Edit: The exit of the maze is only found when there is an empty spot without any walls along the borders of the maze.
Edit2: I've changed the code a bit after receiving the help, but still encountered the same issue.
Could be that I did not use what was mentioned properly.
I've changed the snippet of my code to just one recursion (the moving up portion) to minimize unnecessary repetition when viewing the snippet.
(Still kind of new to coding, so hope you don't mind)
I am attempting to code for a program that prints the movement of a person finding a path through a given maze.
However, I've encountered a problem where after reaching the exit of the maze, the person walks back to the original position where he is at.
The original position is given by an (x, y) coordinate.
Here's a snippet of my function which finds the path through the maze:
bool find_path(char **maze, long **track, long rows, long cols, long x, long y, long steps)
{
bool found;
if (goal_found(maze, rows, cols, x, y)) {
return true;
}
track[y][x] = 2;
if (maze[y - 1][x] == EMPTY && track[y - 1][x] == 1) { // Move Up
steps++;
found = find_path(maze, track, rows, cols, x, y - 1, steps);
if (found) {
return true;
}
++steps;
}
}
Your program finds all paths through the maze; if you want to stop it, you need to change your invocations to stop traversing the maze if you hit success:
found = search_maze(maze, track, rows, cols, x, y + 1, steps);
swap(maze, y, x, y + 1, x);
if (found) {
return found;
}
At all of the recursive invocations.
Related
void put_mandelbrot(int x_left, int x_right, int y_top, int y_bot)
{
pthread_t workers[10];
t_param inputs[10];
int i;
int x;
int y;
i = -1;
y = y_top - 1;
while (++y < y_bot)
{
x = x_left - 1;
while (x < x_right)
{
if (++i < 10)
{
inputs[i] = (t_param){++x, y};
pthread_create(&workers[i], 0, (void *(*)(void *))mandelbrot_thread, (void *)&inputs[i]);
}
else
while (--i > -1)
pthread_join(workers[i], 0);
}
while (--i > -1)
pthread_join(workers[i], 0);
// calling draw_image() here will get the correct result, drawn line by line
}
// if I don't call draw_image() above and for example want to draw
// the whole image here after all of the lines are iterated
// the program will become unresponsive
}
If I have the "draw_image()" function call in between each line, I get the image line by line and the program finishes it (I have checked this). On the other hand, if I only call the "draw image()" function after all of the iterations for all points (x, y) have been completed, the program never draws a single pixel and is unresponsive.
The "mandelbrot_thread" function simply iterates over a pixel (x, y) and sets the color of the pixel to a global image buffer which is always drawn in its entirety with the "draw_image()" call.
Why is this? I can't draw the image after every line as that makes the program insanely slow and makes no sense anyway. At the moment it seems to me that pthreads is working only when I'm actively checking it, almost as if to hide from any debugging attempts.
(edit: my mac also makes a very disturbing sound when I run this program, so there must be something very wrong with this)
I am creating a scrolling shooter for DMG using gbdk, it is based off some youtube tutorials and this example. In fact the link is the skeleton of my program. My issue is that the screen boundary conditions aren't working properly for down and right inputs. For up and left, they work correctly however, and the code for those is basically the exact same. I have also compiled the code from the link above, and it works correctly there. Apologies in advance, I have a childish sense of humor, so the game is penis-based.
The main differences between the skeleton code and mine is that I use a meta-sprite for the player, and an array for the x and y coordinates of the player. I have tried using individual integers for the locations and changing the bounds of the screen, but nothing seems to work.
#include <gb/gb.h>
#include <stdio.h>
#include "gameDicks.c"
#include "DickSprites.c"
#define SCREEN_WIDTH 160
BOOLEAN ishard = TRUE, playing = TRUE;
struct gameDicks flacid;
struct gameDicks hard;
INT8 spritesize = 8, dicklocation[2] = {20, 80};
int i;
void moveGameDicks(struct gameDicks* Dick, UINT8 x, UINT8 y){
move_sprite(Dick->spriteids[0], x, y);
move_sprite(Dick->spriteids[1], x + spritesize, y);
move_sprite(Dick->spriteids[2], x, y + spritesize);
move_sprite(Dick->spriteids[3], x + spritesize, y + spritesize);
}
void setuphard(INT8 dicklocation[2]){
hard.x = dicklocation[0];
hard.y = dicklocation[1];
hard.width = 16;
hard.height = 16;
//load sprites
set_sprite_tile(0,0);
hard.spriteids[0] = 0;
set_sprite_tile(1,1);
hard.spriteids[1] = 1;
set_sprite_tile(2,2);
hard.spriteids[2] = 2;
set_sprite_tile(3,3);
hard.spriteids[3] = 3;
}
void init_screen()
{
SHOW_BKG;
SHOW_SPRITES;
DISPLAY_ON;
}
void init_player()
{
SHOW_SPRITES;
set_sprite_data(0, 8, DickSprites);
setuphard(dicklocation);
}
void input()
{
if (joypad() & J_UP && dicklocation[1])
{
if (dicklocation[1] <= 16){
dicklocation[1] = 16;
}
else{
dicklocation[1]--;
}
}
if (joypad() & J_DOWN && dicklocation[1])
{
if (dicklocation[1] >= 150){
dicklocation[1] = 150;
}
else{
dicklocation[1]++;
}
}
}
void update_sprites()
{
moveGameDicks(&hard, dicklocation[0], dicklocation[1]);
}
int main()
{
init_screen();
init_player();
init_screen();
while(playing)
{
wait_vbl_done(2);
input();
update_sprites();
}
return 0;
}
What I expect is to be able to move the player up to y = 16, and down to y = 150. When it hits these values, it stops moving until you go the other direction. Instead, what I see happen is that the up direction works as expected, but as soon as the down key is pressed - no matter the y-location - the player is immediately sent to the bottom of the screen. From there, pressing up sends it to the very top. Further, the player can only move from the top position to the bottom, and not scroll in between. I'm baffled by this because the conditions are the exact same (except for the y-values), so I don't understand why they behave so differently.
Using an unsigned int may help here as an 8-bit integer will only hold values from -128 to 127, which might cause undefined behaviour when you compare it with over 150, pushing it to a negative value?
You have defined dicklocation as an INT8, when it would be better as a UINT8 or even longer if you plan on ever having a screen size larger than 255 bytes.
I'm working a simple candy crush game for my year 1 assignment.
I am at this stage where I need to show my self-made simple marker( *box made of '|' and '_'* ) on the center of the board ( board[5][5] ) once the program is executed.
Here is the current code:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
//FUNCTION: Draw the Board
int drawBoard()
{
//Declare array size
int board[9][9];
//initialize variables
int rows, columns, randomNumber, flag;
//random number seed generator
srand(time(NULL));
for ( rows = 0 ; rows < 9 ; rows++ )
{
for ( columns = 0 ; columns < 9 ; columns++ )
{
flag = 0;
do
{
//generate random numbers from 2 - 8
randomNumber = rand() %7 + 2;
board[rows][columns] = randomNumber;
//Checks for 2 adjacent numbers.
if ( board[rows][columns] == board[rows - 1][columns] || board[rows][columns] == board[rows][columns - 1] )
{
flag = 0;
continue;
}
else
{
flag = 1;
printf( " %d ", board[rows][columns] );
}
} while ( flag == 0 );
}//end inner for-loop
printf("\n\n");
}//end outer for-loop
//call FUNCTION marker() to display marker around board[5][5]
marker( board[5][5] );
}//end FUNCTION drawBoard
//FUNCTION: Mark the surrounding of the number with "|" and "_" at board[5][5]
void marker( int a )
{
printf( " _ \n" );
printf( "|%c|\n", a );
printf( " _ \n" );
}
int main()
{
drawBoard();
}
At the end of function drawBoard(), I placed the code marker( board[5][5] ).
This should have printed the markers around the number printed at coordinate board[5][5]..but for some reason it displays right after the board has been printed.
So why doesn't it print at that coordinate although I specified it at board[5][5]?
What could be the problem here?
so in your marker function you need to pass the board and the coordinate you want to print at
void marker( int x, int y, int** board )
{
board[x][y-1]="_";
board[x-1][y]="|";
board[x+1][y]="|";
board[x][y+1]="_";
}
then after the call to marker(5,5,board), call drawboard again
my code's a bit off but that's the logic, except you need to check for the case that the marker is at the edge of the board
in other words, you need to keep board around, and any time you make a change to it, clear the screen and print the whole board out again.
There is no persistent drawing in the way that you are doing this. You are just printing straight to the shell/command prompt. The way that you trying to do things will not work. You can't edit something drawn to the prompt after you have drawn it, you need to basically clear the screen and then draw again but with your indicated maker.
I don't know if you are able to use libraries in your assignment, but a very good library that WILL let you do is ncurses
EDIT Full rewrite of answer
Drawing Things On Top of One Another In CMD
Alright, I had some downtime at work, so I wrote a project to do what you need and I'm going to post code and explain what it does and why you need it along the way.
First thin that you are going to need a basically a render buffer or a render context. Whenever you are programming in a graphics API such as OpenGL, you don't just render straight to the screen, you render each object that you have to a buffer that rasterizes your content and turns it into pixels. Once it's in that form, the API shoves the rendered picture onto the screen. We are going to take a similar approach where instead of drawing to a pixel buffer on the GPU, we are going to draw to a character buffer. Think about each character as a pixel on the screen.
Here is a pastebin of the complete source:
Complete Source of Project
RenderContext
Our class to do this will be the RenderContext class. It has fields to hold width and height as well as an array of chars and a special char that we fill our buffer with whenever we clear it.
This class simply holds an array and functions to let us render to it. It makes sure that when we draw to it, we are within bounds. It is possible for an object to try to draw outside of the clipping space (off screen). However, whatever is drawn there is discarded.
class RenderContext {
private:
int m_width, m_height; // Width and Height of this canvas
char* m_renderBuffer; // Array to hold "pixels" of canvas
char m_clearChar; // What to clear the array to
public:
RenderContext() : m_width(50), m_height(20), m_clearChar(' ') {
m_renderBuffer = new char[m_width * m_height];
}
RenderContext(int width, int height) : m_width(width), m_height(height), m_clearChar(' ') {
m_renderBuffer = new char[m_width * m_height];
}
~RenderContext();
char getContentAt(int x, int y);
void setContentAt(int x, int y, char val);
void setClearChar(char clearChar);
void render();
void clear();
};
The two most important functions of this class are setContentAt and render
setContentAt is what an object calls to fill in a "pixel" value. To make this a little more flexible, our class uses a pointer to an array of chars rather than a straight array (or even a two dimensional array). This lets us set the size of our canvas at runtime. Because of this, we access elements of this array with x + (y * m_width) which replaces a two dimensional dereference such as arr[i][j]
// Fill a specific "pixel" on the canvas
void RenderContext::setContentAt(int x, int y, char val) {
if (((0 <= x) && (x < m_width)) && ((0 <= y) && (y < m_height))) {
m_renderBuffer[(x + (y * m_width))] = val;
}
}
render is what actually draws to the prompt. All it does is iterate over all the "pixels" in it's buffer and place them on screen and then moves to the next line.
// Paint the canvas to the shell
void RenderContext::render() {
int row, column;
for (row = 0; row < m_height; row++) {
for (column = 0; column < m_width; column++) {
printf("%c", getContentAt(column, row));
}
printf("\n");
}
}
I_Drawable
Our next class is an Interface that lets us contract with objects that they can draw to our RenderContext. It is pure virtual because we don't want to actually be able to instantiate it, we only want to derive from it. It's only function is draw which accepts a RenderContext. Derived classes use this call to receive the RenderContext and then use RenderContext's setContentAt to put "pixels" into the buffer.
class I_Drawable {
public:
virtual void draw(RenderContext&) = 0;
};
GameBoard
The first class to implement the I_Drawable, thus being able to render to our RenderContext, is the GameBoard class. This is where a majority of the logic comes in. It has fields for width, height, and a integer array that holds the values of the elements on the board. It also has two other fields for spacing. Since when you draw your board using your code, you have spaces between each element. We don't need to incorporate this into the underlying structure of the board, we just need to use them when we draw.
class GameBoard : public I_Drawable {
private:
int m_width, m_height; // Width and height of the board
int m_verticalSpacing, m_horizontalSpacing; // Spaces between each element on the board
Marker m_marker; // The cursor that will draw on this board
int* m_board; // Array of elements on this board
void setAtPos(int x, int y, int val);
void generateBoard();
public:
GameBoard() : m_width(10), m_height(10), m_verticalSpacing(5), m_horizontalSpacing(3), m_marker(Marker()) {
m_board = new int[m_width * m_height];
generateBoard();
}
GameBoard(int width, int height) : m_width(width), m_height(height), m_verticalSpacing(5), m_horizontalSpacing(3), m_marker(Marker()) {
m_board = new int[m_width * m_height];
generateBoard();
}
~GameBoard();
int getAtPos(int x, int y);
void draw(RenderContext& renderTarget);
void handleInput(MoveDirection moveDirection);
int getWidth();
int getHeight();
};
It's key functions are generateBoard, handleInput, and the derived virtual function draw. However, do note that in its constructor it creates a new int array and gives it to its pointer. Then its destructor automatically removes the allocated memory whenever the board goes away.
generateBoard is what we use to actual create the board and fill it with numbers. It will iterate over each location on the board. Each time, it will look at the elements directly to the left and above and store them. Then it will generate a random number until the number it generates does not match either of the stored elements, then it stores the number in the array. I rewrote this to get rid of the flag usage. This function gets called during the construction of the class.
// Actually create the board
void GameBoard::generateBoard() {
int row, column, randomNumber, valToLeft, valToTop;
// Iterate over all rows and columns
for (row = 0; row < m_height; row++) {
for (column = 0; column < m_width; column++) {
// Get the previous elements
valToLeft = getAtPos(column - 1, row);
valToTop = getAtPos(column, row - 1);
// Generate random numbers until we have one
// that is not the same as an adjacent element
do {
randomNumber = (2 + (rand() % 7));
} while ((valToLeft == randomNumber) || (valToTop == randomNumber));
setAtPos(column, row, randomNumber);
}
}
}
handleInput is what deals with moving the cursor around on the board. It's basically a freebie and your next step after getting the cursor to draw over the board. I needed a way to test the drawing. It accepts an enumeration that we switch on to know where to move our cursor to next. If you maybe wanted to have your cursor wrap around the board whenever you reach an edge, you would want to do that here.
void GameBoard::handleInput(MoveDirection moveDirection) {
switch (moveDirection) {
case MD_UP:
if (m_marker.getYPos() > 0)
m_marker.setYPos(m_marker.getYPos() - 1);
break;
case MD_DOWN:
if (m_marker.getYPos() < m_height - 1)
m_marker.setYPos(m_marker.getYPos() + 1);
break;
case MD_LEFT:
if (m_marker.getXPos() > 0)
m_marker.setXPos(m_marker.getXPos() - 1);
break;
case MD_RIGHT:
if (m_marker.getXPos() < m_width - 1)
m_marker.setXPos(m_marker.getXPos() + 1);
break;
}
}
draw is very important because it's what gets the numbers into the RenderContext. To summarize, it iterates over every element on the board, and draws in the correct location on the canvas placing an element under the correct "pixel". This is where we incorporate the spacing. Also, take care and note that we render the cursor in this function.
It's a matter of choice, but you can either store a marker outside of the GameBoard class and render it yourself in the main loop (this would be a good choice because it loosens the coupling between the GameBoard class and the Marker class. However, since they are fairly coupled, I chose to let GameBoard render it. If we used a scene graph, as we probably would with a more complex scene/game, the Marker would probably be a child node of the GameBoard so it would be similar to this implementation but still more generic by not storing an explicit Marker in the GameBoard class.
// Function to draw to the canvas
void GameBoard::draw(RenderContext& renderTarget) {
int row, column;
char buffer[8];
// Iterate over every element
for (row = 0; row < m_height; row++) {
for (column = 0; column < m_width; column++) {
// Convert the integer to a char
sprintf(buffer, "%d", getAtPos(column, row));
// Set the canvas "pixel" to the char at the
// desired position including the padding
renderTarget.setContentAt(
((column * m_verticalSpacing) + 1),
((row * m_horizontalSpacing) + 1),
buffer[0]);
}
}
// Draw the marker
m_marker.draw(renderTarget);
}
Marker
Speaking of the Marker class, let's look at that now. The Marker class is actually very similar to the GameBoard class. However, it lacks a lot of the logic that GameBoard has since it doesn't need to worry about a bunch of elements on the board. The important thing is the draw function.
class Marker : public I_Drawable {
private:
int m_xPos, m_yPos; // Position of cursor
public:
Marker() : m_xPos(0), m_yPos(0) {
}
Marker(int xPos, int yPos) : m_xPos(xPos), m_yPos(yPos) {
}
void draw(RenderContext& renderTarget);
int getXPos();
int getYPos();
void setXPos(int xPos);
void setYPos(int yPos);
};
draw simply puts four symbols onto the RenderContext to outline the selected element on the board. Take note that Marker has no clue about the GameBoard class. It has no reference to it, it doesn't know how large it is, or what elements it holds. You should note though, that I got lazy and didn't take out the hard coded offsets that sort of depend on the padding that the GameBoard has. You should implement a better solution to this because if you change the padding in the GameBoard class, your cursor will be off.
Besides that, whenever the symbols get drawn, they overwrite whatever is in the ContextBuffer. This is important because the main point of your question was how to draw the cursor on top of the GameBoard. This also goes to the importance of draw order. Let's say that whenever we draw our GameBoard, we drew a '=' between each element. If we drew the cursor first and then the board, the GameBoard would draw over the cursor making it invisible.
If this were a more complex scene, we might have to do something fancy like use a depth buffer that would record the z-index of an element. Then whenever we drew, we would check and see if the z-index of the new element was closer or further away than whatever was already in the RenderContext's buffer. Depending on that, we might skip drawing the "pixel" altogether.
We don't though, so take care to order your draw calls!
// Draw the cursor to the canvas
void Marker::draw(RenderContext& renderTarget) {
// Adjust marker by board spacing
// (This is kind of a hack and should be changed)
int tmpX, tmpY;
tmpX = ((m_xPos * 5) + 1);
tmpY = ((m_yPos * 3) + 1);
// Set surrounding elements
renderTarget.setContentAt(tmpX - 0, tmpY - 1, '-');
renderTarget.setContentAt(tmpX - 1, tmpY - 0, '|');
renderTarget.setContentAt(tmpX - 0, tmpY + 1, '-');
renderTarget.setContentAt(tmpX + 1, tmpY - 0, '|');
}
CmdPromptHelper
The last class that I'm going to talk about is the CmdPromptHelper. You don't have anything like this in your original question. However, you will need to worry about it soon. This class is also only useful on Windows so if you are on linux/unix, you will need to worry about dealing with drawing to the shell yourself.
class CmdPromptHelper {
private:
DWORD inMode; // Attributes of std::in before we change them
DWORD outMode; // Attributes of std::out before we change them
HANDLE hstdin; // Handle to std::in
HANDLE hstdout; // Handle to std::out
public:
CmdPromptHelper();
void reset();
WORD getKeyPress();
void clearScreen();
};
Each one of the functions is important. The constructor gets handles to the std::in and std::out of the current command prompt. The getKeyPress function returns what key the user presses down (key-up events are ignored). And the clearScreen function clears the prompt (not really, it actually moves whatever is already in the prompt up).
getKeyPress just makes sure you have a handle and then reads what has been typed into the console. It makes sure that whatever it is, it is a key and that it is being pressed down. Then it returns the key code as a Windows specific enum usually prefaced by VK_.
// See what key is pressed by the user and return it
WORD CmdPromptHelper::getKeyPress() {
if (hstdin != INVALID_HANDLE_VALUE) {
DWORD count;
INPUT_RECORD inrec;
// Get Key Press
ReadConsoleInput(hstdin, &inrec, 1, &count);
// Return key only if it is key down
if (inrec.Event.KeyEvent.bKeyDown) {
return inrec.Event.KeyEvent.wVirtualKeyCode;
} else {
return 0;
}
// Flush input
FlushConsoleInputBuffer(hstdin);
} else {
return 0;
}
}
clearScreen is a little deceiving. You would think that it clears out the text in the prompt. As far as I know, it doesn't. I'm pretty sure it actually shifts all the content up and then writes a ton of characters to the prompt to make it look like the screen was cleared.
An important concept that this function brings up though is the idea of buffered rendering. Again, if this were a more robust system, we would want to implement the concept of double buffering which means rendering to an invisible buffer and waiting until all drawing is finished and then swap the invisible buffer with the visible one. This makes for a much cleaner view of the render because we don't see things while they are still getting drawn. The way we do things here, we see the rendering process happen right in front of us. It's not a major concern, it just looks ugly sometimes.
// Flood the console with empty space so that we can
// simulate single buffering (I have no idea how to double buffer this)
void CmdPromptHelper::clearScreen() {
if (hstdout != INVALID_HANDLE_VALUE) {
CONSOLE_SCREEN_BUFFER_INFO csbi;
DWORD cellCount; // How many cells to paint
DWORD count; // How many we painted
COORD homeCoord = {0, 0}; // Where to put the cursor to clear
// Get console info
if (!GetConsoleScreenBufferInfo(hstdout, &csbi)) {
return;
}
// Get cell count
cellCount = csbi.dwSize.X * csbi.dwSize.Y;
// Fill the screen with spaces
FillConsoleOutputCharacter(
hstdout,
(TCHAR) ' ',
cellCount,
homeCoord,
&count
);
// Set cursor position
SetConsoleCursorPosition(hstdout, homeCoord);
}
}
main
The very last thing that you need to worry about is how to use all these things. That's where main comes in. You need a game loop. Game loops are probably the most important thing in any game. Any game that you look at will have a game loop.
The idea is:
Show something on screen
Read input
Handle the input
GOTO 1
This program is no different. The first thing it does is create a GameBoard and a RenderContext. It also makes a CmdPromptHelper which lets of interface with the command prompt. After that, it starts the loop and lets the loop continue until we hit the exit condition (for us that's pressing escape). We could have a separate class or function do dispatch input, but since we just dispatch the input to another input handler, I kept it in the main loop. After you get the input, you send if off to the GameBoard which alters itself accordingly. The next step is to clear the RenderContext and the screen/prompt. Then rerun the loop if escape wasn't pressed.
int main() {
WORD key;
GameBoard gb(5, 5);
RenderContext rc(25, 15);
CmdPromptHelper cph;
do {
gb.draw(rc);
rc.render();
key = cph.getKeyPress();
switch (key) {
case VK_UP:
gb.handleInput(MD_UP);
break;
case VK_DOWN:
gb.handleInput(MD_DOWN);
break;
case VK_LEFT:
gb.handleInput(MD_LEFT);
break;
case VK_RIGHT:
gb.handleInput(MD_RIGHT);
break;
}
rc.clear();
cph.clearScreen();
} while (key != VK_ESCAPE);
}
After you have taken into consideration all of these things, you understand why and where you need to be drawing your cursor. It's not a matter of calling a function after another, you need to composite your draws. You can't just draw the GameBoard and then draw the Marker. At least not with the command prompt. I hope this helps. It definitely alleviated the down time at work.
This is the code for dfs.
bool processed[MAXV+1]; /* which vertices have been processed */
bool discovered[MAXV+1]; /* which vertices have been found */
int parent[MAXV+1]; /* discovery relation */
#define MAXV 1000 /* maximum number of vertices */
typedef struct {
int y; /* adjacency info */
int weight; /* edge weight, if any */
struct edgenode *next; /* next edge in list */
} edgenode;
typedef struct {
edgenode *edges[MAXV+1]; /* adjacency info */
int degree[MAXV+1]; /* outdegree of each vertex */
int nvertices; /* number of vertices in graph */
int nedges; /* number of edges in graph */
bool directed; /* is the graph directed? */
} graph;
dfs(graph *g, int v)
{
edgenode *p; /* temporary pointer */
int y; /* successor vertex */
if (finished) return; /* allow for search termination */
discovered[v] = TRUE;
time = time + 1;
entry_time[v] = time;
process_vertex_early(v);
p = g->edges[v];
while (p != NULL) {
y = p->y;
if (discovered[y] == FALSE)
{
parent[y] = v;
process_edge(v,y);
dfs(g,y);
}
else if ((!processed[y]) || (g->directed))
process_edge(v,y);
if (finished) return;
p = p->next;
}
process_vertex_late(v);
time = time + 1;
exit_time[v] = time;
processed[v] = TRUE;
}
And for finding the cycles it has modified the process edge function like below
process_edge(int x, int y)
{
if (parent[x] != y) { /* found back edge! */
printf("Cycle from %d to %d:",y,x);
find_path(y,x,parent);
printf("\n\n");
finished = TRUE;
}
}
Now imagine a small tree with just 2 leaf nodes and one root.
When this tree is subjected to this function, I believe it will say that it has cycles.
which is wrong !!
Please correct me if i am wrong.
Thanks.
From the errata corrige, at http://www.cs.sunysb.edu/~skiena/algorist/book/errata:
(*) Page 173, process_edge procedure -- the correct test should be
if (discovered[y] && (parent[x] != y)) { /* found back edge */
I think you're right, and the code is wrong.
It looks to me like the problem is if (parent[x] != y) in process_edge(). In both calls to process_edge(), the supposed parent is passed before the supposed child, i.e. inside process_edge() we expect x to be the parent, so I think that line should be if (parent[y] != x).
Sadly, I think this dfs code is just wrong. It's been a long time since I've studied this stuff in detail, but I think it's clear that the code simply doesn't do what he says it does.
The find cycle code is correct (with the change shown in the errata, as noted by Antonio).
The main problem is that the find cycle routine is in process_edge, but he doesn't process edges to previously discovered nodes! So how will he find the cycle?! If you're interested in cycles (or back edges for any reason), I should think you MUST process all edges.
If you're not interested in back edges and want to avoid processing them, then the code as presented is correct.
It's brilliantly ironic that in the passage immediately preceding the Finding Cycles section in the text he writes:
I find that the subtlety of depth-first search-based algorithms kicks me in the head whenever I try to implement one.
You don't say! :P
The while loop should look something like this:
...
while (p != NULL) {
y = p.y;
process_edge(v,y);
if (discovered[y] == FALSE) {
parent[y] = v;
dfs(g,y);
}
if (finished) {
return;
}
p = p.next;
}
...
In undirected graphs there are only tree and back edges (no forward or cross edges) . The sufficient condition it to be a tree edge is
if (discovered[y] == FALSE)
But in case of the vertices which have been discovered there is a chance that the undirected edge goes back to its parent (instead of its ancestor) . To prevent this case , another condition is added :
if (discovered[y]==TRUE && (parent[x] != y))
For example : y has been discovered by x (say).When the recursive algorithm moves to vertex y (child of x) , in that case vertex X is now vertex y and vertex Y is parent(y) (or x!). If the second condition is removed there is a possibility that the algorithm detects the edge going from child(y) to its parent(x) (in an undirected graph ) as an edge going back to one of the former ancestors which is completely wrong .
I was solving the N Queen problem where we need to place 4 queens on a 4 X 4 chess board such that no two queens can attack each other. I tried this earlier but my approach did not involve backtracking, so I was trying again. The code snippets are
int size=4,i,j;
int arr[4][4];
int lastjindex[4]; // to store the last location which we may need to backtrack
void placeQueen(int i,int j)
{
int availableornot=0;
for(j=0;j<size;j++)
{
if(isAvailable(i,j)==1)
{
availableornot=1;
break;
}
}
if(availableornot==1)
{
arr[i][j]=1;
lastjindex[i]=j;
if((i+1)!=size)
{
placeQueen(i+1,0);
}
}
else
{
// no column was availabe so we backtrack
arr[i-1][lastjindex[i-1]]=0;
placeQueen(i-1,lastjindex[i-1]+1);
}
}
The isAvailable() method returns 1 if arr[i][j] is not under attack, else it returns 0.
int isAvailable(int i,int j)
{
int m,n,flag=0;
for(m=0;m<i;m++)
{
for(n=0;n<size;n++)
{
int k=abs(i-m);
int l=abs(j-n);
if(arr[m][j]==0 || arr[k][l]==0)
{
flag=1;
break;
// means that spot is available
}
}
}
return flag;
}
I call the above method from main as
placeQueen(0,0);
My program compiles successfully but it prints all zeroes.
Is there any problem with my recursion? Please help me correct my code as I am trying to learn how to implement backtracking algorithms!
Also I am not able to decide the base condition to end recursion. How do I choose it here?
There's no printing in the code you posted. If you print after you have backtracked, you will be back to the initial condition of no queens on the board. Print after you have placed N queens, which is also the end condition for recursion. If you only want to print one solution, exit after printing, or set a flag that tells the caller that you're done so you pop all the way out. If you print all solutions, that will include reflections and rotations. You can eliminate one axis of reflection by only placing queens within size/2 in the first level.
Also, there are some clear logic errors in you code, such as
arr[m][j]==0 || arr[k][l]==0
A queen can only be placed if it isn't attacked on the file and it isn't attacked along a diagonal. Use a debugger or add printfs to your code to trace where it is trying to place queens -- that will help you figure out what it is doing wrong.
And aside from being wrong, your isAvailable is very inefficient. You want to know if the [i,j] square is attacked along the file or a diagonal. For that you should have a single loop over the rows of the previous queens for (m = 0; m < i; m++), but you only need three tests, not a loop, to check the file and the diagonals. As soon as you find any previous queen on a file or diagonal, you're done, and the square isn't available -- return false. (And ignore people who tell you that a function should only have one return -- they are wrong, and there are lengthly discussions here at SO and even scientific studies of error rates in code that bear this out.) Only if no previous queen is found is the square available.
Your placeQueen is also wrong. For each available square on a row, you need to place a queen and then recurse, but you're just finding the first available square. And backtracking is achieved simply by removing the queen you placed and then returning ... the previous level of placeQueen will try the next available spot.
Again, trace the code to see what it's doing. And, even more importantly, think through the logic of what is needed. Write your algorithm in words, convince yourself that it will solve the problem, then write the code to carry out the algorithm.
#include <stdio.h>
#define SIZE 4
int size=SIZE;
int arr[SIZE][SIZE] = { 0 };
void placeQueen(int col){
int r,c;
if(col == size){//all queen put!
//print out
for(r = 0;r<size;++r){
for(c = 0;c<size;++c)
printf("%d", arr[c][r]);
printf("\n");
}
printf("\n");
return ;
}
for(r=0;r<size;++r){
if(isAvailable(col, r)==1){
arr[col][r]=1;
placeQueen(col+1);
arr[col][r]=0;//reset
}
}
}
int isAvailable(int col,int row){
int c;
for(c=0;c<col;++c){
int d = col - c;
if(arr[c][row]==1)
return 0;//queen already same row
if(row+d < size && arr[c][row+d]==1 || row-d >= 0 && arr[c][row-d]==1)
return 0;//queen already same slanting position
}
return 1;
}
int main(){
placeQueen(0);
return 0;
}