I am seriously getting rather annoyed with this

*My Ipod went out some time ago after i tried to reload it by deleting all songs on it then putting the songs back on it because it was not playing my songs rite so i did that and it played for a little while but then one morning i woke up and my Ipod turned on showed the apple logo, loaded for 2-3 seconds then it gave me a folder with an exclamation point in the corner and shut off then i connected it to my computer and i get the familiar notification noise stating i connected it but it does not show up on my drive list or on my itunes and i have the latest itunes and everything but i've reset it i've reinstalled itunes restarted my computer and i cant restore it cuz it doesnt show up so i have tried the 5R's and they didnt work so dont talk to me about them and also i have read the page on the error picture i get so that would be a waste of time to post as well if you have any ideas tell me and if it just has to be repaired tell me although if that is the case i would rather hit it with a hammer thatn pay $80 to fix it i'll just buy a brand new one with more space for 30-40 dollars more.........btw its an earlier generation 2GB nano*

It makes posts much easier to read and comprehend if you use proper punctuation, paragraphs, etc. and you get more help that way. It is also a courtesy to include detailed information, like "I replaced the battery" so I wouldn't suggest something you've already tried.
It isn't "pompous" to use proper english if you are asking other people for help. I tried to point out that you post read like the writings of a child or an illiterate, ignorant person in a nice, humorous manner, but I see now I just should have stuck to the facts. And I'll add that you are also rude.
If your battery was put in correctly the only options left are the flash memory or the logic board. And you could still sell it for parts.

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    morgalr wrote:
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            // Add the vertices in the original graph to mst
            // and create the vertex sets at the same time
            for( vertices.first(); vertices.isOnList(); vertices.next() )
                String curName = vertices.get().getName();
                Set<String> curSet = null; // new ArrayBasedSet<Vertex>();
                // Add the name of the current vertex to its set and then
                // add the set to the list that contains the vertex sets
                curSet.add( curName );
                vertexSets.add( curSet );
                // Add the current vertex to the MST graph
                mst.addVertex( curName );
            // Put the edges into a heap which effectively sorts them
            Heap<Edge> edges = new ArrayBasedHeap<Edge>();
            List<Edge> allEdges = g.getEdges();
            for( allEdges.first(); allEdges.isOnList(); allEdges.next() )
                edges.insertHeapNode( allEdges.get() );
            // Setup is complete - run the algorithm
            // There is more than one set left in the list vertex sets
            while( vertexSets.size() > 1 )
                // Get the smallest edge
                Edge cur = edges.getSmallest();
                // Find the sets where these vertices are located
                int sourcePos = findSet( vertexSets, cur.getSource().getName() );
                int destPos = findSet( vertexSets, cur.getDest().getName() );
                // If the vertices are in different sets - add the edge to
                // the MST
                if( sourcePos != destPos )
                    Set<String> sourceSet = vertexSets.get( sourcePos );
                    Set<String> destSet = vertexSets.get( destPos );
                    // Add the edge to the MST
                    mst.addEdge( cur.getSource().getName(),
                            cur.getDest().getName(), cur.getCost() );
                    // Merge the sets
                    sourceSet.union( destSet );
                    vertexSets.remove( destPos );
            // The MST can be read from this graph
            return mst;
         * Return the position of the first set in the list that contains the
         * specified name.
         * @param vertexSets a list of sets to search.
         * @param name the name being searched for.
         * @return the position of the first set in the list that contains the name
         *         or -1 if the name cannot be found.
        private int findSet( List<Set<String>> vertexSets, String name )
            int retVal = -1;
            // Step through the list and examine each set. Stop when you
            // find a set with the name or we fall off the list
            for( int i = 0; retVal == -1 && i < vertexSets.size(); i = i + 1 )
                Set curSet = vertexSets.get( i );
                // Does the current set contain the name we are looking for?
                if( curSet.contains( name ) )
                    retVal = i;
            // Return the position of the set
            return retVal;
         * Perform Dijkstra's Shortest Path algorithm on the given graph, starting
         * at the given vertex.
         * @param g the Graph to traverse.
         * @param name the name of the vertex where the traversal starts.
         * @return an array containing vertex path costs.
        public int[] dijkstraSP( Graph g, String name )
            // The names of the vertices for which the shortest
            // path is not known
            Set<String> u = new ArrayBasedSet<String>();
            // The names of the vertices for which the shortest
            // path is known
            Set<String> s = new ArrayBasedSet<String>();
            // Put the vertices in an array to make things easier
            List<Vertex> vertices = g.getVertices();
            Vertex v[] = new Vertex[ vertices.size() ];
            for( int i = 0; i < vertices.size(); i++ )
                v[ i ] = vertices.get( i );
            // The starting vertex
            Vertex start = g.getVertex( name );
            // The lowest costs seen so far
            int c[] = new int[ v.length ];
            // Temporary edge used by the program
            Edge curEdge = null;
            // Sort the vertices by name so that the costs will
            // appear in order by name
            Heap<String> names = new ArrayBasedHeap<String>();
            // Build the heap
            for( int i = 0; i < v.length; i = i + 1 )
                names.insertHeapNode( v[ i ].getName() );
            // Read out the values
            for( int i = 0; !names.empty(); i = i + 1 )
                v[ i ] = g.getVertex( names.getSmallest() );
            // We "know" the shortest path to the source
            s.add( name );
            // For each vertex, compute the starting cost
            for( int i = 0; i < v.length; i = i + 1 )
                // If this isn't the start node
                if( !v[ i ].getName().equals( name ) )
                    // Put it in the unknown set
                    u.add( v[ i ].getName() );
                    // Compute the initial cost to reach this Vertex
                    curEdge = start.getEdge( v[ i ].getName() );
                    if( curEdge != null )
                        c[ i ] = curEdge.getCost();
                    else
                        // This Vertex is currently unreachable
                        c[ i ] = Integer.MAX_VALUE;
                else
                    // It costs 0 to get to the start vertex
                    c[ i ] = 0;
            // Set is complete - run the algorithm until all of
            // the paths are known
            while( !u.isEmpty() )
                // Find the position of the lowest-cost unknown node
                int min = Integer.MAX_VALUE;
                int minPos = -1;
                for( int i = 0; minPos == -1 && i < c.length; i = i + 1 )
                    if( c[ i ] < min && u.contains( v[ i ].getName() ) )
                        min = c[ i ];
                        minPos = i;
                // We know the shortest path to the vertex
                s.add( v[ minPos ].getName() );
                u.remove( v[ minPos ].getName() );
                // Update the costs based
                for( int i = 0; i < c.length; i = i + 1 )
                    // Get the edge between the new shortest and the
                    // current node in the array
                    curEdge = v[ minPos ].getEdge( v[ i ].getName() );
                    // If there is an edge
                    if( curEdge != null )
                        // If going through the new node is better than
                        // what has been seen update the cost
                        if( c[ i ] > c[ minPos ] + curEdge.getCost() )
                            c[ i ] = c[ minPos ] + curEdge.getCost();
            return c;
        public static void main( String args[] )
            Traversals t = new Traversals();
            Graph g = new LinkedGraph();
            g.addVertex( "A" );
            g.addVertex( "B" );
            g.addVertex( "C" );
            g.addVertex( "D" );
            g.addVertex( "E" );
            g.addVertex( "F" );
            g.addEdge( "A", "B", 5 );             <--------------------------------------------------------------------------------------------Line 311
            g.addEdge( "A", "D", 9 );
            g.addEdge( "A", "C", 10 );
            g.addEdge( "B", "A", 5 );
            g.addEdge( "B", "D", 4 );
            g.addEdge( "C", "A", 10 );
            g.addEdge( "C", "D", 13 );
            g.addEdge( "C", "E", 14 );
            g.addEdge( "D", "B", 4 );
            g.addEdge( "D", "A", 9 );
            g.addEdge( "D", "C", 13 );
            g.addEdge( "D", "E", 7 );
            g.addEdge( "D", "F", 8 );
            g.addEdge( "E", "C", 14 );
            g.addEdge( "E", "D", 7 );
            g.addEdge( "E", "F", 2 );
            g.addEdge( "F", "D", 8 );
            g.addEdge( "F", "E", 2 );
            int costs[] = t.dijkstraSP( g, "A" );
            for( int i = 0; i < costs.length; i = i + 1 )
                System.out.println( costs[ i ] );
    }

    ChainingHashTable.java
    package Graph;
    * An implementation of a HashTable using chaining for collision resolution.
    * Javadoc comments for methods specified in the Table interface have been
    * omitted.
    * This code assumes that the preconditions stated in the comments are true when
    * a method is invoked and therefore does not check the preconditions.
    public class ChainingHashTable<K, V> implements Table<K, V>
        private List<Tuple<K, V>> hashTable[]; // The hashtable
        int size; // The size of the hash table
         * Create a new hash table.
         * @param cardinality the number of elements in the domain.
        public ChainingHashTable( int cardinality )
            // Note that the cast is necessary because you cannot
            // create generic arrays in Java. This statement will
            // generate a compiler warning.
            hashTable = (LinkedList<Tuple<K, V>>[]) new LinkedList[ cardinality ];
            size = 0;
        public void put( K key, V value )
            int bucket = key.hashCode();
            // Do we need to create a new list for this bucket?
            if( hashTable[ bucket ] == null )
                hashTable[ bucket ] = new LinkedList<Tuple<K, V>>();
            hashTable[ bucket ].add( new Tuple<K, V>( key, value ) );
            size = size + 1;
        public void remove( K key )
            int bucket = key.hashCode();
            List<Tuple<K, V>> chain = hashTable[ bucket ];
            boolean found = false;
            // Is there a chain to search?
            if( chain != null )
                // Step through the chain until we fall off the end or
                // find the tuple to delete
                chain.first();
                while( !found && chain.isOnList() )
                    // If this tuple has the key we are looking for
                    // delete it and stop the loop
                    if( chain.get().getKey().equals( key ) )
                        chain.remove();
                        found = true;
                    else
                        chain.next();
        public V get( K key )
            int bucket = key.hashCode();
            List<Tuple<K, V>> chain = hashTable[ bucket ];     <----------------------------------------------------------------Line 76
            V retVal = null;
            // Is there a chain to search?
            if( chain != null )
                // Step through the chain until we find the element or
                // run out of list.
                for( chain.first(); retVal == null && chain.isOnList(); chain
                        .next() )
                    // If this tuple has the key we are looking for,
                    // extract the value
                    if( chain.get().getKey().equals( key ) )
                        retVal = chain.get().getValue();
            return retVal;
        public boolean isEmpty()
            return size == 0;
        public int size()
            return size;
        public int cardinality()
            return hashTable.length;
        public List<K> getKeys()
            List<Tuple<K, V>> chain;
            List<K> keys = new LinkedList<K>();
            // Go through each chain and create a list that contains all
            // of the keys in the table
            for( int i = 0; i < hashTable.length; i++ )
                if( hashTable[ i ] != null )
                    chain = hashTable[ i ];
                    for( chain.first(); chain.isOnList(); chain.next() )
                        keys.add( chain.get().getKey() );
            return keys;
        public List<V> getValues()
            List<Tuple<K, V>> chain;
            List<V> values = new LinkedList<V>();
            // Go through each chain and create a list that contains all
            // of the keys in the table
            for( int i = 0; i < hashTable.length; i++ )
                if( hashTable[ i ] != null )
                    chain = hashTable[ i ];
                    for( chain.first(); chain.isOnList(); chain.next() )
                        values.add( chain.get().getValue() );
            return values;
         * Return a string representation of this hash table.
         * @return a string representation of this hash table.
        public String toString()
            StringBuffer hashAsString = new StringBuffer( "" );
            List chain = null;
            for( int i = 0; i < hashTable.length; i = i + 1 )
                hashAsString.append( "h[" + i + "]==" );
                if( hashTable[ i ] != null )
                    chain = hashTable[ i ];
                    for( chain.first(); chain.isOnList(); chain.next() )
                        hashAsString.append( " " + chain.get() );
                hashAsString.append( "\n" );
            return hashAsString.toString();
    } // ChainingHashTable

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