Hello Hugo,
I have use "notepad" to format the reply and eliminate the special
characters. I hope it is readable now.

In addition, for your program's problem (don't work), maybe firstly
test the simplest planning problem as follows (context) to find the
error, then comparing with the pceudo-code to find the problem. I
haven't start this implementation now.


kind regards/chenyu




;;;;;;;;;;;;;;;;;
;;last message
;;;;;;;;;;;;;;;;;
I have read the pecudo code again give you my thinking here. If you
delete the step 2) (start < s-add < finish), the following problem
will happens, To make the discussion more clearly, I give the context
(To simplify the problem, I will ignore the variables):

Context:

1. actions
	 a)   action1
		 i. name:  start
		 ii. precondition: nothing
		 iii. effect:  nothing
	 b) action2
		 i. name:  finish
		 ii. precondition: predicate
		 iii. effect:  nothing
	 c) action3
		 i. name:  P1action
                 ii. precondition nothing
		 iii. effect:  predicate
2.      initial casual link set bookkeeping
	 a) empty set
3.      initial action order set bookkeeping
	 a) start < finish

Problem and discussion:
1. Problem:
	 a) Is it possible for you to delete the step 2)
(start < s-add < finish)?
2.      discussion:
	 a)       Not correct.
	 b)       My reason:
		 i.              Requirement of Linearization:
			 1.        "POP"'s full name is Parital-
order Planner. It means if needs, it is sure that after
linearizations of POP plan, "Total order plans" can be gained.
			 2.        But if you delete the "step
2)", the order relationship between start and "P1action" will not be
written down in the "action order set bookkeeping", it is difficult
for the above linearizations operation.
			 3.        In last 2nd paragraph of page
350 (AIMA version 1), light arrows in the figure show ordering
constraints¡­.., Also, all causes are constrained to come before
their effects, so you can think of each bold arrow as having a light
arrow underneath it.
		 ii.              Requirement of procedure "RESOLVE-
THREATS" and function "POP"
			 1.        In the function "POP",
procedure "CHOOSE-OPERATOR" and "RESOLVE-THREATS" are closely linked,
that¡¯s, one by one. It means after adding "casual link" and "order
link" in "CHOOSE-OPERATOR", the procedure "RESOLVE-THREATS" will
check the threat has been happened or not. If happens, It will try to
solve it immediately. This immediately check needs the ordering
information. And the solving of threat also needs the ordering
information.
			 2.        If the ordering information
doesn¡¯t exist, The "Resolve-threats" doesn¡¯t work.


I am also interested in the POP implementation for fully
understanding and research later, but I want to use lisp language.


Kind regards/chenyu



--- In aima-talk@yahoogroups.com, "yougooh2000" <hugo.ferreira@m...>
wrote:
> Hi Chenyu,
>
> Thanks for the feedback. Unfortunatelly I could not understand much
of
> your example. I cannot read the characters 8-(. As far as
> linearization goes, you are correct: to check and handle threats I
> need to have the correct orderings. My initial implementation (which
> is still not working), does insert the S-add < Finish because it
> regards Start as a standard step in the plan when selecting a
> sub-goal. As far as Start < S-add I have to check on that and see
how
> it influences threat detection and resolution. Nevertheless because
I
> also handle Finish as a step I do get a Start < S-add, but not in
the
> same point in time.
>
> Regards,
> Hugo F.
>
>
>
> --- In aima-talk@yahoogroups.com, E etech058 <etech058@o...> wrote:
> > Hello,
> > I have read the pecudo code again give you my thinking here. If
you
> delete the step 2) (start < s-add < finish), the following problem
> will happens, To make the discussion more clearly, I give the
context
> (To simplify the problem, I will ignore the variables):
> > Context:
> > 1.      actions
> > a)       action1
> >                         i.              name:
start
> >                       ii.              precondition:      nothing
> >                     iii.              effect:
> nothing
> > b)       action2
> >                         i.              name:
finish
> >                       ii.              precondition:
predicate
> >                     iii.              effect:
> nothing
> > c)        action3
> >                         i.              name:
> P1action
> >                       ii.              precondition       nothing
> >                     iii.              effect:
> predicate
> > 2.      initial casual link set bookkeeping
> > a)       empty set
> > 3.      initial action order set bookkeeping
> > a)       start < finish
> >
> > Problem and discussion:
> > 1.      Problem:
> > a)       Is it possible for you to delete the step 2) (start < s-
add
> < finish)?
> > 2.      discussion:
> > a)       Not correct.
> > b)       My reason:
> >                         i.              Requirement of
> Linearization:
> > 1.        €  ??¼POP€  ??½â'?¹s full name is Parital-order
> Planner. It means
> if needs, it is sure that after linearizations of POP plan,
> €  ??¼Total
> order plans€  ???can be gained.
> > 2.        But if you delete the €  ??¼step 2)€  ??? the order
> relationship between start and €  ??¼P1action€  ???will not be
> written down
> in the €  ??¼action order set bookkeeping€  ??? it is difficult
> for the
> above linearizations operation.
> > 3.        In last 2nd paragraph of page 350 (AIMA version 1),
light
> arrows in the figure show ordering constraints€  ? ¦.., Also, all
> causes
> are constrained to come before their effects, so you can think of
each
> bold arrow as having a light arrow underneath it.
> >                       ii.              Requirement of procedure
> €  ??¼RESOLVE-THREATS€  ???and function €  ??¼POP€  ???> >
1.        In the function €  ??¼POP€  ??? procedure
> €  ??¼CHOOSE-OPERATOR€  ???> and €  ??¼RESOLVE-THREATS€  ???are
closely linked,
> that€  ??¹s, one by one. It
> means after adding €  ??¼casual link€  ???and €  ??¼order
> link€  ???in
> €  ??¼CHOOSE-OPERATOR€  ??? the procedure €  
> ??¼RESOLVE-THREATS€  ???will check
> the threat has been happened or not. If happens, It will try to
solve
> it immediately. This immediately check needs the ordering
information.
> And the solving of threat also needs the ordering information.
> > 2.        If the ordering information doesn€  ??¹t exist, The
> €  ??¼Resolve-threats€  ???doesn€  ??¹t work.
> >
> >
> > I am also interested in the POP implementation for fully
> understanding and research later, but I want to use lisp language.
> >
> >
> > Kind regards/chenyu
> >
> > -----Original Message-----
> > From: yougooh2000 [mailto:hugo.ferreira@m...]
> > Sent: 2003€  å¹?2€  ???2€  ?·¥ 17:18
> > To: aima-talk@yahoogroups.com
> > Subject: [aima-talk] Re: POP Planner Implementation: Choose-
Operator
> >
> > Hi,
> >
> > First versions will (hopefully) be in Prolog. This is for research
> > purposes.
> >
> > Regards,
> > Hugo.
> >
> >
> > --- In aima-talk@yahoogroups.com, E etech058 <etech058@o...>
wrote:
> > > Hello,
> > > Which language will you use for implementation?
> > >
> > > Best regards/chenyu
> > >
> > > -----Original Message-----
> > > From: yougooh2000 [mailto:hugo.ferreira@m...]
> > > Sent: 2003€  ?¤Ã?2€  ?´Ã'?2€  ?¨Ã'?0:06
> > > To: aima-talk@yahoogroups.com
> > > Subject: [aima-talk] POP Planner Implementation: Choose-Operator
> > >
> > > Hi,
> > >
> > > I am attempting to implement the POP planner as is described in
> > AIMA's
> > > 1st version (Chap. 11.). I am having doubts regarding the call
> > > "choose-operator" that is described in the book. The
> > > pseudo-code says
> > > that if the step to add (S-add) is selected from operator's list
> > (not
> > > the plan), then it is necessary to:
> > > 1)Add this new step to the plans already existing steps.
> > > 2)Add the ordering: Start < S-add < Finish
> > >
> > > I cannot understand what is the need for step 2). Say for
example
> I
> > > start the planner with a minimal plan of {Start, Finish} and on
> > > selecting the first operator, I only find an S-add from the
> > operator's
> > > list. I will have to add the ordering S-add < S-need, where
> S-need =
> > > Finish, according to the 5th pseudo-code line of the call. After
> > that
> > > I see that I have to execute steps 1 and 2 as described above.
> Now I
> > > have a _repeated_ attempt to add ordering S-add < Finish.
> > >
> > > I have considered not doing step 2) which would result in a
> missing
> > > ordering Start < S-add. I figure this will eventually be added
to
> > the
> > > orderings when I process step S-add (i.e: attempt to satisfy its
> > > pre-conditions).
> > >
> > > Am I missing something?
> > >
> > > TIA.
> > > Hugo Ferreira.
> > >
> > >
> > >
> > > To unsubscribe from this group, send an email to:
> > > aima-talk-unsubscribe@yahoogroups.com
> > >
> > >
> > >
> > > Your use of Yahoo! Groups is subject to
> > http://docs.yahoo.com/info/terms/
> >
> >
> >
> > To unsubscribe from this group, send an email to:
> > aima-talk-unsubscribe@yahoogroups.com
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
> http://docs.yahoo.com/info/terms/

Hello Imanpreet,
I have read your email carefully, and then read the AIMA book again. My idea is
as follows:
1. A* will find the optimal (best) solution. It is correct.
2. Your idea has one implicit assumption, which is not same as the book. That's
you assume that "GOAL-TEST(NODE)" check will be run immediate after the "EXPAND"
function. The correct one (AIMA books "GENERAL-SEARCH" pceudo-code) is "EXPAND",
then "QUEUE-FN", then "GOAL-TEST".
   2.1 Your above implicit assumption means that you doesn't give the more chance
to "A*" for finding more optimal solution.
   2.2 Explain by your example. After "Node B f(B)=120" is expanded, in the A*
queue, 2 nodes exists. One is "Node D f(D)=180", the other is "Node C f(C)=140".
Then (Please look at "GENERAL-SEARCH" pseudo-code) function "QUEUING-FN" will
work. It will produce list which "Node B f(B)=120" is a header. In the next
loop, "Node B" will be expanded. It means that although the goal of "Node D" has
been founded, the A* will continue to run until it finds a optimal solution.
3. I just think in the actual programming, your implicit assumption will be
used. It means comparing between found pure optimal (A*) solution, the quick and
efficient algorithm is more important.



Kind regards/chenyu



-----Original Message-----
From: Imanpreet Singh Arora [mailto:imanpreet_arora@...]
Sent: 2003Äê12ÔÂ19ÈÕ 3:54
To: aima-talk@yahoogroups.com
Subject: [aima-talk] Re: A* algorithm

--- In aima-talk@yahoogroups.com, E etech058 <etech058@o...> wrote:
> Hi,
> I have used AIMA version 1, do you use version 1 or not?
> I haven't found the inequality of "c(n,a,n') <= h(n) + h(n')" in
page 99.
> In addition, the scenario is not easy to understand, could you
represent it more clearly.
>
>
> Kind regards/chenyu
>

Thanks, I am using version 2.0. Moreover on page 99 you won't find the
given inequality but the following

h(n) <= c(n, a, n') + h(n');

and it was written that it is a triangular inequality, I am wondering
if the triangular inequality is applicable as whole for all the sides
like is it applicable for the inequality

c(n, a, n') <= h(n) + h(n');

I would try to clarify the scenario, though since I know that the
original indentation that I provided in the message seems to have been
lost.

We have a root node A, and the h(A)  is 100.
It has two childern B and C.
g(B) =70 and g(C) = 60
h(B) = 50 and h(C) = 80
so
f(B) = 120
f(C) = 140

Meaning that it would be f(B) that would be expanded, B has a Child
named D and this child is common for the nodes B and C.

Now when B would be expanded the child D would have it's parent set to B.

The cost of getting to D from B and C is respectively 40 and 20 so the
total

g( D through B) is 110
g ( D through C) is  80

Now at this juncture C would be expanded and though D is the child of
C, D's parent would be still set to B because according to section 3.5
the new path would always be rejected.

According to the claims of the authors if h(n) is consitent and
admissible we would certainly find an optimal solution I am not sure
we are able to in this very case.

I belive that if we use the f cost in order to determine the parent of
a node we would certainly find the optimal soution, because for D

f( D through B ) is 190
f ( D through C) is 160

Would the authors and peers be kind to elucidate?  I seem to be unable
to relate the equations and principles to a possible graph space that
can clarify me.

--
Imanpreet Singh Arora



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I'm still intersted in Neural Net Capabilities. See post 253
Can any one provide a constructive proof of the claim from page 744 of
AIMA 2nd edition, "In fact, with a single, sufficiently large hidden
layer, it is possible to represent any continuous function of the
inputs with arbiarary accuracy" or alternatively a net that is able to
learn z = x*y over the range -10 < x, y <10 with an accuracy of
abs(z-xy)<0.1?  or more difficultly compute x,y from r and theta where
x is r * cos(theta) and y is r * sin(theta)?

Chenyu,

Thanks for the feedback.

As far as the problem is concerned, my tests until now show that I do
not need to add the Start < S-add ordering. I need to test additional
worlds, but my initial ideia seems to be valid. I don't know if this
is due to the way in which I implemented the algorithm or some "bug".

Anyway here is the explanation: because the alg. is goal recursive I
only check for preconditions before I regress. If after an
operation/step has been selected I do not add Start < S-add, it does
not matter because I only check for the steps/operators that I have
selected until now with open preconditions. Start will therefore never
be considered, it has no pre-conds.

What about checking for consistency? This doesn't matter either
because I cannot promote anything before Start, nor for that matter
demote it after Finish. In other words consistency only checks for
steps consistency between Start and Finish (exclusive).

What about identifying threats? Just as in the case checking for
consistency, threats may only occur for operators/steps between Start
and Finish. So once again, if I do add a Start < S-add, I don't need
the info to identify threats.
Please note that the book's algo. just checks if an operator/step
"clobbers" anothers pre-conds. This means that after ordering, the
check for threats will still show the threats. I only use ordering
information so as not to repeatedly (and unecessarily) resolve
conflicts.

I hope this explanation is valid. In case I do detect a mistake, I
will post additional information.

Thanks again.
Regards,
Hugo.




--- In aima-talk@yahoogroups.com, "chenyu468" <chenyu468@y...> wrote:
> Hello Hugo,
> I have use "notepad" to format the reply and eliminate the special
> characters. I hope it is readable now.
>
> In addition, for your program's problem (don't work), maybe firstly
> test the simplest planning problem as follows (context) to find the
> error, then comparing with the pceudo-code to find the problem. I
> haven't start this implementation now.
>
>
> kind regards/chenyu
>
>
>
>
> ;;;;;;;;;;;;;;;;;
> ;;last message
> ;;;;;;;;;;;;;;;;;
> I have read the pecudo code again give you my thinking here. If you
> delete the step 2) (start < s-add < finish), the following problem
> will happens, To make the discussion more clearly, I give the
context
> (To simplify the problem, I will ignore the variables):
>
> Context:
>
> 1. actions
>  a)   action1
> 	 i. name:  start
> 	 ii. precondition: nothing
> 	 iii. effect:  nothing
>  b) action2
> 	 i. name:  finish
> 	 ii. precondition: predicate
> 	 iii. effect:  nothing
>  c) action3
> 	 i. name:  P1action
>                 ii. precondition nothing
> 	 iii. effect:  predicate
> 2.      initial casual link set bookkeeping
>  a) empty set
> 3.      initial action order set bookkeeping
>  a) start < finish
>
> Problem and discussion:
> 1. Problem:
>  a) Is it possible for you to delete the step 2)
> (start < s-add < finish)?
> 2.      discussion:
>  a)       Not correct.
>  b)       My reason:
> 	 i.              Requirement of Linearization:
> 		 1.        "POP"'s full name is Parital-
> order Planner. It means if needs, it is sure that after
> linearizations of POP plan, "Total order plans" can be gained.
> 		 2.        But if you delete the "step
> 2)", the order relationship between start and "P1action" will not
be
> written down in the "action order set bookkeeping", it is difficult
> for the above linearizations operation.
> 		 3.        In last 2nd paragraph of page
> 350 (AIMA version 1), light arrows in the figure show ordering
> constraints¡­.., Also, all causes are constrained to come before
> their effects, so you can think of each bold arrow as having a
light
> arrow underneath it.
> 	 ii.              Requirement of procedure "RESOLVE-
> THREATS" and function "POP"
> 		 1.        In the function "POP",
> procedure "CHOOSE-OPERATOR" and "RESOLVE-THREATS" are closely
linked,
> that¡¯s, one by one. It means after adding "casual link" and
"order
> link" in "CHOOSE-OPERATOR", the procedure "RESOLVE-THREATS" will
> check the threat has been happened or not. If happens, It will try
to
> solve it immediately. This immediately check needs the ordering
> information. And the solving of threat also needs the ordering
> information.
> 		 2.        If the ordering information
> doesn¡¯t exist, The "Resolve-threats" doesn¡¯t work.
>
>
> I am also interested in the POP implementation for fully
> understanding and research later, but I want to use lisp language.
>
>
> Kind regards/chenyu
>
>
>
> --- In aima-talk@yahoogroups.com, "yougooh2000"
<hugo.ferreira@m...>
> wrote:
> > Hi Chenyu,
> >
> > Thanks for the feedback. Unfortunatelly I could not understand
much
> of
> > your example. I cannot read the characters 8-(. As far as
> > linearization goes, you are correct: to check and handle threats I
> > need to have the correct orderings. My initial implementation
(which
> > is still not working), does insert the S-add < Finish because it
> > regards Start as a standard step in the plan when selecting a
> > sub-goal. As far as Start < S-add I have to check on that and see
> how
> > it influences threat detection and resolution. Nevertheless
because
> I
> > also handle Finish as a step I do get a Start < S-add, but not in
> the
> > same point in time.
> >
> > Regards,
> > Hugo F.
> >
> >
> >
> > --- In aima-talk@yahoogroups.com, E etech058 <etech058@o...>
wrote:
> > > Hello,
> > > I have read the pecudo code again give you my thinking here. If
> you
> > delete the step 2) (start < s-add < finish), the following problem
> > will happens, To make the discussion more clearly, I give the
> context
> > (To simplify the problem, I will ignore the variables):
> > > Context:
> > > 1.      actions
> > > a)       action1
> > >                         i.              name:
> start
> > >                       ii.              precondition:
nothing
> > >                     iii.              effect:
> > nothing
> > > b)       action2
> > >                         i.              name:
> finish
> > >                       ii.              precondition:
> predicate
> > >                     iii.              effect:
> > nothing
> > > c)        action3
> > >                         i.              name:
> > P1action
> > >                       ii.              precondition
nothing
> > >                     iii.              effect:
> > predicate
> > > 2.      initial casual link set bookkeeping
> > > a)       empty set
> > > 3.      initial action order set bookkeeping
> > > a)       start < finish
> > >
> > > Problem and discussion:
> > > 1.      Problem:
> > > a)       Is it possible for you to delete the step 2) (start <
s-
> add
> > < finish)?
> > > 2.      discussion:
> > > a)       Not correct.
> > > b)       My reason:
> > >                         i.              Requirement of
> > Linearization:
> > > 1.        €  ??¼POP€  ??½â'?¹s full name is
Parital-order
> > Planner. It means
> > if needs, it is sure that after linearizations of POP plan,
> > €  ??¼Total
> > order plans€  ???can be gained.
> > > 2.        But if you delete the €  ??¼step 2)€  ???
the order
> > relationship between start and €  ??¼P1action€
???will not be
> > written down
> > in the €  ??¼action order set bookkeeping€  ??? it is
difficult
> > for the
> > above linearizations operation.
> > > 3.        In last 2nd paragraph of page 350 (AIMA version 1),
> light
> > arrows in the figure show ordering constraints€  ? ¦..,
Also, all
> > causes
> > are constrained to come before their effects, so you can think of
> each
> > bold arrow as having a light arrow underneath it.
> > >                       ii.              Requirement of procedure
> > €  ??¼RESOLVE-THREATS€  ???and function €
??¼POP€  ???> >
> 1.        In the function €  ??¼POP€  ??? procedure
> > €  ??¼CHOOSE-OPERATOR€  ???> and €
??¼RESOLVE-THREATS€  ???are
> closely linked,
> > that€  ??¹s, one by one. It
> > means after adding €  ??¼casual link€  ???and €
??¼order
> > link€  ???in
> > €  ??¼CHOOSE-OPERATOR€  ??? the procedure €
> > ??¼RESOLVE-THREATS€  ???will check
> > the threat has been happened or not. If happens, It will try to
> solve
> > it immediately. This immediately check needs the ordering
> information.
> > And the solving of threat also needs the ordering information.
> > > 2.        If the ordering information doesn€  ??¹t
exist, The
> > €  ??¼Resolve-threats€  ???doesn€  ??¹t work.
> > >
> > >
> > > I am also interested in the POP implementation for fully
> > understanding and research later, but I want to use lisp language.
> > >
> > >
> > > Kind regards/chenyu
> > >
> > > -----Original Message-----
> > > From: yougooh2000 [mailto:hugo.ferreira@m...]
> > > Sent: 2003€  å¹?2€  ???2€  ?·¥ 17:18
> > > To: aima-talk@yahoogroups.com
> > > Subject: [aima-talk] Re: POP Planner Implementation: Choose-
> Operator
> > >
> > > Hi,
> > >
> > > First versions will (hopefully) be in Prolog. This is for
research
> > > purposes.
> > >
> > > Regards,
> > > Hugo.
> > >
> > >
> > > --- In aima-talk@yahoogroups.com, E etech058 <etech058@o...>
> wrote:
> > > > Hello,
> > > > Which language will you use for implementation?
> > > >
> > > > Best regards/chenyu
> > > >
> > > > -----Original Message-----
> > > > From: yougooh2000 [mailto:hugo.ferreira@m...]
> > > > Sent: 2003€  ?¤Ã?2€  ?´Ã'?2€  ?¨Ã'?0:06
> > > > To: aima-talk@yahoogroups.com
> > > > Subject: [aima-talk] POP Planner Implementation:
Choose-Operator
> > > >
> > > > Hi,
> > > >
> > > > I am attempting to implement the POP planner as is described
in
> > > AIMA's
> > > > 1st version (Chap. 11.). I am having doubts regarding the call
> > > > "choose-operator" that is described in the book. The
> > > > pseudo-code says
> > > > that if the step to add (S-add) is selected from operator's
list
> > > (not
> > > > the plan), then it is necessary to:
> > > > 1)Add this new step to the plans already existing steps.
> > > > 2)Add the ordering: Start < S-add < Finish
> > > >
> > > > I cannot understand what is the need for step 2). Say for
> example
> > I
> > > > start the planner with a minimal plan of {Start, Finish} and
on
> > > > selecting the first operator, I only find an S-add from the
> > > operator's
> > > > list. I will have to add the ordering S-add < S-need, where
> > S-need =
> > > > Finish, according to the 5th pseudo-code line of the call.
After
> > > that
> > > > I see that I have to execute steps 1 and 2 as described above.
> > Now I
> > > > have a _repeated_ attempt to add ordering S-add < Finish.
> > > >
> > > > I have considered not doing step 2) which would result in a
> > missing
> > > > ordering Start < S-add. I figure this will eventually be
added
> to
> > > the
> > > > orderings when I process step S-add (i.e: attempt to satisfy
its
> > > > pre-conditions).
> > > >
> > > > Am I missing something?
> > > >
> > > > TIA.
> > > > Hugo Ferreira.
> > > >
> > > >
> > > >
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--- In aima-talk@yahoogroups.com, Fathi Salaymeh <fathi_kazem@y...>
wrote:
> Hi everybody;
> I'm asking if there is any one who can help be in solving  Exercise
3.4 page 89 in our text book.
> please if there is anybody who have the solution, please sent it to
me.
> with regards
> Fathi Salaymeh;
>
>
> ---------------------------------
> Do you Yahoo!?
> Protect your identity with Yahoo! Mail AddressGuard

I have the version 1 of AIMA book, not version 2. In my book,
exercise 3.4 is about missionaries and cannibals and it is in page 88
not in page 89. Do you want this exercise's solution? If yes, I have
written it by "Python", do you need it or not?




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aima-talk-unsubscribe@yahoogroups.com



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Hi,

As per the subject, how do I prove that adding a link to a bayes net
will not decrease the data likelihood?

Thanks!
Yee Seng

Hello everyone,
I have downloaded AIMA version 2 sample chapter 5 and am doing its
exercises.

In exercise 5.7, it requires to compare the algorithms of MRV,
forward checking, FC+MRV, Min-conflicts.
But I wonder is it possible to implement MRV independently without FC?
My impossible reasons are as follows:
1. MRV means "minimium remaining values". It means the unassigned
variable with "minimium remaining values" should be selected firstly.
The 2 implementation ideas are
    1. to filter unassigned variables' domain everytime after
assigning a variable.
    or
    2. calculate the every unassigned variable's legal domains
everytime before select.

The above 2nd implemenation is low efficient for repeat calculation
on legal domains, therefore, it is a bad idea.

So only the 1st implementation is accepted. It seems the exercise 5.7
should be modified to "compare the algorithms of FC,FC+MRV,Min-
conflicts.

What's about your idea?


Thank you for your attention.
Best regards/chenyu

The following aima-talk poll is now closed.  Here are the
final results:


POLL QUESTION: Do you learn AI by yourself for self-
interest, No classmate, No Professor?


CHOICES AND RESULTS
- Yes, I need others to discuss exercises and will reply other's question on my
best., 6 votes, 66.67%
- Yes, I will just ask question but no discuss other's question., 0 votes, 0.00%
- No, but I will join in the discussion., 2 votes, 22.22%
- No, I won't join in the discussion., 0 votes, 0.00%
- Others., 1 votes, 11.11%



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Hello,
Friend has recommended a new IT related website. I have browsed it
and found it contains many useful url links. Of course, it contains
AI related web sites.

But now only Chinese version is available.

it is:
-----------http://www.iturls.com


Thank you for your attention.
Best regards/chenyu

Page 144 of your d/l copy refers to the method of BT + MRV, meaning
that FC is not needed. However, they work like hand in glove. At the
bottom of the same page, the book says FC is an efficient way to
compute MRV. Otherwise, finding the MRV involves many redundant
computations, as you've noticed.

For a more in-depth explanation, see
http://www.cs.toronto.edu/~fbacchus/Papers/BvRCP95.pdf
and look at pages 6 & 7.

Best regards,
Brandon

--- In aima-talk@yahoogroups.com, "chenyu468" <chenyu468@y...> wrote:
> Hello everyone,
> I have downloaded AIMA version 2 sample chapter 5 and am doing its
> exercises.
>
> In exercise 5.7, it requires to compare the algorithms of MRV,
> forward checking, FC+MRV, Min-conflicts.
> But I wonder is it possible to implement MRV independently without
FC?
> My impossible reasons are as follows:
> 1. MRV means "minimium remaining values". It means the unassigned
> variable with "minimium remaining values" should be selected
firstly.
> The 2 implementation ideas are
>    1. to filter unassigned variables' domain everytime after
> assigning a variable.
>    or
>    2. calculate the every unassigned variable's legal domains
> everytime before select.
>
> The above 2nd implemenation is low efficient for repeat calculation
> on legal domains, therefore, it is a bad idea.
>
> So only the 1st implementation is accepted. It seems the exercise
5.7
> should be modified to "compare the algorithms of FC,FC+MRV,Min-
> conflicts.
>
> What's about your idea?
>
>
> Thank you for your attention.
> Best regards/chenyu

I have a couple questions about Figure 4.8, which shows the search cost of
the A* algorithm for the Eight-Puzzle problem.

First, what is search cost?  In the text, it says that "Figure 4.8 gives
the average number of nodes expanded".  However, the search cost for both
heuristicsfor problems with solutions of length 2 is 6.  For both
heuristics, the number of nodes that the expand function is called on
should always be 2 for such problems.  There should be an average of 6
nodes generated for such problems, the ~3 children of the start node and
the ~3 children of the node and the ~3 children of the node whose state is
one away from the goal.  I'd guess that the table actually shows the
number of nodes generated, i.e. number of nodes expanded plus the number
of nodes in the queue.

Second, how were the random problems generated?  I have found that the
number of nodes generated increases with the solution length faster than
the numbers in the table.  My results were
d     Nodes generated                 Nodes expanded
       A*(h1)          A*(h2)          A*(h1)          A*(h2)

2:    6.020000        5.800000        2.000000        2.000000
4:    11.720000       11.620000       4.020000        4.000000
6:    24.690000       18.130000       8.490000        6.290000
8:    55.240000       28.870000       19.610000       10.310000
10:   207.230000      88.260000       74.050000       32.220000
12:   1849.293333     171.324675      661.666667      62.662338

[Note that d = 12 was done with 75 and 79 tests, not 100].  For high d,
these numbers do not compare to those in the table.  For the table, is the
*minimum* solution length of a problem with solution length d actually d?
Has anyone been able to replicate these results?

Thanks,
Kristin

On Fri, 16 Jan 2004, Kristin Mara BRANSON wrote:
>
> Date: Fri, 16 Jan 2004 09:02:05 -0800 (PST)
> From: Kristin Mara BRANSON <kbranson@...>
> To: Charles Elkan <elkan@...>
> Cc: Piotr DOLLAR <pdollar@...>, Anjum GUPTA <a3gupta@...>
> Subject: search cost = number of nodes expanded?
> Parts/Attachments:
>    1 Shown    46 lines  Text
>    2          31 KB     Application, ""
> ----------------------------------------
>
> Although the text says "Figure 4.8 gives the number of nodes expanded", it
> says that the average search cost for problems with solution length 2 is
> 6.  Using either of the two heuristics, you should always expand 2 nodes
> when the solution length is 2.  You should, however, have generated
> somewhere around 6 nodes: the ~3 children of the start node, and the ~3
> children of the node whose state is one away from the goal.  I'd guess
> that the table actually shows the number of nodes generated.
>
> Do you know of any students who have actually got numbers near those in
> the table?  I don't totally trust it.  I implemented the eight puzzle
> stuff this morning, and my numbers grow faster than those in the table:
>
> d Nodes generated 	 Nodes expanded
>  A*(h1)  A*(h2)  A*(h1)  A*(h2)
>
> 2:  6.020000 5.800000 2.000000 2.000000
> 4:  11.720000 11.620000 4.020000 4.000000
> 6:  24.690000 18.130000 8.490000 6.290000
> 8:  55.240000 28.870000 19.610000 10.310000
> 10:  207.230000 88.260000 74.050000 32.220000
> 12:  1849.293333 171.324675 661.666667 62.662338
>
> Note that d = 12 was done with 75 and 79 tests, not 100. I'm not
> guaranteeing that my code is correct, since I wrote it rather fast, but I
> am wondering if maybe the numbers in the book are generated from problems
> that had a solution of length d, but might have had a solution of d' < d
> (i.e. they started from the goal configuration and moved the tiles d times
> randomly, or something a bit smarter than randomly).  I also have a very
> high variance for the search cost for high d.  Here are the variances for
> h1 and nodes generated:
> 2: 1.009697
> 4: 0.789495
> 6: 100.902929
> 8: 1006.527677
> 10: 38149.128384
> 12: 9082827.318198
> Most d = 12 problems that are easy but a few are hard.  Maybe they only
> used easy ones?
>
> I'll let you know if I notice a bug in my code, but otherwise let me know
> if you find anyone who was actually able to reproduce the table, pref
> before section today.
>
> If any of you want to play with my messy, slow code, I attached it.
> There are 2 .mat files with the results.  the variables save_nodes_* have
> the results summarized in the table.
>
>     [ Part 2, ""  Application/OCTET-STREAM (Name: "kristins_astar.tar")  ]
>     [ 42KB. ]
>     [ Cannot display this part. Press "V" then "S" to save in a file. ]
>
>

Hi,

I doubt the solution will involve any learning at all. I guess instead
that it will show how to construct the net directly (or show that
3-layer nets are Turing complete and depend on the problem's answer
being Turing-computable). It's an interesting question though whether
it is possible to learn any such approximation.

I'm not sure if the general solution looks like the following. Sorry
that I don't use more standard notation. I don't quite have the hang of
that.

(I'm somewhat unsure of the math in the next paragraph.)

Assume that the each of the input values x1, x2, ..., xn is drawn from
a bounded interval I1, I2, ... In, each Ii a subinterval of R. Assume
also that you want the output to be within some epsilon > 0 of the
value of some everywhere continuous, n-ary function f: R x R x .. R -->
R. Since the function is everywhere continuous, it's possible to divide
the entire domain (I1 x I2 x .. In) into finitely many sections C1, C2,
... Ck such that for each section there exists some Li such that f is
within epsilon of Li everywhere within the section. (Note each section
Ci is the Cartesian product of n subintervals.)

To build a 3-layer net to approximate f within epsilon: (I assume that
each neuron may have its  own threshold level. If the weighted sum of
the inputs to the neuron is less than this level, it outputs zero;
otherwise, its output is its activation value.)

1) For each of the sections, Ci, with corresponding output value, Li,
add an output node labeled Fi, with activation value Li and threshold
n.

2) For each input variable xi, add an input node labeled Xi. Take the
subintervals corresponding to variable xi in the sections C1, C2, ...,
Ck. Without loss of generality, assume that none of the subintervals
overlaps another. Let ci be the lower bound of the interval
corresponding to xi in Ci. Add a hidden unit labeled Hi with threshold
ci and activation value 1. Add a connection with weight 1 from Xi to
Hi. Add a connection with weight 1 from Hi to Fi.

3) For each hidden unit Hi with threshold ci, add a connection to any
output node Fj if the threshold, cj, of the corresponding hidden unit
Hj is less than ci. Make the weight of this connection -1.

(We could've made all of the neurons in the above have the same
threshold, but the description was complicated enough already.)

I'd also be interested in seeing the solution to the problem posed by
George if someone has it. Thanks.

David

On Dec 20, 2003, at 9:30 AM, georgel360 wrote:

> I'm still intersted in Neural Net Capabilities. See post 253
> Can any one provide a constructive proof of the claim from page 744 of
> AIMA 2nd edition, "In fact, with a single, sufficiently large hidden
> layer, it is possible to represent any continuous function of the
> inputs with arbiarary accuracy" or alternatively a net that is able to
> learn z = x*y over the range -10 < x, y <10 with an accuracy of
> abs(z-xy)<0.1?  or more difficultly compute x,y from r and theta where
> x is r * cos(theta) and y is r * sin(theta)?

I'm taking a graduate course in AI at NYU and am delighted to find the
java demos online.  Much thanks to Ravi.  I've downloaded them and
have walked through the CSPDemo.

The BackTrackingSearch for the CSPDemo worked great.  I have a much
better understading of the AC3 algo. and how the backtracking is
applied to finding the solution.

But the MinConflictsSearch in CSPDemo returned 'No Solution' on
max_step=100000.   Is this right?   I walked through the code and it
appears that the call to 'csp.getConflictedVariables(assignment)'
never returns a value less than 6, even up through 40000 iterations,
if at all.

Also, I had some trouble running the TveDemo.  Has anyone gotten this
demo to work?  First the demo referes to image files such as
clean.gif, and dirty.gif, that don't seem to be included in the
archived download.  Secondly how is the agent for the TveDemo
activated?  I don't see how this is done?

Once again, thanks for all the work put into theses demos. It's a
great learning tool.

sincerely,
Justin

Justin,
  Many thanks for the kind words.
  If you can just hold one for a few more days, there is a new release
of the code coming up which should clear up a lot of these issues.
I plan to make  arelease of teh code on the 9th february (Monday)
which is mainly a cleaned up release of the old code. As i wrote the
jave code for aima, i was initially working off edition 1 of aima and
then switched to edition 2.Also, i was using the code to learn ai
concepts, and the resulting code is a bit grungier than I would like.

Anyway, I have tremendously simplified and refactored the code besides
adding lots of comments (which a  lot of people had asked for ). Also
now the code is fully in synch with edition 2 and there  are afew more
changes like a makefile , removal of circular depenencies , mock
object based tests etc. I am justy waiting for the weekend to put the
finishing touches to the code and send it to Peter.This drop should
clear up the CSP related issues. The TVE demo will probabaly not run
still, since i am not focussing on graphics right now. anyway let me
see if i can squeeze it in .
  Thanks,
  Just hold on till Monday,
Ravi




--- In aima-talk@yahoogroups.com, "jbirtwell" <jbirtwell@y...> wrote:
> I'm taking a graduate course in AI at NYU and am delighted to find the
> java demos online.  Much thanks to Ravi.  I've downloaded them and
> have walked through the CSPDemo.
>
> The BackTrackingSearch for the CSPDemo worked great.  I have a much
> better understading of the AC3 algo. and how the backtracking is
> applied to finding the solution.
>
> But the MinConflictsSearch in CSPDemo returned 'No Solution' on
> max_step=100000.   Is this right?   I walked through the code and it
> appears that the call to 'csp.getConflictedVariables(assignment)'
> never returns a value less than 6, even up through 40000 iterations,
> if at all.
>
> Also, I had some trouble running the TveDemo.  Has anyone gotten this
> demo to work?  First the demo referes to image files such as
> clean.gif, and dirty.gif, that don't seem to be included in the
> archived download.  Secondly how is the agent for the TveDemo
> activated?  I don't see how this is done?
>
> Once again, thanks for all the work put into theses demos. It's a
> great learning tool.
>
> sincerely,
> Justin

Hi all,

Having a graph, is there only one way how to build a tree using
depth first search algorithm?  If there is not, because of cycles,
is there a way how to choose the best tree?

Thanks

I just started learning lisp. For the intelligent program acting in
the vacuum world, I tried to define several agents, but compiler
gives me syntax error.

The agent is defined as :
(defstructure environment
   "The world in which agents exist."
   (agents '())       ;; A list of the agents in the environment
   (step 0)           ;; The number of time steps simulated so far
   (max-steps 1000)   ;; Stop the simulation after this number
   (stream t)         ;; Stream to display output on
   (initialized nil)  ;; Have we run initialize on this environment
yet?
   (state nil)        ;; Current state of the environment; other
subtypes
                      ;; add new slots to hold various state
information
   )

I defined agent like that but not working:
  setq(a (make-agent))
   setq(b (make-agent))
   setq(c (make-agent))

((run-environment (make-vacuum-world
                    (:agents '(a b c))
                    :stream nil
                    :aspec '(reactive-vacuum-agent))))

Hello,
I have run this program before. It also gives some compile error, but after
modification, it can run.
I have used corman-lisp for exercise. Which tools do you use?


Best regard
Chenyu


-----Original Message-----
From: rose_ronglin [mailto:rose_ronglin@...]
Sent: 2004Äê2ÔÂ22ÈÕ 11:08
To: aima-talk@yahoogroups.com
Subject: [aima-talk] how to define several agents

I just started learning lisp. For the intelligent program acting in
the vacuum world, I tried to define several agents, but compiler
gives me syntax error.

The agent is defined as :
(defstructure environment
   "The world in which agents exist."
   (agents '())       ;; A list of the agents in the environment
   (step 0)           ;; The number of time steps simulated so far
   (max-steps 1000)   ;; Stop the simulation after this number
   (stream t)         ;; Stream to display output on
   (initialized nil)  ;; Have we run initialize on this environment
yet?
   (state nil)        ;; Current state of the environment; other
subtypes
                      ;; add new slots to hold various state
information
   )

I defined agent like that but not working:
  setq(a (make-agent))
   setq(b (make-agent))
   setq(c (make-agent))

((run-environment (make-vacuum-world
                    (:agents '(a b c))
                    :stream nil
                    :aspec '(reactive-vacuum-agent))))








Yahoo! Groups Links

"rose_ronglin" <rose_ronglin@...> writes:

> I defined agent like that but not working:
>  setq(a (make-agent))
>   setq(b (make-agent))
>   setq(c (make-agent))

Try the following:

   (setq a (make-agent)
   (setq b (make-agent)
   (setq c (make-agent)


Paolo
--
Why Lisp? http://alu.cliki.net/RtL%20Highlight%20Film

Paolo Amoroso <amoroso@...> writes:

> Try the following:
>
>   (setq a (make-agent)
>   (setq b (make-agent)
>   (setq c (make-agent)

I have forgotten the closing parentheses, sorry:

   (setq a (make-agent))
   (setq b (make-agent))
   (setq c (make-agent))


Paolo
--
Why Lisp? http://alu.cliki.net/RtL%20Highlight%20Film

Hi all,
I have a simple AI quesiton.
On page 73 we read: "...breadth-first search is optimal if the path
cost is nondecreasing function of the depth of the node."
But on page 75 we read: "Breadth-first search is optimal when all step
costs are equal...".
I think both of those conditions are needed. So it seems it's better
we say, Breadth-first search is optimal when both of above conditions
are true.
If I'm wrong please correct me.

Excuse me for my poor English.

Cheers,
Hesham

Hi

I had a couple of doubts while going through Ch. 15 (Probabilistic
Reasoning over Time)

- How do we represent an HMM as a recursive relational probabilistic
model?

- Ex 15.4 asks us to show that for some temporal probability models &
observation sequences, the procedure described on page 547 returns an
impossible state sequence. How do we go about this?

Any pointers would be greatly appreciated.

Thanks
Shraddha

Hi

I had a couple of questions while going through Ch 15 (Probabilistic
Reasoning over Time)

- How do we represent an HMM as a recursive relational probabilistic
model?

- Ex 15.4 asks us to show that the procedure described on page 547
for finding the most likely state sequence returns an impossible
state sequence, for some temporal probability models and observation
sequences. How do we go abt this?

Any pointers to this will be greatly appreciated

Thanks
Shraddha

Hi, i am taking an AI class with AIMA as text and
planning for a mini course project.

Any ideas, suggestions?

aman

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