Detailed mathematical approach

Problem Summary

Let \(p=2m\). We count ordered \(p\)-tuples of frog legs, where each leg starts at node \(1\), ends at node \(n\), and every jump has length \(1\), \(2\), or \(3\). Among the internal nodes \(2,3,\dots,n-1\), at most one node may be missed by every leg. The implementations return the count modulo \(10^9\).

Mathematical Approach

Step 1: Encode One Leg as a Binary Word

Look only at the internal nodes \(2,\dots,n-1\). For one leg, write a binary word of length \(n-2\): put \(1\) when that internal node is visited and \(0\) when it is skipped.

This converts the path problem into a word problem. Because jumps are only \(1\), \(2\), or \(3\), a leg can skip at most two consecutive internal nodes. Therefore valid legs are exactly the binary words with no substring \(000\).

So the problem is not really about geometric paths anymore. It is about \(p=2m\) synchronized binary words of length \(n-2\), each avoiding three consecutive zeros, and we want the union of their \(1\)-positions to miss at most one internal node.

Step 2: A Three-State Automaton for One Leg

Process the internal nodes from left to right. After some prefix has been processed, a single leg is determined by the number of consecutive processed internal nodes that were skipped since its last visit. That number can only be \(0\), \(1\), or \(2\).

Call these states \(S_0,S_1,S_2\), where \(S_r\) means “the current trailing run of skipped internal nodes has length \(r\)”. A visited node resets the state to \(S_0\). A skipped node sends \(S_0 \to S_1\) and \(S_1 \to S_2\). The transition \(S_2 \to S_3\) is impossible, because that would require a jump of length at least \(4\).

This is the key compression: every leg has many possible detailed histories, but for the next step only the current zero-run length matters.

Step 3: Compress All \(p=2m\) Legs into a Triple

Now process all \(p\) legs simultaneously. After a prefix of internal nodes, let

$F_t(a,b,c)$

be the number of ordered \(p\)-tuples of partial legs such that exactly \(a\) legs are in state \(S_2\), exactly \(b\) legs are in state \(S_1\), and exactly \(c\) legs are in state \(S_0\), with

$a+b+c=p.$

The initial state is obvious: before any internal node is processed, no leg has skipped anything yet, so

$F_0(0,0,p)=1,$

and all other triples are zero.

The number of possible triples is

$D=\#\{(a,b,c)\in \mathbb{Z}_{\ge 0}^3:a+b+c=p\}=\binom{p+2}{2}.$

Since \(p=2m\), this is quadratic in \(m\). For the actual target \(m=10\), we get \(p=20\) and \(D=231\).

Step 4: Transition for One More Internal Node

Suppose the old triple is \((A,B,C)\). At the next internal node:

If a leg is in \(S_2\), it must visit the node, otherwise the path would contain three consecutive skips.

If a leg is in \(S_1\), it may either visit the node and reset to \(S_0\), or skip it and move to \(S_2\).

If a leg is in \(S_0\), it may either visit the node and stay effectively in \(S_0\), or skip it and move to \(S_1\).

Therefore, if the new triple is \((i,j,k)\), then \(i\) must be the number of old \(S_1\)-legs that skip, and \(j\) must be the number of old \(S_0\)-legs that skip. The rest visit. In the direct counting basis, the recurrence is

$F_{t+1}(i,j,k)=\sum \binom{B}{i}\binom{C}{j}F_t(A,B,C),$

where the sum runs over old triples satisfying

$k=A+(B-i)+(C-j),\qquad A+B+C=p.$

This already solves the combinatorics, but the implementations use a slightly different basis because it turns the coefficients into the multinomial form used by the matrix construction.

Step 5: Factorial-Scaled Basis and Matrix Entries

Define scaled coordinates

$H_t(a,b,c)=a!\,b!\,c!\,F_t(a,b,c).$

Write the old triple as \((u,v+i,j+w)\), so that \(u+v+w=k\). After substituting the direct recurrence and cancelling factorials, we obtain

$H_{t+1}(i,j,k)=\sum_{u+v+w=k}\binom{k}{u}\binom{k-u}{v}\,H_t(u,v+i,j+w).$

This is exactly the transition rule used by the implementations. The coefficient

$\binom{k}{u}\binom{k-u}{v}=\frac{k!}{u!\,v!\,w!}$

is a multinomial count: among the \(k\) legs that do visit the current node, choose how many came from the old \(S_2\), old \(S_1\), and old \(S_0\) groups.

So the dense matrix is indexed only by triples \((a,b,c)\), not by individual legs. That symmetry reduction is what makes the problem tractable.

Step 6: Mark Internal Nodes Missed by Every Leg

Let \(g_q(n)\) denote the number of ordered \(p\)-tuples of full legs for which exactly \(q\) internal nodes are missed by every leg. The target of the problem is

$g_0(n)+g_1(n).$

Introduce the generating polynomial

$G_n(y)=\sum_{q\ge 0} g_q(n)(1+y)^q.$

At one internal node there are two possibilities: either at least one leg visits it, contributing a factor \(1\), or every leg skips it, contributing a factor \(1+y\).

Let \(T\) be the total transition matrix from Step 5, and let \(Z\) be the submatrix corresponding to the special case “every leg skips the current node”. That special case is deterministic:

$ (0,a,p-a)\longrightarrow (a,p-a,0). $

Hence the weighted one-node transfer is

$M(y)=T+yZ,$

because \(T\) already includes the all-skip case once, and the extra \(yZ\) changes its weight from \(1\) to \(1+y\).

Step 7: Initial Vector, Final Closure, and the Derivative Filter

Let \(e_0\) be the basis vector of the initial triple \((0,0,p)\). After the \(n-2\) internal nodes are processed, we get

$V_{n-2}(y)=M(y)^{n-2}e_0.$

There is still a final jump to node \(n\). A leg finishing in state \(S_0\), \(S_1\), or \(S_2\) must use a final jump of length \(1\), \(2\), or \(3\), respectively, so every terminal automaton state has exactly one legal completion. In the factorial-scaled basis, the corresponding closure row is obtained from the row indexed by \((0,0,p)\) in the total transition matrix. Denote this row by \(L\).

Therefore

$G_n(y)=L\,M(y)^{n-2}e_0.$

Now evaluate at \(y=-1\). Since \((1-1)^q\) vanishes for every \(q\ge 1\), we get

$G_n(-1)=g_0(n).$

Also,

$\left.\frac{d}{dy}(1+y)^q\right|_{y=-1}=\begin{cases}1,&q=1,\\0,&q\neq 1,\end{cases}$

so

$G_n'(-1)=g_1(n).$

Hence the required count is

$\boxed{g_0(n)+g_1(n)=G_n(-1)+G_n'(-1)\pmod{10^9}.}$

Worked Example: \(m=1,\ n=4\)

Here \(p=2\) and the internal nodes are \(2\) and \(3\). A single leg can be one of four paths:

\(1\to4\), \(1\to2\to4\), \(1\to3\to4\), or \(1\to2\to3\to4\).

So there are \(4^2=16\) ordered pairs of legs. The only invalid pair is the pair in which both legs jump directly \(1\to4\), because then both internal nodes are missed and we have \(q=2\).

Therefore

$g_0(4)=9,\qquad g_1(4)=6,\qquad g_2(4)=1,$

and the desired value is

$g_0(4)+g_1(4)=9+6=15.$

This matches the small checkpoint used by the implementations.

How the Code Works

The C++, Python, and Java implementations all follow the same mathematics. They enumerate all triples \((a,b,c)\) with \(a+b+c=2m\), precompute the needed binomial coefficients, build the dense total-transition matrix and the all-skip submatrix, and work entirely modulo \(10^9\).

The hard part is the huge exponent \(n-2\), so the implementation does not iterate over vertices one by one. Instead it evaluates \(M(-1)^{n-2}\) and its derivative at \(y=-1\) by binary matrix exponentiation. If \(U(y)\) and \(V(y)\) are matrix-valued factors, the derivative is updated by the product rule

$\frac{d}{dy}(UV)=U'V+UV'.$

After exponentiation, the implementation applies the terminal closure row, extracts \(G_n(-1)\) and \(G_n'(-1)\), adds them, and reduces the result modulo \(10^9\). The Python version delegates to the same compiled algorithm, so all three language implementations compute the identical transfer-matrix formula.

Complexity Analysis

The state-space dimension is

$D=\binom{2m+2}{2}=O(m^2).$

Dense matrix multiplication dominates the runtime. Binary exponentiation needs \(O(\log n)\) matrix products, so the overall complexity is

$O(D^3\log n)$

time and

$O(D^2)$

memory. The preliminary binomial table is only \(O(m^2)\) and is negligible compared with the matrix work.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=416
2. Generating function: Wikipedia — Generating function
3. Matrix exponentiation / binary exponentiation: cp-algorithms — Binary Exponentiation
4. Multinomial theorem: Wikipedia — Multinomial theorem
5. Transfer-matrix method: Wikipedia — Transfer-matrix method