Voting and Power
Some Mathematics of Political Science
Apportionment: Measuring Unfairness
Introduction
Every ten years when the census is taken the federal government rearranges or
reapportions the delegates in the House of Representatives. This reapportionment is
designed to give equal representation by assuring that all districts with the same population
get equal numbers of representatives. After the 2000 census, the state of Utah sued the
Federal Government arguing that it should be given the delegate that went to the state of
North Carolina. What mathematics is behind the apportionment of the House of
Representatives and did Utah have a case?
Before we begin, let's consider developing our own methods of apportionment.
Suppose there is a township with 5 districts and a total population of 20,000 as shown in
Table 1.
A
B
C
D
E
District
Population 8150 5322 3188 2353 987
Table 1: Population of each district
If the governing body has 20 delegates, how should these delegates be apportioned to the
districts? Give this problem to your students and have them apportion the 20 delegates as
“fairly” as possible.
_________________________________________________________________________
The Hamilton Solution
A common student solution develops as follows: With 20 delegates representing
20,000 people, there should be 1 delegate for each 1000 citizens. Since District A has
8150 citizens, it should be represented by 8.15 delegates. The number 8.15 is called the
quota for district A. We will denote the quota for District A as QA = 8.15 . Similar quotas
can be found for the other districts. If District A is given 9 delegates (or any number of
delegates greater than 8.15 delegates), it is unfair, since District A would be over
represented. If, on the other hand, District A received only 8 delegates, this apportionment
is also unfair, since District A will be underrepresented. The different districts should get
at least the integer part of their quota (shown in the 3rd row of Table 2), that is district A
should have at least 8 delegates, district B at least 5 delegates, district C at least 3
delegates, and district D at least 2 delegates. Additionally, since every district must be
given representation, district E must get at least 1 delegate. However, the total of the
integer parts is only 19. The essential question is which district should receive this last
delegate?
The Hamilton Method gives the last delegate to the district with the largest fraction
of a delegate in its quota. In this case, district D which has 0.353 of a seat as the fractional
part of its quota. So this apportionment plan would apportion 8 seats to A, 5 seats to B, 3
seats to C, 3 seats to D, and 1 seat to E. District B would be somewhat upset, most likely,
since it missed getting the extra delegate by 32 people. This method of apportioning seats
was first proposed by Alexander Hamilton and was used to apportion delegates to the
1
House of Representatives of the U. S. Congress from 1850 to 1900. However, there is a
problem with apportioning seats this way.
District B missed getting the extra delegate by 32 people. If we had 21 delegates,
then District B might be appeased. Consider the same districts with the same populations,
but with 21 seats to apportion. With 20,000 people and 21 delegates, then there should be
1 delegate for every 952 people. To determine the quota for each district divide the
population by 952.
District
Population
Quota (20 Delegates)
Quota (21 Delegates)
A
B
C
D
E
8150
8.15
8.56
5322
5.322
5.59
3188
3.188
3.29
2353
2.353
2.47
987
0.987
1.03
Table 2: Population and Quota for each district
If every district receives at least the integer part of their quota, then District A
should have at least 8 seats, District B at least 5 seats, District C at least 3 seats, District at
least 2 seats, and District E at least 1 seat. Notice that the total is again only 19 seats. If
we apportion the remaining two seats to the districts with the largest fractional parts, then
Districts B and A get the two unassigned delegates since their fractional parts are 0.59 and
0.56, respectively. So the apportionment plan devised by Hamilton would give District A
9 delegates, District B 6 delegates, District C 3 delegates, District D 2 delegates and
District E 1 delegate. District B is satisfied, but what happened to district D? When there
were 20 delegates, District D was apportioned 3 representatives, but with 21 delegates, it
only gets 2. When a seat is added, we expect that some districts will have an increase in
their representation, but we do not expect to find that increasing the number of delegates
would causes a district to lose representation!
The Hamiltonian Method seems a perfectly reasonable way to apportion delegates.
Unfortunately, it leads to unacceptable results. A district may lose representation simply by
having the total delegation increase in size. This is called the Alabama Paradox, since
Alabama would have lost representation in 1880 if the size of the House of Representatives
was increased to 300, the original goal. Instead they increased the total number of
Representatives to 325 to avoid the surprising effect.
The Method of Differences: Apportioning on the Basis of an “Unfairness Index”
One definition of a fair apportionment procedure is one in which the same number
of people in each state are represented by 1 delegate. If 1 delegate is given for every 500
people in one state then 1 delegate should be given for every 500 people in all other states.
Any deviation from this would be unfair to some state.
Consider two states, X and Y, with 100,000 and 60,000 people, respectively.
Suppose state X has 5 representatives while state Y has 4. Is this fair apportionment?
State X has 1 representative for every 20,000 citizens while State Y has 1 representative
for every 15,000 people. This is unfair to state X by 5,000 people/delegate.
If state X has a population PX and is represented by RX delegates while state Y has a
population PY and is represented by RY delegates, then the unfairness of this
P
P
apportionment can be defined by the difference X − Y .
R X RY
2
If
PX PY
−
> 0 , the apportionment is unfair to X, and we call the value of
R X RY
PX PY
−
the unfairness index to X.
R X RY
P
P
If X − Y measures the unfairness to X if state X has R X delegates and state Y
R X RY
P
P
has RY delegates, then X − Y measures how unfair it would be to X if Y got one
R X RY + 1
P
PX
additional delegate. Likewise, Y −
measures how unfair it would be to Y if state
RY R X + 1
X got an additional delegate.
These two expressions are mathematical models of
unfairness. We can use them to develop an apportionment plan that will minimize the
unfairness.
The conditions for which X will get a delegate rather than Y is that the measure of
unfairness to Y when X gets the delegate is smaller than the measure of unfairness to X
when Y gets the delegate. Symbolically, that means that X gets the delegate if
PX
P
P
PX
− Y > Y −
.
R X RY + 1 RY R X + 1
If we rewrite the inequality in our statement by moving all terms with X to the left
of the inequality and all terms with Y to the right side, then we can say to “Give the
P
PX
P
P
delegate to X if X +
> Y + Y .”
R X R X + 1 RY RY + 1
This inequality can be further simplified to
P ( 2 RX + 1) PY ( 2 RY + 1)
“Give the delegate to X if X
.”
>
RX ( RX + 1)
RY ( RY + 1)
To determine which state gets the extra delegate, simply evaluate the expressions
on either side of the inequality above and give the delegate to state X with the larger value.
Moreover, the transitive property of inequalities allows us to include other states. If we
have states A, B, C, D, and E, we compare the values of the expressions
PA ( 2 RA + 1) PB ( 2 RB + 1) PC ( 2 RC + 1) PD ( 2 RD + 1)
P ( 2 RE + 1)
,
,
,
,
, and E
RA ( RA + 1) RB ( RB + 1) RC ( RC + 1) RD ( RD + 1)
RE ( RE + 1)
and give the delegate in question to the district with the largest value.
P ( 2 R + 1)
If we call the expression
the U-value (for unfairness), we should find
R ( R + 1)
the U-values for each district and give the extra seat to the state with the largest U-value.
If we want to apportion 20 delegates to 5 districts, we give each district one delegate and
P ( 2 R + 1)
as R varies from 1 to 10 (no district will receive more
compute the values of
R ( R + 1)
than 10 delegates) for each of the five populations. We then assign additional delegates to
3
the 15 largest U-values. The table below gives each of these values for the five districts.
The values in the table have been rounded to the nearest integer to facilitate the reading.
Notice that the largest number is 12225, so the first new delegate goes to District A.
The second largest is 7983, so the next delegate goes to District B. In Table 2 we indicate
the first 16 assignments in the order in which they were assigned. Again, compare these
tables with the work previously done.
R
1
2
3
4
5
6
7
8
9
10
UA
UB
UC
UD
12225 (1)
7983 (2)
4782 (4)
3530 (8)
6792 (3)
4435 (6)
2657 (11) 1961 (15)
4754 (5)
3105 (9)
1860
1373
3668 (7)
2395 (13)
1435
1059
2988 (10)
1951
1169
863
2523 (12)
1647
987
728
2183 (14)
1426
854
630
1924
1257
753
556
1721
1124
673
497
1556
1016
609
449
Table 3: U-values for R = 1 to 10 with assignments
UE
1481
823
576
444
362
306
264
233
208
188
The final apportionment of 20 delegates by this scheme is district A with 8
delegates, district B with 5 delegates, district C with 3 delegates, district D with 3
delegates, and district E with one. This is the same distribution as the Hamilton Method.
Also notice that if the number of delegates were increased or decreased, we would quickly
be able to apportion the delegates. The apportionment of 21 delegates would have given
district B six delegates. Also notice that District B would receive 7 delegates before E
gets its second, but E would get its second delegate before District D receives its third
delegate.
Method of Equal Proportions
Our first method of apportionment was based on the principle that if two numbers
P
P
P
P
are equal, their difference is zero. If X = Y , then X − Y = 0 and the apportionment
RX RY
RX RY
P
P
between X and Y is fair. The farther the difference X − Y is from zero, the less fair the
RX RY
apportionment. Another way to compare values is to consider their ratio. If two numbers
⎛P ⎞ ⎛P ⎞
P
P
are equal, their ratio is one. If X = Y , then ⎜ X ⎟ ⎜ Y ⎟ = 1 . The larger the ratio
RX RY
⎝ RX ⎠ ⎝ RY ⎠
⎛ PX ⎞ ⎛ PY ⎞
⎜
⎟ ⎜ ⎟ > 1 , the less fair the apportionment is to X. Lets see how this measure of
⎝ RX ⎠ ⎝ RY ⎠
⎛P ⎞ ⎛P ⎞
fairness compares to the previous measure. Using the ratio ⎜ X ⎟ ⎜ Y ⎟ as our measure
⎝ RX ⎠ ⎝ RY ⎠
4
of unfairness to X, we repeat the argument from before. Give the disputed delegate to Y
and compute the unfairness to X, then give the delegate to X and compute how unfair it is
to Y. Give the measure to the district that will produce the smallest computed unfairness.
⎛ PY ⎞
⎛ PX ⎞
⎜ ⎟
⎜
⎟
RY ⎠
RX ⎠
⎝
⎝
<
.
In this case, X will get the extra delegate when
⎛ PX ⎞ ⎛ PY ⎞
⎜
⎟ ⎜
⎟
⎝ RX + 1 ⎠ ⎝ RY + 1 ⎠
Simplifying the inequality, we have
2
2
PX )
PY )
(
(
“Give the extra delegate to X if
.”
>
RX ( RX + 1) RY ( RY + 1)
So District X is assigned the disputed delegate if
PX
PY
>
RX ( RX + 1)
RY ( RY + 1)
Notice that this is totally different from the previous measure and will apportion the
delegates differently as well.
If we compute the unfairness values (denoted U ′ ) for each district with the rule
P
,
U′ =
R ( R + 1)
we generate Table 3.
U A′
U B′
U C′
U D′
U E′
1
5763 (1)
3763 (2)
2254 (5)
1664 (8) 698
2
3327 (3)
2173 (6)
1301 (11)
961
403
3
2353 (4)
1536 (9)
920
679
285
4
1822 (7)
1190 (13)
713
526
221
5
1488 (10)
972 (15)
582
430
180
6
1258 (12)
821
492
363
152
7
1089 (14)
711
426
314
132
8
960
627
376
277
116
9
859
561
336
248
104
10
777
507
304
224
94
Table 4: U ′ - values using ratio method for R = 1 to 10
R
If the total representation is 20, then the delegates will be split so that district A has
8, district B has 6, district C has 3, district D has 2, and district E has 1. This is not the
same representation as the Method of Differences produced. If 21 delegates are used, then
the two apportionments agree.
This Equal Proportions method of apportionment is known as the Huntington-Hill
Method, and is what is presently used to apportion the House of Representatives. The
House now has a fixed number of seats, 435, that are apportioned to the states based on
their population. In the last census, Utah had a population of 2,236,714 and North
5
Carolina a population of 8,067,673. Computing the unfairness indices for each using
2,236,714
8,067,673
′ ( R) =
and U NC
we create the table below.
UU′ ( R ) =
R ( R + 1)
R ( R + 1)
R
Utah
NC
1
1581596
5704706
2
913135
3293613
3
2328937
645684
4
500144
1803987
5
408366
1472949
#
#
#
11
194681
702200
12
179080
645931
13
165796
598016
Table 6: U ′ - values for Utah and North Carolina
North Carolina’s Unfairness Index of 645931 for R = 12 gave North Carolina the
435th seat in the house. It edged out Utah’s Index of 645684 for the last seat. If Utah’s
2,237,571
population had been 857 larger, then Utah’s score of UU′ ( 3) =
= 645931.1
3 ( 3 + 1)
′ (12 ) =
would have been larger than North Carolina’s score of U NC
8,067,673
12 (12 + 1)
= 645930.8 ,
and Utah would have received a 4th delegate instead of North Carolina receiving its 13th.
Utah filed suit claiming that it had many more than 857 citizens of Utah overseas on
Mormon missions and were uncounted in the census. The suit was resolved in North
Carolina’s favor.
Methods of apportionment offer both an interesting mathematical study and a rich
history for students to investigate. The very first presidential veto was Washington’s veto
of the Hamiltonian apportionment in favor of one proposed by Thomas Jefferson. Daniel
Webster proposed a Method of Differences using the measure of delegates per person
RX RY
−
rather than people per delegate.
This leads to an entirely different
PX PY
apportionment. The references below present some of the historical and mathematical
aspects of this intriguing subject.
References:
1.
Balinski, Michael and Peyton Young, Fair Representation⎯Meeting the Ideal of
One Man, One Vote, Yale University Press, 1982.
2.
Burghes, D. N., I. Huntley, and J. McDonald, Applying Mathematics, John Wiley &
Sons, 1982.
3.
Young, H. Peyton (editor), “Fair Allocation”, Proceedings of Symposia in Applied
Mathematics, Volume 33, American Mathematical Society, Providence, Rhode Island,
1985.
6
Voting Power
In ordinary elections, where each person has a single vote, everyone's vote has the
same power. However, in some situations, some voters have more votes than others. The
electoral college, where North Carolina now has 15 votes and Alaska only 3 votes and
shareholders in a company whose votes are equal to the number of shares of stock they
own are two classic examples. On some boards of county commissioners and school
boards that are a union of several districts, the commissioners have votes that are
proportional to the number of people in their district. Does having more votes mean that
these commissioners have more power? To answer this question, we need to create a
mathematical model to measure the power of a single voter with multiple votes.
A few simple examples illustrate the problem.
Example 1: A company has 1,000 shares all owned by one of three people. Person A
owns 499 shares, person B owns 498 shares, and person C owns the remaining 3 shares. A
proposal requires a majority vote (501 shares) for approval. What is the distribution of
power among the 3 shareholders?
After only a little thought, students will realize that all three share power equally.
To have a majority vote, at least two of the shareholders must support the proposal. It
doesn't matter which two. In this situation, even though person A has nearly 150 times the
votes of C, they have the same power. We say that each of the members has 1/3 of the
power. More votes does not necessarily mean more power.
Example 2: A commission is formed with 4 members, A, B, C, and D. Person A has 5
votes, person B has 4 votes, person C has 3 votes, and person D has only 1 vote.
According to the rules of the commission, at least 7 votes are required to pass a particular
proposal. We can describe this situation using the notation 7; 5, 4, 3,1 . How is the power
distributed among the members?
Again, with a little thought, students will realize that commissioners A, B, and C
1
each have of the power while person D has no power. To achieve 7 votes, at least two
3
of A, B, and C must vote for the proposal. Which two of A, B, and C vote for the proposal
does not matter. Person D can never effect the vote. If she votes for a proposal, it will
pass only if it would also pass had she voted against it. Her one vote never is important.
Commissioner D is known as a dummy. Having a vote is not the same as having power!
Example 3: Suppose the commission in Example 2 required 8 votes for passage of a
proposal. In the notation of power indices, this is an 8; 5, 4, 3,1 game. How would the
power be distributed?
In this situation, we see that D does have power, for if A is against the proposal,
and both B and C are for the proposal, then D's vote is necessary for the proposal to pass.
With this example, students usually reason that there is no dummy, that A has the most
power, B and C have the same power with their votes, even though B has twice as many as
C, and that D has the least power. Just exactly how to specify the power is the object of
the assignment that I give.
7
Class Assignment: Given the 8; 5, 4, 3,1 situation described above, find the proportion of
the power for each player. Think about your reasoning in the first two examples. Try to
quantify the principles that allowed you to see that all three shareholders in Example 1
shared power equally, that D was a dummy in Example 2 but not in Example 3, and that B
and C should have equal power, but less power than A in Example 3.
This assignment always leads to some interesting models. Students quickly come
up with reasoned and reasonable methods for measuring the power of each individual.
Solution 1: A common, but incorrect, solution is to list all of the voting coalitions that
will lead to approval of the proposal. In the 8; 5, 4, 3,1 situation, this is
ABCD, ABC, ABD, ACD, BCD, AB, and AC.
Since there are 7 possibilities each of the groupings carries 1/7 power. In each group, the
members share power evening. In the first example, ABCD, there are four members, so
each gets 1/4 of the 1/7 power.
Coalitions
ABCD
ABC
ABD
ACD
BCD
AB
AC
1 1 1
⋅ =
7 3 21
1 1 1
⋅ =
7 3 21
1 1 1
⋅ =
7 3 21
1 1 1
⋅ =
7 3 21
1 1 1
⋅ =
7 3 21
1 1 1
⋅ =
7 2 14
1 1 1
⋅ =
7 2 14
Now, count how much total power each member has.
1
1
1
9
A:
1
+ 3
+ 2
=
≈ 0.3214
28
21
14 28
1
1
1
7
B:
1
+ 3
+ 1
=
≈ 0.25
28
21
14 28
1
1
1
7
C:
1
+ 3
+ 1
=
≈ 0.25
28
21
14 28
1
1
1
5
1
+ 3
+ 0
=
≈ 01786
.
D:
28
21
14 28
While this gives a solution that matches the original student intuition, it doesn't work in the
case of example 2, where D was a dummy. 7; 5, 4, 3,1
bg
bg
bg
bg
Combinations
bg
bg
bg
bg
bg
bg
bg
bg
ABCD
ABC
ABD
ACD
BCD
AB
AC
BC
1 1 1
⋅ =
7 4 28
1 1 1
⋅ =
7 3 21
1 1 1
⋅ =
7 3 21
1 1 1
⋅ =
7 3 21
1 1 1
⋅ =
7 3 21
1 1 1
⋅ =
7 2 14
1 1 1
⋅ =
7 2 14
1 1 1
⋅ =
7 2 14
Students generally understand that D is getting a free ride on the other's majority.
Solution 2: They quickly decide that they should only consider the minimal winning
coalitions. These are the coalitions in which if any member of the coalitions is removed,
the proposal would then fail. There are only three such coalitions, two with two voters and
one with three. Counting as before, we have coalitions AB and AC with a score of
1 1 1
1 1 1
⋅ = and coalition BCD with a score of
⋅ = . Since A was in two winning
3 2 6
3 3 9
8
coalitions, each of which has a power rating of
1
, its power index is given by the sum
6
1 1 1
+ = .
6 6 3
Coalitions
AB
1
3
⋅ 12 =
AC
1
6
1
3
⋅ 12 =
1
6
1
3
BCD
⋅ 13 = 19
1 1 1 6
+ = =
6 6 3 18
1 1 5
B: + =
6 9 18
1 1 5
C: + =
6 9 18
1 2
D: =
9 18
A:
Using this method, A has three times the power of D, while B and C have somewhat more
than twice the power of D.
Solution 3: Another common solution uses Solution 2 as a basis. In that solution case,
each combinations of votes was considered equally likely. However, since AB and AC
only require two members to support the issue, either is more likely to happen than BCD,
which require 3 members. If the probability of getting a individual to vote for passage is p,
then the probability that 2 vote for passage is p 2 and for three is p3 , so for the three
possibilities, we have p 2 + p 2 + p3 = 1. Solving for p, we find that p = 0.618 . The rest of
the solutions follows from before.
Combinations
AB
p = 0.382
2
A:
B:
C:
D:
AC
p = 0.382
BCD
p 3 = 0.236
2
(.382 ) + 12 (.382 ) = 0.382
1
1
2 ( .382 ) + 3 ( .236 ) = 0.270
1
1
2 ( .382 ) + 3 ( .236 ) = 0.270
1
3 ( .236 ) = 0.079
1
2
Solution 4:
Each player is considered according to what must happen for them to be a part of a
winning coalition. For A, we have A and [B or C]. For B we have B and [A or (C and D)].
For C we have C and [A or (B and D)], and for D we have D and [B and C]. Student's then
use the rules of probability of independent events to compare the situations. In this case,
9
and implies multiplication and or implies additions. If the probability of a yes vote is p,
then the situation
A and [B or C] is represented by p p + p .
Similarly, we have both
B and [A or (C and D)] and C and [A or (B and D)] represented by p p + p 2 .
b
c h
g
c
h
Finally, D and [B and C] is represented by p p 2 . Students typically set the sum equal to
c
h
one to find a value of p, then normalize the indices. 2 p 2 + 2 p 2 + p 3 + p 3 = 1 . Solving for
p we find that
Solution 5: Considering Combinations
Another typical solution considers all possible combinations of players. A player
has power in a coalition if the coalition changes from winning to losing when the player is
removed from the coalition. Thus we consider the coalitions:
First, remove from consideration any coalition that is not winning.
Now, consider each player in turn. If they are removed from the coalition, and the
coalition now fails, they have power. The players with power in each coalition have been
circled in the figure below.
Now, count how many time each player has power. Player A has 5 circles, players B and
C both have 3 circles, and player D has one circle. Normalizing, we find that the power of
5
3
3
1
the players is A: , B: , C: , and D: . This method is known as the Banzaf Index
12
12
12
12
and has a rich history in the law. (See references)
Solution 6: Considering Permutations
There are two classics solutions to the measurement of power. The Banzaf Index
above is one that students will often come up with on their own. The second is known as
the Shapley-Shubik index, and is based on permutation rather than combinations. No
student has ever thought of this one on their own.
Consider the vote being taken with each of the four players voting for the proposal. At
some point in the process, enough votes are collected and the measure will pass. If we
consider all possible voting orders, and count how often each member casts the vote that
puts the proposal over the top, the player who is most often in this position will be the one
with the most power. In this situation, we consider the 24 possible orderings.
10
If the votes are cast in each of these orders, when will the necessary 8 votes be obtained?
Of the 24 possibilities, we see that A has been marked 10 time, B and C both 6
times, and D has been marked 2 times. Thus, we measure the power of each player to be
10
6
6
2
, B:
, C:
, and D:
. In this particular example, the Shapley-Shubik and
A:
24
24
24
24
Banzaf indices give the same measure of power. This is not always true.
Calculus Solutions:
Both the Shapely-Shubik and Banzaf indices can be developed through the use of
calculus by forming the "Power Polynomial" for each player. The Power Polynomial for
each player is developed by considering what the other players must do for the player in
question to be important in the coalition. What could the other players do? Since there
are 3 other players, there are 4 possible situations: all three could vote yes, 2 could vote
yes and one could vote no, one could vote yes and two could vote no, or all three could
vote no. If the probability of a player voting yes is p, and assuming independence, then we
have
Scenario
Probability
Number
3 Yes
p3
2 Yes and 1 No
p 2 (1 − p )
1 Yes and 2 No
2
p (1 − p )
3 No
3
(1 − p )
1
3
3
1
Each polynomial is of the form P ( p ) = a p 3 + b p 2 (1 − p ) + c p (1 − p ) + d (1 − p ) , where
2
3
the values of a, b, c, and d are determined by how many of the possibilities the player is
important. For example, consider player A. If all three of the other players vote yes, then
the proposal is passed with or without A's participation. A has no power in this situation,
so a = 0 . If two players vote yes and the other no, A becomes important. If the two yeses
11
are B and C then A is required for passage. If the two yeses are B and D, A is also
required for passage. If the two yeses are C and D, again, A is required for passage. Since
A is necessary in all three cases, b = 3. If there is one yes and two noes, then A may or
may not be important, depending on who votes yes. If the yes is either B or C, then A is
important. If, however, the yes is D, then A's 5 votes are not enough for passage. Thus, A
is important in only two of the cases, and c = 2 . Finally, if all three vote no, then A's vote
is not enough to overcome this, and d = 0 . Player A's Power Polynomial is
2
3
2
PA ( p ) = 0 p 3 + 3 p 2 (1 − p ) + 2 p (1 − p ) + 0 (1 − p ) = 3 p 2 (1 − p ) + 2 p (1 − p ) .
In similar fashion, we can find the polynomials for the other players.
2
3
2
PB ( p ) = PC ( p ) = 0 p 3 + 2 p 2 (1 − p ) + 1 p (1 − p ) + 0 (1 − p ) = 2 p 2 (1 − p ) + 1 p (1 − p )
PD ( p ) = 0 p 3 + 1 p 2 (1 − p ) + 0 p (1 − p ) + 0 (1 − p ) = 1 p 2 (1 − p ) .
2
3
The graphs of these 3 polynomials are shown below.
Solution 7: The average value of the power function over all value of p
Since p varies from 0 to 1, we can compare the power polynomials by finding the
average value of each polynomial over the interval 0,1 . Calculating these three integrals,
1
1
1
5
3
3
,
we find that ∫ PA ( p ) dp = ,
P
p
dp
=
PC ( p ) dp = , and
(
)
B
∫
∫
0
0
0
12
12
12
1
1
∫0 PA ( p ) dp = 12 , the same values as found by the Shapely-Shubik permutation method.
Solution 8: The value of the power function at the average value of p
If we assume the values of p are uniformly distributed on the interval 0,1 , the we
can compare the values of the power polynomials at the "typical' value of p, p = 0.5 .
5
3
1
Evaluating the functions, we have PA ( 12 ) = , PB ( 12 ) = PC ( 12 ) = , and PD ( 12 ) = . Notice
8
8
8
that this sum is 1.5, not 1. If we normalize the values by dividing by the total 1.5, we find
5
3
1
PA ( 12 ) = , PB ( 12 ) = PC ( 12 ) = , and PD ( 12 ) = . These normalized values will always
12
12
12
be the same as the Banzaf index.
12
The Gerrymander Problem: Measuring Compactness
Every ten years, the US House of Representatives is reconfigured according to the
results of the US Census. This reconfiguration comes in two distinct stages. The first is the
reapportionment of the 435 seats in the House of Representatives. The Everybody’s Problems
article “Apportionment: Measuring Unfairness” (Consortium, Number 81, Spring, 2002)
discussed the mathematical model on which this reapportionment is based. After the
appropriate apportionment is determined, states must realign their congressional districts to
account for shifts in population to insure that each voting district has approximately the same
population and to account for either the loss or gain of a new district.
If these districts are created so as to give unfair advantage to one party or group of
people in elections, then the districts are said to have been Gerrymandered. This term arose
from the redistricting of Massachusetts 1812 when Elbridge Gerry was governor (see Figure
1).
The
cartoon
can
be
found
at
http://knight.fcu.edu.tw/~gunning/omancoll/publicch/pcpp6.ppt#323,37,Gerrymandering.
Gerrymandering Cartoon
Figure 1: The Original Gerrymander
What rules have to be followed in redistricting?
Each state has its own rules governing the redistricting process, but the basic guidelines are
similar. In North Carolina (www.ncleg.net), the rules are:
•
One person must equal one vote.
Each district that elects one representative to a legislative body is required to be at
least approximately equal in population to every other such district.
•
Consideration of Minorities
The Voting Rights Act and court cases decided under it forbid drawing districts that
dilute minority voting strength. Section 5 of the Voting Rights Act prohibits
"retrogression," or worsening the position of racial minorities with respect to the
effective exercise of their voting rights, while Section 2 of the Voting Rights Act
13
may require drawing districts which contain a majority minority population if three
threshold conditions are present: 1) a minority group is large enough and lives
closely enough together so that a relatively compact district in which the group
constitutes a majority can be drawn, 2) the minority group has a history of political
cohesiveness, and 3) the white majority has a history of voting as a group sufficient
to allow it to usually defeat the minority group's preferred candidate. These rules
come from Thornburg v. Gingles, a landmark US Supreme Court Voting Rights
Act case arising from North Carolina in the 1980s.
•
Impermissible Consideration of Race
The General Assembly and its redistricting plans are also subject to lawsuits if
considerations of race impermissibly dominate the redistricting process. This may
occur when non-compact majority-minority districts are drawn in such a manner
that traditional redistricting principles, such as compactness, contiguity, respect for
communities of interest, are substantially ignored. These rules come from Shaw v.
Reno, another landmark US Supreme Court case arising from North Carolina in the
1990s. Obviously, abiding by both sets of rules regarding race can be a challenge.
•
Districts Must Be Contiguous
Under the North Carolina Constitution, Senate and House districts must consist of
contiguous territory. By tradition, the contiguity requirement also has been applied
to Congressional districts. Contiguity means that all parts of a district must touch.
• Compactness
Another feature prominent in many state guidelines are a requirement that districts be
as “compact’ as possible, given the natural boundaries imposed by geography and the
present shape of the counties making up the district. One difficulty is that only three
states, Iowa, Colorado, and Michigan actually define compactness. The fundamental
question for students to consider in this issue of Everybody’s Problem is how one
might measure compactness and use that method to evaluate the compactness of
several NC districts.
Properties of Compactness
Rick Gillman (Geometry and Gerrymandering) gives several properties of good
measures of compactness:
1. The measure should be applicable to all geometric shapes, both regular and irregular.
2. The measure should be independent of scale and orientation.
3. The measure should be dimensionless, and preferably be on a scale of 0 to 1, with 1
describing a highly compact region.
4. The measure should not be overly dependent upon one or two extreme points.
5. The measure should correspond with our intuition.
14
Problem Statement: Devise a measure for the compactness and use it to compare the
compactness of the geometric figures shown below. Does your measure correspond to
your intuition about which of the figures is most compact?
Figure 2: Six Geometric Shapes
Once you are satisfied with your measure of compactness, apply it to the two NC districts
below:
NC District 4
NC District 12
Figure 3: NC Districts 4 and 12
To see District maps for your state, go to:
http://nationalatlas.gov/printable/congress.html#list
15
Student methods for measuring compactness
When faced with the task of defining compactness, students generally come
quickly to the conclusion that a circle is the most compact shape, so comparing the known
shapes to circles seems an appropriate approach. Students will often devise methods that
work well with simple geometric shapes like those in Figure 2, but which are difficult to
use one real districts as shown in Figure 3.
1) Area-Perimeter Measure: Since a circle is considered the most compact shape, we will
compare the area of the district to the area of a circle with the same perimeter. Let AD
represent the area of the district (or shape) and let A: represent the area of a circle with the
same perimeter as the district. The ratio of AD to A: is our measure of compactness. If the
shape is a circle, then the ratio is one. The further from one is the ratio, the less compact
the district. If we let p represent the perimeter of the district, the p = 2π r , since we are
p
and
comparing to a circle with circumference equal to the area of the district. So r =
2π
2
⎛ p⎞
p2
A: = π ⎜ ⎟ , or A: =
. Consequently, our measure, M 1 , is
4π
⎝ 2π ⎠
A
A
4π AD
. To use this measure, we simply approximate the area of the
M1 = D = 2 D =
A: p / 4π
p2
district and its perimeter and compute M 1 .
2)
Longest Distance Ratio: For this method we again compute the ratio of the area of
the district to the area of a circle. We will use the distance, L, between the farthest
L
endpoints of the district as the diameter of the circle for comparison. So r = and
2
AD
4A
= D2 . This method is easier than the previous method since measuring
M2 =
2
π ( L / 2)
πL
this distance is often much easier than measuring the perimeter of the district.
3)
Circumcircle Ratio: Make the ratio of Area of the district to Area of the
A
circumscribed circle M 3 = D
. The difficulty here is in finding an appropriate
A:
circumscribed circle. The results of methods 2) and 3) are often (but not always) the same.
4)
Major and Minor Axis Method: Instead of using a circle as the basis for our
measure, some students choose an ellipse. They compare the area of the district to the area
of an ellipse whose major axis is the maximum distance between two points, L, and whose
minor axis is the largest distance across the district perpendicular to this axis. The area of
π Mm
an ellipse with major axis of length M and minor axis of length m is
. So,
4
4 AD
M4 =
.
π Mm
16
5)
Ratio of Major to Minor Axis: A simpler variation on this ellipse method is to
simply compare the length of the longest axis in the district to the length of the longest
m
perpendicular axis. So, M 5 =
. If the district is in the shape of a circle, then this ratio is
M
1. The smaller this ratio, the less dense the district.
6)
Standard Deviation Ratio (Monte Carlo Method): Define a coordinate system on
which to place the district. Choose n points randomly distributed in the district. Find the
center of the district by computing the average x-coordinate and the average y-coordinate
( x , y ) . Calculate the distance, di , of each point from this center. Now compute the
standard deviation of the distances. Do the same for a circle with area equal to that of the
(di − d ) 2
sd :
∑
district. Then M 6 =
, where sd =
. This method can be tedious but
sd D
n −1
works reasonable well for nice geometric shapes. However, the method proves
unworkable (at least for students) on real districts. We will not use it in our computations
that follow.
7)
Sum of the Deviations Ratio: Using polar graph paper on a transparency, select a
point in the center of the region. Measure the radial distance, di , from this center for n
points on the boundary of the region, each separated by a fixed Δθ , say 12 points that are
30 degrees away from each other. Te ratio of this sum to a similar sum for a circle with
area equal to that of the district will be our measure of compactness. The expected sum for
AD
n
AD
the circle is nr , where r =
. So M 7 = n π .
π
∑ di
i =1
8)
Rubber Band Measure: Students notice that a district may have a large perimeter
because its borders are along a river or coast. The students say that this extra length should
not be “counted against” the district, since it is unavoidable (as compared to NC District
12). So, they will argue that this boarder should be approximated with a line. This idea
leads to another method, called the rubber band measure. Compare the area of the district
to the area of the region enclosed by a rubber band stretched around the district. Figure 4
below illustrates this method using NC District 4. Let AD represent the area of the district
(or shape) and let AP represent the area of the surrounding convex polygon. The ratio of
AD to AP is our measure of compactness.
17
Figure 4: Rubber Band Enclosing 4
NC District 4
Figure 5: Approximating Areas with
Graph Paper
Comparison of the Methods
The measures of compactness in the table below are only approximate. While the
lengths and areas for the six geometric shapes can be computed analytically, those for NC
Districts 4 and 12 were estimated using graph paper on a clear transparency placed over
the district (see Figure 5).
The squares on the graph paper were counted with several partial squares making a whole.
Extreme accuracy is not really the issue here, since there is really no difference in
compactness measures of 0.72 and 0.75, but there is between 0.72 and 0.48. Your
student’s approximations will likely be different from what they see in the table, and, of
course, they should use districts from their own state.
Thin
Plus
0.34
Equilateral
Triangle
0.60
Rectangle
Hourglass
Diamond
M 1 (Area/Perimeter)
Fat
Plus
0.59
0.68
NC
4
0.41
NC
12
0.09
0.70
0.30
M 2 (Longest Axis Circle)
0.76
0.53
0.73
0.51
0.28
0.37
0.27
0.29
M 3 (Circumscribed Circle)
0.76
0.53
0.73
M 4 (Major/Minor Axis)
0.76
0.53
0.64
0.51
0.28
0.37
0.27
0.29
0.96
0.64
0.64
0.73
0.35
M 5 (Minor/Major Ratio)
1
1
0.87
0.50
0.43
0.58
0.67
0.28
M 6 (Random Points)
-
-
-
-
-
-
-
-
M 7 (Polar Distances)
0.98
1.11
1.02
1.03
1.10
0.98
1.21
.76
M 8 (Rubber Band)
.086
0.61
1
1
0.50
1
0.78
0.36
Table 1: Approximate Measures of Compactness
Short Commentary on the Various Methods
Measure 1: the Area/Perimeter measure places the figures in an order that generally
matches our intuition about the relative compactness of each region. The difficulty with
this measure is in measuring the perimeter of real districts.
18
Measure 2: the Longest Axis Circle implies that District 12 is more compact than district
4.
This method is flawed.
Measure 3: for the examples given here, the Circumscribed Circle measure is the same as
the
Longest Axis Circle and suffers from the same deficiency.
Measure 4: The Major and Minor Axis measure arranges the figures in an appropriate
order and
is easier to use than the Area/Perimeter method.
Method 5: The Major/Minor Axis Ratio does not distinguish between the two Plus figures,
but
otherwise works well.
Method 6: Unfortunately, the Random Points method is computationally too difficult to
use.
Method 7: The Polar Distances, while appealing in principle, does not give measures
between 0
and 1. There is no easy way to find an upper or lower limit to the measure,
which is a
fatal flaw.
Method 8: The Rubber Band Method works well, except that all convex polygons have a
measure of 1. Since most real districts are not convex polygons, it works better for
real districts than for geometric shapes.
Methods in Use in the US
One of the goals of Everybody’s Problem is to highlight how mathematical models
are used. The models that are actually in use today are no better and often the same as
those the students create. Mathematical modeling is often just a matter of reasoning with
simple mathematical concepts.
The state of Iowa uses what it calls the Average Length-Width measure of
compactness. The absolute difference in miles between the east-west width and the northsouth height (length) of each district, divided by the number of districts to be created. A
lower number indicates better length-width compactness.
(http://www.legis.state.ia.us/Redist/April2001report.htm) If you look at the counties in
Iowa, you will notice that almost all are rectangular with a natural NS and EW orientation,
so this measure make sense for Iowa, where it might not for District 12 in NC. This
measure is similar to our method 5).
According to the report Data for 2001 Redistricting in Texas prepared by the
Research Division of the Texas Legislative Council
(http://www.tlc.state.tx.us/pubspol/red2001data.pdf), the most commonly used and
accepted measure is our method 3), comparing the area of the district to the area of the
smallest circumscribed circle. This is sometimes referred to as the “Reock” or “smallest
circle” measure. This measure penalizes a district for being elongated in any way.
Our measure 1) is also widely used. What we called the Area-Perimeter measure is
actually known as the “Schwartzberg measure”. There are several measures of perimeter
compactness, all of which generally compare to the “Schwartzberg measure”. Again,
according to the Texas Legislative Council, the Area-Perimeter measure is often identified
by scholars as the most useful for ferreting out the kind of districts struck down by the
Supreme Court in the Shaw v. Reno racial gerrymandering cases during the 1990s. The
report also suggests that the Rubber Band Method is also widely used.
19
Final Comments
Typically, given time to brainstorm ideas for measuring compactness and while
working in groups, a class will generate three to five of the ideas presented here. Students
feel empowered and become more confident in their mathematical abilities when they find
the ideas they generate are comparable to those created by professional mathematicians
and used by the states. This problem has wide appeal to students, especially those
interested in politics and government.
References:
Bullard, Floyd, and Dan Teague, “Apportionment: Measuring Unfairness”, Everybody’s
Problems, Consortium, Number 81, COMAP, Inc. Lexington, Massachusetts, Spring, 2002.
Gillman, Rick, “Geometry and Gerrymandering”, Math Horizons, Mathematical Association
of America, September, 2002.
20