Criminal Logic ProblemDate: 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. |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/