HaCkEd By KaSpEr511

General Spiral Numbers : MMM#37

July 22nd, 2009

Here is MMM#37 problem:

Based on the introduction to spiral numbers presented in MMM #36, solve one (or both) of these problems:

Come up with an algorithm that tells what number is at an arbitrary X, Y coordinate.

Come up with an algorithm that tells the X, Y coordinates for an arbitrary positive integer.

Solution (2nd problem):
We can easily check that the coordinate of $1,2,3,4$ are $(0,0),(-1,0),(-1,-1)$ and $(0,-1)$ respectively. Now if $m>4$ then $4n^2< m\leq 4(n+1)^2$ for some $n\geq 1$ . Now consider the following picture

Nested Square

We have proved earlier that the the number $4n^2$ is on the bottom right corner of the $2n\times 2n$ square and its coordinate is $(n-1,-n)$ . It follows that $D=4(n+1)^2$ and the coordinate of $D$ is $(n,-n-1)$ .

If $m=4n^2+k$ where $k=1,2,\cdots, 2n+1$ , then $m$ is on the blue zone which has the same $x$ -coordinate as $D$ . Then clearly the coordinate of $4n^2+k$ is $(n,-n-1+k)$ . In particular $A=4n^2+2n+1$ and its coordinate is $(n, -n-1+(2n+1) )=(n,n)$ .

If $m=4n^2+2n+1+k$ where $k=1,2,\cdots 2n+1$ then $m$ is on the red zone which has the same $y$ -coordinate as $A$ . Hence the coordinate of $4n^2+2n+1+k$ is $(n-k,n)$ . In particular $B=4n^2+4n+2$ and its coordinate is $(n-(2n+1),n)=(-n-1,n)$ .
By using similar argument, if $m=4n^2+4n+2+k$ for $k=1,2,\cdots,2n+1$ this number is on the brown zone and the coordinate of $4n^2+4n+2+k$ is $(-n-1,n-k)$ and in particular $C=4n^2+6n+3$ has coordinate $(-n-1,n-(2n+1) )=(-n-1,-n-1)$ .

The last case, $m=4n^2+6n+3+k$ for $k=1,2,\cdots,2n+1$ has coordinate $(-n-1+k,-n-1)$

In summary for $k=1,2,\cdots, 2n+1$

The coordinate of $m$ is $\begin{cases}(n,-n-1+k)&\text{ if } m=4n^2+k\\ (n-k,n)&\text{ if } m=4n^2+2n+1+k\\(-n-1,n-k)&\text{ if } m=4n^2+4n+2+k\\(-n-1+k,-n-1)&\text{ if } m=4n^2+6n+3+k\end{cases}$ .

One can checks that the coordinate of $1,2,3$ and $4$ satisfy the above formula too for $n=0$ .

Equivalently if $4n^2<m\leq 4(n+1)^2$ for $n\geq 0$ we have that:

The coordinate of $m$ is $\begin{cases}(n,-n-1+m-4n^2)&\text{ if } 4n^2<m\leq 4n^2+2n+1\\(n+4n^2+2n+1-m,n)&\text{ if } 4n^2+2n+1< m\leq 4n^2+4n+2\\(-n-1,n+4n^2+4n+2-m)&\text{ if } 4n^2+4n+2<m\leq 4n^2+6n+3\\(-n-1+m-4n^2-6n-3,-n-1)&\text{ if } 4n^2+6n+3<m\leq 4(n+1)^2\end{cases}$

Note: if you want you can write the coordinate of $m$ completely as a function of $m$ . We have $4n^2<m\leq 4(n+1)^2$ which is equivalent to $n<\frac{1}{2}\sqrt{m}\leq n+1$ . Therefore $n=\left\lceil \frac{1}{2}\sqrt{m}\right\rceil -1$ .

Categories: Math Monday Madness

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General Spiral Numbers : MMM#37

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