The problem with that for miniflak is that you cant perform an operation on every element of the stack because of the lack of random access, so decoding is not really possible
All we have to do is implement the following psuedo code x-- if x == 0 push thing x-- if x == 0 push thing x-- if x == 0 push thing x-- if x == 0 push thing x-- if x == 0 push thing
Ok I'm going to go get lunch with a friend. Hopefully that will give you some time to digest what I have said and give me some time to figure out how I hope to explain the rest.
If you rotate the triangle upside down and attach it to the other triangle, you'd get a polygon of NxN-1 so then you just add N to fill out the top row
YES! ({}({} ({}...) ([{}]))([{}])) allows you to manipulate the third item on the stack. Only problem is it makes the 2nd and third item negative. Fixing that :P
({}([{}] ([{}...]) ([{}]))([{}])) NOW DOES IT. Super happy. It lets you change the 3rd stack item. @WheatWizard
@Riley any ideas on golfing that?
Ofc the new problem is getting the first stack item :P