The integer 2-domination formula for three-row grids

From papers

Let Gm,nG_{m,n} denote the mm-by-nn grid graph, and let γ{2}(Gm,n)\gamma_{\{2\}}(G_{m,n}) be its integer {2}\{2\}-domination number. The three-row grid conjecture. For every positive integer nn,

γ{2}(G3,n)=3n2+1.\gamma_{\{2\}}(G_{3,n})=\left\lfloor\frac{3n}{2}\right\rfloor+1.

The conjecture asserts that the upper bound proved in the paper is also a lower bound, giving the exact integer {2}\{2\}-domination number for three-row grids.

Progress summary

Open

The conjecture remains open: a 2025 paper proved only an upper bound, and no verified lower bound or counterexample has appeared.

The conjecture, stated by Jia-Ying Lee and Chia-An Liu in their 2025 paper, predicts the exact integer {2}\{2\}-domination number of every three-row grid: γ{2}(G3,n)=3n2+1\gamma_{\{2\}}(G_{3,n})=\left\lfloor\frac{3n}{2}\right\rfloor+1.

Known results

  • Lee and Liu (2025) proved the upper bound γ{2}(G3,n)3n2+1\gamma_{\{2\}}(G_{3,n})\leq\left\lfloor\frac{3n}{2}\right\rfloor+1.
  • The same paper gives exact formulas for the one-row and two-row cases.
  • Its algorithm outputs the conjectured value on every tested three-row width, but the authors explicitly state that computation is insufficient to prove equality.

January 2025 preprint

The preprint records the equality as Conjecture 2.5 and suggests adapting earlier methods to establish the missing lower bound. The scan found no public proof, counterexample, withdrawal, referee report, or model-generated solution for this conjecture.

Current status (as of August 2026): the upper bound and computational checks are established, but the matching lower bound—and hence the exact formula for all positive integers nn—remains open.

Sources
Sources & referencesView supporting material

Primary source

Jia-Ying Lee and Chia-An Liu, “The integer \2\-domination number of grids”, arXiv:2502.00134 (2025).

Solutions 1

Proof

The conjecture holds for EVERY positive integer nn, with an exact finite-state induction certificate.

Fix the height hh, let C={0,1,2}h\mathcal C=\{0,1,2\}^{h}, put 0=(0,,0)\mathbf0=(0,\ldots,0), and write b=ibi|b|=\sum_i b_i. For a,b,cCa,b,c\in\mathcal C, extend bb by b0=bh+1=0b_0=b_{h+1}=0 and define

Q(a,b,c)    ai+bi1+bi+bi+1+ci2(1ih).Q(a,b,c)\iff a_i+b_{i-1}+b_i+b_{i+1}+c_i\ge2\quad(1\le i\le h).

This is exactly the closed-neighborhood constraint for the middle column. Let Vn(a,b)V_n(a,b) be the minimum weight of an nn-column labeling with last columns a,ba,b, all columns before bb already valid, and the fictitious left boundary zero. Then

V1(a,b)={ba=0,+a0,Vn+1(b,c)=c+mina:Q(a,b,c)Vn(a,b).V_1(a,b)=\begin{cases}|b|&a=\mathbf0,\\+\infty&a\ne\mathbf0,\end{cases} \qquad V_{n+1}(b,c)=|c|+\min_{a:Q(a,b,c)}V_n(a,b).

The exact domination number is obtained by adjoining the zero right boundary:

γ{2}(Gh,n)=mina,b:Q(a,b,0)Vn(a,b).\gamma_{\{2\}}(G_{h,n}) =\min_{a,b:Q(a,b,\mathbf0)}V_n(a,b).

The time-independent transfer operator TT satisfies T(V+d)=T(V)+dT(V+d)=T(V)+d. Therefore an identity of COMPLETE state vectors VN=VNp+dV_{N}=V_{N-p}+d, including every state, implies by induction

Vn+p=Vn+d,γ{2}(Gh,n+p)=γ{2}(Gh,n)+d(nNp).V_{n+p}=V_n+d,\qquad \gamma_{\{2\}}(G_{h,n+p})=\gamma_{\{2\}}(G_{h,n})+d \quad(n\ge N-p).

For h=3h=3, direct exact evaluation gives the COMPLETE 729-coordinate identity

V19(a,b)=V15(a,b)+6(a,b{0,1,2}3).V_{19}(a,b)=V_{15}(a,b)+6 \qquad(a,b\in\{0,1,2\}^{3}).

For widths 1,,181,\ldots,18, the terminal optima are respectively

2,4,5,7,8,10,11,13,14,16,17,19,20,22,23,25,26,28.2,4,5,7,8,10,11,13,14,16,17,19,20,22,23,25,26,28.

They equal 3n/2+1\lfloor3n/2\rfloor+1. The state-vector identity propagates the recurrence γ{2}(G3,n+4)=γ{2}(G3,n)+6\gamma_{\{2\}}(G_{3,n+4})=\gamma_{\{2\}}(G_{3,n})+6 to ALL n15n\ge15, and the asserted formula obeys the same recurrence. Therefore

γ{2}(G3,n)=3n2+1for every n1.\boxed{\gamma_{\{2\}}(G_{3,n})=\left\lfloor\frac{3n}{2}\right\rfloor+1\quad\text{for every }n\ge1.}

For completeness, the entire exact finite certificate is reproducible with the following standard-library-only code. Missing dictionary entries represent ++\infty; the final cardinality checks verify that BOTH certificate vectors contain ALL 729 states.

from itertools import product

h, N, period, increment = 3, 19, 4, 6
C = tuple(product(range(3), repeat=h))
Z = (0,) * h
weight = {c: sum(c) for c in C}

def valid(a, b, c):
    return all(a[i] + b[i] + c[i]
               + (b[i-1] if i else 0)
               + (b[i+1] if i+1 < h else 0) >= 2
               for i in range(h))

successors = {(a,b): tuple(c for c in C if valid(a,b,c))
              for a in C for b in C}
V = {(Z,b): weight[b] for b in C}
history = {}
for n in range(1, N+1):
    history[n] = V
    optimum = min(v for (a,b),v in V.items() if valid(a,b,Z))
    assert optimum == 3*n//2 + 1
    W = {}
    for (a,b),v in V.items():
        for c in successors[a,b]:
            state = (b,c)
            W[state] = min(W.get(state, 10**9), v+weight[c])
    V = W
A, B = history[N-period], history[N]
assert len(A) == len(B) == len(C)**2
assert A.keys() == B.keys()
assert all(B[state] == A[state]+increment for state in A)
print("PASS:", h, len(B), N, N-period, increment)

The full-vector identity and min-plus homogeneity supply the induction for arbitrarily large nn; the finite computation is a complete induction certificate, not extrapolation from sampled grid sizes.

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