Counting number of quadruples of integers - algorithm

Question

I saw this question today where we need to count the number of
quadruples of integers
(X1, X2, X3, X4), such that Li ≤ Xi ≤ Ri for i
= 1, 2, 3, 4 and X1 ≠ X2, X2 ≠ X3, X3 ≠ X4, X4 ≠ X1.
input:
Li Ri
1 4
1 3
1 2
4 4
output:
8
1 2 1 4
1 3 1 4
1 3 2 4
2 1 2 4
2 3 1 4
2 3 2 4
3 1 2 4
3 2 1 4
My initial thoughts were using
Principle of Inclusion Exclusion
I was able to find number if unrestricted quadruples but I am not able to figure out how can we find the remaining conditions to reach the final solution. Also I came to know this question can be done using DFS .
How can we do this question with Inclusion Exclusion/ DFS

Answer 1

Inclusion/Exclusion will give you the number of quadruples, but won't give you the quadruples themselves.
Let Ai be the set of quadruples satisfying Lj<=Xj<=Rj for all j, with Xi=X(i+1) (where the indices are cyclic, so X5 means X1). In the example you provided,
A1 = { (1114), (1124), (2214), (2224), (3314), (3324) }
A2 = { (1114), (2114), (3114), (4114), (1224), (2224), (3224), (4224) }
A3 = { } (empty set)
A4 = { (4114), (4214), (4314), (4124), (4224), (4324) }
We also need the intersections of pairs of sets:
A1 cap A2 = { (1114), (2224) } (note first three numbers identical)
A1 cap A3 = { }
A1 cap A4 = { } (can't have X4=X1=X2)
A2 cap A3 = { }
A2 cap A4 = { (4114), (4224) }
A3 cap A4 = { }
Intersections of triples of sets:
A1 cap A2 cap A3 = { }
A1 cap A2 cap A4 = { }
A1 cap A3 cap A4 = { }
A2 cap A3 cap A4 = { }
And the intersection of all the sets:
A1 cap A2 cap A3 cap A4 = { }
Inclusion/exclusion in its complementary form tells us that
|intersection of complements of Ai| = |unrestricted quadruples|
- sum of |Ai| + sum of |Ai cap Aj| - sum of |Ai cap Aj cap Ak|
+ sum of |Ai cap Aj cap Ak cap Al|
where none of the indices i,j,k,l are equal. In your example,
|intersection of complements of Ai| = 4x3x2x1 - (6+8+0+6) + (2+0+0+0+2+0) - (0+0+0+0) + 0
= 24 - 20 + 4 - 0 + 0 = 8
In order to find the |Ai| and their intersections, you have to find intersections of intervals [Li,Ri] and multiply the lengths of intersections by the lengths of unrestricted intervals. For example,
|A1| = |[1234] cap [123]| x |[12]| x |[4]| = 3 x 2 x 1 = 6
|A2 cap A4| = |[123] cap [12]| x |[4] cap [1234]| = |[12]| x |[4]| = 2 x 1 = 2
I don't see what depth first search has to do with it in this approach.

Answer 2

It depends if the sets are disjoint or share elements. For n = 4, meaning quadruples, as you asked about, I think I got it down to between 1 and 4 iterations if we commit the ends to four types describing if x_1 is a member of X2 and x_4 a member of X3.
Example with three iterations:
input = {1,2,3}{1,2}{1,2,3}{3,4}
2 * (1)(12)(123)(3) = (1)(2)(1)(3) = 2 * 1 // x_1 ∈ X2, x_4 ∈ X3
2 * (1)(12)(123)(4) = (1)(2)(13)(4) = 2 * 2 // x_1 ∈ X2, x_4 ∉ X3
1 * (3)(12)(123)(4) = (3)(12)(12,3)(4) = 1 * (2 + 2) // x_1 ∉ X2, x_4 ∉ X3
Total = 10
Example with one iteration:
input = {1,2,3,4}{1,2,3,4}{1,2,3,4}{1,2,3,4} // x_1 ∈ X2, x_4 ∈ X3
12 * (1)(1234)(1234)(2) = (1)(2,34)(134)(2) = 12 * (3 + 4)
Total = 84