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Topic: ABS: The Geometry of Congress
Replies: 0

 Kathie Barnes Posts: 21 Registered: 12/3/04
ABS: The Geometry of Congress
Posted: Sep 10, 1992 10:24 AM

Title: Geometric View of Some Apportionment Problems
Source: Mathematics Magazine, February 1992 issue
Summarized by: Kathie Barnes

Representatives that have been suggested. Particular attention is given
to Hamilton's method, and the paradoxes it created. Apportionment laws
are the rules by which states are alloted seats in the House of
Representatives. The author, Brent Bradberry first defines a "fair and
monotone" apportionment with 5 axioms. These axioms ensure that when
populations change, or new states are added, the representation of each
state with respect to the other states is "fair," which Bradberry defines
mathematically through these axioms. For a method to be monotone, no
state can lose a Representative when new states are added to the Union
without a change in population (this would create a larger House of
Representatives, though). Thus, a fair and monotone apportionment is not
biased toward any state or group of states.

Bradberry then describes various methods of apportionment. The basic
form is always the same: find the ratio of the population of state i to
the entire population, call this number the quota, q, of state i, and
then divide the number of seats accordingly. The question that always
arises is what to do with the decimals, as no state can have .67 of a
Representative. Thus the question becomes one of how and when to round
off numbers.

Before embarking on an actual proof, Bradberry defines a geometric
representation of all possible apportionments of h house seats to s
states, called a (s-1)-dimensional simplex. Thus, for 1 state, there is
only 1 possibility: state 1 gets all h seats, which corresponds to the
0-simplex, which is a point. With two states, there are more
possibilities, represented by the one simplex that is a line between the
points which represent apportionments of "all or nothing"; the point
where state 1 has all h seats, and state 2 has 0 seats, and the point
where state 1 has 0 seats, and state 2 has all h seats. Three states get
a bit more complicated, but can be represented by a triangle and its
interior. Again, the vertices correspond to the all or nothing
apportionments. Similarly, the apportionments for four states are
represented by a solid tetrahedron.

While one cannot easily visualize the simplex for 5 states (after all,
it is 4-dimensional), it is an easy generalization from the tetrahedron.
For instance, this simplex would have 5 vertices instead of 4. Keep in
mind that while all possible apportionments of h house seats to s states
are represented in the (s-1)-simplex, not all points in the simplex
represent valid apportionments. In fact, most of them don't. Since states
can't have .67 of a Representative, only the discrete points that have
integer apportionments are allowed.

The need for integer apportionments is why this problem is not a simple
matter of arithmetic. Bradberry proves that it is impossible to have a
fair and monotone method for apportionment. Using geometric simplices,
Bradberry illustrates where each of the five axioms would fail (if they
do) by shading these areas on the simplices. It turns out that the only
types of methods that are monotone are so called divisors methods;
methods where a common divisor is found such that the integer portion of
the quota divided by this common divisor is used as the number of
Representatives for each state. Thus, the integer portions for each state
must add to the total number of seats.

For instance, Hamilton's method is not a divisor method. His method can
be described as follows: Take the quota for each state, and multiply it
by the number of seats h, and take the integer component of this
multiplication. Then parcel out the remaining seats in order of largest
fraction. Obviously, if there are remaining seats, the integer component
does not add to the total number of seats h. A divisor method would be
described as follows: take the quota of each state, and find a common
divisor, e, such that, if all the q/e are rounded up (or down, using a
different e), (summation of q/e) = h (when the summation is taken over
all states). Remember that q depends on i.

However, the need for a divisor method creates a problem, as Bradberry
also proves that divisor methods cannot be fair. Thus, since divisor
methods are the only possible monotone methods, it is impossible to
satisfy all 5 axioms for a fair and monotone apportionment of seats.
Right now, the US uses some combination of the two, in order to create as
just a system as possible.

See Brent Bradberry's "Geometric View of Some Apportionment Problems" in
Mathematics Magazine, February 1992 for further information.