Using the following dataset of year to max drive size in bytes per year [1a][1b]:
{1956,5.00E+06},{1979,5.71E+08},{1980,1.00E+09},{1991,1.00E+09},{1992,2.10E+09},{1997,1.68E+10},{2003,3.70E+10},{2005,4.00E+10},{2006,1.60E+11},{2006,7.50E+11},{2016,5.00E+12}
Linear regression provides the equation:
year = 4.222874399 ln(numBytes) + 1896.534826
Assuming the average bits required to represent a board state is 160 [2], then:
2^155 positions x 160 bits = 9.13e47 bytes
So, a projection shows that by year 2362, we could have a single storage drive holding every permutation of a board state, so it would be less than 2 drives to store the states along with the move to make for that board. Add more drives and you pull that date in.
Linear regression provides the equation: year = 4.222874399 ln(numBytes) + 1896.534826
Assuming the average bits required to represent a board state is 160 [2], then: 2^155 positions x 160 bits = 9.13e47 bytes
4.222874399 ln(9.13e47) + 1896.534826 = 2362.87979
So, a projection shows that by year 2362, we could have a single storage drive holding every permutation of a board state, so it would be less than 2 drives to store the states along with the move to make for that board. Add more drives and you pull that date in.
References:
1a. http://www.pcworld.com/article/127105/article.html 1b. http://www.computerworld.com/article/2473980/data-storage-so... 2. http://codegolf.stackexchange.com/questions/19397/smallest-c...