Data commit

Task/Hailstone-sequence/R/hailstone-sequence-1.r

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+### PART 1:
+makeHailstone <- function(n){
+  hseq <- n
+  while (hseq[length(hseq)] > 1){
+    current.value <- hseq[length(hseq)]
+    if (current.value %% 2 == 0){
+      next.value <- current.value / 2
+    } else {
+      next.value <- (3 * current.value) + 1
+    }
+    hseq <- append(hseq, next.value)
+  }
+  return(list(hseq=hseq, seq.length=length(hseq)))
+}
+
+### PART 2:
+twenty.seven <- makeHailstone(27)
+twenty.seven$hseq
+twenty.seven$seq.length
+
+### PART 3:
+max.length <- 0;  lower.bound <- 1;  upper.bound <- 100000
+
+for (index in lower.bound:upper.bound){
+  current.hseq <- makeHailstone(index)
+  if (current.hseq$seq.length > max.length){
+    max.length <- current.hseq$seq.length
+    max.index  <- index
+  }
+}
+
+cat("Between ", lower.bound, " and ", upper.bound, ", the input of ",
+    max.index, " gives the longest hailstone sequence, which has length ",
+    max.length, ". \n", sep="")

Task/Hailstone-sequence/R/hailstone-sequence-2.r

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+###Task 1:
+collatz <- function(n)
+{
+  lastIndex <- 1
+  output <- lastEntry <- n
+  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.
+    #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) lastEntry <- output[lastIndex <- lastIndex + 1] <- 3 * lastEntry + 1
+    else lastEntry <- output[lastIndex <- lastIndex + 1] <- lastEntry %/% 2
+  }
+  output
+}
+
+###Task 2:
+#Notice how easy it is to access the required elements:
+twentySeven <- collatz(27)
+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(seq_len(99999), function(x) length(collatz(x)))
+longest <- which.max(seqLenghts)
+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).
+#Another good option would be seqLenghts <- lengths(Vectorize(collatz)(1:99999)).