How do you guys find this module, difficult, interesting or both?

Would it be better to do Formal Logic 1 (LGC something) or Formal Logic 2 as a pre-requisite and how would you rate them?]]>

thoughts?]]>

How does this answer looks like?

Which type of search and what is the sequence

of nodes visited? How does the answer looks like??

Regards

Francois du Toit]]>

Anyone have an answer on this one??

Please post example answer that I can work towards

a goal to know if I am correct

What type of search use: breadth first search

and using the four conditions as h(n) function

to know if at the 4rth level if any more moves

exists as their is not a solution at level 4

as shown in example paper.

If node do have more possible nodes and it did

not show as exanded then that node is still in the OPEN

list?? instead of moving it to CLOSED list?

Regards

Francois du Toit]]>

Missionary Cannibals HillClimbing algorithm to make it

with Backtracking

Nowhere in tut202 is the solution -> only states in tut501 that

it can be solved with modified algorithm. cannnot be solved

with Nilssen wrong algorithm.

Has anyone try it out and solve it. If so Please post the steps

or email me the solution at

francoisadt@mweb.co.za

]]>

f = x1 v x2.x3 (Assume x2x3 is multipplication and not logical "and"?

My vectors that is going tobe used below:

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

Use vectors is in the question is the threshold value always 1?

Why they tell use here threshold = 0in the question. Should my last

vector value not be 1 but 0? Is mine below incorrect

Vectors:

========

vectorx1 = <0 1 0 1>

vectorx2 = <1 0 0 1>

vectorx3 = <0 0 1 1>

vectorx4 = <1 1 1 1>

Training:

=========

Step 1:

=======

W1 = (0 0 0 0)

positional then x1 x2 x3 x4 match below to value (positional)

Vectorx1 = (0 1 0 1 )

so when calc d = x1 v x2x3 values in order of the X vector are used

so d = 0 v 1.0 = 0 v 0 = 0

W1.X = (0.0 + 0.1 + 0.0 + 0.1) = 0

So f = 1

Step 2:

=======

W2 = W1 + c(d-f)X1 = (0 0 0 0) + 1(0-1)(0 1 0 1 )

= (0 -1 0 -1)

positional then x1 x2 x3 x4 match below to value

Vector X = (1 0 0 1)

so when calc d = x1 v x2x3 values in order of the X vector are used

so d = 1 v 0.0 = 1 v 0 = 1

W2.X = (0 -1 0 -1).(1 0 0 1) = (0.1 + -1.0 + 0.0 + -1.1) = -1

So f = 0

Step 3:

=======

W3 = W2 + c(d-f)X2 = (0 -1 0 -1) + 1(1-0)(1 0 0 1)

= (0 -1 0 -1) + (1 0 0 1)

= (1 -1 0 0)

positional then x1 x2 x3 x4 match below to value

Vector X = (0 0 1 1)

so when calc d = x1 v x2x3 values in order of the X vector are used

so d = 0 v 0.1 = 0 v 0 = 0

W3.X = (1 -1 0 0).(0 0 1 1)=(1.0 + -1.0 + 0.1 + 0.1) = 0

So f = 1

Step 4:

=======

W4 = W3 + c(d-f)X3 = (1 -1 0 0) + 1(0-1)(0 0 1 1)

= (1 -1 0 0)+(0 0 -1 -1)

= (1 -1 -1 -1)

positional then x1 x2 x3 x4 match below to value

Vector X = (1 1 1 1)

so when calc d = x1 v x2x3 values in order of the X vector are used

so d = 1 v 1.1 = 1 v 1 = 1

W4.X = (1 -1 -1 -1).(1 1 1 1)=(1.1 + -1.1 + -1.1 + -1.1) = -2

So f = 0

Step 5:

=======

W5 = W4 + c(d-f)X4 = (1 -1 -1 -1) + 1(1-0).(1 1 1 1)

= (1 -1 -1 -1)+(1 1 1 1)

= (2 0 0 0)

positional then x1 x2 x3 x4 match below to value

Vector X = (1 0 1 1)

so when calc d = x1 v x2x3 values in order of the X vector are used

so d = 1 v 0.1 = 1 v 0 = 1

W4.X = (2 0 0 0).(1 0 1 1)=(2.1 + 0.0 + 0.1 + 0.1) = 2

So f = 1

Know that both f and d = 1 is this then final vector???

weight vector = w5 = (2 0 0 0) and threshold = 0

How do one know when to stop training??

Step 6:

=======

W6 = W5 + c(d-f)X5 = (2 0 0 0) + 1(1-1).(0 1 1 0)

= (2 0 0 0)

If w6 = w5 then w5 is last weight vector???

Is my answer correct or must I use threshold = 0

in vectors as in q1.4 states threshold = 0

So all my input vectors must have 0 instead of

1 at end ??????

Vectors:

========

vectorx1 = <0 1 0 0> I used above vector <0 1 0 1>

vectorx2 = <1 0 0 0> I used above vector <1 0 0 1>

vectorx3 = <0 0 1 0> I used above vector <0 0 1 1>

vectorx4 = <1 1 1 0> I used above vector <1 1 1 1>

]]>

Is eq x1 v x2x3 => x1 v x2 "and" x3 => ^ (conjunction)

Or is the x2.x3 a dot product to multiply 0(1) = 0

OR eq x1 v x2x3 => x1 v x2 "OR" x3 => V (disjunction)

]]>

For Question 4.1 of Tut 102, How is the answer supposed to be layed out.

Should we only show the states of the OPEN list after the 9 steps of the A* Algorithm, and then the

do we need to show the individual 9 steps of the A* Algorithm on each of the nodes to be expanded ?

Thanks

]]>

How do I determine d => the desired output before dong

the weight change calc.

Nowhere in notes it states when d should be 0 or when equals to 1.

Do when "gyppo" the system and let d be the opposite of f

(actual output). for example if f is 1 let d be 0 If f = 0 let d = 1

untill the weights do not change then you know that d is equal to f

REgards

Francois du Toit]]>

And the answer for Qu7. Why can PSAT be viewed as a constraint satisfaction problem?

Thanks for your help,

Michelle]]>

Hello guys,

Can someone please help me out on this? According to the tutorial letter,

W3 = W2 + c(d2-f2)X2

i.e W3 = [0 0 0 1] + 1(0-1)[0 1 1 1]

which I think should be equal to [0 -1 -1 0].

But W3 in the tutorial letter is [0 -1 -1 -1] .

I would be glad to get a prompt response to this.

Thanks ]]>

I struggling for past week with understanding TLU.

In the testbook and downloaded lecture slides from

other univercities I got some dim light in the tunnel.

but still do not know how the linear separator line

got drawn and got calculated

f(x) = ax + c

so on y-axis and x-axis is the two values x1 and x2.

drawn on the same grid so that x1 is x-axis and x2 is

y-axis?

can you please give example how the XOR and OR

equations got draw up. In some lecture notes they referring

to that c (the value that cut the x-axis is 1.5 ) How and where

they got the c = 1.5 (for x1 AND x2) - cases 0.5 (for x1 OR x2)

How is the x1,x2 grid draw up. Somehow the sepratorline

must have an equation from values x1 and x2. Please

give step by step example

Regards

Francois du Toit]]>

In the explanation we are told that Q11 is true if there is a queen placed there. Then we are asked to say if Q11 then "the queen at Q11 is safe".

Since we are using propositional language we can't even use variables, so how do we link the two.

Question 7.3 has me totally stumped!]]>

If we are going to have to, are we allowed to take caculators into the exam? It will be tricky if we are not allowed to take them and we have to do back propagation!!!]]>

I emailed the lecturer about the correct answer and have had a reply.

The answer is:

Desired output

<0 1 0> 0

<1 0 0> 1

<0 0 1> 0

<1 1 1> 1

Final weight vector (2 -1 0 ) threshold is 1.

Now to double check my workings!]]>

For some reason I managed to figure out the seemingly "difficult" questions, but seem stuck with stuff that I'm sure has a very easy answer but for all my effort cannot seem to find it.

Question 1.1

I've read the textbook so much my brain seems to shut down whenever I open it again. I simply cannot find anything on "space of the input vector" anywhere in chapter 2 or 3. If the answer is in the textbook pls just give me the page number?

Question 1.5

I assume this is the same as figure 2.6? But I have a problem working out the threshold value as I cannot find an example anywhere.

Question 3

I think the statement "Harry has to be present with Tom, or Dick, or both at the same time" is impossible. Here's why:

You cannot send Tom and Dick over together, so it is either Harry and Tom or Harry and Dick. Then once on the other side you cannot send Tom or Dick back and leave Harry at the bank, as Tom and Dick cannot cross together (again). Thus Harry will have to go back to pick up the one left behind, but this leaves Tom and Dick alone on either side of the bank. Surely this is fine? Or is there another way?

And please post more questions / answers - we all appreciate it!

Kind Regards

Michell]]>

I found the following final graph when a node of depth 4 is to be expanded:

G = {1,2,3,6,7,10,11,4,8,12,13,9,5}

As for OPEN = {10,13,6,9,11,12}

and CLOSED = {1,2,3,4,5,7,8}

For Question 5

I found AB(A, -infinity, infinity)=1

and the cut off nodes are node G and hence O and P and it is an alpha cutoff.

I am a bit stuck on question 7 b,c and d????

]]>

for the Exam guidelines question 1.4 I get the final weight vector

[2,0,0] with the threshold value 2.

I can draw the diagram of the input values but do not understand how to plot the separating plane.

Anyone who knows please help?]]>

Thnaks]]>

Anyone got the exam sent out last year? (i.e. 2004's exam).

I might have a few older ones that I downloaded at the end of each year since 2002 if anyone is interested (email me michell.wilkes at gmail.com).

The assignments are all basically the same throughout the years but can also send those along to those interested.

Good luck with the exam preparation!

Kind Regards

Michell]]>

For Question 1 (1.4) I get a final weight vector (2,0,0) and a threshold value of 0.

What do you get?

]]>

If you have to prove by resolution refutation that something, say r, is a logical consequence of the set using the set of support strategy

1. Use the negation of r as the set of support and resolve to nil?

or

2. Use the final clause as the set of support and resolve to r?

Which method??

I am so tired of ploughing through endless stuff on the net just to find one answer! I wish we had been provided with more working examples to practice on.

I am at that stage when the exams are too far but too near!! 8-)

]]>

Just wondering what this course is like. I am thinking about taking it next year. Any advice would be appreciated.

Thanks

Mark]]>

I have a couple of questions regarding chapter 8 of the textbook. If anyone can help me please do so, because i would duly appreciate it!!

Ok..

*pg 131 - Am I correct in saying that a successor function is just a grouping of operators. i.e. It contains many functions?

*pg 132 - Am I correct in saying that if we have reached say:

blank 2 3

1 8 4

7 6 5

by node 5 for example, then we cannot use it again in say node 10, even if we have reached it via expanding of a different node (i.e. both instances are not on the same path?)

*Exercise 8.2 on pg 137

for question 1 - Depth-first search, I get the following nodes:

1-2-5-6-10-11-3-7-12-13-4-8-9

Is this correct?

*could anyone solve Exercise 8.1? {pg 137}. If so, please give me a hint on where to start, because I dont know how to solve this question.]]>