in reply to Re: Pattern enumeration.
in thread Pattern enumeration.

Hm.

2 * 48 * 6 * 6^61 = 1.6890743110126207388309943185817e+50

6^64 = 6.3340286662973277706162286946812e+49

Subtracting the former from the latter is -1.0556714443828879617693714491129e+50?

What I think I have:

Take one row; ignoring the vertical component for the moment, at each iteration:

  1. the first cell has a choice of 6;
  2. the second also has a choice of 6;
  3. the third cell has:
    • a choice of 6

      unless the two preceding cells are the same, which happens 6/36 times

    • when it has a choice of 5

    for an average choice of 5.8333

  4. This figure is true for the other five cells in the row.

Giving ( 6 * 6 * 5.833333333^6 ) ^ 8 = 2.220030530726145590197494583735e+47

However, for the 3rd and subsequent rows, the choice at each position is further constrained by the existing values for the previous two cells in the column.

For the first two columns, 30/36 the choice will be 6; 6/36 it will be 5 giving the familiar 5.8333 average. But for the 3rd and subsequent columns, it gets messy.

I thought that it might be :

28.1944 / 5.8333^2 choice is 6; 5.83333 / 5.8333^2 choice is 5; giving an average of: 5.82857142

For each of the 6x6 cells in the bottom right.

For an overall calculation of:

( 6**2 * 5.833333333333333333333333333333**6 )**2 *

( ( 5.833333333333333333333333333333**2 * 5.8285714285714285714285714285714**6 )**6 )

= 1.13460446277538e+049

But that doesn't make sense because it is bigger than the previous result constrained in only one direction :(.

In either case, it is far to big a number to actually iterate.


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