Recursive functions

We have defined functions in terms of other functions. Besides this essentially the only way to define functions is recursion. In Dependent Type Theory recursion/induction is very powerful.

In particular, unlike imperative programming all loops are implemented in terms of recursion.

The first example is the factorial function:

def fctrl (n : ℕ) : ℕ

The factorial function

Equations

• fctrl 0 = 1
• fctrl (Nat.succ n_2) = (n_2 + 1) * fctrl n_2

We check by evaluating this

#eval fctrl 5 -- 120

Fibonacci numbers:

These are given by the recursive definition

• fib 0 = 1
• fib 1 = 1
• fib (n + 2) = fib n + (fib (n + 1))

def fib : ℕ → ℕ

Equations

• fib 0 = 1
• fib 1 = 1
• fib (Nat.succ (Nat.succ n)) = fib n + fib (n + 1)

The above definition is not efficient as a single computation is called many times.

We can instead define pairs, so if (a, b) = (fib n, fib (n + 1)) then (fib (n + 1), fib (n + 2)) = (b, a+ b)

To define Fibonacci pairs as above, we have two choices. We can view the pair at n as obtained by n iterations of the function g: (a, b) ↦ (b, a + b) starting with the pair (1,1). Note that we can recursively use either g^(n + 1) = g^n ∘ g or g^(n + 1) = g ∘ g^n. The former is more efficient as it means the recursive function is called at the end to give a result, with modified arguments. This allows so called tail call optimization, where new copies of the function are not created.

def fibAux (a : ℕ) (b : ℕ) (n : ℕ) : ℕ × ℕ

Equations

• fibAux a b 0 = (a, b)
• fibAux a b (Nat.succ n_2) = fibAux b (a + b) n_2

def fib' (n : ℕ) : ℕ

Equations

• fib' n = (fibAux 1 1 n).1

The following definition is clearly wrong and will loop infinitely. Indeed if we attempt to define

def silly_fib : ℕ → ℕ
| 0 => 1
| 1 => 1
| n + 2 => silly_fib n + (silly_fib (n + 3))

we get the error message

fail to show termination for
  silly_fib
with errors
argument #1 was not used for structural recursion
  failed to eliminate recursive application
    silly_fib (n + 3)

structural recursion cannot be used

failed to prove termination, use `termination_by` to specify a well-founded relation

partial def silly_fib : ℕ → ℕ

However even if functions terminate we can still get such an error. This is because Lean does not know that the function is terminating. We can fix this by adding the partial keyword but better still we can prove termination.

Thus, the following definition is correct but gives an error.

def hcf (a b : ℕ) : ℕ :=
  if b < a then hcf b a
  else
    if a = 0 then b
    else hcf a (b - a)

#eval hcf 18 12 -- 6

def hcf (a : ℕ) (b : ℕ) : ℕ

Equations

• hcf a b = if c : b < a then hcf b a else if c'' : a = 0 then b else let_fun x := ⋯; hcf a (b - a)

partial def wrong(n: ℕ): Empty :=
  wrong (n + 1)

gives the error message

failed to compile partial definition 'wrong', failed to show that type is inhabited and non empty

Lean has to allow partial definitions due to deep results of Church-Gödel-Turing-..., which say for example that we cannot prove that a Lean interpreter in Lean terminates.