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# Pawn Alice

## Question

The  Queen  and  Alice  traveled  across  the chessboard  dashing
from  square  to  square  without  returning  to  the visited squares.
Starting on  any square.. and if every next  moves are farther than
the  previous  ones,  how  many  squares  shall  they  visit?

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... with 32 jumps visiting 33 squares (including the starting one) for the total path length of approx 174.347

1x0 1.000000
1x1 1.414214
2x0 2.000000
2x1 2.236068
2x2 2.828427
3x0 3.000000
3x1 3.162278
3x2 3.605551
4x0 4.000000
4x1 4.123106
3x3 4.242641
4x2 4.472136
5x0 5.000000
5x1 5.099020
5x2 5.385165
4x4 5.656854
5x3 5.830952
6x0 6.000000
6x1 6.082763
6x2 6.324555
5x4 6.403124
6x3 6.708204
7x0 7.000000
7x1 7.071068
6x4 7.211103
7x2 7.280110
7x3 7.615773
6x5 7.810250
7x4 8.062258
7x5 8.602325
7x6 9.219544
7x7 9.899495

Total distance 174.346982

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thirty-six, not counting the starting square.
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thirty-six, not counting the starting square.
Yes Bonanova! 35 is all the possible distances between 2 squares in a chessboard (center to center).
Got a path ?
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If you move N squares diagonally, do you consider that the same as moving N squares horizontally or vertically because the same number of squares are visited, or longer because you travel a distance that is longer by a factor of sqrt(2)?

Also want to see bonanova's path. I reached a different (smaller) answer, operating under the first condition of having diagonal moves count the same as horizontal and vertical moves.

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Yes, the diagonals are farther. The initial position matters.

They can travel around unrestricted by Queen's regular moves

and not touching any squares along the way.

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Yes, the diagonals are farther. The initial position matters.

They can travel around unrestricted by Queen's regular moves

and not touching any (previously visited?) squares along the way.

32 is trivial without the sqrt(2) differentiator.

29 is my first shot with it, and that probably can be improved with crossing diagonal paths.

Move lengths are 1, 1.4, 2, 2.8, 3, 4, 4.2, 5, 5.6, 6, 7, 7.1, 8, 8.5, 9.9 and 13.3

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@bonanova:

I mean the the moves are non-sliding..more like flying

29 is high for a first shot, i guess you'll max it out anyway..
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...then the highest number of moves may be 9 (10, if a zero distance move is permitted).

One solution is
(0) C1 = 0
(1) C2 = 1
(2) D1 = √2 (≈1.414)
(3) F1 = 2 (en passant: E1)
(4) D3 = 2√2 (≈2.828) (en passant: E2)
(5) G3 = 3 (en passant: E3 and F3)
(6) G8 = 5 (en passant: G4, G5, G6, and G7)
(7) A8 = 6 (en passant: F8, E8, D8, C8, and B8)
(8) A1 = 7 (en passant: A7, A6, A5, A4, A3, and A2)
(9) F6 = 5√2 (≈7.071) (en passant: B2, C3, D4, E5)
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... then the count may be 13 (14, if a zero move is permitted).

(Note: It has been inferred that only a "Queen's move" is permissible, i.e. a diagonal move must be parallel to the diagonal of the square playing field, and otherwise, only orthogonal moves are performed.)
(0) E5 = 0
(1) D5 = 1
(2) E4 = √2 (≈1.414)
(3) C4 = 2
(4) E2 = 2√2 (≈2.828)
(5) H2 = 3
(6) D2 = 4
(7) A5 = 3√2 (≈4.243)
(8) F5 = 5
(9) B1 = 4√2 (≈5.659)
(10) B7 = 6
(11) G2 = 5√2 (≈7.071)
(12) A8 = 6√2 (≈8.485)
(13) H1 = 7√2 (≈9.899)
Edited by DejMar
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...then the highest number of moves may be 9 (10, if a zero distance move is permitted).

One solution is

(0) C1 = 0

(1) C2 = 1

(2) D1 = √2 (≈1.414)

(3) F1 = 2 (en passant: E1)

(4) D3 = 2√2 (≈2.828) (en passant: E2)

(5) G3 = 3 (en passant: E3 and F3)

(6) G8 = 5 (en passant: G4, G5, G6, and G7)

(7) A8 = 6 (en passant: F8, E8, D8, C8, and B8)

(8) A1 = 7 (en passant: A7, A6, A5, A4, A3, and A2)

(9) F6 = 5√2 (≈7.071) (en passant: B2, C3, D4, E5)

Under the interpretation that the total squares 'visited' are to be counted, and not simply the number of moves, the number of squares visited (starting, passing through, and ending in) is much higher than the count of 9 (or 10, the number of moves). Rather, including the beginning square, the 9 squares where the move ended, and the 23 squares passed through, the count would be 33 (or 32, if the starting square is not counted as being visited.)

Edited by DejMar
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@bonanova
Im curious how you found a move length of 13.3. Wouldn't that take you off the playing field? A chessboard is 8x8, but as the initial starting square is not counted*, the longest number of squares moved (as opposed to distance*) is 7.

Edited by DejMar
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...then the highest number of moves may be 9 (10, if a zero distance move is permitted).

One solution is

(0) C1 = 0

(1) C2 = 1

(2) D1 = √2 (≈1.414)

(3) F1 = 2 (en passant: E1)

(4) D3 = 2√2 (≈2.828) (en passant: E2)

(5) G3 = 3 (en passant: E3 and F3)

(6) G8 = 5 (en passant: G4, G5, G6, and G7)

(7) A8 = 6 (en passant: F8, E8, D8, C8, and B8)

(8) A1 = 7 (en passant: A7, A6, A5, A4, A3, and A2)

(9) F6 = 5√2 (≈7.071) (en passant: B2, C3, D4, E5)

Under the interpretation that the total squares 'visited' are to be counted, and not simply the number of moves, the number of squares visited (starting, passing through, and ending in) is much higher than the count of 9 (or 10, the number of moves). Rather, including the beginning square, the 9 squares where the move ended, and the 23 squares passed through, the count would be 33 (or 32, if the starting square is not counted as being visited.)

As TSLF said, Alice and the queen are unrestrained by how a queen normally moves on a chessboard. My understanding of it is that Alice can "jump" from anywhere on the board to any other unvisited square, without touching any of the squares in between.

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I would try this if I had access to a chessboard, but since we're probably trying to make all 32 jumps, I would start with the longest one and then make shorter and shorter jumps. For one, the starting point is obvious (the corner), and there isn't a whole lot of different ways you can make the first few jumps.

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Having some time on my hands, I decided to carry out the strategy I described in the previous post.

But first, I calculated the distances of each move and ordered them by length. What I found was that there are only 33 possible distances between two squares (since 3,4 = 0,5 and 1,7 = 5,5), which is different from what TSLF and bonanova found (35 possible distances)

I went ahead and started laying out a path, starting with the longest jump and then making shorter and shorter jumps. I found that there was only one way to make the first five jumps. However, to my surprise, it was impossible to make the sixth jump.

So contrary to what was established, I think the number of squares visited has to be strictly less than 34 (less than 33 jumps).

Edited by gavinksong
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... with 32 jumps visiting 33 squares (including the starting one) for the total path length of approx 174.347

jumps.png

1x0 1.000000

1x1 1.414214

2x0 2.000000

2x1 2.236068

2x2 2.828427

3x0 3.000000

3x1 3.162278

3x2 3.605551

4x0 4.000000

4x1 4.123106

3x3 4.242641

4x2 4.472136

5x0 5.000000

5x1 5.099020

5x2 5.385165

4x4 5.656854

5x3 5.830952

6x0 6.000000

6x1 6.082763

6x2 6.324555

5x4 6.403124

6x3 6.708204

7x0 7.000000

7x1 7.071068

6x4 7.211103

7x2 7.280110

7x3 7.615773

6x5 7.810250

7x4 8.062258

7x5 8.602325

7x6 9.219544

7x7 9.899495

Total distance 174.346982

You got it.

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... with 32 jumps visiting 33 squares (including the starting one) for the total path length of approx 174.347

1x0 1.000000
1x1 1.414214
2x0 2.000000
2x1 2.236068
2x2 2.828427
3x0 3.000000
3x1 3.162278
3x2 3.605551
4x0 4.000000
4x1 4.123106
3x3 4.242641
4x2 4.472136
5x0 5.000000
5x1 5.099020
5x2 5.385165
4x4 5.656854
5x3 5.830952
6x0 6.000000
6x1 6.082763
6x2 6.324555
5x4 6.403124
6x3 6.708204
7x0 7.000000
7x1 7.071068
6x4 7.211103
7x2 7.280110
7x3 7.615773
6x5 7.810250
7x4 8.062258
7x5 8.602325
7x6 9.219544
7x7 9.899495

Total distance 174.346982

Big Hit K-man,that's the path!

*extra 5.000, 7.071, 8.485 distances excluded

Edited by TimeSpaceLightForce
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Kudos to k-man.

Permitting (real) Queen moves only, this seems optimal.

There are 14 different-length Queen moves for an upper limit of 15 squares touched.

See if that's possible. They are, where D=1.4 signifies diagonal moves:

7D 6D 5D 7 6 4D 5 3D 4 3 2D 2 1D 1

Start at corner a1 (chess notation.) 7D, 6D, 5D takes us to g1.

From here, a move of 7 is not possible. The upper limit is now 14 squares.

Moves 6, 4D and 5 are all constrained, leading to

a1 - h8 - b2 - g7 - g1 - c5 - h5.

Moves 3D and 4 lead to a selection of a2, a8, c4 or c8. Pick
a8.

a1 - h8 - b2 - g7 - g1 - c5 - h5 - e8 - a8 - a5 - c7 - a7 - b6 - b7

(14 squares.)

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@bonanova, I had already demonstrated that in post #9. A difference is that I began at the board's heart instead of a corner.

Oh, and yes, kudos to k-man.

Edited by DejMar

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