Hailstone sequence: Difference between revisions
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</pre>
=={{header|ABC}}==
<syntaxhighlight lang="abc">HOW TO RETURN hailstone n:
PUT {} IN seq
WHILE 1=1:
PUT n IN seq[#seq]
SELECT:
n=1: RETURN seq
n mod 2=0: PUT floor(n/2) IN n
n mod 2=1: PUT 3*n+1 IN n
RETURN seq
PUT hailstone 27 IN h27
WRITE "Length of Hailstone sequence for 27:", #h27/
WRITE "First 4 elements:", h27[0], h27[1], h27[2], h27[3]/
WRITE "Last 4 elements:", h27[#h27-4], h27[#h27-3], h27[#h27-2], h27[#h27-1]/
PUT 0, 0 IN longest, length
FOR n IN {1..100000}:
PUT hailstone n IN hn
IF #hn > length:
PUT n, #hn IN longest, length
WRITE longest, "has the longest hailstone sequence < 100,000, of length:", length/</syntaxhighlight>
{{out}}
<pre>Length of Hailstone sequence for 27: 112
First 4 elements: 27 82 41 124
Last 4 elements: 8 4 2 1
77031 has the longest hailstone sequence < 100,000, of length: 351</pre>
=={{header|ACL2}}==
<syntaxhighlight lang="lisp">(defun hailstone (len)
Line 6,007 ⟶ 6,035:
=={{header|Kotlin}}==
<syntaxhighlight lang="kotlin">
fun hailstone(start: Int) = generateSequence(start) { n ->
when {
n == 1 -> null
n % 2 == 0 -> n / 2
else -> n * 3 + 1
}
}
fun main() {
val hail27 = hailstone(27).toList()
println("The hailstone sequence for 27 has ${hail27.size} elements:\n$hail27")
val (n, length) = (1..100000).asSequence()
.map { it to hailstone(it).count() }
.maxBy { it.second }
println("The number between 1 and 100000 with the longest hailstone sequence is $n, of length $length")
}
</syntaxhighlight>
Alternative, doing it manually:
<syntaxhighlight lang="kotlin">import java.util.ArrayDeque
Line 7,300 ⟶ 7,351:
[351, 77031]
</pre>
=={{header|Onyx (wasm)}}==
<syntaxhighlight lang="C">
use core { * }
hailstone :: (n: u32) -> [..]u32 {
seq: [..]u32;
array.push(&seq, n);
while n > 1 {
n = n/2 if n%2 == 0 else (n*3)+1;
array.push(&seq, n);
}
return seq;
}
Longest :: struct { num, len : u32; }
main :: () {
// -------
// task 1:
// -------
// "Create a routine to generate the hailstone
// sequence for a number."
i := 27;
seq := hailstone(i);
printf("Task 1:\n{}: {}\n\n",
i,
seq
);
// -------
// task 2:
// -------
// "Use the routine to show that the hailstone
// sequence for the number 27 has
// 112 elements starting with
// 27, 82, 41, 124 and ending with 8, 4, 2, 1"
slice_size := 4;
len := seq.length;
slice_first := seq[0..slice_size];
slice_last := seq[seq.length-slice_size..seq.length];
printf("Task 2:\nlength: {}, first: {}, last: {}\n\n",
len,
slice_first,
slice_last
);
// -------
// task 3:
// -------
// "Show the number less than 100,000
// which has the longest hailstone sequence
// together with that sequences length."
l : Longest;
for i in 1..100000 {
seq := hailstone(i);
if l.len < seq.length { l = .{num = i, len = seq.length}; }
}
printf("Task 3:\nLongest Num: {}, Sequence Length: {}\n", l.num, l.len);
}
</syntaxhighlight>
=={{header|ooRexx}}==
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