Indcidence Matrices: Answer to Floor Paths Question is 14, Not 31?

  • Thread starter Thread starter John112
  • Start date Start date
  • Tags Tags
    Matrices
John112
Messages
19
Reaction score
0
Should the right answer to this question(below) be 14 and not 31? because
A_{ij}^{k} means number of paths from i to j of length K. So A_{12}^{8} = 14

We then represent the graph as indcidence matrices and go from there on:

A = { {0,1,0,0}, {1,0,1,0}, {1,1,0,1}, {1,0,0,0} }

A^{8} = { {22,14,13,4}, {31,35,14,13}, {40,31,22,10}, {10,13,4,5} }<br /> <br /> <b>QUESTION:</b><br /> At each step the elevator is able to travel directly from floor to floor as listed below. Suppose we go floor to floor eight times (e.g 1 to 2 then 2 to 3 would be two times). How many different ways can we start at floor 1 and end at floor 2? <br /> <br /> floor 1 to floor 2<br /> floor 2 to floor 3<br /> floor 3 to floor 4<br /> floor 4 to floor 1<br /> floor 2 to floor 1<br /> floor 3 to floor 1<br /> floor 3 to floor 2<br /> <br /> <br /> Correct Answer in the back of the book: 31
 
Last edited:
Physics news on Phys.org
John112 said:
Should the right answer to this question(below) be 14 and not 31? because
A_{ij}^{k} means number of paths from i to j of length K. So A_{12}^{8} = 14

We then represent the graph as indcidence matrices and go from there on:



A^{8} =

QUESTION:
At each step the elevator is able to travel directly from floor to floor as listed below. Suppose we go floor to floor eight times (e.g 1 to 2 then 2 to 3 would be two times). How many different ways can we start at floor 1 and end at floor 2?

floor 1 to floor 2
floor 2 to floor 3
floor 3 to floor 4
floor 4 to floor 1
floor 2 to floor 1
floor 3 to floor 1
floor 3 to floor 2


Correct Answer in the back of the book: 31

I would not call the book's answer "correct"; I get exactly the same ##A^8## answer as you.
 
Last edited by a moderator:
Ray Vickson said:
I would not call the book's answer "correct"; I get exactly the same ##A^8## answer as you.

I guess they mixed it up then, I guess 31 is the answer from floor 2 to floor 1 then. Thanks for the reply.
 
I agree. Quibble: the matrix is the adjacency matrix, not the incidence matrix, right?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top