July 2010 – Page 2 – Coding Friends

Static in c – hide that function

In the c language there is no such thing as a class, private/public setup as such, but if you have a allot of c files that may have allot of the same function names inside e.g. swap(void *data, void *data2). So how do you just want to call your swap function and not another one from another c file. Which in this setup it is similar to private method within a class.

So instead of having access to the private keyword, you can define a function to be a static function as below in this header file that I have called static_test.h

#ifndef STATIC_TEST_H
#define STATIC_TEST_H
 
#include <iostream>
using namespace std;
 
static void PrivateAccessable(char *sting);
 
void PublicAccessable(char *st);
 
#endif // STATIC_TEST_H

the PublicAccessable function is what is *exposed* to other parts of the main program but the static void PrivateAccesable is not.

So if you try and access it via the int main function like so

 
#include <iostream>
#include "static_test.h"
 
int main(int argc, char *argv[])
{
 
    PublicAccessable("hi there");
    // undefined reference to `PrivateAccessable(char*)'
    PrivateAccessable("should not be able to access this!!");
 
    return 0;
}

The compiler will complain with something similar to the error above the PrivateAccessable line, or

undefined reference to `PrivateAccessable(char*)'

because the linker does not know where that function is, since it is “hidden” within the static_test namespace as such.

Just to complete the code example here is the static_test.c

#include "static_test.h"
 
static void PrivateAccessable(char *sting)
{
    cout << sting << endl;
}
 
void PublicAccessable(char *st)
{
    PrivateAccessable(st);
}

Graphs – Part 2

The graphs use two algorithms to either find the shortest path (Dijkstra algorithm) or the minimal spanning tree (Krushal’s method).

This is the second post of the assignment of the Path class and also the Dijkstra / Krushal methods. So to start with the Path class, below is the header of the class with the different methods inside the class, which basically has a Add method to add the arc to the Path, TotalPathDistance which will return the total distance taken in O(1) in big O notation, a toString method which returns a string version of the internal data structure of the arcs added, size method which returns the size of the arcs added to the arcs internal vector array and lastly the GetElementAt method that will return a arc of a element at any given point in the vector array.

class Path {
 
// [TODO: complete the class definition]
 
public:
	Path();
	~Path();
 
/*
 * Function: Add
 * Usage: Add(arcT *arc);
 * ----------------------------------
 * Adds a arc to the internal array of arc(s)
 */
	void Add(arcT *arc);
 
/*
 * Function: TotalPathDistance
 * Usage: TotalPathDistance();
 * ----------------------------------
 * returns the total distance in a O(1) time instead of cycling through the list of arcs
 */
	double TotalPathDistance();
 
 
/*
 * Function: toString
 * Usage: toString();
 * ----------------------------------
 * converts the array of arcs to a string repenstation
 */
	string toString();
 
/*
 * Function: size
 * Usage: size();
 * ----------------------------------
 * returns the size of the internal arcs implemetation e.g. the total amount of arcs already stored.
 */
	int size();
/*
 * Function: GetElementAt
 * Usage: GetElementAt(4);
 * ----------------------------------
 * returns a arcT* at any given element index, as passed by the parameter
 */
	arcT* GetElementAt(int i);
 
private:
	Vector<arcT *> arcs;
	double totalCost;
};

and here is the functions implementation, to start with the Path class is constructed with a total distance of 0 and then every time that a new arc is added the total distance is increased by the value of the cost associated with that arc. The tostring method just loops over the vector array of arcs and builds up a string of there details.

Path::Path()
{
	totalCost = 0;
}
 
Path::~Path()
{
	arcs.clear();
}
 
void Path::Add(arcT *arc)
{
	arcs.add(arc);
	totalCost += arc->cost;
}
 
double Path::TotalPathDistance()
{
	return totalCost;
}
 
string Path::toString()
{
	string returnStr;
	foreach(arcT *arc in arcs)
	{
		returnStr = returnStr + arc->start->name + "->" + arc->finish->name + "(" + RealToString(arc->cost) + ")" +"\n";
	}
	return returnStr;
}
 
int Path::size()
{
	return arcs.size();
}
 
arcT* Path::GetElementAt(int i)
{
	return arcs[i];
}

Because the Dijkstra method was already coded from the course reader handout, here is how I have altered it to use the new Path class to have a big O notation of O(1), I have just basically altered it where it was a Vector types to be a Path type instead and called the relevant methods on that new Path type instead of using the TotalPathDistance function, also with a couple of changes to the bottom part of the function below because there was no = operator, so instead changed to GetElementAt method.

Path FindShortestPath(nodeT *start, nodeT *finish) {
	Path path;
	PQueue< Path> queue;
	Map<double> fixed;
	while (start != finish) {
		if (!fixed.containsKey(start->name)) {
			fixed.put(start->name, path.TotalPathDistance());  
			foreach (arcT *arc in start->arcs) {
				if (!fixed.containsKey(arc->finish->name)) {
					Path newPath = path;
					newPath.Add(arc);
					queue.add(newPath, newPath.TotalPathDistance());
				}
			}
		}
		if (queue.isEmpty()) return Path();
		path = queue.extractMin();
		start = path.GetElementAt(path.size() -1)->finish; 
	}
	return path;
}

and here is the Kruskal function, to start with I only need to store the set of nodes within a array (so using a vector array) with also the just needing to store the arcs that are joining the nodes together (I am using a Set here instead of a vector, but does not really matter as such). So to start with I am first building up a list of nodes (Vector array of Sets of strings) of each node within the graph, whilst also adding to a priority queue of all of the arcs, I could have used a foreach loop and gone over the getNodesSet from thePath variable, but the arcs should have all of the nodes inside it :), and since if the nodes are not attached to the arcs then how could they form a path!. And then the main loop is just going through all of the arcs in order of cost (lowest first to highest) seeing if the start and finish nodes are within the same Set within the vector array, if so they have already been added to the minimal spanning tree (Kruskal) else if they have not been added merge (union) the two sets together and also add the arc to the list of arcs to display to the screen once this function has completed.

void KruskalMethod(PathfinderGraph &thePath)
{
	Set<arcT *> arcs = thePath.getArcSet();
	PQueue<arcT *> path;
	Vector< Set<string> > nodes;
	Set<arcT *> theJoiningArcs;
 
	CleanScreen(thePath);
	// place into a pqueue and also build up a vector of sets of the different node names
	foreach (arcT *arc in arcs)
	{
		path.add(arc, arc->cost);
		bool inAlready = false;
		for (int i = 0; i < nodes.size(); i++)
		{
			if (nodes[i].contains(arc->start->name))
			{
				inAlready = true;
				break;
			}
		}
		if (!inAlready)
		{
			Set<string> newNode;
			newNode.add(arc->start->name);
			nodes.add(newNode);
		}
	}
	while (!path.isEmpty())
	{
		arcT * arc = path.extractMin();
		// start node and end nodes set id
		int startN, endN;
		startN = endN = -1;
		for (int i =0; i < nodes.size(); i++)
		{
			if (nodes[i].contains(arc->start->name))
				startN = i;
			if (nodes[i].contains(arc->finish->name))
				endN = i;
		}
		// if in different sets then
		if (startN != endN)
		{
			nodes[startN].unionWith(nodes[endN]);
			nodes.removeAt(endN);
			theJoiningArcs.add(arc);
//			cout << "Cost : " << arc->cost << " : Start : " << arc->start->name << " : Finish : " << arc->finish->name << endl;
		} /*else
			cout << "Cost : " << arc->cost << " : Start : " << arc->start->name << " : Finish : " << arc->finish->name << " ( not needed) " << endl;*/
	}
	// draw out all of the arcs on the grid in a dim color
	DrawNodes(thePath.getNodeSet(), DIM_COLOR);	
	// draw the minimum arcs in a highlight color
	DrawArcs(theJoiningArcs, HIGHLIGHT_COLOR);
}

Graphs

The graphs use two algorithms to either find the shortest path (Dijkstra algorithm) or the minimal spanning tree (Krushal’s method).

I am going to split up the assignment into 2 posts so that can try and covert as much of the assignment without spanning a massive post, so in this post going to talk about loading in the data from the textual files and also displaying the data on screen (graphics), and how I altered the sublcass to allow for more data to be stored for this assignment. In the next post it will be about the Path class implementation with the Dijkstra and Krushal algorithms.

So to start off with, I altered the type definition for the main/parent class of Graph to be called PathfinderGraphMain so that when I created the subclass I could call it the PathfinderGraph, just liked the look of that :). and since the graph types, nodeT / arcT are capable of holding all of the other data (have included them in the code below) I just need to store the graph image which is what the subclass below is doing for me, was trying to avoid using any global variables, so that is all that my subclass is additional storing and accessing using the get/set functions.

typedef Graph<nodeT, arcT> PathfinderGraphMain;
class PathfinderGraph : public PathfinderGraphMain
{
public:
	string returnJpg() { return jpgimage;}
	void setJpg(string jpg) { jpgimage = jpg;}
 
private:
	string jpgimage;
};
 
// as taken from the graphtypes.h
struct nodeT {
	string name;
	Set<arcT *> arcs;
	pointT loc;
};
struct arcT {
	nodeT *start, *finish;
	double cost;
};

here is the sub part of the method that will read in the data from the textual file, I am basically reading in the nodes to start with and then calling the AddNode function (below the code) and also the AddArc (again further down in the code below) which basically just call the main Graph class addNode / addArc functions.

if (contents=="NODES")
	{
		// end at EOF or ARCS
		while (contents != "ARCS" && !inFile.eof())
		{
			contents = ReadLineFromFile(inFile);
			// add the node into 
			if (contents != "ARCS")
			{
				// need to split the contents string
				pointT location;
				int nameI = contents.find_first_of(" ");
				location.x = StringToInteger(contents.substr(nameI, contents.find_last_of(" ")- nameI));
				location.y = StringToInteger(contents.substr(contents.find_last_of(" ")));
				AddNode(contents.substr(0,nameI), location, thePath);
			}
		}
	}
	// end point is the word ARCS
	if (contents == "ARCS")
	{
		// read till the end
		while (!inFile.eof())
		{
			contents = ReadLineFromFile(inFile);
			if (!inFile.eof())
			{
				string startP, endP;
				int nameI = contents.find_first_of(" ");
				startP = contents.substr(0, nameI);
				nameI++;
				endP = contents.substr(nameI, contents.find_last_of(" ")-nameI);
				double cost = StringToReal(contents.substr(contents.find_last_of(" ")));
				AddArc(startP, endP, cost, thePath);
			}
		}
	}
 
void AddNode(string name, pointT location, PathfinderGraph &thePath)
{
	nodeT *node = new nodeT();
	node->name = name;
	node->loc = location;
	thePath.addNode(node);
}
 
void AddArc(string startP, string endP, double cost, PathfinderGraph &thePath)
{
	nodeT *start = thePath.getNode(startP);
	nodeT *end = thePath.getNode(endP);
 
	arcT *arc = new arcT();
	arc->cost = cost;
	arc->start = start;
	arc->finish = end;
	thePath.addArc(arc);
 
	arcT *arc2 = new arcT();
	arc2->cost = cost;
	arc2->start = end;
	arc2->finish = start;
	thePath.addArc(arc2);
}

To draw to the screen, I have created two methods that will take a set of nodes (nodeT structures) and also the colour that is required to display. Since the Set class has a Iterator I can use the foreach (marco from the CS106B libraries) to loop over the Set class to then just display the nodes/arcs via the CS106B graphics libraries again.

void DrawNodes(Set<nodeT *> nodes, string color)
{	
	foreach (nodeT *node in nodes)
	{
		DrawPathfinderNode(node->loc, color, node->name);
		DrawArcs(node->arcs, color);
	}
	UpdatePathfinderDisplay();
}
 
void DrawArcs(Set<arcT *> arcs, string color)
{
	foreach (arcT *arc in arcs)
		DrawPathfinderArc(arc->start->loc, arc->finish->loc, color);
}

The last part of the graphical aspects is getting the mouse inputs to link to a node on the graphical display, once again using the CS106B libraries for obtaining a point location (x/y coordinates) and then testing to see if this coordinates have clicked on the node location (kinder putting a square around the node of the node radius in size, this NODE_RADIUS is defined in the CS106B libraries and that is what the node will have a circle radius off)

// find the nodes near the mouse cords
// NULL if not found a node, else return a nodeT
nodeT * GetNodeFromMouseCords(PathfinderGraph &thePath, pointT mouseInput)
{
	nodeT * returnNode;
	foreach (nodeT *node in thePath.getNodeSet())
	{
		pointT location = node->loc;
		if (mouseInput.x >(location.x - NODE_RADIUS) && 
			mouseInput.x <(location.x + NODE_RADIUS))
			if (mouseInput.y > (location.y - NODE_RADIUS) &&
				mouseInput.y < (location.y + NODE_RADIUS))
				return node;
	}
	return NULL;
}
 
// need to pull out two places, start and end.
void GetTwoLocationsFromMouse(PathfinderGraph &thePath, nodeT * &startN, nodeT * &finishN)
{
	pointT mouseClick = pointT();
	cout << "Please choice your starting location "<< endl;
	while ((startN = GetNodeFromMouseCords(thePath, mouseClick)) == NULL)
		mouseClick = GetMouseClick();
	cout << "Start : " << startN->name << endl;
	DrawPathfinderNode(startN->loc, HIGHLIGHT_COLOR, startN->name);
	while (true)
	{
		cout << "Please choice your ending location "<< endl;
		mouseClick = pointT();
		while ((finishN = GetNodeFromMouseCords(thePath, mouseClick)) == NULL)
			mouseClick = GetMouseClick();
		if (startN->name == finishN->name)
			cout << "That is silly, you are already there!"<< endl;
		else
			break;
	}
	cout << "Finish : " << finishN->name << endl;
	DrawPathfinderNode(finishN->loc, HIGHLIGHT_COLOR, finishN->name);
}

after the user has selected two nodes it will highlight the nodes with the HIGHLIGHT_COLOR from the CS106B libraries again (which was the colour red). I shall post next on the Path class and also the two breath-search and minimal spanning trees, (Dijkstra and Krushal’s algorithms).

Huffman coding – part 2

The huffman coding is mainly used to compress files that are not already compressed already ( the reason why I say this for is because if you are trying to compress a already compressed file then the assignment 5 will add on more header details onto the file for decompressing the compressed file.

This is a follow on from the part 1 of the assignment 5 this is the decompression of the compressed file.

So to start with, I need to read in the header file of the compressed file. For example if I was going to compress the file

hello there

then the compressed file will be

104=2|101=3|108=2|111=1|32=1|116=1|114=1|-1=1|;u

Huffman coding – part 1

The basics of this assignment, I have included the assignment PDF to the attached file, is to read in the file and place the characters that are read in (1 character at a time) into a mapped of characters that have there associated value of how many times that there are present in the file, so that the most present characters in the file will end up with a very short bit patten in the compressed file (this is how the file will be smaller, e.g. less space taken up on the disk per character within the file) lets say that there is a sentence of “hi there”, then e will be the most used value e.g. 2, so a character is allot bitter in a bit pattern (to store all of the available characters e.g. a-z, smiles etc.) than a bit patten of 2 in length (01) if that was the case.

So to get started this post is going to be doing the compression of the file, and the second post will be the decompression of the file.

Here is how I am reading in that mapped of characters to there occurrences in the file, theMap is a vector of queueI structure, so that we know that will always have the same values inserted into a vector array, thus when it comes to decompressing the file we want to build up the huffman encoding with the same input values.

   struct queueI {
	   int letter;
	   int value;
   };
 
void Encoding::readFile(ibstream &inStream)
{
	theMap.clear();
	char *readIn = new char(1);
	int i;
	queueI insert;
	while (!inStream.eof())
	{
		inStream.read(readIn,1);
		if (!inStream.eof())
		{
			for (i = 0; i < theMap.size(); i++)
			{
				if (theMap[i].letter == readIn[0]) 
					break;
			}
			if (i < theMap.size())
				theMap[i].value++;
			else
			{
				insert.letter = readIn[0];
				insert.value = 1;
				theMap.add(insert);
			}
		}
	}
	// add the EOF
	insert.letter = PSEUDOEOF;
	insert.value = 1;
	theMap.add(insert);
	delete readIn;
}

The last value is the end of file, a pseudo value that will denote that we have reached the end of file value, I then will need to build up a queue(as taken from the last assignment 4 of this course) of the values taken in from the build of theMap above, so to start with I just load all of the values into queue with there order value (value equals the number of occurrences that they are present in the file) and this is how I am loading the values into the queue.

void Encoding::loadPQueue()
{
	theQueue.clear();
	queueT *qt;
	for (int i =0 ; i < theMap.size(); i++)
	{
		qt = new queueT();
		qt->letter = theMap[i].letter; 
		qt->value = theMap[i].value;
		qt->left = NULL;
		qt->right = NULL;
		qt->filled = true;
		theQueue.add(qt, theMap[i].value);
	}	
}

and here is how I build up the huffman encoding list, what it does is to take the lowest two values (the lowest two occurrences, or later on the lowest two joined values) and put them together with a new node that points to them both and takes the value of both of the values added together, so lets say that the character a has a value of 2 and the letter c has the value of 3, then the new node will have a value of 5 and points to both the characters a,c on the left/rights nodes accordlying.

void Encoding::buildTree()
{
	queueT *newQ, *first, *second;
	int valueIn;
	while (!theQueue.isEmpty())
	{
		newQ = new queueT();
		first = theQueue.extractMin();
		valueIn =first->value;
		newQ->left = first;
		if (!theQueue.isEmpty())
		{
			second = theQueue.extractMin();
			valueIn += second->value;
			newQ->right = second;
		}
		newQ->value = valueIn;
		newQ->filled = false;
		if (!theQueue.isEmpty())
			theQueue.add(newQ, valueIn);
	}
	theQueue.add(newQ, valueIn);
}

right that the main part done!, since when we decompress the file we want to have the same order of the loading into the queue we need to write out the header details to the new decompressed file,

void Encoding::outHeaderToFile(obstream &outStream)
{
	string outStr;
	for (int i =0; i < theMap.size(); i++)
	{
		outStr = IntegerToString(theMap[i].letter);
		outStr +="=";
		outStr += IntegerToString(theMap[i].value);
		outStr +="|";
		outStream.write(outStr.c_str(),outStr.length());	
	}
	// end of header details
	outStr = ";";
	outStream.write(outStr.c_str(), 1);
}

the ; is the end of the header details, thus after that we will write out the bit patterns of the compressed characters :). the output of the header details is the character code=number of occurrences so for character a with 10 occurrences then it would output 97=10|

So since we now need to write out the bit patterns of the characters to the file, we need to find out how we placed those characters into the huffman tree, so I wrote a recursive function that will output the bit pattern of where the character is in the huffman tree

string Encoding::printQueue(queueT *node, int letter,string val)
{
	string returnVal;
	if (node != NULL)
	{
		if (node->left != NULL)
		{
			returnVal = printQueue(node->left, letter, val+"0");
			if (returnVal != "")
				return returnVal;
		}
		if (node->filled)
			if (node->letter == letter)
				return val;
		if (node->right != NULL)
		{
			returnVal = printQueue(node->right, letter, val+"1");
			if (returnVal != "")
				return returnVal;
		}
	}
	return "";
}

If we are heading down the left node, then add a 0 to the bit pattern, if we have found the character and it is classed as a filled node (e.g. only nodes that have a character attached to them are filled) then return how we found that character, or try the right node with adding a 1 to the bit pattern.

Then to just write out the uncompressed file to the compressed file we just go through the file again and then write out all of the bit patterns with ending with the pseudo EOF character.

void Encoding::outToFile(ibstream &inStream, obstream &outStream)
{
	queueT *head;
	string foundStr;
	char *readIn = new char(2);
	while (!inStream.eof())
	{
		inStream.read(readIn,1);
		if (!inStream.eof())
		{
			head = theQueue.peek();
			foundStr = printQueue(head, readIn[0], "");
			for (int i =0; i < foundStr.length(); i++)
			{
				int ch = ((int)foundStr[i])-48;
				outStream.writebit(ch);
			}
		}
	}
	// add in the pseudo-EOF
	head = theQueue.peek();
	foundStr = printQueue(head,PSEUDOEOF, "");
	for (int i =0; i < foundStr.length(); i++)
	{
		int ch = ((int)foundStr[i])-48;
		outStream.writebit(ch);
	}
}

that is about it, I shall do a post on how I do the uncompress-ion next.