# Schulze Method

* This document is not part of the formal ASO AC Procedures.

The ASO AC uses the Schulze ballot counting method with two modifications for selection of ICANN Board Seats 9 and 10 Candidates.

1. A tie can result in any point of the counting method.

In the event of a tie, we will randomly select a ballot to break the tie.

If that ballot does one rank one of the tied candidates higher than the other tied candidates then we will pass along the remaining tied candidates to the next randomly selected another ballot, and so on.

If none of the ballots rank one of the tied candidates higher than the others then the tie will be broken by a random process

2. We will run the Schulze process as described successively. This means an election with 5 candidates will have four Schulze counting methods applied. The first round will yield an order list of the 5 candidates. From this list, only the top candidate will be selected, and placed in the top most election ranking. A second round will be held with only the four remaining candidates. The top candidate who has been removed will have her or his votes zeroed out. This second round of Schulze counting will yield an ordered list of four candidates. From this list, only the top candidate will be selected, and placed in the second top most election ranking. This process will be successively repeated with the removed candidatesâ€™ votes zeroed out to fill successive

General information on the Schulze counting method:

https://en.wikipedia.org/wiki/Schulze_method

Math heavy discussion of Schulze:

schulze1.pdf

schulze2.pdf

schulze3.pdf

schulze4.pdf

schulze5.pdf

Example vote, and running the method

This is an example of the Schulze calculations run by the ASO AC members in electing football players and Ron deSilva after the entire board has been removed. In this scenario, Ron has indicated his willingness to continue in service, and Akanori has indicated his preference to not run again.

The link below it to a spreadsheet which records the ballots, calculates the winners of a pairwise comparison of each ballot, calculates the power of winning candidate for each pairwise comparison, and lastly determines the power of the beat path for the winner of each pairwise comparison. Detailed explanation is below the link.

https://docs.google.com/spreadsheets/d/1_MD53cZKVqrli-Rsf3RxSNpXb6sihlCOf_Z0IHY4CY4/edit#gid=0

The candidates are each represented by a two letter code as follows:

CR – Cristiano Ronaldo

LM – Lionel Messi

NJ – Neymar Junior

LS – Luis Suarez

RS – Ron da Silva

Jason marked his ballot as follows:

CR – 4

LM – 3

NJ – 2

LS –

RS – 5

This is recorded in cells A5-F5

This means that Jason prefers Ron the most, followed by Cristiano, followed by Lionel, followed by Neymar. Jason did not vote for Luis Suarez because he prefers all other candidates over him.

A pairwise comparison is done between each candidate on Jasonâ€™s ballot in rows H5 – AF5

You will note H5 is greyed out and has a value of zero. This is a pairwise comparison of a candidate to himself/herself. This cell is meaningless, and is only here for ease of formula cut and pasting. Same for N5, T5, Z5, AF5.

I5 is a comparison between CR > LM. If CR has a higher rank than LM, then this cell will contain a 1. Otherwise it will contain a zero. This process is repeated for each pairwise comparison.

This process is repeated for each ballot.

The number of times a candidate wins a pairwise comparison is summed on B20 – AF20.

The summed pairwise comparison is then transformed into a grid in B22 – G27.

The grid is colored such that the candidate with the most votes in a pairwise comparison is green, and the candidate with the least votes in a pairwise comparison is red. For example cell D23 represents the number of times CR beats LM in a pairwise comparison, 4. Cell C24 represents the number of times LM beats CR in a pairwise comparison, 9. Since 9 is greater than 4, the cell containing 9 is green, and the cell containing 4 is red. The rest of the grid contains all of the pairwise comparisons in the same fashion.

A matrix is then constructed. Since there are five candidates, there will be five points. They are fully meshed by arrows where an arrow points from the winner of the pairwise comparison to the loser of the pairwise comparison. Each arrow is labeled by the number of pairwise comparisons that the winner won (green in the grid).

A beat path is then traced from one candidate to another. Paths may only be traced in the directions of the arrows. Paths may pass through intermediate candidates. The power of the path will be the lowest valued link in the path. When there are multiple paths, the path with the highest value is used. This is then recorded in the strongest path grid in B30 – G35.

For example, RS -> CR has three paths:

RS -> CR

RS -> LM -> CR

RS -> NJ -> CR

RS->CR has a power of 14.

RS -> LM has a power of 12.

RS -> NJ has a power of 13.

LM -> CR has a power of 9.

NJ -> CR has a power of 8.

The path RS -> CR has a power of 14.

The path RS -> LM -> CR has links 12, 9, which means it has a power of 9.

The path RS -> NJ -> CR has links 13, 8, which means it has a power of 8.

Of all the paths, the one with the highest power is 14. This is recorded in C35.

This process is followed for each of the candidates.

The candidate who has the highest power in the most pairwise comparisons is the winner.

The candidate with the highest power in the next most pairwise comparisons is the second place. And so on. This is continued until all the candidates are ranked from highest power to lowest power.

It is possible that the best path between two pairs of candidates may share the same low cost link, and thus have the same power. In this case that link will be removed from the matrix, and a new power will be determined. In the event that they still share a same cost link, it will again be removed from the matrix. This will be repeated until the power of one candidate is greater, or there is no path to either candidate. This will result in a tie. A ballot will be selected at random. If that ballot ranks one of the tied candidates higher than another then the candidate ranked higher will be considered to have the higher power. If neither candidate is ranked higher another randomly selected ballot will be used. If no ballot prefers one candidate over then other than a random process will be used to determine which candidate is considered to have the higher and lower power.

The candidate with the highest power in the first round will take the highest rank.

That candidate will then be zeroed out on all ballots and a second round will be run to find the next highest rank.

Subsequent rounds will be completed in this way, and subsequently lower ranks will be filled, until the final round which will have only two candidate.

Last modified on 17/03/2021