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

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Homework Help Overview

The discussion revolves around a problem involving paths in a graph represented by matrices, specifically focusing on the number of ways to travel between floors using an elevator system over a series of steps. The subject area includes graph theory and matrix operations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are analyzing the interpretation of matrix powers in relation to counting paths between nodes (floors) in a graph. There is a debate about the correct number of paths from floor 1 to floor 2 after eight steps, with references to matrix calculations yielding different results.

Discussion Status

Participants are actively questioning the correctness of the answer provided in the textbook and are sharing their own calculations that suggest an alternative result. There is acknowledgment of potential confusion regarding the terminology used for the matrices involved.

Contextual Notes

There is a noted discrepancy between the calculated results and the answer given in the textbook, leading to discussions about the definitions and types of matrices being used (adjacency vs. incidence matrices). Participants are also considering the implications of the paths counted in relation to the problem's constraints.

John112
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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:
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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?
 

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