Question: Two sheriffs are working on a case to find one culprit. There were initially 8 suspects; through independent work, each sheriff has narrowed this down to a list of 2. Because they are good sheriffs, they can be sure that both of their lists contain the culprit. They plan to make a phone call tonight to see if they can combine their information to find the culprit. However, the locals (who know the 8 suspects, but neither of the narrowed lists) have tapped their phone line. If the locals can figure out the culprit based on this phone call, then they will lynch him before he can be brought to justice. 

How can the sheriffs conduct their conversation so that they can both deduce the culprit, when possible\*, but the locals can't?

\*If the sheriffs have the same list of 2 suspects, then they won't be able to deduce the culprit. As long as their lists are different, they can. The lists cannot be disjoint, since they both contain the unique culprit.

Source: *Mathematical Puzzles, a Connoisseur's Collection*, Peter Winkler

Answer: 
In actual conversation:

Suspects: 1 2 3 4 5 6 7 8

Sheriff A has 1 2

Sheriff B has 1 5

(Phone rings.)

A: Bad news, someone tapped me.

B: OK. Group suspects into 4 pairs with one pair being your shortlist

A: 12 34 56 78

B: Is it either 12 or 34?

A: Yes

B: (so culprit is 1) culprit is either 1 or 3

A: (so culprit is 1) let's go get him

(Both hang up the phone)
The crude answer to this is that they manually perform a [Diffie-Hellman key exchange](http://en.wikipedia.org/wiki/Diffie%E2%80%93Hellman_key_exchange) over the phone. 

Then they can just encrypt their lists with [AES](http://en.wikipedia.org/wiki/Advanced_Encryption_Standard) with the key, and speak it. 
Surely this isn't the most efficient or effective way, but encryption is the catch-all when it comes to questions of communicating secrets in the open :)