# The Pamphlet Collection of Sir Robert Stout: Volume 53

# Algebraic Analysis

## Algebraic Analysis.

Before leaving the case in which there are three candidates only, it may be of interest to give a short algebraical analysis of the question. As before, let the compound symbol AB stand for the number of electors who put A first and B second, and similarly for other cases. Let us suppose, as is clearly possible, that six quantities, *a, b* c, *a, β*, γ, are found from the following equations:

AB = *β + c* BC = γ+a CA = *a+b*

AC = *γ—b* BA= a—c CB = *β*—a

Also let us suppose that 2N denotes the whole number of electors, which is clearly equal to 2 (*a*+*β*+γ), then the states of the poll on the different modes of election which have been discussed are as shown in the following table:—

In the first column is set out an analysis of the votes. In the second is the result of the poll on the single vote method. For instance, in the first line we have the quantity *β+*γ—b+c, which is the sum of AB and AC, *i.e.*, it denotes the number of electors who put A first. In the third column is the result of the poll on the double vote system, in which each elector has two votes. For instance, in the first line we have N + a, or what is the same, 2*a* + *β* + *γ*, and this is equal to AB + AC+BA + CA, *i.e.*, it denotes the number of electors who put A first or second. In the fourth column is the result of the poll on Borda's method. For instance, in the first line we have 2N*—b* + c, and this is equal to 2AB + 2AC+BA + CA, as it ought to be. It is also seen at once that 2N — *b+c* is the sum of the two numbers in the first line in the second and third columns. This shows the truth of what was stated above, *viz.*, that the poll on Borda's method is the aggregate of the polls on the single and double vote systems. In the fifth, sixth, and seventh columns, under the heading Condorcet, are set down the states of the poll on the supposition that each of the candidates, A, B, C, is excluded in turn. Thus, if A be supposed excluded for a moment, we have N + *a* votes for B in preference to C, and consequently N — *a* for C in preference to B. For N + *a* is equal to AB+BC+BA, as it ought to be. Thus it is clear that 2a is the majority for B as against C, so that the letters *a, b*, c, have the same meaning as in the previous part of this paper. It is clear too, as has been proved before, that the number in any row in the column headed Borda, is the sum of the two numbers in the same row in the columns headed Condorcet.

The result of the method of election proposed in this paper depends solely upon the numbers *a, b, c*. The same is true of the method of Borda. On the other hand, the result of the double vote method depends solely on the values of *a, β, γ*. Consequently, whatever be the result of the proposed method or of Borda's method we can clearly construct cases in which the result of the double vote method shall be what we please. The same is true of the single vote method; for although the result of the single vote method depends upon *a, b, c* as well as upon *a, β, γ*, it is easy to see that we can choose *a, β, γ* so as to eliminate the effect of the quantities *a*, 6, c, whatever may be the values of the latter. The results of the Venetian method and of Ware's method depend on the values of a, *b, c* as well as upon those
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of *a, β, γ*, so that although for given values of *a, b, c* we cannot bring about any result we please, still we can choose *a, β, γ* so as to bring about a result different from the true one. This, of course, is to be done by choosing *a, β, γ*, so that the best candidate is thrown out at the first scrutiny. We have already seen that this is possible.

It is clear that no one of the quantities *β* + *γ, γ* + *a, a + β* can be negative. For we have *β* + *γ* = BC + CB, and BC, CB can neither of them be negative. Again, *β + γ* = N — *a*; thus *a* cannot be greater than N. So also *β, γ* can neither of them exceed N. Since *β + γ* cannot be negative, *β* and *γ* cannot both be negative; thus one only of the three *a, β, γ* can be negative. If *a* be negative it is clear that the numerical value cannot exceed N, for *a + β* cannot be negative, and *β* cannot exceed N. So for *β* and *γ*. Thus no one of the three *a, β, γ* can numerically exceed N, and one at most can be negative.

The limits between which *a, b, c* must lie are at once found from the consideration that AB, AC, &c., must none of them be negative. Thus *a* + *γ, β* — *a* can neither of them be negative; thus *a* cannot be less than — *γ* nor greater than *β*. Hence, *a fortiori*, no one of the three *a, b*, c, can be numerically greater than N. This last result is obvious from the fact that no one of the numbers in the columns headed "Condorcet" can be negative.

Formal demonstrations will now be given of a few results.

(i.) If any candidate have less than N votes on the double vote method, he ought not to be elected.

This has already been seen, but the following proof is given. Suppose A has less than N votes; then *a* must be negative, and therefore c must be negative and *b* positive. Thus A is worse than B, and also worse than C.

(ii.) Even if every elector put A in the first or second place it does not follow that A ought to be elected.

For if A has no third places we must have BC = 0 and CB = 0, thus *a = β* = — *γ*. Suppose *β* positive and therefore *γ* negative. Then by preceding case C ought to go out and A or B ought to win as *c* is positive or negative. Now *c* may be negative so that B may win; for the only conditions with reference to *c* are that *c* must be greater than —*β* and less than *a*, and as *β* is positive it is clear that *c* may be negative.

(iii.) It is impossible to arrive at the true result by merely counting the number of first places, the number of page 25 second places, and the number of third places for each candidate.

This result seems obvious enough after what has been given. It may, however, be formally proved as follows.

Let A_{1}, A_{2}, A_{3}, denote the numbers of first, second, and third places respectively for A, and let corresponding meanings be given to B_{1}, &c., C_{1}, &c. Then we have

with corresponding equations for B's and C's. We see at once from these equations that it is impossible to find *a, b, c* even if A_{1}, A_{2}, A_{3}, B_{1}, &c., be all given. We can, however, find *a, β, γ* and the three differences *b* — c, *c* — *a, a* — *b*, viz., the results are

*b* — *c* = A_{3} — A_{1}, *c* — a = B_{3} — B_{1}, a — *b* = C_{3} — C_{1}, where 2N = A_{1} + B_{1} + C_{1} = A_{3} + B_{3} + C_{3} . . . . (i) thus any five of the quantities A_{1}, B_{1}, C_{1}, A_{3}, B_{3}, C_{3}, may be chosen at pleasure; the sixth and N are then determined by the conditions (i) and A_{2}, B_{2}, C_{2} are then given by the equations

(iv.) If there be a demonstrably correct result, say A better than B and B better than C, so that c, *a*, are positive and *b* negative, then if Ware's method be wrong, Venetian method is right, and if Venetian method be wrong, Ware's method is right.

For if Ware be wrong A must be lowest on the single vote method, and therefore we must have

*i.e., a fortiori a* > *γ* because *a, c* are positive and *b* negative. Thus A cannot be lowest on double vote method, so that A will win on the Venetian method. Again, if Venetian be wrong, A must be lowest on double vote method, and therefore we must have *γ > a* and therefore *β* — *γ* — *b* + *c > a + β* — *a + b* because *a, c* are positive and *b* negative. Thus A cannot be lowest on single vote method, so that A will win on Ware's method.

(v.) If we agree to accept the proposed method as correct in all cases, then the conclusions of the last proposition will be true in all cases.

page 26For, in the demonstration of the last proposition, the essential condition is that *a + c*—2*b* should be positive. Now, if We suppose as before that the accepted result is A better than B, and B better than C, we must have *a, b, c* all positive and *b* the smallest of the three, so that it is clear that *a + c* — 2*b* is positive.

Comparing then Ware's method with the Venetian method, we see that both may be right, or one wrong and one right, but both cannot be wrong; so that, if these two methods agree, the result cannot be shown to be wrong. If, however, they do not agree, we cannot tell which is right without in effect having recourse to the proposed method.

(vi.) If *a* = 6 = c, single and double vote methods give different results.

For A's scores on the two methods will be respectively N — *a* and N + *a*. Thus, if *γ > β* > *a*, the candidates are in the order A, B. C on the single vote method, and in the order C, B, A on the double vote method. In this case Borda's method leads to a tie, and consequently the proposed method also. Ware elects A or B as *c* is positive or negative, and Venetian method elects C or B as *a* is negative or positive. Thus, in this case, Ware and Venetian method give different results.

(vii.) If *a* = *β* = *γ*, double vote method, and therefore also Venetian method, gives a tie; single vote method and Borda lead to same result; but Ware and proposed method will not necessarily lead to same result. If one only of the three, *b* — *c _{f} c* —

*a, a*—

*b*, be negative, Ware and proposed method will lead to same result; but if two be negative the results may or may not agree.

(viii.) If AB = AC, BC = BA, CA = CB, all the methods will give the same result, and that result will be demonstrably correct.

This is the case in which the strong supporters of each candidate are equally divided as to the merits of the remaining candidates. In this case we have

and A's scores on the single, double, and Borda's method are respectively 2a, N + a, N + 3a. Thus, if *a* > *β* > *γ*. it is obvious that each of these methods will put A first, B second, and C third, and it is clear that this result is correct, for *a, c* are positive and *b* negative. It is at once seen that all the methods which have been discussed will lead to the same result in this case.

(ix.) If we suppose that

then A's scores on the single, double, and Borda methods will be respectively

Hence we see that

If p < *o* and > — 1, the results of all three methods will be the same.

If *p* — 1, double and Borda methods will give the same result, which will be opposite to that of single method.

If *p* > o, single and Borda methods will give the same result, which will be opposite to that of double method.

Thus, if *p > o* or < — 1, single and double methods will give different results. If we suppose that *b, c* are positive and *a* negative, and also that *2b < c + a*, then it may be shown that these different results will both be wrong.