The Path Of The Wildebeest












7












$begingroup$


Golf a program or function which gives the $n^{text{th}}$ location of the wildebeest who starts at square $1$ on an infinite chessboard which is numbered in an anti-clockwise square spiral, where the wildebeest always visits the lowest numbered square she can reach that she has not yet visited.



Inspiration: The Trapped Knight and OEIS A316667.



The code may produce the $n^{text{th}}$ location, the first $n$ locations, or generate the sequence taking no input.



Feel free to give her location after (or up to) $n$ leaps instead, but if so please state this clearly in your answer and make sure that an input of $n=0$ yields 1 (or [1] if appropriate).



This is code-golf, so the aim is to produce working code in as few bytes as possible in your chosen language.



Note: I believe that the path of the wildebeest is not finite (that she never becomes trapped, like the knight does at his $2016^{text{th}}$ location, square $2084$, and the camel does at his $3723^{text{rd}}$, square $7081$), but have not proven this - if this is not the case then the behaviour may be undefined for $n$ larger than the step at which she does get trapped. (I also believe the path will turn out to be a permutation of the natural numbers, but this is a pure hunch.)*



Detail



The board looks like the below, and continues indefinitely:



101 100  99  98  97  96  95  94  93  92  91

102  65  64  63  62  61  60  59  58  57  90

103  66  37  36  35  34  33  32  31  56  89

104  67  38  17  16  15  14  13  30  55  88

105  68  39  18   5   4   3  12  29  54  87

106  69  40  19   6   1   2  11  28  53  86

107  70  41  20   7   8   9  10  27  52  85

108  71  42  21  22  23  24  25  26  51  84

109  72  43  44  45  46  47  48  49  50  83

110  73  74  75  76  77  78  79  80  81  82

111 112 113 114 115 116 117 118 119 120 121


A wildebeest is a fairy chess piece - a non-standard chess piece which may move both as a knight (a $(1,2)$-leaper) and as a camel (a $(1,3)$-leaper).

As such she could move to any of these locations from her starting location of $1$:



  .   .   .   .   .   .   .   .   .   .   .

  .   .   .   .  35   .  33   .   .   .   .

  .   .   .   .  16   .  14   .   .   .   .

  .   .  39  18   .   .   .  12  29   .   .

  .   .   .   .   .  (1)  .   .   .   .   .

  .   .  41  20   .   .   .  10  27   .   .

  .   .   .   .  22   .  24   .   .   .   .

  .   .   .   .  45   .  47   .   .   .   .

  .   .   .   .   .   .   .   .   .   .   .


The lowest of these is $10$ and she has not yet visited that square, so $10$ is the second term in the sequence.



Next she could move from $10$ to any of these locations:



  .   .   .   .   .   .   .   .   .   .   .

  .   .   .   .   .   .  14   .  30   .   .

  .   .   .   .   .   .   3   .  29   .   .

  .   .   .   .   6   1   .   .   .  53  86

  .   .   .   .   .   .   . (10)  .   .   .

  .   .   .   .  22  23   .   .   .  51  84

  .   .   .   .   .   .  47   .  49   .   .

  .   .   .   .   .   .  78   .  80   .   .

  .   .   .   .   .   .   .   .   .   .   .


However, she has already visited square $1$ so her third location is square $3$, the lowest she has not yet visited.




The first $100$ terms of the path of the wildebeest are:



1, 10, 3, 6, 9, 4, 7, 2, 5, 8, 11, 14, 18, 15, 12, 16, 19, 22, 41, 17, 33, 30, 34, 13, 27, 23, 20, 24, 44, 40, 21, 39, 36, 60, 31, 53, 26, 46, 25, 28, 32, 29, 51, 47, 75, 42, 45, 71, 74, 70, 38, 35, 59, 56, 86, 50, 78, 49, 52, 80, 83, 79, 115, 73, 107, 67, 64, 68, 37, 61, 93, 55, 58, 54, 84, 48, 76, 43, 69, 103, 63, 66, 62, 94, 57, 87, 125, 82, 118, 77, 113, 72, 106, 148, 65, 97, 137, 91, 129, 85


The first $11$ leaps are knight moves so the first $12$ terms coincide with A316667.




* There are over one hundred and fifty million terms. After one million terms, numbering the layers of the spiral from the centre out, she has visited $512$ layers and the ($(512times 2-1)^2-10^6=46529$) unvisited squares by layer are: [(1,0),(2,0),...,(494,0),(495, 2), (496, 53), (497, 204), (498, 662), (499, 1208), (500, 1899), (501, 2459), (502, 2921), (503, 3133), (504, 3288), (505, 3397), (506, 3538), (507, 3712), (508, 3868), (509, 3978), (510, 4046), (511, 4074), (512, 4087)], so it's not entirely clear that all squares will be visited or that the wildebeest will never get trapped. Proofs of whether she will ever become trapped or whether she will visit all squares would, of course, be gratefully received!






code-golf sequence integer chess







share|improve this question











$endgroup$












  • $begingroup$
    I suspect the wildebeest gets trapped eventually -- it just takes longer since she has more movement options.
    $endgroup$
    – lirtosiast
    4 hours ago










  • $begingroup$
    "There are over ten million terms." LOL do you have a reference implementation?
    $endgroup$
    – Erik the Outgolfer
    3 hours ago










  • $begingroup$
    @EriktheOutgolfer yes, but she does get stuck somewhere - I need to rerun as was not catching the StopIteration :(
    $endgroup$
    – Jonathan Allan
    3 hours ago
















7












$begingroup$


Golf a program or function which gives the $n^{text{th}}$ location of the wildebeest who starts at square $1$ on an infinite chessboard which is numbered in an anti-clockwise square spiral, where the wildebeest always visits the lowest numbered square she can reach that she has not yet visited.



Inspiration: The Trapped Knight and OEIS A316667.



The code may produce the $n^{text{th}}$ location, the first $n$ locations, or generate the sequence taking no input.



Feel free to give her location after (or up to) $n$ leaps instead, but if so please state this clearly in your answer and make sure that an input of $n=0$ yields 1 (or [1] if appropriate).



This is code-golf, so the aim is to produce working code in as few bytes as possible in your chosen language.



Note: I believe that the path of the wildebeest is not finite (that she never becomes trapped, like the knight does at his $2016^{text{th}}$ location, square $2084$, and the camel does at his $3723^{text{rd}}$, square $7081$), but have not proven this - if this is not the case then the behaviour may be undefined for $n$ larger than the step at which she does get trapped. (I also believe the path will turn out to be a permutation of the natural numbers, but this is a pure hunch.)*



Detail



The board looks like the below, and continues indefinitely:



101 100  99  98  97  96  95  94  93  92  91

102  65  64  63  62  61  60  59  58  57  90

103  66  37  36  35  34  33  32  31  56  89

104  67  38  17  16  15  14  13  30  55  88

105  68  39  18   5   4   3  12  29  54  87

106  69  40  19   6   1   2  11  28  53  86

107  70  41  20   7   8   9  10  27  52  85

108  71  42  21  22  23  24  25  26  51  84

109  72  43  44  45  46  47  48  49  50  83

110  73  74  75  76  77  78  79  80  81  82

111 112 113 114 115 116 117 118 119 120 121


A wildebeest is a fairy chess piece - a non-standard chess piece which may move both as a knight (a $(1,2)$-leaper) and as a camel (a $(1,3)$-leaper).

As such she could move to any of these locations from her starting location of $1$:



  .   .   .   .   .   .   .   .   .   .   .

  .   .   .   .  35   .  33   .   .   .   .

  .   .   .   .  16   .  14   .   .   .   .

  .   .  39  18   .   .   .  12  29   .   .

  .   .   .   .   .  (1)  .   .   .   .   .

  .   .  41  20   .   .   .  10  27   .   .

  .   .   .   .  22   .  24   .   .   .   .

  .   .   .   .  45   .  47   .   .   .   .

  .   .   .   .   .   .   .   .   .   .   .


The lowest of these is $10$ and she has not yet visited that square, so $10$ is the second term in the sequence.



Next she could move from $10$ to any of these locations:



  .   .   .   .   .   .   .   .   .   .   .

  .   .   .   .   .   .  14   .  30   .   .

  .   .   .   .   .   .   3   .  29   .   .

  .   .   .   .   6   1   .   .   .  53  86

  .   .   .   .   .   .   . (10)  .   .   .

  .   .   .   .  22  23   .   .   .  51  84

  .   .   .   .   .   .  47   .  49   .   .

  .   .   .   .   .   .  78   .  80   .   .

  .   .   .   .   .   .   .   .   .   .   .


However, she has already visited square $1$ so her third location is square $3$, the lowest she has not yet visited.




The first $100$ terms of the path of the wildebeest are:



1, 10, 3, 6, 9, 4, 7, 2, 5, 8, 11, 14, 18, 15, 12, 16, 19, 22, 41, 17, 33, 30, 34, 13, 27, 23, 20, 24, 44, 40, 21, 39, 36, 60, 31, 53, 26, 46, 25, 28, 32, 29, 51, 47, 75, 42, 45, 71, 74, 70, 38, 35, 59, 56, 86, 50, 78, 49, 52, 80, 83, 79, 115, 73, 107, 67, 64, 68, 37, 61, 93, 55, 58, 54, 84, 48, 76, 43, 69, 103, 63, 66, 62, 94, 57, 87, 125, 82, 118, 77, 113, 72, 106, 148, 65, 97, 137, 91, 129, 85


The first $11$ leaps are knight moves so the first $12$ terms coincide with A316667.




* There are over one hundred and fifty million terms. After one million terms, numbering the layers of the spiral from the centre out, she has visited $512$ layers and the ($(512times 2-1)^2-10^6=46529$) unvisited squares by layer are: [(1,0),(2,0),...,(494,0),(495, 2), (496, 53), (497, 204), (498, 662), (499, 1208), (500, 1899), (501, 2459), (502, 2921), (503, 3133), (504, 3288), (505, 3397), (506, 3538), (507, 3712), (508, 3868), (509, 3978), (510, 4046), (511, 4074), (512, 4087)], so it's not entirely clear that all squares will be visited or that the wildebeest will never get trapped. Proofs of whether she will ever become trapped or whether she will visit all squares would, of course, be gratefully received!






code-golf sequence integer chess







share|improve this question











$endgroup$












  • $begingroup$
    I suspect the wildebeest gets trapped eventually -- it just takes longer since she has more movement options.
    $endgroup$
    – lirtosiast
    4 hours ago










  • $begingroup$
    "There are over ten million terms." LOL do you have a reference implementation?
    $endgroup$
    – Erik the Outgolfer
    3 hours ago










  • $begingroup$
    @EriktheOutgolfer yes, but she does get stuck somewhere - I need to rerun as was not catching the StopIteration :(
    $endgroup$
    – Jonathan Allan
    3 hours ago














7












7








7





$begingroup$


Golf a program or function which gives the $n^{text{th}}$ location of the wildebeest who starts at square $1$ on an infinite chessboard which is numbered in an anti-clockwise square spiral, where the wildebeest always visits the lowest numbered square she can reach that she has not yet visited.



Inspiration: The Trapped Knight and OEIS A316667.



The code may produce the $n^{text{th}}$ location, the first $n$ locations, or generate the sequence taking no input.



Feel free to give her location after (or up to) $n$ leaps instead, but if so please state this clearly in your answer and make sure that an input of $n=0$ yields 1 (or [1] if appropriate).



This is code-golf, so the aim is to produce working code in as few bytes as possible in your chosen language.



Note: I believe that the path of the wildebeest is not finite (that she never becomes trapped, like the knight does at his $2016^{text{th}}$ location, square $2084$, and the camel does at his $3723^{text{rd}}$, square $7081$), but have not proven this - if this is not the case then the behaviour may be undefined for $n$ larger than the step at which she does get trapped. (I also believe the path will turn out to be a permutation of the natural numbers, but this is a pure hunch.)*



Detail



The board looks like the below, and continues indefinitely:



101 100  99  98  97  96  95  94  93  92  91

102  65  64  63  62  61  60  59  58  57  90

103  66  37  36  35  34  33  32  31  56  89

104  67  38  17  16  15  14  13  30  55  88

105  68  39  18   5   4   3  12  29  54  87

106  69  40  19   6   1   2  11  28  53  86

107  70  41  20   7   8   9  10  27  52  85

108  71  42  21  22  23  24  25  26  51  84

109  72  43  44  45  46  47  48  49  50  83

110  73  74  75  76  77  78  79  80  81  82

111 112 113 114 115 116 117 118 119 120 121


A wildebeest is a fairy chess piece - a non-standard chess piece which may move both as a knight (a $(1,2)$-leaper) and as a camel (a $(1,3)$-leaper).

As such she could move to any of these locations from her starting location of $1$:



  .   .   .   .   .   .   .   .   .   .   .

  .   .   .   .  35   .  33   .   .   .   .

  .   .   .   .  16   .  14   .   .   .   .

  .   .  39  18   .   .   .  12  29   .   .

  .   .   .   .   .  (1)  .   .   .   .   .

  .   .  41  20   .   .   .  10  27   .   .

  .   .   .   .  22   .  24   .   .   .   .

  .   .   .   .  45   .  47   .   .   .   .

  .   .   .   .   .   .   .   .   .   .   .


The lowest of these is $10$ and she has not yet visited that square, so $10$ is the second term in the sequence.



Next she could move from $10$ to any of these locations:



  .   .   .   .   .   .   .   .   .   .   .

  .   .   .   .   .   .  14   .  30   .   .

  .   .   .   .   .   .   3   .  29   .   .

  .   .   .   .   6   1   .   .   .  53  86

  .   .   .   .   .   .   . (10)  .   .   .

  .   .   .   .  22  23   .   .   .  51  84

  .   .   .   .   .   .  47   .  49   .   .

  .   .   .   .   .   .  78   .  80   .   .

  .   .   .   .   .   .   .   .   .   .   .


However, she has already visited square $1$ so her third location is square $3$, the lowest she has not yet visited.




The first $100$ terms of the path of the wildebeest are:



1, 10, 3, 6, 9, 4, 7, 2, 5, 8, 11, 14, 18, 15, 12, 16, 19, 22, 41, 17, 33, 30, 34, 13, 27, 23, 20, 24, 44, 40, 21, 39, 36, 60, 31, 53, 26, 46, 25, 28, 32, 29, 51, 47, 75, 42, 45, 71, 74, 70, 38, 35, 59, 56, 86, 50, 78, 49, 52, 80, 83, 79, 115, 73, 107, 67, 64, 68, 37, 61, 93, 55, 58, 54, 84, 48, 76, 43, 69, 103, 63, 66, 62, 94, 57, 87, 125, 82, 118, 77, 113, 72, 106, 148, 65, 97, 137, 91, 129, 85


The first $11$ leaps are knight moves so the first $12$ terms coincide with A316667.




* There are over one hundred and fifty million terms. After one million terms, numbering the layers of the spiral from the centre out, she has visited $512$ layers and the ($(512times 2-1)^2-10^6=46529$) unvisited squares by layer are: [(1,0),(2,0),...,(494,0),(495, 2), (496, 53), (497, 204), (498, 662), (499, 1208), (500, 1899), (501, 2459), (502, 2921), (503, 3133), (504, 3288), (505, 3397), (506, 3538), (507, 3712), (508, 3868), (509, 3978), (510, 4046), (511, 4074), (512, 4087)], so it's not entirely clear that all squares will be visited or that the wildebeest will never get trapped. Proofs of whether she will ever become trapped or whether she will visit all squares would, of course, be gratefully received!






code-golf sequence integer chess







share|improve this question











$endgroup$




Golf a program or function which gives the $n^{text{th}}$ location of the wildebeest who starts at square $1$ on an infinite chessboard which is numbered in an anti-clockwise square spiral, where the wildebeest always visits the lowest numbered square she can reach that she has not yet visited.



Inspiration: The Trapped Knight and OEIS A316667.



The code may produce the $n^{text{th}}$ location, the first $n$ locations, or generate the sequence taking no input.



Feel free to give her location after (or up to) $n$ leaps instead, but if so please state this clearly in your answer and make sure that an input of $n=0$ yields 1 (or [1] if appropriate).



This is code-golf, so the aim is to produce working code in as few bytes as possible in your chosen language.



Note: I believe that the path of the wildebeest is not finite (that she never becomes trapped, like the knight does at his $2016^{text{th}}$ location, square $2084$, and the camel does at his $3723^{text{rd}}$, square $7081$), but have not proven this - if this is not the case then the behaviour may be undefined for $n$ larger than the step at which she does get trapped. (I also believe the path will turn out to be a permutation of the natural numbers, but this is a pure hunch.)*



Detail



The board looks like the below, and continues indefinitely:



101 100  99  98  97  96  95  94  93  92  91

102  65  64  63  62  61  60  59  58  57  90

103  66  37  36  35  34  33  32  31  56  89

104  67  38  17  16  15  14  13  30  55  88

105  68  39  18   5   4   3  12  29  54  87

106  69  40  19   6   1   2  11  28  53  86

107  70  41  20   7   8   9  10  27  52  85

108  71  42  21  22  23  24  25  26  51  84

109  72  43  44  45  46  47  48  49  50  83

110  73  74  75  76  77  78  79  80  81  82

111 112 113 114 115 116 117 118 119 120 121


A wildebeest is a fairy chess piece - a non-standard chess piece which may move both as a knight (a $(1,2)$-leaper) and as a camel (a $(1,3)$-leaper).

As such she could move to any of these locations from her starting location of $1$:



  .   .   .   .   .   .   .   .   .   .   .

  .   .   .   .  35   .  33   .   .   .   .

  .   .   .   .  16   .  14   .   .   .   .

  .   .  39  18   .   .   .  12  29   .   .

  .   .   .   .   .  (1)  .   .   .   .   .

  .   .  41  20   .   .   .  10  27   .   .

  .   .   .   .  22   .  24   .   .   .   .

  .   .   .   .  45   .  47   .   .   .   .

  .   .   .   .   .   .   .   .   .   .   .


The lowest of these is $10$ and she has not yet visited that square, so $10$ is the second term in the sequence.



Next she could move from $10$ to any of these locations:



  .   .   .   .   .   .   .   .   .   .   .

  .   .   .   .   .   .  14   .  30   .   .

  .   .   .   .   .   .   3   .  29   .   .

  .   .   .   .   6   1   .   .   .  53  86

  .   .   .   .   .   .   . (10)  .   .   .

  .   .   .   .  22  23   .   .   .  51  84

  .   .   .   .   .   .  47   .  49   .   .

  .   .   .   .   .   .  78   .  80   .   .

  .   .   .   .   .   .   .   .   .   .   .


However, she has already visited square $1$ so her third location is square $3$, the lowest she has not yet visited.




The first $100$ terms of the path of the wildebeest are:



1, 10, 3, 6, 9, 4, 7, 2, 5, 8, 11, 14, 18, 15, 12, 16, 19, 22, 41, 17, 33, 30, 34, 13, 27, 23, 20, 24, 44, 40, 21, 39, 36, 60, 31, 53, 26, 46, 25, 28, 32, 29, 51, 47, 75, 42, 45, 71, 74, 70, 38, 35, 59, 56, 86, 50, 78, 49, 52, 80, 83, 79, 115, 73, 107, 67, 64, 68, 37, 61, 93, 55, 58, 54, 84, 48, 76, 43, 69, 103, 63, 66, 62, 94, 57, 87, 125, 82, 118, 77, 113, 72, 106, 148, 65, 97, 137, 91, 129, 85


The first $11$ leaps are knight moves so the first $12$ terms coincide with A316667.




* There are over one hundred and fifty million terms. After one million terms, numbering the layers of the spiral from the centre out, she has visited $512$ layers and the ($(512times 2-1)^2-10^6=46529$) unvisited squares by layer are: [(1,0),(2,0),...,(494,0),(495, 2), (496, 53), (497, 204), (498, 662), (499, 1208), (500, 1899), (501, 2459), (502, 2921), (503, 3133), (504, 3288), (505, 3397), (506, 3538), (507, 3712), (508, 3868), (509, 3978), (510, 4046), (511, 4074), (512, 4087)], so it's not entirely clear that all squares will be visited or that the wildebeest will never get trapped. Proofs of whether she will ever become trapped or whether she will visit all squares would, of course, be gratefully received!






code-golf sequence integer chess




code-golf sequence integer chess






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 16 mins ago







Jonathan Allan

















asked 5 hours ago









Jonathan AllanJonathan Allan

51.4k535166




51.4k535166












  • $begingroup$
    I suspect the wildebeest gets trapped eventually -- it just takes longer since she has more movement options.
    $endgroup$
    – lirtosiast
    4 hours ago










  • $begingroup$
    "There are over ten million terms." LOL do you have a reference implementation?
    $endgroup$
    – Erik the Outgolfer
    3 hours ago










  • $begingroup$
    @EriktheOutgolfer yes, but she does get stuck somewhere - I need to rerun as was not catching the StopIteration :(
    $endgroup$
    – Jonathan Allan
    3 hours ago


















  • $begingroup$
    I suspect the wildebeest gets trapped eventually -- it just takes longer since she has more movement options.
    $endgroup$
    – lirtosiast
    4 hours ago










  • $begingroup$
    "There are over ten million terms." LOL do you have a reference implementation?
    $endgroup$
    – Erik the Outgolfer
    3 hours ago










  • $begingroup$
    @EriktheOutgolfer yes, but she does get stuck somewhere - I need to rerun as was not catching the StopIteration :(
    $endgroup$
    – Jonathan Allan
    3 hours ago
















$begingroup$
I suspect the wildebeest gets trapped eventually -- it just takes longer since she has more movement options.
$endgroup$
– lirtosiast
4 hours ago




$begingroup$
I suspect the wildebeest gets trapped eventually -- it just takes longer since she has more movement options.
$endgroup$
– lirtosiast
4 hours ago












$begingroup$
"There are over ten million terms." LOL do you have a reference implementation?
$endgroup$
– Erik the Outgolfer
3 hours ago




$begingroup$
"There are over ten million terms." LOL do you have a reference implementation?
$endgroup$
– Erik the Outgolfer
3 hours ago












$begingroup$
@EriktheOutgolfer yes, but she does get stuck somewhere - I need to rerun as was not catching the StopIteration :(
$endgroup$
– Jonathan Allan
3 hours ago




$begingroup$
@EriktheOutgolfer yes, but she does get stuck somewhere - I need to rerun as was not catching the StopIteration :(
$endgroup$
– Jonathan Allan
3 hours ago










1 Answer
1






active

oldest

votes


















2












$begingroup$


JavaScript (Node.js),  191 ... 170  169 bytes



Returns the $N$th term.





n=>(g=(x,y)=>n--?g(Buffer('<3!3(*8*A1C1<:,:').map(m=c=>(i=(a=4*((x+=c-48>>3)*x>(y+=c%8-2)*y?x:y)**2)-(x>y||-1)*(a**.5+x+y))>m|g[i]||(H=x,V=y,m=i))&&H,V,g[m]=1):m+1)(1,2)


Try it online!






share|improve this answer











$endgroup$













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    $begingroup$


    JavaScript (Node.js),  191 ... 170  169 bytes



    Returns the $N$th term.





    n=>(g=(x,y)=>n--?g(Buffer('<3!3(*8*A1C1<:,:').map(m=c=>(i=(a=4*((x+=c-48>>3)*x>(y+=c%8-2)*y?x:y)**2)-(x>y||-1)*(a**.5+x+y))>m|g[i]||(H=x,V=y,m=i))&&H,V,g[m]=1):m+1)(1,2)


    Try it online!






    share|improve this answer











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      2












      $begingroup$


      JavaScript (Node.js),  191 ... 170  169 bytes



      Returns the $N$th term.





      n=>(g=(x,y)=>n--?g(Buffer('<3!3(*8*A1C1<:,:').map(m=c=>(i=(a=4*((x+=c-48>>3)*x>(y+=c%8-2)*y?x:y)**2)-(x>y||-1)*(a**.5+x+y))>m|g[i]||(H=x,V=y,m=i))&&H,V,g[m]=1):m+1)(1,2)


      Try it online!






      share|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$


        JavaScript (Node.js),  191 ... 170  169 bytes



        Returns the $N$th term.





        n=>(g=(x,y)=>n--?g(Buffer('<3!3(*8*A1C1<:,:').map(m=c=>(i=(a=4*((x+=c-48>>3)*x>(y+=c%8-2)*y?x:y)**2)-(x>y||-1)*(a**.5+x+y))>m|g[i]||(H=x,V=y,m=i))&&H,V,g[m]=1):m+1)(1,2)


        Try it online!






        share|improve this answer











        $endgroup$




        JavaScript (Node.js),  191 ... 170  169 bytes



        Returns the $N$th term.





        n=>(g=(x,y)=>n--?g(Buffer('<3!3(*8*A1C1<:,:').map(m=c=>(i=(a=4*((x+=c-48>>3)*x>(y+=c%8-2)*y?x:y)**2)-(x>y||-1)*(a**.5+x+y))>m|g[i]||(H=x,V=y,m=i))&&H,V,g[m]=1):m+1)(1,2)


        Try it online!







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 56 mins ago

























        answered 3 hours ago









        ArnauldArnauld

        74.1k690310




        74.1k690310






























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