We continue our discussion on the shortest
path problem. In the previous lecture, I explained the shortest path problem using this numerical
example. The problem was to find out the shortest distance
and the shortest path from 1 to 7. We have already said that this arc could represent
a cost or a distance or a time. We would like to find out the shortest path and the corresponding
sum of weights associated with the shortest path. We started with a 0 here, and then we
first found out the minimum distance that is reachable from this. 2, is reachable at
15, 3 was reachable at 20 so we picked the minimum and labeled this as 15. From 2 we
proceeded to subsequently label 3 at 20, 4 at 25, 5 and 6 at 40 and 7 at 50. We found
out that the shortest path was 1 to 2, 2 to 5, 5 to 7 with total cost or distance equal
to 50. We then proceeded to explain a tabular form
or a labeling algorithm to do the same thing that we did on the network. This rudimentary
algorithm is the Dijkstra's algorithm and we now present a labeling version of the Dijkstra's
algorithm. Here 1 was the source node so we start with 1 we write the rest of the nodes
here from 2 to 7. We go back to this network to see which ones are directly reachable from
1. They happen to be 1 to 2 with 15 and 1 to 3 with 20. We write 15 and 20 here, 1 4,
1 5, 1 6, and 1 7 do not exist. Therefore, we write a dash here or we can write infinity,
saying that at present the distance is infinity. Among these values, there are only two of
them, between these two we pick the one which has the smallest numbers which is 15 and mark
it with the star or with some form of a label. Since node 2 is marked we write 2 here and
then we proceed to this network to find out what are all the other nodes that are reachable
from 2. They happen to be 4 and 5. Now 2 to 4 with 15 plus 10 is equal to 25 and 2 to
5 with 15 plus 25 is equal to 40. We look at these values. If you look at 4, 4 is reachable
with 15 plus 10 is equal to 25. Earlier, the distance was infinity, 25 being smaller than
infinity, we changed this value to 25. Similarly, 40 being smaller than infinity, the value
are updated here from infinity
to 40. Now, 20, that was a distance from 1 to 3, remains as 20 because 3 is not reachable
from 2, so 20 remains as 20. We have three values 20, 25 and 40. Out of these three values,
we pick the minimum one, which happens to be for 20 and then we mark a label here.
Since 3 is marked, we write 3 here and then go back to this network to see what are all
the things that are reachable from 3. They turn out to be 3 to 4 and 3 to 6. 3 to 4 is
20 plus 15, is 35. 35, happens to be more than this 25, so we retain this 25. Now, 5,
is not reachable from 3, so 40 remains as 40. 6 is reachable from 3, 20 plus 20 is equal
to 40, 40 being less than infinity, infinity value is changed to 40. Again, there are three
values; choose the minimum which happens to be at 25, which also happens to be for node
number 4. So, node number 4 comes here and this process is repeated till node number
7, which is the destination node; it is labeled and the value is equal to 50.
What we have done here is we have systematically written this. This labeling version of Dijkstra's
algorithm is far easier to understand and implement, if one was to write a computer
program to find out the shortest path on the network. The same 50 which happens to be the
shortest distance, obviously, the same 50 we have got here. While for this, we said
it is 1 to 2, 2 to 5, and 5 to 7. The motivation came from the fact that 0 plus 15 is 15, 15
plus 25 is 40, 40 plus 10 is equal to 50. We have a path, such that every label for
every i j in that path, the label corresponding j is the sum of the label corresponding to
i, plus this Cij. For example, you could, if we call this as
some w2, which is the label corresponding to 2, then w2 is w1 plus c12, w5 is w2 plus
c25 and w7 is w5 plus c27. There is a path from 1 to 7 and we got this 50. How can we
identify this path from this table? This is something we need to do. What happens is the
final value is 50. There is a change, move upwards, so this change is happened from 55
to 50, because we passed through node number 5. The first thing, we have to do is write
7, which is the destination. Since, there is a change here from 55 to 50, that change
has happened because we have passed through this 5, so, we write 5 here.
Now, come back to this, we will be at 5, now, move upwards again. Here, there is a change
from infinity to 40 and this change has happened, because we had moved from 2, so we write 2
here. Then this is the change, and then come back here to 2, we moved here, there is a
change and this change happened, because we labeled this 2, so we write 2 here and then
we write 1 here. To repeat, we got this 50, there is a change from 55 to 50. We represent
the first destination node here. Then there is a change from 55 to 50. This change has
happened because we moved from 5. So, 5 come here, so automatically, we go to node number
5, which is 40. Move upwards, there is a change between this infinity and 40, because we moved
from here to the labeled node 2, so we write 2 here. Then, we come back to this 2, we realise
there is no change; therefore, it is 1 2. The path becomes 1 to 2, 2 to 5, and 5 to
7, with weight equal to 50. This is how we explain the labeling algorithm for the Dijkstra's
and the way to get both the shortest distance as well as the shortest path. What else can this Dijkstra's algorithm
do? When we look at shortest path so far we have
seen shortest path from a given source to a given destination. The second type of problem
that we can think of is from a given source to any destination. Let us see if the Dijkstra's
algorithm can solve this problem as well. If we assume that our given source is 1 we
can easily show that these label is that we have written here, not only helps us in getting
the shortest distance from 1 2 7, we have actually solved the shortest path from 1 to
2, 1 to 3, 1 to 4, 1 to 5, 1 to 6, 1 to 7. We can show that the shortest distance from
1 to 2 is 15, 1 to 3 is 20, which is obvious for 4. It is 25 for 5 and 6 it is 40 and for
7, it is 50. The Dijkstra's algorithm not only solves
the first problem whereas it helps us to find the shortest path and the shortest distance
from a given source to a given destination. It also solves the second problem where it
does it for a given source to all destinations. With one pass of the algorithm, we have actually
solved both these problems. There is a third problem which is like from any source to a
given destination can be seen as a third problem. This can be done in more than one way, one
of which is to do a backward pass of the algorithm from here to here.
Remember, what we did first is a forward pass from 1 to 7. We can actually do a backward
pass from 7 to back, if we do that, we will be able to find out the shortest distance
from every one of this to 7, because a backward pass implies that we are actually trying to
find out the distance from 7 to every other node in the network. Backward pass of the
Dijkstra's algorithm would help us to find the shortest path from every node to a destination
or from destination to every node on the network. The fourth problem is between every source
to every destination, which means, we wish to find out something from 2 to 6 or 3 to
5 or 4 to 7 and so on. One of the ways of using this algorithm is to do this, keep every
vertex as the source and run it n times if there are n vertices. Running the Dijkstra's
algorithm n times, in this case 7 times, would help us get the shortest path from every source
to every destination in this network. Effectively, the algorithm that we have seen actually solves
all four of them, but with certain increasing amount of computation and complexity. Effectively,
it solves both these in a single stroke, given 1 as source and 7 as destination. In a single
pass of the algorithm, either through this method or through this method, we can find
out the shortest distance from 1 to 7, which is from a given source to a given destination.
We can also find the shortest distance from a given source 1 to every other node as destination
using the same algorithm. This is one more pass of the algorithm, which is a backward
pass; whereas, this is a little more complex where we have to run this algorithm 7 times
or (n minus 1) times, 6 times as the case may be. We can find out the shortest distance
from every source node to every destination node on this network. There are other algorithms
which try to handle this problem a little more efficiently where they specifically address
this problem and where we find out the shortest distance from every node to every other node
in the network. There is a more popular algorithm called the
Floyd's algorithm which we are not going to look at in this course. Nevertheless, we
understand that we can run the Dijkstra's algorithm (n minus 1) times with each of these
nodes as the starting nodes. Keeping each of these nodes as a starting node and solving
this problem to find out the shortest path from a given source node to every destination
node we will be able to handle the fourth one. Effectively, we understand that we can
use this algorithm to solve all these four problems though there are slightly more efficient
algorithms to solve this. With this, let us come back and try to understand
how the Dijkstra's algorithm is optimal. Or how do we confidently say that, whatever
we have done here or the labels essentially represent the correct shortest path and the
corresponding shortest distance. In order to do that, we first formulate the shortest
path problem as a linear programming problem and then try to write the dual of the linear
programming problem to understand how these labels work. Let us
write the linear programming formulation for
this. The formulation will be like this: Xij equal to 1, if arc i - j is in the shortest
path and is equal to 0, otherwise. The
objective function will be to minimize CijXij,
so Xij will be equal to 1, if arc i - j is in the shortest path and 0 otherwise.
Now, if we take this node 1, which is a starting node, we should have at least one arc that
leaves this node. We should write a constraint that X12 plus X13 should be equal to 1, because
we have to leave this node in order to reach this destination. We can leave this node either
through this or through this. We say X12 plus X13 is equal to 1. If we take this node, this
is an intermediate node. The shortest path may pass through this, may not pass through
this. This being an intermediate node the only thing we need to do is to model what
is called the flow balance equation. If this lies in the shortest path, then you enter
and then you leave. If it does not lie in the shortest path you do not worry about it.
That is written in the form of a constraint which is like this: minus X12 plus X24 plus
X25 is equal to 0. What does this constraint do if this lies in the shortest path? Then,
you have to enter this node which means X12 should be equal to 1, which also means that
you should leave this node by either to 4 or to 5, so the constraint will take care
of it, X12 is 1. Then this contribution is minus 1, so one of them will take a plus 1,
so that you get a 0. Therefore, if it enters 2 through X12, it has to leave either through
X24 or through X45. If it does not lie in the shortest path then
this will be 0, automatically these two will be 0 and therefore, this constraint will take
care of that requirement. For any intermediate node, all we need to do is to model the flow
balance equation, so whatever comes in should go out. For this node, we will have minus
X13 plus X34 plus X36 is equal to 0. For this 1, we will have minus X24 minus X34 plus X45
plus X46 plus X47 is equal to 0. Remember, there are 5 arcs associated with
this node. Two of them enter this node and three of them leave this node. For 5, there
are three arcs associated, so you get minus X25 minus X45 plus X57 is equal to 0. For
6, we get minus X36 minus X46 plus X67 is equal to 0 and for 7, 7 being the destination
node, you have to reach 7. So, we write X57 or X47 or X67, sum of them should be equal
to 1. So the constraint will be X57 plus X47 plus X67 is equal to 1. We write it in a slightly
different way. We write it as minus X47 minus X57 minus X67 is equal to minus 1. Please
note that this constraint should ordinarily read as X47 plus X57 plus X67 is equal to
1, but we have multiplied this by minus 1. We have written minus X47 minus X57 minus
X67 is equal to minus 1. We have Xij is equal to 0 1. This completes
the linear or linear integer programming formulation of the shortest path problem, because Xij
is equal to 0 1, makes Xij a binary variable. Therefore this formulation is a linear integer
formulation where all the variables are restricted to be binary variables.
Let us go back and try to understand that this problem is a network problem. We have
seen some characteristics of network problems earlier when we have looked at transportation
and assignment problems. We also said the transportation and assignment problems have
an important property called unimodularity and because of this unimodularity property,
even if you solve the linear integer problem by relaxing this binary variable and by writing
it as continuous variable and by relaxing this we would still get solutions which are
either 0 or 1. We have already seen a little bit about this
unimodularity property. Unimodularity property would simply mean the rank of the constraint
coefficient matrix. We can actually show the way this matrix is structure simple addition
on the left-hand side and simple addition on the right-hand side, would give us 0 equal
to 0. In fact, one of the reasons we have written this in this form, particularly here,
is to facilitate that, if you add all the left-hand sides and all the right-hand sides,
you will see this 1 and minus 1 cancel out. Every Xij will cancel out to give a 0 is equal
to 0 which tells us that this is a linearly dependent system of equations.
Before that we also observe that this formulation will have as many constraints as the number
of nodes or vertices and as many variables as a number of arcs or edges. There are n
vertices, we have a linearly dependent system and we can also show that the rank is (n minus
1). Then, we also realise that the determinant of the corresponding matrix will have a plus
1 or a minus 1. Because of that property, B inverse will be integer value and therefore,
the actual values of Xij will only be integers. In this case 0 1, even if we relax the binary
assumption and treat it as a continuous variables because of this unimodularity property, we
can ignore this 0 1, treat it as a continuous variables, solve it as a linear programming
problems. We would still get values for 0 1, for these variables.
Another way of showing unimodularity is that, if we have a constraint coefficient matrix,
such that it can be written in this form, where every variable appears in only two constraints,
whether plus 1 and minus 1, then we could show that, this one is unimodularity. In more
than one way, we can establish the unimodularity characteristic of this matrix. Therefore,
we can relax this 0 1 assumption and then treat it as a linear programming problem.
Once we treated as a linear programming problem, it is easy to write the dual. We start writing
the dual of this problem and therefore we introduce dual variables w1 to w7, call them
w1, w2, w3 and so on, till w7. Now, we write the dual of this problem. From
duality, once we have defined dual variables w1 to w7, primal being a minimization problem,
dual is a maximization problem. We will get maximize w1 minus w7, 1 into w1 minus 1 into
w7; will be the objective function, maximize w1 minus w7. Each variable Xij appears in
the ith and the jth constraint here. We look at this w1 minus w2. If we look at variable
X12 w1 minuss w2, we have a c12 here. Remember, the primal is a minimization problem; dual
is a maximization problem. So minimization problem with all greater than or equal to
constraint would give us some maximization problem with less than or equal to this. wi
minus wj is less than or equal to Cij. In our case, w1 minus w2 will be less than or
equal to c12 which is 15. Similarly, we can write w1 minus w3 is less
than or equal to 20. From this, we now start writing w2 minus w4 is less than or equal
to 10; w2 minus w5 is less than or equal to 25; w3 minus w4 is less than or equal to 15;
w3 minus w6 is less than or equal to 20; w4 minus w5 is less than or equal to 20; w4 minus
w6 is less than or equal to 15; w4 minus w7 less than or equal to 3;. w5 minus w7 less
than or equal to 10; w6 minus w7 less than or equal to 20; and importantly, wj is unrestricted
in size. The unrestricted in size comes because all the primal constraints are equations,
so dual variables are unrestricted in size. This completes the dual of the problem. If
one can also see that the dual has seven variables corresponding to seven constraints of the
primal, which correspond to seven nodes in this network. The primal has twelve variables,
which correspond to
eleven arcs. There are eleven arcs in this network. There
are eleven variables here. These correspond to 11 constraints in the dual. Dual will have
as many variables as the number of nodes and as many constraints as the number of arcs.
Where, the primal will have as many constraints as a number of nodes and as many variables
as the number of arcs. With this primal and with this dual, let us try and get a solution
to this problem. Let us start with w7 equal to 0 that we like
to put. So, w7 is equal to 0, would actually give us from here w6 is 20 and w5 is 10 out
of this. This would give us w4 as less than or equal to 30. This would give us w4 less
than are equal to 35. This would give w4 less than or equal to 30. We write w4 is equal
to 30. From this, using this 20, w3 is less than equal to 40, so, w3 is less than equal
to 45. You get w3 is 40. From here w2 is 35, w2 minus w4 would give us 40, this would give
us w2 is equal to 35 and from here, we have w1 is less than or equal to 60. w1 is less
than or equal to 50, so, w1 is equal to 50. If we start with the solution w7 equal to
0 and work backwards, we get w7 equal to 0, w6 is 20, w5 is 10, w4 is 30, w3 is 40, w2
is 35, and w1 is 50; w1 minus w7, so this w value is 50. We realise that this is a feasible
solution to this dual. This is feasible to the dual in the sense that now if we look
at this value w1 minus w2 is equal to 15, this is satisfied as an equation. w1 minus
w3 is 50 minus 40 is less than, so this is an inequality; w2 minus w4 is 5, so inequality;
w2 minus w5, so, equation; w3 minus w4 is 10 inequality; w3 minus w6, 20, inequality;
w4 minus w5, 20, equation; w4 minus w6 inequality; w4 minus w7 equation; w5 minus w7 equation
and w6 minus w7 equation. Now, all these dual feasible solution satisfies
all these unrestricted inside. We also realise that, this is satisfied as an equation, this
is satisfied as an equation, this is satisfied as an equation, this is satisfied as an equation.
If we apply complementary slackness to this dual feasible solution, wherever it is satisfied
as an equation, the corresponding variable is a basic variable in the primal. X12 is
a basic variable. I am just circling them so X12 is a basic variable. 2 5 is a basic
variable. I am again doing this, now 4 5 is a basic variable. You may have 4 7 as a basic
variable, I have 5 7 and I have 6 7. These are my basic variables.
There are six basic variables, which is also understandable. There are seven constraints.
It is a linearly dependent system, so we will have (n minus 1) basic variable. There are
six basic variables. Now, corresponding to these basic variables, can we define a solution
to the primal? This would give us X12 is equal to 1; this is satisfied because X12 is 1,
X12 plus X25 is equal to 0. Remember, these are non-basic and therefore they take value
0. Therefore minus X12 plus X25 is equal to 0 would give us X25 is equal to 1. As far
as this is concerned, X13, X34, X36. We go back to this 1 3, 1 is 50, 3 is 40 so inequality.
3 is 40, 30 inequality. 4 6 is 30 plus 20, again inequality. We come back to this, 4
5 and 4 7 right now, we put X25 is equal to 1.
This would give us a contribution of minus 1. We have four minus X45 and X57 is equal
to 0. This would give us X57 is equal to 1. From here, you would have minus X57 is equal
to minus 1. You will have X47 is equal to X67 equal to anyone from here. X13 or X34
or X36 is equal to 0. We would have one more basic variable to represent this equation.
But then that basic variable would also take a value 0. So corresponding to this feasible
solution, we observe that X12 is equal to 1, X25 is equal to 1, X57 is equal to 1, satisfies
the complementary slackness condition. It also provides a basic feasible solution to
the primal. This would give us Z is equal to 50, 1 to 2 is 15, 2 to 5 is 25, 5 to 7
is 10 so this would give us Z is equal to 50.
We realize two things: one is, having written the dual, and we have got a dual feasible
solution by inspection. Corresponding to this dual feasible solution we applied complementary
slackness. We first found out which are the ones that are satisfied as equation and which
are the ones that are satisfied as inequalities. Those that are satisfied as equation, we go
back here; we try to identify the corresponding basic variable for those that are satisfied
as an equation. From that we observe that 1 to 2, 2 to 5, 5 to 7 is equal to 1 is primal
feasible. In more than one way we can show the optimality. One is that, we have a primal
a dual feasible solution. We apply complementary slackness and we obtained primal feasible
solution. Another way is to show that we have a dual
feasible solution with value equal to 50; we have a primal feasible solution, with value
equal to 50. They automatically, the values being equal they satisfy the complementary
slackness condition, therefore they are optimum. This is a way to understand that we get the
same 50 here, as the optimal solution, both from the primal and the dual of the linear
programming problem. We have got the same 50 here, which represents the optimal solution.
This is a way to show from the linear programming formulation that the answer you have got here,
which is 50, is indeed the optimal solution. We need to discuss a little bit more on what
do these w s represent? This w's effectively representing the labels that we have actually
given here. There are seven values of w. We also realise that one of them is 0. Then we
progressively move towards getting this 50. Alternately, if we look at this carefully,
the seven labels, that we have given here for w1 to w7 are actually different from the
labels that we obtained there. it is because we have w1 is 0 here, w1 is 50 there, w2 is
15 here, w2 35 there, w3 is 20 here, w3 was 40 there, w4 is 25 here, w4 was 30. You got
different values of these labels. Is there any relationship between these labels
and these values that we actually got? Yes, there is a relationship between these values
that we have here, as well as these labels and the relationship looks from this point
of view, this is 0 and that was 50, from 7 it 50 and 0. Effectively, we add up to 50,
which is the value of Z. More importantly, the difference lies in the
fact that, if we have done a backward pass of this algorithm, starting from 0 here, then
the labels that we would have got would be exactly those labels that we have got here.
The only difference is when I have described this part of the algorithm; I have described
a forward pass starting from 0 for the source and 50 to the destination. When I describe
the dual here, I started with destination equal to 0 and the source getting the value
50. Effectively, these labels represent the labels corresponding to a backward pass of
the Dijkstra's algorithm, whereas here we have done the forward pass. We have already
seen whether you do a forward pass or a backward pass, you will still get the shortest distance
from the source 1 to the destination. The Dijkstra's algorithm is actually a very
nice primal dual implementation of the linear programming formulation of the shortest path
problem. There is also an interesting relationship
between these forward passes and the backward passes. The dual was w1 minus w7 subject to wi minus
wj is less than or equal to Cij, then we said wj unrestricted. What we had written here
are the eleven constraints corresponding to the eleven arcs which satisfy wi minus wj
is less than or equal to Cij. We can do one more thing, because this is unrestricted.
I can always replace wj by minus wj and therefore we will write this as maximize minus w1 plus
w7, subject to minus wi plus wj is still less than or equal to Cij. Then we say minus wj
or wj unrestricted. This will lead us to a situation where you want to minimize w7 minus
w1, subject to w minus wi plus wj is less than or equal to Cij and wj unrestricted.
We start writing a dual in this form and then expand it, and then get the labels. Then,
the corresponding labels that we would got here would exactly be these labels which are
0, 15, 20, etc respectively. There is a nice trick that lies in the fact that, because
this wj is unrestricted, we can simply replace wj by a minus wj and yet retain it.
So, starting now with w7 equal to 0, would actually give us the forward pass of the Dijkstra's
algorithm. The way we actually did it here we did the backward pass of the algorithm;
whereas, in this place, we did the forward path of the algorithm. Nevertheless, we have
seen a framework through which we will be able to show that the labels generated by
the Dijkstra's algorithms were indeed optimal. That is the most important thing to learn
from the shortest path point of view The shortest path problem, even though formulated
this way, because of unimodularity property has a continuous variable. We can write the
dual and then we can get a dual solution by inspection and prove that the corresponding
dual solution is indeed optimal. Provided, we generate the dual solution based on a certain
principle and that principle was the same principle that we followed when we generated
these labels. If each of these labels represent some dual variables and represents some wj,
then, what we did was, we made sure that wi minus wj is less than or equal to Cij, then
we then picked that wj that was minimum. If we are able do that then we are intuitively
generating a feasible solution to the dual of the shortest path problem and by complementary
slackness conditions, it will be optimal. Therefore, the labeling algorithm that we
have seen for the Dijkstra's algorithm is indeed optimal. The next thing that can happen
is we make one important assumption which was told in the earlier lecture. For the optimality
of the Dijkstra's algorithm, all the Cijs have to be greater than or equal to 0. We
get into the next question, can sum of these Cij be negative or can they take negative
values? Let us see what happens if some of these Cijs can take negative values. Let us take a simple network to explain this.
Consider a simple network where the distance here is 20. We call this as 10, we call this
as minus 15, call this as 10, and we call this as 5. Let these be the distances that
we have here. If we start Dijkstra's algorithm for this, we will label this with 0. We have
1 to 2 is 20, 1 to 3 is 10, so we label this as 10. Then we will do this 1 to 2 is 20,
3 to 4 is 10 plus 10 is equal to 20. Let us say we label here as 20, now will do 10 plus
10 is 20, 20 plus 10, 25, so we would label this as 20 and say that the Dijkstra's algorithm
terminates like this. The single pass of the Dijkstra's algorithm would give us a shortest
distance from 1 to 4 as 20; whereas, we know that if we take the path 1 to 2, 2 to 3, 3
to 4 the distance is 20 minus 15 plus 5, which is actually 10 and not 20.
The reason we did not get the correct answer here is because we violated one of the assumptions
that Cij has to be greater than or equal to 0. We modify the shortest path algorithm,
if we have these negatives. Such problems are called shortest path problems with arbitrary
cost. Arbitrary cost would mean either a positive or a 0 or a negative cost. We can modify this
algorithm through this. What we do there is this, instead of these
labels, let me simply call them as w1. One being the source, we call this as w1 is equal
to 0 will have w2 equal to infinity, w3 equal to infinity and w4 is equal to infinity. What
is the dual of the shortest path problem? Dual of the shortest path problem is wi minus
wj is less than or equal to Cij. So, first thing is if we are able to find wi and wj
such that if we write the dual this way by changing this, we get wj minus wi is less
than or equal to Cij. If we replace this unrestricted variable by the negative of that, you will
get wj minus wi is less than or equal to cij If you are able to get an ij such that wj
minus wi is greater than Cij, then we can actually correct the wj. We look at this pair
1 2, wj minus wi is greater than Cij. Please observe that we are looking at this
particular form of the dual. This is the first form of the dual. Here, we got wi minus wj
less than equal to Cij, wj, unrestricted. This form comes because wj is an unrestricted
variable. I am replacing this wj by minus wj which would continue to be unrestricted.
So, wj unrestricted, but you will have minus i plus wj is less than or equal to Cij. We
are using this form of the dual. Here, you have an ij pair such that wj minus wi is greater
than Cij, so update this one. So, pass 1 of the algorithm. We have w2 is updated to 20.
Now again, go back to wi minus wj infinity and 0, wj minus wj is greater than Cij, upgrade
w3 to 10. Here, you have wj minus wi is infinity minus 20 is greater than Cij, upgrade w4 to
25 such that you have wj minus wi is equal to Cij in this case. Now, the values are updated
from the original values till first pass you get 25. Go to the second pass and check whether for
all ij, wj minus wj is less than or equal to Cij. 20 minus 0 is equal to 20, satisfied.
10 minus 0 is equal to 10, satisfied. Come back to this wj minus wi is minus 10, is greater
than Cij, so, update this
to 5, now this one is satisfied. Again in pass 3, verify for all ij, whether wj minus
wi is less than or equal to Cij. wj minus wi is equal to Cij, wj minus wi is less than
or equal to Cij wj is 5 minus 20 is less than or equal to 15, 25 minus 5 is less than or
equal to 20. wj minus wi is greater than Cij, 25 minus 5 is 20, 20 is greater than 10. Now
change w4 to 15, so w4 becomes 15. Once again go back to the algorithm to check whether
for all ij's, wj minus wj is less than or equal to Cij, wj minus wj is equal to Cij,
wj minus wj is less than Cij. wj minus wj is equal to Cij, wj minus wj is less than
Cij, wj minus wj is equal to Cij. So this set of wjs satisfies the condition: wj minus
wj less than or equal to Cij. Therefore, this value is optimal.
The only difference is if we have arbitrary cost, if we have some Cij less than 0. It
is necessary to run or modify the Dijkstra's algorithm, so that the dual condition is satisfied
by the resultant once. The first type of thing which is a Dijkstra's algorithm, they come
under the category of what is called label setting algorithm where you have Cij greater
than or equal to 0. Once the label has set, the label will not change. So these come under
the category of label setting algorithms. The optimality principle is that, once a label
is fixed, which means I have determined the shortest path from the source to that node,
I will not change the label. But when we have arbitrary cost like a negative coming here,
then it may be necessary to correct the label as we move along and we apply this duality
condition. These algorithms come under the category of order called label correcting
algorithms. The
labels that are set here 0, 20, etc., are now corrected. They are not fixed as it was
done in the earlier case. So, they come under the category of order called label correcting
algorithms. This way we can handle shortest path problems, even if some of them have arbitrary
cost. There is another important thing that can happen. Let us look at this same network with the
slight change. Let us say this cost is minus 40. Once again, it is very similar to the
earlier example, except that one Cij is negative and it is minus 40. Most importantly, it is
going to create a big problem for us. Let us make another interesting change here. Instead
of this 5, let me put this 5 and this arc here.
Let us look at a network like this. Remember the changes that I have made. One change is
this is minus 40; here. I have changed the direction and it goes like this. The disadvantage
with a network like this is, if we look at 2 to 3, 3 to 4, 4 to 2, we have a minus 40,
10 and 5. If we look at this cycle, this cycle has a total length of minus 25, minus 40 plus
10 minus 5. So, this has a total length of minus 25. This is a negative length cycle. When we have a negative length cycle and we
are interested in finding the shortest path from 1 to 4, 1 will always say that I can
start with 0. I come back to 20, I complete one cycle. I get 20 minus 25, which is minus
5. I complete another cycle, I get minus 30, I get minus 55 and it goes on. The algorithm
will not terminate here. One of the things is whether you use the basic Dijkstra's
algorithm; we ensure that all Cij's are greater than or equal to 0.
There is no question of negative sign whereas in the modified label correcting algorithm,
we said if we have arbitrary cost, we can still handle it, provided, there are no negative
cycles. If there is a negative cycle in the network, then the label correcting algorithm
will keep on correcting the labels till it starts getting negative values and it gets
smaller and smaller, wj's become much smaller and smaller as we proceed.
The other question that comes is how do I know that there is a negative cycle in a network?
This is a very small network. This has been created only to explain this principle. By
inspection, one can say that this is a negative sign, but if we are dealing with a very large
network, how would somebody know that there is a negative cycle here? The answer is, if
we apply the label correcting algorithm and then we realize that these labels are becoming
smaller and smaller, we fix a particular value like say, if w4 happens to be less than equal
to some minus big M or minus 1000 or some large value. Look at some arbitrarily defined
large value and when one of the dual variables becomes less than such a value, then the algorithm
will terminate saying that there is a negative length cycle.
The label correcting algorithm does two things. One is, if there are arbitrary costs and there
are no negative length cycles, then the algorithm will terminate by giving the optimum value.
If there are negative length cycles, then the algorithm will terminate by saying that
there is a negative length cycle. This is because somewhere, as we move along in this,
we would started with 0, 20 so we would start with 20. After one cycle, we will get minus
5 because the length of this cycle is minus 25. After another cycle, we get minus 30 minus
55 and so on. Somewhere, the value keeps reducing and it will go to less than this 1000 and
therefore the algorithm will terminate at that point saying there is a negative length
cycle. The algorithm will not find the shortest path
from 1 to M. What we have effectively done is, we have now shown, first of all, the primary
dual relationship that is associated with the Dijkstra's algorithm. We formulated
the problem as a linear programming problem. We wrote the dual of that problem. Then, we
derived the optimal solution to this problem using some ideas from duality. We first created
a dual feasible solution here, based on an idea that we kept this as small as possible,
so that we finally got to this 50. Then, we showed that applying the complementary slotness
conditions to this dual feasible solution, we were able to get a primary feasible solution
which was optimal. We also showed that there is relationship between both of them, while
this represented the forward pass that represented backward pass.
We also said the underlying idea is to keep wj minus wj less than or equal to Cij and
then try to keep it as small as possible; small as possible comes by labeling here.
The moment we know this as 15, we labeled it as the smallest possible value of 15. We
labeled this or the smallest possible value of 25 such that wj minus wj is less than or
equal to Cij. We also showed why this is true by redefining the dual this way, so that we
actually end up minimizing this, satisfying this condition. Then, we extended this idea
to look at shortest path problems with arbitrary cost, some places where you have some negative
values. Then, we modified that algorithm based on this duality procedure. We also had showed
a way by which, if we have negative length cycle the algorithm will terminate without
giving the optimal solution. We will continue in a subsequent class on
order called successive shortest path problem.
