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Useful Formulae

Useful Formulae. Analysis of Algorithms.  for (i = 0; i <= n; i++) { j = i; do j = j - 2 while (j > 0); }  for (i = 0; i <= n; i++) { j = i; do j = j div 2 while (j > 0); }. T(n) = O(n 2 ). T(n) = O(n log n). Recurrence Relations.

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Useful Formulae

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  1. Useful Formulae

  2. Analysis of Algorithms  for (i = 0; i <= n; i++) { j = i; do j = j - 2 while (j > 0); }  for (i = 0; i <= n; i++) { j = i; do j = j div 2 while (j > 0); } T(n) = O(n2) T(n) = O(n log n)

  3. Recurrence Relations • A sequence of numbers can be defined by a recurrence relation and a set of initial conditions • Examples: • an = an-1 + 2 ; a0 = 5 gives the sequence a0 = 5, a1 = 7, a2 = 9, a3 = 11 • an = an-1 + an-2 ; a0 = 1 a1 = 1 gives the sequence a0 = 1, a1 = 1, a2 = 2, a3 = 3, a4 = 5, a5 = 8

  4. How to solve a recurrence relation? • Ad Hoc Method -- Calculate the first few values of the recurrence -- Look for regularity -- Guess a suitable general form -- Prove by mathematical induction e.g., T(n) = 2T(n - 1) + 1, T(1) = 1 • Iterated Method -- Converts the recurrence into a summation -- Use bounded summation to solve the recurrence e.g., T(n) = 3T(n/4) + n, T(1) = 1  Characteristic Equation or Annihilator Method

  5. Linear Recurrence Relations with Constant Coefficients

  6. Superposition of Solutions

  7. Obtaining a General Solution • A general solution is of the form g(n)=anwhere a is some constant. To determine a, substitute g(n)=an into the recurrence. c0an - c1an-1 - ... - ckan-k = 0, divide by an-k to get c0ak - c1ak-1 - ... - ck = 0,   • Solving this polynomial equation gives all possible a's. (This is called the characteristic equation.)

  8. Example

  9. Multiple Roots • If a is the root of the characteristic equation with multiplicity m, there is a general solution

  10. Getting a Particular Solution • Mainly try and error. However, some very good suggestions do exist: • Sometimes your initial guess "degenerates" and gives contradictory conditions. Then try a solution of higher degree.

  11. Examples

  12. A Total Example • Once the general and particular solutions have been found, they are added together to give the total solution. • The initial conditions are then used to determine the constants of the general solution.

  13. A Total Example

  14. Pseudo Nonlinear Recurrence • Range Transformations: on the values of the sequences • Domain Transformations: on the value of indices

  15. Example

  16. A Practical Example •  Mergesort • Split list in half • Sort both halves • Merge the two halves • Analysis: The recurrence relation is T(0) = 0

  17. Example Continued T(n) = 2T(n/2) + (n – 1)

  18. Example Continued

  19. Example Continued (3) Impose the initial conditions on A2n + n  2n + 1 (n=0) A + 0 + 1 = U(0) = 0  A = -1 So U(n) = n  2n + 2n + 1 = T(2n) Replace n by log n gives T(n) = nlog n + n + 1 = Q(n  log n)

  20. How to deal with this? • What if the master theorem cannot be applied? • Extend the recurrence and compute the sum of the terms.

  21. Polygon Triangulation • Given a polygon in the plane, add diagonals so that each face is a triangle None of the diagonals are allowed to cross. • Triangulation is an important first step in many geometric algorithms.

  22. Polygon Triangulation • The simplest algorithm might be to try each pair of points and check if they see each other. If so, add the diagonal and recur on both halves, for a total of O(n3). • However, Chazelle gave an algorithm which runs in T(n) =2T(n/2)+O( ) time. Since = O(n1-e) by case 1 of the Master Theorem, Chazelle's algorithm is linear, i.e., T(n) =O(n).

  23. Application to Algorithm Design • To improve the efficiency of an algorithm • if logba > c, try to decrease the ratio of a/b (i.e., split the problem into fewer subproblems with larger size. • if logba ≤ c, try to find a better way for splitting/merge subproblems.

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