Haskell Basics

Question 1

17. .
23. . Note that * has higher precedence than +.
1. Exponentiation has a higher precedence than modulo (non-operator functions like mod that are called in an infix manner can have a well-defined operator precedence level).
24.25. Regular division of integers gives a Fractional type.
24. The div function is similar to // in Python.
1. First we evaluate the condition let x = 3 in x + 3 evaluates to 3 + 3 which therefore is 6. Clearly 6 /= 5 is true, so we need to also evaluate 3 < 4, which is also true. && is the same as and in Python, so True and True is therefore True. Thus, the whole expression evaluates to the if branch, which is 1.
False. otherwise is actually just True by definition, so not True becomes False.
It actually causes a compile-time error since it is a type error. fst and snd receive pairs, so these functions do not work on triples.
1.5. The succ function returns the successor of any enumerable type. For numbers, this would be one more than the number.
1.4142135623730951. Straightforward. Notice that Haskell’s Prelude (the built-in stuff) comes with many math functions.
True. The elem function is similar to in in Python.
4. When writing let bindings in a single line, we can separate multiple definitions with ;. Therefore, we have defined two functions f and g which add one and multiply by 2 respectively. The . operator is function composition, where , so (g . f) 1 is the same as g (f 1), which evaluates to 4.
[1, 2, 3, 4, 5, 6]. This is straightforward, since ++ concatenates two lists.
1. head returns the first element of the list.
[2, 3]. tail returns the suffix of the list without the first element.
[1, 2]. init returns the list without the last element.
1. !! is indexing.
True. null checks whether a list is empty.
3. Obvious.
[3]. drop n drops the first n elements of a list.
[-1, 0, 1, 2, 3]. take n takes the first n elements of a list. The range [-1..] is an infinitely long range from -1 to infinity.
[5, 1, 2 ,3]. dropWhile f will drop elements from a list until f returns false for an element.
30. The easiest way to see this is by converting this to the equivalent Python expression:
Python interactive session
```
>>> sum([x[0] for x in
            [(i, j) for i in range(1, 5)
                    for j in range(-1, 2)]])
```
Going back to Haskell land, let us evaluate the inner list first. [(i, j) | i <- [1..4], j <- [-1..1]] gives [(1, -1), (1, 0), (1, 1), (2, -1), ..., (4, 1)] then, [fst x | x <- ...] would therefore give [1,1,1,2,2,2,3,3,3,4,4,4] which sums to 30.

Question 2

Idea: take the last elements of both lists, and check for equality. For this, we can use the last function.

1
eqLast xs ys = last xs == last ys

Question 3

Idea: reverse the string, and check if the string and its reverse are equal. For this, we can use the reverse function.

1
isPalindrome w = w == reverse w

Question 4

1
taxiFare f r d = f + r * d

Question 5

There are several ways to approach this problem. Let us first define the ingredientPrice function which should be straightforward to do.

1
ingredientPrice i
2
  | i == 'B' = 0.5
3
  | i == 'C' = 0.8
4
  | i == 'P' = 1.5
5
  | i == 'V' = 0.7
6
  | i == 'O' = 0.4
7
  | i == 'M' = 0.9

Then we can define burgerPrice recursively. If the string is empty then the price is 0. Otherwise, take the price of the first ingredient and add that to the price of the remaining burger.

8
burgerPrice burger
9
  | null burger = 0
10
  | otherwise =
11
      let first = ingredientPrice (head burger)
12
          rest  = burgerPrice (tail burger)
13
      in  first + rest

Of course, we know that we can do the following in Python quite nicely:

1
def burger_price(burger):
2
    return sum(ingredient_price(i) for i in burger)

This can be done in Haskell too as follows:

8
burgerPrice burger = sum [ingredientPrice i | i <- burger]

We can also replace the comprehension expression in Python using map:

1
def burger_price(burger):
2
    return sum(map(ingredient_price, burger))

Haskell also has a map (or fmap) function that does the same thing:

8
burgerPrice burger = sum $ map ingredientPrice burger

The $ sign is just regular function application, except that $ binds very weakly. So sum $ map ingredientPrice burger is basically sum (map ingredientPrice burger).

Finally, notice that burgerPrice x = sum ((map ingredientPrice) x), so effectively we can finally define our function this way:

1
burgerPrice = sum . map ingredientPrice
2
  where ingredientPrice i
3
          | i == 'B' = 0.5
4
          | i == 'C' = 0.8
5
          | i == 'P' = 1.5
6
          | i == 'V' = 0.7
7
          | i == 'O' = 0.4
8
          | i == 'M' = 0.9

To see this, let be burgerPrice, be sum and be map ingredientPrice. We have shown that

By definition,

This style of writing functions is known as point-free style, where functions are expressed as a composition of functions.

Question 6

Again, there are several ways to solve this. To do so numerically, we can define our function recursively:

1
sumDigits n
2
  | n < 10    = n
3
  | otherwise = n `mod` 10 + sumDigits (n `div` 10)

Alternatively, we may convert n into a string, convert each character into integers, then obtain the sum. This might be expressed in Python as:

1
def sum_digits(n):
2
    return sum(map(int, str(n)))

Converting n into a string can be done by show:

ghci> show 123
"123"

Converting back into an integer can be done with read (you have to explicitly state the output type of the read function since this can be ambiguous):

ghci> read "123" :: Int
123

However, we can’t read from characters since the read function receives strings. Good thing that strings are lists of characters, so by putting the character in a list, we now obtain the ability to read a digit (as a character) as an integer.

ghci> read '1' :: Int
-- error!
ghci> read ['1'] :: Int
1

To put things into lists, we can use the return function!

ghci> return '1' :: String
"1"
ghci> (read . return) '1' :: Int
1

Thus, the read . return function allows us to parse each character into an integer. Combining this with what we had before, we can obtain the list of the digits (as integers) from n using:

ghci> [(read . return) digit | digit <- show 123] :: [Int]
[1, 2, 3]

Again, we can use map instead of list comprehension.

ghci> map (read . return) (show 123) :: [Int]
[1, 2, 3]

Obtaining the sum of this list gives us exactly what we want. Thus, our sumDigits function is succinctly defined as follows:

sumDigits = sum . map (read . return) . show

Question 7

Idea: drop the first start elements, then take the stop - start elements after that.

1
ls @: (start, stop) = take (stop - start) (drop start ls)