In 1996, a 19-year-old man was arrested, and charged with murder. Although, the man claimed he was innocent, he was convicted of the crime, and sentenced to 25 years in prison. In 2006, after serving ten years of his sentence, a man came forward with an incredible story; that he, in fact, was the person that committed the crime, and the wrong man had been in prison for the past ten years.

After an intense investigation, it was concluded that the man was, in fact, telling the truth; he committed the crime. He had too many details of the crime that were never released to the public, and was able to answer questions, that, until now, remained unanswered.

After a short trial, this man was convicted of the crime, and was sentenced to 25 years in prison. The 19-year-old man (now 29), that was wrongly imprisoned, was released the next day. He was issued a formal apology, and was given a check for a measly $300, to help him start a new life, and find a job.

Shortly after the man’s release from prison, a third man came to the police department, and confessed that he, and the now 29-year-old man, committed a robbery, in 1993. So, after being released for only 6 days, the 29-year-old man found himself arrested once again, and back in the court room. After a trial that lasted 5 weeks, and all the evidence was presented, along with a confession from both men, they were both convicted of the robbery.

The third man was sentenced to 17 years in prison for his part, and the 29-year-old man was sentenced to 10 years in prison, for his part in the crime. Now, the 29-year-old man obtains another lawyer, and asks to have his case dismissed. His lawyer presents the argument that he should be released from prison. His sentence is ten years for the robbery; however, he has already spent ten years in prison for a crime that he was acquitted of committing.

The victims of the robbery seek justice, and want both men placed in prison.

What does the judge do? How does he decide the fate of the 29-year-old man? Should the man remain in prison, and serve his ten-year sentence, or should the previous ten years that he served for a crime that he didn’t commit, be applied to his current sentence? It was a sentence for a different crime, but he was wrongly convicted because of a failure in the justice system.

Should the man be required to give away another ten years of his life, or should the system give him a break, and set him free, with time paid?

• Should the judge give him a jail pass?
• Would it even be legal, for the judge to give him a jail pass?

This image, provided by PETA, is a court exhibit in a case that states the photographer should be granted copyrights to the image, and therefore, receive monetary damages, because the photograph was used on the cover of a wildlife book.

The catch? The photographer is a macaque monkey (Naruto) who lives on the Indonesian island of Sulawesi. Reportedly, the monkey took this selfie using a camera that belonged to David Slater, a wildlife photographer.

A lawsuit was filed by PETA in San Francisco, on September 22, 2015.  Slater states that he should be the copyright holder of the photograph, because (1) he setup the camera so that the photograph could be taken, and (2), he was the owner of the camera, and the “intellect” behind the photograph.

PETA claims that Naruto should own the copyrights to the photograph and should receive any benefits that are gained from the photograph, which was used on the cover of a wildlife book. PETA stated that macaque monkeys are a critically endangered species, and that their numbers have decreased by around 90%, over the past 25 years.

PETA also sought a court order that would grant both PETA and Dr. Antje Engelhardt (a primatologist), the rights to administer Naruto’s benefits, on the condition that all benefits would be used specifically for the benefit of Naruto, his family, and community. The benefits would be used to help preserve Naruto’s habitat.

Can a monkey be granted photographic copyrights?

There is an island, upon which a tribe resides. The residents either have blue eyes, or brown eyes. Yet, it is taboo to talk about eye color in any way. Thus, one resident can see the eye colors of all other residents but has no way of discovering his own (there are no reflective surfaces).  If a resident does discover his or her own eye color, then he or she must commit ritual suicide, at 12-noon, the following day, in the village square, for all to witness.

Note: Everyone on the island is perfectly logical. On this island there are 100 blue-eyed people, and an unknown number of brown-eyed people.

On day zero, a traveler comes to the island, and says aloud for all residents to hear: “On this island, there is at least one person with blue eyes.”

Having said that, he departs; on day 100, all 100 blue-eyed people commit suicide simultaneously. Explain the logic behind these suicides.

Corollary: Why is the traveler important?

Identification of the problem

One crux to this problem, could be the fact that the people on the island are perfectly logical. If the people weren’t logical, they wouldn’t be able to make conclusions about other people’s eye colors, or, about their own eye color. This would make the traveler completely irrelevant.

The traveler is important, because he or she starts the count, and gets things rolling. If the traveler hadn’t come to the island and made the statement about at least one person having blue eyes, none of the people on the island would have been able to figure out their own eye color, and they would all still be living happily ever after.

If either of the two situations above weren’t true or didn’t happen; everyone on the island would still be alive. The people on the island only died because they were both perfectly logical, and the traveler came.

Analysis

This problem can be solved using multiple cases, with smaller, more manageable numbers. It’s hard to consider the entire problem at once, but if we break the problem down, it becomes much easier to understand.

Case 1: To start off, let’s assume that there’s only one blue-eyed person on the island. Let’s name this person “Mark”. This will make the problem a little easier to understand. The traveler comes to the island, and states that he sees at least one person on the island with blue eyes. Mark, the person with blue eyes, looks around the island, and notices that everyone else on the island has brown eyes. From this, Mark concludes that he must have blue eyes, and, therefore, kills himself at noon the next day.

Mark can make this conclusion because the traveler stated there was at least one person with blue eyes on the island; however, he can tell that everyone else on the island has brown eyes by looking at them.

The other people on the island notice that Mark killed himself, and therefore, must have figured out his eye color. When the other people on the island see that Mark has blue eyes, they put that together with what the traveler stated. Since they are perfectly logical, they also realize that this means Mark must not have seen anyone else on the island with blue eyes.

This information leads everyone else on the island to conclude that they have brown eyes. At noon the next day, everyone remaining on the island kills themselves.

Case 2: In case two, we will have two blue-eyed people. We’ll name them Mark and Bob. After the traveler comes to the island, and makes his statement, both Mark and Bob look around the island at everyone else.

Mark looks at Bob, and notices that Bob has blue eyes, but everyone else on the island has brown eyes. Mark assumes that Bob will kill himself at noon the next day, but when Bob fails to kill himself, Mark deduces that Bob must see someone else on the island with blue eyes. Because Mark is perfectly logical, he knows that this means he must have blue eyes.

At the same time Bob notices that Mark has blue eyes, but everyone else on the island has brown eyes. Bob also assumes that Mark will kill himself at noon the next day. When Mark fails to kill himself, Bob concludes that Mark must see someone else on the island with blue eyes. Bob realizes that he must be the other person with blue eyes, and both men kill themselves at noon the next day.

When the rest of the people living on the island notice Mark and Bob kill themselves on day 2, they realize that they must all have brown eyes, and, therefore, kill themselves at noon on day 3.

Case 3: We’ll use three people in this case. At this point we can continue to use the previous case to help us solve the problem. We already know that Mark was waiting for Bob to kill himself, and vice versa. Let’s add the third person into the problem. We’ll name him, Jim.

In this case, Jim is also waiting for Mark and Bob to kill themselves; however, just as in previous cases, when Mark and Bob fail to kill themselves on day 2, Jim realizes that there must be another person on the island with blue eyes. He can conclude this because neither Mark, nor Bob, have killed themselves, and, therefore, they must see another person on the island with blue eyes.

Just as in previous cases, Jim looks around and notices that everyone else on the island has brown eyes, and, therefore, he must be the other person with blue eyes. There are a couple of things that are happening here, Mark is waiting for Bob and Jim to commit suicide. Bob is waiting for Mark and Jim to commit suicide, and Jim is waiting for Mark and Bob to commit suicide.

Because none of the men are killing themselves, they conclude that they all have blue eyes, and, therefore, they all kill themselves at noon on day 3. When the three men kill themselves, the rest of the people living on the island instantly know that they all have brown eyes. After learning their eye color, the rest of the people on the island commit suicide at noon the next day.

Case N: From the previous cases you should be able to notice the patterning that is arising. This information can be used to construct an algorithm to help solve the larger problem.

In case 1, we had one blue-eyed person. That person killed themselves on day one, and everyone else killed themselves on day two. In case 2, the two blue-eyed people killed themselves on day two, and everyone else killed themselves on day three. Similarly, in case 3, the three blue-eyed people killed themselves on day three, and everyone else on day four.

We’ve used a minimum of three cases in order to prove that a pattern exists. Now that we know there is a pattern, we can solve the finial problem. We could list out every case up until day one hundred, or we could take a more logical approach to reach the solution.

The case-number itself, directly reflects the number of blue-eyed people in the problem. Case 1 had one blue-eyed person; Case 2 had two blue-eyed people, and so on. This means we can substitute the letter N (meaning number of blue-eyed people in this case) for the number of blue-eyed people in our problem.

Case (N): We can break Case N down into two sub-cases: N₀, will be the day in which all the blue-eyed people will die, and N₁, will be the day in which everyone else on the island dies. If we substitute N, for the number of blue-eyed people, we can figure out when everyone is going to die.

The sum of N plus its sub-case will provide the answer we need:

Case N = {x | x ϵ N and x is the number of blue-eyed people}.

Case N+₀ = Blue-eyed people die.

Case N+₁ = Brown-eyed people die.

Heuristics

• We used multiple cases to help break a large problem down, into more manageable pieces.
• We constructed a simple algorithm to help solve the problem.

Looking Back

I think this is a pretty silly problem in the first place. First of all, it states that everyone is perfectly logical. Why would anyone that is perfectly logical kill themselves over eye color? If the people really were perfectly logical, they would use their logic to put together a raft, and sail away from the island, while waving their fists, and telling the rest of the inhabitants to “stuff” their crazy laws.

But on the bright side of things, if you happened to be lucky enough to have brown eyes, you get to live one day longer than everyone that had blue eyes. On the downside, you had to watch 100 blue-eyed people commit suicide, which is probably the real reason why all the brown-eyed people killed themselves the following day. Either that, or they all died from heat stroke while attempting to dig 100 graves.

Mr. & Mrs. Anbouba recently attended a party, at which, there were three other couples. Various handshakes took place. No one shook hands with their spouse. No one shook his, or her, own hand. No one shook hands with the same person twice. At the end of the evening, Mrs. Anbouba asked the seven other people how many hands he, or she, had shaken. Each person gave her a different answer. How many hands did Mr. Anbouba shake?

Identification of the problem

The crux of this problem is to alternate extremes, so that we can eliminate possibilities. One way we can solve this problem, is to use a few diagrams to assist us in obtaining a solution.

Analysis

For the first part of the problem, we need to figure out the maximum number of handshakes that are going to be possible. We know that there are two people in a couple. The problem states that Mr. & Mrs. Anbouba recently attended a party at which there were three other couples. From this, we can conclude that there must be four couples total. Mr. & Mrs. Anbouba being one couple, and three other couples, whose names we don’t know. This proves that we have a total of eight people.

The problem also states that no one shook hands with their spouse, and no one shook hands with themselves. Now that we have this information, we can form a simple math equation, that will tell us the maximum number of handshakes we will have.

We’ll take the total number of people at the party and remove the two people (yourself and your spouse) that you cannot shake hands with. Now we have a formula we can use to solve the first part of the problem.

Formula: total people – (self + spouse) = max handshakes, Or 8 – (1 + 1) = 6

That formula tells us that the maximum number of handshakes any one person can have will be six. Now, we’ll need to figure out the minimum number of handshakes a person could have.

If you read the problem closely, you’ll notice that there is no constraint that requires every person to shake someone else’s hand. From this information, we can deduce that it’s possible for someone to have zero handshakes. This leaves us with zero for the minimum number of handshakes. We can use diagram 1, to represent the information we already know.

For this problem, we can have two different types of responses, one possible, and one absolute. I’ve listed them below:

Responses

1. Possible: 0, 1, 2, 3, 4, 5, or 6, handshakes
2. Absolute: All of the possible responses must be used at least once.

We now have enough information to draw the initial diagram that we will use to solve this problem. The circles will represent the people, and the double-sided arrows will represent the couples. Colored lines will represent the handshakes.

We know that someone is going to have six handshakes, so I placed a number six inside one of the circles, as a starting point. It should be noted that I chose this circle at random. I could have chosen any of the circles as a starting point, and the end result would have been the same. I’ve also crossed-off the number six, in the list, to make it clear that the number has already been used.

The next thing we want to do, is draw a line between person number six, and all of the people they will shake hands with. We know that person number six can’t shake hands with their spouse, so we won’t be drawing a line to that person. Diagram 4 shows the result.

We can tell that person number six, has six handshakes. The spouse of person number six doesn’t have any handshakes, and everyone else has one handshake.

We can figure out the next part of the problem, by alternating extremes. In other words, we’ll try to figure out what person will have zero handshakes. By looking at the diagram we can tell that only one person can have zero handshakes. We know this because person number six already has six handshakes, and everyone else already has one handshake. We can identify that person in the next diagram.

Now, someone needs to have five handshakes. It doesn’t matter at this point who we pick to have five handshakes. So, I’ll pick another person at random. Once I’ve identified that person, I’ll go ahead and draw the handshake lines, and mark the number five as being used. You can see the result in the next diagram.

The blue lines indicate the four additional people that person number five has shaken hands with. Notice that they did not shake hands with their spouse, or the spouse of person number six, because we’ve already stated that person won’t have any handshakes.

We can now figure out who will have only one handshake. We can tell from the next diagram, that it’s only possible for the spouse of person number five to have one handshake, because everyone else already has more than one. This is represented in the next diagram.

For person number four, we can pick any of the unlabeled people, because they all have less than four handshakes. I’ll place person number four at the top and draw the handshake lines to the two people on the left, as those are the only two people that person number four can shake hands with. Remember, that they can’t shake hands with their spouse, and everyone else is already labeled with their corresponding handshakes.

If we look at the diagram now; we can see that the unlabeled person (right) on the top, currently has two handshakes. So, we can mark the remaining person on the top, as being person number two. If you look at the two unlabeled people on the left side, you’ll notice that both of them have three handshakes. Let’s update the diagram with the changes for person number two and take a closer look at the remaining two people on the left.

As stated above, both of the people on the left, currently have three handshakes, that are identified by a red, a blue, and a green line. We can also tell by looking at our number diagram (diagram 1), that we only have one number left to use: the number three. Let’s label the person on the top-left as person number three and see what happens.

Looking at the diagram, we can tell that there is only one unlabeled person left. We are also out of numbers to use. If we count the lines on the unlabeled person, we can tell they have three lines, one red, one blue, and one green. This means that this person also has three handshakes. We can now complete our diagram.

The final diagram shows us that we have two people at the party that will have three handshakes. If we look back at the original question, it states that; at the end of the evening, Mrs. Anbouba asked the seven other people how many hands he or she had shaken, and each person gave her a different answer.

When we look at the diagram, we can tell that Mrs. Anbouba must have been one of the people on the left. We know this because if she was any other person, then two of the people at the party would have reported the same number of handshakes when she asked at the end of the evening. Two of the people would have reported they shook hands three times.

The original problem also states that she asked the seven other people at the party how many times they shook hands. This tells us that she did not respond to the question. Otherwise, there would have been eight responses to the question.

We don’t know what person on the left is Mrs. Anbouba, but it doesn’t matter. We only need to know who her spouse was, because the question is asked how many times Mr. Anbouba shook hands. Since we know that Mrs. Anbouba was one of the people on the left, we also know that Mr. Anbouba must have been one of the people on the left, because he was her spouse.

With this information, we can conclude that the total number of handshakes Mr. Anbouba received, was three.

Heuristics

• We made several diagrams.
• We listed the possible responses.
• We calculated the minimum and maximum responses.
• We alternated extremes.
• We eliminated possibilities.
• We created a formula to calculate the max handshakes.

Looking Back

When looking at a problem like this, the solution can seem impossible. It can be hard to picture in your mind, what the final outcome might be. To get around this, we first calculated the minimum, and maximum, responses, and listed what the possible response types were.

We talked through the problem while using diagrams to help break the problem down into more manageable pieces. When using these diagrams, we were able to get a clearer understanding of what was going on at the party. We were able to conclude that Mrs. Anbouba was, in fact, a duplicate.

We also used the method of alternating extremes, to make solving the problem easier. This helped us eliminate possibilities while completing the diagram. If we would have worked from highest to lowest, (or lowest to highest) we would have run into complications solving the problem. Try doing this problem again, without alternating extremes, and you’ll see that it becomes a lot harder to eliminate the possibilities.

While it’s possible to solve this problem without the use of diagrams, it would have been a lot harder to come up with the correct answer, and there wouldn’t have been an easy way to test that the answer was correct.