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Subject: Re: Number of six men tablebases.

Author: blass uri

Date: 12:29:44 08/09/99

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On August 09, 1999 at 13:46:37, José de Jesús García Ruvalcaba wrote:

>	I finally sat down and make all the computations from scratch and then
>rechecked the results.
>Number of 5+1 tablebases: 40
>Number of 4+2 tablebases: 200
>Number of 3+3 tablebases: 120
>Total number of six men tablebases: 360
>	If there is interest I can explain how I got the results, they are simple
>combinatorics problems.
>José.
Your calculation is wrong.
For example the first number is 70

if I represent  P=1,N=2,B=3,R=4,Q=5 then the I can represent every 5+1 material
configuration(4+0 without kings) as 4 numbers and the number of possible
configuration is 70

I can represent pppp configuration by 1111
pppn by 1112 pppb by 1113

You can count non decreasing sequence of 4 numbers(out of numbers 1,2,3,4,5)
1111 1112 1113 1114 1115
1222 1223 1224 1225 1233
1234 1235 1244 1245 1255
1333 1334 1335 1344 1345
1355 1444 1445 1455 1555
2222 2223 2224 2225 2233
2234 2235 2244 2245 2255
2333 2334 2335 2344 2345
2355 2444 2445 2455 2555

are only 45 options.

You can translate every non decreasing sequence by adding 1 to the second
number, adding 2 to the 3th number and adding 3 to the 4th number.

The problem is how many increasing sequences you can choose from the numbers 1-8
 and the answer is 8!/(4!*4!)=70


Uri



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