Subfield: Combinatorics


Question: Fix an integer $n \geq 2$. An $n \times n$ sieve is an $n \times n$ array with $n$ cells removed so that exactly one cell is removed from every row and every column. A stick is a $1 \times k$ or $k \times 1$ array for any positive integer $k$. For any sieve $A$, let $m(A)$ be the minimal number of sticks required to partition $A$. Find all possible values of $m(A)$, as $A$ varies over all possible $n \times n$ sieves.

Solution:
Given $A, m(A)=2 n-2$, and it is achieved, for instance, by dissecting $A$ along all horizontal (or vertical) grid lines. It remains to prove that $m(A) \geq 2 n-2$ for every $A$.



By holes we mean the cells which are cut out from the board. The cross of a hole in $A$ is the union of the row and the column through that hole.



Arguing indirectly, consider a dissection of $A$ into $2 n-3$ or fewer sticks. Horizontal sticks are all labelled $h$, and vertical sticks are labelled $v ; 1 \times 1$ sticks are both horizontal and vertical, and labelled arbitrarily. Each cell of $A$ inherits the label of the unique containing stick.



Assign each stick in the dissection to the cross of the unique hole on its row, if the stick is horizontal; on its column, if the stick is vertical.



Since there are at most $2 n-3$ sticks and exactly $n$ crosses, there are two crosses each of which is assigned to at most one stick in the dissection. Let the crosses be $c$ and $d$, centred at $a=\left(x_{a}, y_{a}\right)$ and $b=\left(x_{b}, y_{b}\right)$, respectively, and assume, without loss of generality, $x_{a}<x_{b}$ and $y_{a}<y_{b}$. The sticks covering the cells $\left(x_{a}, y_{b}\right)$ and $\left(x_{b}, y_{a}\right)$ have like labels, for otherwise one of the two crosses would be assigned to at least two sticks. Say the common label is $v$, so each of $c$ and $d$ contains a stick covering one of those two cells. It follows that the lower (respectively, upper) arm of $c$ (respectively, $d$ ) is all- $h$, and the horizontal arms of both crosses are all- $v$, as illustrated below.



![](https://cdn.mathpix.com/cropped/2023_12_21_5f10889eb78c8b64e212g-1.jpg?height=651&width=651&top_left_y=1302&top_left_x=680)



Each of the rows between the rows of $a$ and $b$, that is, rows $y_{a}+1, y_{a}+2, \ldots, y_{b}-1$, contains a hole. The column of each such hole contains at least two $v$-sticks. All other columns contain at least one $v$-stick each. In addition, all rows below $a$ and all rows above $b$ contain at least one $h$-stick each. This amounts to a total of at least $2\left(y_{b}-y_{a}-1\right)+\left(n-y_{b}+y_{a}+1\right)+\left(n-y_{b}\right)+\left(y_{a}-1\right)=2 n-2$ sticks. A contradiction.
We provide a different proof that $m(A) \geq 2 n-2$.


Call a stick vertical if it is contained in some column, and horizontal if it is contained in some row; $1 \times 1$ sticks may be called arbitrarily, but any of them is supposed to have only one direction. Assign to each vertical/horizontal stick the column/row it is contained in. If each row and each column is assigned to some stick, then there are at least $2 n$ sticks, which is even more than we want. Thus we assume, without loss of generality, that some exceptional row $R$ is not assigned to any stick. This means that all $n-1$ existing cells in $R$ belong to $n-1$ distinct vertical sticks; call these sticks central.



Now we mark $n-1$ cells on the board in the following manner. ( $\downarrow$ ) For each hole $c$ below $R$, we mark the cell just under $c$; $(\uparrow)$ for each hole $c$ above $R$, we mark the cell just above $c$; and $(\bullet)$ for the hole $r$ in $R$, we mark both the cell just above it and just below it. We have described $n+1$ cells, but exactly two of them are out of the board; so $n-1$ cells are marked within the board. A sample marking is shown in the figure below, where the marked cells are crossed.



![](https://cdn.mathpix.com/cropped/2023_12_21_8706eea02666b23df93fg-1.jpg?height=414&width=466&top_left_y=752&top_left_x=772)



Notice that all the marked cells lie in different rows, and all of them are marked in different columns, except for those two marked for $(\bullet)$; but the latter two have a hole $r$ between them. So no two marked cells may belong to the same stick. Moreover, none of them lies in a central stick, since the marked cells are separated from $R$ by the holes. Thus the marked cells should be covered by $n-1$ different sticks (call them border) which are distinct from the central sticks. This shows that there are at least $(n-1)+(n-1)=2 n-2$ distinct sticks, as desired.
In order to prove $m(A) \geq 2 n-2$, it suffices to show that there are $2 n-2$ cells in $A$, no two of which may be contained in the same stick.



To this end, consider the bipartite graph $G$ with parts $G_{h}$ and $G_{v}$, where the vertices in $G_{h}$ (respectively, $G_{v}$ ) are the $2 n-2$ maximal sticks $A$ is dissected into by all horizontal (respectively, vertical) grid lines, two sticks being joined by an edge in $G$ if and only if they share a cell.



We show that $G$ admits a perfect matching by proving that it fulfils the condition in Hall's theorem; the $2 n-2$ cells corresponding to the edges of this matching form the desired set. It is sufficient to show that every subset $S$ of $G_{h}$ has at least $|S|$ neighbours (in $G_{v}$, of course).



Let $L$ be the set of all sticks in $S$ that contain a cell in the leftmost column of $A$, and let $R$ be the set of all sticks in $S$ that contain a cell in the rightmost column of $A$; let $\ell$ be the length of the longest stick in $L$ (zero if $L$ is empty), and let $r$ be the length of the longest stick in $R$ (zero if $R$ is empty).



Since every row of $A$ contains exactly one hole, $L$ and $R$ partition $S$; and since every column of $A$ contains exactly one hole, neither $L$ nor $R$ contains two sticks of the same size, so $\ell \geq|L|$ and $r \geq|R|$, whence $\ell+r \geq|L|+|R|=|S|$.



If $\ell+r \leq n$, we are done, since there are at least $\ell+r \geq|S|$ vertical sticks covering the cells of the longest sticks in $L$ and $R$. So let $\ell+r>n$, in which case the sticks in $S$ span all $n$ columns, and notice that we are again done if $|S| \leq n$, to assume further $|S|>n$.



Let $S^{\prime}=G_{h} \backslash S$, let $T$ be set of all neighbours of $S$, and let $T^{\prime}=G_{v} \backslash T$. Since the sticks in $S$ span all $n$ columns, $|T| \geq n$, so $\left|T^{\prime}\right| \leq n-2$. Transposition of the above argument (replace $S$ by $T^{\prime}$, shows that $\left|T^{\prime}\right| \leq\left|S^{\prime}\right|$, so $|S| \leq|T|$.
Yet another proof of the estimate $m(A) \geq 2 n-2$. We use the induction on $n$. Now we need the base cases $n=2,3$ which can be completed by hands.



Assume now that $n>3$ and consider any dissection of $A$ into sticks. Define the cross of a hole as in Solution 1, and notice that each stick is contained in some cross. Thus, if the dissection contains more than $n$ sticks, then there exists a cross containing at least two sticks. In this case, remove this cross from the sieve to obtain an $(n-1) \times(n-1)$ sieve. The dissection of the original sieve induces a dissection of the new array: even if a stick is partitioned into two by the removed cross, then the remaining two parts form a stick in the new array. After this operation has been performed, the number of sticks decreases by at least 2 , and by the induction hypothesis the number of sticks in the new dissection is at least $2 n-4$. Hence, the initial dissection contains at least $(2 n-4)+2=2 n-2$ sticks, as required.



It remains to rule out the case when the dissection contains at most $n$ sticks. This can be done in many ways, one of which is removal a cross containing some stick. The resulting dissection of an $(n-1) \times(n-1)$ array contains at most $n-1$ sticks, which is impossible by the induction hypothesis since $n-1<2(n-1)-2$.

Final answer: $m(A) \geq 2n-2$ for every $A$