The Hare quota is the sum of all the states' populations, divided by 435, the number of House seats.
A state's number of Hare quotas is its population divided by the Hare quota. Cycle 0 contains all the states with fewer than 1 Hare quota. Cycle 1 contains all the states with at least 1 Hare quota, and fewer than 2. Cycle 2 contains all the states with at least 2 Hare quotas and fewer than 3. Cycle n contains all the states with at least n Hare quotas, and fewer than n+1. Any cycle containing no states isn't included in the process described in this message. A cycle's q is the sum of its states' Hare quotas. A cycle's P is the sum of its states' populations. [I use Hare quotas with Webster and Cycle Webster, but one could use P instead, so I mention both] Use Webster to apportion the 435 seats to the cycles exactly as one would, in ordinary Webster, apportion the 435 seats to the states. Each cycle gives its seats to its states so that, in that cycle, a state with more population never has fewer seats than a state with less population. [end of Cycle-Webster definition] Suggested Webster implementation for Cycle-Webster: (If there were no 1-seat-minimum requirement): 1. All cycles initially have 0 seats. 2. Calculate each cycle's score by dividing its q (or P) by the rounding point immediately above its current number of seats. Give a seat to the cycle with highest score. Repeat 2 till 435 seats have been given. (Of course really the only scores needing re-calculation are those of cycles that have received a seat since their score was last calculated). (With the 1-seat-minimum requirement, each cycle initially has a number of seats equal to its number of states that it contains. Instruction #2 is unchanged. Mike _________________________________________________________________ >From photos to predictions, The MSN Entertainment Guide to Golden Globes has it all. http://tv.msn.com/tv/globes2007/?icid=nctagline1 ---- election-methods mailing list - see http://electorama.com/em for list info
