Solution for problem 3:

Part a:
heuristic = 2 * the least # of doors between kitchen and living room (+ 1 if orange not picked up)

this would give you: (h,f)

	5,5

	3,5

5,9	3,7	3,8

3,9	1,7

3,11	0,7	1,10

Part b:

go LR to FR
go FR to DR
go DR to K
pickup orange

total cost = 7. It is optimal because h is admissible and A* is optimal with
admissible h.

Part c:

go K to DR
go DR to FR
go FR to LR

total cost = 6

Part d:

go LR to FR
open door
go FR to K
pickup orange
go K to FR
go FR to LR

total cost = 12, the combined plans for b and c would cost 13. The reason for the difference
is that now our overall goal is different and the cost of opening the door is amortized because
we dont have to go through the DR anymore.

Part a:
	- not defining the heuristic clearly. this would cost you 5-8 points.
	- h is not admissible.. almost correct but h(goal) = 0... 4 points
	- a lot of people defined h as : "estimated cost of the cheapest path". 
	well this is true but stating a heuristic for a problem involves giving 
	a computable function where given a state s, it will give a value v. (-5)

	- if your definition was not adequate BUT you gave a laydown of h values for
	each room, then you got full points.

	the things that we were looking for in the definition of heuristic was

	- some notion of "shortest" (3 points)
	- mentioning of "rooms" or doors (3 points)

	the best answer was "least # of doors between kitchen and living room" 
	or "least # of rooms".. people only saying the shortest path got deducted 
	3 points.. and people only saying # of rooms got taken off 3 points as well..

	- not filling in the h-f values. 7 points
	- just filling in the h-f values of node x's. 5 points 
	(because the point of other h-f values is to show that A* with your heuristic 
	will take the expansion path shown in the figure)

	- incorrect f function.. some people thought f = g.. 4 points

part b:
	- not saying whether the path is optimal -1
	- not saying why it is optimal -2
	- basically what we were looking for "why" was that A* with admissible 
	h is optimal.if you missed this, you missed 1 point.

part c:
	- no problems here

part d:

	- not saying why and how the new path is different -2