LL]]>

Congrats to everybody that passed, I wish I were you guys. I am only writing an aegro in January.

Can you please help me here with criptarithmetic, I have tried to understand it through the notes in the phorum, they area great and the websites but I don't seem to get it clearly.

Your input will be highly appreciated

Thanks]]>

I am still waiting in anticipation for all my results!!

So stressed out...

Was wondering how many results you got and when you wrote it!!

Anyway!

Hope all your results are what you wished for!!

(:P)]]>

This is actually shocking!!

But maybe I will pass now!!]]>

1) TreeDBNotes

2) Paint.NET

See my blog for an example screenshot and link to the TreeDBNotes site where the make available a free version for private (non-commercial use)

Ian's Blog]]>

I got the tree right, but only calculated entropy in my rough work. Used the ratios at the back to calculate information gain. Hope they're lenient :/]]>

now the waiting for the results.....]]>

Thank you for being so helpful it certainly made a difference to me. GOOD LUCK with the exam, hope this year's paper will be better.

LL>:D<]]>

]]>

Once again, the exam guidelines are vague :/]]>

Inc Oct 2007, the CSP given in the exam was:

HOCUS +POCUS ------ PRESTO

How would you go about solving that during the exam other then brute force guessing ?

Using the text book, i can get to these conclusions:

S+S = O + 10*X1 X1 + U+U = T + 10*X2 X2 + C+C = S + 10*X3 X3 + O+O = E + 10*X4 X4 + H+P = R + 10*X5 X5 = P

constraint graph:

]]>

Where is this Mock exam you are referring to ? i've gone through the tuts for this module and couldn't find it.]]>

Variables: T, A, B, P

Domain(s): {1, 2, 3, 4}

Constraint(s): (note I'm using <> for not equal to)

T <> A

T <> B

T <> P

A <> B

A <> P

B <> P

A = 1 or 2

T <> 4

That's it, let me know what you think please. ;-)]]>

The definition is a bit unclear, so here it is again:

Refutation Completeness = resolution can always be used to either confirm or refute a sentence.

http://ian-coetzer.blogspot.com/]]>

http://ian-coetzer.blogspot.com/2008/11/cos351-d-mock-exam-question-4-proposed.html]]>

========================================

Variables = {T, A, B, P}

Domain for all variables = {1, 2, 3, 4}

Constraints:

- Alldiff(T, A, B, P)

- A < P

- A < B

- T != 4

1. Arc consistency: remove 4 from A (because of A<P and A<B ) and 4 from T (T!=4).

- T: {1, 2, 3}

- A: {1, 2, 3}

- B: {1, 2, 3, 4}

- P: {1, 2, 3, 4}

2. MRV states we use T or A. The degree heuristic then tells us to use A (because it is highly constrained). The least-constraining value for A is 1. Arc consistency then removes 1 from P, B and T.

- T: {2, 3}

- A: 1

- B: {2, 3, 4}

- P: {2, 3, 4}

3. MRV states we assign T next since it has the fewest remaining values. Values 2 and 3 rule out the same values for the domains of B and P so the least-constraining-value heuristic cannot be used. The value 2 will be used because it is before 3 in the domain's value list. Arc consistency removes 2 from B and P.

- T: 2

- A: 1

- B: {3, 4}

- P: {3, 4}

4. The MRV and degree heuristics cannot distinguish between B and P (they are equally constrained and have the same number of legal values remaining). Hence, B will be used since it is first in the variable list. Value 3 will be used since it is first in the domain list. Arc consistency removes value 3 from P's domain.

- T: 2

- A: 1

- B: 3

- P: {4}

5. P = 4.

- T: 2

- A: 1

- B: 3

- P: 4]]>

Any assistance would be very much appreciated.]]>

Please check the correctness and quality of my answer. Thanks.

We can represent each state as two sets, like so: {...}{...}. The first set contains those people who havn't crossed the river, and the second contains those who have crossed. Tom, Dick, and Harry are represented by T, D, and H respectively.

Initial state: {T, D, H}{}

Successor function: Returns states that move one or two people only from one side of the river to the other. Tom and Dick are not allowed to be on the same side of the river without Harry being present.

Goal test: {}{T, D, H}

Path cost: 1 per boat trip (i guess??)

State space:

{T,D,H}{},

{T}{D,H},

{D},{T,H},

{T,H}{D},

{D,H}{T},

{}{T,D,H}]]>

Here is my answer for question3. I'm not entirely sure if my queue orders are correct when f is the same amoungst nodes. Please check my answer!!

f(1) = g(1) + h(1) = 1 + 4 = 5

Open Queue = {1}

Expand node 1:

f(2) = g(2) + h(2) = 2 + 4 = 6

f(3) = g(3) + h(3) = 2 + 4 = 6

f(4) = g(4) + h(4) = 2 + 4 = 6

f(5) = g(5) + h(5) = 2 + 4 = 6

Open Queue = {2, 3, 4, 5}

Expand node 2:

No fringe

Expand node 3:

f(6) = g(6) + h(6) = 3 + 4 = 7

f(7) = g(7) + h(7) = 3 + 3 = 6

Open Queue = {7, 4, 5, 6}

Expand node 7:

f(10) = g(10) + h(10) = 4 + 2 = 6

f(11) = g(11) + h(11) = 4 + 3 = 7

Open Queue = {10, 4, 5, 11, 7}

Expand node 10:

Stop: Node with depth = 4 selected.

Cheers]]>

http://www.emunix.emich.edu/~evett/AI/AlphaBeta_movie/sld001.htm]]>

According to the exam guidelines, we only need to know the basics from 5.1 (variables, domains and constraints), 5.3 and 5.4.

Does this mean no backtracking search???]]>

They don't explicitly mention that we have to study belief state spaces (as with Assign2, Q3).

In the guidelines it says we must study "Graphical representations of searches" - does this refer to search trees like in figure 3.12 and 3.15?]]>