Making a Directed Acyclic Graph - c

I'm working on a task, where the final step is to take an array of pairs (where a pair is essentially an edge in a graph) and make an Acyclic graph from it. If a pair happens to create a cycle in the graph, then it should be skipped. The DAG is to be stored as an adjacency matrix (the edges are unweighted, hence it is of type bool matrix[][] )
I tried to implement a modified DFS, based on what I read online. The task is in C, and I'm new to the language, so sorry for the crudeness of the code.
The point is that it doesn't skip the cycle-forming pairs and I'm stuck at this point. Any advice or help is appreciated.
int MAX; //number of nodes in the graph
int player; //a node in the graph
typedef struct
{
int winner;
int loser;
} pair; //a directed edge from player 'winner' to player 'loser'
pair pairs[MAX * (MAX - 1) / 2]; //array of all valid pairs
int main(void)
{
/* Get input from console for:
MAX - desired number of players, <= 9,
. . .
int results[MAX][MAX]; - a table (2D array), where each element
results[A][B] shows the number of wins that player A
has over player B, and vice versa.
Element results[X][X] = 0 always.
A new pair is only added when two players have unequal
number of wins over each other: results[A][B] != results[B][A],
so that pairs[i].winner is the one having more wins than losses
against pairs[i].loser .
pairs[] is then sorted in descending order according to
the value of pairs[i].winner .
The goal is to create another 2D array
bool matrix[MAX][MAX];
by adding each pair in pairs[] sequentially,
so that matrix[][] is the adjacency matrix of a
Directed Acyclic Graph. (a.k.a. if a pair happens to create
a cycle, it must not be added)
*/
DAG();
}
void DAG(void)
{
int from, to;
for (int i = 0; i < pair_count; i++)
{
//Create an edge in graph
from = pairs[i].winner;
to = pairs[i].loser;
matrix[from][to] = true;
//Check if this edge made a cycle
bool visited[MAX];
bool onStack[MAX];
if (cyclicGraph(visited, onStack))
{
matrix[from][to] = false;
}
//Here we should have the DAG in locked
return;
}
bool cyclicGraph(bool visited[], bool onStack[])
{
for (int k = 0; k < MAX; k++)
{
//Run the hasCycle-DFS only from unvisited vertices
if (!visited[k] && hasCycle(k, visited, onStack))
{
//if it forms a cycle,
return true;
}
}
return false;
}
bool hasCycle(int x, bool visited[], bool onStack[])
{
// we push this 'x' node onto the stack
onStack[x] = true;
int child;
for (int i = 0; i < MAX; i++)
{
if (locked[x][i]) //x's children are only i's holding True values in array "locked[x][i]"
{
child = i;
if (onStack[child])
{
return true;
}
if (!visited[child] && hasCycle(child, visited, onStack))
{
return true;
}
}
}
//we pop the 'x' from the stack and mark it as visited
onStack[x] = false;
visited[x] = true;
return false;
}

I went back to this problem after a while and I found the bug. The two arrays bool visited[MAX]; bool onStack[MAX]; holding the information for the nodes being visited or being on the recursion stack during the DFS hadn't been initialized. A simple solution is to initialize them with false values:
memset(visited, false, sizeof visited);
memset(onStack, false, sizeof onStack);
Conclusion: always make sure to initialize your variables.

Related

How do I interact with a grid using a 2D array (Proce55ing)

For a project I need to create Connect Four using Processing, I have created the grid which the game will be played in however I am lost when it comes to players interacting with it.
I have been told to use a 2D array however my knowledge of arrays is very limited. I am currently trying to code the bit where the program detects where a player has clicked and spawning a coin there.
int c = 8;
int r = 10;
int[][] grid = new int [c][r];
int CoinSpawn = -1;
void setup () {
size(1000, 800);
}
void draw () {
background(1, 1, 1);
translate(100, 100);
noStroke();
drawColumns();
drawRows();
}
void keyPressed () {
for (int i=0; i<grid.length-1; i++) {
grid[i][i] = grid[i+1][i+1];
}
}
void drawRows(){
for (int i=0; i < r; i++){
int x = 80;
x = x * i;
translate(x,0);
drawColumns();
translate(-x,0);
}
}
void drawColumns(){
for (int i=0; i < c; i++){
int y = 80;
y = y * i;
translate(0,y);
drawCell();
translate(0,-y);
}
}
void drawCell(){
fill(0,0,255);
rect(0,0,80,80);
noFill();
fill(0);
ellipseMode(CENTER);
translate(40,40);
ellipse(0,0,75,75);
noFill();
translate(-40,-40);
}
Would I be able to assign the 2D array to the grid? so that each slot in the grid represents an element from the array? That is the best option I can see at the moment however I have no idea how to do it.
I really appreciate any replies as I am completely lost at the moment.
You have a pretty good start, I made a lot more changes than I had planned... got a little into it!
Let me know if you have any questions, I did use one simple OOP class called cell that tracks the value and the cell's position, as well as provides a display function, I converted a lot of your variables and hard coded numbers to constants (starts with final and has the same value for the entire program)
you will notice I left the win conditions to you!
I hope this is helpful, I feel like seeing 2D arrays used in a context you are familiar with might help you understand them!
My normal process using 2D arrays for processing:
use Setup() to set initial array values
use Draw() to call each of the item's display function
use other functions to modify the array data, which the display function of the cell knows how to display
Main File
// Grid Size constants
final int GRID_COLUMNS = 10;
final int GRID_ROWS = 8;
// Display constants
final int CELL_SCALE = 80;
final int COIN_RADIUS = 75;
final int BOARD_PADDING = 100;
final color[] PLAYER_COLORS = new color[]{color(0), color(200, 10, 10), color(200, 200, 10)};
// cell values
final int EMPTY_CELL = 0;
final int PLAYER_1 = 1;
final int PLAYER_2 = 2;
// game data
Cell[][] grid = new Cell [GRID_COLUMNS][GRID_ROWS];
int currentPlayer;
void setup () {
size(1000, 800);
ellipseMode(CENTER); // only needs to be set once per sketch
setupBlankBoard();
}
// method to populate the array with blank cells, used to initiate and reset the game
void setupBlankBoard() {
currentPlayer = PLAYER_1; // reset game and start with first player
// populate array
for (int y=0; y < GRID_ROWS; y++) { // for every vertical row,
for (int x=0; x < GRID_COLUMNS; x++) { // for every horizontal cell in that row
// rather than translates I prefer to calculate the actual x,y position and just display it there,
// we add the padding offset to every cell, and then take the column/row times the width of the square cell
grid[x][y] = new Cell(BOARD_PADDING+x*CELL_SCALE, BOARD_PADDING+y*CELL_SCALE);
}
}
}
void changePlayers() {
if (currentPlayer == PLAYER_1) {
currentPlayer = PLAYER_2;
} else {
// already was player 2, so change to 1
currentPlayer = PLAYER_1;
}
}
boolean placeCoin(int column) {
boolean coinPlaced = false;
// now we know the column, we need to find the lowest cell in that column that is empty
for (int y = GRID_ROWS-1; y >= 0; y-- ) { // let's start at bottom and move up to reduce computations
// for every cell, test if it is empty
if (grid[column][y].isEmpty()) {
// if found a valid cell, place current players token and exit the loop (break)
grid[column][y].setValue(currentPlayer);
coinPlaced = true;
break;
}
}
return coinPlaced;
}
void checkCoins() {
// scan the array for 4 of the same value in a row
for (int y=0; y < GRID_ROWS; y++) { // for every vertical row,
for (int x=0; x < GRID_COLUMNS; x++) { // for every horizontal cell in that row
//grid[x][y]
// I will leave this to you ;)
// keep in mind to check neighbors all you need to do is add or subtract 1 from the x or y value
// however when you are at the edge of the board be careful not to try and look at a neighbor that is off the edge, you will get a null pointer or an array out of bounds exception
if (x+1<GRID_COLUMNS) {
Cell toTheRight = grid[x+1][y];
}
// for each cell you could look at the 3 above the current, 3 below the current, 3 to the right and 3 to the left, 3 diagnal in each direction and then manually try and find 4 adjacent same color cells
// or a bit more complicated is a recursive solution that checks its 8 immediate neighbor and for each that match the center cell run the same function to test its 8 neighbors keeping track of the current inARowCount and returning true when it is found.
// would be a bit hard because you would need to make sure the second cell doesn't follow back to the original cell, and start an endless loop
}
}
}
void draw () {
background(1, 1, 1);
noStroke();
// draw all cells
for (int y=0; y < GRID_ROWS; y++) { // for every vertical row,
for (int x=0; x < GRID_COLUMNS; x++) { // for every horizontal cell in that row
grid[x][y].display(); // draw this cell
}
}
// draw next coin floating above the board, contrain its positon to above the board
fill(PLAYER_COLORS[currentPlayer]);
int currentMouseX = constrain(mouseX, BOARD_PADDING+COIN_RADIUS/2, BOARD_PADDING+(GRID_COLUMNS*CELL_SCALE)-COIN_RADIUS/2);
//currentMouseX = 40*(ceil(abs(currentMouseX/40)));
ellipse(currentMouseX, BOARD_PADDING/2, COIN_RADIUS, COIN_RADIUS);
}
// press any key to rest the game, probably want to test for a certain key here!
void keyPressed () {
setupBlankBoard();
}
// on press attempt to place a coin
void mousePressed() {
int currentMouseX = constrain(mouseX, BOARD_PADDING+COIN_RADIUS/2, BOARD_PADDING+(GRID_COLUMNS*CELL_SCALE)-COIN_RADIUS/2);
// determine what column we are over
int column = (currentMouseX - BOARD_PADDING)/CELL_SCALE;
// if choice is a valid coin placement
if (placeCoin(column)) {
// toggle players if a coin was placed
changePlayers();
// after each placement check win conditions
checkCoins();
}
}
Cell Class
class Cell {
int x, y;
int value; // 0 for empty, 1 for team 1, 2 for team 2 (constants defined at top of main file)
Cell(int x, int y) {
// default constructor to create empty cell
this(x, y, EMPTY_CELL);
}
// allows cell value to be set at creation
Cell(int x, int y, int value) {
this.x = x;
this.y = y;
this.value = value;
}
boolean setValue(int value){
value = constrain(value, EMPTY_CELL,PLAYER_2); // keeps it from setting a value outside of our range
if(this.value == EMPTY_CELL){
this.value = value;
return true; // placed
}
return false; // was not able to place it as there is already a value
}
boolean isEmpty(){
return this.value == EMPTY_CELL;
}
void display() {
fill(0, 0, 255);
rect(this.x, this.y, CELL_SCALE, CELL_SCALE);
// Draw Circle color based on current value, could simply just put fill(playerColors[this.value]); but this seems a bit more clear
if(this.value == EMPTY_CELL){
fill(PLAYER_COLORS[EMPTY_CELL]);
}else if(this.value == PLAYER_1){
fill(PLAYER_COLORS[PLAYER_1]); // red
}else if(this.value == PLAYER_2){
fill(PLAYER_COLORS[PLAYER_2]); // yellow
}
ellipse(this.x + CELL_SCALE/2, this.y + CELL_SCALE/2, COIN_RADIUS, COIN_RADIUS);
}
}

Transform an array to another array by shifting value to adjacent element

I am given 2 arrays, Input and Output Array. The goal is to transform the input array to output array by performing shifting of 1 value in a given step to its adjacent element. Eg: Input array is [0,0,8,0,0] and Output array is [2,0,4,0,2]. Here 1st step would be [0,1,7,0,0] and 2nd step would be [0,1,6,1,0] and so on.
What can be the algorithm to do this efficiently? I was thinking of performing BFS but then we have to do BFS from each element and this can be exponential. Can anyone suggest solution for this problem?
I think you can do this simply by scanning in each direction tracking the cumulative value (in that direction) in the current array and the desired output array and pushing values along ahead of you as necessary:
scan from the left looking for first cell where
cumulative value > cumulative value in desired output
while that holds move 1 from that cell to the next cell to the right
scan from the right looking for first cell where
cumulative value > cumulative value in desired output
while that holds move 1 from that cell to the next cell to the left
For your example the steps would be:
FWD:
[0,0,8,0,0]
[0,0,7,1,0]
[0,0,6,2,0]
[0,0,6,1,1]
[0,0,6,0,2]
REV:
[0,1,5,0,2]
[0,2,4,0,2]
[1,1,4,0,2]
[2,0,4,0,2]
i think BFS could actually work.
notice that n*O(n+m) = O(n^2+nm) and therefore not exponential.
also you could use: Floyd-Warshall algorithm and Johnson’s algorithm, with a weight of 1 for a "flat" graph, or even connect the vertices in a new way by their actual distance and potentially save some iterations.
hope it helped :)
void transform(int[] in, int[] out, int size)
{
int[] state = in.clone();
report(state);
while (true)
{
int minPressure = 0;
int indexOfMinPressure = 0;
int maxPressure = 0;
int indexOfMaxPressure = 0;
int pressureSum = 0;
for (int index = 0; index < size - 1; ++index)
{
int lhsDiff = state[index] - out[index];
int rhsDiff = state[index + 1] - out[index + 1];
int pressure = lhsDiff - rhsDiff;
if (pressure < minPressure)
{
minPressure = pressure;
indexOfMinPressure = index;
}
if (pressure > maxPressure)
{
maxPressure = pressure;
indexOfMaxPressure = index;
}
pressureSum += pressure;
}
if (minPressure == 0 && maxPressure == 0)
{
break;
}
boolean shiftLeft;
if (Math.abs(minPressure) > Math.abs(maxPressure))
{
shiftLeft = true;
}
else if (Math.abs(minPressure) < Math.abs(maxPressure))
{
shiftLeft = false;
}
else
{
shiftLeft = (pressureSum < 0);
}
if (shiftLeft)
{
++state[indexOfMinPressure];
--state[indexOfMinPressure + 1];
}
else
{
--state[indexOfMaxPressure];
++state[indexOfMaxPressure + 1];
}
report(state);
}
}
A simple greedy algorithm will work and do the job in minimum number of steps. The function returns the total numbers of steps required for the task.
int shift(std::vector<int>& a,std::vector<int>& b){
int n = a.size();
int sum1=0,sum2=0;
for (int i = 0; i < n; ++i){
sum1+=a[i];
sum2+=b[i];
}
if (sum1!=sum2)
{
return -1;
}
int operations=0;
int j=0;
for (int i = 0; i < n;)
{
if (a[i]<b[i])
{
while(j<n and a[j]==0){
j++;
}
if(a[j]<b[i]-a[i]){
operations+=(j-i)*a[j];
a[i]+=a[j];
a[j]=0;
}else{
operations+=(j-i)*(b[i]-a[i]);
a[j]-=(b[i]-a[i]);
a[i]=b[i];
}
}else if (a[i]>b[i])
{
a[i+1]+=(a[i]-b[i]);
operations+=(a[i]-b[i]);
a[i]=b[i];
}else{
i++;
}
}
return operations;
}
Here -1 is a special value meaning that given array cannot be converted to desired one.
Time Complexity: O(n).

Merging two arraylists without creating third one

Here is one task, i was trying to solve. You must write the function
void merge(ArrayList a, ArrayList b) {
// code
}
The function recieves two ArrayLists with equal size as input parameters [a1, a2, ..., an], [b1, b2, ..., bn]. The execution result is the 1st ArrayList must contain elements of both lists, and they alternate consistently ([a1, b1, a2, b2, ..., an, bn]) Please read the bold text twice =)
Code must work as efficiently as possible.
Here is my solution
public static void merge(ArrayList a, ArrayList b) {
ArrayList result = new ArrayList();
int i = 0;
Iterator iter1 = a.iterator();
Iterator iter2 = b.iterator();
while ((iter1.hasNext() || iter2.hasNext()) && i < (a.size() + b.size())) {
if (i % 2 ==0) {
result.add(iter1.next());
} else {
result.add(iter2.next());
}
i++;
}
a = result;
}
I know it's not perfect at all. But I can't understand how to merge in the 1st list without creating tmp list.
Thanks in advance for taking part.
Double ArrayList a's size. Set last two elements of a to the last element of the old a and the last element of b. Keep going, backing up each time, until you reach the beginnings of a and b. You have to do it from the rear because otherwise you will write over the original a's values.
In the end i got this:
public static void merge(ArrayList<Integer> arr1, ArrayList<Integer> arr2) {
int indexForArr1 = arr1.size() - 1;
int oldSize = arr1.size();
int newSize = arr1.size() + arr2.size();
/*
decided not to create new arraylist with new size but just to fill up old one with nulls
*/
fillWithNulls(arr1, newSize);
for(int i = (newSize-1); i >= 0; i--) {
if (i%2 != 0) {
int indexForArr2 = i%oldSize;
arr1.set(i,arr2.get(indexForArr2));
oldSize--; // we reduce the size because we don't need tha last element any more
} else {
arr1.set(i, arr1.get(indexForArr1));
indexForArr1--;
}
}
}
private static void fillWithNulls(ArrayList<Integer> array, int newSize) {
int delta = newSize - array.size();
for(int i = 0; i < delta; i++) {
array.add(null);
}
}
Thanks John again for bright idea!

Sudoku in Java. Index out of Bounds Exception

I've got an IndexOutOfBounds exception in the following program. It consists of three files:
Important are only two of them, the GUI is working fine. Here is the first one:
interface SudokuObserver {
public void modified(int i, int j);
}
public class SudokuData
{
public int[][] feld = new int[9][9];
public SudokuObserver obs = null;
public SudokuData()
{
int i,j;
for (i=0; i<9; i++) {
for (j=0; j<9; j++) {
feld[i][j] = 0;
}
}
}
public int getNumber(int x, int y)
{
return feld[x][y];
}
public void setNumber(int x, int y, int v)
{
feld[x][y] = v;
if (obs != null)
obs.modified(x, y);
}
public void setObserver(SudokuObserver o)
{
obs = o;
}
So the Sudoku field is allocated as a 9x9 integer array. The following file is called SudokuSolver and has an algorithm to write the possible numbers for each square into an ArrayList. Then the second algorithm works as following: He finds the square which has the minimum of possible numbers, sets the first of the numbers saved in the ArrayList on that square and does this recursive, so he starts again at defining the possible numbers for each square, taking the one with the smallest number of possibilities and picks the first one to put it into that field. A for-loop runs over the possible Numbers for each square while doing that.
import java.util.*;
public class SudokuSolver
{
SudokuData data;
public SudokuSolver(SudokuData d)
{
data = d;
}
{
/*Pseudoalgorithm:
- Inserts the numbers 1-9 into a Collection called res
- Looks at line x, which numbers are in there and erases them out of the
collection
- Looks at column y, which numbers are in there and erases them out of the
collection
- Looks in the 3x3 Square (x,y) which numbers are already in there and erases
them out of the collection
- Gives back the possible candidates for that field
*/
Here i initialize my ArrayList.
public ArrayList<Integer> offen(int x, int y)
{
ArrayList<Integer> res = new ArrayList<Integer>();
/* The collection is saved in an ArrayList */
int k = 0;
Here I just fill in the numbers 1-9 in my ArrayList.
for (int i=1;i<10;i++)
{
res.add(i);
}
Now comes the difficult part: I loop over j from zero to nine, then over k. The line is constant with the given x, the j runs over the columns, so i got every square in the given line, and in every square i check for every number from 1-9. Care: the index goes from 0-9 while the elements go from 1-9 so k has to be 0-9 cause the get()-method takes an index as input. If there is any compliance I remove the element from the ArrayList.
for (int j=0;j<9;j++)
{
for (k=0;k<9;k++)
{
if (this.data.feld[x][j] == (res.get(k)))
res.remove(k);
}
Same stuff as above for the columns, constant column and j loops.
for (k=0;k<9;k++)
{
if (this.data.feld[j][y] == res.get(k))
res.remove(k);
}
}
Now i get my inputs in two new variables, just because i had typed the code part below before with wrong variable names.
int m = x;
int n = y;
Here is the part for the 3x3 squares, i do this with if conditions, so this is just one of the 9 parts, I didn't want to post them all here, cause they just differ in a few constants. I check in which square my input x,y is, and then I loop over the square and check which numbers are there, which are also still in my ArrayList and remove them.
if (m<=2 && n<=2)
{
for (m=0;m<3;m++)
{
for (n=0;n<3;n++)
{
for (k=0;k<9;k++)
{
if (this.data.feld[m][n] == res.get(k))
res.remove(k);
}
}
}
}
Now I return the ArrayList
return res;
}
//findSolution() finds a Solution
public boolean findSolution()
{
/*Possible Strategy:
- Find the square, which has the fewest possible candidates
- If there are more than one candidates, who have the minimum of candidates,
take any of them
- If there are no more open candidates, there is a solution found. Return
true
- Loop over the candidates of this square and by setting the first possible
candidate into this square[x][y]
- Call the method findSolution() recursive to find in dependence of the set
value the values for the other fields
If there is a blind alley, take the next possible candidate (Backtracking!)
*/
int j = 0;
int k = 0;
int x = 0; // x coordinate of the field with the fewest open candidates
int y = 0; // y coordinate of the field with the fewest open candidates
int counter_offene_felder = 0; // counts the number of open fields
int min = 9;
I'm looping over j and k, looking if the number of possible candidates is more than 0, that means I'm running through the whole sudoku field and count the number of open fields.
for (j=0;j<9;j++)
{
for (k=0;k<9;k++)
{
if ( this.offen(j,k).size() >= 0)
{
counter_offene_felder += 1;
}
If the number is < than min = 9 possible candidates, i take it as the min and save the coordinates of that field
if ( (this.offen(j,k)).size() < min )
{
x = j;
y = k;
}
}
}
now i initialize and ArrayList for the field with the fewest possible candidates and put them into this ArrayList with my offen-method
ArrayList<Integer> candidate_list = this.offen(x,y);
for (k=0;k<this.offen(x,y).size();k++)
{ // runs over candidates
int v = this.offen(x,y).get(k); // takes the first candidate
this.data.setNumber(x,y,v); // writes the first candidate into square [x][y]
this.findSolution(); // calls findSolution() recursive
}
If there are no more open fields, I've found a solution
if (counter_offene_felder == 0)
{
return true;
}
else return false;
}
}
The problem is, that I get an IndexOutOfBounds Exception at line 39, at Index 8 Size 8. But I don't know why. :(
Not positive that this is where you are getting your error... but you could run into an issue when you do something like this.
for (k=0;k<9;k++)
{
if (this.data.feld[j][y] == res.get(k))
res.remove(k);
}
For instance, say that at k=1 the if statement evaluates to true. Then you will remove an element from the ArrayList. Then when k=8, and IndexOutOfBounds exception will be thrown because the ArrayList only contains 8 elements (0-7)
Assuming that no other threads will be modifying this.data.feld[][], you will only ever get one match when going through this loop.. so you could do something like this...
int match = -1;
for (k=0;k<res.size();k++) {
if (this.data.feld[j][y] == res.get(k)){
match = k;
break;
}
}
if(match != -1)
res.remove(match);
I think the contains() method will help eliminate your exceptions for this loop.
Try replacing your code with this:
for (m=0;m<3;m++)
{
for (n=0;n<3;n++)
{
if (res.contains(this.data.field[m][n]))
res.remove(res.indexOf(this.data.field[m][n]));
}
}
It will iterate over the data.field, and check the ArrayList to see if it contains the value at m,n. If it does, it will remove it.

replace/remove cycle in graph

I'm trying to replace cycle in the graph to a group of the vertexes (remove this cycle and put there once vertex with maximum number)
struct group {
int master; // representative of cycle
};
struct vertex {
int *to; // neighbor list
int nb; // how many neighbor
int p; // parent
short v; // was visited? 0 = false, 1 = true
struct group *cycle; // is part of cycle? NULL = no, else pointer to group
};
I'm running dfs on each vertex
void searchCycles() {
int i;
for (i = 0; i < MAX_VER; i += 1)
if (ver[i].v == 0 && ver[i].nb > 0)
dfs(i);
}
dfs:
void dfs(int v) {
ver[v].v = 1;
int i;
for (i = 0; i < ver[v].nb; i += 1) {
ver[ver[v].to[i]].p = v;
if (ver[ver[v].to[i]].v == 0)
dfs(ver[v].to[i]);
else
// cycle found
replaceCycle(ver[v].to[i]);
}
}
and replace function shout print what vertexes are in cycle
void replaceCycle(int v) {
struct group *g = &gr[usedGroup++];
g->master = -1;
printf("\nCYKL: %d ", v);
int p = ver[v].p;
while (p != v) {
printf("%d(%d) ", p, v);
p = ver[p].p;
}
printf("\n");
}
Generally it's works, but sometimes it get a infinity loop. I tried to debug it and if there are two or more cycles, parents (p in vertex struct) are lost, it's means it works fine but there is wrong number. I'm learning C and algorithms, so I don't know a lot of it.
It's not a homework, it's a spoj problem
Once you replace a cycle, restart your dfs.
Basically, the visited flag might be set for your first cycle, but you'd want that cleared to test your second cycle. (And third, and fourth, etc.)

Resources