Detailed mathematical approach

Problem Summary

Problem 907 works with cup sizes \(1,2,\dots,n\), but each size can appear in two oriented states, written \((s,0)\) and \((s,1)\). A legal stack is not an arbitrary sequence: from the current oriented cup you may move only along a small list of allowed transitions, and by the end you must have used every size exactly once. The path therefore has length \(n\), not \(2n\), because using size \(s\) in one orientation already forbids using the same size again in the opposite orientation.

If \(S(n)\) denotes the number of legal full-length paths, the implementations aim at \(S(10^7)\bmod (10^9+7)\). A brute-force search over all admissible paths is feasible only for very small \(n\), so the solution strategy has two layers: an exact state-space count for small cases and a fixed linear recurrence for large \(n\).

Mathematical Approach

The mathematical content of the implementations is completely driven by three problem-specific objects: the directed graph on oriented cups, the invariant that only sizes matter for the visited-set, and the order-\(8\) recurrence that takes over once the initial values have been certified.

Oriented cups form a local directed graph

For each size \(s\in\{1,\dots,n\}\) and orientation \(o\in\{0,1\}\), introduce a vertex \((s,o)\). The allowed transitions are

$ (s,0)\to(s-1,0),\qquad (s,0)\to(s-2,1),\qquad (s,0)\to(s+2,1), $

$ (s,1)\to(s+1,1),\qquad (s,1)\to(s-2,0),\qquad (s,1)\to(s+2,0), $

whenever the target size still lies between \(1\) and \(n\). So every state has outdegree at most \(3\). There are two kinds of moves:

same-orientation moves, which change the size by \(1\), and orientation-flipping moves, which change the size by \(2\).

This already explains why the problem is local. A move never jumps arbitrarily far; it only interacts with neighboring sizes. That locality is exactly what later collapses the large-\(n\) problem into a short linear recurrence.

The correct counting state is a set of sizes, not a set of oriented vertices

A natural first thought is to count paths on the \(2n\)-vertex directed graph. But that is not the real combinatorial state space, because the rule is "use each size once", not "visit each oriented vertex once". The implementations encode this by remembering only which sizes have already appeared.

Define

$ F(M,s,o) $

to be the number of legal completions from current state \((s,o)\) when the set of used sizes is exactly \(M\subseteq\{1,\dots,n\}\), with \(s\in M\). Then the terminal condition is

$ F(\{1,\dots,n\},s,o)=1, $

because a path that has already used all sizes is complete. Otherwise, every legal next move must go to an allowed neighbor \((t,p)\) whose size \(t\) is not yet in \(M\), so

$ F(M,s,o)=\sum_{\substack{(s,o)\to(t,p)\\ t\notin M}} F(M\cup\{t\},t,p). $

The total number of valid full stacks is therefore

$ S(n)=\sum_{s=1}^{n}\sum_{o=0}^{1} F(\{s\},s,o). $

This recurrence is not a generic template; it encodes the central invariant of this problem: opposite orientations of the same size compete for a single slot in the path.

Worked exact example: starting from \((1,1)\) when \(n=3\)

For \(n=3\), consider the start state \((1,1)\). The legal outgoing moves are to \((2,1)\) and \((3,0)\), so

$ F(\{1\},1,1)=F(\{1,2\},2,1)+F(\{1,3\},3,0). $

From \((2,1)\) with used sizes \(\{1,2\}\), the only unused legal continuation is \((3,1)\), giving

$ F(\{1,2\},2,1)=F(\{1,2,3\},3,1)=1. $

From \((3,0)\) with used sizes \(\{1,3\}\), the only unused legal continuation is \((2,0)\), so

$ F(\{1,3\},3,0)=F(\{1,2,3\},2,0)=1. $

Hence

$ F(\{1\},1,1)=2. $

The two complete stacks are

$ (1,1)\to(2,1)\to(3,1) $

and

$ (1,1)\to(3,0)\to(2,0). $

Summing over all possible starts gives \(S(3)=6\), which matches the certified initial table used later by the fast method.

Locality forces a fixed-order recurrence

The exact recurrence above is still exponential because \(M\) can be any subset. For large \(n\), the implementations exploit a second structural fact: every move changes the size by only \(1\) or \(2\). When one extends the problem from sizes \(\{1,\dots,n-1\}\) to \(\{1,\dots,n\}\), only a bounded fringe near the largest sizes can interact with the new cup. That means one can compress the effect of the unfinished top end into finitely many boundary patterns.

After eliminating those auxiliary pattern counts, the implementations use the scalar recurrence

$ S_n=2S_{n-1}-3S_{n-2}+5S_{n-3}-4S_{n-4}+4S_{n-5}-3S_{n-6}+S_{n-7}-S_{n-8}, \qquad n\ge 10. $

The exact derivation of the auxiliary boundary system is not encoded explicitly in the production code; what the implementations store is the reduced one-dimensional recurrence above. Its correctness is anchored by exact counts for small \(n\), not taken on faith.

Verified seeds and a concrete recurrence evaluation

The initial values used by the recurrence are

$ S(1)=2,\ S(2)=2,\ S(3)=6,\ S(4)=12,\ S(5)=16,\ S(6)=22,\ S(7)=36,\ S(8)=58,\ S(9)=82. $

Applying the recurrence once gives

$ \begin{aligned} S_{10}&=2\cdot 82-3\cdot 58+5\cdot 36-4\cdot 22+4\cdot 16-3\cdot 12+6-2\\ &=164-174+180-88+64-36+6-2\\ &=114. \end{aligned} $

The exact small-\(n\) search confirms \(S(10)=114\), so the rolling recurrence starts from values that have already been checked against the real combinatorial count.

How the Code Works

The C++, Python, and Java implementations all compute the same mathematical sequence, but the C++ version also contains a built-in exact verifier for small \(n\).

Exact verification on small instances

The C++ implementation first builds the directed transition graph on the \(2n\) oriented states. It then memoizes the function \(F(M,s,o)\) using three pieces of information: a bitmask for the used sizes, the current size, and the current orientation. This is the crucial implementation decision dictated by the mathematics: the mask tracks sizes only, because choosing one orientation of size \(s\) consumes size \(s\) completely.

That verifier checks the exact values for \(n=1,\dots,10\), so the base table \(2,2,6,12,16,22,36,58,82,114\) is not guessed. It is computed directly from the path-counting definition.

Rolling the order-\(8\) recurrence up to \(10^7\)

Once the initial values are known, all three implementations switch to the fast recurrence. They keep only the latest eight sequence values, form the next one with the fixed coefficients, reduce modulo \(10^9+7\), normalize after the subtractions, and slide the window forward by one position.

The C++ implementation also checks several checkpoints, including \(S(4)=12\), \(S(8)=58\), and \(S(20)=5560\). The Python and Java implementations keep only the production part, since the recurrence alone is enough once its seed values are trusted.

Complexity Analysis

The exact verifier has exponentially many subset states. In the worst case there are \(O(n2^n)\) memo states for the pair \((M,s)\), doubled by the orientation bit, and each state examines at most three outgoing edges. So the exact method is exponential and intended only for small certification cases.

The production computation is much smaller. Each new term \(S_n\) requires a constant number of modular additions and subtractions, so computing up to \(S(10^7)\) takes \(O(n)\) time and \(O(1)\) extra memory.

Footnotes and References

1. Problem page: Project Euler 907: Stacking Cups
2. Directed graph: Wikipedia - Directed graph
3. Hamiltonian path: Wikipedia - Hamiltonian path
4. Dynamic programming: Wikipedia - Dynamic programming
5. Recurrence relation: Wikipedia - Recurrence relation
6. Modular arithmetic: Wikipedia - Modular arithmetic