From 98d1bc59d87c3c25daecb3ada94ba761f2ff7477 Mon Sep 17 00:00:00 2001 From: jwarren-scottlogic Date: Wed, 23 Oct 2024 16:46:16 +0100 Subject: [PATCH] Neatening up --- ...building-an-assignment-algorithm-1.markdown | 18 +++++++++--------- 1 file changed, 9 insertions(+), 9 deletions(-) diff --git a/_posts/2024-10-21-building-an-assignment-algorithm-1.markdown b/_posts/2024-10-21-building-an-assignment-algorithm-1.markdown index 6b6ed9e88..2b7dfd49b 100644 --- a/_posts/2024-10-21-building-an-assignment-algorithm-1.markdown +++ b/_posts/2024-10-21-building-an-assignment-algorithm-1.markdown @@ -73,15 +73,15 @@ An intergalactic 3-talk conference for universal problems has 5 attendees: The three talks run concurrently (denoted T1, T2, T3): - - The Dark Side of Work-Life Balance: Avoiding Burnout in the Empire (T1) - - Lightsaber Practise and Etiquette: Health and safety in the workplace (T2) - - Parenting 101: how to tell a son that you’re his father (T3) + - T1: The Dark Side of Work-Life Balance: Avoiding Burnout in the Empire + - T2: Lightsaber Practise and Etiquette: Health and safety in the workplace + - T3: Parenting 101: how to tell a son that you’re his father T1 has a maximum capacity of 3 attendees, the other two talks have a maximum capacity of 1. Each attendee makes an ordered list of the talks. For example Anakin’s first choice is “Parenting 101”, his second choice is “The Dark Side of Work-Life Balance” and third choice is “Lightsaber Practise and Etiquette”. We can represent this as [1st: T3, 2nd: T1, 3rd: T2], or even more simply [3,1,2]. If we do the same for the other attendees, we have: -
A [3,1,2] | B [1,3,2] | C [1,2,3]
D [1,3,2] | E [3,1,2]
+
A [3,1,2]  |  B [1,3,2]  |  C [1,2,3]
D [1,3,2]  |  E [3,1,2]

![fig1: The attendees with their different choices]({{ site.github.url }}/jwarren/assets/assignment-algorithm-1/characterChoices.png) *fig1: the attendees' choices represented graphically.* @@ -98,11 +98,11 @@ Let’s begin with a simple solution. We go through the list of attendees alphab In summary: -1. A - T3 (1st) -2. B - T1 (1st) -3. C - T1 (1st) -4. D - T1 full, gets 2nd choice T3 -5. E - T3 full, 2nd choice T3 full, 3rd choice T2 +1. A - T3 (1st) +2. B - T1 (1st) +3. C - T1 (1st) +4. D - T1 full, gets 2nd choice T3 +5. E - T3 full, 2nd choice T3 full, gets 3rd choice T2 A second choice and a third choice is not ideal among 5 attendees. Of course the ordering here is contrived, but how would we process the assignments for an optimal result, no matter the situation?