CSE 311 Lecture 17: Strong Induction

Topics

Induction

A brief review of Lecture 16.

Strong induction

Induction with a stronger hypothesis.

Using strong induction

An example proof and when to use strong induction.

Recursively defined functions

Recursive function definitions and examples.

A template for proofs by induction

A template for proofs starting at any integer $b\in\Z$

① Let $P(n)$ be [ definition of $P(n)$ ].

We will show that $P(n)$ is true for every integer $n\geq b$ by induction.

② Base case ($n = b$):

[ Proof of $P(b)$. ]

③ Inductive hypothesis:

Suppose that $P(k)$ is true for an arbitrary integer $k\geq b$.

④ Inductive step:

We want to prove that $P(k+1)$ is true.

[ Proof of $P(k+1)$. This proof must invoke the inductive hypothesis. ]

⑤ The result follows for all $n\geq b$ by induction.

 

Induction with a stricter hypothesis

If you can’t prove $P(k) \to P(k+1)$ then you need a stricter hypothesis!

Find a $Q(n)$ such that $Q(n) \to P(n)$ and $Q(k) \to Q(k+1)$.

Prove $Q(n)$ by induction to conclude that $P(n)$ holds (by MP).

There is no recipe for finding $Q(n)$.

But looking at examples can help.

And also thinking about why the attempted proof of $P(k) \to P(k+1)$ fails.

For a proof like this, you must argue that $Q(n) \to P(n)$.

If it’s not obvious that $P(n)$ follows from $Q(n)$, you have to prove it.

Otherwise, it’s enough to state that you are using a stricter hypothesis.

There is no recipe for determining what is obvious :)

Recall how induction works

Note that we have $P(0), \ldots, P(k)$ when proving $k+1$.
So we can safely assume all of them, rather than just $P(k)$.

The strong induction rule of inference

Strong inductive proofs for any base case $b\in\Z$

① Let $P(n)$ be [ definition of $P(n)$ ].

We will show that $P(n)$ is true for every integer $n\geq b$ by strong induction.

② Base case ($n = b$):

[ Proof of $P(b)$. ]

③ Inductive hypothesis:

Suppose that for some arbitrary integer $k\geq b$, $P(j)$ is true for every integer $b \leq j \leq k$.

④ Inductive step:

We want to prove that $P(k+1)$ is true.

[ Proof of $P(k+1)$. The proof must invoke the strong inductive hypothesis. ]

⑤ The result follows for all $n\geq b$ by strong induction.

 

Example: the fundamental theorem of arithmetic

Examples

$48= 2\cdot 2\cdot 2\cdot 2\cdot 3$

$591 = 3 \cdot 197$

$45,523 = 45,523$

$321,950 = 2 \cdot 5 \cdot 5 \cdot 47 \cdot 137$

$1,234,567,890 = 2 \cdot 3 \cdot 3 \cdot 5 \cdot 3,607 \cdot 3,803$

We use strong induction to prove that a factorization into primes exists (but not that it is unique).

Prove that every integer $\geq 2$ is a product of primes

Strong induction is particularly useful when …

We need to reason about procedures that given an input $k$ invoke themselves recursively on an input different from $k-1$.

Example:

Euclidean algorithm for computing $\gcd{a}{b}$.

// Assumes a >= b >= 0.
public static int gcd(int a, int b) {
  if (b == 0)
    return a;             // GCD(a, 0) = a
  else  
    return gcd(b, a % b); // GCD(a, b) = GCD(b, a mod b)
}

We will use strong induction to reason about this algorithm and other functions with recursive definitions.

Giving a recursive definition for a function

To define a recursive function $f$ over $\N$, give its output in two cases:

Base case: the value of $f(0)$.

Recursive case: the value of $f(n+1)$, given in terms of $f(n)$.

When the recursive case refers only to $f(n)$, as in these examples, we can prove properties of $f(n)$ easily using ordinary induction.

Example: prove $n! \leq n^n$ for all $n\geq 1$

Fun: can we verify $n! \leq n^n$ for all natural numbers?

// x^y where 0^0 = 1
function expt(x : nat, y: nat) : nat {
   if y == 0 then 1 else x * expt(x, y-1)
}

// n!
function fact(n : nat) : nat {
  if n == 0 then 1 else n * fact(n-1)   
}

// n! <= n^n for all natural numbers
lemma factLemma(n : nat)
  ensures fact(n) <= expt(n, n)
{ }

// x^y where 0^0 = 1
function expt(x : nat, y: nat) : nat {
   if y == 0 then 1 else x * expt(x, y-1)
}

// n!
function fact(n : nat) : nat {
  if n == 0 then 1 else n * fact(n-1)   
}

// n! <= n^n for all natural numbers
lemma factLemma(n : nat)
  ensures fact(n) <= expt(n, n)
{
  if (n == 0) {           // Base case
    assert fact(0) <= expt(0, 0);
  } else {               // Inductive step
    factLemma(n-1);      // Inductive hypothesis
    exptLemma(n-1, n-1); // (n-1)^(n-1) <= n^(n-1)
    assert fact(n) == n * fact(n-1);               // by fact defn
    assert n * fact(n-1) <= n * expt(n-1, n-1);    // by IH
    assert n * expt(n-1, n-1) <= n * expt(n, n-1); // by exptLemma
    assert fact(n) <= expt(n, n);                  // qed.
  }
}

// x^y <= (x+1)^y for all natural numbers.
lemma exptLemma(x: nat, y: nat)
  ensures expt(x, y) <= expt(x + 1, y)
{}

Defining a recursive function with multiple base cases

A recursive function can have more than one base case.

Base cases give the value of $f(0), \ldots, f(m)$ where $m\geq 0$.

Recursive case defines $f(n+1)$ in terms of $f(n-m), \ldots, f(n-1), f(n)$ for all $n\geq m+1$. Or it defines $f(n)$ in terms of $f(n-1-m), \ldots, f(n-1)$.

Strong inductive proofs with base cases $b, \ldots, b+m$

Example: prove $f_n < 2^n$ for all $n\geq 0$

Fun: can we verify $f_n < 2^n$ for all $n\geq 0$?

// 2^n
function pow2(n : nat) : nat {
   if n == 0 then 1 else 2 * pow2(n-1)
}

// Fibonacci function f_n
function fib(n: nat): nat {
  if n == 0 then 0
  else if n == 1 then 1
  else fib(n-2) + fib(n-1)
}

// f_n < 2^n
lemma fibLemma(n : nat)
  ensures fib(n) < pow2(n)
{ }