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)
> 
> they number the suspects in binary, then each officer tells the other which digit he needs to incriminate the right suspect. the other officer can either give him that digit right away if he has it, or request a digit himself.
> 
> 
> 

example

officer A has  

011  

101 

officer B has  

000  

011 

officer A needs either the first or second digit.  

officer B needs either the second or third digit.  

officer A tells B the third digit it 1.  

officer B tells A the first digit is 0.