Recurrences

Separable form for recurrence relation:

$$\begin{gather*} T(n) = G(n, T(n_1), ..., T(n_k)) \end{gather*}$$

where $G(x_0, x_1, ..., x_k)$ is a function in $k+1$ variables and each $n_i (i=1,...,k)$ is a function of $n$ that is strictly less than $n$.

Fibonacci sequence is defined by the recurrence relation:

$$\begin{gather*} F(n) = F(n-1) + F(n-2) \end{gather*}$$

For this, we have $k=2, n_1=n-1, n_2=n-2$

$$\begin{gather*} G(n, x, y) = x + y \end{gather*}$$

Merge sort complexity:

$$\begin{gather*} T(n) = n + 2T(n/2) \end{gather*}$$

For this, we have $k=1, n_1=n/2$

$$\begin{gather*} G(n, x) = n + 2x \end{gather*}$$

Rote Method

Related homework questions: h2-Q7

EGVS method

4 stages: Expand, Guess, Verify and Stop-and-Sum.

Example:

$T(n) = 2T(n/2) + n$

1. Expand:

   $$\begin{align*} T(n) &= 2T(n/2) + n \\ &= 2 (2 T(n/4) + (n/2)) + n \\ &= 4 T(n/4) + 2n \\ &= 4 (2T(n/8) + (n/4)) + 2n \\ &= 8T(n/8) + 3n \\ &= \dots \end{align*}$$
2. Guess:

   $$\begin{gather*} T(n) = 2^i T({n\over{2^i}}) + i n \end{gather*}$$
3. Verify: (use natural induction)

   Base case: $T(n)_{i=1} = 2T(n/2) + n$ holds

   Inductive step: Assume $T(n)_{i=k}$ holds, the goal is to prove $T(n)_{i=k+1}$ also holds.

   $$\begin{align*} T(n) &= 2^k T({n\over{2^k}}) + k n \\ &= 2^k (2 T({n\over{2^{k+1}}}) + {n\over{2^k}}) + k n \\ &= 2^{k+1} T(n/2^{k+1}) + （k+1) n \end{align*}$$

   Therefore, this proves $T(n)_{i=k+1}$ also holds.

   Combining the base case and the inductive step we verified that $T(n) = 2^i T({n\over{2^i}}) + i n$.
4. Stop:

   We can pick $i = \lfloor {\lg n} \rfloor$ for stopping, then $0 < {n\over{2^i}} \le 2$.

   By DIC, choose $T(n) = 0$ for all $n \le 2$.

   Therefore, for $n > 1$,

   $$\begin{gather*} T(n) = \lfloor {\lg n} \rfloor n \end{gather*}$$

Basic Sums

Other kinds of sums are often reduces to the following forms.

Arithmetic Sums

$$\begin{gather*} S^k_n := \sum^n_{i=1}{i^k} \end{gather*}$$

Solution: $S^k_n = \Theta(n^{k+1})$

Geometric Sums

When $x \ne 1$,

$$\begin{gather*} S_n(x) := \sum^{n-1}_{i=0}{x^i} \end{gather*}$$

Solution: $S_\infty = {{x^n - 1}\over{x - 1}}$

Infinite Geometric Series

When $\lvert{x}\rvert < 1$,

$$\begin{gather*} S_{\infty}(x) := \sum^{\infty}_{i=0}{x^i} \end{gather*}$$

Solution: $S_{\infty}={1\over{1 - x}}$

Harmonic Series

$$\begin{gather*} H_n := 1 + {1\over{2}} + {1\over{3}} + \dots + {1\over{n}} \end{gather*}$$

Solution: $H_n = ln(n) + g(n)$ where $0 < g(n) < 1$.

Summation Techniques

Growth Types

Related homework questions: h3-Q1

Polynomial Type

A real function $f$ is polynomial-type if $f$ is non-decreasing (ev.) and there is some $C > 1$ such that:

$$\begin{gather*} f(x) \le C f(x/2)\ (ev.) \end{gather*}$$

Increasing Exponential Type

$f$ increases exponentially if there exists real numbers $C > 1$ and $k > 0$ such that:

$$\begin{gather*} f(x) \ge C f(x - k)\ (ev.) \end{gather*}$$

Decreasing Exponential Type

$f$ decreases exponentially if there exists real numbers $C > 1$ and $k > 0$ such that:

$$\begin{gather*} f(x) \le f(x - k) / C\ (ev.) \end{gather*}$$

Lemma 8: Closed Properties

(a) Polynomial-type functions are closed under addition, multiplication, and raising to any positive power $a > 0$.

(b) Exponential-type functions $f$ are closed under addition, multiplication, and raising to any power $a$. In case $a > 0$, the function $f^a$ will not change its subtype (increasing or decreasing). In case $a < 0$, the function $f^a$ will change its subtype.

(c) If $f$ is polynomial-type and $\lg f$ is non-decreasing then $\lg f$ is also polynomial-type. If $f$ is exponential-type and $a > 1$ then so is $a ^ f$.

Summation Rules by Growth Type

Related homework questions: h3-Q2

Theorem 6: Summation Rules

$$S_f(n) = \Theta \begin{cases} n f(n) & \text{if f is polynomial-type,} \\ f(n) & \text{if f is increasing exponentially,} \\ 1 & \text{if f is decreasing exponentially.} \end{cases}$$

Examples:

• Polynomial Sums:

  $$\sum^n_{i\ge{1}}{i \log(i)} = \Theta(n^2 \log(n)), \sum^n_{i\ge{1}}{\log(i)} = \Theta(n \log(n)), \sum^n_{i\ge{1}}{i^a} = \Theta(n^{a+1})(a\ge{0}).$$
• Exponentially Increasing Sums:

  $$\sum^n_{i\ge{1}}{b^i} = \Theta(b^n)(b\ge{0}), \sum^n_{i\ge{1}}{i^{-5}2^{2^i}} = \Theta(n^{-5}2^{2^n}), \sum^n_{i\ge{1}}{i!} = \Theta(n!).$$
• Exponentially Decreasing Sums:

  $$\sum^n_{i\ge{1}}{b^{-i}} = \Theta(1)(b\ge{0}), \sum^n_{i\ge{1}}{i^2 i^{-i}} = \Theta(1), \sum^n_{i\ge{1}}{i^{-i}} = \Theta(1).$$

More Examples

Transformation Techniques

Domain Transformation

Related homework questions: h3-Q3 and h3-Q4(combining range transformation)

Example:

Consider:

$$\begin{gather*} T(N) = T(N/2) + N \end{gather*}$$

We define $t(n) := T(2^n) \text{or} N = 2^n$.

This transforms the original N-domain into the n-domain:

$$\begin{align*} T(2^n) &= T(2^n/2) + 2^n \\ T(2^n) &= T(2^{n-1}) + 2^n \\ t(n) &= t(n-1) + 2^n \end{align*}$$

We get this standard form. By DIC, we choose the boundary condition $t(n) = 0$ for all $n \le 0$.

We get sum:

$$\begin{gather*} t(n) = \sum^n_{i\ge{0}}{2^n}=\Theta(2^n) \end{gather*}$$

Because we have $n = \lg N$,

$$\begin{gather*} T(N) = \Theta(N) \end{gather*}$$

On Lecture II Page 46, L1120, we get the exact sum by transforming the descending sum to the ascending sum.

Range Transformation

Related homework questions: h3-Q4(combining domain transformation)

Example:

$$\begin{gather*} T(n) = 2T(n-1) + n \end{gather*}$$

We define $t(n) := {T(n)\over{2^n}}$, then divide the both sides of the original equation by $2^n$.

$$\begin{align*} {T(n)\over{2^n}} &= 2{T(n-1)\over{2^n}} + {n\over{2^n}} \\ {T(n)\over{2^n}} &= {T(n-1)\over{2^{n-1}}} + {n\over{2^n}} \\ t(n) &= t(n-1) + {n\over{2^n}} \end{align*}$$

We get this standard form. By DIC, we choose the boundary condition $t(n) = 0$ for all $n \le 0$.

We get sum: ($\sum^n_{i\ge{0}}{n\over{2^n}}$ is exponentially decreasing)

$$\begin{gather*} t(n) = \sum^n_{i\ge{0}}{n\over{2^n}}=\Theta(1) \end{gather*}$$

Because $t(n) = {T(n)\over{2^n}}$,

$$\begin{gather*} T(n) = \Theta(2^n) \end{gather*}$$

Master Theorem

Related homework questions: h2-Q8

Master recurrence:

$$\begin{gather*} T(n) = a T(n/b) + d(n) \end{gather*}$$

where $a > 0$, $b > 0$ are real constants and $d(n)$ the driving/forcing function.

Watershed constant

$w = \log_b(a)$

Watershed function

$W(n) = n^w = n^{\log_b(a)}$

Useful link: Online basic master theorem solver

Master Theorem

Theorem 10: Master Theorem on Lecture II Page 51, L1207

$$\begin{gather*} T(n) = \Theta \begin{cases} n^w, & \text{if } d(n) = O(n^{w-\epsilon}) \text{ for some } \epsilon > 0, & CASE(-) \\ n^w \log(n), & \text{if } d(n) = \Theta(n^w), & CASE(0) \\ d(n), & \text{if } a\cdot d(n/b) \le c\cdot d(n) \text{ for some } 0 < c < 1. & CASE(+) \\ \end{cases} \end{gather*}$$

Extended Master Theorem

Theorem 12: Extended Master Theorem on Lecture II Page 54, L1282

$$\begin{gather*} T(n) = \Theta \begin{cases} d(n), & \text{if } a\cdot d(n/b) \le c\cdot d(n) \text{ for some } 0 < c < 1. & CASE(+) \\ W(n)\log\log n, & \text{if } d(n) = \Theta(W(n)\log^c(n)) \text{ for } c = -1, & CASE(1) \\ W(n)\log^{c+1}(n), & \text{if } d(n) = \Theta(W(n)\log^c(n)) \text{ for } c > -1, & CASE(0) \\ W(n), & \text{if } d(n) = O(W(n)\log^c(n)) \text{ for } c < -1. & CASE(-) \\ \end{cases} \end{gather*}$$

Multi-term Master Recurrences

Related homework questions: h3-Q5

$$\begin{gather*} T(n) = d(n) + \sum_{i=1}^{k}{a_i T(n/b_i)} \end{gather*}$$

where $a_i > 0$ and $b_1 > b_2 > \cdots > b_k > 1$ are real constants.

Watershed constant

It is the real number $\alpha$ such that:

$$\begin{align*} 1 = \sum_{i=1}^{k}{a_i\over{b_i^{\alpha}}} \end{align*}$$

It clearly exists and is unique.

Multi-term Master Theorem

Theorem 15: Multi-term Master Theorem on Lecture II Page 62, L1483

$$\begin{gather*} T(n) = \Theta \begin{cases} n^{\alpha}, & \text{if } d(n) = O(n^{w-\epsilon}) \text{ for some } \epsilon > 0, & CASE(-) \\ n^w, & \text{if } d(n) = \Theta(n^w), & CASE(0) \\ d(n), & \text{if } \sum^k_{i=1}{a_i\cdot d(n/{b_i})} \le c\cdot d(n), & CASE(+) \\ \end{cases} \end{gather*}$$

Real Induction

Related homework questions: h2-Q9