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Practice English Speaking&Listening with: Lecture - 10 Constraint Satisfaction Problems 2

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Today we start with the lecture ten of the course Artificial Intelligence. In the last

class we introduced what you mean by constraint satisfaction problems and we

looked at how we cast such problems as search problems and solve them by depth first search

with backtracking. Today we will explore some more efficient

techniques for solving constraint satisfaction problems.

The instructional objectives of today's class are as follows:

In this class the student will be introduced to more efficient search for constraint satisfaction

problem. We will talk about the following strategies:

Forward checking and then we will look at constraint propagation algorithms like the

AC-3 algorithm. Then we will briefly talk about intelligent backtracking or

backjumping. On taking these topics the student should be able to apply these techniques to

constraint satisfaction problems.

Before we start let us recapitulate the formulation we did for constraint satisfaction problems.

If you remember, the sort of constraint satisfaction problems we had

a number of variables and each variable has a domain. So v1 is a variable and v1 has a

domain it can take a set of values. v2 is another variable it can take some

set of values. So we have different variables along with their domains. Our objective is

to assign variables to values and the value for a particular variable must

come from its domain and the constraints must be satisfied.

We especially looked at a most common type of constraint called binary constraints. Binary

constraints involve a pair of variables. Suppose we can have a

constraint saying that v1 and v2 cannot have the same value this is one type of constraint.

We can say that the constraint between the variables v2 and v3 is that

the value of v3 must be greater than the value of v2. So we have a set of variables, each

variable has a domain. The variables must be assigned values from the

domains such that the constraints are satisfied.

This is a general class of constraint satisfaction problems. These constraint satisfaction problems

can be cast as state space search problems and we can do the

search in the following manner: We pick up a variable, so this is the start, when none

of the variables are assigned any values so v1 is not assigned a value, v2 is

not assigned a value, v3 is not assigned a value and so on. So this is the empty state

where none of the variables are assigned any values. In the next step we look

at the assignment of different values to variable v1. Suppose variable v1 can take the values

1, 2, 4, 7 we have 4 branches corresponding to all the possible values

that v1 can take.

In the next state we have to pick up another variable. Let us say we pick up v2 and we

have to look at all possible assignments of values to v2 that are consistent

with the assignment that we have so far. For example, if v1 has the domain 1, 2, 4, 7 and

v2 has the domain 2, 3, 4 then if v1 is already one v2 can be either 2 or 3

or 4. If v1 is 2 then v2 can be either 3 or 4. We cannot have v2 is equal to 2 because

any further assignment after we have v1 is equal to 2, v2 is equal to 2 will not

undo this constraint. So if we have an assignment which is inconsistent we should not proceed

further in that direction.

Next, suppose we have another variable v3 which has a domain 1 and 5 and we have the

constraint that the value of v3 must be greater than the value of v2 then

in this case v3 can only take the value of 5, in this case v3 again can only take the

value of 5 and so on. So, in this way we can construct the search tree

corresponding to the constraint satisfaction problem and we note that a solution to the

constraint satisfaction problem is a leaf in the search tree which corresponds

to all variables being assigned to specific values that do not violate any constraints.

Therefore, if there are n variables all the leaves at level n are the different solutions

to the search problem. And the strategy which is appropriate for solving such

search problems is depth first search. Whenever we find that a constraint is violated we do

backtracking. This backtracking is called chronological backtracking

because we backtrack to the previous choice point. So, depth first search with chronological

back tracking is a very appropriate method for solving constraint

satisfaction search problems.

We also discussed that we can try to make this search efficient by looking at a proper

order in which we choose the next variable that we select next for

assignment. So if you take a proper order of the variables we might end up in reaching

the first solution much faster. And then once we pick a variable to assign

values too we will be considering in what order should we assign values to these variables.

Then we will look at other questions like, can we detect inevitable failure earlier?

So, in the general search problem we said that whenever we see that a constraint is

violated we terminate the search at that point and then backtrack to the

previous choice point. But it may be the case that future problems can be detected even

earlier on. Let us see some techniques to handle that and then there is a

question whether we can take advantage of the problem structure to further prune the

search specific to a particular problem. One heuristic to choose the next

variable is the minimum remaining value heuristic. In the minimum remaining value heuristic the

variable with the fewest remaining values is chosen next.

What do you mean by variable with the fewest remaining values?

Every variable has a domain. So when we have a partial search tree some of the variables

have been already assigned values and as a result of assigning values to

v2 and v3 the domain of v1 has become restricted, the domain of v4 has become restricted, domain

of v5 has become restricted. So we find out the size of the

domains of the remaining variables and we pick the variable for which the remaining

domain is smallest in size. The intuition behind this is, since in the constraint

satisfaction solutions or constraint satisfaction problems all variables have to be assigned

values, the variable which is most constrained will have the fewest

remaining values. So we need less of backtracking in order to consider the effect of that on

the search. Once you have chosen a variable we have to choose the

order of value assignment. And here the heuristic is, choose the least

constraining value. All variables have to be assigned value and the solution but each

variable has to be assigned only one

value. So our intuition is to choose a value which is most likely to take you to a solution.

And consider that value which constraints the domains of the remaining

variables the least. That is the value that rules out the fewest values in the remaining

variables. Apart from the depth first search formulation we are considering we

can propagate constraints earlier on.

Instead of considering variables one at a time we look at the constraints early on to

reduce the search space. And there are several ideas which people have

formulated which gave rise to different algorithms for constraint satisfaction search and the

simplest of these ideas is forward checking.

What is forward checking? We keep track of the remaining legal values

for the unassigned variables. So, when we assign values to a variable we suitably constrain

the domains of the

remaining variables. And then if we detect that there is a variable which does not have

any legal values left then we can immediately terminate that path because

finally we have to assign values to every variable along a correct solution path. So,

if you find that there is a variable which does not have any legal values left we

will not be able to assign values to the variables so that search path is bound to fail. If we

can detect this early on we can terminate search at that point of the

search tree. This is an example of a graph coloring problem where we have five nodes

and there are three colors red green and blue and these are the links in the

graph.

Corresponding to this problem suppose we show how a constraint satisfaction search proceeds.

Initially suppose we pick the node MH we can paint is red green or blue we paint MH is

equal to red then if we pick KN next KN is a neighbor of MH so KN can

be either green or blue. If we pick KN is equal to green and then we pick KE next KE

is a neighbor of both MH and KN so KE is a neighbor of KN so it has to

have a different color than KN so it can be either red or blue. So this way we can construct

the search tree and we can do a depth first search.

Now, if we do forward checking let us see how we will proceed. Initially we have five

variables MH KN KE AP and TN. Each of these variables can take each of

the three values red green and blue. When we set MH is equal to red KN and AP are neighbors

of MH so they can be either G or B, the others can be RGB. If we

pick now KN is equal to green then AP which is a neighbor cannot be green so it can only

be blue, KE and TN can be either RB or RB they are also neighbors of

KN. Then we pick KE is equal to red as a result AP and TN can be only blue. Now if you take

AP blue we find that the domain of TN is empty because TN is a

neighbor of AP. So we detect here that there is a problem with the assignment and we backtrack.

This is what forward checking does.

Let us look back at the figure. At the 4th step we found that KE and TN have only blue

left in their domains. Now KE has a legal value left, TN has a legal value

left but we also know that there is a constraint between KE and TN. KE and TN cannot be of

the same color. We are talking about AP and TN. They cannot be of

the same color because they are neighbors. So in this step you see that AP and TN have

only one legal value left. Even though they are individually legal together

AP is equal to blue and TN is equal to blue cannot happen. So a smart algorithm might

detect this early on and terminate search. This is a limitation of forward

checking. Forward checking checks that every remaining

variable has at least one legal value but it does not explore further whether these

values can be assigned to this

variable subject to the constraints between these variables. Forward checking detects

some of the inconsistencies but it does not detect all inconsistencies. For

example, in the previous example it does not detect that AP and TN cannot be blue simultaneously.

Now let us look at another constrained graph on which we will run the next algorithm. Again

this is a graph coloring problem. We have 6 nodes which we have

numbered 1, 2, 3, 4, 5 and 6 and this is the connectivity information between the nodes.

Each of the nodes can be colored red, green or blue but two neighboring

nodes cannot be assigned the same color. Now, if we do forward checking let us see what

happens.

Suppose we first pick up 1 and we assign 1 to red. Now, if I assign 1 to red 2 and 3

cannot be red so we reduce the domains of 2 and 3 we have blue or green. So,

for each variable which is in constraint with 1 we eliminate the conflicting values from

the domain. Next is, suppose we assign green to 4 then 2, 3 and 6 cannot be

green so we remove green from the domains of 2, 3 and 6.

Now 2 can be only blue, 3 can be blue, 6 can be red or blue. Next we assign blue to 5 and

as a result it affects the domains of 3 and 6. So we remove blue from the

domains of 3 and the domains of 6.

Now as a result we see that the domain of 3 becomes empty. So this is impossible so

we have to back track at this point. Let us now find out the weakness of

forward checking. When we assigned for green we saw that 2 and 3 have only one legal value

left which is blue but these two legal values cannot be assigned

together so we should have detected this problem early on, forward checking does not detect

this.

So this is not detected by forward checking. We should have backtracked at this step.

Now let us see what further improvements we can do. There are several things that we could

do. One is we could try to propagate the implications of the constraint

on one variable on to the other variables. We propagate not just values but constraints

between these variables. This is the idea of constraint propagation we will

explore next.

Secondly what we will see after that is, we can backtrack intelligently which we call

conflict directed backtracking. So we will save possible conflicts for a variable

value when we assign that variable value and we will back jump to that assignment which

gave rise to this conflict.

In chronological backtracking on normal depth first search we backtrack to just the previous

choice of point.

In backjumping when we detect a constraint violation we backtrack to that assignment

which led to the conflict which could have taken several decisions earlier.

So first, let us explore constraint propagation.

Arc consistency is a fast method for constraint propagation. This is how arc consistency works:

Given the current domains of v1 and v2 the arc from v1 to v2 is consistent if, for every

value x of v1 there is some value y of v2 that is consistent with x. In

forward checking check if the domain of in the remaining legal the legal domain of v1

is empty or not. Here we are not only checking whether the domain has a

value but we are just checking whether for every value left in the domain there is some

value y in the domain of another variable so that these two assignments of

values are together consistent. This is what constraint propagation does.

There are several constraint propagation algorithms. We will look at AC-3 and some algorithms like

that. Let us see an example of the same graph that we

discussed earlier.

We assign green to 4. After we assigned green to 4 we check all the arcs from 4, 2, 3 and

6. First we check 2. So when we check 4 is equal to 2 we eliminate green

from the domain of 2. But we have to check back later with the variables with which 2

has some constraint.

Next we check the constraint 4 to 3 and eliminate green from the domain of 3. After that we

check the constraint 4 to 6 and eliminate green from the domain of 6.

After we have finished checking the neighbors of 4 we go back to those variables whose domains

have been changed to check if their current remaining values are

consistent with the values of their neighbors.

We now go back to 2 and check the neighbors of 2. First we check 2 with 1 there is no

violation, there is no problem. Then we check 2 with 3 and then we notice

that if 2 is blue 3 cannot be blue so we eliminate blue from 3.

If we do that the domain of 3 becomes empty which cannot happen so we backtrack at this

point. This is the arc consistency algorithm and we can backtrack when

we detect that there are variables which do not have domains which are consistent together

because of the constraints between those variables. This is the idea of

arc consistency.

Whenever the domain of a variable is revised the other arcs may need further revisions

until no more inconsistencies remain. One implementation of the arc

consistency algorithm is the AC-3 algorithm. The AC-3 algorithm uses a queue to keep track

of the arcs that need to be checked for consistency because when we

assign 4 to green it is not enough that we check the domains of 2, 3 and 6 which is neighbors

of 4 but if these domains did change we have to check 2 with its

neighbors and do this recursively until there is no further constraint propagation. So AC-3

does this constraint propagation so whenever the domain of a variable is

reduced according to this algorithm you propagate the constraints to the neighbors of that variable.

This is done recursively in the AC-3 algorithm. Now let us look

at an implementation of the AC-3 algorithm.

We have a data structure of a queue. Initially the queue contains all the arcs of the graph.

So we have all the Vi Vj that are the edges of the graph such that i0 is

equal to j. Then while q is not empty we select one of the arcs from this queue that is Vk,

Vm and we delete Vk, Vm from queue. Then we check if revise Vk, Vm

is true. That is, when we delete this arc we see if the domains get reduced. When we

consider this constraint between Vk, Vm we check if the domains of these

variables get reduced. So as a result of assignment of Vm if the domain of Vk gets changed then

we find all the arcs in the graph of the form Vi, Vk. That is, if Vi is

a neighbor of Vk then we have to check those constraints. So we append to the queue all

the arcs Vi, Vk. If as a result of an assignment to Vm the domain of Vk

gets changed.

Now let us see what is the complexity of this algorithm AC-3?

Now, if we have a binary constraint satisfaction problem then the constraints are only between

pair of variables. So there are utmost n square arcs in the

constrained graph.

In the algorithm every arc can be inserted utmost d times if the domain of Vi has d values.

So, when we are putting an arc involving a variable into the queue we

are only doing it when the domain of the variable is reduced. Suppose initially the domain of

a variable Vk has b values we consider this arc for revision only when

this domain is getting reduced. Corresponding to every Vk the arc can be put into the queue

utmost d times. Now, checking the consistency of an arc can be done

in order d square times because we check it against all the arcs in the graph. So the

worst case time complexity of this algorithm is n square times d times d square

which is O(n square d cube). So O(n square d cube) is the worst case complexity of the

AC-3 algorithm. This is what constraint propagation is about.

In constraint propagation as a result of reduction in the domain of a variable we look at the

implications in terms of the constraints of the variables with other

variables. And there are a set of algorithms which do constraint propagation. One of the

well known algorithms is AC-3 which we discussed. Next we will consider

another technique which we mentioned as intelligent backtracking.

We have already seen backtracking which has a very important role in solving the constraint

satisfaction problems. So what you have in plain depth first search is

chronological backtracking. When the branch of a search fails we need to backtrack to

the most recent choice point. This is called depth first search. In depth first

search we backtrack to the previous choice point but intelligent backtracking involves backtracking

to the point because of which the conflict has arisen.

Suppose we have a search tree like this and we have a constraint violation here, in chronological

backtracking what we would do is, if there is a violation here we

will go back here and then go back here which has a choice point. So we will go back to

this node which is the next choice point. But in chronological backtracking

if we find that this violation is because of an assignment here we will backtrack directly

to this place. This is called backjumping or conflict directed backjumping.

In chronological backtracking consistency check is performed in the order in which the

variables were instantiated. If consistency check fails we try the next value

of the current variable. If the current variable has no more values we backtrack to the most

recent variable that has a choice left. This is what is done is normal

depth first search. Backjumping is also a type of backtracking but it is more efficient

when there is no consistent instantiation to be found for the current variable.

And in this case we go to the deepest variable which was constraint checked against the current

variable. So, when we find that the current variable is inconsistent

and has no possible instantiation left instead of backtracking only to the previous choice

point we find out what is the last variable which constrain the current

variable. The current variable has no legal values left. In order that the current variable

does get some legal value some constraint which participated with the

current variable has to be changed. So we find the deepest node in the path from the

current variable to the root where a variable was assigned a value and that

constrains the domain of the current variable, we have to find that and that is what intelligent

backtracking does.

Backjumping reverts to the deepest variable which was checked against the current variable

in which introduce the constraints. Now let us look at this example. So

here we have the 6 queens problem and we have put queen 1 in row 2, queen 2 in row 5, queen

3 in row 3 and queen 4 in row 6. After that we have put queen 5 in

row 4. Now we see that, after we have placed these 4 queens the last queen cannot be placed

in any of the rows. When we cannot put queen 5 we really do not

need to look at any further assignment to the value of queen 5.

We have looked at some efficient constraint satisfaction problems. We have revised forward

checking then we have looked at constraint propagation for which we

looked at arc consistency algorithms and then we looked at backjumping.

What is the idea behind these algorithms? In forward checking when we assign values

to some of the variables we look forward and then we update the domains of the remaining

variables. And in forward

checking we just checked whether the domain of any of the remaining variables become empty.

If they do become empty then we terminate search at that point. In constraint propagation,

in the general constraint satisfaction class of problems what we do is

that, we not only propagate the constraints to restrict the domains of the unassigned

variables but we also do a consistency check on those variables which have

constraints between them. That is we check whether every variable whose domain gets affected,

whether it has a legal value for a legal value of its neighbor with

which it has some constraints. AC-3 is a constraint propagation algorithm which does this recursively.

Whenever it reduces a domain of a variable it takes up those

variables and looks at its constraints with its neighbors.

If there are constraints with its neighbors which reduce the domains of the neighbors

then again it checks those with their neighbors and so on. So AC-3 is quite an

efficient algorithm and is able to detect constraints earlier. As you make your constraint

checking algorithms very intelligent you are reducing your reducing the

number of nodes in the constraint satisfaction tree that you are going to search but you

are spending more time at every step. So you have to select a right trade off

about whether these intelligent processing does give you benefit.

Does this take up time? Whether the benefit that you get in terms of reduced search space

if they warrant that sort of benefit? Finally we looked at backjumping. What backjumping

does is that, when in your search tree some constraint or some conflict is discovered

that is the domain of

this variable has become empty. In backjumping we find the earliest variable which had been

checked with this variable in which we introduced " ". Now we must

undo those effects in order to get out of this situation. So instead of going to the

most recent choice point we jump to a conflict point.

Now we look at the questions for this lecture: The first question: Apply the different constraint

satisfaction algorithms we discussed in today's class on the 5 queens problem. In each case

you should find out

the number of nodes that are expanded. So first you will try ordinary depth first search.

Second; you will try variable ordering heuristic along with ordinary depth

first search. In the third instance you try forward checking. The 4th instance you work

out AC-3 and finally you work with backjumping.

Let us look back briefly at the questions we have set for lecture nine. If you remember

the first question asked you to do the class room scheduling, to look at the

class room scheduling problem and formulate it as a constraint satisfaction problem. So

let us briefly look at the class room scheduling problem. We have a number

of class rooms. Suppose cl1, cl2, cl3 are the class rooms and we have some times slots

Let T1, T2, T3, T4 be the time slots and we have some teachers. So let P1, P2, P3, P4

etc be the teachers. Now for each teacher we have a set of classes which the

teacher should be able to teach. So let us say P1 should be able to teach cl1 and cl2,

P2 should be able to teach let us say cl2 and cl3, P3 should be able to teach

cl4, P4 should be able to teach cl5 and cl1. So we have a set of teachers and each teacher

should be able to teach a set of classes. We have to assign courses so we

have a set of courses.

Now this is a problem which involves a large number of variables of different types. Now

let us look at the basic types of constraints which are involved. Now, if a

teacher has to teach both the courses namely course 1 and course 2 then it means that course

1 and course 2 must be in different time slots. If a teacher is teaching

c1 and c5 it means c1 and c5 must be in different time slots. Now each class room at a slot

can have only one class. This is something you have to follow. And if a

teacher is teaching two classes those two classes cannot be of the same time slot. Now

what we have to do is, we have to assign classes to class rooms and time

slots.

So we can start with picking class c1 and assigning it to class room 1 time slot T1.

Now, if c2 and c5 cannot have the same time slot c2 domain will not have this

time slot T1 and c5's domain will also not have T1. So when we consider the next instantiation,

when we look at c2 we have to look at the choices for c2 in time

slots other than T1.

When we look at c3, if after c1 we look at c3, c3 can be in time slot T1 but in time

slot T1 it cannot be in a class room cl1. So any two courses cannot share the

same room and slot. So both rooms and slots cannot be the same for two courses. And two

courses that can be taught by the same professor cannot share a time

slot. So, based on these constraints we have to do the assignment of values. You can work

out these for a small set of values.

The second problem we said was a crypt arithmetic problem. Specifically we were asked to work

out the values of the different letters for this crypt arithmetic

problem. I will discuss briefly how you go about the

solution to this problem. Sometimes for these problems if we are working out manually you

can use certain heuristics

from your knowledge. The value of M in this case cannot be more than 1. So even if S and

M have the highest values this can be utmost 1. This is not 0 because if

this value is 0 we would not be putting this here so this is very easy to see that M is

equal to 1. So what we can do is, we can simplify this problem by putting 1 in

this position. Now we have some other variables left.

What are the variables we have? The variables are D, E, Y, N, R, O and S so

we have seven variables in this problem. Let us say we first pick up D.

Now what are the values that D can take? D can be 1 or 2 or 3 or 4 or 5 or 6 or 7 or

8 or 9 or 0 so we try the different values of D. There are many possible values of D

so we try all possible values of D.

Now let us say first we try D is equal to 1. Now if D is equal to 1 we have to try the

different values of E. We first try the value E is equal to 1. Now if E is equal

to 1 then Y has to be is equal to 2 if D is 1 and E is 1 and Y is 2. So Y is fixed and

here we have E is 1, this E is 1, these are the values we have assigned and Y has

become 2.

Now we have to choose N and R so that we have 1 here. Suppose so we choose the different

possible values of N, suppose you first choose N is equal to 1, if N is

equal to 1 then R has to be 0 to satisfy this constraint so we have this assignment. And

then E is 1 along this path and we do not know the value of O and N is 1 so

E is 1, O is not known, N is 1 so it must be is equal to 0. So here we have o is equal

to 0 and we have this is 1 so S must be is equal to 9. Therefore a solution to this

problem is S is equal to 9, E is equal to 1, N is equal to 1 and D is equal to 1, M

is 1, O is 0, R is 0, E is 1 and M is 1, O is 0, N is 1, E is 1 and Y is 2 and you can

verify whether this is correct.

It is 1 plus 1 is equal to 2, 1 plus 0 is equal to 1, 1 plus 0 is equal to 1, 9 plus

1 is equal to 10. So this is a solution to the crypt arithmetic problem send plus more

is

equal to money. And we have solved this problem by simple depth first search and in this case

this is a simple problem with many solutions so we could get this

problem by only exploring one path. so we will stop here today in the class we will

first briefly discuss the answers to today's problem and we will start on the next

topic which is on machine learning thank you very much

The Description of Lecture - 10 Constraint Satisfaction Problems 2