Enumerative problem (A1425)2(A2315)2 = 2 on Fl(2,4;6)
Experimental data
Related Problems
Number of Real Solutions
Necklaces
0
2
ABAB
0
10003
AABB
790
9213
Projections
Variety
Problem
#
Fl(2;6)
(W)2(W)2
1
Key
Condition
Name
Symbol
Codimension
1425
A1425
A
3
2315
A2315
B
3
Point Selection
Total time of computation: 5 106.98 GHz-seconds or 1.42 GHz-hours on Schubert
20 006 Polynomial systems solved
The coefficients of a typical eliminant had 25 digits.
The typical eliminant had size 84 bytes.
This table automatically generated from the data in This File using This Maple ScriptCreated: Fri Jul 15 15:54:48 CDT 2005
)2(W
)2