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[Max White]

Because I am too hazy right now to think clearly.
A person is surrounded by eight nodes.
Each node may be turned either on or off.
How many different combinations of nodes being on or off are there?

Show working, so that I can recall how to do this next time. Used to know, but for the life of me I can't remember right now.

[Lectorog]

Each node has two modes: on or off. There are eight. Thus, there are 2^8 combinations: two times itself eight times. That's 256. (Which is also the number of possibilities of storage in a byte, for the same reason.)

There's also a Mathematics Help Thread, just so you know.

[Bauglir]

2^8, or 256; this is basically asking what the largest number you can represent in 8 bits is, if you start at 1.

0000 0000
to
1111 1111

More generally, the number of possible values of X digits with Y possible values each is Y^X. I'm not quite sure how to explain that, though, because I suck. I am really bad at moving very far to the right on that curve, you see, so even though this isn't particularly obvious on its own, the fact that I already know it and find it intuitive makes it difficult for me to articulate. Which probably means I don't actually understand it, but oh well.

EDIT: Damn, I've been ninja'd.

[Vector]

Yes, well, that would be why we threw that under the bus called "Oh fuck it, it's axiomatic."

[Leatra]

Combinations. One of the rare maths topic which I can study without my stomach burning with pain.

[askovdk]

I hate to complicate matters, but is it significant that the person is surrounded by the nodes?
If the problem is about high school level, then I agree with the other posters, but if the question is at college level, then you may need to consider rotation symmetries.
I.e.  10111000  is the same as 11000101. 

[Vector]

Yeah, you'd probably want to use the Burnside formula.  Fortunately, that's relatively easy.

[Neonivek]

I hate to complicate matters, but is it significant that the person is surrounded by the nodes?
If the problem is about high school level, then I agree with the other posters, but if the question is at college level, then you may need to consider rotation symmetries.
I.e.  10111000  is the same as 11000101. 

DANG IT! I used to remember how to answer that one too.

[Max White]

I don't think 255 is the answer I am looking for, so I must have misinterpreted the problem. This is neither for school nor collage or anything like that, this is practical application. I need a quick tileset for testing a program, and each block has a different tile depending on those around it. In the end I had 47 different tiles, as shown below

Spoiler (click to show/hide)

Top = grass
Bottem = Key
Each tile is 8*8

But I'm still not sure if that is every combination covered.

[IronyOwl]

I think the issue here is that not every combination is unique, or else you would indeed need a different tile for all grass, all grass but one tile to the north, all grass but north and northeast, all grass but north, northeast, and east, etc. What really matters for what tile is displayed?

[Max White]

Well, the tiles around it matter... I have a tile for when a block is surrounded, and when it is surrounded from all sides but the north, and when it is surrounded by all but the north east, and so on and so forth.

Come to think of it, as a design choice there is an exception, and that is that corners do not count unless both adjacent sides are true. So, the northeast corner doesn't matter unless both the north and east sides are occupied. That should lower the tile count a bit.

[IronyOwl]

So... you've got four nodes, and when two "adjacent" nodes are both active, it opens up another node?

That gets a bit more complicated than raw multiplication. Hm...

[Nivim]

 The four adjacents sum to 16 combinations (if you aren't rotating and reusing tiles), and then if corners are considered when two adjacents that are adjacent to eachother are the same that adds 56 combinations, for 72 total. I guess you've probably already tested it for yourself by now, though.