(*

* Solution to Project Euler problem 106

* Copyright (c) Project Nayuki. All rights reserved.

*

* https://www.nayuki.io/page/project-euler-solutions

* https://github.com/nayuki/Project-Euler-solutions

*)

(*

* Lemma to confirm denominator:

* For each natural number n >= 2, any set of size n has exactly

* (3^n + 1) / 2 - 2^n unordered pairs of non-empty disjoint subsets.

* Proof:

* 0. Let A be an arbitrary set of size n. We want to count its subset pairs.

* 1. Suppose we can label each element of A with a letter from the set {B, C, N}.

* A label is defined as an n-tuple of letters from the set {B, C, N}.

* For example, the set {a,b,c,d} can be labeled as (B,B,C,N) or (N,C,N,N).

* 2. Let L be the set of all possible labels on A. We know that |L| = 3^n.

* 3. If a label doesn't contain any B's, then it is equivalent to saying that the

* label is composed of any number of C's and N's, so there are 2^n of them.

* 4. If a label doesn't contain any C's, then it is equivalent to saying that the

* label is composed of any number of B's and N's, so there are 2^n of them.

* 5. If a label doesn't contain any B's or C's, then it is the singleton of all N's.

* 6. How many labels contain at least one B and at least one C?

* Using the inclusion-exclusion principle, we have:

* |Have B and Have C| = |All labels| - |Lacks B or Lacks C|

* = |All labels| - (|Lacks B| + |Lacks C| - |Lacks B and Lacks C|)

* = |All labels| - |Lacks B| - |Lacks C| + |Lacks B and Lacks C|

* = 3^n - 2^n - 2^n + 1.

* 7. For an arbitrary label that has at least one B and at least one C,

* what actually matters is which elements of A are put into one subset

* and which elements of A are put into the other disjoint subset, but

* the names of these subsets as B and C are interchangeable. Hence we

* divide by 2 to remove this degree of freedom, giving us a final count

* of (3^n - 2^n - 2^n + 1) / 2 = (3^n + 1) / 2 - 2^n.

* Corollary:

* This confirms some values mentioned in the problem statement:

* - Set size n = 7 has 966 subset pairs.

* - Set size n = 12 has 261625 subset pairs.

*

*

* Main theorem:

* Let A be an arbitrary set such that all of the following hold:

* - Its size is n.

* - It consists only of positive integers.

* - It satisfies the property (ii) given in the problem statement.

* - a0 < a1 < ... < a_{n-1} are all the elements of A listed in ascending order.

* To verify property (i), we would need to test some number of unordered pairs

* of non-empty disjoint subsets of A to see that the subsets have unequal sums

* Then we claim that exactly this many pairs need to be tested for equality:

* The summation of (n choose 2k) * [(2k choose k) / 2 - (2k choose k) / (k + 1)]

* for k from 2 to floor(n / 2) (inclusive).

*

* Proof:

* We begin by arguing about what subset pairs don't need to be tested,

* then progress to counting which subset pairs do and don't need to be tested.

*

* Let B and C be an arbitrary pair of non-empty disjoint subsets of A.

* Furthermore, restrict the pair (and prevent duplicate counting) so that

* the smallest element of B is smaller than the smallest element of C.

* Assume that for each element of B and C, we know what index of A

* it comes from, but we don't look at the actual element's value.

*

* If sets B and C have different sizes, then property (ii) implies that

* either S(B) < S(C) or S(B) > S(C), which in both cases imply S(B) != S(C).

* Hence we only care about the cases where |B| = |C|.

*

* If |B| = |C| = 1, then we know by disjointness that S(B) != S(C).

* (Namely because each set has a different singleton element.)

*

* For the interesting case, we have |B| = |C| >= 2. If we can match each

* element of B to a unique larger element in C, then we know for sure that

* S(B) < S(C), which further implies that S(B) != S(C). When we find such

* a matching, the ordering of the elements of A already implies inequality,

* without the need to examine the actual element values.

*

* To illustrate with a concrete example, suppose B = {a0,a1} and C = {a2,a3}

* are subsets of some set A. For each element of the set B or C, we are

* assumed to know what index of A the element came from. Because we know

* a0 < a2 and a1 < a3, we add these inequalities to get S(B) = a0 + a1

* < a2 + a3 = S(C), implying S(B) != S(C). Similarly, with B = {a0,a2} and

* C = {a1,a3}, we have a0 < a1 and a2 < a3, leading to a0 + a2 < a1 + a3.

* But in the case of B = {a0,a3} and C = {a1,a2}, we cannot conclude

* whether S(B) equals S(C) without examining the actual element values.

*

* Let's imagine scanning the elements of A in ascending order. When we

* encounter an element a_i that is:

* - In B, then we push it onto a stack, remembering that we need to

* pair it with a later (and thus larger) element that is in C.

* - In C, then we consult the stack. If the stack is empty, then this

* element cannot be paired with an earlier (and thus smaller) element that

* is in B, so no match exists. Otherwise we pop one element from the stack.

* - Not in B or C, then we ignore it because it plays no role in the sums.

*

* Suppose we lay out the elements of A as a sequence, and label each element

* according to the subsets B and C in a particular way. If element a_i is in B,

* then label it as ( (left parenthesis). If element a_i is in C, then label it

* as ) (right parenthesis). Otherwise label it as nothing/space.

*

* To illustrate, suppose A has size 5, B = {a0,a1}, and C = {a2,a3}.

* Then this pair of subsets corresponds with the label "(( ))" on A.

*

* We can see that a pair of subsets doesn't need to be tested if

* its label corresponds to a string of balanced parentheses. This is

* a well-known combinatorics problem (which I won't try to prove):

* The Catalan number C_k = (2k choose k) / (k + 1) represents the

* number of ways that k pairs of parentheses can be arranged in a

* sequence to produce a proper expression (i.e. no prefix contains

* more right parentheses than left parentheses).

*

* There are (n choose 2k) ways to choose elements from A that will

* then be split among the subsets B and C. Focus on one arbitrary choice.

*

* Subsequently, there are (2k choose k) / 2 ways to put k of those

* elements into B and k of those elements into C, but without

* regards to the ordering of B and C.

*

* However, C_k = (2k choose k) / (k + 1) of those choices of subset

* pairs will necessarily have unequal sums due to the ordering of

* elements, hence they don't need to be tested.

*

* This means that for a given k >= 2, we need to test (n choose 2k)

* * ((2k choose k) / 2 - (2k choose k) / (k + 1)) subset pairs.

*

* Finally, we sum this term for k = 2, 3, 4, ... as long as 2k <= n,

* so that we consider all the sizes of k where we can take a pair

* of k-sized disjoint subsets of A.

*

* Corollary:

* Although the derivation/justification is long, the amount of

* arithmetic is small enough to be doable by hand calculation.

*)

setSize = 12;

CatalanNum[n_] := Binomial[n * 2, n] / (n + 1)

Sum[Binomial[setSize, i * 2] * (Binomial[i * 2, i] / 2 - CatalanNum[i]),

{i, 2, Floor[setSize / 2]}]