Proving Theorems

1
Proving Theorems

• Direct proof
• An implication p?q can be proved by showing that
  if p is true, then q is also true.
• Example Give a direct proof of the theorem If
  n is odd, then n2 is odd.
• Idea Assume that the hypothesis of this
  implication is true (n is odd). Then use rules of
  inference and known theorems to show that q must
  also be true (n2 is odd).

2
Proving Theorems

• n is odd.
• Then n 2k 1, where k is an integer.
• Consequently, n2 (2k 1)2.
• 4k2 4k 1
• 2(2k2 2k) 1
• Since n2 can be written in this form, it is odd.

3
Proving Theorems

• Indirect proof
• An implication p?q is equivalent to its
  contra-positive ?q ? ?p. Therefore, we can prove
  p?q by showing that whenever q is false, then p
  is also false.
• Example Give an indirect proof of the theorem
  If 3n 2 is odd, then n is odd.
• Idea Assume that the conclusion of this
  implication is false (n is even). Then use rules
  of inference and known theorems to show that p
  must also be false (3n 2 is even).

4
Proving Theorems

• n is even.
• Then n 2k, where k is an integer.
• It follows that 3n 2 3(2k) 2
• 6k 2
• 2(3k 1)
• Therefore, 3n 2 is even.
• We have shown that the contrapositive of the
  implication is true, so the implication itself is
  also true (If 2n 3 is odd, then n is odd).

5
Induction

• The principle of mathematical induction is a
  useful tool for proving that a certain predicate
  is true for all natural numbers.
• It cannot be used to discover theorems, but only
  to prove them.

6
Induction

• If we have a propositional function P(n), and we
  want to prove that P(n) is true for any natural
  number n, we do the following
• Show that P(0) is true. (basis step)
• Show that if P(n) then P(n 1) for any n?N.
  (inductive step)
• Then P(n) must be true for any n?N.
  (conclusion)

7
Induction

• Example
• Show that n lt 2n for all positive integers n.
• Let P(n) be the proposition n lt 2n.
• 1. Show that P(1) is true.(basis step)
• P(1) is true, because 1 lt 21 2.

8
Induction

• 2. Show that if P(n) is true, then P(n 1) is
  true.(inductive step)
• Assume that k lt 2k is true.
• We need to show that P(k 1) is true, i.e.
• k 1 lt 2k1
• We start from k lt 2k
• k 1 lt 2k 1 2k 2k 2k1
• Therefore, if n lt 2n then n 1 lt 2n1

9
Induction

• Then P(n) must be true for any positive
  integer.(conclusion)
• n lt 2n is true for any positive integer.
• End of proof.

10
Induction

• Another Example (Gauss)
• 1 2 n n (n 1)/2
• Show that P(0) is true.(basis step)
• For n 0 we get 0 0. True.

11
Induction

• Show that if P(n) then P(n 1) for any n?N.
  (inductive step)
• 1 2 n n (n 1)/2
• 1 2 n (n 1) n (n 1)/2 (n 1)
• (2n 2 n (n 1))/2
• (2n 2 n2 n)/2
• (2 3n n2 )/2
• (n 1) (n 2)/2
• (n 1) ((n 1) 1)/2

12
Induction

• Then P(n) must be true for any n?N. (conclusion)
• 1 2 n n (n 1)/2 is true for all n?N.
  
• End of proof.

13
Complexity of Algorithms

• Algorithm - A finite set of precise instructions
  for performing a computation or for solving a
  problem. Examples
• - Find the largest integer in a set
• - Locate a particular element in a set
• - Sort a list of elements
• Pseudocode - Intermediate steps between an
  English language description of the procedure and
  an actual specification using a programming
  language.
• Method Analyze the pseudocode and determine the
  number of steps required, f(n), as a function of
  the problem size n

14
Time Complexity

• Consider
• a 7
• x 3
• for i 1 to n
• y ax
• Complexity?

15
Time Complexity

• Consider
• a 7
• b 2
• c 1
• x 3
• for i 1 to n
• for j 1 to n
• y abbc(abc)/(ab) 4
• Complexity?

16
Time Complexity

• Assume an algorithm takes
• 3n 3 steps Then it is a O(n) Algorithm
• 5n2 5n 105 steps Then it is a O(n2)
  Algorithm
• 2n3 2n 10 steps Then it is a O(n3)
  Algorithm
• f(n) is O(g(n)) iff positive constants c and n0
  exist such that f(n) lt cg(n) for all n, n gt k.
• 3n 3 lt 4n for all n, n gt 3
• 5n2 5n 105 lt 10n2 for all n, n gt
  6
• 2n3 2n 10 lt 3n3 for all n, n gt 3

17
Time Complexity
Typical Complexities O(1) O(log n) O(n) O(n
log n) O(n2) O(n3) NP

• cg(n)

f(n)
n
k
f(n) is O(g(n)) iff positive constants c and k
exist such that f(n) lt cg(n) for all n, n gt k.
18
Example Find the Maximum Element in a Finite
Sequence

• Procedure max(a1, a2,..., an integers)
• max a1
• for i 2 to n
• if max lt ai then max ai
• max is the largest element

a
6
5
7
1
10
8
1 2 3 4 5 6
19
ExampleBinary Search Algorithm

• Procedure binary search(x integer, a1,...,an
  increasing integers)
• i 1 left endpoint of search interval
• j n right endpoint of search interval
• while (i lt j )
• m ?(i j )/2?
• if x gt am then i m 1
• else j m
• end
• if x ai then location i
• else location 0
• location is the subscript of the term that
  equals x location is 0 of x is not found

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25
27
31
35
38
3
5
7
9
10
18
47
51
60
68
20
Homework

• Weiss p. 35-38 2.1, 2.2, 2.18
• Levitin P. 51-52 3, 8, 9
• P. 59-60 1, 2, 3

21
Programming Assignment 1

• Weiss p. 35-36 2.6
• Instructions
• Implement each routine in C
• Turn in with a report stating clearly for each
  algorithm what the running time is
• Provide a bar chart showing the number of
  iterations (or the running time) for various
  values of n