2 ms·

> On the other hand, a so-called mountain peak would be a 5 surrounded by 4’s or lower. The odds for having this happen in 10D are 0.2*(1-0.8^10) = 0.18. Then t

> On the other hand, a so-called mountain peak would be a 5 surrounded by 4’s or lower. The odds for having this happen in 10D are 0.2*(1-0.8^10) = 0.18. Then the total density of mountain peaks, in a 10D hyperlattice with 5 potential values, is only 18%. I believe the odds are actually 0.2 (odds of it being a 5) × 0.8^10 (odds of each of the neighbors being ∈ {1,2,3,4}) which is ~0.021 or around 2%. This makes much more sense, since 18% of the nodes being peaks doesn't sound like they are rare.

ted_dunning 1y ago
Even if you assume that they are asking about when given a node is a 5 what is the probability it will not have any 5 as a neighbor the result is about 10% which is your result without the leading 0.2. That's still not what they got.
jlokier 1y ago
It's not 0.8^10 anyway, because there aren't 10 neighbours. There are neighbours in both directions in each dimension, i.e. 20 neighbours in 10 dimensions if you're allowed to wrap around the edges of the lattice. With wrapping, I think the probability is 0.2 × (0.8^20) ≈ 0.002306 ≈ 0.23%. If you're not allowed to wrap at the edges, a unformly random point has 2/N probability of having one neighbour on each dimension independently, and (N-2)/N probability of having two. With D dimensions, that's on average D(2N-2)/N neighbours. With D = 10, N = 5, I think it's 16 neighbours on average, with a distribution from 10 to 20. That makes the probability of landing on a mountain peak lower than ~2% and higher than ~0.23%. (Not 0.2 × (0.8^16) though, due to the distribution.)
- MarkusQ 1y ago
  You are correct about the first part (I actually came back here to add a comment to mine saying that I'd goofed and the exponent should be 20, not 10). Also, I suspect that the width (your N) should be large enough in most of the cases where we'd be wondering about peaks vs. ridge lines that the large-N limit of (2N-2)/N could safely be taken (that is, 2). [The whole motivation for peaks vs. ridges is the question of whether there is a connected path that passes "near" all the points in the space, which is generally taken to mean within a ball of radius r, with 1 < r << N; since the grid is discrete, this implies N is >> 1.]

ted_dunning 1y ago

Even if you assume that they are asking about when given a node is a 5 what is the probability it will not have any 5 as a neighbor the result is about 10% which is your result without the leading 0.2. That's still not what they got.

jlokier 1y ago

It's not 0.8^10 anyway, because there aren't 10 neighbours. There are neighbours in both directions in each dimension, i.e. 20 neighbours in 10 dimensions if you're allowed to wrap around the edges of the lattice. With wrapping, I think the probability is 0.2 × (0.8^20) ≈ 0.002306 ≈ 0.23%. If you're not allowed to wrap at the edges, a unformly random point has 2/N probability of having one neighbour on each dimension independently, and (N-2)/N probability of having two. With D dimensions, that's on average D(2N-2)/N neighbours. With D = 10, N = 5, I think it's 16 neighbours on average, with a distribution from 10 to 20. That makes the probability of landing on a mountain peak lower than ~2% and higher than ~0.23%. (Not 0.2 × (0.8^16) though, due to the distribution.)

MarkusQ 1y ago
You are correct about the first part (I actually came back here to add a comment to mine saying that I'd goofed and the exponent should be 20, not 10). Also, I suspect that the width (your N) should be large enough in most of the cases where we'd be wondering about peaks vs. ridge lines that the large-N limit of (2N-2)/N could safely be taken (that is, 2). [The whole motivation for peaks vs. ridges is the question of whether there is a connected path that passes "near" all the points in the space, which is generally taken to mean within a ball of radius r, with 1 < r << N; since the grid is discrete, this implies N is >> 1.]