Detailed mathematical approach

Problem Summary

We are given a 48-bit linear congruential generator with state update

$a_{n+1}=A a_n + C \pmod{2^{48}},\qquad A=25214903917,\quad C=11.$

Each state emits one letter from a 52-symbol alphabet by taking the upper 32 bits, reducing them modulo \(52\), and mapping \(0,\dots,25\) to a through z and \(26,\dots,51\) to A through Z. The problem asks for the exact step distance between the two target words that appear in this sequence. Since the generator has period \(2^{48}\), direct simulation is not realistic; the solution instead reconstructs the unique starting state of each target word and then computes the distance algebraically.

Mathematical Approach

Let the emitted code at time \(n\) be

$u_n=\left\lfloor\frac{a_n}{2^{16}}\right\rfloor \bmod 52.$

For a target word with code sequence \(t_0,t_1,\dots,t_{m-1}\), we must find all starting states \(a_0\) such that

$u_i=t_i\qquad (0 \le i \lt m).$

Step 1: Split the State into High and Low Words

Write every state as

$a_n=2^{16}q_n+\ell_n,\qquad 0\le \ell_n \lt 2^{16},\qquad 0\le q_n \lt 2^{32}.$

Substituting this into the recurrence separates the dynamics into a low 16-bit part and a high 32-bit part. The low word evolves by

$\ell_{n+1}\equiv A\ell_n+C \pmod{2^{16}},$

and the carry into the high word is

$\kappa_n=\left\lfloor\frac{A\ell_n+C}{2^{16}}\right\rfloor.$

The high word therefore satisfies

$q_{n+1}\equiv A q_n+\kappa_n \pmod{2^{32}}.$

The output condition depends only on \(q_n\):

$q_n\equiv t_n \pmod{52}.$

This decomposition is the key observation: the entire carry sequence is determined by the starting low word \(\ell_0\), while the starting high word \(q_0\) only has to satisfy a linear recurrence modulo \(2^{32}\).

Step 2: Use Adjacent Characters to Filter the Low Word

The implementation first exploits a very cheap necessary condition. Since \(A\equiv 1 \pmod 4\), reducing the high-word recurrence modulo \(4\) gives

$q_{n+1}-q_n\equiv \kappa_n \pmod 4.$

Because \(q_n\equiv t_n \pmod{52}\), this implies

$\kappa_n\equiv t_{n+1}-t_n \pmod 4.$

So each adjacent pair of target characters prescribes the carry residue modulo \(4\). The low word \(\ell_0\) can take only \(2^{16}\) possible values, and for each one the low recurrence produces a deterministic carry sequence. Any starting low word whose carry residues fail even one of these congruences is impossible and can be discarded immediately.

This turns the first stage into a complete scan of only \(65536\) candidates, which is tiny compared with the full \(2^{48}\)-state space.

Step 3: Recover the High Word from an Arithmetic Progression

Once a starting low word survives the modulo-\(4\) filter, the exact carries \(\kappa_0,\kappa_1,\dots,\kappa_{m-2}\) are known. The starting high word must satisfy

$q_0\equiv t_0 \pmod{52},$

so we can write

$q_0=t_0+52r,\qquad 0\le q_0 \lt 2^{32}.$

For each such candidate, we iterate

$q_{i+1}\equiv A q_i+\kappa_i \pmod{2^{32}}$

and check whether every step still satisfies \(q_i\equiv t_i \pmod{52}\). Every successful \(q_0\) yields a full starting state

$a_0=2^{16}q_0+\ell_0.$

For the two actual target words of the problem, this reconstruction returns exactly one valid starting state for each word.

Step 4: Jump Ahead by Many Steps with an Affine Power

The one-step transition is affine: \(x\mapsto A x+C\). Composing affine maps keeps the same form, so for every nonnegative integer \(s\) there exist coefficients \(\alpha_s\) and \(\beta_s\) such that

$a_{n+s}=\alpha_s a_n+\beta_s \pmod{2^{48}}.$

If two affine maps are written as \(x\mapsto \alpha_1 x+\beta_1\) and \(x\mapsto \alpha_2 x+\beta_2\), then their composition is

$x\mapsto \alpha_1\alpha_2 x+(\alpha_1\beta_2+\beta_1).$

This means binary exponentiation works exactly as it does for ordinary modular powers: repeated squaring gives \((\alpha_s,\beta_s)\) in \(O(\log s)\) time. The implementation uses this both for verification and for checking the few candidate distances that remain after the discrete-log step.

Step 5: Convert the Affine Recurrence into Multiplication

Define a transformed quantity

$b_n=(A-1)a_n+C \pmod{2^{48}}.$

Then

$\begin{aligned} b_{n+1} &=(A-1)(A a_n+C)+C\\ &=A\bigl((A-1)a_n+C\bigr)\\ &\equiv A b_n \pmod{2^{48}}. \end{aligned}$

So the affine recurrence becomes a purely multiplicative recurrence. This is the central algebraic simplification.

Moreover, every \(b_n\) is odd. Indeed, \(A-1\) is divisible by \(4\), while \(C=11\) is odd, so \((A-1)a_n+C\equiv 3 \pmod 4\). Therefore \(b_n\) is invertible modulo \(2^{48}\).

If \(s\) and \(t\) are the two reconstructed starting states and \(t\) occurs \(\Delta\) steps after \(s\), then

$b_t\equiv A^{\Delta} b_s \pmod{2^{48}},$

hence

$A^{\Delta}\equiv b_t\,b_s^{-1}\pmod{2^{48}}.$

The ratio on the right is congruent to \(1 \pmod 4\), so it lies in the relevant cyclic \(2\)-power subgroup generated by \(A\).

Step 6: Recover the Distance Modulo \(2^{46}\) and Lift It

Modulo \(2^{48}\), the multiplier \(A\) has order \(2^{46}\) inside the subgroup used above. Therefore the exponent \(\Delta\) is first determined only modulo \(2^{46}\).

The implementation extracts this exponent bit by bit. Suppose the current ratio is \(A^e\). Raising it to \(2^{45-i}\) tells whether bit \(i\) of \(e\) is zero or one: if the result is not \(1\), then bit \(i\) must be \(1\), and the corresponding factor \(A^{2^i}\) is divided out before moving to the next bit. This recovers the base solution

$\Delta\equiv \Delta_0 \pmod{2^{46}}.$

However, the map \(a\mapsto (A-1)a+C\) is not injective modulo \(2^{48}\): since \(\gcd(A-1,2^{48})=4\), the transformed sequence identifies four LCG states at a time. Therefore the true distance must be one of

$\Delta_0,\qquad \Delta_0+2^{46},\qquad \Delta_0+2\cdot 2^{46},\qquad \Delta_0+3\cdot 2^{46}.$

The affine jump formula from Step 4 checks these four lifts directly and selects the exact step count.

Worked Example: Reconstructing a Short Prefix

Take a target prefix of length \(2\) with codes \(t_0\) and \(t_1\). A starting low word \(\ell_0\) produces a carry

$\kappa_0=\left\lfloor\frac{A\ell_0+C}{2^{16}}\right\rfloor.$

A necessary condition for this low word to work is

$\kappa_0\equiv t_1-t_0 \pmod 4.$

Most of the \(2^{16}\) low-word candidates fail this immediately. For each survivor, the high word must lie on the arithmetic progression

$q_0=t_0+52r.$

Testing these candidates with the exact 32-bit recurrence either rejects them or produces a genuine starting state. Longer prefixes do the same thing repeatedly, one carry constraint and one high-word update at a time.

How the Code Works

The C++, Python, and Java implementations follow the same mathematical pipeline. First they encode letters into numbers \(0\) through \(51\). Then they enumerate all \(2^{16}\) possible low 16-bit starts, simulate only the low-word recurrence, and compare the carry residues modulo \(4\) against the differences forced by the target word.

For every surviving low-word candidate, the implementation recomputes the exact carries and tests the arithmetic progression \(t_0+52r\) of possible high words. Each successful pair of high and low words is assembled into a 48-bit starting state, and duplicate states are removed after sorting.

After both target words have been converted into their unique starting states, the implementation transforms them into odd residues modulo \(2^{48}\), computes a modular inverse, and solves the exponent in the order-\(2^{46}\) subgroup one bit at a time. It then checks the four possible lifts by fast affine jumping, which is much cheaper than iterating the generator step by step.

The language-specific arithmetic details differ slightly. C++ uses a 128-bit intermediate product and masks the result back to 48 bits. Python uses arbitrary-precision integers with the same masking rule. Java avoids overflow in signed 64-bit arithmetic by splitting each 48-bit factor into two 24-bit halves and recombining the partial products before applying the mask.

Complexity Analysis

Let \(m\) be the target-word length, and let \(R\) be the number of starting low words that survive the carry-residue filter. The low-word scan costs \(O(2^{16}m)\). Recomputing the exact carry sequences costs \(O(Rm)\). The high-word search is \(O\!\left(R\left\lceil\frac{2^{32}}{52}\right\rceil m\right)\) in the pessimistic worst case, because each surviving low word is paired with an arithmetic progression of possible 32-bit high words.

In practice, the carry constraints are extremely restrictive, so \(R\) is tiny for the fixed words in this problem, and the reconstruction stage is dominated by the initial \(2^{16}\) scan. The distance phase needs only \(O(\log 2^{48})\) modular multiplications, plus at most four affine-jump verifications. Memory usage is \(O(R+m)\), which is effectively constant for the given input sizes.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=803
2. Linear congruential generator: Wikipedia — Linear congruential generator
3. Modular multiplicative inverse: Wikipedia — Modular multiplicative inverse
4. Discrete logarithm: Wikipedia — Discrete logarithm
5. Multiplicative group of integers modulo \(n\): Wikipedia — Multiplicative group of integers modulo n