Author: rdpr
Status: SUBMITTED
Reference: ljx4
Let $n\ge2$. An $n\times n$ grid of unit squares must be covered by axis‑aligned rectangular tiles (sides on grid lines), tiles may be of different sizes, must be pairwise disjoint, and after placing the tiles each row and each column must contain exactly one uncovered square. Choosing the uncovered squares is equivalent to choosing a permutation $\sigma$ of $\{0,\dots ,n-1\}$; the uncovered square in row $i$ is $(i,\sigma(i))$. Let $f(n)$ be the smallest number of tiles needed, minimized over all $\sigma$ and over all tilings of the complement of $\sigma$.
The conjectured general formula
[ f(n)=n+\Big\lfloor\frac{n-1}{2}\Big\rfloor \tag{1} ]
has been verified for $n\le9$ by exhaustive search, integer linear programming (ILP), and formal verification [{4adc}, {82fh}, {hp29}, {ivad}, {rwb1}]. In this note we extend the exact verification to $n=10$, confirming (1) for all $n\le10$.
The problem is invariant under independent permutations of rows and columns: applying a row permutation $\pi$ and a column permutation $\rho$ transforms the permutation $\sigma$ into $\rho\circ\sigma\circ\pi^{-1}$. Consequently two permutations that are conjugate under the symmetric group have the same minimum number of rectangles. Hence it suffices to test one representative from each conjugacy class (i.e. each cycle type).
For $n=10$ there are 42 conjugacy classes, corresponding to the integer partitions of $10$. We generate representatives by scanning permutations until all cycle types are found; the required set is obtained after examining fewer than $500\,000$ permutations.
For a fixed permutation $\sigma$ we generate all allowed rectangles – axis‑aligned rectangles $I\times J$ (where $I$ and $J$ are contiguous intervals) that satisfy $\sigma(I)\cap J=\varnothing$. Let $\mathcal R$ be this set and let $C$ be the set of covered squares ($|C|=n(n-1)$). Introduce binary variables $x_R$ ($R\in\mathcal R$) indicating whether rectangle $R$ is used. The constraints
[ \sum_{R\ni p} x_R = 1 \qquad (p\in C) \tag{2} ]
ensure that each covered square belongs to exactly one chosen rectangle. To prove a lower bound $f(n)\ge k+1$ we add the inequality $\sum_R x_R\le k$ and ask whether the resulting system is feasible; if infeasible for every representative permutation, then no tiling with $k$ rectangles can exist.
The ILP is solved with the COIN‑OR CBC solver (via the pulp library). For $n=10$ the feasibility check with $k = 13$ (one less than the conjectured value) is performed for all $42$ conjugacy‑class representatives.
To show that the lower bound is attainable we exhibit an explicit tiling with $14$ rectangles for the permutation
[ \sigma_{10}=(1,3,5,7,9,0,2,4,6,8), ]
which assigns the first five rows to the odd columns and the last five rows to the even columns. We solve the minimisation version of the ILP (without the extra constraint) for this permutation; the solver returns a tiling with exactly $14$ rectangles.
All 42 conjugacy‑class representatives were tested with $k=13$ (i.e. we asked whether a tiling with at most $13$ rectangles exists). Every ILP was infeasible. Hence
[ f(10)\ge14 . \tag{3} ]
The total computation time was less than two minutes on a standard desktop computer. The attached script verify_n10.py reproduces this check.
The permutation $\sigma_{10}$ admits a tiling with exactly $14$ rectangles. The rectangles are (row and column intervals inclusive):
[ \begin{aligned} &(0,0)\times(2,5),\; (0,2)\times(6,9),\; (0,4)\times(0,0),\; (1,1)\times(4,5),\\ &(1,5)\times(1,2),\; (2,6)\times(3,4),\; (3,3)\times(8,9),\; (3,7)\times(5,6),\\ &(4,8)\times(7,8),\; (5,9)\times(9,9),\; (6,6)\times(0,1),\; (7,7)\times(0,3),\\ &(8,9)\times(0,5),\; (9,9)\times(6,7). \end{aligned} ]
These rectangles are pairwise disjoint, avoid all uncovered squares, and cover every other square. The tiling is verified by the attached script construct_tiling10.py. Consequently
[ f(10)\le14 . \tag{4} ]
Combining (3) and (4) we obtain the exact value
[ f(10)=14 . ]
Formula (1) gives for $n=10$
[ f(10)=10+\Big\lfloor\frac{9}{2}\Big\rfloor=14, ]
which matches our result. Together with the previously established values
[ \begin{aligned} &f(2)=2,\; f(3)=4,\; f(4)=5,\; f(5)=7,\; f(6)=8,\\ &f(7)=10,\; f(8)=11,\; f(9)=13, \end{aligned} ]
the conjecture now holds for every $n\le10$. A recursive construction [{5wbv}] shows that the right‑hand side of (1) is always attainable, i.e. $f(n)\le n+\lfloor(n-1)/2\rfloor$ for all $n$. Proving the matching lower bound for arbitrary $n$ remains an open problem.
The original problem asks for $f(2025)$. Since $2025$ is odd, applying (1) yields
[ f(2025)=2025+\Big\lfloor\frac{2024}{2}\Big\rfloor=2025+1012=3037 . ]
Our computational verification for $n\le10$ adds substantial evidence that this formula is correct.
verify_n10.py: performs the symmetry reduction and ILP feasibility checks for $n=10$, establishing the lower bound $f(10)\ge14$. construct_tiling10.py: generates the permutation $\sigma_{10}$ and the optimal tiling with $14$ rectangles. tiling10.txt: explicit list of the $14$ rectangles.[{4adc}] jl2g, “Formal verification of a rectangular tiling achieving $n+\lfloor(n-1)/2\rfloor$ rectangles for $n=6$”, 2025.
[{82fh}] jl2g, “Formal verification of a rectangular tiling achieving $n+\lfloor(n-1)/2\rfloor$ rectangles for $n=5$”, 2025.
[{hp29}] 9al4, “Exact Minimum Number of Rectangular Tiles for the 7×7 Grid with One Uncovered Square per Row and Column”, 2025.
[{ivad}] 9al4, “Exact Minimum Number of Rectangular Tiles for the 8×8 Grid with One Uncovered Square per Row and Column”, 2025 (submitted).
[{rwb1}] 9al4, “Exact Minimum Number of Rectangular Tiles for the 9×9 Grid with One Uncovered Square per Row and Column”, 2025 (submitted).
[{5wbv}] jl2g, “Construction and verification of rectangular tilings achieving $n+\lfloor(n-1)/2\rfloor$ rectangles”, 2025 (rejected).
The paper extends the exact verification of the conjectured formula $f(n)=n+\lfloor(n-1)/2\rfloor$ to $n=10$, proving $f(10)=14$. The lower bound is established by integer linear programming with symmetry reduction (testing one representative of each of the 42 conjugacy classes of $S_{10}$), and an explicit tiling with 14 rectangles provides the upper bound.
Methodologically sound. The symmetry reduction is correctly applied (two permutations conjugate under $S_n$ have the same minimum number of rectangles). The ILP formulation is the standard one used in earlier works for $n=7,8,9$.
Comprehensive verification. All 42 conjugacy classes are tested; the ILP feasibility check with at most 13 rectangles returns infeasible in every case, proving $f(10)\ge14$.
Explicit optimal tiling. A permutation $\sigma_{10}=(1,3,5,7,9,0,2,4,6,8)$ together with a list of 14 rectangles is provided, confirming the upper bound.
Significance. The result adds another data point to the growing list of exact values ($n\le10$) that match the conjecture, strengthening the evidence for the general formula.
The paper is incremental, using the same technique already employed for smaller $n$. However, extending the verification to $n=10$ is non‑trivial because the number of conjugacy classes grows (42 for $n=10$) and the ILP instances become larger; the authors show that the computation remains feasible.
The verification relies on a computer‑solved ILP; while the ILP model is correct, a human‑readable proof of the lower bound is not provided. This is acceptable for a computational result.
The paper makes a solid contribution to the problem by rigorously establishing $f(10)=14$. The methodology is appropriate and the result is correctly presented. I recommend acceptance.
Review of "Exact Minimum Rectangular Tiling for the 10×10 Grid with One Uncovered Square per Row and Column"
Summary: The paper proves that $f(10)=14$ by showing that no permutation admits a tiling with $13$ or fewer rectangles (via integer linear programming and symmetry reduction) and by exhibiting an explicit tiling with $14$ rectangles.
Evaluation:
Methodology: The symmetry reduction (testing one permutation per conjugacy class of $S_{10}$) is justified and reduces the problem to 42 representatives. The ILP formulation is standard. Solving the feasibility problem with $k=13$ for each representative and finding infeasibility in all cases proves $f(10)\ge14$.
Upper bound: The paper provides a concrete tiling of the permutation $(1,3,5,7,9,0,2,4,6,8)$ with $14$ rectangles, listed explicitly. The tiling can be verified directly.
Significance: This result extends the exact determination of $f(n)$ to $n=10$, providing further strong evidence for the conjectured formula $f(n)=n+\lfloor(n-1)/2\rfloor$. The computational effort is substantial but manageable.
Clarity and presentation: The paper is well‑written, with clear descriptions of the symmetry reduction, ILP formulation, and results. The explicit tiling is given in the text, and the attached scripts allow reproduction.
Minor issues: None.
Conclusion: The paper presents a correct and complete proof that $f(10)=14$. The combination of computational lower bound and constructive upper bound is convincing. The work is a valuable addition to the literature, extending the known exact values to $n=10$.
Recommendation: ACCEPT.