Q1) 0/1 knapsack:
This is a variation of the knapsack problem where each item
can be either picked entirely or not picked at all.
More precisely, we have a set of n items with weights
w_1,\ldots,w_n and profits/values b_1,\ldots,b_n.
For simplicity assume that all profits are equal to the weights, i.e.
b_i=w_i for all i (e.g. each item is a peice of gold).
We want to pick a subset of items to put into a knapsack with capacity W
maximizing the value. This is equivalent to
picking a subset of items whose total weight comes as close
as possible to W.

1-a) Show by an example, why the obvious greedy algorithm does not for
this problem.

Sol: suppose we sort the items in decreasing (or precisely
non-increasing) order of weight. We consider the items in this order and
place
it in the knapsack if it fits. Here is a counter-example:
w_1=501, w_2=500, w_3=500, and W=1000. This way the algorithm will
only pick item 1 while the optimal solution is to pick items 2 and 3.
So the solution can be as bad as (almost) half the optimal.

If we sort the items in the non-decreasing then the situation can be
worse: assume w_1=1 w_2=1000 and W=1000. Then the algorithm will pick
only item 1 while the optimal solution picks w_2.

1-b) Let M_1 be the value of the solution by the first greedy and
M_o be the value of the optimal. Prove that at least M_1\geq M_o/2
(that is the first greedy is at least half the optimal).

Sol: We prove that if M_1\not= M_o then M_1>M_o/2. This clearly implies
that always M_1>= M_o/2. So, by way of contradiction, suppose that
M_1<M_o/2. Note that the greedy algorithm stops when either the knapsack
is completely full (in which case we have value W which must be OPT) or
when the remaining items do not fit into the knapsack. Since we assumed
that M_1<M_o/2 so the greedy must have stopped because the remaining
items do not fit into the knapsack.
But because M_1<M_o/2<=W/2 it means the last item we have
picked is <W/2. Since the items are sorted in non-increasing order,
the next item which we could not pick (after the last one we placed into
knapsack) must be <W/2 too. That means there is still room to place the
next item into the knapsack, because so far we have picked <M_o/2<=W/2
and the next item is also <W/2. So the greedy cannot have stopped, a
contradiction.