@vinicity oh my this already looks epic!.

Love that jumping action! The character reminds me of the robots from the old Smash advert

I am so sad not to be working on this. It looks so much fun!

I made some layer cake tiles (size 16x16x16) in the Fuze image editor and written the code to draw them. This is what they look like from above and from the side. tiles combined views.jpg

That looks very intriguing. Can’t wait to get started on that next week!

ah ok thanks

My friend code is sw-6941-1875-1015

@monkeee Now, voxel art! 🤩 That is just geat! I love the way of thinking around here.

@andyg1985 Correct!

For Task #2, I did an implementation of the shunting yard method. It can evaluate valid infix expressions and it will evaluate invalid ones too since there is no validity checking.

Function Eval(Expr) var OpChar = [ "(", "*", "+", "-", ")" ] var OpPrec = [ 4, 3, 2, 2, 1 ] float VS[0] int TopVS = 0 // Value Stack int OS[0] int TopOS = 0 // Operator Stack Expr += "+0" // This makes it finish up the stuff on int I // the stack when it gets to the end For I = 0 To Len(Expr) Loop int J = ChrVal(Expr[I]) - ChrVal("(") If J < 64 And 1 << J & 0x2F Then // Expr[I] ~ "()+-*" J = [ 0, 4, 1, 2, -1, 3 ][J] // Remap the character code While TopOS > 0 Loop If OpPrec[J] > OpPrec[OS[TopOS - 1]] Then Break EndIf If OpChar[OS[TopOS - 1]] == "(" Then TopOS -= OpChar[J] == ")" // Pop matching '()' Break EndIf TopOS -= 1 // Evaluate the operator on top If OpChar[OS[TopOS]] == "+" Then TopVS -= 1 VS[TopVS - 1] += VS[TopVS] EndIf If OpChar[OS[TopOS]] == "-" Then TopVS -= 1 VS[TopVS - 1] -= VS[TopVS] EndIf If OpChar[OS[TopOS]] == "*" Then TopVS -= 1 VS[TopVS - 1] *= VS[TopVS] EndIf Repeat If OpChar[J] != ")" Then OS[TopOS] = J TopOS += 1 EndIf Else int J For J = I To Len(Expr) Loop int N = ChrVal(Expr[J]) - 46 // Expr[J] !~ ".0123456789" If N > 63 Or 1 << N & 0xFFD == 0 Then Break EndIf Repeat If J > I Then // It is a Number J -= 1 VS[TopVS] = Float(Expr[I: J]) TopVS += 1 I = J EndIf EndIf Repeat Return VS[0]

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.