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Lect 15 - November 14, 2008 - ECS 20 - Fall 2008 - Phil Rogaway
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Today: o Induction, recursion and recurrence relations, cont.
(special Friday 50-minute in-discussion-section lecture)

--------------------------
1. Induction and recursion
--------------------------

EXAMPLE 1. Sam's Dept Store sells enveloped in packages of 5 and 12. 
        Prove that, for any n\ge 44, the store can sell you exactly n envelopes.
       [GP, p.147]

Try it:  44 = 2(12) + 4(5)
         45 =         9(5)
         46 = 3(12) + 2(5)
         ?...?

SUPPOSE: it is possible to buy k envelopes for some k\ge44.
SHOW:    it is possible to buy k+1 envelopes

If purchasing at least seven packets of 5, trade them for three packets of 12:
  7(5) -> 3(12)
  35       36

If <7 packets of 5, ie <=6 fewer packets of 5, so at most
30 of the envelopes are in packets of 5; so there are >= 44-30 = 14 envelopes
being bought in packets of 12, so >= 2 two packets of twelve.
So take two of the packets of 12 and trade them for 5 packets of 5:
  2(12) -> 5(5)
   24      25


-----------------------------------
Principle of mathematical induction
-----------------------------------
To prove a proposition P(n) for all integers n\ge n_0:
1) Prove P(n_0)    (Basis)
2) Prove that P(k) --> P(k+1) for all k > n_0    (inductive step)
    (Inductive hypothesis)

(initially make n_0=1)

And that's as far as we got today!!

Sometimes convenient to strengthen this:

-----------------------------------
Principle of mathematical induction
-----------------------------------
To prove a proposition P(n) for all integers n\ge n_0:
1) Prove P(n_0)    (Basis)
2) Prove that (P(n_0) and P(k+1) and ... and P(k))  --> P(k+1) 
for all k > n_0    (inductive step)
    (Inductive hypothesis)


EXAMPLE 2: Show that you can tile any "punctured" n x n grid with
"trominos" 

           #
           ##      (may be rotated)

Illustrate and prove, dividing board in into 2^k x 2^k to prove
2^{k+1} x 2^{k+1}, puncturing three near-center center points (for 
the three 2^k} x 2^k pieces that lacking the puncture), and recurse.

EXAMPLE 3:  Prove that the sum of the the odd integers 2 .. 2n-1 is n^2
    1 + 3 + ... + (2n-1) = n^2.


Basis:   n=1, check

Inductive step:   


 4*2^{2k}-1 = (2^{2k} - 1) + (2^{2k} - 1) + (2^{2k} - 1)  + (2^{2k} - 1) + 3
               3 divides      3 divides      3 divides       3 divides     3 divides


suppose 1 + 3 + ... (2n-3) = (n-1)^2
                                    + 2n-1   = (n-1)^2 + 2n-1
                                             = n^2 - 2n + 1 + 2n - 1 = 1 
                                             = n^2 

(For HW p. 158, p. 30?) 

EXAMPLE 4: Cake cutting.  
See http://www.cs.berkeley.edu/~daw/teaching/cs70-s05/Notes/cakecutting.pdf
for a nice writeup

n people. want to divide equally.
n=2: known case.
n>=3:


1. Persons 1 .. n-1 people divide the cake 
   into n-1 pieces (using a recursive call to this procedure).

2. Persons 1 .. n-1 divide their piece into n equal shares.

3. Person n takes the largest piece among the pieces held by
   each person 1..n-1. 

4. Persons 1..n-1 keep their remaining n-1 pieces for themselves


Remaining n-1 people keep their remaining portion of cake.

Running time:
T_n = T_{n-1} + n
T_n = Theta(n!)

Yuck!