A $n$ x $n$ square (where n is an odd positive integer) is divided into $n^2$ unit squares . Each unit square is colored either black or white . Probability of a square being black or white is equal . The square is rotated $90^o$ with respect to the center . If a white square overlaps on a square which was previously black , then the white square is colored black . What is the probability of the new square being entirely black ?
[This problem is inspired from another problem in a math olympiad , organized by Ramanujan Gonit Songho a few days ago. there, it was a $3$ x $3$ square.]
rotating a colored square
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Re: rotating a colored aquare
Is it allowed to rotate the square in both sides? Or just one specific side?
Why so SERIOUS?!??!
Re: rotating a colored aquare
you can rotate to any side , it doesn't matter for calculation .
Try not to become a man of success but rather to become a man of value.-Albert Einstein
- asif e elahi
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Re: rotating a colored aquare
How many times is the square rotated ?
Re: rotating a colored square
the question said $90^{\circ}$.asif e elahi wrote:How many times is the square rotated ?
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- Tahmid Hasan
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Re: rotating a colored square
So let me get this straight $\mathbb B \rightarrow \mathbb B=\mathbb B, \mathbb B \rightarrow \mathbb W=\mathbb B, \mathbb W \rightarrow \mathbb B=\mathbb B, \mathbb W \rightarrow \mathbb W=\mathbb W$.
WLOG we assume the square is rotated anti-clockwise. For simplicity, we assume the centre of a square corresponds to the whole square i.e. the centre and the whole square has the same colour.
We use cartesian co-ordinates. Let the centre be $(0,0)$. We can denote each point as a unique co-ordinate $(i,j)$.
Now notice that $(i,j) \rightarrow (-j,i) \rightarrow (-i,-j) \rightarrow (j,-i) \rightarrow (i,j)$ where $(i,j) \neq (0,0)$.
Call these four points a chunk. A chunk can be coloured $2^4=16$ ways. Of these colourings $7$ produce a complete black chunk.
So the probability of a chunk being black is $\frac{7}{16}$. There are $\frac{n^2-1}{4}$ chunks.
So the probability of the whole board being black is $\frac 12(\frac{7}{16})^{\frac{n^2-1}{4}}$.
WLOG we assume the square is rotated anti-clockwise. For simplicity, we assume the centre of a square corresponds to the whole square i.e. the centre and the whole square has the same colour.
We use cartesian co-ordinates. Let the centre be $(0,0)$. We can denote each point as a unique co-ordinate $(i,j)$.
Now notice that $(i,j) \rightarrow (-j,i) \rightarrow (-i,-j) \rightarrow (j,-i) \rightarrow (i,j)$ where $(i,j) \neq (0,0)$.
Call these four points a chunk. A chunk can be coloured $2^4=16$ ways. Of these colourings $7$ produce a complete black chunk.
So the probability of a chunk being black is $\frac{7}{16}$. There are $\frac{n^2-1}{4}$ chunks.
So the probability of the whole board being black is $\frac 12(\frac{7}{16})^{\frac{n^2-1}{4}}$.
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