[EM] Divisor method equivalence

[Dan Bishop]

Kristofer Munsterhjelm wrote:
> Hello all,
>
> Could someone show me a proof of how the Sainte-Laguë and Webster's 
> method are equivalent? On the surface of it, each method has an 
> approach that's quite different from the other: Sainte-Laguë divides 
> subsequently by (1, 3, 5, 7, etc), while Webster's finds the correct 
> divisor/multiplier for all in one go.
Yes, but because Webster doesn't suffer from the Alabama Paradox, it can 
also be thought of as assigning seats in order.

For example, given the state populations / party-list votes

A: 3112
B: 1771
C: 910

successive Webster apportionments are:

1 seat: A=1, B=0, C=0 (divisors 3543-6224)
2 seats: A=1, B=1, C=0 (divisors 2075-3542)
3 seats: A=2, B=1, C=0 (divisors 1821-2074)
4 seats: A=2, B=1, C=1 (divisors 1245-1820)
5 seats: A=3, B=1, C=1 (divisors 1181-1244)
6 seats: A=3, B=2, C=1 (divisors 890-1180)
7 seats: A=4, B=2, C=1 (divisors 709-889)
8 seats: A=4, B=3, C=1 (divisors 692-708)
9 seats: A=5, B=3, C=1 (divisors 607-691)
10 seats: A=5, B=3, C=2 (divisors 566-606)
11 seats: A=6, B=3, C=2 (divisors 507-565)
12 seats: A=6, B=4, C=2 (divisors 479-506)
13 seats: A=7, B=4, C=2 (divisors 415-478)
14 seats: A=8, B=4, C=2 (divisors 394-414)
15 seats: A=8, B=5, C=2 (divisors 367-393)
16 seats: A=9, B=5, C=2 (divisors 365-366)
17 seats: A=9, B=5, C=3 (divisors 328-364)
18 seats: A=10, B=5, C=3 (divisors 323-327)
19 seats: A=10, B=6, C=3 (divisors 297-322)
20 seats: A=11, B=6, C=3 (divisors 273-296)
21 seats: A=11, B=7, C=3 (divisors 271-272)
22 seats: A=12, B=7, C=3 (divisors 261-270)
23 seats: A=12, B=7, C=4 (divisors 249-260)
24 seats: A=13, B=7, C=4 (divisors 237-248)

This is equivalent to saying:

Seat #1 goes to A when divisor <= 6224
Seat #2 goes to B when divisor <= 3542
Seat #3 goes to A when divisor <= 2074
Seat #4 goes to C when divisor <= 1820
Seat #5 goes to A when divisor <= 1244
Seat #6 goes to B when divisor <= 1180
Seat #7 goes to A when divisor <= 889
Seat #8 goes to B when divisor <= 708
Seat #9 goes to A when divisor <= 691
Seat #10 goes to C when divisor <= 606
Seat #11 goes to A when divisor <= 565
Seat #12 goes to B when divisor <= 506
Seat #13 goes to A when divisor <= 478
Seat #14 goes to A when divisor <= 414
Seat #15 goes to B when divisor <= 393
Seat #16 goes to A when divisor <= 366
Seat #17 goes to A when divisor <= 364
Seat #18 goes to C when divisor <= 327
Seat #19 goes to B when divisor <= 322
Seat #20 goes to A when divisor <= 296
Seat #21 goes to B when divisor <= 272
Seat #22 goes to A when divisor <= 270
Seat #23 goes to C when divisor <= 260
Seat #24 goes to A when divisor <= 248

or using the priority numbers

A: 6224, 2074, 1244, 889, 691, 565, 478, 414, 366, 364, 296, 270, 248, ...
B: 3542, 1180, 708, 506, 393, 322, 272, ...
C: 1820, 606, 327, 260, ...

Each list is the fractions 2, 2/3, 2/5, 2/7, 2/9, 2/11, 2/13, etc. of 
the population/votes.

To see why, note that Webster requires at least (N+1)/2 quotas to get N 
seats.  That is, if P is population/votes and Q is the quota, then P >= 
Q*(N+1)/2.  But this can be rearranged to Q <= 2P/(N+1).

> In general, what is the relation between the divisors in a 
> highest-average method and the rounding in the equivalent "find 
> divisor so that the rounded numbers sum up to what we want" method? 
The divisors are the minimum numbers of quotas needed to get a seat.  
That is, the rounding cutoffs of the method.  So, for Huntington-Hill, 
the divisors are sqrt(0*1), sqrt(1*2), sqrt(2*3), sqrt(3*4), etc.

And for Webster, they're 0.5, 1.5, 2.5, 3.5, etc.  This is equivalent to 
using 1, 3, 5, 7, etc.: Multiplying by a constant doesn't disturb the 
relative order of priority numbers.