minor style updates

modules/CS260/Joes_Notes/Greedy_Algorithms_Scheduling.md

@@ -22,7 +22,7 @@ Given a set of jobs with a start and end time. Find the largest set of jobs such
 * Sort the jobs by end time in ascending order
 * Create an empty set of selected jobs
 * Iterate though the set of jobs
-* * If a job is compatible with the current selected jobs then add it
+  * If a job is compatible with the current selected jobs then add it
 
 ##### Checking For compatibility
 
@@ -38,9 +38,9 @@ jobs <- The available jobs with .start and .end
 
 jobs = jobs.sort(job.end) // Sort the job by end time in increasing order O(nlogn)
 selected = [] // empty set
-for( job : jobs ){
+for ( job : jobs ) {
     // the first job is added by default
-    if (selected.length ==0 || last.finish <= job.start){
+    if (selected.length == 0 || last.finish <= job.start){
         selected.add(job)
         last=job
     }
@@ -58,7 +58,7 @@ proof by contradiction
 * The most efficient set ($$j_1,j_2,j_3...j_m$$) also exists such that the largest $$r$$ can be found
 * Select an $$r$$ such that $$\forall v < r : i_v = j_v \land i_r \neq j_r$$ (r is the largest value such that all pairs before r are equal)
 
-As $i_r$ is selected by the finish first algorithm; no job exists that is compatible with $$\{ i_1...i_{r-1} \}$$ and finishes before $$i_r$$ so $$j_r$$ finishes after or with $$i_r$$
+As $$i_r$$ is selected by the finish first algorithm; no job exists that is compatible with $$\{ i_1...i_{r-1} \}$$ and finishes before $$i_r$$ so $$j_r$$ finishes after or with $$i_r$$
 
 This implies that $$i_r$$ can replace $$j_r$$ this is contradictory as it goes against the maximality of $$r$$ 
 
@@ -78,7 +78,7 @@ Given a set of jobs (intervals) schedule the jobs into the least number of rooms
 * Going through each job in turn
 * * check if the new schedule and add the job
 
-#### **Checking For compatibility
+#### Checking For compatibility
 * Store the schedules in a priority queue
 * use the end time of the last job added as the key
 * Check compatibility with queue.Find_Min 
@@ -90,18 +90,18 @@ Given a set of jobs (intervals) schedule the jobs into the least number of rooms
 ```java
 rooms= new PriorityQueue
 jobs = jobs.sort(job.start)
-for (job in jobs){
+for (job : jobs) {
     // the first job is added by default
-    if (rooms.length>0 && rooms.find_Min_Key <= job.start){
+    if (rooms.length > 0 && rooms.find_Min_Key <= job.start) {
         room=rooms.pop_Min()
         room.add(job)
-        rooms.add(room,job.end)
-    }else{
-        room=[job]
-        rooms.add(room,job.end)
+        rooms.add(room, job.end)
+    } else {
+        room = [job]
+        rooms.add(room, job.end)
     }
 }
-return rooms.values();
+return rooms.values()
 ```
 <div id="screen1"></div>
 
@@ -219,4 +219,4 @@ Show that at every stage of the the greedy algorithm produces a solution that is
 Discover structural bounds asserting that every solution must have a set value. Then show your algorithm achieves this bound.
 
 #### Exchange 
-Gradually transform any optimal solution to the one found by the greedy algorithm without hurting the solution quality.
+Gradually transform any optimal solution to the one found by the greedy algorithm without hurting the solution quality.

modules/CS260/Joes_Notes/Greedy_Algorithms_Shortest_Path.md

@@ -10,15 +10,20 @@ part: true
 
 ### Single-Pair Shortest path problem
 **Input**:
+
 * A weighted graph/digraph
 * A source node(start)
 * A destination node(end)
+
 **Output**:
 * A shortest path through the graph from the source to the destination 
 
 ### Single-Source shortest path problem
+**Input**:
+
 * A weighted graph/digraph
 * A source node(start)
+
 **Output**:
 * A set of shortest paths through the graph from the source to every node in the graph
 * Can be represented as a tree with the source as the node
@@ -36,10 +41,7 @@ initialize $$S \leftarrow \{s\}$$ , $$d[u] \leftarrow 0$$
 repeatedly:
 * choose unexplored node $$v\notin S$$ that minimises $$\pi(v)$$ where:
 
-$$
-\pi(v) = \min_{e = (u,v):u \in S}
-
-$$
+$$\pi(v) = \min_{e = (u,v):u \in S}$$
 
 * add $$v$$ to $$S$$ , $$d[v]\leftarrow \pi(v)$$
 * prev[v] $$\leftarrow e$$