How to determinate the convergence the start and the finish points?











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I have found the Gridpatterns page.



One can apply the next algoritm and obtaine the grid pattern "1-2-3".




  1. On square grid paper start in the middle.

  2. Draw a line 1-unit long.

  3. Turn a right angle clockwise.

  4. Draw a line 2-unit long.

  5. Turn a right angle clockwise.

  6. Draw a line 3-unit long.

  7. Repeat steps 1-6 four times.


Result is below on the left figure. As you can see we returned to the start point (red).



Edit. After the lesnik's answer. Your positions after step 6 will be four points forming a square (right figure).



enter image description here



Question. Is there a rule by which it can be determined that the pattern "a-b-c" will return (convergence) to the starting point? Here a, b, c are integer.










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    up vote
    1
    down vote

    favorite












    I have found the Gridpatterns page.



    One can apply the next algoritm and obtaine the grid pattern "1-2-3".




    1. On square grid paper start in the middle.

    2. Draw a line 1-unit long.

    3. Turn a right angle clockwise.

    4. Draw a line 2-unit long.

    5. Turn a right angle clockwise.

    6. Draw a line 3-unit long.

    7. Repeat steps 1-6 four times.


    Result is below on the left figure. As you can see we returned to the start point (red).



    Edit. After the lesnik's answer. Your positions after step 6 will be four points forming a square (right figure).



    enter image description here



    Question. Is there a rule by which it can be determined that the pattern "a-b-c" will return (convergence) to the starting point? Here a, b, c are integer.










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I have found the Gridpatterns page.



      One can apply the next algoritm and obtaine the grid pattern "1-2-3".




      1. On square grid paper start in the middle.

      2. Draw a line 1-unit long.

      3. Turn a right angle clockwise.

      4. Draw a line 2-unit long.

      5. Turn a right angle clockwise.

      6. Draw a line 3-unit long.

      7. Repeat steps 1-6 four times.


      Result is below on the left figure. As you can see we returned to the start point (red).



      Edit. After the lesnik's answer. Your positions after step 6 will be four points forming a square (right figure).



      enter image description here



      Question. Is there a rule by which it can be determined that the pattern "a-b-c" will return (convergence) to the starting point? Here a, b, c are integer.










      share|cite|improve this question















      I have found the Gridpatterns page.



      One can apply the next algoritm and obtaine the grid pattern "1-2-3".




      1. On square grid paper start in the middle.

      2. Draw a line 1-unit long.

      3. Turn a right angle clockwise.

      4. Draw a line 2-unit long.

      5. Turn a right angle clockwise.

      6. Draw a line 3-unit long.

      7. Repeat steps 1-6 four times.


      Result is below on the left figure. As you can see we returned to the start point (red).



      Edit. After the lesnik's answer. Your positions after step 6 will be four points forming a square (right figure).



      enter image description here



      Question. Is there a rule by which it can be determined that the pattern "a-b-c" will return (convergence) to the starting point? Here a, b, c are integer.







      geometry algorithms vectors invariant-theory






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      edited 10 hours ago

























      asked 11 hours ago









      Nick

      294112




      294112






















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          Looks like for any $a$, $b$ and $c$ following your algorithm you will return to the starting point.



          After steps 1-6 you shift from original position by some vector $vec{x_1}$ and turn by 90 degrees. Now you repeat steps 1-6, and shift from your current position by some vector $vec{x_2}$. Which is exactly like $vec{x_1}$, but turned by 90 degrees. And so on. Sum of vectors $vec{x_1}$, $vec{x_2}$, $vec{x_3}$ and $vec{x_4}$ is zero because all of them have the same length, and each one is turned by 90 degrees.



          Your positions after step 6 will be four points forming a square.






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            up vote
            1
            down vote













            Looks like for any $a$, $b$ and $c$ following your algorithm you will return to the starting point.



            After steps 1-6 you shift from original position by some vector $vec{x_1}$ and turn by 90 degrees. Now you repeat steps 1-6, and shift from your current position by some vector $vec{x_2}$. Which is exactly like $vec{x_1}$, but turned by 90 degrees. And so on. Sum of vectors $vec{x_1}$, $vec{x_2}$, $vec{x_3}$ and $vec{x_4}$ is zero because all of them have the same length, and each one is turned by 90 degrees.



            Your positions after step 6 will be four points forming a square.






            share|cite|improve this answer

























              up vote
              1
              down vote













              Looks like for any $a$, $b$ and $c$ following your algorithm you will return to the starting point.



              After steps 1-6 you shift from original position by some vector $vec{x_1}$ and turn by 90 degrees. Now you repeat steps 1-6, and shift from your current position by some vector $vec{x_2}$. Which is exactly like $vec{x_1}$, but turned by 90 degrees. And so on. Sum of vectors $vec{x_1}$, $vec{x_2}$, $vec{x_3}$ and $vec{x_4}$ is zero because all of them have the same length, and each one is turned by 90 degrees.



              Your positions after step 6 will be four points forming a square.






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                Looks like for any $a$, $b$ and $c$ following your algorithm you will return to the starting point.



                After steps 1-6 you shift from original position by some vector $vec{x_1}$ and turn by 90 degrees. Now you repeat steps 1-6, and shift from your current position by some vector $vec{x_2}$. Which is exactly like $vec{x_1}$, but turned by 90 degrees. And so on. Sum of vectors $vec{x_1}$, $vec{x_2}$, $vec{x_3}$ and $vec{x_4}$ is zero because all of them have the same length, and each one is turned by 90 degrees.



                Your positions after step 6 will be four points forming a square.






                share|cite|improve this answer












                Looks like for any $a$, $b$ and $c$ following your algorithm you will return to the starting point.



                After steps 1-6 you shift from original position by some vector $vec{x_1}$ and turn by 90 degrees. Now you repeat steps 1-6, and shift from your current position by some vector $vec{x_2}$. Which is exactly like $vec{x_1}$, but turned by 90 degrees. And so on. Sum of vectors $vec{x_1}$, $vec{x_2}$, $vec{x_3}$ and $vec{x_4}$ is zero because all of them have the same length, and each one is turned by 90 degrees.



                Your positions after step 6 will be four points forming a square.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 10 hours ago









                lesnik

                1,375611




                1,375611






























                     

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