Subfield: Combinatorics


Question: For an integer $m \geq 1$, we consider partitions of a $2^{m} \times 2^{m}$ chessboard into rectangles consisting of cells of the chessboard, in which each of the $2^{m}$ cells along one diagonal forms a separate rectangle of side length 1 . Determine the smallest possible sum of rectangle perimeters in such a partition.

Solution:
For a $k \times k$ chessboard, we introduce in a standard way coordinates of the vertices of the cells and assume that the cell $C_{i j}$ in row $i$ and column $j$ has vertices $(i-1, j-1),(i-$ $1, j),(i, j-1),(i, j)$, where $i, j \in\{1, \ldots, k\}$. Without loss of generality assume that the cells $C_{i i}$, $i=1, \ldots, k$, form a separate rectangle. Then we may consider the boards $B_{k}=\bigcup_{1<i<j<k} C_{i j}$ below that diagonal and the congruent board $B_{k}^{\prime}=\bigcup_{1<j<i<k} C_{i j}$ above that diagonal separately because no rectangle can simultaneously cover cells from $B_{k}$ and $B_{k}^{\prime}$. We will show that for $k=2^{m}$ the smallest total perimeter of a rectangular partition of $B_{k}$ is $m 2^{m+1}$. Then the overall answer to the problem is $2 \cdot m 2^{m+1}+4 \cdot 2^{m}=(m+1) 2^{m+2}$.

First we inductively construct for $m \geq 1$ a partition of $B_{2^{m}}$ with total perimeter $m 2^{m+1}$. If $m=0$, the board $B_{2^{m}}$ is empty and the total perimeter is 0 . For $m \geq 0$, the board $B_{2^{m+1}}$ consists of a $2^{m} \times 2^{m}$ square in the lower right corner with vertices $\left(2^{m}, 2^{m}\right),\left(2^{m}, 2^{m+1}\right),\left(2^{m+1}, 2^{m}\right)$, $\left(2^{m+1}, 2^{m+1}\right)$ to which two boards congruent to $B_{2^{m}}$ are glued along the left and the upper margin. The square together with the inductive partitions of these two boards yield a partition with total perimeter $4 \cdot 2^{m}+2 \cdot m 2^{m+1}=(m+1) 2^{m+2}$ and the induction step is complete.

Let

$$
D_{k}=2 k \log _{2} k
$$

Note that $D_{k}=m 2^{m+1}$ if $k=2^{m}$. Now we show by induction on $k$ that the total perimeter of a rectangular partition of $B_{k}$ is at least $D_{k}$. The case $k=1$ is trivial (see $m=0$ from above). Let the assertion be true for all positive integers less than $k$. We investigate a fixed rectangular partition of $B_{k}$ that attains the minimal total perimeter. Let $R$ be the rectangle that covers the cell $C_{1 k}$ in the lower right corner. Let $(i, j)$ be the upper left corner of $R$. First we show that $i=j$. Assume that $i<j$. Then the line from $(i, j)$ to $(i+1, j)$ or from $(i, j)$ to $(i, j-1)$ must belong to the boundary of some rectangle in the partition. Without loss of generality assume that this is the case for the line from $(i, j)$ to $(i+1, j)$.

Case 1. No line from $(i, l)$ to $(i+1, l)$ where $j<l<k$ belongs to the boundary of some rectangle of the partition.

Then there is some rectangle $R^{\prime}$ of the partition that has with $R$ the common side from $(i, j)$ to $(i, k)$. If we join these two rectangles to one rectangle we get a partition with smaller total perimeter, a contradiction.

Case 2. There is some $l$ such that $j<l<k$ and the line from $(i, l)$ to $(i+1, l)$ belongs to the boundary of some rectangle of the partition.

Then we replace the upper side of $R$ by the line $(i+1, j)$ to $(i+1, k)$ and for the rectangles whose lower side belongs to the line from $(i, j)$ to $(i, k)$ we shift the lower side upwards so that the new lower side belongs to the line from $(i+1, j)$ to $(i+1, k)$. In such a way we obtain a rectangular partition of $B_{k}$ with smaller total perimeter, a contradiction.

Now the fact that the upper left corner of $R$ has the coordinates $(i, i)$ is established. Consequently, the partition consists of $R$, of rectangles of a partition of a board congruent to $B_{i}$ and of rectangles of a partition of a board congruent to $B_{k-i}$. By the induction hypothesis, its total



perimeter is at least

$$
2(k-i)+2 i+D_{i}+D_{k-i} \geq 2 k+2 i \log _{2} i+2(k-i) \log _{2}(k-i)
\tag{1}
$$

Since the function $f(x)=2 x \log _{2} x$ is convex for $x>0$, JENSEN's inequality immediately shows that the minimum of the right hand sight of (1) is attained for $i=k / 2$. Hence the total perimeter of the optimal partition of $B_{k}$ is at least $2 k+2 k / 2 \log _{2} k / 2+2(k / 2) \log _{2}(k / 2)=D_{k}$.
We start as in Solution 1 and present another proof that $m 2^{m+1}$ is a lower bound for the total perimeter of a partition of $B_{2^{m}}$ into $n$ rectangles. Let briefly $M=2^{m}$. For $1 \leq i \leq M$, let $r_{i}$ denote the number of rectangles in the partition that cover some cell from row $i$ and let $c_{j}$ be the number of rectangles that cover some cell from column $j$. Note that the total perimeter $p$ of all rectangles in the partition is

$$
p=2\left(\sum_{i=1}^{M} r_{i}+\sum_{i=1}^{M} c_{i}\right)
$$

No rectangle can simultaneously cover cells from row $i$ and from column $i$ since otherwise it would also cover the cell $C_{i i}$. We classify subsets $S$ of rectangles of the partition as follows. We say that $S$ is of type $i, 1 \leq i \leq M$, if $S$ contains all $r_{i}$ rectangles that cover some cell from row $i$, but none of the $c_{i}$ rectangles that cover some cell from column $i$. Altogether there are $2^{n-r_{i}-c_{i}}$ subsets of type $i$. Now we show that no subset $S$ can be simultaneously of type $i$ and of type $j$ if $i \neq j$. Assume the contrary and let without loss of generality $i<j$. The cell $C_{i j}$ must be covered by some rectangle $R$. The subset $S$ is of type $i$, hence $R$ is contained in $S$. $S$ is of type $j$, thus $R$ does not belong to $S$, a contradiction. Since there are $2^{n}$ subsets of rectangles of the partition, we infer

$$
2^{n} \geq \sum_{i=1}^{M} 2^{n-r_{i}-c_{i}}=2^{n} \sum_{i=1}^{M} 2^{-\left(r_{i}+c_{i}\right)}
\tag{2}
$$

By applying JENSEN's inequality to the convex function $f(x)=2^{-x}$ we derive

$$
\frac{1}{M} \sum_{i=1}^{M} 2^{-\left(r_{i}+c_{i}\right)} \geq 2^{-\frac{1}{M} \sum_{i=1}^{M}\left(r_{i}+c_{i}\right)}=2^{-\frac{p}{2 M}}
\tag{3}
$$

From (2) and (3) we obtain

$$
1 \geq M 2^{-\frac{p}{2 M}}
$$

and equivalently

$$
p \geq m 2^{m+1}
$$
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