Data update

Task/Hailstone-sequence/ABC/hailstone-sequence.abc

@@ -0,0 +1,22 @@
+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/

Task/Hailstone-sequence/Onyx-(wasm)/hailstone-sequence.onyx

@@ -0,0 +1,61 @@
+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);
+}