On 2013-04-17, at 09:31, DASDBILL2 wrote: > I tried your algorithm with 13 multiplied by 81 and produced the correct > answer. This algorithm is undoubtedly how the microcode for the M (multiply > fullword) instruction does its math. > It has a lot to do with where the 1-bits are in the binary representation of the multiplier, yes.
GIYF. "Wallace tree" PDP-6 et al. inspected two bits of the multiplier at each iteration and mixed adds and subtracts to get a 2s complement product without a restoring step. > ----- Original Message ----- > From: "Gerhard Postpischil" > Sent: Wednesday, April 17, 2013 10:19:22 AM > > Simple - you write the two numbers with the larger on the left. In the > next row, double the number on the left, and halve the number on the > right, discarding any fraction. Upon reaching 1 on the right, cross out > any row where the right number is even. Add the remaining rows on the left. -- gil
