Jump to content

  • Log In with Google      Sign In   
  • Create Account


Finding a path of length n in a graph


Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.

  • You cannot reply to this topic
14 replies to this topic

#1 warnexus   Prime Members   -  Reputation: 1379

Like
0Likes
Like

Posted 30 October 2013 - 09:33 AM

path_zpsb4b2abdd.png

 

K4_zpse47064df.png

 

I am doing part a of Ex 19. The answer has 2 but I can list at least 11 paths.

 

Here are the paths:

1) {a,b,d}

2) {a,d,c}

3) {a,c,d}

4) {b,a,c}

5) {b,d,c}

6) {b,c,d}

7) {c,a,b}

8) {c,d,b}

9) {d,b,a}

10) {d,c,a}

11) {d,b,c}

 

The complete graph k4 graph does not have labeled vertices a,b,c,d.

So I am not sure if I can label a,b,c,d anywhere I want.


Edited by warnexus, 30 October 2013 - 09:34 AM.


Sponsor:

#2 Waterlimon   Crossbones+   -  Reputation: 2349

Like
1Likes
Like

Posted 30 October 2013 - 09:44 AM

Assuming the diagonals count as 1, if you pick any 2 vertices, there will be only 2 paths of length 2 between them. This is what i assume you were supposed to do.


Waterlimon (imagine this is handwritten please)


#3 Oxyd   Members   -  Reputation: 592

Like
0Likes
Like

Posted 30 October 2013 - 09:57 AM

Sure you can label them as you want.

 

I think you are misunderstanding the problem, though. I would interpret it as, “Pick two fixed vertices. Tell me how many paths of length 2 there are between them.” The beauty here is that you should get the same answer no matter which two vertices you pick – this is thanks to your graph being a complete one; the question would be ill-formed if it were for a different kind of graph.



#4 warnexus   Prime Members   -  Reputation: 1379

Like
0Likes
Like

Posted 30 October 2013 - 10:03 AM

Assuming the diagonals count as 1, if you pick any 2 vertices, there will be only 2 paths of length 2 between them. This is what i assume you were supposed to do.

I think I get what you saying. So the 11 paths I listed are not correct.



#5 warnexus   Prime Members   -  Reputation: 1379

Like
0Likes
Like

Posted 30 October 2013 - 10:04 AM

Sure you can label them as you want.

 

I think you are misunderstanding the problem, though. I would interpret it as, “Pick two fixed vertices. Tell me how many paths of length 2 there are between them.” The beauty here is that you should get the same answer no matter which two vertices you pick – this is thanks to your graph being a complete one; the question would be ill-formed if it were for a different kind of graph.

Well there are 2 edges between any two vertices you pick so 2 paths of length. Now that I am doing part b. I do not get the same answer as the textbook so I think I am struggling with what the question is actually saying



#6 Álvaro   Crossbones+   -  Reputation: 11846

Like
0Likes
Like

Posted 30 October 2013 - 10:09 AM

I count

a) 2

b) 7

c) 20

d) 61

 

That's assuming that it's OK to repeat nodes and edges in the paths: I am not sure what precise definition you are using.

 

If you want more terms,

 

A(0) = 0, A(1) = 1, A(n) = 3*A(n-2) + 2*A(n-1) if n>1


Edited by Álvaro, 30 October 2013 - 10:10 AM.


#7 warnexus   Prime Members   -  Reputation: 1379

Like
0Likes
Like

Posted 30 October 2013 - 10:12 AM

I count

a) 2

b) 7

c) 20

d) 61

 

That's assuming that it's OK to repeat nodes and edges in the paths: I am not sure what precise definition you are using.

 

If you want more terms,

 

A(0) = 0, A(1) = 1, A(n) = 3*A(n-2) + 2*A(n-1) if n>1

You did get them all correct. You mention you counting the edges, so you do not need to write them down on a piece of paper?

 

Can you explain how you get 7 for part b?

 

I think if I understand how 7 was acheived I should be able to get part c and d intuitively. part a seems to do a poor job of figuring out how the thought process should be approached.


Edited by warnexus, 30 October 2013 - 10:12 AM.


#8 N.I.B.   Members   -  Reputation: 1027

Like
1Likes
Like

Posted 30 October 2013 - 10:33 AM

A(0) = 0, A(1) = 1, A(n) = 3*A(n-2) + 2*A(n-1) if n>1

To find a path with length n you need to find a path with length n-1 to any of the other nodes. You have 2 options:
1. Find a path to a node other than the start node. So 2 possible nodes, and you need a path of length n-1 to get to them. This is the 2*A(n-1) term.

2. Find a n-1 path to the start node. To do that, you need a n-2 path to any of the other 3 nodes. That's the 3*A(n-2) term.

#9 Álvaro   Crossbones+   -  Reputation: 11846

Like
0Likes
Like

Posted 30 October 2013 - 10:56 AM

Yup, satanir's explanation is correct.



#10 warnexus   Prime Members   -  Reputation: 1379

Like
0Likes
Like

Posted 30 October 2013 - 11:00 AM

 

A(0) = 0, A(1) = 1, A(n) = 3*A(n-2) + 2*A(n-1) if n>1

To find a path with length n you need to find a path with length n-1 to any of the other nodes. You have 2 options:
1. Find a path to a node other than the start node. So 2 possible nodes, and you need a path of length n-1 to get to them. This is the 2*A(n-1) term.

2. Find a n-1 path to the start node. To do that, you need a n-2 path to any of the other 3 nodes. That's the 3*A(n-2) term.

 

Oh I get it now. I was doing it wrong the whole time. Thanks for the explaination.



#11 Álvaro   Crossbones+   -  Reputation: 11846

Like
1Likes
Like

Posted 30 October 2013 - 11:06 AM

Oh, also

 

A(n) = (pow(3,n)-pow(-1,n))/4

 

Magic! :)



#12 N.I.B.   Members   -  Reputation: 1027

Like
0Likes
Like

Posted 30 October 2013 - 11:25 AM

Oh, also

 

A(n) = (pow(3,n)-pow(-1,n))/4

 

Magic! smile.png

I was just driving my car and thinking about the non-recursive solution...



#13 warnexus   Prime Members   -  Reputation: 1379

Like
0Likes
Like

Posted 30 October 2013 - 07:25 PM

Oh, also

 

A(n) = (pow(3,n)-pow(-1,n))/4

 

Magic! smile.png

Wow! Holy cow! That is awesome! Strange how my textbook does not have that formula! How did you come up with that formula? That's ingenious.



#14 Álvaro   Crossbones+   -  Reputation: 11846

Like
0Likes
Like

Posted 30 October 2013 - 07:38 PM

You start at `a' and then you have three choices at each step. Of the possible paths, only 1/4 of them will end at the right spot, so the formula is pow(3,n)/4. Well, powers of 3 are not usually divisible by 4, so you need to do some rounding. </mostly kidding>

Once you have the recursive formula, you can find the explicit formula by trying to find the numbers x such that pow(x,n) satisfies the recursive formula.

pow(x,n) = 3*pow(x,n-2) + 2*pow(x,n-1)

Dividing by pow(x,n-2), you get

x^2 = 3 + 2*x => x^2 - 2*x - 3 = 0

Solving the quadratic equation, you find out that 3 and -1 are the roots. Now all the sequences that satisfy the recursive formula form a vector space of dimension 2 (because the one that starts (1,0,...) and the one that starts (0,1,...) are a basis), and you just found two independent vectors. If you don't know what on Earth I am talking about, don't worry. It means that any sequence that satisfies the recursive formula can be expressed as a*pow(3,n) + b*pow(-1,n), for appropriate values of a and b. Since we know A(0)=0 and A(1)=1,

a*pow(3,0) + b*pow(-1,0) = 0 => a + b = 0
a*pow(3,1) + b*pow(-1,1) = 1 => 3*a - b = 1

Solving that system of linear equations, you get a=1/4, b=-1/4, and that gives us the explicit formula.

This may look like magic if you haven't seen it before, but it's just a procedure you learn.


Edited by Álvaro, 31 October 2013 - 06:12 AM.


#15 Paradigm Shifter   Crossbones+   -  Reputation: 5112

Like
0Likes
Like

Posted 30 October 2013 - 07:46 PM

It's called a recurrence relation http://en.wikipedia.org/wiki/Recurrence_relation

 

They are also called "difference equations". They are the discrete version of differential equations.


"Most people think, great God will come from the sky, take away everything, and make everybody feel high" - Bob Marley




Old topic!
Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.



PARTNERS