CS432, Fall 2005
Solutions for HW5

Problem 1:
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Part A:
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The question is:  Explain in a sentence or two why the optimal solution
would be found by the final step(s) when we do the continuous knapsack?
What leads to the difference in these results?

Answer:  The greedy solution is optimal for the *continuous* knapsack
problem because it can add a fraction of the next best item in order
to fill up the knapsack to its capacity.  But for the 0/1 knapsack,
it cannot add a franction of an item, but only all of the item or
nothing of the item.

Part B:
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The table should look like this:

	w = 0	w=1	w=2	w=3	w=4
k = 0	0	0	0	0	0
k = 1	0	3	3	3	3
k = 2	0	3	5	8	8
k = 3	0	3	5	8	9



Problem 2:
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Solution:
You must show that a non-deterministic algorithm can solve the 
0/1 knapsack problem in polynomial time.

Pseudocode:

check-knapSack(v[], w[], n, W, k) {
  // phase 1: generate certificate (a set if items to be chosen)
  s = GetItems(v, w, n, W, k); // non-deterministic!
  // assume s is an array s[1..m] of unique int values,
  //   where each is a index of an item to be chosen.

  // phase 2: check s to see if it is a subset that meets the 
  //    problem constraints
  weight = value = 0
  for (i=1; i < m; ++i) { // for each item in certificate
     if ( s[i] > n || s[i] < 1 ) return false; // valid item?
     weight = weight + w[ s[i] ];
     value  = value  + v[ s[i] ];
  }
  if ( weight > W ) return false;  // too heavy
  if ( value  < k ) return false;  // didn't meet value-target
  return true; // else OK!
}

Is this polynomial?
Phase 1 is, since we assume s[] is of length m<=n.
Cost Theta(n) to write it out.
Phase 2 is too, since m<=n.  This is O(n).