Re: OT: CSES Number Spiral algorithm

On 20/03/2025 16:47, DFS wrote:

I don't have a C question, but rather I'd like input about algorithmic approaches.
 https://cses.fi/problemset/task/1071
 It's an 'infinite grid'.  You have to find the value at rows-columns from 1 to 10^9.
 First thing you notice about the 5x5 grid is the values in odd-numbered columns begin with the square of the column number, and the values in even-numbered rows begin with the square of the row number.
 I followed the number pattern and built a grid in Excel and expanded it to 10x10 for more testing.
 https://imgur.com/x4VymmA
 Then coded 4 conditions      solution
1. row <= col and col odd    (col * col) - row + 1
2. row <= col and col even   ((col-1) * (col-1)) + row
3. row >  col and row odd    ((row-1) * (row-1)) + col
4. row >  col and row even   (row * row) - col + 1
 My full C code submission was accepted the first time.
 How would you have done it?

I see that you've already solved it, so that's good.
My first observation was the main diagonal, which goes:
1 3 7 13 21...
Differences are 2 4 6 8... respectively
Consider the triangle numbers (n(n+1)/2):
0 1 3 6 10 15 21 28...
Double and add 1:
1 3 7 13 21 31...
So the main diagonal is readily calculable - either (row, row) or (col, col).
I'd then have picked the biggest out of (row, col) and calculated its triangle number: n(n+1)/2, doubled (so don't divide) and add 1, for (n(n+1))+1 and then either added the column or subtracted the row (whichever of them is smaller).
Reviewing your solution after the fact, I don't see the need to distinguish between odd and even. What did I miss?
--
Richard Heathfield
Email: rjh at cpax dot org dot uk
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