws