Enumerative problem (A1432)2XYZ = 2 on Fl(2,3,4;5)

Experimental data Related Problems
Number of Real Solutions
  Necklaces     0     2  
  AaAbc     0     1000  
  AacAb     0     1000  
  AabAc     327     673  
  AAacb     354     646  
  AAabc     649     351  
  AAbac     649     351  
Projections
VarietyProblem#
Fl(2,3;5) (A143)2XY 2
Problems fibred over (A1432)2XYZ
VarietyProblem#
Fl(1,2,3,4;5) (A1432)2WXYZ 2
Key
Condition Name Symbol Codimension
 1432   A1432   A   3 
 1324   X   a   1 
 1243   Y   b   1 
 1235   Z   c   1 
Point Selection
Total time of computation: 247.16 GHz-seconds or 4.12 GHz-minutes on Noether
6 000 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 15:49:22 CDT 2005