Working Out My Desired Income in Haskell

1. desiredIncome x = x > $3000 per month after tax and superannuation contributions
2. Tax = $4650 + 30% of per-annum income between $37,000 and $80,000

Eventually I'd like something that takes all of my expenses into account and calculates the desired income as well, so it's for general use not just personal.  For now, this is a good start.

Floor Mats and Haskell

This is a problem that I'd like to solve in Haskell.

The other day I went to Clark's Rubber to purchase some carpet vinyl. I'm also interested in mats for doing rolls and handstands on a danceFloorArea.

There are two sizes:

smallMatLength = (600mm)
smallMatArea = smallMatLength^2
smallMatUnitPrice = $30

largeMatLength = (1000mm ^2)
largeMatArea = largeMatLength^2
largeMatUnitPrice = $50

I have a danceFloorArea in mind:

danceFloorWidth = 2200mm
danceFloorLength = 3200mm
danceFloorArea = danceFloorWidth * danceFloorLength

I want to cover as much danceFloorArea as possible for the lowest possible price, i.e. I want the highest possible possible danceFloorArea/[small/large]MatUnitPrice ratio:

Should I use smallMats, largeMats, or a combination of both?

1) How do I calculate (danceFloorArea / smallMatArea) without a remainder?
2) How I calculate (danceFloorArea / largeMatArea) without a remainder?
3) How do I calculate danceFloorCoverage = (a * smallMatArea) + (b * largeMatArea) without a remainder OR with the smallest possible remainder

Other questions, I guess...

1) How many smallMats can I fit into the area?
2) How much would (1) cost?
3) How much of the area would I cover?
4) How many largeMats can I fit into the area?
5) How much would (4) cost?
6) How much of the area would I cover?

Haskell Notes 2011-07-23

This is a note about Haskell

Haskell Notes

I tried to write the Douglas Adams joke but got an error.

*Main> let "The Answer" = 42

:1:20:
    No instance for (Num [Char])
      arising from the literal `42'
    Possible fix: add an instance declaration for (Num [Char])
    In the expression: 42
    In a pattern binding: "The Answer" = 42

Starting Out In Haskell (Lists)

I'm going through a delightful introduction to the programming language Haskell called Learn You A Haskell For Greater Good.

So far I like it a lot.
I'm no expert programmer, but I have a few neat bits of Python in my pastebin, and have lost many hours of sleep tapping away in Python in a mildly addicted manner.

The great appeal for me is that I can read something in Haskell (at least basic stuff) without being confronted by a wall of punctuation and brackets.  It also differs from Python in that once you define a function:

doubleMe x = x +x

Then that function will never change.  You can rest assured a thousand lines later that "doubleMe 20" will return 40.

Haskell generates lists in a really neat way.

Let's say I want a list of all the integers between 1 and 1000:

Prelude> [1..100]
[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100]

That's it.  Haskell's got it covered.

For my next trick, let's say I want a list of every second odd integer between 1 and 1000:

Prelude>[1,3..100]
[1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,49,51,53,55,57,59,61,63,65,67,69,71,73,75,77,79,81,83,85,87,89,91,93,95,97,99]

Neat, eh?

I thought, "OK, let's get crazy with this.  I want the difference between each integer on the list to be one greater than the previous difference, i.e...

1,3,6,10,15

And so forth.

So I tried:

Prelude> [1,3,6..100]

And got this error message:
:1:7: parse error on input `..'

I wonder how I can convey this in Haskell?  Oh well, something to look forward to.  For now, it's a lot of fun, but it's time for me to get to bed.