Kontraktbaseret Programmering 1 Induction and Recursion Jens Bennedsen.
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Transcript of Kontraktbaseret Programmering 1 Induction and Recursion Jens Bennedsen.
Kontraktbaseret Programmering 1Induction and Recursion
Jens Bennedsen
Ingeniørhøjskolen i ÅrhusSlide 2
Agenda
Syllabus for this week• Recursive definitions of sets• Recursive definitions of algorithms• Mathematical Induction – First Principle• Mathematical Induction – Second Principle
Ingeniørhøjskolen i ÅrhusSlide 3
Syllabus for this week
• Discrete Structures Web Course Material – Section 5A: Recursive Definitions– Section 5D: Proof by Induction– But these web pages also contain material about
• Propositional logic• Predicate logic• Sets • Relations and• Functions
– Including on-line interactive exercises– May be handy for repetition before exam
Ingeniørhøjskolen i ÅrhusSlide 4
Agenda
Syllabus for this week Recursive definitions of sets• Recursive definitions of algorithms• Mathematical Induction – First Principle• Mathematical Induction – Second Principle
Ingeniørhøjskolen i ÅrhusSlide 5
Recursive Definition of a set
• The basis clause:– What are the basic starting elements of the set.
• The inductive clause:– In which ways can existing elements be combined to
produce new elements.
• The extremal clause:– Unless an object can be shown to be a member of the
set by applying the first two clauses it is not a member of the set.
Ingeniørhøjskolen i ÅrhusSlide 6
Example: The natural numbers
• The set of the natural numbers can be defined as:– Basis clause: 0 N.– Inductive clause: For any element x in N, x + 1 is in N.
– Extremal clause: Nothing is in N unless it is obtained from either the basis or the inductive clauses.
• The “x + 1” is also called “succ(x)” in the literature• This was formalized by the mathematician Peano
Ingeniørhøjskolen i ÅrhusSlide 7
How to define the even natural numbers?
• The set of the even natural numbers (NE) can be defined as:– Basis clause: 0 NE.– Inductive clause: For any element x in NE, x + 2 is in
NE.– Extremal clause: Nothing is in NE unless it is obtained
from either the basis or the inductive clauses.
Ingeniørhøjskolen i ÅrhusSlide 8
How to define the even integers?
• The set of the even integers (EI) can be defined as:– Basis clause: 0 EI.– Inductive clause: For any element x in EI, x + 2 and x -2
are in EI.– Extremal clause: Nothing is in EI unless it is obtained
from either the basis or the inductive clauses.
Ingeniørhøjskolen i ÅrhusSlide 9
How to define propositional forms?• Let V = {p,q,r,…} be a set of propositional
variables, where V does not contain any of the following symbols: (, ), , , , , , T and F. Then the set of the propositional forms can be defined as:– Basis clause: T and F are propositional forms and if x V
then x is a propositional form.– Inductive clause: If E1 and E2 are propositional forms
then (E1), (E1 E2), (E1 E2), (E1 E2), (E1 E2) are all propositional forms.
– Extremal clause: Nothing is a propositional form unless it is obtained from either the basis or the inductive clauses.
Ingeniørhøjskolen i ÅrhusSlide 10
Agenda
Syllabus for this weekRecursive definitions of sets Recursive definitions of algorithms• Mathematical Induction – First Principle• Mathematical Induction – Second Principle
Ingeniørhøjskolen i ÅrhusSlide 11
Recursive Thinking
• Recursion is a problem-solving approach that can be used to generate simple solutions to certain kinds of problems that would be difficult to solve in other ways
• Recursion splits a problem into one or more simpler versions of itself
Ingeniørhøjskolen i ÅrhusSlide 12
Recursion in practice
Ingeniørhøjskolen i ÅrhusSlide 13
Recursion Recipe
1. Handle the “base case”, where a recursive call is not made.
2. For the other cases, make a recursive call (the recursive “leap of faith”), which must make progress towards a goal (towards the base case)
3. Typically no need for an extremal clause
If the base case is not handled, or the recursive call does not progress towards the base case,
then the method will call itself infinitely (it will “diverge”).
Ingeniørhøjskolen i ÅrhusSlide 14
Recursive Definitions of Algorithms
• The factorial N!, for any positive integer N, is defined to be the product of all integers between 1 and N inclusive
• This definition can be expressed recursively as:
1! = 1 N! = N * (N-1)!
• A factorial is defined in terms of another factorial
• Eventually, the base case of 1! is reached
Ingeniørhøjskolen i ÅrhusSlide 15
Recursive Definitions
5!
5 * 4!
4 * 3!
3 * 2!
2 * 1!
1
2
6
24
120
Ingeniørhøjskolen i ÅrhusSlide 16
A recursive C++ algorithm for factorial
// Precondition n >= 1
// Postcondition result is n!
int recursiveFac(int n)
{
if(n==1) return 1;
else return n*recursiveFac(n-1);
}
Base case
Recursive case
Ingeniørhøjskolen i ÅrhusSlide 17
Fibonacci numbers
• Developed by Leonardo Pisano in 1202.– Investigating how fast rabbits could breed under
idealized circumstances.– Assumptions
• A pair of male and female rabbits always breed and produce another pair of male and female rabbits.
• A rabbit becomes sexually mature after one month, and that the gestation period is also one month.
– Pisano wanted to know the answer to the question how many rabbits would there be after one year?
Ingeniørhøjskolen i ÅrhusSlide 18
Fibonacci Numbers
• The Fibonacci Numbers are defined by– The first two numbers are 1 and 1.– Each subsequent number is the sum of the preceding 2
numbers.
• 1,1,2,3,5,8,13,21,34,55,etc.• Recursively it is defined as:• Fibonachi(n) = 1 if n = 0 • Fibonachi(n) = 1 if n = 1• Fibonachi(n) = Fibonachi(n-2) + Fibonachi(n-1) otherwise• Non-recursively:
n
nn
nF25
5151)(
Ingeniørhøjskolen i ÅrhusSlide 19
The fibonacci Method
// Precondition: n >= 0
// Postcondition: The corresponding fibonachi number is
// returned
long fibonacci(long n){
int fib;
if (n <= 1)
return 1;
else
return fibonacci(n-1) + fibonacci(n-2);
}
Base case
Recursive case
Ingeniørhøjskolen i ÅrhusSlide 20
Execution Trace (decomposition)
fibonacci(3)
fibonacci(2) fibonacci(1)
Ingeniørhøjskolen i ÅrhusSlide 21
Execution Trace (decomposition)
fibonacci(3)
fibonacci(2) fibonacci(1)
fibonacci(0)fibonacci(1)
Ingeniørhøjskolen i ÅrhusSlide 22
Execution Trace (composition)
fibonacci(3)
fibonacci(2) fibonacci(1)
fibonacci(0)->1fibonacci(1)->1
+
+
Ingeniørhøjskolen i ÅrhusSlide 23
Execution Trace (composition)
fibonacci(3)
fibonacci(2)->2 fibonacci(1)->1
+
Ingeniørhøjskolen i ÅrhusSlide 24
Execution Trace (composition)
fibonacci(3)->3
Ingeniørhøjskolen i ÅrhusSlide 25
Towers of Hanoi
In the great temple of Brahma in Benares, on a brass plate under the dome that marks the center of the world, there are 64 disks of pure gold that the priests carry one at a time between these diamond needles according to Brahma's immutable law: No disk may be placed on a smaller disk. In the beginning of the world all 64 disks formed the Tower of Brahma on one needle. Now, however, the process of transfer of the tower from one needle to another is in mid course. When the last disk is finally in place, once again forming the Tower of Brahma but on a different needle, then will come the end of the world and all will turn to dust.
R. Douglas Hofstadter. Metamagical themas. Scientific American,
248(2):16-22, March 1983.
Ingeniørhøjskolen i ÅrhusSlide 26
Towers of Hanoi Problem with Three Disks
Ingeniørhøjskolen i ÅrhusSlide 27
Towers of Hanoi: Three Disk Solution
Ingeniørhøjskolen i ÅrhusSlide 28
Towers of Hanoi: Three Disk Solution
Ingeniørhøjskolen i ÅrhusSlide 29
Towers of Hanoi: Recursive FunctionThe general algorithm to solve this problem is
1. Move the top n-1 disks from needle 1 to needle 2 using needle 3 as the intermediate needle
2. Move disk number n from needle 1 to needle 3
3. Move the top n-1 disks from needle 2 to needle 3 using needle 1 as the intermediate needle
Ingeniørhøjskolen i ÅrhusSlide 30
Recursive Towers of Hanoi in C++
// Precondition: count >= 0// Postcondition: count disks are moved from from_needle// to to_needle using via_needle for helpvoid moveDisks(int count, int from_needle, int to_needle,
int via_needle){ if(count > 0) { moveDisks(count - 1, from_needle, via_needle, to_needle); cout<<"Move disk "<<count<<“ from "<<from_needle <<“ to "<<to_needle<<"."<<endl; moveDisks(count - 1, via_needle, to_needle, from_needle); }}
Recursive call
Recursive call
Ingeniørhøjskolen i ÅrhusSlide 31
General problem solving strategy
• Divide and conquersolve(p: Problem) {
if <p is a simple problem>
solution = <solve it directly>
else
<split p in two disjoint problems p1 and p2>
res1 = solve(p1)
res2 = solve(p2)
solution = join(p1, p2)
Ingeniørhøjskolen i ÅrhusSlide 32
Example: search
private bool search(int m, List<int> p) { if (p.Count == 0) return false; else if (p.Count == 1) return p[0] == m; else { List<int> p1 = new List<int>(); List<int> p2 = new List<int>(); for (int i = 0; i < p.Count / 2; i++) p1[i] = p[i]; for (int j = (p.Count / 2) + 1; j < p.Count; j++) p2[j - ((p.Count / 2) + 1)] = p[j]; return search(m, p1) || search(m, p2); } }
Ingeniørhøjskolen i ÅrhusSlide 33
Agenda
Syllabus for this weekRecursive definitions of setsRecursive definitions of algorithms Mathematical Induction – First Principle• Mathematical Induction – Second Principle
Ingeniørhøjskolen i ÅrhusSlide 34
Mathematical Induction
• In the mathematical induction we have the law already formulated. We must prove that it holds generally
• The basis for mathematical induction is the property of the well-ordering principle for the natural numbers
Ingeniørhøjskolen i ÅrhusSlide 35
1. principle of Mathematical Induction
• Also called weak induction• Suppose P(n) is a statement involving an integer n• Then to prove that P(n) is true for every n ≥ n0 it is
sufficient to show that these two things hold:
1. P(n0) is true
2. For any k ≥ n0, if P(k) is true, then P(k+1) is true
– In general we have a basis step– Followed by an inductive step.
Ingeniørhøjskolen i ÅrhusSlide 36
Example Induction Proof
• Prove that for any natural number n it holds that: 0+1+…+n = n(n + 1)/2– Basis step: If n = 0 then LHS = 0 and RHS = 0(0+1)/2=0– Induction step: Let the induction hypothesis be:
For an arbitrary k it holds that 0+1+…+k = k(k + 1)/2
Let us try to express LHS for (k+1). Using the induction hypothesis the LHS = k(k+1)/2 + (k+1)
If we factor out (k+1) we can rewrite this to: (k+1)(k+2)/2
This is identical to the RHS for (k+1)
QED.
Ingeniørhøjskolen i ÅrhusSlide 37
Example Induction Proof
• Prove that for any natural number n it holds that: 1+3+…+(2n+1) = (n + 1)2
– Check for n = 0: 2*0 + 1 = (0 + 1)2
– Assume that:For an arbitrary k it holds that 1+3+…+(2k+1) = (k + 1)2
Then for k+1 we need to prove: 1+3+…+(2k+1)+(2(k+1)+1) = ((k+1)+1)2
Using the induction hypothesis we get: 1+3+…+(2k+1)+(2(k+1)+1) = (k+1)2 + (2(k+1) +1)Since (k+1)2 + 2(k+1)+1 = (k+1+1)2 we have proved our objective.
QED.
Ingeniørhøjskolen i ÅrhusSlide 38
More on Inductive Proofs
General strategy:
Clarify on which variable you are going to do the induction
Calculate some small cases n=0,1,2,3,…
(Come up with your conjecture)
Make clear what the induction step nn+1 is
Prove it, and say that you proved it.
Ingeniørhøjskolen i ÅrhusSlide 39
Agenda
Syllabus for this weekRecursive definitions of setsRecursive definitions of algorithmsMathematical Induction – First Principle Mathematical Induction – Second Principle
Ingeniørhøjskolen i ÅrhusSlide 40
2. principle of Mathematical Induction
• Also called strong induction• Suppose P(n) is a statement involving an integer n
• Then to prove that P(n) is true for every n ≥ n0 it is sufficient to show that this holds:
1. P(n0) is true
2. For any k ≥ n0, if P(x) is true for all x smaller than k, then P(k) is true
Ingeniørhøjskolen i ÅrhusSlide 41
Strong induction
• Weak mathematical induction assumes P(k) is true, and uses that (and only that!) to show P(k+1) is true
• Strong mathematical induction assumes P(1), P(2), …, P(k) are all true, and uses that to show that P(k+1) is true.
)1P()P(...)3P()2P()1P( kk
Ingeniørhøjskolen i ÅrhusSlide 42
Strong induction example
• Show that any number > 1 can be written as the product of primes
• Base case: P(2)– 2 is the product of 2 (remember that 1 is not prime!)
• Inductive hypothesis: P(1), P(2), P(3), …, P(k) are all true
• Inductive step: Show that P(k+1) is true
Ingeniørhøjskolen i ÅrhusSlide 43
Strong induction example
• Inductive step: Show that P(k+1) is true• There are two cases:
– k+1 is prime• It can then be written as the product of k+1
– k+1 is composite• It can be written as the product of two composites, a and b,
where 2 ≤ a ≤ b < k+1• By the inductive hypothesis, both P(a) and P(b) are true
– QED
Ingeniørhøjskolen i ÅrhusSlide 44
Induction in Computer Science
Inductive proofs play an important role in computer science because of their similarity with recursive algorithms.
Analyzing recursive algorithms often require the use of recurrent equations, which require inductive proofs.
Also, recursive algorithms are constructive such that we often get a solution “for free”.
Ingeniørhøjskolen i ÅrhusSlide 45
Summary
• The notion of Recursion and Induction– Recursive definitions of sets– Recursive definitions of algorithms– Mathematical Induction
• Read Section 5A: Recursive Definitions and Section 5D: Proof by Induction from the web pages on slide 3
• Carry out as many exercises as possible from the web pages