please tell me if another way to show nonEquivalence of loops
for(int i = 0; i < 10; i++){
if (i % 2 == 0)
continue;
System.out.println(i);
}
//non equivalent statement. is there any like this(without continue)
int i = 0;
while(i < 10){
if (i % 2 == 0)
continue;
System.out.println(i);
}
From what i gather from you code, you want to display only odd number in your loop. Try this
for(int i = 0; i < 10; i++){
if (i % 2 != 0)
System.out.println(i);
}
Instead of finding when i is even and doing the continue statement, simply "reverse" the condition. The same goes for the while
Related
I am having an issue where when my program prints (base)^0=, it doesn't print the answer (1)
(I shorend down the out put examples as I'm only having an issue with the first line of the output)
expected output:
2^0=1
2^1=2
2^2=2*2=4
2^3=2*2*2=8
2^4=2*2*2*2=16
actual output:
> 2^0=
> 2^1=2=2
> 2^2=2*2=4
> 2^3=2*2*2=8
> 2^4=2*2*2*2=16
code:
else if(option == 2){
base = Input.nextInt();
for(int i = 0; i<10; i+=1){
System.out.print(base+"^"+i+"=");
for(int j = 0; j < i; j+=1){
if(j != i -1){
System.out.print(base+"*");
}else{
System.out.format(base+"="+"%.0f",Math.pow(base,i));
}
}
System.out.println("");
}
}
The first round when i = 0, you don't enter the inner for loop as the condition to enter is j < i, that is 0 < 0 => false
I am to write a program that prints ONLY the NON-BOUNDARY AND CORNER elements of an (n*n) array, for my assignment, and this is the main part of the code:
The output I am getting is this:
As you can see, the non-boundary elements (6,7,10,11) are not in their correct positions, which I believe, is because of incorrect printing of tab spaces within the loop. (My code is totally a mess) I would like some help or suggestions to fix this. Thanks!
I generally find that flattening things (the if-conditions in particular), and putting conditions into boolean-returning methods helps. Try something like
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++ {
if (isCorner(i,j,n) || !isEdge(i,j,n)) {
//...
} else {
//...
}
}
System.out.println();
}
where isCorner(i,j,n) and isEdge(i,j,n) are defined something like
public boolean isCorner(int row, int column, int gridSize) {
//...
}
A you got a solution, just missing spaces, I'll add some smart things:
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
boolean visible = (i % (n - 1) == 0) == (j % (n - 1) == 0);
if (visible) {
System.out.printf(" %4d", a[i][j]);
} else {
System.out.print(" ");
}
}
System.out.println();
}
No longer any problems with tabs "\t", though I used spaces here.
Keep it simple, too many cases just cause problems - as you experienced.
The trick here is to consider whether to print or not. Hence I started with
a variable visible.
The border condition
i == 0 || i == n - 1
could also be written with modulo as
i % (n - 1) == 0
If this is "too smart", hard to grasp reading:
boolean iOnBorder = i % (n - 1) == 0;
boolean jOnBorder = j % (n - 1) == 0;
boolean visible = iOnBorder == jOnBorder;
The "X" pattern checks the _equivalence of i-on-border and j-on-border.
For the rest: formatted printf allows padding of a number.
Try this i have optimized your if condition
No need to again check for i == 0 or i == n-1
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
if(i==0 || i==n-1){
if(j==0 || j==n-1){
System.out.print(a[i][j]);
}
}else{
if(j != 0 && j!= n-1){
System.out.print(a[i][j]);
}
}
System.out.print("\t");
}
System.out.println();
}
Just gave a try in case you might find it helpful.
public static void main(String[] args) throws ParseException {
int[][] ar = new int[4][4];
int[] input = new int[]{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16};
int pointer=0;
int imin=0,jmin=0,imax=3,jmax=3;
for(int i=0;i<4;i++){
for(int j=0;j<4;j++){
ar[i][j]=input[pointer];
pointer++;
}
}
for(int i=0;i<4;i++){
for(int j=0;j<4;j++){
if(!((i==imax && j==jmin)||(i==imin && j==jmax)||i==j) && //For skipping the corners
(i == imin || j == jmin || i == imax || j == jmax)){// Not to print the borders
continue;
}
else {
System.out.println(ar[i][j]);
}
}
}
}
Is there a quicker way to increment a counter or break from an outer loop?
while(myArrayList.get(i) > myNumber) {
// some operations
if(i + 1 < myArrayList.size())
i++;
else
break;
}
A better way to write your code is definitely there:
for (int i = 0; i < myArrayList.size(); i++) {
if (myArrayList.get(i) <= myNumber) break;
//Some operations...
}
You need to check the size first to avoid running off the end of the list and getting an error.
int i;
for (i = 0; i < myArrayList.size(); i++)
if (myArrayList.get(i) <= myNumber)
break;
With the Streams API you could do
int n = IntStream.range(0, myArrayList.size())
.filter(i -> myArrayList.get(i) <= myNumber)
.findFirst()
.orElse(myArrayList.size()); // or -1
A for-loop is more appropriate here:
int counter = 0
for(int i = 0; i < myArrayList.size(); i++) {
if(!(myArrayList.get(i) > myNumber)) {
break;
}
counter++;
}
If the intention is count values that are greater than myNumber, then break will possibly exclude some values (unless you know that myArrayList is sorted), and the loop should rather be:
for(int i = 0; i < myArrayList.size(); i++) {
if(myArrayList.get(i) > myNumber) {
counter++;
}
}
I think you want to start counting from some index i,
int count=0;
for( ;((i < myArrayList.size()-1) && (myArrayList.get(i) > myNumber));i++){
count++;
}
I am trying to make the following number line is Java
2,3,5,7,11,13,17 (Prime Numbers)
I tried this code
for(int i =0; i <= 100; i++) {
if(i < 2) {
continue;
}
for(int j = 2; j < 1; j++) {
if(i % j == 0) {
break;
} else {
System.out.print(i + ",");
}
}
}
But it doesn't work
Anyone help please?
Your code is quite poor, but the minimal amount of changes needed to make it work yields this code:
outerLoop:
for(int i = 0; i <= 100; i++) {
if(i < 2) {
continue;
}
for(int j = 2; j < i; j++) {
if(i % j == 0) {
continue outerLoop;
}
}
System.out.print(i + ",");
}
But a further improvement would be to start the first loop at 2 right away:
outerLoop:
for(int i = 2; i <= 100; i++) {
for(int j = 2; j < i; j++) {
// and so on...
EDITED
There are a lot of errors in that code, first of all that second loop is a infinite loop and second one that the System.out.println line should not be in second loop it should be at end of first loop! If you place it in second it will print numbers hundreds of time.
This is the correct code :
for(int i = 2; i <= 100; i++)//begin loop from 2 instead of 0
{
boolean flag = true;
for(int j = 2; j < i; j++)
{
if(i % j == 0)
{
flag = false;
break;
}
}
if(flag)System.out.print(i + ",");
}
You need to set a flag to check if a factor was found outside the loop.
Hi so I'm building a program to create the classical minesweeper game in Java, and have almost everything down but I cant figure out how to check the sqaures around a mine and to write in the numbers (i.e. if there are one, two, three mines next to it). I only included the method it's under, but I can post the rest of the program if necessary. What should my approach be? Thanks!
private void countAdjacentMines()
{
for (int i = 0; i < mineField.length; i++)
{
for (int j = 0; j < mineField.length; j++)
{
if (!(mineField[i][j].getIsMine()))
{
mineField[i-1][j-1];
mineField[i-1][j];
mineField[i-1][j+1];
mineField[i][j-1];
mineField[i][j+1];
mineField[i+1][j-1];
mineField[i+1][j];
mineField[i+1][j+1];
mineField[i][j].setAdjacentMines(0);
}
} // end for loop rows
} // end for loop columns
} // end countAdjacentMines
Something like this:
private void countAdjacentMines()
{
for (int i = 0; i < mineField.length; i++)
{
for (int j = 0; j < mineField.length; j++)
{
if (!(mineField[i][j].getIsMine()))
{
int count = 0;
for (int p = i - 1; p <= i + 1; p++)
{
for (int q = j - 1; q <= j + 1; q++)
{
if (0 <= p && p < mineField.length && 0 <= q && q < mineField.length)
{
if (mineField[p][q].getIsMine())
++count;
}
}
}
mineField[i][j].setAdjacentMines(count);
}
} // end for loop rows
} // end for loop columns
} // end countAdjacentMines
For each item in that "list", do something like:
if ((i-1) >= 0 && (j-1) >= 0 && mineField[i-1][j-1].getIsMine()) {
numAdjacentMines++;
}
It's probably worth writing a helper function to do all of this, and then you just need to call it 8 times.
You're on the right track. You should be keeping a counter, that represents the count of adjacent mines that return true for .getIsMine().
if (!(mineField[i][j].getIsMine()))
{
counter = 0;
if (i-1 >= 0)
{
if (j-1 >=0 && mineField[i-1][j-1].getIsMine()) counter++;
if (mineField[i-1][j].getIsMine()) counter++;
if (j+1 < mineField.length && mineField[i-1][j+1].getIsMine()) counter++;
}
if (j-1 >=0 && mineField[i][j-1].getIsMine()) counter++;
if (j+1 < mineField.length && mineField[i][j+1].getIsMine()) counter++;
if (i+1 < mineField.length)
{
if (j-1 >=0 && mineField[i+1][j-1].getIsMine()) counter++;
if (mineField[i+1][j].getIsMine()) counter++;
if (j+1 < mineField.length && mineField[i+1][j+1].getIsMine()) counter++;
}
mineField[i][j].setAdjacentMines(counter);
}
You also need to be checking that all of those values (i-1, j-1, i+1, j+1) don't go outside the bounds of your array (i - 1 > -1, etc)
EDIT:: I think I covered all of the checks.
I'm not sure what you're trying to do. This statement, mineField[i-1][j-1]; and the ones like it just access data and do nothing with it.
I'm assuming that the array stores something that includes a boolean that tells you whether or not there's a mine in that space. If that's true, just make a counter and change those statements to something like if(mineField[i-1][j-1].hasMine()) counter++;. And then change the last statement to mineField[i][j].setAdjacentMines(counter);.
int sum = 0;
sum += mineField[i-1][j-1].getIsMine() ? 1 : 0;
sum += mineField[i-1][j].getIsMine() ? 1 : 0;
sum += mineField[i-1][j+1].getIsMine() ? 1 : 0;
sum += mineField[i][j-1].getIsMine() ? 1 : 0;
sum += mineField[i][j+1].getIsMine() ? 1 : 0;
sum += mineField[i+1][j-1].getIsMine() ? 1 : 0;
sum += mineField[i+1][j].getIsMine() ? 1 : 0;
sum += mineField[i+1][j+1].getIsMine() ? 1 : 0;
mineField[i][j].setAdjacentMines(sum);
This is count up how many are mines and use it properly. It may be more clean to create a loop to perform this calculation, since it is the same thing for each adjacent field.
Try something like:
private static int countAdjacentMines(int x, int y) {
int adjacentMines = 0;
for(int i = -1; i <= 1; i++) {
if((x + i < 0) || (x + i >= width)) {
continue;
}
for(int j = -1; j <= 1; j++) {
if((y + j < 0) || (y + j >= height)) {
continue;
}
if(mineField[x + i][y + j].getIsMine()) {
adjacentMines++;
}
}
}
return adjacentMines;
}
This should count the number of mines neighbouring a block at (x, y).