Detailed mathematical approach

Problem Summary

In the standard optimal Tower of Hanoi transfer with \(n\) disks, after each move we record the peg populations \((a,b,c)\). The position is losing in normal Nim exactly when

$a\oplus b\oplus c=0.$

The required quantity is the sum of all move numbers \(t\in\{1,\dots,2^n-1\}\) for which the Hanoi position is Nim-losing. The C++, Python, and Java implementations compute this value modulo \(10^9+7\) without simulating the full exponential move sequence.

Mathematical Approach

The key reduction is to count how many losing times correspond to each admissible ordered population triple, and only at the very end convert that count into a sum of move indices.

Step 1: Turn the index sum into a counting problem

The optimal Hanoi path has length \(2^n-1\). Reversing time sends move \(t\) to move \(2^n-1-t\), and this reversal only permutes peg roles, so the Nim condition is preserved. Therefore losing times come in complementary pairs whose indices add to \(2^n-1\).

If \(\Lambda(n)\) denotes the total number of losing times, then

$S(n)=\frac{2^n-1}{2}\,\Lambda(n)\pmod{10^9+7}.$

So the real task is to compute \(\Lambda(n)\), the number of Nim-losing moments along the Hanoi path.

Step 2: Describe all admissible losing triples

Any Hanoi state satisfies

$a+b+c=n.$

For a losing Nim state we also need

$a\oplus b\oplus c=0.$

Bit by bit, xor zero means that at every binary position there are either zero 1s or exactly two 1s among \(a,b,c\). So the contribution of bit \(2^r\) to the sum \(a+b+c\) is either \(0\) or \(2^{r+1}\). This immediately shows that \(n\) must be even; if \(n\) is odd, the answer is \(0\).

Write \(n=2m\), and let the binary expansion of \(m\) be

$m=\sum_{r\in R}2^r.$

For each \(r\in R\), exactly one peg omits the bit \(2^r\), so the local contribution is one of

$\left(0,2^r,2^r\right),\qquad \left(2^r,0,2^r\right),\qquad \left(2^r,2^r,0\right).$

Adding these choices independently over all set bits of \(m\) generates every ordered triple \((a,b,c)\) with \(a+b+c=n\) and \(a\oplus b\oplus c=0\), and it generates each such triple exactly once. Hence the number of admissible losing triples is

$3^{s_2(m)}=3^{s_2(n/2)},$

where \(s_2(\cdot)\) is the number of 1s in the binary expansion.

Step 3: Introduce the auxiliary generating function

The implementation evaluates an auxiliary coefficient extracted from

$\Psi(X,Y,Z)=\frac{1}{1-X^2-Y^2-Z^2-2XYZ}.$

Define

$\mathcal{A}(\alpha,\beta,\gamma)=\left[X^\alpha Y^\beta Z^\gamma\right]\Psi(X,Y,Z).$

Using the geometric-series expansion,

$\Psi(X,Y,Z)=\sum_{i\ge 0}\left(X^2+Y^2+Z^2+2XYZ\right)^i.$

So in the \(i\)-th layer we choose some copies of \(X^2\), some of \(Y^2\), some of \(Z^2\), and some of \(2XYZ\). That multinomial expansion is exactly the algebra encoded by the implementations.

Step 4: Closed form for the auxiliary coefficient

Fix nonnegative \(\alpha,\beta,\gamma\) and set

$m=\alpha+\beta+\gamma.$

In one term of degree \(i\), suppose \(k\) factors contribute \(2XYZ\). Then

$k=m-2i,$

and the remaining exponents must come from squares, so

$r_X=\frac{\alpha-k}{2},\qquad r_Y=\frac{\beta-k}{2},\qquad r_Z=\frac{\gamma-k}{2}.$

These must be nonnegative integers. Equivalently, with

$A=\frac{\beta+\gamma}{2},\qquad B=\frac{\alpha+\gamma}{2},\qquad C=\frac{\alpha+\beta}{2},$

we have

$r_X=i-A,\qquad r_Y=i-B,\qquad r_Z=i-C.$

Therefore the \(i\)-th contribution is

$Q_i(\alpha,\beta,\gamma)=\frac{i!\,2^{m-2i}}{(m-2i)!\,(i-A)!\,(i-B)!\,(i-C)!},$

and the full auxiliary term is

$\mathcal{A}(\alpha,\beta,\gamma)=\sum_{i=\ell}^{\lfloor m/2\rfloor} Q_i(\alpha,\beta,\gamma),$

where

$\ell=\max\left(\left\lceil\frac{m}{3}\right\rceil,A,B,C\right).$

If the parity conditions fail, meaning \(\alpha,\beta,\gamma\) do not all have the same parity as \(m\), then \(\mathcal{A}(\alpha,\beta,\gamma)=0\).

Step 5: Recover the exact multiplicity of an ordered triple

The actual number of losing occurrences of a specific ordered Hanoi population triple is obtained by multiplying the generating function by a correction polynomial. Define

$M(a,b,c)=\left[X^aY^bZ^c\right]\left(1+X+Z+XY+YZ-Y^2\right)\Psi(X,Y,Z).$

Extracting coefficients gives the six-term identity

$\begin{aligned} M(a,b,c)=&\ \mathcal{A}(a,b,c)+\mathcal{A}(a-1,b,c)+\mathcal{A}(a,b,c-1)\\ &+\mathcal{A}(a-1,b-1,c)+\mathcal{A}(a,b-1,c-1)-\mathcal{A}(a,b-2,c). \end{aligned}$

This expression is intentionally not symmetric in \(a,b,c\). The standard Hanoi transfer treats the three pegs as source, auxiliary, and target in a fixed order, so the middle peg receives a different correction term.

Step 6: Sum over all losing triples

Let \(\mathcal{T}_n\) be the set of ordered triples satisfying

$a+b+c=n,\qquad a\oplus b\oplus c=0.$

Then the total number of losing times is

$\Lambda(n)=\sum_{(a,b,c)\in\mathcal{T}_n} M(a,b,c).$

Combining this with Step 1 yields the implemented formula

$\boxed{S(n)=\frac{2^n-1}{2}\sum_{\substack{a+b+c=n\\a\oplus b\oplus c=0}} M(a,b,c)\pmod{10^9+7}.}$

Worked Example: \(n=4\)

Here \(n/2=2\) has one set bit, so there are exactly three admissible losing triples:

$\left(0,2,2\right),\qquad \left(2,0,2\right),\qquad \left(2,2,0\right).$

For \((2,2,0)\), only one summation level contributes to the auxiliary coefficient, giving

$\mathcal{A}(2,2,0)=\frac{2!}{0!\,1!\,1!\,0!}=2.$

Also \(\mathcal{A}(2,0,0)=1\), and the other shifted terms in the six-term identity vanish, so

$M(2,2,0)=2-1=1.$

Similarly,

$M(0,2,2)=1,\qquad M(2,0,2)=2.$

Therefore

$\Lambda(4)=1+1+2=4.$

The required sum of move indices is then

$S(4)=\frac{2^4-1}{2}\cdot 4=\frac{15}{2}\cdot 4=30,$

which matches the checkpoint used by the implementations.

How the Code Works

The C++, Python, and Java implementations all follow the same mathematical pipeline. First they reject odd \(n\), because Step 2 proves that no losing triple can exist in that case. Next they precompute factorials, inverse factorials, powers of two, and modular inverses up to about \(n/2\), so each multinomial term can be evaluated quickly modulo \(10^9+7\).

Then the implementation enumerates every admissible triple by scanning the set bits of \(n/2\) and, for each such bit, choosing which peg does not receive that contribution. For one ordered triple, the auxiliary coefficient is not recomputed from scratch for every \(i\); instead the code starts at the smallest feasible \(i\) and advances with the exact ratio

$\frac{Q_{i+1}}{Q_i}=\frac{(i+1)(m-2i)(m-2i-1)}{4(i+1-A)(i+1-B)(i+1-C)}.$

That turns the inner sum into a linear sweep with \(O(1)\) modular work per step. After the auxiliary values are known, the implementation combines six shifted coefficients to obtain \(M(a,b,c)\), sums these values over all admissible triples to get \(\Lambda(n)\), and finally multiplies by \((2^n-1)/2\). The C++ version parallelizes the outer triple loop; the Python and Java versions keep the same mathematics in serial form.

Complexity Analysis

If \(n\) is odd, the answer is returned immediately. For even \(n\), the number of admissible ordered triples is exactly

$T=3^{s_2(n/2)}.$

For each triple, the auxiliary summation runs over at most \(\lfloor n/2\rfloor+1\) indices \(i\), and every transition uses only \(O(1)\) modular operations because the tables are precomputed. Therefore the worst-case running time is

$O\!\left(n\cdot 3^{s_2(n/2)}\right),$

with memory usage

$O(n).$

Since \(s_2(n/2)\le \lfloor\log_2 n\rfloor\), this is dramatically smaller than brute-force Hanoi simulation, which would require \(\Theta(2^n)\) moves and states.

Footnotes and References

1. Project Euler problem page: https://projecteuler.net/problem=806
2. Tower of Hanoi: Wikipedia — Tower of Hanoi
3. Nim: Wikipedia — Nim
4. Gray code: Wikipedia — Gray code
5. Multinomial theorem: Wikipedia — Multinomial theorem
6. Generating function: Wikipedia — Generating function