Detailed mathematical approach

Problem Summary

The quantity \(C(n,s)\) counts losing positions in the card-stacking game with \(s\) independent suits and \(n\) cards per suit. The real target is \(C(10^7,10^7)\) modulo \(M=10^9+7\), so brute force is impossible. The implementation therefore does not enumerate positions directly: it first computes the Grundy-value distribution for a single suit, then combines \(s\) identical suit distributions by xor convolution.

Mathematical Approach

Let \(a_g\) be the number of one-suit positions whose Grundy value is \(g\). The implementation shows that only \(0\le g\le n-1\) occur, and that these frequencies partition all one-suit states:

$\sum_{g=0}^{n-1} a_g = 2^n.$

Once this one-suit distribution is known, the multi-suit count follows from standard impartial-game xor rules.

Step 1: Reduce the Full Game to One-Suit Grundy Values

If the \(j\)-th suit contributes Grundy value \(x_j\), then the total position has value

$x_1\oplus x_2\oplus \cdots \oplus x_s.$

A position is losing exactly when this xor is \(0\). Therefore

$C(n,s)=\sum_{x_1\oplus\cdots\oplus x_s=0}\prod_{j=1}^{s} a_{x_j}.$

This formula says that the whole problem is controlled by the frequency table \(a_g\) for one suit.

Step 2: Rewrite the Linear-Time Recurrence in Closed Form

The one-suit distribution is generated by a recurrence indexed by \(m=\lfloor g/2\rfloor\). After simplifying the ratio updates used by the implementation, the auxiliary quantities become ordinary binomial coefficients:

$\beta_m=\binom{n-m-2}{m},\qquad \alpha_m=\begin{cases} 0,&m=0,\\ \binom{n-m-2}{m-1},&m\ge 1. \end{cases}$

Applying Pascal's identity gives two more useful expressions:

$\pi_m=\alpha_m+\beta_m=\binom{n-m-1}{m},$

$\chi_m=\pi_m\frac{n-2m-1}{m+1}=\binom{n-m-1}{m+1}.$

The power-of-two baseline that appears in the code starts at \(2^{n-2}\) and is halved at every step, so at stage \(m\) it is exactly

$2^{n-2-m}.$

Step 3: Recover the One-Suit Distribution

The remaining correction term is another linear recurrence:

$\tau_0=1,\qquad \tau_{m+1}=\frac{\tau_m+\alpha_{m+1}}{2}+\beta_{m+1}.$

With this notation the one-suit frequencies are

$a_0=2^{n-2}+2,$

$a_{2m+1}=2^{n-2-m}+\chi_m-\tau_m \qquad \left(0\le m\le \left\lfloor\frac{n-2}{2}\right\rfloor\right),$

$a_{2m}=2^{n-2-m}+\pi_m+\beta_m-\tau_m \qquad \left(1\le m\le \left\lfloor\frac{n-1}{2}\right\rfloor\right).$

This produces every nonzero \(a_g\) in \(O(n)\) time, without traversing any of the \(2^n\) one-suit states individually.

Step 4: Use Structural Identities as Sanity Checks

The implementation verifies two identities that are mathematically informative and computationally useful:

$\sum_{g=0}^{n-1} a_g = 2^n,$

$a_{n-1}=1.$

The first identity confirms that the recurrence accounts for every one-suit position exactly once. The second shows that the largest attainable Grundy value appears exactly once. Both checks are strong consistency tests before the xor-convolution stage starts.

Step 5: Combine the \(s\) Suits with XOR Convolution

For arrays \(f\) and \(g\), define xor convolution by

$ (f *_\oplus g)(t)=\sum_{x\oplus y=t} f(x)g(y). $

Then \(C(n,s)\) is the coefficient at \(t=0\) of the \(s\)-fold xor convolution of the one-suit distribution with itself. Let

$P=2^{\lceil \log_2 n\rceil}.$

Pad \((a_0,\dots,a_{n-1})\) with zeros to length \(P\), apply the Walsh-Hadamard transform, and write the transformed vector as \(\widehat{a}\). The xor-convolution theorem yields

$C(n,s)=\frac{1}{P}\sum_{i=0}^{P-1}\widehat{a}_i^s \pmod{M},\qquad M=10^9+7.$

So the \(s\)-suit aggregation becomes a pointwise power in transform space.

Worked Example: \(n=3,\ s=2\)

For one suit, the base value is

$a_0=2^{1}+2=4.$

At \(m=0\), we have \(\beta_0=1\), \(\alpha_0=0\), \(\pi_0=1\), \(\chi_0=2\), and \(\tau_0=1\). Therefore

$a_1=2^1+\chi_0-\tau_0=2+2-1=3.$

Next, \(\alpha_1=1\) and \(\beta_1=0\), so

$\tau_1=\frac{1+1}{2}+0=1,$

and hence

$a_2=2^0+\pi_1+\beta_1-\tau_1=1+1+0-1=1.$

Thus the one-suit distribution is

$a=(4,3,1).$

Pad it to \(P=4\):

$a=(4,3,1,0).$

Its Walsh-Hadamard transform is

$\widehat{a}=(8,2,6,0).$

For two suits, the losing count is

$C(3,2)=\frac{8^2+2^2+6^2+0^2}{4}=26,$

which matches the checkpoint used by the implementation.

How the Code Works

The C++, Python, and Java implementations all follow the same mathematics. They first precompute modular inverses up to \(n\), so that every rational-looking factor in the recurrence can be evaluated modulo \(10^9+7\). They then run one forward pass over \(m\), updating the auxiliary sequences and filling the one-suit frequency table for all Grundy values below \(n\).

After that, the distribution is padded to the next power of two and transformed in place with the Walsh-Hadamard transform. In the transformed domain, each spectral entry is raised to the \(s\)-th power, the results are summed, and the final multiplication by \(P^{-1}\) extracts the zero-xor coefficient. The C++ and Java implementations also parallelize the large transform and exponentiation loops on big inputs, while the Python implementation delegates to the same compiled arithmetic core, so all three language versions evaluate the same recurrence and the same transform formula.

Complexity Analysis

Building the modular inverses and the one-suit distribution costs \(O(n)\) time and \(O(n)\) memory before padding. Let \(P=2^{\lceil\log_2 n\rceil}\), so \(n\le P<2n\). The Walsh-Hadamard transform costs \(O(P\log P)\), and exponentiating all transformed entries to the \(s\)-th power costs \(O(P\log s)\). Therefore the total running time is \(O(n + P\log P + P\log s)\), which is effectively \(O(n\log n + n\log s)\), and the working memory is \(O(P)\).

Footnotes and References

1. Problem page: https://projecteuler.net/problem=798
2. Sprague-Grundy theorem: Wikipedia - Sprague-Grundy theorem
3. Nim and xor sums: Wikipedia - Nim
4. Hadamard transform: Wikipedia - Hadamard transform
5. Walsh-Hadamard transform for xor convolution: cp-algorithms - Walsh-Hadamard transform