[seqfan] Re: Mobile phone security!

[M. F. Hasler]

Rob,

I'm not sure the rules stated there are correct:
not only neighbours may be connected, you can connect e.g. the "1"
with the "6" on a 3x3 grid (labelled as a numeric keypad)(*), but you
cannot directly connect the 1 with the 3 or the 9: if you try, the 2
resp. 5 is added to the path.
(I did not check whether your calculation does use these rules.)

(*) Here's a screenshot, not sure whether you can access:
https://lh6.googleusercontent.com/-s7VaC2EXKiA/VB7ViGkqC5I/AAAAAAAABzk/KMos3nWrcko/w346-h519/Screenshot_2014-09-21-08-54-34.png

-- 
Maximilian

On Sat, Sep 20, 2014 at 4:17 PM, Rob Arthan <rda at lemma-one.com> wrote:
> I have been playing with a two-dimensional sequence inspired by the
> security patterns that you can use like passwords on Android phones.
> See:
>
> http://math.stackexchange.com/questions/37167/combination-of-smartphones-pattern-password
>
> But I don’t think any of the answers there except (possibly) the one I added this
> afternoon can be right.
>
> Ignoring the minor detail that Android requires the patterns to have at least
> four points, this suggests a two-dimensional sequence a(m, n) defined as follows.
> Let G(m, n) be the graph whose vertices are the integer lattice points (p, q)
> with 0 <= p < m and 0 <= q < n. The graph has an edge between v
> and w iff the line segment [v, w] does not contain any other
> integer lattice points (equivalently, iff v - w = (i, j) with i and j coprime).
> a(m, n) is the number of acyclic eulerian paths in G(m, n).
>
> I implemented a brute force search and got the following values of:
>  a(m, n) for 1 <= m <= 3 and 1 <= n <= 4:
>
> 0, 2, 6, 12
> 2, 60, 1058, 25080
> 6, 1058, 140240, 58673472
>
> I have only been able to verify these results independently
> in the cases when one of m or n is 1 (which is easy because
> the paths are uniquely determined by their end-points in that
> case and in the case m = n = 2 (which is easy because G(2, 2)
> is the complete graph on 4 vertices).  I would be very grateful
> for confirmation of (or corrections to) my results and for any
> thoughts on an efficient way of calculating the sequence.
>
> My OEIS searches haven’t come up with anything like this.
> Do people think this sequence is worth submitting to OEIS?
>
> Regards,
>
> Rob.