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elements in an ArrayList

I have been given an algorithm which i have to implement which is doing a breadth first traversal. In the one sections it says For each car C2 that is adjacent to C Begin so my question is, how do you use a for loop to find the adjacent elements in that arrayList? the only ways i know of using for loops is the following (which i've tried and it doesn't work). for(Car C2 : C) {}; and then within that for loop i have to other things.

I think you might be misunderstanding the concept of graph data structures vs a list. You can implement a graph by using an adjacency list(I suspect this is what you mean by "in that arrayList").Each node in the graph will contain a list of adjacent nodes. By definition every node in the adjacency list of the current node is an adjacent node.
Another implementation is an adjacency matrix.
Do some research on graph structures.

Is this a proper implementation of a Graph for DFS, BFS

I'm learning about search algorithms BFS and DFS. I plan to implement both but before I do that, I need to implement my graph structure. Here's my idea:
A graph of connecting cities: Each city is represented by a Node.
Our graph will simply be an ArrayList of Nodes added as they're created, and each Node will have a list of it's neighbors, and a parent which will let us know where we came from (for path retrieval). I haven't coded anything up yet, I wanted to get some feedback on my ideas before spending the time writing up something that won't work. Here's some pseudocode-ish, code. One potential problem I can see is how we're going to deal with Nodes that we can get to from multiple places (multiple parents). If anyone has any suggestions on dealing with that, feel free to share.
public class Node{
String name;
Node parent;
ArrayList<Node> neighbors;
public addNeighbor(Node n);
public setParent(Node n);
public getNeighbors()
...
}
public static void main(String[] args){
ArrayList<Node> graph = new ArrayList<Node>();
//build node
Node node = new Node(String name);
//add neighbors
node.addNeighbor(neighbor1);
node.addNeighbor(neighbor2);
//set parent
node.setParent(parent1);
//add to graph
graph.add(node);
path = dfs(graph, startNode, goalNode);
System.out.print(path);
}
Edit: I know I could look online and find implementations of this pretty easily, but I'd prefer to come up with my own solutions.

Your implementation look good. It's the classic implentation of a graph structure (a node with a list of neighbors). Some points:
You can use backtracking to deal with multiples paths that reach the same node. If the dfs method have a recursive implementation, you need to avoid the recursive call if the Node has already a parent. But if the new path is better that the old one, then you discard the old parent, and set the new one.
Your implementation is a directional graph. In other words, you can build a graph that has a path from A to B, but has no path from B to A. I don't know if this is ok for you.
I recommend you encapsulate the building of the graph inside a wrapper, that build both paths automatically, whith a unique call to a method. That way, always you build bidirectional paths.
You can use a Set to store the neighbors. That way, there is no duplicates. Of course, you need to implements the "equals" method in the Node class.

should the adjacency information be stored in the graph object or in the vertex itself?

For purpose of the demonstrating the concept of adjacency list, I would assume that it is easy to represent the list as a list of lists and put the vertices as numbers and put them in an array where we can reference them by index directly in a graph. Once we get the index of a vertex, we can just get the respective list array.
However, for actual objects, where the vertex can't be just referenced by an array index, I would assume a separate vertex object would need to be created. In this case, my question is where should the edge information be implemented.
In OO based programming languages, such as java, I have see implementations of the adjacency list based graph implementation where the adjacency information is stored in the vertex object itself with a data member of some type of data structure, such as arrays, list, and etc. However, I have also seen people implement the adjacency edge information in the graph object itself by maintaining a list of edges.
Is it more of a personal preference as to where the edge information should be stored?

I think it depends on how you want the graph API to look like.
Suppose for example that the graph works with an arbitrary generic type representing a vertex. In this case, instead of defining a vertex wrapper for internal use (and managing adjacency in it), you can simply have an adjacency data structure as a member of the graph class:
public class Graph<V> {
private Map<V, Set<V>> adjacencies;
..
}
If on the other hand you do wish to expose a Vertex object together with its adjacencies (I don't find a good reason why), then you should be careful about the data consistency. The adjacency data must be read only for the caller (or defensively copied), otherwise the caller may alter an edge in such way that it's not symmetric anymore.
To conclude, the adjacency data maintained by the graph class seems a better option to me. It simplifies the API and the implementation, and it doesn't require special measures to protect the class invariants if you decide to expose the vertex object.

How to create a simple unordered tree(not BST) in java with given node pairs(u,v)?

Problem is I don't understand how to create a tree. I have gone through many code examples on trees but I don't even know how to work with/handle a node and hence I don't understand how the node class works(that was present in all the program examples ). When i try to use methods such as appendChild(as mentioned in java docs),I get an error,and I am asked to create one such appendChild method inside that node class within the main program. Couldn't understand why that happened.
I am given integer pairs((u,v) meaning there is an edge between u & v) of nodes and I also need to know if any Element-to-node conversion is required for using u and v(of type integer) as nodes.
Please bear with me since my basics are weak. Little explanation on how the entire thing works/functions would be very helpful.
Thank you.
EDIT 1: I went through the following links:(hardly found anything on just unordered trees) http://www.cs.cmu.edu/~adamchik/15-121/lectures/Trees/code/BST.java http://www.newthinktank.com/2013/03/binary-tree-in-java/ . Tried to modify these codes to meet my own purpose but failed.
I only got a blurry idea and that is not enough for implementation. I am trying to make a simple unordered tree,for which i am given u v pairs like:
(4,5) (5,7) (5,6). I just need to join (4<--5),(5<--7) and (5<--6). So how do I write a node class that only joins one node to the prev node? Besides,to do only this,do I need to bother myself with leftchild,rightchild? If not,how will I be able to traverse the tree and do similar operations such as height diameter calculation etc later?
Thank you for your patience.

Well Its not entirely clear whether you want an explanation on tree creation in general, on some tree implementation you have found, or you have a basic code already that you cannot get working. You might want to clarify that :).
Also tree creation in general:
Most explanation and implementation you will find might be "overly" elegant :). So try to imagine a simple linked list first. In that list you have nodes(the elements of the list). A node contains some valuable data and a reference to an other node object. A very simple tree is different only in that a node have more than one reference to other nodes. For example it has a Set of references, its "children".
Here is a VERY crude code only as an example for addChild() or appendChild in your case:
public class Node {
private String valueableData;
private Set<Node> children;
public Node(){
this.children=new HashSet<Node>();
}
public Node (String valueableData){
this.valueableData=valueableData;
this.children=new HashSet<Node>();
}
public void addChild(Node node){
children.add(node);
}
}
Now this implementation would be quite horrible (I even deleted the setter/getters), also it would be wise to keep the root nodes reference in some cases, and so on. But you might get the basic idea.
You might wanna create a cycle or recursion to go over the (u,v) integer pairs. You create the root node first then you just addChild() all the other nodes recursively, or create every node first then setChild() them according to your rules.

How to build in Java a Weighted Directed Acyclic Graph

I did a search on similar topics, but the answers are too vague for my level of understanding and comprehension, and I don't think they're specific enough to my question.
Similar threads:
Tree (directed acyclic graph) implementation
Representing a DAG (directed acyclic graph)
Question:
I have formatted a text file which contains data of the following format...
Example dataset:
GO:0000109#is_a: GO:0000110#is_a: GO:0000111#is_a: GO:0000112#is_a: GO:0000113#is_a: GO:0070312#is_a: GO:0070522#is_a: GO:0070912#is_a: GO:0070913#is_a: GO:0071942#part_of: GO:0008622
GO:0000112#part_of: GO:0000442
GO:0000118#is_a: GO:0016581#is_a: GO:0034967#is_a: GO:0070210#is_a: GO:0070211#is_a: GO:0070822#is_a: GO:0070823#is_a: GO:0070824
GO:0000120#is_a: GO:0000500#is_a: GO:0005668#is_a: GO:0070860
GO:0000123#is_a: GO:0005671#is_a: GO:0043189#is_a: GO:0070461#is_a: GO:0070775#is_a: GO:0072487
GO:0000126#is_a: GO:0034732#is_a: GO:0034733
GO:0000127#part_of: GO:0034734#part_of: GO:0034735
GO:0000133#is_a: GO:0031560#is_a: GO:0031561#is_a: GO:0031562#is_a: GO:0031563#part_of: GO:0031500
GO:0000137#part_of: GO:0000136
I'm looking to construct a weighted directed DAG from this data (the above is just a snippet). The whole dataset of 106kb is here: Source
--------------------------------------------------
Taking into consideration line-by-line, the data of each line is explained as follows...
First line as an example:
GO:0000109#is_a: GO:0000110#is_a: GO:0000111#is_a: GO:0000112#is_a: GO:0000113#is_a: GO:0070312#is_a: GO:0070522#is_a: GO:0070912#is_a: GO:0070913#is_a: GO:0071942#part_of: GO:0008622
'#' is the delimeter/tokenizer for the line data.
The First term, GO:0000109 is the node name.
The subsequent terms of is_a: GO:xxxxxxx OR part_of: GO:xxxxxxx are the nodes which are connected to GO:0000109.
Some of the subsequent terms have connections too, as depicted in the dataset.
When it is is_a, the weight of the edge is 0.8.
When it is part_of, the weight of the edge is 0.6.
--------------------------------------------------
I have Google-d on how DAGs are, and I understand the concept. However, I still have no idea how to put it into code. I'm using Java.
From my understanding, a graph generally consists of nodes and arcs. Does this graph require an adjacency list to determine the direction of the connection? If so, I'm not sure how to combine the graph and adjacency list to communicate with each other.
After constructing the graph, my secondary goal is to find out the degree of each node from the root node. There is a root node in the dataset.
For illustration, I have drawn out a sample of the connection of the first line of data below:
Image Link
I hope you guys understand what I'm trying to achieve here. Thanks for looking through my problem. :)

Because it's easier to think about, I'd prefer to represent it as a tree. (Also makes it easier to traverse the map and keep intermediate degrees.)
You could have a Node class, which would have a Collection of child Node objects. If you must, you could also represent the child relationships as a Relationship object, which would have both a weight and a Node pointer, and you could store a Collection of Relationship objects.
Then you could do a walk on the tree starting from the root, and mark each visited node with its degree.
class Node{
String name;
List<Relationship> children;
}
class Relationship{
Node child;
double weight;
}
class Tree{
Node root;
}
Here, Tree should probably have a method like this:
public Node findNodeByName(String name);
And Node should probably have a method like this:
public void addChild(Node n, double weight);
Then, as you parse each line, you call Tree.findNodeByName() to find the matching node (and create one if none exists... but that shouldn't happen, if your data is good), and append the subsequent items on the line to that node.
As you've pointed out, DAGs cannot really be converted to trees, especially because some nodes have multiple parents. What you can do is insert the same node as the child of multiple parents, perhaps using a hash table to decide if a particular node has been traversed or not.

Reading the comments, you seem confused by how a Node can contain Relationships which each in turn contains a Node. This is quite a common strategy, it is in general called the Composite pattern.
The idea in the case of trees is that the tree can be thought of as consisting of multiple subtrees - if you were to disconnect a node and all its ancestors from the tree, the disconnected nodes would still make a tree, though a smaller one. Thus, a natural way to represent a tree is to have each Node contain other Nodes as children. This approach lets you do many things recursively, which in the case of trees is often, again, natural.
Letting a Node keep track of its children and no other parts of the tree also emulates the mathematical directed graph - each vertex is "aware" only of its edges and nothing else.
Example recursive tree implementation
For instance, to search for an element in a binary search tree, you would call the root's search method. The root then checks whether the sought element is equal, less or greater than itself. If it is equal, the search exits with an appropriate return value. If it is less or greater, the root would instead call search on the left or right child, respectively, and they would do exactly the same thing.
Analogously, to add a new Node to the tree, you would call the root's add method with the new node as a parameter. The root decides whether it should adopt the new node or pass it on to one of its children. In the latter case, it would select a child and call its add method with the new Node as a parameter.