# Gratuitous Mathematics

robin Monday, August 23 2004 @ 08:46 PM BST Views: 4,612

Here's a good one for all you maths fiends...

Map the five values `1..5`

to the range `10..50`

. Easy, huh? (Just in case you're not clear the answer I'm looking for is `10, 20, 30, 40, 50`

!)

What if the range starts at zero? i.e. map `0..5`

to `10..50`

? Hmmm, a little more tricky but do-able. Answer: `10, 18, 26, 34, 42, 50`

.

Now, how about mapping the values `0..40`

to the range `0..1`

? Erm...

Actually, it's just a case of finding a generic formula to calculate the new values from the old values. To map a series of consecutive integer values Imin .. Imax to a range of continous real values Rmin .. Rmax, I came up with the following formula:

( Rmax - Rmin )( Ii - Imin ) Ri = ---------------------------- ( Imax - Imin )where:

`Ri`

is the ith value in the range`Rmax`

is the maximum value of the range`Rmin`

is the minimum value of the range`Ii`

is the ith integer in the series`Imin`

is the 1st integer in the series`Imax`

is the last integer in the series

`0..40`

to the range `0..1`

.In this example:

- Rmax = 1
- Rmin = 0
- Imax = 40
- Imin = 0

R1 = (1 - 0)( 0 - 0) / (40 - 0) = 0 R2 = (1 - 0)( 1 - 0) / (40 - 0) = 1/40 = 0.025 .. R40 = (1 - 0)(39 - 0) / (40 - 0) = 39/40 = 0.975 R41 = (1 - 0)(40 - 0) / (40 - 0) = 40/40 = 1So, now you know.

Gratuitous Mathematics| 0 comments