415 lecture note #3

[Ch 3] Algorithms (1)

3.1 Introduction & Notation

• An algorithm is a list of step-by-step instructions on how to accomplish a task.  Required properties include:
  • Input/Output -- An algorithm takes some input and produces an output.
  • Precision -- Each step is clearly indicated and unambiguous.
  • Finiteness -- The algorithm stops after a finite number of steps.

  Example: Find the maximum of three numbers, a, b, and c

  • Non-algorithm
    "Compare the three (input) numbers and return (or output) the largest one."
  • Algorithm
    • Step 1:  Let x (another number) to have the value of a.
    • Step 2:  If b > x, then set x's value to b.
    • Step 3:  If c > x, then set x's value to c.
    • Step 4:  Return x.

• Algorithms are often written as pseudocode, which is sort of the middle ground between a formal programming language and plain English.
• Some of the common elements of pseudocode:
  • Each algorithm starts with "procedure name(input values)".
  • There must be at least one "return value" statement.
    NOTE:  A procedure can only return ONE value only.
  • Operators are:
    • +, -, *, /, mod
    • <, <=, >, >=
    • ==, !=
  • The assignment operator is := (e.g. "x := 2").

• The basic constructs are
  • "if cond then action1 else action2"
  • "for var := initial_value to final value do action"
  • "while cond do action"
  • "repeat action until cond"
  • "begin statement₁ .. statement_n end"

• Example:  Algorithm max which finds the maximum of three numbers, a, b, and c.

  Input: Three numbers a, b, c
  Output: A number which is the largest of the tree input numbers

  procedure max(a, b, c)
  1. x := a           // x is a local/temporary variable
  2. if b > x then
  3.   x := b         // update x
  4. if c > x then
  5.   x := c         // update x
  6. return x
  

3.2 Example Algorithms

1. Algorithm find_largest which finds the largest number in the sequence s₁,..,s_n.

   Input:  A list of integers s = {s₁,..,s_n}, and the length n of the list
   Output: A number which is the largest in s.
   
   

    

    

    

    

    

    

    

    

2. Algorithm Div which calculate, for a given integer (n) and a positive integer (d),  the quotient (q) and remainder (r) where q and r ate integers.  Division is done by repeated subtraction, not by arithmetic division.

   Input:  n (an integer), and d (a positive integer)
   Output:  q and r (integers)

   procedure Div(n, d, q, r)
   
   
   

    

    

    

    

    

    

    

    

3. Algorithm is_even which tests whether a positive integer m is even

   Input:  A positive integer m
   Output:  true if m is even; false if m is odd

    

3.3 The Euclidean Algorithm

• An algorithm which finds the greatest common divisor (gcd) of two integers.

  Example:  gcd(30, 105) = 15

  divisor of 30
  divisor of 105
  common

• "divides"
  • If a, b and q are integers (where b != 0) satisfying a = bq, we say that "b divides a", noted b | a. The value q is called the quotient, and b a divisor of a.

    Examples:  15 | 30, and the quotient is 2.

• Propoeties of divisors:

  Let m, n and c be integers.

  1. If c | m and c | n, then c | (m + n)
  2. If c | m and c | n, then c | (m - n)
  3. If c | m, then c | mn

• Theorem:  For all integers a, b, q and r, such that a = bq + r, and a >= 0, b > 0 and 0 <= r < b, then gcd(a, b) = gcd(b, r).

  First, let's check some cases:

  • 105 = 30 * 3 + 15 -- gcd(105, 30) = gcd(30, 15) = 15
  • 504 = 396 * 1 + 108 -- gcd(504, 396) = gcd(396, 108) = 36

  Proof:

  Let C1 be the set of common divisors of a and b, and C2 be the set of common divisors of b and r.  We show C1 Ê C2, and C2 Ê C1.  Then we can conclude C1 º C2. 

  1) Show C1 Ê C2
  Let c Î C1 be a common divisor of a and b, that is, c | a and c | b.  
  From the third property of divisors, we get that c | bq. 
  Then  from the second property of divisors, we get that c | (a - bq) = r. ... (A)
  From (A) and hypothesis, we have that c | b and c | r.  Therefore, c is a common divisor of b and r, that is, c Î C2.

  2) Show C2 Ê C1
  Let c Î C2 be a common divisor of b and r, that is, c | b and c | r.  
  From the third property of divisors, we get that c | bq. 
  Then  from the first property of divisors, we get that c | ( bq + r) =a.  ... (B)
  From (B) and hypotehsis, c | a and c | b.  Therefore, c is a common divisor of a and b, that is, c Î C1.

  From 1) and 2), we showed that all common divisor of a and b are also common divisors of b and r. In particular, this is true for the greatest common divisor, i.e., gcd(a, b) = gcd(b, r).

• Algorithm

  procedure gcd(a, b)
  1. if a < b, then
  2.   swap(a, b)     // make a the larger of the two
  3. while b != 0 do
  4.   begin
  5.     r := a mod b
  6.     a := b
  7.     b := r
  8.   end
  9. return a
  

  iteration		a	b	r
  1	gcd(105, 30)	105	30	15
  2
  3

   

• Example:  gcd(110, 273) -- Section 3.3, question #2

  iteration		a	b	r
  1
  2
  3