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 \(C(n)\).

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TopNewestI 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. – Renjith Joshua · 2 years, 3 months ago

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– Tan Li Xuan · 2 years, 3 months 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.Log in to reply

– Renjith Joshua · 2 years, 3 months ago

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

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). – Ivan Koswara · 2 years, 3 months ago

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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. – Ben Frankel · 2 years, 3 months ago

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jee advance2015 paper 2 ellipse question – Rahul Jain · 2 years, 3 months ago

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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. – Calvin Lin Staff · 2 years, 3 months ago

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– Tan Li Xuan · 2 years, 3 months ago

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

– Ivan Koswara · 2 years, 3 months ago

I don't see why \(C(n) = n-1\); a staircase of order \(n\) has \(n\) rows.Log in to reply

Though, given that he says the image has order 3, I guess he did pick the natural ordering. – Calvin Lin Staff · 2 years, 3 months ago

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– Ivan Koswara · 2 years, 3 months 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".Log in to reply

– Shenal Kotuwewatta · 2 years, 3 months ago

That is what I meant.Log in to reply