3 Jan
2006
3 Jan
'06
1 p.m.
Happy new year! I found out that if we do (notation simplified to be more readable but less compilable): (a^n*b^m*b^p).subs($0^$1*$0^$2==$0^($1+$2),subs_options::algebraic) nothing happens because a^n already matches a factor in the pattern and in that case no other matches are attempted. I fixed this. Also, I allowed algebraic substitution to match multiple times on different factors with the same pattern. E.g. the above .subs() can be used to simplify a^n*a^m*b^p*b^q -> a^(n+m)*b^(p+q) These changes are in CVS. Bye, Chris