Aryaman · I wasn't

CS 213

Lecture 3

Analyse Code

Work Cost
Assignment 1 unit
Operation 1 unit
Array reference 2 units
Comparison 1 unit
if-then-else cost(condition) + cost(then) + cost(else) + 2
  [2 for jumps]
for loop body once = 3 + cost(body)
function call cost(body) + 10
  [10 for calling and returning]

Note that we’re only looking at time, so we’ve ignored declarations (space).

Example

unsigned short search(int a[], unsigned short size, int elt){  
	unsigned short pos = -1;			// cost 1  
	for(unsigned short i = 0; i < size; i++)	//3  
		if(a[i] == elt)				//2 + 1, array + comparison  
			return pos = i;			//6  
	return pos;  
}

Execution of search over an array of n elements: \(9n + 1 \sim c_1n + c_2\)

Big-O Notation

Let \(f, g: \mathbb{Z} \to \mathbb{R}\).
The function \(f(x)\) is said to be of \(O(g(x))\) if there are constants \(C\) and \(k\) such that \(|f(x)| \le C|g(x)|\) for all \(x > k\).

The constants \(C\) and \(k\) are called witnesses to the relationship of \(f(x)\) being of \(O(g(x))\).

E.g. \(f(x) = 9x + 1\), \(g(x) = 10\)
\(C = 1\), \(k = 1\) works and thus, \(f(x)\) is of \(O(10x)\) and also of \(O(x)\).

E.g. \(f_1(x) = log_2(x)\), \(f_2(x) = \sqrt{x}\)
Have to find \(C\) and \(k\) to claim \(f_1(x)\) is of \(O(f_2(x))\).

Results

\(f(x) = \sum_{i = 0}^{i = n}a_ix^i\), then \(f(x) = O(x^n)\). (Assuming \(a_n \neq 0\).)
\(n! = O(n^n)\)
\(\log n! = O(n \log n)\)

Numerical application

Polynomial = p(x) = \(\sum_{i = 0}^{n} a_i x^i\)

Given a polynomial, the following are natural operations to do with it.

  • Evaluate it at a point, \(P(x_0)\)
  • Find the roots
  • Extrema in domain
  • \(P_1(x) + P_2(x)\), \(P_1(x)P_2(x)\), \(P_1(x) \% P_2(x)\)
  • Given roots (data), construct the polynomial satisfying that stuff. (Clearly, something is missing. Maybe assume monic.)

How do you represent a polynomial?

  • Array, possibly. \(4x^5 - 3x^3 + x^2 + 250 \to [250, 0, 1, -3, 0, 4]\) ** Not good for sparse polynomials like \(9x^{200} - 5x^{100} + 10\)
    \(\{10, 0\},\; \{-5, 100\},\; \{9, 200\}\) But this is not v good for the cases where we have contiguous exponents.

Evaluating at a point. Cost?

  • Blind way: \(a_0 + a_1 x + ... + a_n x^n\), do every operation
    \(n\) additions
    \(1 + 2 + \cdots + n\) multiplications = \(n(n+1)/2\) multiplications
    ~ \(O(n^2)\)

    \(4x^5 - 3x^3 + x^2 + 250 = ((4x^2 - 3)x + 1)x^2 + 250\)
    ~ \(O(n)\)

Can we do better than \(O(n)\)?
No. In worst case scenario, we have \(n\) different coefficients (as input) and we do need to look at them at least once.

Matrix: \(n^2\) elements

A * B
Output: \(n^2\) elements Should expect algorithm of \(O(n^2)\). → No one has gotten close to this.
However, we don’t have a better (proven) lower bound.
Has gotten down to \(O(n^{2.49})\).
In class, we’ll look at one algorithm of complexity \(O(n^3)\).

All roots of a polynomial of degree \(n\)

  • Newton Raphsson
  • Interval half
  • Graephe’s Root square -> Gives all n roots (approx, of course) simultaneously
Next

CS 213