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Criminal Logic Problem

Date: 02/19/2005 at 18:37:53
From: Jay
Subject: logical reasoning

Five criminals appeared before the court for sentence.  Their names,
strange to say, were Fraud, Libel, Blackmail, Theft and Murder--each
the namesake of the crime with which one of the others was charged.

The namesake of the crime with which Blackmail was charged was himself
charged with the crime of which the namesake was charged with murder.

The namesake of the crime with which Murder was charged was himself
charged with the crime of which the namesake was charged with fraud.

All the prisoners were found guilty and sentenced.  Theft, for 
example, got seven years.  The criminal who was found guilty of murder 
was placed on death row.  Who committed the murder?  Explain your 
reasoning.

I don't know how to figure this out.  I get confused reading the
problem itself!



Date: 02/20/2005 at 11:04:34
From: Doctor Vogler
Subject: Re: logical reasoning

Hi Jay,

Thanks for writing to Dr Math.

That's a very good question.  Here's how I analyze it:  We have five
(charged and convicted) criminals, whom I will denote F, L, B, T, and
M.  I will write

  x -> y

to mean that x was charged with the crime y.  Of course, crimes and
criminals have the same names, so that

  x -> y -> z

means that x was charged with the crime called y, and the criminal y
was charged with the crime z.  Notice that

  x -> x

is never true, as was stated in the problem.  The next sentence of the
problem is a little hard to work out, but it ultimately comes down to

  B -> x -> y -> M

and the sentence following is exactly parallel

  M -> z -> w -> F.

Now, if we consider the five crimes/criminals, starting with, say, B,
and looking at the sequence

  B -> x1 -> x2 -> x3 -> x4 -> ...

we should find that the sequence repeats.  You see, there are only
five crimes (or five criminals) so some two x's will be the same. 
Then the sequence between them will repeat.  Furthermore, since each
crime was only charged to one criminal, the sequence repeats backward
to the B as well.  We mathematicians call this a "cycle."  As I
already pointed out, the length of the cycle cannot be one, since no
criminal was charged with his namesake.

That leaves two possibilities:  Either two criminals were charged with
one another's namesake (a cycle of length two) and the other three
form a cycle of length three, or there is only one cycle of length
five.  (Of course, we can't have a cycle of length four, since the
leftover criminal would have to be charged with his namesake.)  But

  B -> x -> y -> M
  M -> z -> w -> F

means that B, M, and F cannot be in a cycle of length three.  There
are only two other criminals left, which isn't enough for a cycle of
length three.  That means that the first possibility is not possible,
so there is only one cycle of length five.  So that means that we have

  B -> x -> y -> M -> z -> B

with the cycle being (B, x, y, M, z).  But then

  M -> z -> w -> F

means that w = B, and F = x, so that our cycle is

  B -> F -> y -> M -> z -> B.

Now it only remains to determine y and z.  We want to find out who was
charged with murder (and therefore convicted), which means who is y? 
Well, y was placed on death row, which means that y is not Theft,
because Theft got seven years.  So that means that z is Theft, and the
only remaining criminal is Libel, who must be the murderer y.

  B -> F -> L -> M -> T -> B

And we conclude that

  Blackmail was charged with fraud,
  Fraud was charged with libel,
  Libel was charged with murder,
  Murder was charged with theft, and
  Theft was charged with blackmail.

If you have any questions about this, please write back, and I will
try to explain further.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 02/20/2005 at 12:35:25
From: Jay
Subject: Thank you (logical reasoning)

Thank You Doctor Vogler,

This detailed explanation really helped... thank you so much.
Associated Topics:
High School Logic
High School Puzzles
Middle School Logic
Middle School Puzzles

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