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Math-1042 · mathematics · Graph theory

Pebbling a cartesian product

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Statement

Definitions and mathematical statement

Let $G$ be a finite nonempty connected simple undirected graph. A pebble configuration is a function $C:V(G)\to\mathbb Z_{\ge0}$, with total size $\sum_v C(v)$. A legal pebbling move chooses adjacent vertices $u,v$ with $C(u)\ge2$, removes two pebbles from $u$, and adds one pebble at $v$. One pebble is therefore lost at each move.

For a chosen target vertex $t$, a configuration is solvable at $t$ if some finite sequence of legal moves leaves at least one pebble there. The pebbling number $p(G)$ is the least integer $N$ such that every configuration of total size $N$ is solvable at every target vertex. The sequence of moves may depend on both the configuration and the target; simultaneous delivery to all targets is not required.

For two such graphs, their Cartesian product $G_1\mathbin\Box G_2$ has vertex set $V(G_1)\times V(G_2)$. Two distinct pairs are adjacent if one coordinate is equal and the other coordinate is an adjacent pair in its factor graph. Graham's conjecture asserts \[ p(G_1\mathbin\Box G_2)\le p(G_1)p(G_2) \] for every pair of finite connected simple graphs.

Thus the product bound must handle arbitrary, possibly highly uneven, placements on the product graph. A construction that succeeds only for placements distributed uniformly among factor fibres is insufficient. Connectedness ensures that pebbling numbers are finite; the single-vertex graph is permitted and has pebbling number one.

Short English statement

Is the pebbling number of a Cartesian product at most the product of the pebbling numbers of its factors?

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Math-1042-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-1042

Pebbling a cartesian product

Definitions and mathematical statement

Let $G$ be a finite nonempty connected simple undirected graph. A pebble configuration is a function $C:V(G)\to\mathbb Z_{\ge0}$, with total size $\sum_v C(v)$. A legal pebbling move chooses adjacent vertices $u,v$ with $C(u)\ge2$, removes two pebbles from $u$, and adds one pebble at $v$. One pebble is therefore lost at each move.

For a chosen target vertex $t$, a configuration is solvable at $t$ if some finite sequence of legal moves leaves at least one pebble there. The pebbling number $p(G)$ is the least integer $N$ such that every configuration of total size $N$ is solvable at every target vertex. The sequence of moves may depend on both the configuration and the target; simultaneous delivery to all targets is not required.

For two such graphs, their Cartesian product $G_1\mathbin\Box G_2$ has vertex set $V(G_1)\times V(G_2)$. Two distinct pairs are adjacent if one coordinate is equal and the other coordinate is an adjacent pair in its factor graph. Graham's conjecture asserts \[ p(G_1\mathbin\Box G_2)\le p(G_1)p(G_2) \] for every pair of finite connected simple graphs.

Thus the product bound must handle arbitrary, possibly highly uneven, placements on the product graph. A construction that succeeds only for placements distributed uniformly among factor fibres is insufficient. Connectedness ensures that pebbling numbers are finite; the single-vertex graph is permitted and has pebbling number one.

Short English statement

Is the pebbling number of a Cartesian product at most the product of the pebbling numbers of its factors?

Research attempt

Literature check and attempted route.
Chung's original pebbling work [A1] and the original problem page [S2] were checked on 1 October 2026. Products of paths are known positive cases; the present proof does not claim those results as new. We test the most direct fibre-by-fibre strategy for general factors, derive a rigorous bound with an explicit loss, and show on four vertices why rounding away the residues cannot prove Graham's sharp product bound.
Collecting disjoint packets in a fibre.
Write $a=p(G)$, $b=p(H)$ and $h=|V(H)|$, and target $(g_0,h_0)$. For each $y\in V(H)$ let $c_y$ be the number of pebbles in the $G$-fibre $V(G)\times\{y\}$. Partition any $a\lfloor c_y/a\rfloor$ of those pebbles into packets of size $a$. By definition of $p(G)$, each packet can supply a pebble to $(g_0,y)$ using moves inside the fibre. Treat packets as coloured only for bookkeeping: process one packet at a time, never consuming the pebbles reserved from another. Every move legal for the selected packet is legal for the actual uncoloured configuration. Thus at least $\lfloor c_y/a\rfloor$ pebbles can be collected at $(g_0,y)$. If their sum over $y$ is at least $b$, the definition of $p(H)$ finishes the job along the fibre $\{g_0\}\times V(H)$.
A general upper bound with a controlled error.
Write $c_y=aq_y+r_y$, $0\le r_y\le a-1$. If $\sum_yq_y\le b-1$, then \[ N=\sum_yc_y\le a(b-1)+h(a-1). \] Consequently every configuration of size \[ a(b-1)+h(a-1)+1=ab+(a-1)(h-1) \] is solvable. Interchanging the two factors gives the proved estimate \[ p(G\mathbin\Box H)\le p(G)p(H)+ \min\{(p(G)-1)(|V(H)|-1),(p(H)-1)(|V(G)|-1)\}. \] The single-vertex case makes the error zero. In general the error is genuine in this argument, even though it is not evidence that the conjecture is false. It represents packets that cannot be completed using only their initial fibre.
A four-vertex stress test of the sharp inference.
For $G=H=K_2$, each factor has pebbling number two: a target already occupied needs no move, and otherwise both pebbles at its neighbour give one move. Use coordinates $(i,j)\in\{0,1\}^2$, target $(0,0)$, and initial counts \[ C(0,0)=0,\ C(1,0)=1,\ C(0,1)=1,\ C(1,1)=2. \] There are four pebbles, exactly the conjectural product. The two fibre totals are one and three, and their rounded packet counts sum to only $\lfloor1/2\rfloor+\lfloor3/2\rfloor=1<2$. Nevertheless the configuration is solvable: move the two pebbles from $(1,1)$ to $(0,1)$, which now has two, then move them to $(0,0)$. The successful route uses a residue already sitting on the collecting fibre. Discarding residues loses precisely that interaction. The example does not refute fibre methods; it refutes the claim that the packet lower bound alone always supplies $b$ pebbles at size $ab$.
Verification and remaining gap.
The packet proof checks legality, order of processing, and arbitrary uneven initial distributions. The algebraic threshold follows from an integer contrapositive, so the final $+1$ is necessary. A finite exact search over all 35 distributions of four pebbles on the four-cycle verifies solvability at a fixed target; symmetry covers all targets. It also verifies the displayed two-move certificate directly. For general factors a sharp proof must exploit residues while preserving the ability to gather enough pebbles; the current bound deliberately leaves their possible contribution unproved.
Reproducibility.
The finite calculations above are implemented in scripts/check_attempt_batch_2026_10_01.py; run it with Python 3 from the repository. It uses exact integer arithmetic and no external packages.
Final status.
UNRESOLVED. A general bound with an additive error and an exact obstruction to a naive sharp proof are established. The product inequality itself is not proved for arbitrary factors. Completion estimate: 5%.

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