3.1. Examples Chapter 3 Programming with Recursion Factorial - - PDF document

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3.1. Examples Chapter 3 Programming with Recursion Factorial - - PDF document

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Ch.3: Programming with Recursion Ch.3: Programming with Recursion Plan 3.1. Examples 3.1. Examples Chapter 3 Programming with Recursion Factorial (revisited from Section 2.13) 1. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . .

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SLIDE 1

Ch.3: Programming with Recursion 3.1. Examples

3.1. Examples

Factorial (revisited from Section 2.13)

Program (fact.sml) fun fact n = if n < 0 then error "fact: negative argument" else if n = 0 then 1 else n ∗ fact (n−1) Useless test of the error case at each recursive call! Hence we introduce an auxiliary function, and can then use pattern matching in its declaration: fun factAux 0 = 1 | factAux n = n ∗ factAux (n−1) fun fact1 n = if n < 0 then error "fact1: negative argument" else factAux n In fact1: pre-condition verification (defensive programming) In factAux: no pre-condition verification

Sven-Olof Nystr¨

  • m/IT Dept/Uppsala University

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3.2

Ch.3: Programming with Recursion Plan

Chapter 3 Programming with Recursion

  • 1. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2
  • 2. Correctness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5
  • 3. Construction methodology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.13
  • 4. Forms of recursion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.16

Sven-Olof Nystr¨

  • m/IT Dept/Uppsala University

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3.1

Ch.3: Programming with Recursion 3.1. Examples

Triangle

Specification function triangle a b TYPE: int → int → int PRE: (none) POST:

  • a ≤ i ≤ b

i Construction Base case: a > b : return 0 General case: a ≤ b : return

  • a≤i≤b

i = a +

  • a+1≤i≤b

i = a + triangle (a+1) b Program (triangle.sml) fun triangle a b = if a > b then 0 else a + triangle (a+1) b

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  • m/IT Dept/Uppsala University

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3.4

Ch.3: Programming with Recursion 3.1. Examples

Exponentiation

Specification function expo x n TYPE: real → int → real PRE: n ≥ 0 POST: xn Construction Error case: n < 0: produce an error message Base case: n = 0 : return 1 General case: n > 0 : return xn = x ∗ xn−1 = x ∗ expo x (n−1) Program (expo.sml) fun expoAux x 0 = 1.0 | expoAux x n = x ∗ expoAux x (n−1) fun expo x n = if n < 0 then error "expo: negative argument" else expoAux x n

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  • m/IT Dept/Uppsala University

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3.3

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SLIDE 2

Ch.3: Programming with Recursion 3.2. Correctness

Correctness (cont) What does this function compute? fun f n = if n = 0 then 0 else 1 + f(n-1) Answer: f n = n, if n>=0 Seems reasonable, but how do we prove it?

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3.6

Ch.3: Programming with Recursion 3.2. Correctness

3.2. Correctness

How do we know that a recursive program computes what we want? Lets try a very simple program. fun f n = if n = 0 then 0 else 1 + f(n-1) What does f compute?

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3.5

Ch.3: Programming with Recursion 3.2. Correctness

Another example, slightly harder What does this function compute? fun g(0) = 0 | g(n) = n + g(n-1)

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3.8

Ch.3: Programming with Recursion 3.2. Correctness

Proof by induction For all n >= 0, f(n) = n. Base case: n = 0. f(n) = 0 = n follows immediately. Induction step: Hypothesis: f(k) = k, for all k such that 0 <= k < n We have f(n) = 1 + f(n − 1), by definition 1 + f(n − 1) = 1 + (n − 1), by induction hypothesis 1 + (n − 1) = n, by algebraic transformations Thus f(n) = n It follows that f(n) = n, for all n >= 0. Question: What can we say about f(−1)?

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  • m/IT Dept/Uppsala University

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3.7

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SLIDE 3

Ch.3: Programming with Recursion 3.2. Correctness

If you think these examples where too simple Exercises:

  • Prove that the factorial given earlier produces the intended results.
  • Prove the same for the gcd function.

What do the functions below compute? Make an educated guess (try on an SML system if you like) and show by induction that your guess was correct. fun h(n) = if n = 0 then 0 else h(n-1)+2*n-1; fun m(a,b) = if a = 0 then 0 else b+m(a-1,b);

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3.10

Ch.3: Programming with Recursion 3.2. Correctness

Another example (cont) What does this function compute? fun g(0) = 0 | g(n) = n + g(n-1) Answer: g(n) = (n ∗ n + n)/2 Proof by induction. For all n >= 0, g(n) = (n ∗ n + n)/2 Base case: n = 0. g(n) = (n ∗ n + n)/2 by definition (n ∗ n + n)/2 = (0 ∗ 0 + 0)/2 = 0 Induction step Hypothesis: g(k) = (k ∗ k + k)/2, for all k such that 0 <= k < n We have g(n) = n + g(n − 1), by definition = n + ((n − 1) ∗ (n − 1) + (n − 1))/2, by induction hypothesis = n + (n ∗ n − 2 ∗ n + 1 + n − 1)/2, by algebraic transformations = n + (n ∗ n − n)/2 − n, simplification = (n ∗ n + n)/2 Thus g(n) = (n ∗ n + n)/2 It follows that g(n) = (n ∗ n + n)/2, for all n >= 0.

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  • m/IT Dept/Uppsala University

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3.9

Ch.3: Programming with Recursion 3.2. Correctness

Example (factorial): fun fac(x) = if x = 0 then 1 else x*fac(x-1) Assume that fac(x) terminates for all x >= 0. Show that P(x, fac(x)) <=> fac(x) = 1 ∗ 2 ∗ . . . ∗ x holds. Assuming property holds for recursive calls... By the definition of the function fac we have fac(x) = if x = 1 then 0 else x ∗ (1 ∗ 2 ∗ . . . ∗ (x − 1)) For x = 0 the property is obvious. For x <> 0 we prove the property by algebraic manipulations.

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  • m/IT Dept/Uppsala University

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3.12

Ch.3: Programming with Recursion 3.2. Correctness

Correctness of functional programs Suppose we have a recursive function fun f(x) = ... f(..) ... and we want to show that it satisfies some property P(x, f(x)) Solution:

  • Show that the function terminates (for all values of x that we are

interested in)

  • Assume that all recursive calls satisfy the property P(. . . , f(. . .)).

Show P(x, f(x)). Then we have shown that P(x, f(x)) holds for all values of x we are interested in. (This is just an induction proof in disguise.) (By the way, showing that if an answer is returned it will be correct is also known as partial correctness.)

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  • m/IT Dept/Uppsala University

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3.11

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SLIDE 4

Ch.3: Programming with Recursion 3.4. Variants in general

3.4. Variants in general

Generally, a variant may be any expression whose set of possible values is some set A.

  • For each recursive call, the variant should decrease strictly.
  • There should be no infinite descending chain x0 > x1 > x2 > . . . in A.

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3.14

Ch.3: Programming with Recursion 3.3. Construction methodology

3.3. Construction methodology

Objective Construction of an SML program computing the function: f(x) : D → R given its specification S Methodology

  • 1. Case analysis

Recognize

  • base case(s), and
  • recursive case(s).
  • 2. Partial correctness

Check that the base case returns correct result. Show that the recursive case gives correct result, assuming that all recursive calls give correct result.

  • 3. Termination

Choose a variant. A variant is typically an integer-valued expression on the input parameters together with a lower bound b so that

  • for any recursive call, the variant is ≥ b, and
  • for any recursive call, the variant of the recursive call is strictly less

than the current call.

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  • m/IT Dept/Uppsala University

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3.13

Ch.3: Programming with Recursion 3.6. Forms of recursion

3.6. Forms of recursion

Up to now:

  • One recursive call
  • Some variant is decremented by one

That is: simple recursion (construction process by simple induction) Forms of recursion

  • Simple recursion
  • Complete recursion
  • Multiple recursion
  • Mutual recursion
  • Nested recursion
  • Recursion on a generalised problem

Sven-Olof Nystr¨

  • m/IT Dept/Uppsala University

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3.16

Ch.3: Programming with Recursion 3.5. Tips and hints

3.5. Tips and hints

Superfluous or overlapping cases Try to keep your code short and compact. Avoid redundant cases! fun fact n = if n = 0 then 1 else if n = 1 then 1 else n ∗ fact (n−1) Variants Variants are usually simple; an integer given by a parameter or the size

  • f some input data. But watch out for the difficult cases!

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  • m/IT Dept/Uppsala University

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3.15

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SLIDE 5

Ch.3: Programming with Recursion 3.6. Forms of recursion

Program (intDiv.sml) fun intDivAux a b = if a < b then (0,a) else let val (q1,r1) = intDivAux (a−b) b in (q1+1,r1) end fun intDiv a b = if a < 0 orelse b <= 0 then error "intDiv: invalid argument" else intDivAux a b Necessity of the induction hypothesis not only for a−1, but actually for all values less than a: complete induction!

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  • m/IT Dept/Uppsala University

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3.18

Ch.3: Programming with Recursion 3.6. Forms of recursion

Complete recursion

Example 1: integer division (quotient and remainder) Specification function intDiv a b TYPE: int → int → (int ∗ int) PRE: a ≥ 0 and b > 0 POST: (q,r) such that a = q ∗ b + r and 0 ≤ r < b Construction Variant: a Error case: a < 0 or b ≤ 0 : produce an error message Base case: a < b : since a = 0 ∗ b + a, return (0,a) This covers more than the minimal value of a (namely 0)! General case: a ≥ b : since a = q ∗ b + r iff (a−b) = (q−1) ∗ b + r, the call intDiv (a−b) b will give q−1 and r

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3.17

Ch.3: Programming with Recursion 3.6. Forms of recursion

Program (expo.sml) fun fastExpo x n = let fun fastExpoAux x 0 = 1.0 | fastExpoAux x n = let val r = fastExpoAux x (n div 2) in if even n then r ∗ r else x ∗ r ∗ r end in if n < 0 then error "fastExpo: negative argument" else fastExpoAux x n end Complete recursion, but the size of the input is divided by 2 each time! Complexity Let C(n) be the number of multiplications made (in the worst case) by fastExpo: C(0) = 0 C(n) = C(n div 2) + 2 for n > 0 One can show that C(n) = O(log n)

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3.20

Ch.3: Programming with Recursion 3.6. Forms of recursion

Example 2: exponentiation (revisited from Section 3.1) Specification function fastExpo x n TYPE: real → int → real PRE: n ≥ 0 POST: xn Construction Variant: n Error case: n < 0: produce an error message Base case: n = 0 : return 1 General case: n > 0 : if n is even, then return xn div 2 ∗ xn div 2

  • therwise, return x ∗ xn div 2 ∗ xn div 2

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  • m/IT Dept/Uppsala University

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3.19

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SLIDE 6

Ch.3: Programming with Recursion 3.6. Forms of recursion

Mutual recursion

Example: recognising even integers and odd integers Specification function even n TYPE: int → bool PRE: n ≥ 0 POST: true if n is even false

  • therwise

function odd n TYPE: int → bool PRE: n ≥ 0 POST: true if n is odd false

  • therwise

Program (even.sml) Variant: n fun even 0 = true | even n = odd (n−1) and odd 0 = false | odd n = even (n−1)

  • Simultaneous declaration of the functions
  • Global correctness reasoning

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  • m/IT Dept/Uppsala University

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3.22

Ch.3: Programming with Recursion 3.6. Forms of recursion

Multiple recursion

Example: the Fibonacci numbers Definition fib (0) = 1 fib (1) = 1 fib (n) = fib (n−1) + fib (n−2) for n > 1 Specification function fib n TYPE: int → int PRE: n ≥ 0 POST: fib (n) Program (fib.sml) Variant: n fun fib 0 = 1 | fib 1 = 1 | fib n = fib (n−1) + fib (n−2)

  • Double recursion
  • Inefficient: multiple recomputations of the same values!

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  • m/IT Dept/Uppsala University

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3.21

Ch.3: Programming with Recursion 3.6. Forms of recursion

Example 2: Graham’s number, the “largest” number Definition Operator ↑n (invented by Donald Knuth): a ↑1 b = ab a ↑n b = a ↑n−1 (b ↑n−1 b) for n > 1 Program (graham.sml) Variant: n fun

  • pKnuth 1 a b = Math.pow (a,b)

|

  • pKnuth n a b = opKnuth (n−1) a (opKnuth (n−1) b b)
  • pKnuth 2 3.0 3.0 ;

val it = 7.62559748499E12 : real

  • pKnuth 3 3.0 3.0 ;

! Uncaught exception: Overflow Graham’s number is opKnuth 63 3.0 3.0 It is in the Guiness Book of Records!

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  • m/IT Dept/Uppsala University

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3.24

Ch.3: Programming with Recursion 3.6. Forms of recursion

Nested recursion and lexicographic order

Example 1: the Ackermann function Definition For m, n ≥ 0: acker (0,m) = m+1 acker (n,0) = acker (n−1, 1) for n > 0 acker (n,m) = acker (n−1, acker (n, m−1)) for n, m > 0 Program (acker.sml) Variant: the pair (n,m) fun acker 0 m = m+1 | acker n 0 = acker (n−1) 1 | acker n m = acker (n−1) (acker n (m−1)) where (n′, m′) <lex (n, m) iff n′ < n or (n′ = n and m′ < m) This is called a lexicographic order

  • The function acker always terminates
  • It can be shown that Ackermann’s function can not be expressed as a

primitive-recursive function.

Sven-Olof Nystr¨

  • m/IT Dept/Uppsala University

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3.23

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SLIDE 7

Ch.3: Programming with Recursion 3.6. Forms of recursion

Specification of the generalised function function divisors n low up TYPE: int → int → int → bool PRE: n, low, up ≥ 1 POST: true if n has no divisors in {low, . . . , up} false otherwise Construction of a program for the generalised function Variant: up − low + 1, which is the size of {low, . . . , up} Base case: low > up : return true because the set {low, . . . , up} is empty General case: low ≤ up : if n is divisible by low, then return false

  • therwise, return whether n has a divisor in {low+1, . . . , up}

Construction of a program for the original function The function prime is a particular case of the function divisors, namely when low is 2 and up is n−1 One can also take up as ⌊√n⌋, and this is more efficient

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3.26

Ch.3: Programming with Recursion 3.6. Forms of recursion

Recursion on a generalised problem

Example: recognising prime numbers Specification function prime n TYPE: int → bool PRE: n > 0 POST: true if n is a prime number false

  • therwise

Construction It is impossible to determine whether n is prime via the reply to the question “is n − 1 prime”? It seems impossible to directly construct a recursive program We thus need to find another function:

  • that is more general than prime, in the sense

that prime is a particular case of this function

  • for which a recursive program can be constructed

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  • m/IT Dept/Uppsala University

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3.25

Ch.3: Programming with Recursion 3.6. Forms of recursion

Program (prime.sml) fun divisors n low up = low > up

  • relse

(n mod low) <> 0 andalso divisors n (low+1) up fun prime n = if n <= 0 then error "prime: non-positive argument" else if n = 1 then false else divisors n 2 (floor (Math.sqrt (real n)))

  • The function divisors may be useful for other problems
  • The discovery of divisors requires imagination and creativity
  • There are some standard methods of generalising problems:

– let the recursive function take more parameters, so that the problem we want to solve is a special case – let the recursive function return more information than is required in the problem statement – tupling generalisation: replace a parameter by a list of parameters of the same type

  • Sometimes the only way to solve the problem at hand is to consider a

more general problem.

  • Often, a generalized problem may be more efficient regarding time

and/or space consumption. Exercises:

  • 1. Which standard method of generalization did we apply in the

implementation of the primes function?

  • 2. (Harder) Implement a function that computes fib(n) in time

proportional to n. Hint: You will need to define a function that solves a more general problem.

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  • m/IT Dept/Uppsala University

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3.27