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--- courses:cs211:winter2014:journals:deirdre:chapter3 [2014/01/29 04:48] – [Section 3.3 - Implementing Graph Traversal Using Queues and Stacks] tobind
+++ courses:cs211:winter2014:journals:deirdre:chapter3 [2014/02/12 04:40] (current) – [Section 3.6 - Directed Acyclic Graphs and Topological Ordering] tobind
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 ====== Section 3.1 - Basic Definitions and Applications ======
+The reading wasn't as helpful as the past few weeks because I felt like I already knew a lot of it from class.
 Recall from Chapter 1 that a graph //G// is simply a way of encoding pariwise relationships among a set of objects: it consists of a collection //V// of nodes and a collection //E// of edges, each of which "joins" two of the nodes. We thus represent an edge //e ∈ E// as a two-element subset of //V: e = {u,v}// for some //u,v ∈ V//, where we call //u// and //v// the ends of //e//. Edges indicated a symmetric relationsihp between their ends. WE use a directed graph when we want to show asymmetric relationships. A directed graph //G'// consists of a set of nodes //V// and a set of directed edges //E'//. Each //e' ∈ E'// is an ordered pair (//u,v//). We call //u// the tail of the edge and //v// the head. We also say that the edge leaves node //u// and enters node //v//. By default, graph will mean an undirected graph. Two notes: an edge should be written as a set of nodes {//u,v//}, but will frequently be written as (//u,v//); a node is often called a vertex.
@@ Line 18: / Line 20: @@
        *G does not contain a cycle.
        *G has n - 1 edges.
+I give this an 8. It was pretty straightforward to read.
 ====== Section 3.2 - Graph Connectivity and Graph Traversal ======
@@ Line 59: / Line 64: @@
 ==The Set of All Connected Components==
 For any two nodes //s// and //t// in a graph, their connected components are either identical or disjoint.
+I give this an 8.
 ====== Section 3.3 - Implementing Graph Traversal Using Queues and Stacks ======
@@ Line 67: / Line 74: @@
 Linear time = //O(m+n)//, //n = |V|, m = |E|//
-The simplest way to represent a graph is by an adjacency matrix, which is an //n// x //n// matrix //A// where //A[u,v]// is equal to 1 if the graph contains the edge (//u,v//) and 0 otherwise. If the graph is undirected, the matrix //A// is symmetric. The adjacency matrix representation allows us to check in //O(1) time if a given edge (//u,v//) is present in the graph. However, it has two disadvantages:
+The simplest way to represent a graph is by an adjacency matrix, which is an //n// x //n// matrix //A// where //A[u,v]// is equal to 1 if the graph contains the edge (//u,v//) and 0 otherwise. If the graph is undirected, the matrix //A// is symmetric. The adjacency matrix representation allows us to check in //O(1)// time if a given edge (//u,v//) is present in the graph. However, it has two disadvantages:
   * Takes θ(n^2) space
-  * If you need to examin all edges incident to a node //v//, you have to consider all other nodes //w// and checking the matrix entry //A[v,w]// to see whether the edge (//v,w//) is present and this takes θ(n^2) time.
+  * If you need to examine all edges incident to a node //v//, you have to consider all other nodes //w// and checking the matrix entry //A[v,w]// to see whether the edge (//v,w//) is present and this takes θ(n^2) time.
 In the adjacency list representation, there is a record for each node //v//, containing a list of the nodes to which //v// has edges.
@@ Line 83: / Line 90: @@
 ==Implementing BFS==
 BFS(//s//):
-  * Set Discovered[s] = true and Discovered[v] = false for all other v
+   Set Discovered[s] = true and Discovered[v] = false for all other v
-  * Initialize L[0] to consist of the single element s
+   Initialize L[0] to consist of the single element s
-  * Set the layer counter i = 0
+   Set the layer counter i = 0
-  * Set the current BFS tree T = ~0
+   Set the current BFS tree T = ~0
-  * While L[i] is not empty
+   While L[i] is not empty
-        * Initialize an empty list L[i+1]
+        Initialize an empty list L[i+1]
-        * For each node u ∈ L[i]
+        For each node u ∈ L[i]
-              * Consider each edge (u,v) incident to u
+              Consider each edge (u,v) incident to u
-              * If Discovered[v] = false then
+              If Discovered[v] = false then
-                    * Set Discovered[v] = true
+                    Set Discovered[v] = true
-                    * Add edge (u,v) to the tree T
+                    Add edge (u,v) to the tree T
-                    * Add v to the list L[i+1]
+                    Add v to the list L[i+1]
-              * Endif
+              Endif
-        * Endfor
-        * Increment the layer counter i by one
+        Endfor
-  * Endwhile
+        Increment the layer counter i by one
+  Endwhile
+The above implementation of the BFS algorithm runs in time //O(m + n)// if the graph is given by the adjacency list representation.
+==Implementation DFS==
+DFS(//s//):
+   Initialize S to be a stack with one element s
+   While S is not empty
+      Take a node u from S
+      If Explored[u] = false then
+         Set Explored[u] = true
+         For each edge (u,v) incident to u
+            Add v to the stack S
+         Endfor
+      Endif
+   Endwhile
+The above algorithm implements DFS in the sense that it visits the nodes in exactly the same order as the recursive DFS procedure in the previous section.
+The above implementation of the DFS algorithm runs in time //O(m+n)// if the graph is given by the adjacency list representation.
+(Is there a way to get to the left of the "=="Headings"=="?)
+I give this section a 9 on the topic, but only a 7 on the interesting/helpful. I feel like I got a good grasp on this from class/my notes.
+====== Section 3.4 - Testing Bipartiteness: An Application of BFS ======
+**The Problem**
+How do we figure out if a graph is bipartite?
+- If a graph G is bipartite, it cannot contain an odd cycle.
+**Designing the Algorithm**
+In fact, there is a very simple procedure to test for bipartiteness, and its analysis can be used to show that odd cycles are the only obstacle. First, we assume the graph is //G// is connected, since otherwise we can first compute its connected components and analyze each of them separately. Next, we pick any node //s ∈ V// and color it red. It follows that all the neighbors of //s// must be colored blue. And then their neighbors must be red. Etc. At the end, we either have a valid coloring or we don't. If we don't, it's not bipartite. We can see that we can implement BFS to do this!
+**Analyzing the Algorithm**
+Let //G// be a connected graph and let L1, L2 be the layers produced by BFS starting at node s. Then exactly one of the following two things must hold.
+  * There is no edge of //G// joining two nodes of the same layer. In this case //G// is a bipartite graph in which the nodes in even-numbered layers can be colored red and the nodes in odd-numbered layers can be colored blue..
+  * There is an edge of G joining two nodes of the same layer. In this case, G contains an odd-length cycle and so it cannot be bipartite.
+See book for proof (p96). This part was a 8 for interesting, but had minimal knowledge gain I feel like? It might have just been that I thought it was relatively simple when we talked briefly about it during class.
+====== Section 3.5 - Connectivity in Directed Graphs ======
+Remember: in directed graphs, the edge //(u,v)// has a direction: it goes from //u// to //v//. (relationship is asymmetric)
+To represent a dg, we use aversion of the adjacency list representation. Now, instead of each node having a single list of neighbors, each node has two lists associated with it: one list consists of nodes to which it has edges and a second list consists of nodes from which it has edges.
+**The Graph Search Algorithm**
+BFS starts at node //s//, defines first layer, second layer, etc. The nodes in layer //j// are precisely those for which the shortest path from //s// has exactly //j// edges. running time = //O(m+n)//. DFS also runs in linear time.
+**Strong Connectivity**
+(strongly connected = u -> v ^ v -> u)
+Two nodes //u// and //v// in a dg are mutually reachable if there is a path from //u// to //v// and also a path from //v// to //u//. If //u// and //v// are mutually reachable and //v// and //w// are m.r., then //u// and //w// are m.r.
+There is a simple linear time algorithm to test if a directed graph is strongly connected. We pick any node //s// and run BFS starting from //s//. we then also run BFS starting from //s// in G^rev. If one of the two searches fail to reach every node, G is not strongly connected.
+For any two nodes //s// and //t// in a dg, their strong components are either identical or disjoint.
+====== Section 3.6 - Directed Acyclic Graphs and Topological Ordering ======
+If an undirected graph has no cycles, then each of its connected components is a tree. But it's possible for a dg to have no cycles and still "have a very rich structure". If a dg has no cycles, we call it a DAG (directed acyclic graph).
+**The Problem**
+.18 If G has a topological ordering, then //G// is a DAG.
+Does every DAG have a topological ordering? How do we find one efficiently?
+**D and A the Algorithm**
+(Spoiler alert: The converse of 3.18 is true.) Which node do we put at the beginning of the topological ordering? Such a node would need to have no incoming edges. (In every DAG G, there is a node v with no incoming edges)
+Algorithm:
+ To compute a topological ordering of G:
+ Find a node v with no incoming edges and order it first
+ Delete v from G
+ Recursively compute a topological ordering of G-{v} and append this order after v
+We can achieve a running time of //O(m+n)// by iteratively deleting nodes with no incoming edges. We can do this efficiently by declaring nodes "active" and maintaining:
+ - for each node //w//, the number of incoming edges that //w// has from active nodes;
+ - the set S of all active nodes in //G// that have no incoming edges from other active nodes.
+At the start, all nodes are active. This allows us to keep track of nodes that are eligible for deletion.
+This section was an 8 to read. I must have been tired in class this day or else we didn't cover it very much because the acyclic stuff makes way more sense now.