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1,790 charsAssume 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:
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it samples i at the first attempt,
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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.