## Simple Puzzle, Stunning Solution

So I stumbled upon one of those logic puzzles that I love.

Six boys are accused of stealing apples. Exactly two are guilty. Which two? When the boys are questioned, Harry names Charlie and George, James names Donald and Tom, Donald names Tom and Charlie, George names Harry and Charlie, and Charlie names Donald and James. Tom can’t be found. Four of the boys who were questioned named one guilty boy correctly and one incorrectly, and the fifth lied outright. Who stole the apples?

And I got the solution correct (in a few mninutes).

But how a mathematician proved the solution in a few seconds was a revelation to me.

Hat-Tip:Futility Closet.

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I also solved this problem graphically, but used a different method. For each accusation, I drew a directed link from the accuser to the accused, for a total of 10 directed links. Since exactly 4 of the accusations are correct and there are exactly 2 guilty boys, I need to find 2 boys who have a total of 4 links pointing toward them. Thus, the two boys must each have 2 accusations or one must have 3 and the other 1. Donald and Tom are the only two boys accused twice, but James accused both of them, so they can’t both be guilty. Thus, Charlie, the only boy accused 3 times, must be guilty and the other guilty boy must have only one accusation. The 3 boys who accused Charlie must have selected an innocent boy with their other accusation, so that eliminates Harry, George, and Tom. James is the only remaining boy with only one accusation, so he’s the other guilty party.

Once I drew the graph, this took me about 30 seconds. This might actually be equivalent to the proof provided, but it isn’t obvious to me that it is.

huzonfirstFebruary 24, 2014 at 9:54 pm

This strikes me as isomorphic to the solution provided.

taogamingFebruary 24, 2014 at 10:33 pm

Thanks for posting this. I’ve done a lot of these puzzles, and I’ve never once thought of using a graphical approach to solving them, so I found the mathematician’s description very interesting. I’ll have to keep it in mind in the future.

My solution was the non-graphical version of Larry’s. I noticed right away that Charlie was listed three times, realized that that meant he had to be guilty, and then solved for James.

Lou WFebruary 26, 2014 at 7:47 am

I don’t think that being accused 3 times guarantees guilt, Lou. Charlie could have been accused incorrectly by 2 boys (so that the other boys accused by them are guilty), as well as by the boy who lied twice. Digging a little deeper probably nails him, but I don’t think you can start with the premise that Charlie is guilty because he was listed 3 times.

huzonfirstFebruary 26, 2014 at 10:49 pm

I didn’t mean to oversimplify my solution, because you are right, three namings alone isn’t definitive enough, although it’s a strong indication. Here was my actual solution process, which did take that into account.

1) Count the names, notice Charlie shows up three times

2) Assume that Charlie is guilty, and check if I can get a valid unique solution [Yes, James]

3) Assume that Charlie is innocent and ensure that there isn’t a solution [No. Two of Harry, George or Tom would have to be guilty and Harry and George aren’t named again].

I’ve always been a big believer when doing puzzles of this type of using the experimental approach of testing the solution that feels the most likely, right at the beginning. It is often faster to say “Assume [X} is guilty” and iterate through the candidates, than to try to envision all the relationships so that the unique valid solution is apparent (which is what the graphical approach did). That said, it is a matter of scale, sufficiently hard puzzles are likely better attacked through that graphing method, which is why I was happy to learn about it and have access to a new tool.

Lou WFebruary 27, 2014 at 7:23 am