Detailed mathematical approach

Problem Summary

Problem 910 asks for the value of a deeply nested L-expression modulo \(10^9\). The solution is built around the quotient ring

$M=10^9,\qquad P(x)=\prod_{k=1}^{40}(x+k),\qquad A=(\mathbb{Z}/M\mathbb{Z})[x]/(P(x)).$

Rather than expanding enormous symbolic expressions, the computation keeps every intermediate object as a single polynomial class in \(A\). For the official tuple \((12,345678,9012345,678,90)\), the target quantity is obtained by constructing the final polynomial layer, evaluating it at \(x=678\), adding \(90\), and reducing modulo \(10^9\).

The hard part is therefore not the final substitution. It is finding a representation in which repeated multiplication, repeated composition, and repeated nesting stay finite and algorithmically manageable.

Mathematical Approach

The implementations succeed because the whole L-expression recursion can be rewritten as arithmetic on polynomials of degree at most \(39\). That degree bound is not a heuristic; it is a strict invariant forced by the ring \(A\).

Why the Quotient Ring Is the Correct State Space

The modulus polynomial \(P(x)\) is monic and has degree \(40\). Therefore every class in \(A\) has a unique remainder of the form

$r(x)=r_0+r_1x+\cdots+r_{39}x^{39}.$

So every gigantic symbolic L-expression is compressed to a coefficient vector in the fixed basis

$1,x,x^2,\dots,x^{39}.$

This remains true even though the coefficient ring is \(\mathbb{Z}/10^9\mathbb{Z}\), which is not a field. No division or interpolation is needed; the whole method uses only ring operations that are valid modulo a composite number.

The Three Nested Polynomial Layers

The recursive structure used by the implementations is

$D_1(\ell)=x^\ell+x^{\ell+1},$

$D_2(m,U)=U^{\circ m}\circ(xU),$

$D_3(0,m,\ell)=D_1(\ell)\circ D_2(m,D_1(\ell)),$

$D_3(n,m,\ell)=D_2(m,D_3(n-1,m,\ell)).$

Here \(xU\) is ordinary multiplication in \(A\), while \(U^{\circ m}\) means \(m\)-fold self-composition. The final answer is then

$\operatorname{Ans}(n,m,\ell,t,s)=\bigl(D_3(n,m,\ell)(t)+s\bigr)\bmod M.$

The mathematical task is to carry out all three layers without ever leaving the degree-\(39\) basis.

Folding \(x^{40}\) Back into the Basis

Write

$P(x)=x^{40}+p_{39}x^{39}+\cdots+p_1x+p_0.$

Because \(P(x)\equiv 0\) in the quotient ring, we have the reduction rule

$x^{40}\equiv-\bigl(p_{39}x^{39}+\cdots+p_1x+p_0\bigr)\pmod{P(x)}.$

If

$q(x)=q_0+q_1x+\cdots+q_{39}x^{39},$

then multiplication by \(x\) becomes

$xq(x)\equiv \sum_{i=0}^{38} q_i x^{i+1}-q_{39}\bigl(p_0+p_1x+\cdots+p_{39}x^{39}\bigr).$

This identity is the central reduction step. Once it is available, every overflow term is folded back immediately, so no polynomial of degree \(40\) or more ever has to be stored explicitly.

From Shifts to Products, Powers, and Composition

The previous formula gives multiplication by \(x\). General ring multiplication follows from linearity:

$\left(\sum_{i=0}^{39} a_i x^i\right)q(x)=\sum_{i=0}^{39} a_i\bigl(x^i q(x)\bigr).$

So one can build a product from repeated shift-and-reduce steps. Ordinary powers such as \(x^\ell\) are then computed by binary exponentiation inside the ring \(A\).

Composition is handled in the same basis by

$p\circ q=p(q(x))=\sum_{i=0}^{39} a_i q(x)^i,$

where all products \(q(x)^i\) are again reduced in \(A\). The same repeated-squaring idea also works for iterated composition, because iterates of one polynomial satisfy

$U^{\circ a}\circ U^{\circ b}=U^{\circ(a+b)}.$

That is exactly what makes the huge parameter \(m=345678\) tractable.

Worked Checkpoint

A small example from the built-in validation set is \((n,m,\ell,t,s)=(0,1,1,1,0)\). First,

$D_1(1)=x+x^2.$

Since \(m=1\), the middle layer is just one composition:

$D_2(1,U)=U\circ(xU).$

Substituting \(U=D_1(1)\) gives

$xU=x(x+x^2)=x^2+x^3,$

$D_2(1,D_1(1))=(x+x^2)\circ(x^2+x^3)=(x^2+x^3)+(x^2+x^3)^2,$

with every multiplication reduced modulo \(P(x)\) and modulo \(10^9\). If we call this intermediate polynomial \(V\), then

$D_3(0,1,1)=D_1(1)\circ V=V+V^2.$

Evaluating at \(t=1\) yields

$\bigl(D_3(0,1,1)(1)+0\bigr)\bmod 10^9=42.$

This example is small enough to inspect by hand, but it already shows the full mechanism: repeated ring reduction prevents the symbolic nesting from exploding.

How the Code Works

Fixed-Size Polynomial Arithmetic

The C++, Python, and Java implementations store one polynomial as exactly \(40\) coefficients. Before the main recurrence starts, they expand \(P(x)\), keep its lower-degree coefficients, negate them modulo \(10^9\), and reuse that data whenever an \(x^{40}\) term appears. As a result, multiplying by \(x\) is only a coefficient shift plus one overflow fold-back.

Using that primitive, the implementation builds full polynomial multiplication from shifted copies, computes ordinary powers by binary exponentiation, and computes composition by accumulating \(1,q,q^2,\dots,q^{39}\) in the quotient ring. Iterated self-composition is then obtained by the same binary idea, but with composition replacing multiplication.

Building the Nested Expression

The implementation first constructs \(x^\ell\) and then forms the base layer \(x^\ell+x^{\ell+1}\). Next it applies the middle operator \(U\mapsto U^{\circ m}(xU)\) to obtain the first nontrivial nested polynomial. After that, it repeats the same operator \(n\) more times to build the outer layer.

When the final degree-\(39\) representative is ready, the implementation evaluates it at the integer point \(t\) by accumulating the powers \(1,t,t^2,\dots,t^{39}\) modulo \(10^9\). The last operation is simply to add \(s\) and reduce once more.

Complexity Analysis

Let \(D=40\). One ring multiplication costs \(O(D^2)\), because it combines \(D\) shifted-and-reduced copies of a degree-\(39\) polynomial. One composition costs \(O(D^3)\), because it builds the powers \(1,q,q^2,\dots,q^{D-1}\) and combines them with the outer coefficients.

Computing \(x^\ell\) costs \(O(D^2\log \ell)\). Computing \(U^{\circ m}\) costs \(O(D^3\log m)\). Since the outer layer applies the same middle operator \(n+1\) times, the full running time is

$O\bigl(D^2\log \ell+(n+1)D^3\log m\bigr).$

With \(D=40\) fixed, this is effectively logarithmic in \(\ell\) and \(m\), and linear in the number of outer nestings. The polynomial storage is \(O(D)\), with only a small constant number of temporaries, plus the shallow recursion depth used for the outer layer.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=910
2. Quotient ring: Wikipedia - Quotient ring
3. Polynomial ring: Wikipedia - Polynomial ring
4. Function composition: Wikipedia - Function composition
5. Exponentiation by squaring: Wikipedia - Exponentiation by squaring
6. Polynomial evaluation: Wikipedia - Polynomial evaluation