There's not really any way around the sub issue, unless I don't use strings at all. But then what's the alternative, matrices? I'd rather not have a matrix that big here

:"XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX                                  X   XX       X          X                   XX  X                   X      X        XX        X     X    X       X          XX    XX  X       X                 X   XX   X                     X       X    XX                       X      X     X XX                                      XX      X          X                    XX                       X      X       XX         X                    X   X   XX                X       X             XX                   X    X        X    XX                          X           XX           X     X        X       X   XX    X             X     X             XX        X        X         X  X       XX         X    X     X          X      XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX→Str1
:20→L
:40→W
:10→R
:20→C
:7→P
:13→Q
:For(A,1,8
:Output(A,1,sub(Str1,WP-2W+Q+WA,16
:End
:Output(4,8,"π
:Repeat K=45
:Repeat max(Ans={24,25,26,34,45
:getKey→K
:End
:C-(K=24 and sub(Str1,WR-W-1+C,1)=" ")+(K=26 and sub(Str1,WR-W+1+C,1)=" →B
:R-(K=25 and sub(Str1,WR-2W+C,1)=" ")+(K=34 and sub(Str1,WR+C,1)=" →A
:min(L-7,max(1,P+Ans-R+(K=25 and R-P>4)-(K=34 and R-P<3→P
:min(W-15,max(1,Q+B-C+(K=24 and C-Q>8)-(K=26 and C-Q<7→Q
:A→R
:B→C
:For(A,1,8
:Output(A,1,sub(Str1,WP-2W+Q+WA,16
:End
:Output(R-P+1,Ans-Q+1,"π
:End

And with that, I think I'm done with this random mini-project.

I'll never cease to be amazed with the things you people can do with a graphing calculator. If this is what we can do ten years after the 83+ came out, what kinds of things will we be doing after twenty years?

Practically on an impulse, I decided to try and make a typical movement-with-a-string-map routine, but this map is 20x40 and scrolls. I also wanted it to stop scrolling when it gets to the edge of the map (domain error prevention, plus I think it's a nice effect). So when that happens, the player who has been so far stuck in the middle of the screen becomes off-centered. What I want to do next though is when you move away from the edge of the map, I want to re-center the player (one step at a time, of course). I'm probably not explaining this adequately...

Also, I'd prefer if no one gave me an answer in code. Just the theory behind it, so I can still get my practice.

:"a huge string implementation of a 20x40 map→Str1
:10→R
:20→C
:7→P
:13→Q
:For(A,1,8
:Output(A,1,sub(Str1,40P-80+Q+40A,16
:End
:Output(4,8,"π
:Repeat K=45
:Repeat max(Ans={24,25,26,34,45
:getKey→K
:End
:R-(K=25 and sub(Str1,40R-80+C,1)=" ")+(K=34 and sub(Str1,40R+C,1)=" →A
:C-(K=24 and sub(Str1,40R-41+C,1)=" ")+(K=26 and sub(Str1,40R-39+C,1)=" →B
:P+A-R→P
:Q+B-C→Q
:A→R
:B→C
:min(13,max(1,P→P
:min(25,max(1,Q→Q
:For(A,1,8
:Output(A,1,sub(Str1,40P-80+Q+40A,16
:End
:Output(R-P+1,C-Q+1,"π
:EndDisregard how it's not optimized right now. I'll do that after getting it all to work.

All I know of is xLib, but I'm sure there are others, so what are they and which ones are the best/most used/still frequently updated?

I forgot most animes are that short. Most of the shows I've marathoned through have at least 100 hour-long episodes. Then again, that's why I switch between typical show and anime whenever I finish a series

I tend to like things with deep plots... Maybe I'll give this anime a watchthrough. How long is it?

Well I just decided to use a turn counter in my C4 program, but this is still an interesting problem.

Your pseudo-recursive solution is nice, albeit big, but it can easily be extended to find the min of any matrix in general:
assuming the dimensions of the matrix are {R,C}...

R->dim(L1
For(B,1,R
[A](B,1
For(A,2,C
min(Ans,[A](B,A->L1(B
End
End
min(L1

But I still hate how there doesn't look to be an elegant way to do this

While trying to finish my simple Connect 4 game to detect tied games (aka the board matrix is all nonzero), I realized that min( isn't overloaded to handle matrices, so I'm in need of some ingenuity. Since I realized I only have to care about the min of the top row, my current solution is this:
Matr>list([A]^T,L₁
min(L₁

But I want to know if there's a better way to do this. And I also want to know how to do the general case of finding the minimum of an entire matrix rather than just the top row like I did.

EDIT: Actually my current idea doesn't work either. Now I'm utterly clueless.

EDIT2: lol it's min(L1, not min(Ans. Disregard my last edit.