Comparison chain

Categories: Mathematics | FAQ | Idioms | Code

Contents 1 Problem 2 Solutions 2.1 simple 2.2 complex

1 Problem

Question: The compiler doesn't accept a <= x <= b. Why?

Answer: The expression cannot be parsed, because the infix symbol <= has no (left or right) associativity.

In languages like C the expression is parsed as (a <= x) <= b which is even worse. The first part is evaluated to a boolean value, which is then compared with b. (For C "boolean" and "integer" are the same type.)

2 Solutions

2.1 simple

• You must be aware, that the mathematical notation is shorthand for . Consequently a possible Haskell solution is a <= x && x <= b.
• Another fine mathematical notation is . You can roll your own function

isInRange :: Ord a => a -> a -> a -> Bool
isInRange lower upper x = lower <= x && x <= upper

to capture this notation (isInRange a b x), or

(<?) :: Ord a => a -> (a,a) -> Bool
(<?) = flip (uncurry isInRange)

(x <? (a,b)). In case of integers you can use the inRange function from Ix class.

• You can easily write a function, which checks if a list of numbers increases monotonicly.

monotonicIncreasing :: Ord a => [a] -> Bool
monotonicIncreasing xs = and (zipWith (<=) xs (tail xs))

You can use that for the initial problem by monotonicIncreasing [a,x,b].

2.2 complex

• For more complex checks of whether an element is contained in some ranges, you should have a look at the ranged sets library.
• If you want to program more complex chains with different kinds of comparisons, try the following code.

module ChainRelation where
 
{- * chains of relations (comparison, subsets, logical implications etc.) -}
 
infixr 4 &-, -&
 
type Rel   a = (a -> a -> Bool)
type Chain a = [(Rel a, a)]
 
endChain :: Chain a
endChain = []
 
-- separate comparison and operand
(&-) :: Rel a -> (a, Chain a) -> Chain a
rel &- (x,xs) = (rel,x):xs
 
-- separate operand and comparison
(-&) :: a -> Chain a -> (a, Chain a)
(-&) = (,)
 
-- check if all comparisons are true
check :: (a, Chain a) -> Bool
check (x,chain) =
   let (rels,xs) = unzip">unzip chain
   in  and (zipWith3 id rels (x:xs) xs)
 
 
example1 :: Bool
example1 =
   check (1 -& (<) &- 5 -& (==) &- 5 -& (<=) &- 10 -&
      (endChain :: Chain Integer))
 
 
{- * specialised infix operators for comparison -}
 
infixr 4 ==:, /=:, <:, >:, <=:, >=:
 
(==:), (/=:), (<:), (>:), (<=:), (>=:) :: Ord a =>
   a -> (a, Chain a) -> (a, Chain a)
(==:) = lift (==)
(/=:) = lift (/=)
(<:)  = lift (<)
(>:)  = lift (>)
(<=:) = lift (<=)
(>=:) = lift (>=)
 
lift :: Rel a -> a -> (a, Chain a) -> (a, Chain a)
lift f x (y,chain) = (x, (f,y):chain)
 
example2 :: Bool
example2 =
   check (1 <: 5 ==: 5 <=: 10
             -& (endChain :: Chain Integer))

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