# The staircase-covering problem

Remove the unit squares lying on a main diagonal of a $(n+1) \times (n+1)$ square grid. You get two congruent shapes, each containing $\frac{n(n+1)}{2}$ unit squares. Call such a shape a staircase of order $n$. (For example, the above image shows a staircase of order $3$)

Let $C(n)$ be the minimum number of rectangles one needs to cover a staircase of order $n$ such that all unit squares in the staircase are covered, no two rectangles overlap, and no part of a rectangle is outside the staircase.

Find an explicit formula for $C(n)$.

Note by Shenal Kotuwewatta
4 years ago

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I think C(n) should be equal to n itself. I guess you could give a straightforward statement based proof where you show separately that it cannot be greater than n, nor can it be less than n.

- 4 years ago

We can easily show that $n$ rectangles is sufficient by covering each row with one rectangle. Then, we observe that a staircase of order $n$ has $n$ "stairs", and each of the rectangles can cover at most one stair. So, we need at least $n$ rectangles.

- 4 years ago

Precisely what I meant. For some reason, the words 'necessary' and sufficient' eluded me when I made my previous comment :)

- 4 years ago

More interesting is the number of ways to cover the staircase with $C(n)$ rectangles (there are $C_n$ ways where $C_n$ is the $n$-th Catalan number).

- 4 years ago

I claim that $C(n) = n$.

$C(n) \le n$: We can always tile a staircase with as many rectangles as steps by covering each row with a long rectangle.

$C(n) \ge n$: We can never tile a staircase with less rectangles than steps. Consider some step's top-left corner: this corner must coincide with the top-left corner of a rectangle in the tiling in order for the step to be covered. Thus we must have at least $n$ top-left corners of rectangles in our tiling, or, at least $n$ rectangles.

- 4 years ago

Claim: $C(n) = n-1$ (due to the way you indexed it).

There is a one-line proof of that claim. It is left as an exercise to the reader.

Staff - 4 years ago

Then wouldn't C(1) = 0? How would we cover a staircase of order 1 with 0 rectangles?

- 4 years ago

I don't see why $C(n) = n-1$; a staircase of order $n$ has $n$ rows.

- 4 years ago

I didn't understand the "divided by one of it's main diagonals". Won't it be more of a $n \times (n+1)$ rectangle instead?

Though, given that he says the image has order 3, I guess he did pick the natural ordering.

Staff - 4 years ago

I take it as "cut the square with one diagonal, and remove the squares that are also cut apart with the diagonal; the rest is two congruent pieces, each piece is a staircase".

- 4 years ago

That is what I meant.

- 4 years ago

jee advance2015 paper 2 ellipse question

- 4 years ago