CS 411 Fall 2025  >  Outline for September 8, 2025

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CS 411 Fall 2025
Outline
for September 8, 2025

Outline

Analysis Examples [L 2.5, Appendix B]

• More on Recurrences
  • Given a recurrence and initial conditions, the general solution consists of all functions that satisfy the recurrence, ignoring initial conditions.
  • Procedure for solving recurrences:
    1. Find general solution of the recurrence. This will usually involve unknown constants.
    2. Use initial conditions to determine the values of these constants. Result: specific solution to the recurrence and initial conditions.
• Fibonacci Numbers
  • Sequence \(0,1,1,2,3,5,8,13,21,34,55,89,144,\dots\). Property: sum of two consecutive values is next value.
  • Recurrence: \(F(0) = 0\), \(F(1) = 1\), and \(F(n) = F(n-1)+F(n-2)\) for \(n\ge 2\).
  • Slow algorithm, based on recurrence.
• Second-order linear homogeneous recurrences
  • Order of a recurrence: how far back its dependence on previous values goes.
    • First-order recurrence: \(x(n) = 5 \bigl[x(n-1)\bigr]^2\).
    • Second-order recurrence: \(x(n) = 5 \left[x(n-1)\right]^2 - \sin\bigl[x(n-2)\bigr]\).
  • Linear recurrence: constants times values, added up, is equal to ...
    • Example linear recurrence: \(x(n) - 2 x(n-1) + 3 x(n-2) = 15\).
    • Equivalently: \(x(n) = 2 x(n-1) - 3 x(n-2) + 15\).
  • A linear recurrence is homogeneous if the right-hand side (in the first form, above) is the constant zero.
    • Example linear homogeneous recurrence: \(6 x(n) - 2 x(n-1) + 3 x(n-2) = 0\).
  • Finding the general solution of a second-order linear homogeneous recurrence
    • Recurrence: \(a\,x(n) + b\,x(n-1) + c\,x(n-2) = 0\), where \(a\), \(b\), \(c\) are constants.
    • Characteristic equation: \(ar^2 + br + c = 0\).
    • Find general solution to recurrence based on roots of characteristic equation:
      • 2 real roots \(r_1\), \(r_2\). General solution: \(x(n) = \alpha r_1^n + \beta r_2^n\), where \(\alpha\), \(\beta\) are real numbers.
      • 1 real root \(r\). General solution: \(x(n) = \alpha r^n + \beta n r^n\), where \(\alpha\), \(\beta\) are real numbers.
      • No real roots (we will not need to use this case). Roots are distinct complex numbers \(r_1, r_2 = u \pm vi\). General solution: \(x(n) = \gamma^n\left[\alpha \cos n\theta + \beta \sin n\theta\right]\), where \(\alpha\), \(\beta\), \(\gamma\) are real numbers.
• Analyze Slow Fibonacci
  • Size: given \(n\). Basic operation: base-case return. Result: second-order linear homogeneous recurrence.
  • Basic operations required to compute \(F(n)\): \(\Theta\left(\left[\frac{1+\sqrt{5}}{2}\right]^n\right)\). Exponential time!
• Fast Fibonacci
  • Fibonacci number \(F(n)\) can easily be computed using \(\Theta(n)\) integer arithmetic operations.
  • Both recursive and iterative algorithms demonstrated.
• Super-Fast Fibonacci
  • The text mentions that there is an algorithm based on matrix multiplication that computes \(F(n)\) using only \(\Theta(\log n)\) integer arithmetic operations.
  • This allows for the computation of very large Fibonacci numbers, if such numbers can be represented by the numeric types used.
  • Python implementation demonstrated.