Arranging bricks safely

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The goal is to arrange bricks safely so that there is no line that can be cut through.



For example



example1



This arrangement is considered unsafe



example2



because you can cut through the red line.



example3



This is a safe arrangement because you cannot cut through any line.



So the question is, What is the condition for the length and the width so that there is at least one possible way to arrange bricks safely?



Obviously, one of the length and the width should be even number, for obvious reason.



I also figured out that the length and width should be bigger than 3



figure4



because it will be one of these cases, which already creates line.



Also $ 1times2 $, which is just a single brick will also be one of the possible way.



I also know that $ 6times11 $ is possible.



So what would the general condition be?




puzzle




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  • 4




    These are called fault-free rectangles. Mathworld has a short article that gives the conditions you seek.
    – Ross Millikan
    Jul 26 at 3:15










  • These notes outline the solution in the case of the $1times2$ bricks the OP asked about.
    – saulspatz
    Jul 26 at 3:24














up vote
5
down vote

favorite
2












The goal is to arrange bricks safely so that there is no line that can be cut through.



For example



example1



This arrangement is considered unsafe



example2



because you can cut through the red line.



example3



This is a safe arrangement because you cannot cut through any line.



So the question is, What is the condition for the length and the width so that there is at least one possible way to arrange bricks safely?



Obviously, one of the length and the width should be even number, for obvious reason.



I also figured out that the length and width should be bigger than 3



figure4



because it will be one of these cases, which already creates line.



Also $ 1times2 $, which is just a single brick will also be one of the possible way.



I also know that $ 6times11 $ is possible.



So what would the general condition be?




puzzle




share|cite|improve this question















  • 4




    These are called fault-free rectangles. Mathworld has a short article that gives the conditions you seek.
    – Ross Millikan
    Jul 26 at 3:15










  • These notes outline the solution in the case of the $1times2$ bricks the OP asked about.
    – saulspatz
    Jul 26 at 3:24












up vote
5
down vote

favorite
2









up vote
5
down vote

favorite
2






2





The goal is to arrange bricks safely so that there is no line that can be cut through.



For example



example1



This arrangement is considered unsafe



example2



because you can cut through the red line.



example3



This is a safe arrangement because you cannot cut through any line.



So the question is, What is the condition for the length and the width so that there is at least one possible way to arrange bricks safely?



Obviously, one of the length and the width should be even number, for obvious reason.



I also figured out that the length and width should be bigger than 3



figure4



because it will be one of these cases, which already creates line.



Also $ 1times2 $, which is just a single brick will also be one of the possible way.



I also know that $ 6times11 $ is possible.



So what would the general condition be?




puzzle




share|cite|improve this question











The goal is to arrange bricks safely so that there is no line that can be cut through.



For example



example1



This arrangement is considered unsafe



example2



because you can cut through the red line.



example3



This is a safe arrangement because you cannot cut through any line.



So the question is, What is the condition for the length and the width so that there is at least one possible way to arrange bricks safely?



Obviously, one of the length and the width should be even number, for obvious reason.



I also figured out that the length and width should be bigger than 3



figure4



because it will be one of these cases, which already creates line.



Also $ 1times2 $, which is just a single brick will also be one of the possible way.



I also know that $ 6times11 $ is possible.



So what would the general condition be?





puzzle





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 26 at 3:11









Pizzaroot

1056




1056







  • 4




    These are called fault-free rectangles. Mathworld has a short article that gives the conditions you seek.
    – Ross Millikan
    Jul 26 at 3:15










  • These notes outline the solution in the case of the $1times2$ bricks the OP asked about.
    – saulspatz
    Jul 26 at 3:24












  • 4




    These are called fault-free rectangles. Mathworld has a short article that gives the conditions you seek.
    – Ross Millikan
    Jul 26 at 3:15










  • These notes outline the solution in the case of the $1times2$ bricks the OP asked about.
    – saulspatz
    Jul 26 at 3:24







4




4




These are called fault-free rectangles. Mathworld has a short article that gives the conditions you seek.
– Ross Millikan
Jul 26 at 3:15




These are called fault-free rectangles. Mathworld has a short article that gives the conditions you seek.
– Ross Millikan
Jul 26 at 3:15












These notes outline the solution in the case of the $1times2$ bricks the OP asked about.
– saulspatz
Jul 26 at 3:24




These notes outline the solution in the case of the $1times2$ bricks the OP asked about.
– saulspatz
Jul 26 at 3:24










1 Answer
1






active

oldest

votes

















up vote
9
down vote



accepted










These are called fault-free rectangles. Mathworld has a short article that gives the conditions you seek. For $1 times 2$ tiles you can tile a $p times q$ rectangle if



  • At least one of $p$ and $q$ is even

  • $x +2y=p$ and $x+2y=q$ each have two distinct solutions in positive integers

  • $p$ and $q$ are not both $6$





share|cite|improve this answer





















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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    9
    down vote



    accepted










    These are called fault-free rectangles. Mathworld has a short article that gives the conditions you seek. For $1 times 2$ tiles you can tile a $p times q$ rectangle if



    • At least one of $p$ and $q$ is even

    • $x +2y=p$ and $x+2y=q$ each have two distinct solutions in positive integers

    • $p$ and $q$ are not both $6$





    share|cite|improve this answer

























      up vote
      9
      down vote



      accepted










      These are called fault-free rectangles. Mathworld has a short article that gives the conditions you seek. For $1 times 2$ tiles you can tile a $p times q$ rectangle if



      • At least one of $p$ and $q$ is even

      • $x +2y=p$ and $x+2y=q$ each have two distinct solutions in positive integers

      • $p$ and $q$ are not both $6$





      share|cite|improve this answer























        up vote
        9
        down vote



        accepted







        up vote
        9
        down vote



        accepted






        These are called fault-free rectangles. Mathworld has a short article that gives the conditions you seek. For $1 times 2$ tiles you can tile a $p times q$ rectangle if



        • At least one of $p$ and $q$ is even

        • $x +2y=p$ and $x+2y=q$ each have two distinct solutions in positive integers

        • $p$ and $q$ are not both $6$





        share|cite|improve this answer













        These are called fault-free rectangles. Mathworld has a short article that gives the conditions you seek. For $1 times 2$ tiles you can tile a $p times q$ rectangle if



        • At least one of $p$ and $q$ is even

        • $x +2y=p$ and $x+2y=q$ each have two distinct solutions in positive integers

        • $p$ and $q$ are not both $6$






        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 26 at 3:19









        Ross Millikan

        275k21186351




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