Now, let try another approach. We construct the distribution for the smallest enclosing angle incrementally. The case for the smallest enclosing angle for 2 duck is just the uniform distribution in [0,π].
Solving the distribution function for 3 ducks is key, because the same approach is likely useful for 4 ducks, or even n ducks.
Let 2a be the smallest enclosing angle for the first 2 ducks
Let b be the smallest enclosing angle for the first 3 ducks.
For simplicity, we let duck 1 has polar angle a , duck 2 has polar angle −a.
For duck 3, let it polar angle be t.
If t∈[a,−a], then b=2a, in other words, P(b=2a)=2a2π=aπ.
Now suppose t∈[a,π], we need to compare all choices:
(Option 1) duck 2 -> duck 1 -> duck 3 will have angle t+a
(Option 2) duck 1 -> duck 3 -> duck 2 will have angle 2π−2a
(Option 3) duck 3 -> duck 2 -> duck 1 will have angle 2π−(t−a)=2π+a−t
We know (option 1) is always as good as than (option 3) because (option 3) - (option 1) = (2π+a−t)−(t+a)=2π−2t≥0 (Remember the maximum of t is π). So we can simply ignore option 3.
It is not so simple for (option 2), we see
(option 1) - (option 2) = (t+a)−(2π−2a)=t+3a−2π
We know if the difference is greater than 0, then (option 2) is better. The condition for that to be true is t≥2π−3a, which means if the range of t=[a,π] contains 2π−3a, then we need to worry about splitting it so that we choose the options wisely.
For that to happen, a≤2π−3a⟹4a≤2π⟹a≤π2, which is automatically true, we also need 2π−3a≤π⟹π≤3a⟹a≥π3
So we have 4 cases to worry about for positive t.
Case 1: t≤a - in that case b=2a.
Case 2: t≥a and a≤π3 - in that case b=t+a.
Case 3: t≥a, a≥π3 and t≤2π−3a - in that case b=t+a, and at last
Case 4: t≥a, a≥π3 and t>2π−3a - in that case b=2π−2a.
The cases for negative t is completely symmetric so we can skip it. Now, if we are given a and t, we can calculate b.
But, that's not our goal. We wanted to know the distribution of b, To get there, the plan is to compute F(b|a), the conditional cumulative density function (conditional CDF) for b first.
Case 1 is easy to encode, P(b≤t|a)=0 if t∈[0,2a).
Case 2 means P(b≤t+a|a)=2t2π=tπ, a≤π3 and t≥a, as the diagram would indicate.
An algebraic manipulation of the above is to let s=t+a, that would give P(b≤s)=s−aπ, a≤π3 and s≥2a.
Case 1 and Case 2 form the complete distribution for b (condition on a), in this case, we get a curve like this:
Case 3 is getting complicated, as long as we stay within the 2π−3a region, the probability is pretty much the same as it was for case 2.
P(b≤s)=s−aπ, a≥π3,s≥2a and t≤2π−3a⟹s≤2π−2a
Case 4 seems complicated but in fact it is not. It merely says that the s is capped above by 2π−2a, so the final CDF look like this in that case.
It is too late at night for me now, so I will stop here with the conditional CDF. If time allows I will continue and work on the PDFs.
Solving the distribution function for 3 ducks is key, because the same approach is likely useful for 4 ducks, or even n ducks.
Let 2a be the smallest enclosing angle for the first 2 ducks
Let b be the smallest enclosing angle for the first 3 ducks.
For simplicity, we let duck 1 has polar angle a , duck 2 has polar angle −a.
For duck 3, let it polar angle be t.
If t∈[a,−a], then b=2a, in other words, P(b=2a)=2a2π=aπ.
Now suppose t∈[a,π], we need to compare all choices:
(Option 1) duck 2 -> duck 1 -> duck 3 will have angle t+a
(Option 2) duck 1 -> duck 3 -> duck 2 will have angle 2π−2a
(Option 3) duck 3 -> duck 2 -> duck 1 will have angle 2π−(t−a)=2π+a−t
We know (option 1) is always as good as than (option 3) because (option 3) - (option 1) = (2π+a−t)−(t+a)=2π−2t≥0 (Remember the maximum of t is π). So we can simply ignore option 3.
It is not so simple for (option 2), we see
(option 1) - (option 2) = (t+a)−(2π−2a)=t+3a−2π
We know if the difference is greater than 0, then (option 2) is better. The condition for that to be true is t≥2π−3a, which means if the range of t=[a,π] contains 2π−3a, then we need to worry about splitting it so that we choose the options wisely.
For that to happen, a≤2π−3a⟹4a≤2π⟹a≤π2, which is automatically true, we also need 2π−3a≤π⟹π≤3a⟹a≥π3
So we have 4 cases to worry about for positive t.
Case 1: t≤a - in that case b=2a.
Case 2: t≥a and a≤π3 - in that case b=t+a.
Case 3: t≥a, a≥π3 and t≤2π−3a - in that case b=t+a, and at last
Case 4: t≥a, a≥π3 and t>2π−3a - in that case b=2π−2a.
The cases for negative t is completely symmetric so we can skip it. Now, if we are given a and t, we can calculate b.
But, that's not our goal. We wanted to know the distribution of b, To get there, the plan is to compute F(b|a), the conditional cumulative density function (conditional CDF) for b first.
Case 1 is easy to encode, P(b≤t|a)=0 if t∈[0,2a).
Case 2 means P(b≤t+a|a)=2t2π=tπ, a≤π3 and t≥a, as the diagram would indicate.
Case 1 and Case 2 form the complete distribution for b (condition on a), in this case, we get a curve like this:
Case 3 is getting complicated, as long as we stay within the 2π−3a region, the probability is pretty much the same as it was for case 2.
P(b≤s)=s−aπ, a≥π3,s≥2a and t≤2π−3a⟹s≤2π−2a
Case 4 seems complicated but in fact it is not. It merely says that the s is capped above by 2π−2a, so the final CDF look like this in that case.
It is too late at night for me now, so I will stop here with the conditional CDF. If time allows I will continue and work on the PDFs.
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