21 | 22 | 23 | 24 | 25 |

20 | 7 | 8 | 9 | 10 |

19 | 6 | 1 | 2 | 11 |

18 | 5 | 4 | 3 | 12 |

17 | 16 | 15 | 14 | 13 |

The "**Spiral Matrix**" is a recurring exercise in computer science classes and occasionally programming job interviews.
I've come across the problem in various settings a number of times over the years, but only had occasion to develop a solution
yesterday morning *(circa June 2011)*. The matrix at right illustrates
the 'spiral' nature of the exercise:

The first 95% of the solution is entirely straightforward. But depending on approach, the thought-provoking task is the final step of calculating each cell value on the fly.

I decided on a client-side JavaScript solution which followed these assumptions:

- I quickly determined to avoid an exercise of 'chasing' the spiral through the matrix with a series of nested
*for-next*loops, since this is both convoluted and unnecessary (and, furthermore, "Yougly" -- as Jar Jar Binks might say ;-) - I found it equally redundant and unnecessary to separate the tasks of populating nested arrays with the solution, then subsequently displaying it in the page,
- Consequently, I simply build the matrix (a table) dynamically in a document fragment, calculating the value of each cell in
*row, column*sequence as I went, and, finally, add the completed fragment to the document's array container (i.e. div) when done; - In order to do this, I adopted the notion (although not the Ruby code) that a spiral actually consists of a series of concentric 'rings', each of which presents a fixed starting value ("offset") at its northeast origin. This is well-explained at www.rubyquiz.com/quiz109.html;
- The example below is designed to work for all possibilities of
**odd**numbers of horizontal cells. It would be a minor extension to make it work for even numbers as well, but that's beyond the purpose of this example. - Finally, the cell value calculations are mine. They simply calculate the number of cells which have preceded the current cell in the ring and subtract it from the offset. Note from the ternary operator in the code that the final leg of each ring is an exception since it decrements in ascending (reversed) direction.

I've encapsulated the JavaScript code here if you'd like to download it. The calculations are in mathematically simplified form, so the algorithm may not be entirely obvious from the code. Perhaps I can put the expansions here as time permits. Meanwhile, try the example below to tour the results.

### Build a Spiral or view the Rings and Offsets: