Snake Game Combinatorics












2












$begingroup$


I am doing an algorithmic analysis of the video game snake. It would be helpful to know the total amount of game positions in a $10times10$ board or an approximation of it.



There is another question on this site that uses a $5times5$ board and a snake of length $3$ and can solve this problem, but what I have realized is that since the snake is not allowed to overlap itself, it gets much more complicated to compute its possibilities after length $4$, and I need to know all the way up to length $100$ on a $10times10$ board.



Does anyone know how to do this?



Thanks










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$endgroup$

















    2












    $begingroup$


    I am doing an algorithmic analysis of the video game snake. It would be helpful to know the total amount of game positions in a $10times10$ board or an approximation of it.



    There is another question on this site that uses a $5times5$ board and a snake of length $3$ and can solve this problem, but what I have realized is that since the snake is not allowed to overlap itself, it gets much more complicated to compute its possibilities after length $4$, and I need to know all the way up to length $100$ on a $10times10$ board.



    Does anyone know how to do this?



    Thanks










    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I am doing an algorithmic analysis of the video game snake. It would be helpful to know the total amount of game positions in a $10times10$ board or an approximation of it.



      There is another question on this site that uses a $5times5$ board and a snake of length $3$ and can solve this problem, but what I have realized is that since the snake is not allowed to overlap itself, it gets much more complicated to compute its possibilities after length $4$, and I need to know all the way up to length $100$ on a $10times10$ board.



      Does anyone know how to do this?



      Thanks










      share|cite|improve this question











      $endgroup$




      I am doing an algorithmic analysis of the video game snake. It would be helpful to know the total amount of game positions in a $10times10$ board or an approximation of it.



      There is another question on this site that uses a $5times5$ board and a snake of length $3$ and can solve this problem, but what I have realized is that since the snake is not allowed to overlap itself, it gets much more complicated to compute its possibilities after length $4$, and I need to know all the way up to length $100$ on a $10times10$ board.



      Does anyone know how to do this?



      Thanks







      combinatorics number-theory






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      edited Dec 13 '18 at 22:54







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      asked Dec 13 '18 at 22:38









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          $begingroup$

          I do not think anyone is going to have an exact answer, but here is an upper bound.



          There are $100$ positions for the snake's head, at most $4$ possibilities for the body segment next to the head, and at most $3$ for each subsequent body part. Therefore, there are at most $400cdot 3^{k-2}$ possibilities for a snake of length $k$. Summing from $k=2$ to $100$ gives $$#text{ snake positions}le 400(3^{99}-1)/2.$$I suppose you should multiply that by $100$ to account for the position of the food piece.






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            $begingroup$

            I do not think anyone is going to have an exact answer, but here is an upper bound.



            There are $100$ positions for the snake's head, at most $4$ possibilities for the body segment next to the head, and at most $3$ for each subsequent body part. Therefore, there are at most $400cdot 3^{k-2}$ possibilities for a snake of length $k$. Summing from $k=2$ to $100$ gives $$#text{ snake positions}le 400(3^{99}-1)/2.$$I suppose you should multiply that by $100$ to account for the position of the food piece.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              I do not think anyone is going to have an exact answer, but here is an upper bound.



              There are $100$ positions for the snake's head, at most $4$ possibilities for the body segment next to the head, and at most $3$ for each subsequent body part. Therefore, there are at most $400cdot 3^{k-2}$ possibilities for a snake of length $k$. Summing from $k=2$ to $100$ gives $$#text{ snake positions}le 400(3^{99}-1)/2.$$I suppose you should multiply that by $100$ to account for the position of the food piece.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                I do not think anyone is going to have an exact answer, but here is an upper bound.



                There are $100$ positions for the snake's head, at most $4$ possibilities for the body segment next to the head, and at most $3$ for each subsequent body part. Therefore, there are at most $400cdot 3^{k-2}$ possibilities for a snake of length $k$. Summing from $k=2$ to $100$ gives $$#text{ snake positions}le 400(3^{99}-1)/2.$$I suppose you should multiply that by $100$ to account for the position of the food piece.






                share|cite|improve this answer









                $endgroup$



                I do not think anyone is going to have an exact answer, but here is an upper bound.



                There are $100$ positions for the snake's head, at most $4$ possibilities for the body segment next to the head, and at most $3$ for each subsequent body part. Therefore, there are at most $400cdot 3^{k-2}$ possibilities for a snake of length $k$. Summing from $k=2$ to $100$ gives $$#text{ snake positions}le 400(3^{99}-1)/2.$$I suppose you should multiply that by $100$ to account for the position of the food piece.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 13 '18 at 22:57









                Mike EarnestMike Earnest

                23.9k12051




                23.9k12051






























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