Review

Generalizing Recursive Math Operations

Last class I posed the question: Can we generalize the pattern of defining (*) in terms of (+), (^) in terms of (*), and so on?

Generalizing Recursive Math Operations

First let me remind you of the definition of (***) and (^^^):

(***) :: Int -> Int -> Int
m *** 0 = 0
m *** n = m + (m *** (n - 1))

(^^^) :: Int -> Int -> Int
m ^^^ 0 = 1
m ^^^ n = m *** (m ^^^ (n - 1))

Exercise: Generalize the Pattern

Try to generalize this pattern by first defining a function hyper which takes three arguments:

• x the level of recursion (level 1 is addition, level 2 is multiplication and so on)
• m is the “base”
• n is the “exponent”

hyper 2 2 3 = 6     = 2 * 3     = 2 + 2 + 2
hyper 3 2 3 = 8     = 2 ^ 3     = 2 * 2 * 2
hyper 4 2 3 = 16    = 2 ^!^ 3   = 2 ^ (2 ^ 2)
hyper 5 2 3 = 65536 = 2 ^!!^ 3  = 2 ^!^ (2 ^!^ 2)
hyper 6 2 3 = BIG#  = 2 ^!!!^ 3 = 2 ^!!^ (2 ^!!^ 2)

Exercise: Generalize the Pattern

Step 1: Define the type

hyper :: Int -> Integer -> Integer -> Integer

Exercise: Generalize the Pattern

Step 2: Enumerate the cases

Ask your self what are we recursing on?

hyper 1 m n = ...
hyper 2 m n = ...
hyper 3 m n = ...
hyper x m 0 = ...
hyper x m 1 = ...
hyper x m n = ...

Exercise: Generalize the Pattern

Step 3: Define the simple cases

hyper 1 m n = m + n
hyper 2 m n = m * n
hyper 3 m n = m ^ n
hyper x m 0 = 1
hyper x m 1 = m
hyper x m n = ...

Exercise: Generalize the Pattern

Step 4: Define the other cases

hyper 1 m n = m + n
hyper 2 m n = m * n
hyper 3 m n = m ^ n
hyper x m 0 = 1
hyper x m 1 = m
hyper x m n = hyper (x - 1) m (hyper x m (n - 1))

Exercise: Generalize the Pattern

Step 4: Define the other cases

hyper 1 m n = m + n
hyper 2 m n = m * n
hyper 3 m n = m ^ n
hyper x m 0 = 1
hyper x m 1 = m
hyper x m n = hyper (x - 1) m   (hyper x m   (n - 1))
              m `hyper (x - 1)` (m `hyper x` (n - 1))
              m ***             (m ^^^       (n - 1))

Exercise: Generalize the Pattern

Step 5: Generalize and simplify

You could generalize the type to use Num, Integral and Eq type classes.

hyper :: (Num a, Integral b, Eq a, Eq b) => a -> b -> b -> b

Exercise: Generalize the Pattern

Step 6: Test the function

hyper 1 2 3 === 5
hyper 2 2 3 === 6
hyper 3 2 3 === 8
hyper 4 2 3 === 16
hyper 5 2 3 === 65536
hyper 6 2 3 === BIG#

Aside: Graham’s Number

Graham’s number is an extremely large number which can be expressed with the following function:

$$ g(n) = \left\{ \begin{array}{ll} 3 \uparrow\uparrow\uparrow\uparrow 3 & \text{if } n = 1 \\ 3 \uparrow^{g(n-1)} 3 & \text{if } n > 1, n \in \mathbb{N} \end{array} \right. $$

Aside: Graham’s Number

How do we express it with our hyper function?

Aside: Graham’s Number

Let’s try to define it:

graham :: Integer -> Integer
graham 1 = hyper 6 3 3
graham n = ...

Aside: Graham’s Number

What about the other case?

graham :: Integer -> Integer
graham 1 = hyper 6 3 3
graham n = hyper (graham (n - 1)) 3 3

Aside: Graham’s Number

Let’s try it:

> graham 1

Aside: Graham’s Number

Let’s try to get a sense for just how big even the base case of this is:

$$ \begin{array}{lllllll} \footnotesize{3 \uparrow 3} & \footnotesize{=} & \footnotesize{3 * (3 * 3)} & \footnotesize{=} & \footnotesize{3 * 9} & \footnotesize{=} & \footnotesize{27} \\ \footnotesize{3 \uparrow\uparrow 3} & \footnotesize{=} & \footnotesize{3 ^ {3 ^ 3}} & \footnotesize{=} & \footnotesize{3 ^ {27}} & \footnotesize{=} & \footnotesize{7625597484987} \\ \footnotesize{3 \uparrow\uparrow\uparrow 3} & \footnotesize{=} & \footnotesize{3 \uparrow\uparrow (3 \uparrow\uparrow 3)} & \footnotesize{=} & \footnotesize{3 ^ {7625597484987}} & \footnotesize{=} & \footnotesize{\text{BIG}} \\ \footnotesize{3 \uparrow\uparrow\uparrow\uparrow 3} & \footnotesize{=} & \footnotesize{3 \uparrow\uparrow\uparrow (3 \uparrow\uparrow\uparrow 3)} & \footnotesize{=} & \footnotesize{3 \uparrow\uparrow\uparrow \text{BIG}} & \footnotesize{=} & \footnotesize{\text{REALLY BIG}} \\ \end{array} $$