# What is the probability that $n$ random points in $d$ dimensions are linearly separable?

Given $n$ data points, each with $d$ features, $n/2$ are labeled as $0$, the other $n/2$ are labeled as $1$. Each feature takes a value from $[0,1]$ randomly (uniform distribution). What's the probability that there exists a hyperplane that can split the two classes?

Let's consider the easiest case first, i.e. $d = 1$.

• This is a really interesting question. I think this might be able to be reformulated in terms of whether or not the convex hulls of the two classes of points intersect or not - though I don't know if that makes the problem more straightforward or not. – Don Walpola Sep 10 '18 at 2:55
• This will clearly be a function of the relative magnitudes of $n$ & $d$. Consider the easiest case w/ $d=1$, if $n=2$, then w/ truly continuous data (ie, no rounding to any decimal place), the probability they can be linearly separated is $1$. OTOH, $\lim n\to \infty\ \ \text{Pr(linearly separable)} \to 0$. – gung Sep 10 '18 at 14:08
• You should also clarify if the hyperplane needs to be 'flat' (or if it could be, say, a parabola in a $2d$-type situation). It seems to me that the question strongly implies flatness, but this should probably be stated explicitly. – gung Sep 10 '18 at 14:13
• @gung I think the word "hyperplane" unambiguously implies "flatness", that's why I edited the title to say "linearly separable". Clearly any dataset without duplicates can is in principle nonlinearly separable. – amoeba Sep 10 '18 at 15:30
• @gung IMHO "flat hyperplane" is a pleonasm. If you argue that "hyperplane" can be curved, then "flat" can also be curved (in an appropriate metric). – amoeba Sep 10 '18 at 18:07

## 1 Answer

Assuming no duplicates exist in the data.

If $$n\leq d+1$$, the probability is $$\text{Pr}=1$$.

For other combinations of $$(n,d)$$, see the following plot: I generated this plot simulating input and output data as specified in the OP. Linear separability was defined as failure of convergence in a logistic regression model, due to the Hauck-Donner effect.

We can see the probability decreases for increasing $$n$$. In fact, we could fit a model relating $$n, d$$ to $$p$$, and this was the result:

$$P(n,d)={ 1 \over {1 + e^ {-(5.82944-4.58261\times n + 1.37271 \times d -0.0235785 \times n \times d)} } }$$ Code for the plot (in Julia):

using GLM

ds = 10; #number of dimensions to be investigated
ns = 100 #number of examples to be investigated
niter = 1000; #number of iterations per d per n
P = niter * ones(Int64, ds, ns); #starting the number of successes

for d in 1:ds
for n in (d+1):ns
p = 0 #0 hits
for i in 1:niter
println("Dimensions: $$d; Samples:$$n; Iteration: $i;") try #we will try to catch errors in the logistic glm, these are due to perfect separability X = hcat(rand((n,d)), ones(n)); #sampling from uniform plus intercept Y = sample(0:1, n) #sampling a binary outcome glm(X, Y, Binomial(), LogitLink()) catch p = p+1 #if we catch an error, increase the count end end P[d,n] = p end end using Plots gui(heatmap(P./niter, xlabel = "Number of Samples", ylabel = "Number of Dimensions", title = "Probability of linear separability"))  Code for the model relating $$(n,d)$$ to $$p$$ (in Julia): probs = P./niter N = transpose(repmat(1:ns, 1, ds)) D = repmat(1:ds, 1, ns) fit = glm(hcat(log.(N[:]), D[:], N[:].*D[:], ones(ds*ns)), probs[:], Binomial(), LogitLink()) coef(fit) #4-element Array{Float64,1}: # -4.58261 # 1.37271 # -0.0235785 # 5.82944 gui(heatmap(reshape(predict(fit), ds, ns), xlabel = "Number of Samples", ylabel = "Number of Dimensions", title = "Fit of probability of linear separability"))  • +1. Why log(n) and not n? The yellow-black boundary looks like a straight line to me on the top figure, but appears bent on the second figure. Is it maybe because of the log(n)? Not sure. – amoeba Sep 26 '18 at 13:50 • @amoeba I changed it. I also included the interaction, because it could explain the gradual broadening of the border between$p=1$and$p=0\$ (which was the reason I had tried the logarithm before). – Firebug Sep 26 '18 at 16:54