Friday Puzzler -- Handshakes

Just to be clear

Alexy

And I hope this hasn't thrown anyone off.

The only people at this event were the 5 couples mentioned in the puzzle. Obviously, if there were other people we were shaking hands with, there would be no possible solution.

Re: Friday Puzzler -- Handshakes clarification

Larry Barrett

To be sure we understand, suppose there is just one other couple, A1 and A2, in addition to your wife and you, W and Y.

So A1 could shake hands with W and Y, A2 could also shake hands with W and Y.

Then, when you stop and ask, A1 would say 2, A2 would say 2, and W would say 2.

Another scenario, which matches the result you postulated, is A1 shakes hands with W an Y, A2 shakes hand with no one. Then, when you stop and ask, A1 would say 2, A2 would say 0, and W would say 1.

You don't count your handshakes as part of the answer. Is this correct?

Re: Friday Puzzler -- Handshakes

Larry Barrett

Assuming my simple example is correct, I think the answer to the 4 couples problem is that your wife shook hands with 4 people.

Here is my reasoning:

In the simple example with one couple, A1 and A2, A1 could have shaken hands with W and Y, and A2 with no one, or vice versa. In either case, W shook hands wth one person.

Extend this example to 2 couples, A1, A2, B1, B2, plus W and Y.

One solution is for A1 to shake hands with B1, B2, W and Y (4 people), A2 to shake hands with no one, B1 to shake hands with W and Y (2 people, plus A1), B2 to shake hands with no one else (plus A1).

So A1 shakes hands with 4 people, A2 with 0, B1 with 3, B2 with 1, and W with 2. I think A1, A2, B1, B2 could be various permutations, but W will always be 2.

Extend this to 3 couples and one solution is

A1 with 6, A2 with 0, B1 with 5, B2 with 1, C1 with 4, C2 with 2 and W with 3. Again, various permutations of A1, A2, B1, B2, C1, C3 but W always with 3.

Extend this to 4 couples and W will shake hands with 4 people.

Correct!

Alex Y

Nice reasoning from smaller case to larger!

Another way to look at it: There are 9 people I asked. The most hands anyone could shake is 8, so the answers I got were 0,1,2,...,7,8

Call the person who shook hands with 8 people A1. He or she shook hands with both members of couples B, C, D, and E. Since they all shook hands with him, the only one left to have shaken hands 0 times is A2.

Then call the person who shook hands with 7 people B1. He shook hands with A1 and with both members of C, D, and E. Since C, D, and E all shook hands with both A1 and B1, that leaves only B2 as the person that shook hands with 1.

Similarly, we find that the couples shook hands with

A:9,0

B:7,1

C:6,2

D:5,3

E:4,4

Since everyone I asked gave a different answer, I obviously didn't ask both E1 and E2, so one of those is me and the other is my wife.