Enumerative problem A1425A2315(A1326)2 = 2 on Fl(2,4;6)

Experimental data
Number of Real Solutions
  Necklaces     0     2  
  ABCC     0     2000  
  ACBC     0     2000  
Key
Condition Name Symbol Codimension
 1425   A1425   A   3 
 2315   A2315   B   3 
 1326   A1326   C   3 
Point Selection
Total time of computation: 676.65 GHz-seconds or 11.28 GHz-minutes on Schubert
4 000 Polynomial systems solved
The coefficients of a typical eliminant had 19 digits.
The typical eliminant had size 66 bytes.
This table automatically generated from the data in This File using This Maple Script
Created: Fri Jul 15 15:54:43 CDT 2005