Patrick added the fact Alain Hertz had alerted me to, namely that the independence number of skew-starfree (-free) graphs can be computed efficiently. He then added a test for skew-start free graphs. The last difficult graph I?bbrr[ko was indeed skew-star free, the independence numbers of the remaining graphs of order 10 can be computed efficiently, and now Patrick is running graphs of order 11. So I?bbrr[ko is solved. On to n=11!