Experiencing Michael Fenton’s Rectangle World
Check out the simulation. Trust me. It’s cool.
]]>In summary, imagine an m x n rectangle where n is greater than or equal to m. This solution involves taking the solution for an m x m square and adding to it the number of additional squares that are created by adding the additional rows required to make it an m x n rectangle.
]]>I scribbled some notes into a Dropbox Paper document while I watched the Warriors game. 🙂
Two objectives here: Give the Paper app a thorough test drive (I started using it yesterday) and generate something I could easily share in the comments of the blog.
Voilà: https://paper.dropbox.com/doc/How-Many-Squares-v8kjPTUcKVpDWlrw6y9Mo
As for the test drive… I like this app. Super easy to use. I might just ditch my other note taking app for this one. We’ll see. Anyway, I assume no one actually cares about that except for me. 🙂
]]>It should be three columns, the first being mxn = number of Squares (for example 1X2 = 2 S)
The second column is the product (for example prod 2)
The third column is the difference between product and number of squares.
2X3=8 S prod 6 2
2X4 = 11 S prod 8 3
2X5 = 14 S prod 10 4
3X4 = 20 S prod 12 8
3X5 = 26 S prod 15 11
3X6 = 32 S prod 18 14
4X5 = 40 S prod 20 20
4X6 = 50 S prod 24 26
4X7 = 60 S prod 28 32
I started looking for patterns and found the product of each mxn (second column) and then the difference between product and the number of sqares (3rd column). Then I noticed that the differences were the same as the number of squares in an (n-1)X(m-1) rectangle. So they just build on themselves.
It’s like a little Russian doll situation!
So, to find the number of squares in an mxn rectangle, you can find mxn + (m-1)*(n-1) + (m-2)*(m-2) + … until one of those factors is zero. That sum will give you the number of rectangles.
So my question is: Is there a better way to generalize that formula, other than just finding the sum of a lot of products (if m and n are bigger numbers)?
]]>