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?
Take one row; ignoring the vertical component for the moment, at each iteration:
unless the two preceding cells are the same, which happens 6/36 times
for an average choice of 5.8333
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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