1 / 43

Function Definition by Cases and Recursion

Function Definition by Cases and Recursion. Lecture 2, Programmeringsteknik del A. Definitions Revisited. A definition double :: Int -> Int double x = 2*x makes a true statement about the function defined, (whatever x is, then double x and 2*x are equal)

meli
Download Presentation

Function Definition by Cases and Recursion

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Function Definition by Cases and Recursion Lecture 2, Programmeringsteknik del A

  2. Definitions Revisited • A definition • double :: Int -> Int • double x = 2*x • makes a true statement about the function defined, • (whatever x is, then double x and 2*x are equal) • gives a way of computing calls of the function.

  3. Quiz Given the definition x :: Int x*x = 4 Is x equal to 2?

  4. Quiz Given the definition x :: Int x*x = 4 Is x equal to 2? NO! This is not a valid Haskell definition. It makes a true statement about x, but it does not give a way of computing x.

  5. Computing with Definitions • A function call is computed by • replacing the call with a copy of the right hand side, • with the argument names replaced by the actual arguments. double :: Int -> Int double x = 2*x double 8 2*8 16

  6. Evaluation Order There may be more than one way to evaluate an expression: double 8 double (3+5) 2*8 16 2*(3+5) You can use any order of evaluation; they all give the same result. Haskell chooses a suitable one; you don’t need to know which.

  7. Sharing Evaluation double :: Int -> Int double x = x+x double (3*5) Is it more work to evaluate the expression in this order? (3*5)+(3*5) double 15 15+(3*5) 15+15 30

  8. Sharing Evaluation double :: Int -> Int double x = x+x double (3*5) NO! Haskell `remembers´ that both occurrences of 3*5 are really the same, and evaluates both in one step. (3*5)+(3*5) double 15 15+15 30

  9. Definition by Cases Often programs must make decisions, and compute different results in different cases. Example: Define max x y to return the maximum of its two arguments. If x <= y, then max x y should be y. If x>y, then max x y should be x.

  10. The Type of Booleans We make a decision by asking: does a condition hold? (e.g. Does x<=y hold?) A condition is either true or false: this is a piece of data, a value! We introduce a new basic type with two values, named after the mathematician George Boole: True, False :: Bool Constants begin with a capital letter.

  11. Some Operators Producing Booleans Note two equals signs, to avoid confusion with a definition. 2 <= 3 True 2 > 3 False 2 < 3 True 2 == 3 False 2 /= 3 True Not equals.

  12. Functions Returning Booleans Functions can return boolean results (or any other type). Example: inOrder :: Int -> Int -> Int -> Bool inOrder x y z = x <= y && y <= z a && b is True if both a and b are True.

  13. Using Booleans to Define Functions by Cases max :: Int -> Int -> Int max x y | x <= y = y max x y | x > y = x max :: Int -> Int -> Int max x y | x <= y = y | x > y = x OR A guard: an expression of type Bool. If the guard is True, the equation applies.

  14. Evaluation with Guards • To evaluate a function call, • evaluate each guard in turn until one is True, • replace the call with a copy of the right hand side following the true guard. max 4 2 ?? 4 <= 2 False ?? 4 > 2 True 4 max :: Int -> Int -> Int max x y | x <= y = y | x > y = x

  15. Is max Correct? Programming is a very error prone process; programs are rarely correct `first time´. A large part of the cost of software development goes on finding and correcting errors. It is essential to test software: try it on a variety of inputs and see if the output is correct.

  16. Choosing Test Data Test data should be chosen carefully, to include `difficult´ cases that might induce a failure. The max function should be tested at least with x<y, x==y, x>y, and probably combinations of positive and negative arguments. Choose enough test examples so that every case in your program is used at least once!

  17. Dijkstra on Testing ”Testing can never demonstrate the absence of errors in software, only their presence” Edsger W. Dijkstra (but it is very good at the latter).

  18. Specifications What do we mean by `max is correct´? A specification formulates properties we expect max to satisfy. Property: x <= max x y Property: y <= max x y

  19. Why Formulate Specifications? • Helps us clarify what max is supposed to do. • Can help in testing. • Enables us to prove programs correct.

  20. Specifications and Testing We can define function to check whether properties hold. prop_Max :: Int -> Int -> Bool prop_Max x y = x <= max x y && y <= max x y If prop_Max always returns True, then the specification is satisfied. We can test max on many inputs without needing to inspect the results by hand.

  21. Testing with QuickCheck QuickCheck is a tool to help you test your programs. Main> quickCheck prop_Max OK, passed 100 tests quickCheck generates random values to test your property thoroughly.

  22. Testing with QuickCheck (2) • What if we make a mistake? • max x y | x <= y = x • | x > y = y Main> quickCheck prop_Max Falsifiable, after 0 tests 1 0

  23. Specifications and Proofs From the definition of max: x <= y ==> max x y = y x > y ==> max x y = x Theorem: x <= max x y Proof: Consider two cases: Case x <= y: y = max x y, so x <= max x y. Case x > y: max x y = x and x <= x, so x <= max x y.

  24. Formal Methods • Proofs are costly and also error-prone, but can guarantee correctness. • Thorough testing is the most common method today. • Customers for safety critical software demand proofs today. • Proofs of correctness will play a growing role, thanks to • automatic tools to help with proving, • demand for better quality software.

  25. Quiz • Define • abs x to return the absolute value of x (e.g. abs 2 = 2, abs (-3) = 3. • sign x to return 1 if x is positive, and -1 if x is negative. • State (and prove?) a property relating abs and sign.

  26. Quiz Answer abs x | x <= 0 = -x | x > 0 = x sign x | x < 0 = -1 | x > 0 = 1 | x == 0 = 0 Property: x == sign x * abs x Did you consider this case? This can also be written sign 0 = 0

  27. Recursion Problem: define fac :: Int -> Int fac n = 1 * 2 * … * n What if we already know the value of fac (n-1)? Then fac n = 1 * 2 * … * (n-1) * n = fac (n-1) * n

  28. A Table of Factorials Must start somewhere: we know that fac 0 = 1. n fac n 0 1 1 1 2 2 3 6 4 24 ... So fac 1 = 1 * 1. So fac 2 = 1 * 2. So fac 3 = 2 * 3.

  29. A Recursive Definition of Factorial Base case. fac :: Int -> Int fac 0 = 1 fac n | n > 0 = fac (n-1) * n Recursive case.

  30. Evaluating Factorials fac :: Int -> Int fac 0 = 1 fac n | n > 0 = fac (n-1) * n fac 4 ?? 4 == 0 False ?? 4 > 0 True fac (4-1) * 4 fac 3 * 4 fac 2 * 3 * 4 fac 1 * 2 * 3 * 4 fac 0 * 1 * 2 * 3 * 4 1 * 1 * 2 * 3 * 4 24

  31. There is No Magic! What if we define fac :: Int -> Int fac n = div (fac (n+1)) (n+1) ? fac 4 div (fac 5) 5 div (div (fac 6) 6) 5 div (div (div (fac 7) 7) 6) 5 ... A true statement. Not a useful definition.

  32. Primitive Recursion • Define • f n in terms of f (n-1), for n > 0. • f 0 separately. • What if I already know the value of f (n-1)? • Can I compute f n from it?

  33. Quiz Define a function power so that power x n == x * x * … * x n times (Of course, power x n == x^n, but you should define power without using ^).

  34. Quiz Define a function power so that power x n == x * x * … * x n times Don’t forget the base case! power x 0 = 1 power x n | n > 0 = power x (n-1) * x Since this equals (x * x * … * x) * x n-1 times

  35. General Recursion What if I know the values of f x for all x less than n? Can I compute f n from them? Example x^(2*n) == (x*x)^n x^(2*n+1) == (x*x)^n * x

  36. Power Using General Recursion Base case is still needed. power :: Int -> Int -> Int power x 0 = 1 power x n | n `mod` 2 == 0 = power (x*x) (n `div` 2) | n `mod` 2 == 1 = power (x*x) (n `div` 2) * x Two recursive cases. Why might this definition of power be preferred?

  37. Comparing the Versions First Version power 3 5 power 3 4 * 3 power 3 3 * 3 * 3 power 3 2 * 3 * 3 * 3 power 3 1 * 3 * 3 * 3 * 3 power 3 0 * 3 * 3 * 3 * 3 * 3 1 * 3 * 3 * 3 * 3 * 3 243 Second Version power 3 5 power 9 2 * 3 power 81 1 * 3 power 81 0 * 81 * 3 1 * 81 * 3 243 4 function calls, 4 multiplications. 6 function calls, 5 multiplications.

  38. A More Difficult Example Define prime :: Int -> Bool, so that prime n is True if n is a prime number. What if we know whether (n-1) is prime? What if we know whether each smaller number is prime? NO HELP!

  39. Generalise the Problem! n is prime means No k in the range 2<=k<n divides n. Generalisation Replace 2 by a variable. Define factors m n == True if Some k in the range m<=k<n divides n. So prime n = not (factors 2 n) not x is True if x is False, and vice versa.

  40. Recursive Decomposition Problem: Does any k in the range m<=k<n divide n? What if we know whether any k in a smaller range divides n? Some k in the range m<=k<n divides n if m divides n, or some k in the range m+1<=k<n divides n.

  41. Recursive Solution factors :: Int -> Int -> Bool factors m n | m == n = False | m < n = divides m n || factors (m+1) n divides :: Int -> Int -> Bool divides m n = n `mod` m == 0 There is no k in the range n<=k<n. x || y is True if x is True or y is True.

  42. What is Getting Smaller? The range m<=k<n contains n-m elements. Call this the problem size. factors m n | m == n = False | m < n = divides m n || factors (m+1) n Base case: n-m == 0 Recursive case: n-(m+1) == (n-m)-1 The problem size gets smaller in each call, until it reaches zero. So recursion terminates.

  43. Lessons • Recursion lets us decompose a problem into smaller subproblems of the same kind -- a powerful problem solving tool in any programming language! • A more general problem may be easier to solve recursively than a `simpler´ one, because the recursive calls can do more. • To ensure termination, define a `problem size´ which must be greater than zero in the recursive cases, and decreases by at least one in each recursive call.

More Related