How to iterate over pixels on edge of a square in 1 iteration [closed]

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Hi is there a way of looping around the edges of a square using one for loop? The square will be aligned with the x-axis and y-axis. The square will also have a known length and center position.

edited Nov 22 at 14:17

Ishaan

7711417

asked Nov 22 at 14:07

Richard8365

closed as too broad by idursun, user6910411, Max Vollmer, Skynet, Rob Nov 22 at 18:06

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

3

It's hard to answer without having some more details e.g. code.
– Rene Knop
Nov 22 at 14:10

1

The answer is most likely "Yes" but we need to see your code.
– Stefan
Nov 22 at 14:22

so far i'm just planning the function that this is for and haven't got any code yet but the image will be a binary image in a 2d x and y array of 1's and 0's. the location of the center of the square is known and so is the distance from its center to the edge.
– Richard8365
Nov 22 at 14:27

add a comment |

up vote
-1
down vote

favorite

Hi is there a way of looping around the edges of a square using one for loop? The square will be aligned with the x-axis and y-axis. The square will also have a known length and center position.

edited Nov 22 at 14:17

Ishaan

7711417

asked Nov 22 at 14:07

Richard8365

closed as too broad by idursun, user6910411, Max Vollmer, Skynet, Rob Nov 22 at 18:06

3

It's hard to answer without having some more details e.g. code.
– Rene Knop
Nov 22 at 14:10

1

The answer is most likely "Yes" but we need to see your code.
– Stefan
Nov 22 at 14:22

so far i'm just planning the function that this is for and haven't got any code yet but the image will be a binary image in a 2d x and y array of 1's and 0's. the location of the center of the square is known and so is the distance from its center to the edge.
– Richard8365
Nov 22 at 14:27

add a comment |

up vote
-1
down vote

favorite

Hi is there a way of looping around the edges of a square using one for loop? The square will be aligned with the x-axis and y-axis. The square will also have a known length and center position.

edited Nov 22 at 14:17

Ishaan

7711417

asked Nov 22 at 14:07

Richard8365

Hi is there a way of looping around the edges of a square using one for loop? The square will be aligned with the x-axis and y-axis. The square will also have a known length and center position.

java image geometry coordinates

edited Nov 22 at 14:17

Ishaan

7711417

asked Nov 22 at 14:07

Richard8365

edited Nov 22 at 14:17

Ishaan

7711417

asked Nov 22 at 14:07

Richard8365

edited Nov 22 at 14:17

Ishaan

7711417

edited Nov 22 at 14:17

Ishaan

7711417

edited Nov 22 at 14:17

Ishaan

7711417

asked Nov 22 at 14:07

Richard8365

asked Nov 22 at 14:07

Richard8365

asked Nov 22 at 14:07

Richard8365

closed as too broad by idursun, user6910411, Max Vollmer, Skynet, Rob Nov 22 at 18:06

3

It's hard to answer without having some more details e.g. code.
– Rene Knop
Nov 22 at 14:10

1

The answer is most likely "Yes" but we need to see your code.
– Stefan
Nov 22 at 14:22

so far i'm just planning the function that this is for and haven't got any code yet but the image will be a binary image in a 2d x and y array of 1's and 0's. the location of the center of the square is known and so is the distance from its center to the edge.
– Richard8365
Nov 22 at 14:27

add a comment |

3

It's hard to answer without having some more details e.g. code.
– Rene Knop
Nov 22 at 14:10

1

The answer is most likely "Yes" but we need to see your code.
– Stefan
Nov 22 at 14:22

so far i'm just planning the function that this is for and haven't got any code yet but the image will be a binary image in a 2d x and y array of 1's and 0's. the location of the center of the square is known and so is the distance from its center to the edge.
– Richard8365
Nov 22 at 14:27

It's hard to answer without having some more details e.g. code.
– Rene Knop
Nov 22 at 14:10

The answer is most likely "Yes" but we need to see your code.
– Stefan
Nov 22 at 14:22

so far i'm just planning the function that this is for and haven't got any code yet but the image will be a binary image in a 2d x and y array of 1's and 0's. the location of the center of the square is known and so is the distance from its center to the edge.
– Richard8365
Nov 22 at 14:27

add a comment |

1 Answer
1

active

oldest

votes

up vote
1
down vote

accepted

You certainly can do it in one loop, but I think you are probably looking for something like this:

//start at top-left

int x = center_x - (len/2);

int y = center_y - (len/2);



//point to the right

int dx=1;

int dy=0;



for (int side=0; side<4; ++side) {

    for (int i=1; i<len; ++i) {

        do_something(x,y);

        x+=dx;

        y+=dy;

    }

    //turn right

    int t=dx;

    dx=-dy;

    dy=t;

}

answered Nov 22 at 14:26

Matt Timmermans

18.2k11532

Thanks, this should work
– Richard8365
Nov 22 at 14:30

this can be transferred to one loop but no reason to do that because the number of iterations will be the same - the number of points around the square are the same
– Veselin Davidov
Nov 22 at 14:35

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
1
down vote

accepted

You certainly can do it in one loop, but I think you are probably looking for something like this:

//start at top-left

int x = center_x - (len/2);

int y = center_y - (len/2);



//point to the right

int dx=1;

int dy=0;



for (int side=0; side<4; ++side) {

    for (int i=1; i<len; ++i) {

        do_something(x,y);

        x+=dx;

        y+=dy;

    }

    //turn right

    int t=dx;

    dx=-dy;

    dy=t;

}

answered Nov 22 at 14:26

Matt Timmermans

18.2k11532

Thanks, this should work
– Richard8365
Nov 22 at 14:30

this can be transferred to one loop but no reason to do that because the number of iterations will be the same - the number of points around the square are the same
– Veselin Davidov
Nov 22 at 14:35

add a comment |

up vote
1
down vote

accepted

You certainly can do it in one loop, but I think you are probably looking for something like this:

//start at top-left

int x = center_x - (len/2);

int y = center_y - (len/2);



//point to the right

int dx=1;

int dy=0;



for (int side=0; side<4; ++side) {

    for (int i=1; i<len; ++i) {

        do_something(x,y);

        x+=dx;

        y+=dy;

    }

    //turn right

    int t=dx;

    dx=-dy;

    dy=t;

}

answered Nov 22 at 14:26

Matt Timmermans

18.2k11532

Thanks, this should work
– Richard8365
Nov 22 at 14:30

this can be transferred to one loop but no reason to do that because the number of iterations will be the same - the number of points around the square are the same
– Veselin Davidov
Nov 22 at 14:35

add a comment |

up vote
1
down vote

accepted

You certainly can do it in one loop, but I think you are probably looking for something like this:

//start at top-left

int x = center_x - (len/2);

int y = center_y - (len/2);



//point to the right

int dx=1;

int dy=0;



for (int side=0; side<4; ++side) {

    for (int i=1; i<len; ++i) {

        do_something(x,y);

        x+=dx;

        y+=dy;

    }

    //turn right

    int t=dx;

    dx=-dy;

    dy=t;

}

answered Nov 22 at 14:26

Matt Timmermans

18.2k11532

You certainly can do it in one loop, but I think you are probably looking for something like this:

//start at top-left

int x = center_x - (len/2);

int y = center_y - (len/2);



//point to the right

int dx=1;

int dy=0;



for (int side=0; side<4; ++side) {

    for (int i=1; i<len; ++i) {

        do_something(x,y);

        x+=dx;

        y+=dy;

    }

    //turn right

    int t=dx;

    dx=-dy;

    dy=t;

}

answered Nov 22 at 14:26

Matt Timmermans

18.2k11532

answered Nov 22 at 14:26

Matt Timmermans

18.2k11532

answered Nov 22 at 14:26

Matt Timmermans

18.2k11532

answered Nov 22 at 14:26

Matt Timmermans

18.2k11532

Thanks, this should work
– Richard8365
Nov 22 at 14:30

this can be transferred to one loop but no reason to do that because the number of iterations will be the same - the number of points around the square are the same
– Veselin Davidov
Nov 22 at 14:35

add a comment |

Thanks, this should work
– Richard8365
Nov 22 at 14:30

this can be transferred to one loop but no reason to do that because the number of iterations will be the same - the number of points around the square are the same
– Veselin Davidov
Nov 22 at 14:35

Thanks, this should work
– Richard8365
Nov 22 at 14:30

this can be transferred to one loop but no reason to do that because the number of iterations will be the same - the number of points around the square are the same
– Veselin Davidov
Nov 22 at 14:35

add a comment |

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