I’m working to bone up on my python skills so I decided to spend my Sunday doing problems 1-10 from Project Euler. I’ve done them before with C or Java but this was my first time with Python. Here are the problems and my commented code for each one in case it interests anybody.
Last week a podcast I listen to, the 404, discussed a math problem where you roll six 20-sided die and count how often you get a situation where at least one dice matches another dice. They discussed the math a little and came to the conclusion that it happens far more than you’d think. I thought it’d make a good monte carlo programming exercise so I’ve done just that. Below, you’ll find my C code (though it’s not great) and results for 2-20 dice.
Here’s my solution to this week’s Riddler Classic puzzle FiveThirtyEight. Here’s the question: ” You are the only sane voter in a state with two candidates running for Senate. There are N other people in the state, and each of them votes completely randomly! Those voters all act independently and have a 50-50 chance of voting for either candidate. What are the odds that your vote changes the outcome of the election toward your preferred candidate? More importantly, how do these odds scale with the number of people in the state? For example, if twice as many people lived in the state, how much would your chances of swinging the election change?”
Here’s my solution to this weeks Riddler Classic puzzle from FiveThirtyEight. Here’s the question: “While traveling in the Kingdom of Arbitraria, you are accused of a heinous crime. Arbitraria decides who’s guilty or innocent not through a court system, but a board game. It’s played on a simple board: a track with sequential spaces numbered from 0 to 1,000. The zero space is marked “start,” and your token is placed on it. You are handed a fair six-sided die and three coins. You are allowed to place the coins on three different (nonzero) spaces. Once placed, the coins may not be moved. Continue reading “Riddler Classic – 10-14-2016 – Arbitraria”
Here’s my solution to this week’s Riddler Express question from FiveThirtyEight. Here’s the question: “You place 100 coins heads up in a row and number them by position, with the coin all the way on the left No. 1 and the one on the rightmost edge No. 100. Next, for every number N, from 1 to 100, you flip over every coin whose position is a multiple of N. For example, first you’ll flip over all the coins, because every number is a multiple of 1. Then you’ll flip over all the even-numbered coins, because they’re multiples of 2. Then you’ll flip coins No. 3, 6, 9, 12 … And so on. What do the coins look like when you’re done? Specifically, which coins are heads down?”