L
Lew Pitcher
In said:I do not have TAOCP.
Can you quote the algorithm instead?
D1. [Iterate on k.] Do step D2 for k - m-n, m-n-1, ...,0; then the algorithm
terminates with u[n-1] == r[n-1],...,u[0] == r[0]
D2. [Division loop] Set q[k] to u[n+k] / v[n], and then set u[j] to
u[j] - q[k] * v[j-k] for j == n + k - 1, n + k - 2, ..., k
--
Lew Pitcher
Master Codewright & JOAT-in-training | Registered Linux User #112576
http://pitcher.digitalfreehold.ca/ | GPG public key available by request
---------- Slackware - Because I know what I'm doing. ------