Sunday 18 July 2021

Why getting a very good voting system will bring dictator ship

 Why getting a very good voting system will bring dictatorship - Arrow's impossibility theorem


Premise

Imagine that we want to improve how the world works, and as a start, we look at our voting system and decide that it does not represent the real preferences of people. So we decide that our voting system should have two conditions:
  1. Pareto Efficiency : If every voter says that the party A is more preferred that the party B, then in that case, the result, after voting should also have A ranked above over B.
  2. Independence of irrelevant alternatives: Let's say there are two parties, A and B, and there is an output A > B, in the results. Now let's shuffle the other preferences of voters for every other party than A and B. Then in that case, the relative ranking of A and B shouldn't change in the result, even though there absolute position in the output ranking can change.

So what? Our system already has these two conditions!

So these two perfectly reasonable conditions are the ones that we want to have in our voting system. 

Now let's say our system is a majoritarian system, that is, the one person getting the maximum amount of votes will be the winner. I will try to prove, that in this voting system, there are cases when the second condition, independence of irrelevant alternatives (IIA), will be broken. 

Let's say we have a total of 21 voters, and 3 parties in the election, with the following configuration:

  • 10 voters have the preference : $ A > B > C $
  • 2 voters have the preference : $ C > B > A $
  • 9 voters have the preference : $ B > C > A $
Now in accordance to this voting system, A would win. But let's remove C from this. In that case, this would be the situation :

  • 10 voters : $ A > B $
  • 2 voters : $ B > A $
  • 9 voters : $ B > A $
That is, 11 votes to B, and 10 votes to A, which would mean that B would win.

This means that the presence of C, actually changes the results, and the relative position between A and B. Which means that this system doesn't follow the IIA condition.

Then let's improve things?

Okay. So this means let's go towards a system which actually has these two conditions followed. But there is a mathematical proof which says that if both of these conditions are followed, there will be dictatorship. Wow. Pretty dark theorem.

Less go.

Why can't we improve things?

Let's prove one important theorem, which is a secret tool that will help us later.

Theorem 1: If some preference B is either at the top or bottom of everyone's preference sequence, then it would be either on the top, or at the bottom of the result's sequence.

Let's prove this using contradiction. Let's have $n$ voters with different preferences, each represented by a line. As shown in the  diagram. Now, let's say that in the output, the output $B$, is in the middle. It's neither at the top, nor at the bottom. So this would mean that it has some output $A$ above $B$, and $C$ below $B$. 



Now as you can see in the output $ A > B $, and $ B > C $. Now IIA says that only the relative positions of A, and B determine $A > B$, and the relative positions of B and C, determine $B > C$. So let's just change the positions of A and C in each of our voter's preferences, and put C above A. This would mean that $C > A$. 

This neither changes the relative position of A and B, nor the relative positions of B and C, since A and C are either both above or below B.

Now the condition looks some thing like:



Now using Pareto efficiency (PE)  which was our first condition : Since everyone agrees that $C > A$, therefore, in the results, we should also see that $C >_{Result}  A$.

Now let's go to our original conditions, that is : $A > B$, and $B > C$. Due to transitivity, $A > C$. 

So we have a contradiction, and $B$ can never be in the middle. It can either be at the top, or at the bottom of the result.


[X] (The box that comes in math books, after the "hence proved").


Theorem 2: In the voting system where PE and IIA both are present, there always will be a dictator.

Slow down death. 

Let's say initially we have $B$ at the bottom of every voter's preference. And gradually we move $B$ from the bottom to the top of the preference sequence. Let's say there is some voter $n^*$ who, as soon as moves $B$ from bottom to the top of his preference sequence, the result also puts $B$ back to the top. 

Due to theorem 1, we can say that $B$ will go directly to the top, because $B$'s are either at the top or at the bottom of the preferences of every voter.

The process can be described here:

Now let's prove that this guy, $n^*$ is a dictator for every pair A and C (excluding B). This means that, $n^*$ decides the result directly for every pair of outputs.

Let's put A above B for $n^*$ only, and let all the preferences for all the other voters remain same. For all the other preferences the relationship between A and B's positions is random.

The output should look something like this:


Now, let's compare this to the time, when $n^*$ had decided to put $B$ at the bottom of their preferences. 

In that case, the result had $B$ at the bottom. That is $A > B$. But notice that in this case too, the relative positions between A and B are the same. A is above B, for $n^*$'s preferences too. 

So using IIA we can say that for this case too, $A>B$.

Now let's see the case, when $n^*$ has shifted B to the top of their preferences. In that case, C is below B for $n^*$, and also in the result.

In this case too, C is below B. Moreover, the relative positions of B and C are the same, so we can directly say that $B > C$.

Using this, we can say that $n^*$ is a dictator for all pairs A and C.

Now let's see whether this works for B too. Well obviously! Since B's position at the top, for $n^*$'s preferences determines that the result has B at the top, and vice-versa, we can say that $n^*$ is a dictator for B too.

Using this information, we can say that $n^*$ is a dictator for all pairs.

1295D - Same GCDs

 1295D - Same GCDs Link to the problem :  https://codeforces.com/problemset/problem/1295/D I had to see the tutorial to understand this. $$ ...