Lemon Candy and Dynamic Programming

Over at TheDailyWtf, hidden among some comments was an interesting dynamic programming problem:

Consider this problem:

George bought a sack of 100 pieces of candy at the store. 90 of the pieces are lemon flavored and ten are cherry flavored. Of the two, George prefers the lemon flavored candies.

Every day George randomly picks a piece of candy out of the bag. If it is lemon flavored, he eats it and puts the bag away for the next day.

But if the candy he chose is cherry flavored, he puts it back in the bag and then randomly picks a candy out of the bag and eats it regardless of the flavor. In other words, he’ll only put a piece of candy back at most once per day.

What are the odds that when one piece of candy remains, it will be lemon flavored?

I posed the problem at a company where I used to work. All but one person tried to do it recursively. The remaining person tried to do it using an Excel spreadsheet!!!

Maybe I (and some of the other posters on that thread) are morons, but Excel (or in my case, OpenOffice) seemed like a fine way to solve it, so I did.

Read more about the Lemon-Cherry problem, and download the speadsheets used to solve it

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The 100 Doors Puzzle in R at Rosetta Code

From time to time I see a puzzle at Rosetta Code that interests me, and I post an R solution for it.  This time it was the 100 doors puzzle.

Problem: You have 100 doors in a row that are all initially closed. You make 100 passes by the doors. The first time through, you visit every door and toggle the door (if the door is closed, you open it; if it is open, you close it). The second time you only visit every 2nd door (door #2, #4, #6, …). The third time, every 3rd door (door #3, #6, #9, …), etc, until you only visit the 100th door.

Question: What state are the doors in after the last pass? Which are open, which are closed?

The code for this in R is pretty simple:

# UNOPTIMIZED
doors_puzzle <- function(ndoors=100,passes=100) {     doors <- rep(FALSE,ndoors)     for (ii in seq(1,passes)) {         mask <- seq(0,ndoors,ii)         doors[mask] <- !doors[mask]        }     return (which(doors == TRUE)) } doors_puzzle() ## optimized version... we only have to to up to the square root of 100 seq(1,sqrt(100))**2 [/sourcecode]


Monty Hall in R

Inspired by paddy3118, I decided to write up a Monty Hall simulation in R for Rosetta Code.   Enjoy!

… The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it…

R-project code follows