Here's the problem for those who haven't seen it before:

*There are 5 rational pirates A, B, C, D and E. They find 100 gold coins. They must decide how to distribute them.*

*The pirates have a strict order of seniority: A is superior to B, who is superior to C, who is superior to D, who is superior to E.*

*The pirate world's rules of distribution are thus: that the most senior pirate should propose a distribution of coins. The pirates, including the proposer, then vote on whether to accept this distribution. In case of a tie vote the proposer has the casting vote. If the distribution is accepted, the coins are disbursed and the game ends. If not, the proposer is thrown overboard from the pirate ship and dies, and the next most senior pirate makes a new proposal to begin the system again.*

*Pirates base their decisions on three factors. First of all, each pirate wants to survive. Second, given survival, each pirate wants to maximize the number of gold coins each receives. Third, each pirate would prefer to throw another overboard, if all other results would otherwise be equal.The pirates do not trust each other, and will neither make nor honor any promises between pirates apart from a proposed distribution plan that gives a whole number of gold coins to each pirate.*

Five pirates vying for 10 coins is an interesting problem but what if there were one hundred pirates each trying to get the most loot out of 100 coins? What if there were more pirates than coins? If there

*were*more pirates than coins would the first pirate be able to get

*any*coins? Is there a strategy where the first pirate would be able to stay onboard? Can you devise a formula for the number of coins any given pirate in such a situation would get where the number of pirates (N) is greater than the number of coins (G)? Use the backwards induction model like we used with the first version of the game!

Speaking of formulating your answer using variables, we continued to derive the pythagorean theorem in Geometry, replicating the work of the indian mathematician Bhakarsa in his famous proof. Walking the same path he did in 1114 CE!

In graph theory we built off of our definitions of days one through three to find isomorphic graphs amidst a maelstrom of interconnected nodes and edges. Mathematicians then used this fresh skill set to draw their own planar graphs.

During lunch I noticed a young mathematician who was playing a game of Cat's Cradle. I wondered, if they were to chop up the string could they make a planar graph with the same number of nodes and edges as we see in the picture below?

*If you were to play cat's cradle without chopping up the string, how many different non-planar graphs can you make? How many of these are identical?*

In computer science we talked about Boolean logic and the special definitions of

*and*or

*or*in programming.

Here's a joke about the

*inclusive or*:

A logician's wife is in labor. The logician is in the waiting room.

The doctor comes out of the delivery room and says to the logician, "Congratulations, your wife gave birth to a beautiful baby!"

"Is it a boy or a girl?" asks the logician.

"Yes", says the doctor.

In art, mathematicians continued to use their budding architectural acumen to make platonic solids. We're eagerly shedding the templates that we made in the beginning of the week as tetrahedrons, octahedrons, dodecahedrons and more take form.

We're disappointed that C.A.M.P. is discrete and not continuous. The last update (until next summer of course) is tomorrow, stay tuned!

-- Eliana