Cs 6515 Exam 1 Complete Review Notes

CS 6515 EXAM 1 COMPLETE REVIEW NOTES

◉ DP: Types of Subproblems (4).
Answer: Input = x1, x2, ..., xn
1) Subproblem = x1, x2, ..., xi ; O(n)
2) Subproblem = xi, xi+1, ..., xj ; O(n^2)


Input = x1, x2, ..., xn; y1, y2, ..., ym
1) Subproblem = x1, x2, ..., xi; y1, y2, ..., yj ; O(mn)


Input = Rooted Binary Tree
1) Subproblem = Smaller rooted binary tree inside the Input.


◉ DC: Geometric Series.
Answer: Given r = common ratio and a = first term in series
=> a + ar + ar^2 + ar^3 + ... + ar^(n-1)


=> a * [(1 - r^n) / (1-r)]


◉ DC: Arithmetic Series.
Answer: Given d = common difference and a = first term in series =>
a + (a + d) + (a + 2d) + ... + (a + (n-1)d

,Sum = n/2 * [2*a + (n-1)d]


◉ DC: Solving Recurrences - Master Theorem.
Answer: If T(n) = aT([n/b]) + O(n^d) for constants a>0, b>1, d>=0:


T(n) = {
O(n^d) if d > logb(a)
O((n^d)logn) if d = logb(a)
O(n^(logb(a))) if d < logb(a)
}


◉ Nth roots of Unity.
Answer: (1, 2*PI*j/n) for j = 0, 1, ..., n-1
*Around the Unit Circle!


◉ Steps to solve for FFT.
Answer: 1) Write out Matrix Coefficient Form based on n (size of
input) Mn(w) = [ 1 1 ... 1
1 w ... w^n-1
...
1 w^n-1 ... w^((n-1)*(n-1)) ]

, 2) Find value for w = e^(2*PI*i)/n, Substitute in Mn(w).


3) For the input coefficients into nx1 matrix. I.E. [4 0 1 1], let known
as B.


4) Evaluate FFT:
a) FFT of Input = Mn(w) x B
b) Inverse FFT of Input = 1/n * Mn(w^-1) x B


◉ Euler's Formula.
Answer: e^ix = cosx + isinx


◉ Imaginary Number Multiples.
Answer: i = i, i^2 = -1, i^3 = -i, i^4 = 1
i = -i, i^2 = -1, i^3 = i, i^4 = 1


◉ Omega(w).
Answer: w = (1, 2*PI / n) = e^(2*PI*i/n)


◉ DC Algorithms and Runtimes (6).