Mathematical Induction

Transcription

1 Mathematical Induction COM1022 Functional Programming Techniques Professor Steve Schneider University of Surrey Semester 2, 2010 Week 10 Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week 10 1 / 22

2 Outline 1 Mathemtical induction exercises 2 List Induction 3 Structural Induction over algebraic types Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week 10 2 / 22

3 Mathemtical induction exercises Review: The principle of induction If P(0) and P(n) P(n + 1) for all n Then we can conclude n.p(n) An infinite sequence of dominoes... Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week 10 3 / 22

4 Exercises Mathemtical induction exercises These exercises were set last week in place of the lab: Prove each of the following by induction: n = 2 (n+1) 1 for all n n 2 = (n(n+1)(2n+1)) 6 for all n 4 n + 6n 1 is divisible by 9 for all n Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week 10 4 / 22

5 List Induction Proving statements about lists How do we understand what a function on lists does? [Thompson Chapter 8]. Example: length [] = 0 length (x:xs) = 1 + length xs Is it true that length (xs ++ ys) = length xs + length ys If so, how do we prove it? Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week 10 5 / 22

6 List Induction List induction In order to prove a logical property P(xs) holds for all finite lists xs we have to prove two things: Base case: prove P([]) Inductive step: Prove P(x:xs) on the assumption that P(xs) holds. The idea is similar to mathematical induction, (which allows proofs for finite numbers, i.e. positive integers) Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week 10 6 / 22

7 List Induction Setting up the induction For our example, we will use P(xs) to be length (xs++ys) = length xs + length ys for any ys. We will prove this by induction on xs Also need to use the definition of ++: [] ++ xs = xs (x:xs) ++ ys = x:(xs ++ ys) Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week 10 7 / 22

8 Base case: P([]) List Induction We need to prove length ([]++ys) = length [] + length ys for any ys. By the Haskell definitions, this reduces to length ys = 0 + length ys for any ys. and this is clearly true. Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week 10 8 / 22

9 List Induction Inductive Step: P(xs) P(x:xs) Assume that length (xs++ys) = length xs + length ys for any ys Now we need to consider whether P(x:xs) is true. length((x : xs) + +ys) = length (x:(xs++ys)) = 1 + length (xs++ys) = 1 + (length xs + length ys) (since P(xs) is assumed) = (1 + length xs) + length ys = length (x:xs) + length ys Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week 10 9 / 22

10 List Induction Exercise Prove by list induction that: xs ++ [] = xs for all xs Base case Inductive step You will need to use the definitions for ++ Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week / 22

11 List Induction Another example The function reverse is defined as follows: reverse [] = [] reverse (x:xs) = (reverse xs) ++ [x] Prove by list induction over xs that: reverse (xs ++ ys) = reverse ys ++ reverse xs Base case Inductive step Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week / 22

12 Structural Induction over algebraic types Recursive algebraic types A recursive type is defined in terms of itself. For example, a type for trees can be defined as follows: data NTree = NilT Node Int NTree NTree An NTree is defined to be either the empty tree NTree or else a node containing an integer and two subtrees. [see Thompson 14.2] We will use tt NTree as the running example, but the principles hold for any recursive data type. Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week / 22

13 Question Structural Induction over algebraic types Draw some NTrees. For each tree, how many nodes does it contain? How many occurrences of NilT? Can you identify a relationship between the number of nodes and the number of occurrences of NilT? Is this true for all trees? How would you prove that you have got it right? Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week / 22

14 Structural Induction over algebraic types Structural induction over trees In order to prove a logical property P(tr) holds for all trees tr we have to prove two things: Base case prove P(NilT) Inductive step Prove P(Node x tr1 tr2), on the assumption that P(tr1) and P(tr2) hold. Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week / 22

15 Structural Induction over algebraic types Exercise: Characterising the property Define the number of nils and the number of nodes by completing the following functions: nils (NilT) =... nils (Node x tr1 tr2) =... nodes (NilT) =... nodes (Node x tr1 tr2) =... Express the property in terms of nils and nodes: P(tr) =??? Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week / 22

16 Structural Induction over algebraic types Base case Prove the property P(NilT). Use the definitions of nils NilT and nodes NilT Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week / 22

17 Structural Induction over algebraic types Inductive step Assume P(tr1) and P(tr2) are true. Prove P(Node x tr1 tr2) using the definitions of nils and nodes Once the base case and the inductive step are both established, then the property is proven for all trees. Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week / 22

18 Structural Induction over algebraic types Exercise: height of tree and number of nodes height is defined as follows: height NilT = 0 height (Node x tr1 tr2) = 1 + max(height tr1, height tr2) Prove that height tr nodes tr for all trees tr What is the base case? Prove the inductive step Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week / 22

19 Structural Induction over algebraic types Example: height of tree and number of nodes Prove that nodes tr 2 height tr 1 for all trees tr Base case: nodes NilT = 0 height NilT = , which establishes the base case. Inductive step: assume true for tr1 and tr2 See next slide... Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week / 22

20 Inductive step Structural Induction over algebraic types nodes(node x tr1 tr2) = 1 + (nodes tr1) + (nodes tr2) 1 + (2 height tr1 1) + (2 height tr2 1) 1 + (2 max(height tr1)(height tr2) 1) + (2 max(height tr1)(height tr2) 1) (2 2 max(height tr1)(height tr2) ) 1 (2 max(height tr1)(height tr2)+1 ) 1 = 2 height (Node x tr1 tr2) 1 Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week / 22

21 Structural Induction over algebraic types Summary Structural induction (for lists, trees, and any other recursive data types) to prove a property P(x) for all objects x of that type: Base case: prove P(x) holds for all clauses of the type which are not recursive, i.e. which do not have components of that type. Inductive step: for all clauses of the type which do refer recursively to the type, prove that P holds for any instance x of that case under the assumption that it holds for the components of x which are of that type. This is also the principle for list induction. Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week / 22

22 Structural Induction over algebraic types This week s exercise data STree = Leaf Int Node STree STree elements (Leaf a) = 1 elements (Node tr1 tr2) = elements tr1 + elements tr2 elementlist (Leaf a) = [a] elementlist (Node tr1 tr2) = (elementlist tr1) ++ (elementlist tr2) Prove that elements tr = length (elementlist tr) for all STrees tr Professor Steve Schneider Mathematical Induction Semester 2, 2010 Week / 22