🚀 go-pugleaf

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.