Enumerative problem (A1425)3A2315 = 2 on Fl(2,4;6)
Experimental data
Related Problems
Number of Real Solutions
Necklaces
0
2
AAAB
0
2000
Projections
Variety
Problem
#
Fl(2;6)
(W)3W
1
Key
Condition
Name
Symbol
Codimension
1425
A1425
A
3
2315
A2315
B
3
Point Selection
Total time of computation: 1 001.07 GHz-seconds or 16.68 GHz-minutes on Schubert
2 000 Polynomial systems solved
The coefficients of a typical eliminant had 24 digits.
The typical eliminant had size 83 bytes.
This table automatically generated from the data in This File using This Maple ScriptCreated: Fri Jul 15 15:54:55 CDT 2005
)3W