Problem 910: L-expressions II

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Project Euler Problem 910 Solution

EulerSolve provides an optimized solution for Project Euler Problem 910, L-expressions II, with C++, Python, Java, and a step-by-step mathematical explanation.

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 \(40\)....

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

Problem 910 source code

C++

#include <array>
#include <cstdint>
#include <iomanip>
#include <iostream>

using namespace std;

static constexpr int DEG = 40;
static constexpr uint32_t MOD = 1000000000u;

struct Poly {
    array<uint32_t, DEG> c{};

    uint32_t eval(uint32_t x) const {
        uint64_t res = 0;
        uint64_t p = 1;
        for (int i = 0; i < DEG; ++i) {
            res += (p * c[i]) % MOD;
            res %= MOD;
            p = (p * x) % MOD;
        }
        return static_cast<uint32_t>(res);
    }
};

static Poly exponent_poly;

static inline Poly add_poly(Poly a, const Poly& b) {
    for (int i = 0; i < DEG; ++i) {
        uint32_t v = a.c[i] + b.c[i];
        if (v >= MOD) {
            v -= MOD;
        }
        a.c[i] = v;
    }
    return a;
}

static inline Poly lambda_poly(Poly a, uint32_t k) {
    for (int i = 0; i < DEG; ++i) {
        a.c[i] = static_cast<uint32_t>((static_cast<uint64_t>(a.c[i]) * k) % MOD);
    }
    return a;
}

static inline Poly shift_poly(Poly p) {
    uint32_t last = p.c[DEG - 1];
    for (int i = DEG - 1; i > 0; --i) {
        p.c[i] = p.c[i - 1];
    }
    p.c[0] = 0;
    return add_poly(p, lambda_poly(exponent_poly, last));
}

static inline Poly mul_poly(const Poly& p, Poly q) {
    Poly res;
    for (int i = 0; i < DEG; ++i) {
        res = add_poly(res, lambda_poly(q, p.c[i]));
        q = shift_poly(q);
    }
    return res;
}

static Poly pow_poly(Poly p, uint32_t n) {
    Poly res;
    res.c[0] = 1;
    while (n > 0) {
        if (n & 1u) {
            res = mul_poly(res, p);
        }
        n >>= 1u;
        if (n > 0) {
            p = mul_poly(p, p);
        }
    }
    return res;
}

static inline Poly compose_poly(const Poly& p, const Poly& q) {
    Poly res;
    Poly prod;
    prod.c[0] = 1;
    for (int i = 0; i < DEG; ++i) {
        res = add_poly(res, lambda_poly(prod, p.c[i]));
        prod = mul_poly(prod, q);
    }
    return res;
}

static Poly iter_compose(Poly p, uint32_t n) {
    if (n == 0) {
        Poly id;
        id.c[0] = 1;
        return id;
    }

    Poly res;
    bool init = false;
    while (n > 0) {
        if (n & 1u) {
            if (!init) {
                res = p;
                init = true;
            } else {
                res = compose_poly(p, res);
            }
        }
        n >>= 1u;
        if (n > 0) {
            p = compose_poly(p, p);
        }
    }
    return res;
}

static inline Poly d1(uint32_t n) {
    Poly p;
    p.c[1] = 1;
    Poly q = pow_poly(p, n);
    return add_poly(q, shift_poly(q));
}

static inline Poly d2(uint32_t n, const Poly& u) {
    Poly w = iter_compose(u, n);
    return compose_poly(w, shift_poly(u));
}

static Poly d3(uint32_t n, uint32_t m, uint32_t l) {
    if (n == 0) {
        Poly u = d1(l);
        Poly v = d2(m, u);
        return compose_poly(u, v);
    }
    Poly u = d3(n - 1, m, l);
    return d2(m, u);
}

static void init_exponent_poly() {
    exponent_poly.c.fill(0);
    exponent_poly.c[0] = 1;

    for (uint32_t i = 1; i <= DEG; ++i) {
        for (int j = DEG - 1; j > 0; --j) {
            exponent_poly.c[j] = static_cast<uint32_t>((exponent_poly.c[j - 1] + static_cast<uint64_t>(i) * exponent_poly.c[j]) % MOD);
        }
        exponent_poly.c[0] = static_cast<uint32_t>((static_cast<uint64_t>(exponent_poly.c[0]) * i) % MOD);
    }

    for (int i = 0; i < DEG; ++i) {
        exponent_poly.c[i] = (MOD - exponent_poly.c[i]) % MOD;
    }
}

static uint32_t solve(uint32_t a, uint32_t b, uint32_t c, uint32_t d, uint32_t e) {
    Poly p = d3(a, b, c);
    return (p.eval(d) + e) % MOD;
}

static bool run_validation() {
    struct T {
        uint32_t a;
        uint32_t b;
        uint32_t c;
        uint32_t d;
        uint32_t expected;
    };

    const T tests[] = {
        {0, 1, 1, 1, 42},
        {0, 2, 1, 2, 599882556},
        {1, 1, 1, 2, 707063360},
        {1, 2, 2, 3, 600000000},
        {2, 1, 2, 2, 0},
    };

    for (const auto& t : tests) {
        uint32_t got = solve(t.a, t.b, t.c, t.d, 0);
        if (got != t.expected) {
            cerr << "Validation failed for a=" << t.a << " b=" << t.b << " c=" << t.c
                 << " d=" << t.d << " expected=" << t.expected << " got=" << got << '\n';
            return false;
        }
    }

    return true;
}

int main() {
    const uint32_t a = 12;
    const uint32_t b = 345678;
    const uint32_t c = 9012345;
    const uint32_t d = 678;
    const uint32_t e = 90;

    init_exponent_poly();

    if (!run_validation()) {
        return 1;
    }

    uint32_t ans = solve(a, b, c, d, e);
    cout << setw(9) << setfill('0') << ans << '\n';
    return 0;
}

Python

MOD = 1000000000
DEG = 40

class Poly:
    def __init__(self):
        self.c = [0] * DEG

    def eval(self, x):
        res = 0
        p = 1
        for i in range(DEG):
            res = (res + p * self.c[i]) % MOD
            p = (p * x) % MOD
        return res

def add_poly(a, b):
    res = Poly()
    for i in range(DEG):
        res.c[i] = (a.c[i] + b.c[i]) % MOD
    return res

def lambda_poly(a, k):
    res = Poly()
    for i in range(DEG):
        res.c[i] = (a.c[i] * k) % MOD
    return res

exponent_poly = Poly()

def shift_poly(p):
    res = Poly()
    last = p.c[DEG - 1]
    for i in range(DEG - 1, 0, -1):
        res.c[i] = p.c[i - 1]
    res.c[0] = 0
    return add_poly(res, lambda_poly(exponent_poly, last))

def mul_poly(p, q):
    res = Poly()
    for i in range(DEG):
        res = add_poly(res, lambda_poly(q, p.c[i]))
        q = shift_poly(q)
    return res

def pow_poly(p, n):
    res = Poly()
    res.c[0] = 1
    while n > 0:
        if n & 1:
            res = mul_poly(res, p)
        n >>= 1
        if n > 0:
            p = mul_poly(p, p)
    return res

def compose_poly(p, q):
    res = Poly()
    prod = Poly()
    prod.c[0] = 1
    for i in range(DEG):
        res = add_poly(res, lambda_poly(prod, p.c[i]))
        prod = mul_poly(prod, q)
    return res

def iter_compose(p, n):
    if n == 0:
        id_poly = Poly()
        id_poly.c[0] = 1
        return id_poly

    res = Poly()
    init = False
    while n > 0:
        if n & 1:
            if not init:
                res = p
                init = True
            else:
                res = compose_poly(p, res)
        n >>= 1
        if n > 0:
            p = compose_poly(p, p)
    return res

def d1(n):
    p = Poly()
    p.c[1] = 1
    q = pow_poly(p, n)
    return add_poly(q, shift_poly(q))

def d2(n, u):
    w = iter_compose(u, n)
    return compose_poly(w, shift_poly(u))

def d3(n, m, l):
    if n == 0:
        u = d1(l)
        v = d2(m, u)
        return compose_poly(u, v)
    u = d3(n - 1, m, l)
    return d2(m, u)

def init_exponent_poly():
    global exponent_poly
    exponent_poly = Poly()
    exponent_poly.c[0] = 1

    for i in range(1, DEG + 1):
        for j in range(DEG - 1, 0, -1):
            exponent_poly.c[j] = (exponent_poly.c[j - 1] + i * exponent_poly.c[j]) % MOD
        exponent_poly.c[0] = (exponent_poly.c[0] * i) % MOD

    for i in range(DEG):
        exponent_poly.c[i] = (MOD - exponent_poly.c[i]) % MOD

def solve():
    a = 12
    b = 345678
    c = 9012345
    d = 678
    e = 90

    init_exponent_poly()
    p_poly = d3(a, b, c)
    ans = (p_poly.eval(d) + e) % MOD
    return f"{ans:09d}"

if __name__ == "__main__":
    print(solve())

Java

public class Euler910 {
    static final int DEG = 40;
    static final long MOD = 1000000000L;

    static class Poly {
        long[] c = new long[DEG];

        long eval(long x) {
            long res = 0;
            long p = 1;
            for (int i = 0; i < DEG; ++i) {
                res = (res + p * c[i]) % MOD;
                p = (p * x) % MOD;
            }
            return res;
        }
    }

    static Poly exponentPoly = new Poly();

    static Poly addPoly(Poly a, Poly b) {
        Poly res = new Poly();
        for (int i = 0; i < DEG; ++i) {
            res.c[i] = (a.c[i] + b.c[i]) % MOD;
        }
        return res;
    }

    static Poly lambdaPoly(Poly a, long k) {
        Poly res = new Poly();
        for (int i = 0; i < DEG; ++i) {
            res.c[i] = (a.c[i] * k) % MOD;
        }
        return res;
    }

    static Poly shiftPoly(Poly p) {
        Poly res = new Poly();
        long last = p.c[DEG - 1];
        for (int i = DEG - 1; i > 0; --i) {
            res.c[i] = p.c[i - 1];
        }
        res.c[0] = 0;
        return addPoly(res, lambdaPoly(exponentPoly, last));
    }

    static Poly mulPoly(Poly p, Poly q) {
        Poly res = new Poly();
        Poly curQ = new Poly();
        System.arraycopy(q.c, 0, curQ.c, 0, DEG);
        for (int i = 0; i < DEG; ++i) {
            res = addPoly(res, lambdaPoly(curQ, p.c[i]));
            curQ = shiftPoly(curQ);
        }
        return res;
    }

    static Poly powPoly(Poly p, long n) {
        Poly res = new Poly();
        res.c[0] = 1;
        Poly curP = new Poly();
        System.arraycopy(p.c, 0, curP.c, 0, DEG);
        while (n > 0) {
            if ((n & 1) != 0) {
                res = mulPoly(res, curP);
            }
            n >>= 1;
            if (n > 0) {
                curP = mulPoly(curP, curP);
            }
        }
        return res;
    }

    static Poly composePoly(Poly p, Poly q) {
        Poly res = new Poly();
        Poly prod = new Poly();
        prod.c[0] = 1;
        for (int i = 0; i < DEG; ++i) {
            res = addPoly(res, lambdaPoly(prod, p.c[i]));
            prod = mulPoly(prod, q);
        }
        return res;
    }

    static Poly iterCompose(Poly p, long n) {
        if (n == 0) {
            Poly id = new Poly();
            id.c[0] = 1;
            return id;
        }

        Poly res = new Poly();
        boolean init = false;
        Poly curP = new Poly();
        System.arraycopy(p.c, 0, curP.c, 0, DEG);
        while (n > 0) {
            if ((n & 1) != 0) {
                if (!init) {
                    System.arraycopy(curP.c, 0, res.c, 0, DEG);
                    init = true;
                } else {
                    res = composePoly(curP, res);
                }
            }
            n >>= 1;
            if (n > 0) {
                curP = composePoly(curP, curP);
            }
        }
        return res;
    }

    static Poly d1(long n) {
        Poly p = new Poly();
        p.c[1] = 1;
        Poly q = powPoly(p, n);
        return addPoly(q, shiftPoly(q));
    }

    static Poly d2(long n, Poly u) {
        Poly w = iterCompose(u, n);
        return composePoly(w, shiftPoly(u));
    }

    static Poly d3(long n, long m, long l) {
        if (n == 0) {
            Poly u = d1(l);
            Poly v = d2(m, u);
            return composePoly(u, v);
        }
        Poly u = d3(n - 1, m, l);
        return d2(m, u);
    }

    static void initExponentPoly() {
        exponentPoly = new Poly();
        exponentPoly.c[0] = 1;

        for (long i = 1; i <= DEG; ++i) {
            for (int j = DEG - 1; j > 0; --j) {
                exponentPoly.c[j] = (exponentPoly.c[j - 1] + i * exponentPoly.c[j]) % MOD;
            }
            exponentPoly.c[0] = (exponentPoly.c[0] * i) % MOD;
        }

        for (int i = 0; i < DEG; ++i) {
            exponentPoly.c[i] = (MOD - exponentPoly.c[i]) % MOD;
        }
    }

    public static String solve() {
        long a = 12;
        long b = 345678;
        long c = 9012345;
        long d = 678;
        long e = 90;

        initExponentPoly();
        Poly pPoly = d3(a, b, c);
        long ans = (pPoly.eval(d) + e) % MOD;
        return String.format("%09d", ans);
    }

    public static void main(String[] args) {
        System.out.println(solve());
    }
}