Answer to The Usual Suspects

This is the solution to the puzzle The Usual Suspects which I published today.  If you haven't read it yet, don't self-spoil - go have a read of it and see how far you get.  Then come back here.

So, there are two ways to answer this puzzle.  The first is with some very, very careful reasoning.  After a great deal of time, and a not of expended energy, you can get there.  It is possible.  But its not very elegant so I'll not go in to it here.

The second way, the elegant way, is using Boolean Algebra.

So, we have 6 suspects, which to simplify matters I'll call A, B, C, D, E and F.

We know that two of them are guilty, and only two.

We also have the brainwave of the detective, that each of the suspects when interrogated will give the names of one guilty man, and one innocent man.

So lets set up a table of the various claims and counter claims.

And here's a logic table of all possible combinations.  There are actually far fewer than the puzzle might have implied, as we're dealing with combinations of two.

Now, for each line in the table of all possible combinations, we're checking against the claims and counterclaims table whether.  So,

  1. Take each line from the table of all possible combinations.  This will give a different pair each time.  E.g. the first line gives AB, the second line gives AC.
  2. Using the two letters from step one, check each row of the claims table.  
    1. If that row has one of the two letters as the accused, move on to the next line.
    2. If that row does not have either of the two letters, then that accusation is false.  Move to the next line in the table of all possible combinations and go back to Step One.
  3. When you get to the last row of the claims table, having matched at one of the letters in each line, then that's your answer.  Those are the two perpetrators.
 So to illustrate.

Therefore, the solution to the puzzle is:  It was Charlie and Donald what dunnit.


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