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Assume spawn(e) spawns points uniformly within an element e, i.e., for any i∈e, P⁢(i∣e)=1v⁢(e), where P⁢(i∣e) is the probability density function describing the likelihood of sampling a point i within the element e, and v⁢(e) is the volume of e. Let 𝒮 be the set of all solutions satisfying constraints 𝒞. The probability of sampling i∈(ℐ∪𝒪) involves selecting an element e∋i and spawning i uniformly within e. The selection probability of e is proportional to its volume:

P⁢(e)=v⁢(e)∑e′∈ℐ∪𝒪v⁢(e′).

Thus, the joint probability of selecting e and spawning i is:

P(i is spawned)=P(e)⋅P(i∣e)=v⁢(e)∑e′∈ℐ∪𝒪v⁢(e′)⋅1v⁢(e)=1∑e′∈ℐ∪𝒪v⁢(e′)⋅

Now, a point sampled with this process, without proper rejection, may not belong to 𝒮. If e∈ℐ, all points i∈e satisfy 𝒞, so i is always accepted. If e∈𝒪, i is accepted only if i∈𝒮, and otherwise we repeat the sampling process. It follows that Algorithm 3 returns a particular solution i∈𝒮 either if:

  • it samples i at the first attempt,

  • or if the first attempt fails to produce a solution and i is sampled in a subsequent attempt.

The probability of sampling a particular solution i∈𝒮 is thus

P⁢(i) =1∑e′∈ℐ∪𝒪v⁢(e′)+P⁢(fail)⁢P⁢(i)
--- --- --- ---
=1(1−P⁢(fail))⁢∑e′∈ℐ∪𝒪v⁢(e′)

where P⁢(fail) is the probability that a uniform element in ℐ∪𝒪 does not satisfy 𝒞. This is constant for all i∈𝒮, proving uniformity.