My time as a grad student will soon draw to a close. With this comes the terrifying realisation that I'm going to start applying for jobs and, hopefully, interviewing soon, forever leaving my comfortable security blanket of academia.

With that horrible thought in mind, I've been doing some poking around to see what various kinds of technical interviews are like. Apparently, it is not entirely uncommon in such interviews to ask for solutions to little toy programming problems. These are viewed as a way of weeding out the hoi polloi, and as a way of determining how you approach a novel problem under stress.

One such problem that has been discussed at length is the fizzbuzz problem. The idea is simple:

Write a program that prints the numbers from 1 to 100. But for multiples of three print “Fizz” instead of the number and for the multiples of five print “Buzz”. For numbers which are multiples of both three and five print “FizzBuzz”.

Sounds simple enough, but if you've never though about this problem before, then I encourage you to do so before reading on. One way we might try to solve the problem is as follows:

# fizzbuzz with a loop for (i in 1:100){ if (i%%3==0) if (i%%5==0) print("fizzbuzz") else print("fizz") else if (i%%5==0) print("buzz") else print(i) }

The above is very "C-ish" in spirit, or as Patrick Burns says in the R Inferno (which is necessary reading for anyone considering putting R on his/her resume) this is "speaking R with a C accent." So we might want to think about vectorizing our code. My philosophy on vectorizing is that you win when it's completely incomprehensible. Unfortunately, I don't believe I've won here, but I did concoct this:

# vectorized fizzbuzz m <- 1:100 m[which(m%%5==0)][which(m%%3==0)] <- "fizzbuzz" m[which(as.numeric(m)%%3==0)] <- "fizz" m[which(as.numeric(m)%%5==0)] <- "buzz" print(m)

which will give you all kinds of warnings, but you can safely ignore them. These are just a few ways one can proceed. The problem has somewhat captivated internet nerds everywhere, with people coming up with solutions where no conditionals are used or trying to solve fizzbuzz with as few bytes of code as possible (which I first heard of through this R-bloggers blog). Why would anyone do this? One word: nerdcred.

Another toy problem I found while looking for job interview advice involves taking a fair 5-sided die and turning it into a fair 7-sided die. I originally found the question from a link pointing over to a stackoverflow page. The problem statement given there is

Given a function which produces a random integer in the range 1 to 5, write a function which produces a random integer in the range 1 to 7.

Since the problem does not explicitly say that the original function produces the numbers uniformly randomly, or that the new function should itself be under this restriction, there are several cute solutions you could concoct which fit the letter, though not the spirit of the problem. At that very stackoverflow page, user DrStalker gives this solution (with translation to R)

rand7 <- function() rand5()

which I think is my favorite such solution. My best friend (who otherwise wishes to remain nameless) gave this solution:

rand7 <- function() rand5() + floor(rand5() / 2)

which is more sneaky, since it is actually theoretically possible to generate all digits 1, 2, 3, 4, 5, 6, and 7 with this, though not uniformly. Again, if you haven't thought about this problem before, give it a shot. In my opinion, it's not all that trivial. What follows is the first solution I could come up with:

rand5 <- function() sample(1:5, 1) # the given # Translate 5-sided die to fair coinflips coin.flip <- function(){ n <- rand5() if (n==1 || n==2) return(0) else if (n==3 || n==4) return(1) else coin.flip() } # Fair 7-sided die rand7 <- function(){ x <- coin.flip() + coin.flip()*2 + coin.flip()*4 while(x == 0) x <- coin.flip() + coin.flip()*2 + coin.flip()*4 return(x) }

A few things about this solution before moving on. First, I think this has a kind of mathematical elegance, since it's clear from this solution how to easily translate this solution into creating other fair dice. Second, it's computationally inelegant in more ways than one. First of course, note that this could technically run forever. If you wanted to build your fair 7-sided die this way, you would probably want to do something about that, even if all you do is stop after "too many" tries. I didn't do any such optimization because who the hell cares; that line of thought completely misses the point of the exercise. Lastly, both in terms of source code and in terms of actual running, the above solution is costly.

After satisfying myself in producing that solution, I decided to see what kinds of solutions others were coming up with. I've already mentioned what is, I think, my favorite solution. But there is a much more computationally elegant "real" solution provided by Adam Rosenfield (and translated into R):

rand7 <- function(){ x <- rand5() + (rand5() - 1)*5 while (x > 21) x <- rand5() + (rand5() - 1)*5 return(x%%7 + 1) }

Now, it's fairly obvious that these solutions are producing fair dice, assuming the original function is itself fair. However, suppose we wanted numerical verification. This is a good simple exercise in Monte Carlo simulation. And while we're at it, let's parallelize our simulation.

We'll be using the built-in R library (as of 2.14) called parallel. This handy little guy merges the two former powerhouse parallel R libraries, namely multicore and snow. If you're using Windows, then you have to use the snow side of parallel--which frankly, I find to be a huge, annoying mess. If you're using a POSIX-like OS such as Linux or Mac OS X, then first pat yourself on the back for being awesome. Also, unlike the unwashed Windows peasants (only kidding!), you get your choice of multicore or snow functions. For the sake of inclusion, we'll talk a bit about both approaches.

First, the snow way of doing things:

# The SNOW way of doing things -- necessary for Windows users library(parallel) # Setting up the workers. The function detectCores() used below is a # function from library(parallel) which finds the total number of # available cores. You can change the call below to a smaller number # if you don't want to use all of your cores for some reason. cl <- makeCluster(detectCores()) # Send our functions to the workers clusterEvalQ(cl, rand5 <- function() sample(1:5, 1)) clusterEvalQ(cl, coin.flip <- function(){ n <- rand5() if (n==1 || n==2) return(0) else if (n==3 || n==4) return(1) else coin.flip() } ) # Our many die rolls, stored as a list results <- clusterApply(cl, 1:1e5, rand7) table(unlist(results))

which gives output

1 2 3 4 5 6 7 14279 14489 14210 14289 14279 14238 14216

Which looks exactly like we had hoped it would. We can also look at a quick summary

summary(unlist(results)) Min. 1st Qu. Median Mean 3rd Qu. Max. 1.000 2.000 4.000 3.994 6.000 7.000

We can also flex our mighty R muscles and give a pretty histogram using Hadley Wickham's incredible ggplot2 package

which looks exactly like we expect it to.

Ok, so that's the snow way of doing things; now for my personal favorite, the multicore way of doing things. The function we will be using is mclapply, which is exactly what it looks like. It's a **m**ulti**c**ore**lapply** function.

library(parallel) results <- mclapply(X=1:1e5, FUN=rand7, mc.cores=detectCores()) table(unlist(results))

Note the difference. And for those who think I'm just being cranky for no good reason, here's a little speed test

time.capply <- system.time({results <- clusterApply(cl=cl, x=1:1e5, fun=rand7)})[3] time.capplylb <- system.time({results <- clusterApplyLB(cl=cl, x=1:1e5, fun=rand7)})[3] time.mclapply <- system.time({results <- mclapply(X=1:1e5, FUN=rand7, mc.cores=detectCores())})[3] cat(sprintf("\nWith %d cores, the times are:\nclusterApply():\t\t%.3f seconds\nclusterApplyLB()\t%.3f seconds\nmclapply():\t\t%.3f seconds\n", detectCores(), time.capply, time.capplylb, time.mclapply))

which gives output

With 4 cores, the times are: clusterApply(): 17.119 seconds clusterApplyLB() 14.849 seconds mclapply(): 0.800 seconds

Again, note the difference.

Finally, it is worth mentioning that there is another way we could parallelize things, namely by using the foreach package. I'm not a huge fan of foreach, but you might find it useful depending on what you are trying to do.

As for more "toy" problems, the motherload is over at Project Euler. These problems hastily go from being a fun, simple way to practice R to being fairly cumbersome and challenging. I've posted several Project Euler in R solutions on this blog, and I have a bunch more that I'm too lazy to write up and post. Should you get bitten by the Project Euler bug, once you get an initial solution to a problem there, try your best to get the run time to below a minute (or below a second). This can sometimes be a deeply frustrating (given how R performs on loops), though ultimately rewarding process. Have a look.

I was thinking about a more intuitive solution in order to generate n uniformly distributed values in range 1..7 starting from the rand5 function (which returns n uniformly distributed values in 1..5):

rand7b() :

. generate 7n uniformly distributed values in 1..35

. uniformly pick n values of the 7n

. translate 1..35 into 1..7 with x%%5 +1

Here is my code (sorry for not managing the highlighting):

## Returns n random integers in the range 1..5.

rand5 <- function(n=1) sample(1:5, n, replace=TRUE)

## Returns n random integers in the range 1..7.

## FAIL !!

rand7a <- function(n=1) {

round(rand5(n) * 7 / 5)

}

## Returns n random integers in range 1..7

rand7b <- function(n=1) {

## x will have 7*n uniformly distributed values in range 1..35

x <- c()

for (i in 0:6)

x <- c(x, i*5 + rand5(n))

## Uniformly pick n elements in range 1..35

## Just for fun, from each group of 7 elements, pick 1 of the first 5 :)

# i <- 0:(n-1) * 7 + rand5(n)

## ... or just do it quickly

i <- 0:(n-1) * 7

## Translate 1..35 into 1..7

x[i] %% 7 + 1

}

## Adam Rosenfield's solution.

## Good, but slower then rand7b.

rand7c1 <- function(dummy=1) {

x 21) x <- rand5() + (rand5() - 1) * 5

return (x %% 7 + 1)

}

rand7c <- function(n=1) {

sapply(1:n, FUN=rand7c1)

}

## Pick the function to use.

rand7 <- rand7b

## Generate (large) samples.

x5 <- rand5(35000)

x7b <- rand7b(35000)

x7c <- rand7c(35000)

## Print the "histogram".

print(table(x5))

print(table(x7b))

print(table(x7c))

In the first problem you can avoid the warnings as follows:

m <- 1:100

is.mod.3 <- m%%3 == 0

is.mod.5 <- m%%5 == 0

m[is.mod.3] <- "fizz"

m[is.mod.5] <- "buzz"

m[is.mod.3 & is.mod.5] <- "fizzbuzz"

print(m)

Another way to avoid the warnings with fizzbuzz:

n <- 1:100

m <- n

m[n%%3==0] <- "fizz"

m[n%%5==0] <- "buzz"

m[(n%%5==0)&(n%%3==0)] <- "fizzbuzz"

print(m)