Q1: Show that \sum_{i=1}^n i^2 is in O(n^3)

Sol: let's prove by induction on n that \sum_{i=1}^n i^2 <= n^3 for n>=1
- do the basis (easy)
- Ind. Hyp: true for all values up to k>=1.
  for n=k+1: \sum_{i=1}^{k+1} i^2 = \sum_{i=1}^k i^2 + (k+1)^2
                                 <=  k^3 + k^2 + 2k + 1
                                 <=  k^3 + 3k^2 + 3k + 1
                                  =  (k+1)^3

Q2: Suppose you are given a set of small boxes, numbered 1 to n, identical
in every aspect except that each of the first i contains a pearl whereas
the remaining n-i are empty. You can have two magic wands that can each
test if a box is empty or not in a single touch, except that a wand 
disappears if you test it on a box that is empty. Show that, without
knowing the value of i, you can use the two wands to determine all
the boxes containing pearls using at most no more than 2\sqrt{n} wand
touches.

Sol: We will fix number k later.
The idea is first use one wand on boxes 1, k, 2k, 3k, ...
The smallest i for which the wand burns on box i*k indicates
that the first empty box is among (i-1)*k+1....i*k
Now we use the second wand sequentially from (i-1)*k+1 to i*k-1
to find it.
The total number of toches will be at most: n/k + k
n/k is the number of boxes for the first wand and k for the second one

If we now choose k to be (about) \sqrt{n} then we have n/\sqrt{n}+\sqrt{n}
which is in O(\sqrt{n}) touches.

Q3: LOOP INVARIANT EXAMPLE
Prove the following pseudocode is correct, i.e., returns,
max_{i in 1..length(A)-1} |A[i] - A[i+1]|

DIFF(A)
   d = |A[1] - A[2]|
   for i = 2 to length(A)-1 do
     if (|A[i] - A[i+1]| > d)
        d = |A[i] - A[i+1]|
   return d

Proof:

Loop invariant is,
   d = max_{i in 1..i-1} |A[i] - A[i+1]|

We proved initialization, maintenance, and showed the loop invariant at
termination proves correctness