Enumerative problem A143(X)3(Y)2 = 5 on Fl(2,3;5)

Experimental data
Number of Real Solutions
  Necklaces     1     3     5  
  Aaaabb     0     0     100000  
  Abbaaa     0     0     100000  
  Abaaab     0     19108     80892  
  Abaaba     0     41023     58977  
  Aabaab     0     41288     58712  
  Aabbaa     1582     30987     67431  
  Aaabba     1589     31071     67340  
  Ababaa     13514     36861     49624  
  Aaabab     13637     36410     49953  
  Aababa     18049     43609     38342  
Key
Condition Name Symbol Codimension
 143   A143   A   3 
 132   X   a   1 
 124   Y   b   1 
Point Selection
Total time of computation: 70 056.24 GHz-seconds or 19.46 GHz-hours on TAMU
1 000 000 Polynomial systems solved
The coefficients of a typical eliminant had 24 digits.
The typical eliminant had size 168 bytes.
This table automatically generated from the data in This File using This Maple Script
Created: Fri Jul 15 15:50:23 CDT 2005