

Title: Knight on an infinite chessboard Post by NickH on Apr 15^{th}, 2003, 12:01pm Consider a knight on an infinite chessboard. How many squares can it reach after precisely n moves? 

Title: Re: Knight on an infinite chessboard Post by towr on Apr 15^{th}, 2003, 1:09pm it might be [hide]2^(2n+1)+n1[/hide] but it's probably not.. (I only tried n=0,1,2 so far ) 

Title: Re: Knight on an infinite chessboard Post by NickH on Apr 15^{th}, 2003, 2:08pm I would say f(0) = 1. After no moves, it can only be on one square  the square it starts from. The knight can move back to a square it has moved from. 

Title: Re: Knight on an infinite chessboard Post by cho on Apr 15^{th}, 2003, 2:22pm It takes a few rounds to stabilize. It starts 1,8,33,76,129,196. From that point on each round grows by a number 14 greater than the previous. I'm not sure how to turn the whole sequence into one equation. 

Title: Re: Knight on an infinite chessboard Post by THUDandBLUNDER on Apr 15^{th}, 2003, 4:20pm Quote:
I'm not sure what you mean there, cho. Do you mean f(n)  f(n1) = 14n + some constant ? OR Do you mean f(n)  f(n1) = 14 ? OR Perhaps you mean something else? Why 14? How did you get your sequence? 

Title: Re: Knight on an infinite chessboard Post by cho on Apr 15^{th}, 2003, 6:13pm Okay, I'm going to show my utter lack of math skills here. By instability I mean that moves 1 and 2 don't fit the pattern. If they did the sequence would be 1,12,37,76,129,196,277,372,481... Notice that the difference between adjacent values is 11,25,39,53,67... There's the 14. The pattern is obvious, but my last math class was a basic algebra class in high school 30 years ago. How I got my sequence was by counting (very little math required). I took a piece of graph paper and marked out the possible locations for one quadrant and multiplied by 4: 0 3212343456567878 214323454567678 8 3232343456567878 232343454567678 8 3434345456567878 434345456567678 8 5454545656767878 454545656767878 8 56565656767878 8 6565656767878 8 767676767878 8 67676767878 8 7878787878 8 878787878 8 8 8 8 8 8 8 8 8 8 8 

Title: Re: Knight on an infinite chessboard Post by cho on Apr 15^{th}, 2003, 6:20pm I'm also assuming the question seeks the number of squares you could be sitting on on the Nth move, and not all the squares it has touched so far (only the same colored squares as the original square after an even number of moves, eg) 

Title: Re: Knight on an infinite chessboard Post by Chronos on Apr 17^{th}, 2003, 4:51pm Quote:
Which is exactly what cho found, so this post is rather redundant now... 

Title: Re: Knight on an infinite chessboard Post by Netman on Apr 23^{rd}, 2003, 6:59pm For the version where the knight cannot revisit squares... after wasting 12 CPU minutes bruteforcing the problem, I came to the realization that the only square that restriction eliminates is the starting square, all other squares can be reached via some sequence of moves. So the equation: f(n) = 7n^{2} + 4n + 1, if n > 2 holds for all odd n, and for even n, just subtract one for the starting square. Chris 

Title: Re: Knight on an infinite chessboard Post by THUDandBLUNDER on Apr 24^{th}, 2003, 10:36am But the original square is already included. That is, f(0) = 1. Let's generalize the question: Consider a knight moving on an infinite chessboard. How many squares can it arrive at after precisely n moves given that: 1) it can double back and revisit squares? 2) it cannot double back and revisit squares? 3) we also count every square that it has visited? Also, 4) how many for n or less moves? 1) is the original question. You are suggesting that 1) and 2) are the same sets, which seems doubtful to me. Have you written a program? However, 3) and 4) would appear to be identical. [i][/i] 

Title: Re: Knight on an infinite chessboard Post by cho on Apr 24^{th}, 2003, 11:27am In #2, of course, you can never end up back at square one because it's been visited, but any other square that can be reached can be reached without using a square twice. For example: Move 1 could put you at 2,1 from the start. Move 3 could put you there by going 2,1; 4,0; 2,1. For any square you want to reach 2 or 4 or 6 or 8 moves after the move that would most quickly put it there, you just move 1 or 2 or 3 or 4 moves beyond it and come back on a parallel route. Your #3 is just each number added to the previous term in the series. Even I can do that math: 14x^{2}6x+5. 

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