Second round of Chris' comments

_posts/2024-10-21-building-an-assignment-algorithm-2.markdown

@@ -109,7 +109,7 @@ Let’s introduce a new attendee, Bob, with exactly the same votes as Alice (to
 ![fig2: The play off between Alice and Bob's aggregate compromise.]({{ site.github.url }}/jwarren/assets/assignment-algorithm-2/table2.JPG)
 
 - Slot 1: Bob is given his 2nd choice (chosen arbitrarily), so his aggregate compromise rises to 2. However, Alice is given her 3rd choice and so has had to compromise more in this first round. In slot 2 Alice will take precedence over Bob.  
-- Slot 2: Alice gets her first choice, Bob gets his 2nd choice. Even though Bob gets a worse score than Alice in the second slot, Alice’s aggregate compromise is higher - accounting for previous slots. This is because she had to get a 3rd choice in the first slot, which - we consider 3rd choices worse than 2x 2nd choices, as justified by the formula for slot compromise.  
+- Slot 2: Alice gets her first choice, Bob gets his 2nd choice. Even though Bob gets a worse score than Alice in the second slot, Alice’s aggregate compromise is higher. This is because she was given her 3rd choice in the first slot and we consider one 3rd choice worse than two 2nd choices - refer to the slot compromise formula under the first expanding more+ section (under the Compromise heading).  
 - Slot 3: Alice gets her second choice and Bob gets his 3rd.  
 - Slot 4: Bob’s aggregate compromise score has surpassed Alice’s, and so he will take precedence in choice here.  
 
@@ -133,11 +133,11 @@ If Chewbacca’s aggregate compromise was greater than the emperors, then the so
 
 ![fig3: Prioritising sorting by compromise: the play off between Chewbacca and The Emperor]({{ site.github.url }}/jwarren/assets/assignment-algorithm-2/table3.JPG)
 
-However, if Chewbacca’s aggregate compromise was 5 and the emperor’s compromise was 4, we would want a different outcome. Since the difference between the two aggregate compromises is small, we would rather Chewbacca move and have his 2nd choice (and have an aggregate compromise of 7) rather than the Emperor to take his 3rd choice, shooting his aggregate compromise from 4 up to 9. I would argue therefore that Chewbacca should take his 2nd choice shown in the slideshow, meaning Chewbacca would have an aggregate compromise of 7 and the Emperor have an aggregate compromise of 4. This is prioritising sorting by surplus difference. All of this is to say, we can’t sort one way after the other, we need to sort compromise and surplus difference simultaneously to cover all circumstances. 
+However, if Chewbacca’s aggregate compromise was 5 and the emperor’s compromise was 4, we would want a different outcome. Since the difference between the two aggregate compromises is small, we would rather Chewbacca move and have his 2nd choice (and have an aggregate compromise of 7) rather than the Emperor to take his 3rd choice, shooting his aggregate compromise from 4 up to 9. I would argue therefore that Chewbacca should take his 2nd choice shown in the slideshow, meaning Chewbacca would have an aggregate compromise of 7 and the Emperor have an aggregate compromise of 4. This is prioritising sorting by surplus difference.
 
 ![fig4: Prioritising sorting by surplus difference: the 2nd play off between Chewbacca and The Emperor]({{ site.github.url }}/jwarren/assets/assignment-algorithm-2/table4.JPG)
 
-Capturing these nuances in an algorithm however is easier said than done. Aggregate compromise, by nature, increases in size every slot so it is hard to compare with surplus difference, which remains roughly within the same range. 
+All of this is to say, we can’t sort one way after the other, we need to sort compromise and surplus difference simultaneously to cover all circumstances. Capturing these nuances in an algorithm however is easier said than done. Aggregate compromise, by nature, increases in size every slot so it is hard to compare with surplus difference, which remains roughly within the same range. 
 
 <details><summary>For an example click the 'more' button.</summary>
 In slot 2, aggregate compromise per attendee could range from 0-5 (1st choice = 0, 3rd choice = 5), but in slot 10, the aggregate compromise per attendee could range between 0 and 50. Ignoring the fact that the algorithm would not be working very well if one person had 10x 3rd choices (giving an aggregate compromise score of 50)!  
@@ -228,7 +228,7 @@ standardisedCompromiseScore \; = \quad \;
     The rationale behind this was as follows: 
 </h4>
 <p>
-    The \(\text{standardisedSurplusScore}\) should be in comparison to the maximum value, otherwise the compromise would give an extreme value. We want the compromise to be in the same range of values as the \(\text{standardisedSurplusScore}\), except for the outlying compromise, and therefore \(\frac{\text{mean surplus difference}}{\text{max surplus}}\) brings the \(\text{standardisedCompromiseScore}\) into the relative range of values, and \(\frac{\text{attendee Z score}}{2.72}\) should be in the range of \(\pm 1.3\), with the larger values being extremal. When this overtakes the \(\text{standardisedSurplusScore}\), (surpassing the value just greater than 1), we want this to occur quite rapidly because extremal compromise is much more important to deal with. Therefore we cube it. Cubing not only rises quickly, but unlike squaring, it maintains the \(\pm\), which is important for capturing whether the value is above or below the median. After some fine tuning, it also appears to give an optimal result.  
+    The \(\text{standardisedSurplusScore}\) should be in comparison to the maximum value, otherwise the compromise would give an extreme value. We want the compromise to be in the same range of values as the \(\text{standardisedSurplusScore}\), except for the outlying compromise, and therefore \(\frac{\text{mean surplus difference}}{\text{max surplus}}\) brings the \(\text{standardisedCompromiseScore}\) into the relative range of values, and \(\frac{\text{attendee Z score}}{2.72}\) should be in the range of \(\pm 1.3\), with the larger values being extremal. When this overtakes the \(\text{standardisedSurplusScore}\), (surpassing the value just greater than 1), we want this to occur quite rapidly because extremal compromise is much more important to deal with. Therefore we cube it. Cubing not only rises quickly, but unlike squaring, it maintains the sign, which is important for capturing whether the value is above or below the median. After some fine tuning, it also appears to give an optimal result.  
 </p>
 <p>
 The value of 2.72 comes from the fact that for a normal distribution, 95.4% of values are found within 2 standard deviations of the average and 99.7% of values are found within 3 standard deviations of the average. This gave a rough range between 2-3 and after some fine tuning, 2.72 gave the optimal result.