CMSC250

Recall Weak Induction

Goal: $\forall n \in \mathbb{Z}^{\ge 0}, \sum\limits_{i=1}^n i = \frac{n(n+1)}{2}$

Goal: $\forall n \in \mathbb{Z}^{\ge 0}, \sum\limits_{i=1}^n i = \frac{n(n+1)}{2}$

$P(0),P(1),P(2),\ldots$

Instead what if we proved:
$P(0),P(0) \Rightarrow P(1), P(1) \Rightarrow P(2), P(2) \Rightarrow P(3), \ldots$

Goal: $\forall n \in \mathbb{Z}^{\ge 0}, \sum\limits_{i=1}^n i = \frac{n(n+1)}{2}$

Instead what if we proved:
$P(0),P(0) \Rightarrow P(1), P(1) \Rightarrow P(2), P(2) \Rightarrow P(3), \ldots$

(If we generalize this: $P(0),P(k) \Rightarrow P(k+1), k\ge 0$)

Strong induction is slightly different

Strong induction is slightly different

 public int strong(int n){
  if(n==0) return 3;
  if(n==1) return 4;
  return -n + strong(n-1) + strong(n-2);
}

 public int strong(int n){
  if(n==0) return 3;
  if(n==1) return 4;
  return -n + strong(n-1) + strong(n-2);
}

claim: $\text{strong}(n) = n + 3$

 public int strong(int n){
  if(n==0) return 3;
  if(n==1) return 4;
  return -n + strong(n-1) + strong(n-2);
}

Weak induction would not work

strong$(n) \nRightarrow \text{ strong}(n+1)$

instead: strong$(n) \land \text{strong}(n -1) \Rightarrow \text{ strong}(n+1)$

strong$(n) \nRightarrow \text{ strong}(n+1)$

instead: $\text{strong}(n) \land \text{strong}(n -1) \Rightarrow \text{ strong}(n+1)$

Recall the weak induction assumption:$k \ge n_0, P(k)$

We need to modify for Strong: $\forall i, k \ge i \ge n_0, P(i)$

 public int strong(int n){
  if(n==0) return 3;
  if(n==1) return 4;
  return -n + strong(n-1) + strong(n-2);
}

\[ s(x) = \begin{cases} 3 & x = 0 \\ 4 & x = 1 \\ -n + s(x-1) + s(x-2) & x \ge 2 \end{cases} \]

\[ s(x) = \begin{cases} 3 & x = 0 \\ 4 & x = 1 \\ -x + s(x-1) + s(x-2) & x \ge 2 \end{cases} \]

Claim: s(x) = x + 3

\[ s(x) = \begin{cases} 3 & x = 0 \\ 4 & x = 1 \\ -x + s(x-1) + s(x-2) & x \ge 2 \end{cases} \]

Explicit Base cases

Not always obvious

Suppose I have 3¢ and 10¢ coins

Prove I can pay anything 18¢ or more

Question: How many Base Cases?

How many things we rely on?

Suppose I have 3¢ and 10¢ coins

18¢ = 3¢ + 3¢ + 3¢ + 3¢ + 3¢ + 3¢

19¢ = 10¢ + 3¢ + 3¢ + 3¢

20¢ = 10¢ + 10¢

21¢ = 18¢ + 3¢

22¢ = 19¢ + 3¢

23¢ = 20¢ + 3¢

3 Base cases: P(18), P(19), P(20)

Suppose I have 3¢ and 10¢ coins

3 Base cases: P(18), P(19), P(20)

$P(n) = P(n-3) + 3, n\ge 21$

$P(n+1) = P(n-2) + 3, n\ge 20$

Prove: All integers $\ge 2$ has a prime factor