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Subject: Re: Number of six men tablebases.

Author: José de Jesús García Ruvalcaba

Date: 14:02:21 08/09/99

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On August 09, 1999 at 15:29:44, blass uri wrote:

>
>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
>

	You are right. I forgot to add the tables with two equal pieces and the other
two different (there are exactly 30 of them).
	There can be more mistakes, of course. Now that I use your techniques to count
the 4+2 tables, I get 175 (it is five times the number of 4+1 tables), I wonder
where I was wrong. My new numbers are:
5+1: 70
4+2: 175
3+3: 120
Total: 365.

>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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