November 2011 Exam Question 1 Incorrect??

I posted this onto the UNISA discussion forum, just circulating it here...

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The question says that jackals must not outnumber coyotes at any time:

Some scenarios:

Start state: Assume that one jackal crosses. There will be one jackal on the opposite riverbank. A case will exist where jackals outnumber coyotes, so this cannot be correct.
Start State: Assume that one coyote crosses. There will be three jackals and two coyotes on the first riverbank. A case will exist where jackals outnumber coyotes, so this cannot be correct.
Start State: Assume that both a jackal and a coyote cross. All okay in this step. Next step, and a similar problem happens as mentioned above.

I my mind there is no way that this is feasible?

Any thoughts??
Thanks
MIke


(2012-05-23 22:59)

Now see below:

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There is a problem with the way this question is and was worded in the exam which changed the meaning completely!!! "Three jackals and three coyotes come to a river. There is a boat on THEIR side of the river...." This leads me to believe that all the animals are on the same side of the river. But the link below shows that this is not the case!!!

http://answers.yahoo.com/question/index?qid=20100307182303AA3ebpk



(2012-05-23 23:10)

I answered it in in the MyUnisa forum, but I will repeat it here as well:

Actually there is a solution for the problem as it is stated (it took me a while to figure it out though, I really hope they do not give one this difficult in the exam, it just seem silly to me to even include this type of question at all in an exam):

State space: s = {A(Coyotes, Jackals, Boat?), B(Coyotes, Jackals, Boat?)}
Initial State s = {A(3, 3, Y), B(0, 0, N)}
Goal State g = {A(0, 0, N), B(3, 3, Y)}

{A(3, 3, Y), B(0, 0, N)}
{A(2, 2, N), B(1, 1, Y)}
{A(3, 2, Y), B(0, 1, N)}
{A(3, 0, N), B(0, 3, Y)}
{A(3, 1, Y), B(0, 2, N)}
{A(1, 1, N), B(2, 2, Y)}
{A(2, 2, Y), B(1, 1, N)}
{A(0, 2, N), B(3, 1, Y)}
{A(0, 3, Y), B(3, 0, N)}
{A(0, 1, N), B(3, 2, Y)}
{A(0, 2, Y), B(3, 1, N)}
{A(0, 0, N), B(3, 3, Y)}

Hi, yes I agree there is a solution and yes, it takes forever to draw the whole state space (for 8 marks is seems silly to have such a question).

Regarding the first post here:

Start state: Assume that one jackal crosses. There will be one jackal on the opposite riverbank. A case will exist where jackals outnumber coyotes, so this cannot be correct.
Start State: Assume that one coyote crosses. There will be three jackals and two coyotes on the first riverbank. A case will exist where jackals outnumber coyotes, so this cannot be correct.
Start State: Assume that both a jackal and a coyote cross. All okay in this step. Next step, and a similar problem happens as mentioned above.

The first move above is legal (i think their wording was just wrong). Having a single jackal on the opposite side is fine (there are no coyotes to worry about). I think the rule only applies when you actually have coyotes and jackals (then jackals may not out number).
The same rules as with the cannibals/missionaries question in the book.

Danie van Eeden
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Thanks guys...if i followed the question completely literally I would be correct. Thanks for the clarification.