The more I think about it, the less sure I am it’s just a metaphor.
It is obviously a metaphor in the sense that I meant it as one. I use surface area metaphors all the time because I find them conceptually useful. (I often think of plans as having “fuck-up targets”, which is the union of the probabilities of all major modes of failure, and then think what I’d have to do to minimise the target of any plan I’m implementing).
I’m thinking it might not be a metaphor in the sense that the properties of large surface area solids might be isomorphic with the properties of search spaces that are easy to systematically search. For example, for an object made up of a finite number of points, the maximal surface area is in effect a binary tree.
I am now no longer sure whether I’ve come up with a useful way of thinking about problem-solving, or if I’m just digging myself into a tautological hole.
The most obvious reason, to me, why it’s just a metaphor is that there’s no rigorous definition of what the space actually is. So surface area and volume of problems are in fact not defined. If we let volume be unique to a problem as you seem to be thinking, then we could say that it has units of knowledge. And then surface area has units of knowledge^(2/3). So… I would guess not.
I assumed it was obvious that the volume of a problem is its solution space and the attack surface is the affordances given to it by whatever method of analysis you’re using.
I guess if that isn’t obvious, it probably reads very differently.
Why would you assume a solution space to be 3-dimensional? That said, I don’t see how the possible approaches could be regarded as the surface area even to an n-sphere (even with noninteger n). A triangulation might be more accurate.
Fair enough. You get my point, though. The units don’t work for a literal interpretation. Instead, what seems to be talked about as attack angles is simply which solutions have a decent chance of working. But there’s no reason why high-likelihood avenues should correspond to the edges—in fact, they should tend to represent the center.
The more I think about it, the less sure I am it’s just a metaphor.
It is obviously a metaphor in the sense that I meant it as one. I use surface area metaphors all the time because I find them conceptually useful. (I often think of plans as having “fuck-up targets”, which is the union of the probabilities of all major modes of failure, and then think what I’d have to do to minimise the target of any plan I’m implementing).
I’m thinking it might not be a metaphor in the sense that the properties of large surface area solids might be isomorphic with the properties of search spaces that are easy to systematically search. For example, for an object made up of a finite number of points, the maximal surface area is in effect a binary tree.
I am now no longer sure whether I’ve come up with a useful way of thinking about problem-solving, or if I’m just digging myself into a tautological hole.
The most obvious reason, to me, why it’s just a metaphor is that there’s no rigorous definition of what the space actually is. So surface area and volume of problems are in fact not defined. If we let volume be unique to a problem as you seem to be thinking, then we could say that it has units of knowledge. And then surface area has units of knowledge^(2/3). So… I would guess not.
I assumed it was obvious that the volume of a problem is its solution space and the attack surface is the affordances given to it by whatever method of analysis you’re using.
I guess if that isn’t obvious, it probably reads very differently.
Ah, so the units of surface area are solutions^(2/3) then.
Why would you assume a solution space to be 3-dimensional? That said, I don’t see how the possible approaches could be regarded as the surface area even to an n-sphere (even with noninteger n). A triangulation might be more accurate.
Fair enough. You get my point, though. The units don’t work for a literal interpretation. Instead, what seems to be talked about as attack angles is simply which solutions have a decent chance of working. But there’s no reason why high-likelihood avenues should correspond to the edges—in fact, they should tend to represent the center.
Yes, I agree with that.