Looping Numbers

pHghost

You can use the mod function to loop through a sequence of integers, like 0-8. Is there any method which would do the same, but you could start at a non-zero point -- eg. 5-17, 200-230?

Socks

@Hoodloc said:
That's really simple. Looping between a and b = mod(0,(b-a)+1)+a.

5-17 = mod(0,13)+5
200-230 = mod(0,31)+200

I think you should have an attribute (or equation) where you have a zero ?

eg. . . . mod(X,13)+5 . . . or . . . mod(X+1,13)+5 . . . etc

The '0' might be a bit confusing ?

birdboy

@Socks said:

@Hoodloc said:
That's really simple. Looping between a and b = mod(0,(b-a)+1)+a.

5-17 = mod(0,13)+5
200-230 = mod(0,31)+200

I think you should have an attribute (or equation) where you have a zero ?

eg. . . . mod(X,13)+5 . . . or . . . mod(X+1,13)+5 . . . etc

The '0' might be a bit confusing ?

Already deleted my post.
Both of our solutions are wrong.

Edit: more precisely, my solution was wrong, it's still wrong when you include an attribute.

Socks

@Hoodloc said:

Already deleted my post.
Both of our solutions are wrong.

Are you sure ? I'm not at my computer to check, but your approach looks fine to me ?

If X is always rising (let's imagine it's game.time), then mod(X,13)+5 should give you 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 5 - 6 - 7 - 8 . . . etc

?

birdboy

@Socks said:

@Hoodloc said:

Already deleted my post.
Both of our solutions are wrong.

Are you sure ? I'm not at my computer to check, but your approach looks fine to me ?

If X is always rising (let's imagine it's game.time), then mod(X,13)+5 should give you 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 5 - 6 - 7 - 8 . . . etc

?

Yeah, that was my initial thought too. Our solution results in 5 - 10 - 15 - 7 - 12 - ... because of the second addend.

Here's how you do it @pHghost :

For looping x between a and b -> x = mod(x-(a-1),(b-a)+1)+a
There's an example attached under the next post.

Socks

@Hoodloc said:
Yeah, that was my initial thought too. Our solutions results in 5 - 10 - 15 - 7 - 12 - ... because of the addend.

That depends on exactly how you have it set up.

For example in my suggestion above (using a rising value like game.time) the values would loop correctly.

What setup results in 5 - 10 - 15 . . etc ?

birdboy

@pHghost Added Attachment.

@Socks Yes, if you're using game.time or self.time, you're right. But for a more universal solution you need to use your own attribute.

pHghost

Perfect, thank you guys!

Makes sense and is quite straightforward, once you put the numbers together -- next time I need to be a bit less lazy.

The basic premise behind mod(X,13)+5 is logically correct (which I was trying previously, but it wasn't working, which led to this thread), but is only halfway to the solution. The thing I was missing was that you need to counteract the +5, which causes a big jump. So, if you want to do +1 steps, you counteract the +5 with a -4, so you get: mod(X-4,13)+5. The same way, you would counteract the +200 in the second example with -199.

The awesome thing about this breakdown (thanks for the formula @Hoodloc!) is that it also answers the second part of my question (which I hand't asked yet). The question was to be: how can you use the same mod function to count down in a set range?

Well, the answer is easy, you counteract the final offset with a number one higher! So, for example mod(X-6,13)+5 will count down from 17 to 5.

So yeah:

For a counting up loop between a and b -> x = mod(x-(a-1),(b-a)+1)+a

For a counting down loop between a and b -> x = mod(x-(a+1),(b-a)+1)+a