Posted by: Gary Ernest Davis on: November 30, 2010

translation servicesTranslate | French translation Spanish translation translate German translate Chinese

On the rectangular grid shown below we can imagine walking from the red spot to a yellow spot, by taking unit steps up, down, left or right, to adjacent spots:

There are many ways to walk from the red spot to the purple spot shown below, but only 2 of those paths – the ones show in color – are shortest paths:

All other paths from the red spot to the purple spot take more than two steps.

We call a shortest path from the red spot to a yellow spot a *geodesic*.

The *length* of a geodesic is the number of yellow spots it passes through.

The geodesics shown aboveÂ in color have length 2.

I want to think about the idea of a geodesic having offspring.

Suppose a geodesic goes from the red dot to yellow dot 1, then to yellow dot 2, and to yellow dot 3:

This geodesic, which we will name (Greek “gamma”) has length 3: it passes through 3 yellow dots.

It is also a part of 2 geodesics of length 4, namely those geodesics that after reaching yellow spot 3, go to yellow spot A, or to yellow spot B.

These length 4 geodesics we call *daughters* of the geodesic : they are geodesics of length one greater than and they consist of the path plus one extra step.

The geodesic has 2 daughters, and this is typical of most geodesics.

However, some geodesics have 3 daughters. These are the geodesics that travel directly up, or directly down, or directly left, or directly right, from the red spot:

The blue geodesic (of length 2) has 3 daughters (of length 3).

Let’s call a geodesic of type A if it has 2 daughters, and of type B if it has 3 daughters.

Let denote the number of geodesics of type A that have length , and let denote the number of geodesics of type B that have length .

There are 4 geodesics of length 1, all of which are of type B.

So .

A geodesic of type A of length has 2 daughters, both of which are of type A, and of length .

A geodesic of type B of length has 3 daughters, all of length ,Â two of which are of type A and the other of which is of type B.

So, .

This means

The recurrence together with the initial condition allows us to calculate how many geodesics there are of type A of length , for any .

The total number of geodesics of length we denote by , so .

From the recurrence for and the fact that we get the same recurrence for as for , but with a different initial condition:

.

A spreadsheet can calculate the values of for small from the recurrence formula:

n | Gamma(n) | Gamma(n)/Gamma(n-1) |

1 | 4 | |

2 | 12 | 3 |

3 | 28 | 2.333333333 |

4 | 60 | 2.142857143 |

5 | 124 | 2.066666667 |

6 | 252 | 2.032258065 |

7 | 508 | 2.015873016 |

8 | 1020 | 2.007874016 |

9 | 2044 | 2.003921569 |

10 | 4092 | 2.001956947 |

11 | 8188 | 2.000977517 |

12 | 16380 | 2.00048852 |

13 | 32764 | 2.0002442 |

14 | 65532 | 2.000122085 |

15 | 131068 | 2.000061039 |

16 | 262140 | 2.000030519 |

17 | 524284 | 2.000015259 |

18 | 1048572 | 2.000007629 |

19 | 2097148 | 2.000003815 |

20 | 4194300 | 2.000001907 |

21 | 8388604 | 2.000000954 |

22 | 16777212 | 2.000000477 |

23 | 33554428 | 2.000000238 |

24 | 67108860 | 2.000000119 |

25 | 134217724 | 2.00000006 |

Also shown in this table is the ratio .

This ratio seems to approach 2 as increases.

This numerical evidence suggests that the total number of geodesics of length is increasing exponentially, by a multiplication factor of about 2.

To see *why* approaches a limit as increases, as distinct from observing from a table that it seems to, we use the fact that so .

Because increases without bound, we see that for large , as the numerical evidence suggested.

The rectangular grid could be replaced by a triangular grid:

or a hexagonal grid:

Venturing into 3 dimensions, we could consider a cubical grid:

How does the number of geodesics from a given starting spot vary with the length of the geodesic in each of these cases?

1 | Dave Radcliffe

December 1, 2010 at 1:10 am

Square grid: 2^n – 4

Triangular grid: 6*(2^n – 1)

Hexagonal grid: 6*(2^(n/2)-1) if n is even; 9*[2^((n-1)/2)]-6 if n is odd.

Cubical grid: 8*3^n – 12*2^n + 6

The formula for the hexagonal grid can be found here:

http://oeis.org/A061776

The last two sequences are missing from the OEIS. All of the above formulas assume n>0. The formula for the cubical grid can be generalized to higher dimensions.

Gary Davis

December 1, 2010 at 6:04 am

Thanks for the information and links, Dave