Hailstone sequence: Difference between revisions

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The previous solution is entirely satisfactory and may be more efficient than the following solution. However, problems like these are a great chance to show off the strength of R's vectorization. Also, this lets us show off how the <- syntax can do multiple variable assignments in one line. Observe how short the following code is:

<lang rsplus>###Task 1:

collatz<-function(n)

collatz <- function(n)

{

lastIndex<-1

lastIndex <- 1

output<-lastEntry<-n

output <- lastEntry <- n

while(lastEntry!=1)

while(lastEntry != 1)

{

#Each branch updates lastEntry, lastIndex, and appends a new element to the end of output.

#Note that the return value of lastIndex<-lastIndex+1 is lastIndex+1.

#Note that the return value of lastIndex <- lastIndex + 1 is lastIndex + 1.

#You may be surprised that output can be appended to despite starting as just a single number.

#If so, recall that R's numerics are vectors, meaning that output<-n created a vector of length 1.

#It's ugly, but efficient.

if(lastEntry%%2~~==0~~){lastEntry<-output[lastIndex<-lastIndex+1]<-lastEntry~~%/%2}~~

if(lastEntry %% 2) lastEntry <- output[lastIndex <- lastIndex + 1] <- 3 * lastEntry + 1

else{lastEntry<-output[lastIndex<-lastIndex+1]<-3*lastEntry~~+1}~~

else lastEntry <- output[lastIndex <- lastIndex + 1] <- lastEntry %/% 2

}

output

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###Task 2:

#Notice how easy it is to access the required elements:

twentySeven<-collatz(27)

twentySeven <- collatz(27)

cat("The first four elements are:", twentySeven[1:4],"and the last four are:", twentySeven[length(twentySeven)-3:0])

cat("The first four elements are:", twentySeven[1:4], "and the last four are:", twentySeven[length(twentySeven) - 3:0], "\n")

###Task 3:

#Notice how a several line long loop can be avoided with R's sapply or Vectorize:

seqLenghts<-sapply(1:99999,function(x) length(collatz(x)))

seqLenghts <- sapply(seq_len(99999), function(x) length(collatz(x)))

longest<-which.max(seqLenghts)

longest <- which.max(seqLenghts)

~~print(paste0~~("The longest sequence before the 100000th is found at n=",longest,". It has length ",seqLenghts[longest],"."))

cat("The longest sequence before the 100000th is found at n =", longest, "and it has length", seqLenghts[longest], "\n")

#Equivalently, line 1 could have been: seqLenghts<-sapply(Vectorize(collatz)(1:99999),length).

#Equivalently, line 1 could have been: seqLenghts <- sapply(Vectorize(collatz)(1:99999), length).

#Another good option would be seqLenghts<-lengths(Vectorize(collatz)(1:99999)).</lang>

#Another good option would be seqLenghts <- lengths(Vectorize(collatz)(1:99999)).</lang>

⚫

<pre>The first four elements are: 27 82 41 124 and the last four are: 8 4 2 1

<pre>

⚫

The longest sequence before the 100000th is found at n = 77031 and it has length 351</pre>

> twentySeven<-collatz(27)

> cat("The first four elements are:", twentySeven[1:4],"and the last four are:", twentySeven[length(twentySeven)-3:0])

⚫

The first four elements are: 27 82 41 124 and the last four are: 8 4 2 1

> seqLenghts<-sapply(1:99999,function(x) length(collatz(x)))

> longest<-which.max(seqLenghts)

> print(paste0("The longest sequence before the 100000th is found at n=",longest,". It has length ",seqLenghts[longest],"."))

⚫

~~[1] "~~The longest sequence before the 100000th is found at n=77031. It has length 351."

</pre>

=={{header|Racket}}==