ดูต้นฉบับสำหรับ พูดคุย:Computing the volume of partitions in boolean cubes

ทำไมบางที font math มันก็ตัวเล็ก เดี๋ยวก็ตัวใหญ่ฟระ -- [[ผู้ใช้:Jung|Jung]] 08:25, 28 มีนาคม 2007 (ICT)

==comments from P' manow==
Could P' manow explain in more details for these points?
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* But I just looked at Elekes's paper... it said that a volume of an m-point convex hull of m points in an n-dimensional ball with volume V  is at most V * m / 2^n. 

:That means in our case, the number of points would be very very large.... so that the fact that you can compute the volume, given the points, doesn't really help us.

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:* Maybe a better way to distribute the volume might be to look at the subspace spanned by points in ''S'' closest to ''u'', and compute the volumes of each partition (recursively). 
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== from Ed's mail==
What if we start with N= empty set and add v in S to N one by one. Adding the first vertex gives a partition that occupies the whole cube. The second vertex, when added, will take half volume from the first, resulting in each having half the volume. When we add the third vertex, can we say anything about volume from the coordinates of these three vertices? Can you find a formula (just for k=3 first)? 

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I made straightforward experiments for these stuffs in 2-dim and 3-dim problems...
in all cases, the summation is 0.99 not 1.00 as i ignore an empirical uncertainty ...
(These cases are simple, 100,000 points should be enough)
===2-dim===
There is only one distinguish case:
(0,0) (0,1) (1,1) wil have volumes, respectively, about

.37 .25 .37

===3-dim===
I can find only three distinguish cases:
(0,0,0)    (1,1,1)    (0,0,1)

.35          .39         .24

(0,0,0)    (1,1,0)    (0,1,0)

.37          .37         .25

(0,0,0)    (1,1,0)    (0,1,1)  ['''symmetry!''']

.33          .33         .33


[[ผู้ใช้:Jung|Jung]] 03:14, 31 มีนาคม 2007 (ICT)
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