Hailstone sequence: Difference between revisions

=={{header|CLU}}==

<lang clu>% Generate the hailstone sequence for a number

hailstone = iter (n: int) yields (int)

while true do

yield(n)

if n=1 then break end

if n//2 = 0 then

n := n/2

else

n := 3*n + 1

end

end

end hailstone

% Make an array from an iterator

iter_array = proc [T,U: type] (i: itertype (U) yields (T), s: U) returns (array[T])

arr: array[T] := array[T]$[]

for item: T in i(s) do array[T]$addh(arr, item) end

return(arr)

end iter_array

start_up = proc ()

po: stream := stream$primary_output()

% Generate the hailstone sequence for 27

h27: array[int] := iter_array[int,int](hailstone, 27)

lo27: int := array[int]$low(h27)

hi27: int := array[int]$high(h27)

stream$putl(po, "The hailstone sequence for 27 has "

|| int$unparse(array[int]$size(h27)) || " elements.")

stream$puts(po, "The first 4 elements are:")

for i: int in int$from_to(lo27, lo27+3) do

stream$puts(po, " " || int$unparse(h27[i]))

end

stream$puts(po, ", and the last 4 elements are:")

for i: int in int$from_to(hi27-3, hi27) do

stream$puts(po, " " || int$unparse(h27[i]))

end

stream$putl(po, "")

% Find whichever sequence < 100 000 has the longest sequence

maxnum: int := 0

maxlen: int := 0

for i: int in int$from_to(1, 99999) do

len: int := array[int]$size(iter_array[int,int](hailstone, i))

if len > maxlen then

maxnum, maxlen := i, len

end

end

stream$putl(po, int$unparse(maxnum)

|| " has the longest hailstone sequence < 100000: "

|| int$unparse(maxlen))

end start_up</lang>

{{out}}

<pre>The hailstone sequence for 27 has 112 elements.

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

77031 has the longest hailstone sequence < 100000: 351</pre>