Hailstone sequence: Difference between revisions

==Non-cached version==

Calculates sequence without using previous calculated sequences.

==Cached version==

Calculations are stored for used in later calculations.

<syntaxhighlight lang="miniscript">

cache = {}

calc = function(n)

if cache.hasIndex(n) then return

items = [n]

origNum = n

while n > 1 and not cache.hasIndex(n)

if n % 2 then

n = 3 * n + 1

else

n = n /2

end if

items.push(n)

end while

cache[origNum] = {"len": items.len,"items":items}

end function

getLen = function(n)

if not cache.hasIndex(n) then calc(n)

if n == 1 then return 1

return cache[n].len + getLen(cache[n].items[-1]) - 1

end function

getSequence = function(n)

if not cache.hasIndex(n) then calc(n)

if n == 1 then return [1]

return cache[n].items[:-1] + getSequence(cache[n].items[-1])

end function

h = getSequence(27)

print "The hailstone sequence for 27 has " + h.len + " elements starting with"

print h[:4]

print "and ending with"

print h[-4:]

print

longSeq = 0

longSeqVal =0

for i in range(2, 100000)

seq = getLen(i)

if longSeq < seq then

longSeq = seq

longSeqVal = i

end if

end for

</syntaxhighlight>

{{out}}Output is the same as the non-cached version above.

<pre>The hailstone sequence for 27 has 112 elements starting with

[27, 82, 41, 124]

and ending with

[8, 4, 2, 1]

The number < 100,000 which has the longest hailstone sequence is 77031.

This sequence has 351 elements.</pre>