Modular Arithmetic in Haskell

Question

How would I go about making a function so that x has a range of values from x=0 to x=19 and if the x value exceeds 19 or is below zero how can I get it to wrap around

From: x=20, x=21, x=22 and x=(-1), x=(-2), x=(-3)

To: x=0, x=1, x=2 and x=19, x=18, x=17 respectively?

I've heard of modular arithmetic which is apparently the way I should deal with it.

score 7 · Accepted Answer · answered May 20 '11 at 13:11

7

Usually you would use the built-in functions mod and rem, but I assume they are off-limits for homework. So you can write your own function, e.g.

mod20 x | x < 0 = ...
        | x > 19 = ...
        | otherwise = x

There are different things you can try to fill in the ...s. One of the easiest is repeated addition or subtraction, but I don't want to spoil all the fun.

Once you have this function, you can "rescale" the values after every "normal" arithmetic operation, e.g. mod20 (12 + 17).

answered May 20 '11 at 13:11

Landei

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are `mod` and `rem` exactly the same things? They seem to be. – maclunian May 20 '11 at 13:18
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`mod` and `rem` behave differently when fed negative numbers; for positive numbers they're the same. – dave4420 May 20 '11 at 13:31

score 3 · Answer 2 · edited May 20 '11 at 22:06

3

Try using the mod function:

(-5) `mod` 20 ==> 15
5 `mod` 20 ==> 5
20 `mod` 20 ==> 0
25 `mod` 20 ==> 5

See also wikipedia on the topic.

edited May 20 '11 at 22:06

luqui

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answered May 20 '11 at 13:08

aleator

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`mod (-5) 20 = 15`, `rem (-5) 20 = 5` – augustss May 20 '11 at 13:10
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25 `mod` 25 is not equal to 5 – maclunian May 20 '11 at 13:20

score 1 · Answer 3 · answered May 20 '11 at 13:09

1

Use

x `mod` 20

(This is a filler to make the answer 30 characters.)

answered May 20 '11 at 13:09

augustss

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Modular Arithmetic in Haskell

3 Answers3