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- From: jhaukioj@sara.cc.utu.fi (Jussi Haukioja)
- Newsgroups: sci.logic
- Subject: Re: The poor parachutist...
- Message-ID: <1992Aug20.114621.1@sara.cc.utu.fi>
- Date: 20 Aug 92 08:56:53 GMT
- References: <1992Aug18.131259.14667@wisipc.weizmann.ac.il> <1992Aug19.090212.1658@gacvx2.gac.edu>
- Sender: news@polaris.utu.fi (Usenet News admin)
- Organization: University of Turku, Finland
- Lines: 54
-
- Maybe a group ?.logic.puzzles would be in order?
-
- In article <1992Aug19.090212.1658@gacvx2.gac.edu>, logic@gacvx2.gac.edu writes:
- >
- > Now for a riddle of a different flavor.. I first saw this in an interview
- > article with Paul Halmos in the Mathematical Intelligencer.. I offer it here
- > because it is (1) not as widely known as the riddle above, (2) more difficult
- > than the riddle above, and (3) its solution offers some surprises.. Here we
- > go. My wife and I go to a party.. There are 4 other couples there for a total
- > of 5 couples. (No funny business here.. a couple is husband and wife..) As
- > is usual at a social gathering a certain amount of handshaking takes place..
- > After the handshaking has completed, I interview all of the people at the party
- > including my wife. I ask them how many hands they shook. (note: spouses don't
- > shake hands and no one shakes their own hand).. Now each person, therefore,
- > could have shook as few as 0 hands or as many as 8.. Now, in fact, that range
- > of answers is exactly what i get for answers! I.e., the first person I
- > interview responds 0, the second 1, the third 2, ..., the ninth 8.. The riddle
- > is "How many hands did my wife shake?".. I'll even give you the answer.. It
- > is 4.. The real riddle, of course, is explaining why this is the correct
- > answer.. If "pigeon hole" suddenly occurs to you, you're on the right track..
- > Subriddle: Who is the "meanest" person at this party married to?
- >
- > enjoy,
- > al hart
- >
-
- First, lets call the nine people you (represented here by U) interviewed by the
- letters A-I (A shook 0 hands, B shook 1, ... ,I shook 8 hands).
-
- Now drawing a simple graph helps (actually, the graph could be omitted; it's
- just nice to see how things turn out and to be assured that the puzzle is
- soluble, that is, that U has not shaken hands with E): we see that the only way
- to have the required numbers of handshakes for A-I is:
- Person The people with whom s/he shook hands
- A
- B I
- C H,I
- D G,H,I
- E F,G,H,I
- F E,G,H,I,U
- G D,E,F,H,I,U
- H C,D,E,F,G,I,U
- I B,C,D,E,F,G,H,U
- which shows us:
- U F,G,H,I
-
- Now we see that I has to be married to A: I has shaken hands with everyone else
- (subriddle)
- By the same reasoning, H is married to B: H has shaken hands with everyone but A
- and B, and A is already married to I
- And so on: G&C are married to each other, so U is left with E, who shook 4
- hands. Q.E.D.
-
- Jussi.
-