I will post a little selection sort implementation I have handy:

function sortList(list) sorted = list imin=0 swap = 0 /* assume smallest element value */ for i=0 to len(sorted)-1 loop imin=i for j=i+1 to len(sorted) loop if sorted[j] < sorted[imin] then imin=j endif repeat if imin!= i then swap = sorted[j] sorted[i] = sorted[imin] sorted[imin] = swap endif repeat return sorted

I've made a little scene/greetings card thing :) Download ID: 24K73WK5KR

Hi guys! Since it's getting close to the next challenge. I thought I'd post up my solution for last month's challenge.

table = [1, 2, 4, 8, 16, 32, 64, 128] rtable = [128, 64, 32, 16, 8, 4, 2, 1] //TASK #1 task1_n = random(255) function _abs(x) r = 0 if x < 0 then r = (~x) + 1 else r = x endif return r function extract_bin(n) r = "00000000" for i = 0 to 8 loop if n & rtable[i] then r[i] = "1" else r[i] = "0" endif repeat return r function reverse_bin(n) r = 0 for i = 0 to 8 loop if n & table[i] then r = r | rtable[i] endif repeat return r print(str(task1_n) + " " + extract_bin(task1_n) + " -> " + str(reverse_bin(task1_n)) + " " + extract_bin(reverse_bin(task1_n))) //TASK #2 task2_n1 = random(255) task2_n2 = random(255) is_negative1 = random(2) is_negative2 = random(2) function mul(a, b) r = 0 sign = 0 loops = 0 if a == 0 or b == 0 then r = 0 else if a == 1 then r = b else if a == -1 then r = -b else if b == 1 then r = a else if b == -1 then r = -a else if a < 0 or b < 0 then sign = 1 endif if a < 0 and b < 0 then sign = 0 endif a = _abs(a) b = _abs(b) for i = 0 to 8 loop if b & table[i] then r += a endif a = a << 1 loops += 1 repeat endif endif endif endif endif if sign then r = -r endif res = [r, loops] return res if is_negative1 then task2_n1 = -task2_n1 endif if is_negative2 then task2_n2 = -task2_n2 endif task2_res = mul(task2_n1, task2_n2) print(str(task2_n1) + " x " + str(task2_n2) + " = " + str(task2_res[0]) + " in " + str(task2_res[1]))

These are great! Very cool to see you venturing into 3D, Vinicity :D Here's my little scene also using some of KayKit's assets

2021103020590100-86FC867622BE2C2315A009D470F52DAE.jpg

(Download code is N4N4VMND9H)

I wanted to give this a try. But Scrubz solution cant be beaten, its nice. But maybe diversity is welcome.

A bad game can still have a good story, so, here is the story of Task 1: I call this "lazy recursion", the code uses recursion, but it just hands over an index, and always the same text variable. So, it does use a lot of memory - maybe you could use this for memory testing (somehow) ;-). By the way I probably introduced a new concept: the measurement of numerics: it answers the question "how numeric is a variable containing text?" (I had no clue how to name this recursive function, because it tests only one char but also sums up all found numeric values).

Task 1

function howNumericIsCharAtIndexAndAfter(index, text) isNumeric = false if index >= len(text) then else c = chrVal(text[index]) if c >= chrVal("0") and c <= chrVal("9") then isNumeric = true endif index += 1 isNumeric += howNumericIsCharAtIndexAndAfter(index, text) endif return isNumeric t = "ab12cd34" print(howNumericIsCharAtIndexAndAfter(0, t))

output:

4

A recursive solution for TASK #1:

B6388EEC-9A97-4589-A9B5-2AA67D843C14.jpeg

Recursive solutions are probably not the best way to do things, but they are so much fun to write. Elegant solutions for a more civilised age!

I’ll skip the other task, since it seems to me that it’s hard to come up with any solution that isn’t just loops and counters. But I would be happy to be proved wrong about that! ;)

Reading through the posts, I feel like my SVG approach is not fitting the core of this challenge. I will move it to a separate WIP thread. So, this submission fits better: download code NQ6HVMNDN8
2020122702493900-86FC867622BE2C2315A009D470F52DAE.jpg 2020122703120200-86FC867622BE2C2315A009D470F52DAE.jpg
The dark/light green structure in the tree, was a typo, but I decided that it looks nice 😄 so I left it.