## Description

This article is from the Puzzles FAQ,
by Chris Cole chris@questrel.questrel.com and Matthew Daly
mwdaly@pobox.com with numerous contributions by others.

# 228 induction/handshake.p

A married couple organizes a party. They only invite other married

couples. At least one person of an invited couple is acquainted to

at least the host or the hostess (so between sets {host,hostess} and

{male of invited couple, female of invited couple} there exists at

least one relation, but two, three or four relations is also possible).

Upon arrival at the party, each person shakes hands with all other

guests he/she doesn't know yet (it is assumed everybody knows

him/herself and his/her partner).

When all couples have arrived and all the handshaking has been done,

the host mingles between the guests and ask everybody (including his

wife): "How many hands did you shake?". To his surprise, all responses

are different.

With the above information, you must be able to determine how many

hands the host and hostess each shook.

induction/handshake.s

Assume there were 2n people (including host and hostess)

in the party.

1. When the host asked the question he must have got

2n-1 responses (including from his wife).

2. All of the responses were different.

The responses have to be (0, 1, 2, 3, ..., 2n-2)

to satisfy the above requirements. As 2n-2 is the maximum

possible handshakes any person in this party could have made.

/** Below,
P{x} - means a person who shook x hands.
H - means the host
**/
H: <-------->2n-2 0
2n-3 1
2n-4 2
2n-5 3
. .
. .
. .
n n-1 n-2

(There are 2n-1 on the circle.)

P{2n-2} must have handshaked with H (because in the circle he

can handshake with only 2n-3. He has to exclude himself also.)

P{2n-3} must have handshaked with H (because in the circle he

can handshake with only 2n-4.)

P{2n-4} must have handshaked with H (because in the circle he

can handshake with only 2n-5.)

P{n} must have handshaked with H (because in the circle he

can handshake with only n-1.)

from P{n-1} to P{0} nobody handshakes with H, because, for them

the handshake numbers are satisfied on the circle itself.

This leaves H with (n-1) handshakes.

----------------------------------

P{0} must be the spouse of P{2n-2} (since P{2n-2} handshakes with

everbody else.)

Continue to: