Detailed mathematical approach

Problem Summary

We are given a list of symbolic expressions built from variables and a binary constructor. For each unordered pair of positions containing different expressions, we ask whether a substitution can make the two expression trees identical. If no such substitution exists, that pair contributes \(0\). If a substitution does exist, the resulting common tree is evaluated numerically, and the overall total is taken modulo

$M=10^9.$

A brute-force search over all substitutions is hopeless, so the solution works directly with tree structure, recursive unification, and a cache of repeated pair computations.

Mathematical Approach

The key observation is that each pair of expressions can be handled in two stages: first solve the symbolic equation by unification, then evaluate the unified term by a fixed arithmetic rule.

Step 1: Model the expressions as rooted binary trees

In abstract form, the expressions follow the grammar

$E ::= v \mid I(E,E),$

where \(v\) denotes a variable. Thus every expression is either a leaf carrying a variable name or an internal node with exactly two children. Structural equality means same tree shape and same variable labels in the corresponding positions.

This viewpoint turns the problem into a question about tree equations rather than about string manipulation.

Step 2: Solve one pair by recursive unification

For two expressions \(E_1\) and \(E_2\), we seek a substitution \(\sigma\) such that

$\sigma(E_1)=\sigma(E_2).$

A substitution maps variables to expressions and extends recursively:

$\sigma\bigl(I(A,B)\bigr)=I\bigl(\sigma(A),\sigma(B)\bigr).$

The recursive cases are straightforward. If both current nodes are binary constructor nodes, then their left children must unify and their right children must unify. If one side is a variable, we may bind that variable to the other side, provided the binding is legal. If neither rule applies, the pair cannot be unified.

Before each comparison, already known bindings are followed until the current node is fully resolved. This is why chains such as \(a \mapsto b\) and \(b \mapsto I(c,d)\) behave exactly as the direct binding \(a \mapsto I(c,d)\).

Step 3: Prevent cyclic substitutions with the occurs-check

The only dangerous kind of binding is a self-referential one. If a variable \(v\) is to be replaced by a term \(T\), we must require

$v\notin \operatorname{Vars}(T).$

This is the classical occurs-check. Without it, an equation such as

$v=I(v,w)$

would force an infinite cyclic object instead of an ordinary finite term. The implementations explicitly reject such cases, so any successful unification always produces an acyclic expression tree.

Step 4: Evaluate the unified tree modulo \(10^9\)

Once unification succeeds, the common term is interpreted numerically. Unbound variables contribute \(0\). For the binary constructor, the recursive valuation rule is

$J(x,y)=\bigl(1+x+y\bigr)^2+(y-x)\pmod{M}.$

Equivalently, if \(V_\sigma(E)\) denotes the value of expression \(E\) under the successful substitution \(\sigma\), then

$V_\sigma(v)=0\quad\text{for an unresolved variable},$

$V_\sigma\bigl(I(A,B)\bigr)=J\bigl(V_\sigma(A),V_\sigma(B)\bigr).$

Therefore the pair contribution is

$P(E_1,E_2)= \begin{cases} V_\sigma(E_1), & \text{if }E_1\text{ and }E_2\text{ unify},\\ 0, & \text{otherwise}. \end{cases}$

Because the two expressions become identical after applying \(\sigma\), evaluating either side gives the same value.

Step 5: Compress duplicates and reconstruct the final sum

The full input may contain many repeated expressions. After converting every parsed tree to one canonical textual form, equal expressions can be merged into unique representatives \(U_0,U_1,\dots,U_{U-1}\).

For those representatives, define the symmetric pair table

$C_{i,j}=P(U_i,U_j),\qquad C_{i,j}=C_{j,i}.$

If the original list has \(N\) positions and the expression at position \(p\) corresponds to representative index \(a_p\), then the required total is

$S=\sum_{\substack{1\le p\lt q\le N \\ a_p\ne a_q}} C_{a_p,a_q}\pmod{M}.$

The condition \(a_p\ne a_q\) matters: even if the same structural expression appears at different positions, that positional pair is excluded by the problem statement.

Worked Example

Take

$E_1=I(a,b),\qquad E_2=I(c,I(c,d)).$

The roots have the same binary shape, so we unify their children. From the left children we get

$a\mapsto c.$

From the right children we get

$b\mapsto I(c,d).$

This second binding is legal because \(b\) does not occur inside \(I(c,d)\). Hence the pair unifies, and the common resolved tree is

$I(c,I(c,d)).$

Now evaluate it. Since \(c\) and \(d\) remain unbound, both contribute \(0\). The inner node gives

$J(0,0)=(1+0+0)^2+(0-0)=1.$

The outer node then gives

$J(0,1)=(1+0+1)^2+(1-0)=4+1=5.$

So this pair contributes \(5\) modulo \(10^9\). By contrast, the pair \(a\) and \(I(a,b)\) fails immediately, because binding \(a\) to a term that already contains \(a\) violates the occurs-check.

How the Code Works

The C++, Python, and Java implementations first parse the entire input into expression trees while preserving the original order and all duplicates. They then serialize each tree into a canonical form so that structurally identical expressions receive the same unique index.

Next, the implementation computes the pair value for every unique-expression pair. Each pair is processed by recursive unification with an occurs-check; if unification succeeds, the resolved tree is evaluated recursively modulo \(10^9\). During evaluation, already resolved subtrees are memoized inside the current pair computation so that repeated branches are not recomputed. The C++ implementation additionally distributes the unique-pair precomputation across several worker tasks, while the Python and Java implementations perform the same logic in direct nested loops.

Finally, the implementations scan all original input positions \((p,q)\) with \(p<q\), skip the pairs whose two positions refer to the same canonical expression, and add the cached value \(C_{a_p,a_q}\) for the remaining pairs.

Complexity Analysis

Let \(N\) be the total number of input expressions, \(U\) the number of distinct structural expressions after deduplication, and \(L\) the total size of all parsed trees. Parsing plus canonicalization is linear in the input size, so it costs \(O(L)\) time.

The dominant symbolic phase is the unique-pair table. Its cost is

$O\bigl(U^2\cdot T_{\text{pair}}\bigr),$

where \(T_{\text{pair}}\) is the cost of one unification and, when successful, one recursive evaluation. On ordinary acyclic inputs this is close to linear in the number of visited nodes, although the occurs-check can revisit large subtrees in harder cases. The final positional summation costs \(O(N^2)\) time. Memory usage is \(O(L+U^2)\) for the stored trees and the pair cache. Parallelism in the C++ implementation improves wall-clock time but not the asymptotic bound.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=674
2. Unification in computer science: Wikipedia - Unification (computer science)
3. Occurs check: Wikipedia - Occurs check
4. Substitution in logic: Wikipedia - Substitution (logic)
5. Abstract syntax tree: Wikipedia - Abstract syntax tree