meteor-contest.go

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    notice, this list of conditions and the following disclaimer in the
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    * Neither the name of "The Computer Language Benchmarks Game" nor the
    name of "The Computer Language Shootout Benchmarks" nor the names of
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    from this software without specific prior written permission.

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IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
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*/

/* The Computer Language Benchmarks Game
 * http://shootout.alioth.debian.org/
 *
 * contributed by The Go Authors.
 * based on meteor-contest.c by Christian Vosteen
 */

package main

import (
	"flag"
	"fmt"
)

var max_solutions = flag.Int("n", 2100, "maximum number of solutions")


func boolInt(b bool) int8 {
	if b {
		return 1
	}
	return 0
}

/* The board is a 50 cell hexagonal pattern.  For    . . . . .
 * maximum speed the board will be implemented as     . . . . .
 * 50 bits, which will fit into a 64 bit long long   . . . . .
 * int.                                               . . . . .
 *                                                   . . . . .
 * I will represent 0's as empty cells and 1's        . . . . .
 * as full cells.                                    . . . . .
 *                                                    . . . . .
 *                                                   . . . . .
 *                                                    . . . . .
 */

var board uint64 = 0xFFFC000000000000

/* The puzzle pieces must be specified by the path followed
 * from one end to the other along 12 hexagonal directions.
 *
 *   Piece 0   Piece 1   Piece 2   Piece 3   Piece 4
 *
 *  O O O O    O   O O   O O O     O O O     O   O
 *         O    O O           O       O       O O
 *                           O         O         O
 *
 *   Piece 5   Piece 6   Piece 7   Piece 8   Piece 9
 *
 *    O O O     O O       O O     O O        O O O O
 *       O O       O O       O       O O O        O
 *                  O       O O
 *
 * I had to make it 12 directions because I wanted all of the
 * piece definitions to fit into the same size arrays.  It is
 * not possible to define piece 4 in terms of the 6 cardinal
 * directions in 4 moves.
 */

const (
	E = iota
	ESE
	SE
	S
	SW
	WSW
	W
	WNW
	NW
	N
	NE
	ENE
	PIVOT
)

var piece_def = [10][4]int8{
	[4]int8{E, E, E, SE},
	[4]int8{SE, E, NE, E},
	[4]int8{E, E, SE, SW},
	[4]int8{E, E, SW, SE},
	[4]int8{SE, E, NE, S},
	[4]int8{E, E, SW, E},
	[4]int8{E, SE, SE, NE},
	[4]int8{E, SE, SE, W},
	[4]int8{E, SE, E, E},
	[4]int8{E, E, E, SW},
}


/* To minimize the amount of work done in the recursive solve function below,
 * I'm going to allocate enough space for all legal rotations of each piece
 * at each position on the board. That's 10 pieces x 50 board positions x
 * 12 rotations.  However, not all 12 rotations will fit on every cell, so
 * I'll have to keep count of the actual number that do.
 * The pieces are going to be unsigned long long ints just like the board so
 * they can be bitwise-anded with the board to determine if they fit.
 * I'm also going to record the next possible open cell for each piece and
 * location to reduce the burden on the solve function.
 */
var (
	pieces       [10][50][12]uint64
	piece_counts [10][50]int
	next_cell    [10][50][12]int8
)

/* Returns the direction rotated 60 degrees clockwise */
func rotate(dir int8) int8 { return (dir + 2) % PIVOT }

/* Returns the direction flipped on the horizontal axis */
func flip(dir int8) int8 { return (PIVOT - dir) % PIVOT }


/* Returns the new cell index from the specified cell in the
 * specified direction.  The index is only valid if the
 * starting cell and direction have been checked by the
 * out_of_bounds function first.
 */
func shift(cell, dir int8) int8 {
	switch dir {
	case E:
		return cell + 1
	case ESE:
		if ((cell / 5) % 2) != 0 {
			return cell + 7
		} else {
			return cell + 6
		}
	case SE:
		if ((cell / 5) % 2) != 0 {
			return cell + 6
		} else {
			return cell + 5
		}
	case S:
		return cell + 10
	case SW:
		if ((cell / 5) % 2) != 0 {
			return cell + 5
		} else {
			return cell + 4
		}
	case WSW:
		if ((cell / 5) % 2) != 0 {
			return cell + 4
		} else {
			return cell + 3
		}
	case W:
		return cell - 1
	case WNW:
		if ((cell / 5) % 2) != 0 {
			return cell - 6
		} else {
			return cell - 7
		}
	case NW:
		if ((cell / 5) % 2) != 0 {
			return cell - 5
		} else {
			return cell - 6
		}
	case N:
		return cell - 10
	case NE:
		if ((cell / 5) % 2) != 0 {
			return cell - 4
		} else {
			return cell - 5
		}
	case ENE:
		if ((cell / 5) % 2) != 0 {
			return cell - 3
		} else {
			return cell - 4
		}
	}
	return cell
}

/* Returns wether the specified cell and direction will land outside
 * of the board.  Used to determine if a piece is at a legal board
 * location or not.
 */
func out_of_bounds(cell, dir int8) bool {
	switch dir {
	case E:
		return cell%5 == 4
	case ESE:
		i := cell % 10
		return i == 4 || i == 8 || i == 9 || cell >= 45
	case SE:
		return cell%10 == 9 || cell >= 45
	case S:
		return cell >= 40
	case SW:
		return cell%10 == 0 || cell >= 45
	case WSW:
		i := cell % 10
		return i == 0 || i == 1 || i == 5 || cell >= 45
	case W:
		return cell%5 == 0
	case WNW:
		i := cell % 10
		return i == 0 || i == 1 || i == 5 || cell < 5
	case NW:
		return cell%10 == 0 || cell < 5
	case N:
		return cell < 10
	case NE:
		return cell%10 == 9 || cell < 5
	case ENE:
		i := cell % 10
		return i == 4 || i == 8 || i == 9 || cell < 5
	}
	return false
}

/* Rotate a piece 60 degrees clockwise */
func rotate_piece(piece int) {
	for i := 0; i < 4; i++ {
		piece_def[piece][i] = rotate(piece_def[piece][i])
	}
}

/* Flip a piece along the horizontal axis */
func flip_piece(piece int) {
	for i := 0; i < 4; i++ {
		piece_def[piece][i] = flip(piece_def[piece][i])
	}
}

/* Convenience function to quickly calculate all of the indices for a piece */
func calc_cell_indices(cell []int8, piece int, index int8) {
	cell[0] = index
	for i := 1; i < 5; i++ {
		cell[i] = shift(cell[i-1], piece_def[piece][i-1])
	}
}

/* Convenience function to quickly calculate if a piece fits on the board */
func cells_fit_on_board(cell []int8, piece int) bool {
	return !out_of_bounds(cell[0], piece_def[piece][0]) &&
		!out_of_bounds(cell[1], piece_def[piece][1]) &&
		!out_of_bounds(cell[2], piece_def[piece][2]) &&
		!out_of_bounds(cell[3], piece_def[piece][3])
}

/* Returns the lowest index of the cells of a piece.
 * I use the lowest index that a piece occupies as the index for looking up
 * the piece in the solve function.
 */
func minimum_of_cells(cell []int8) int8 {
	minimum := cell[0]
	for i := 1; i < 5; i++ {
		if cell[i] < minimum {
			minimum = cell[i]
		}
	}
	return minimum
}

/* Calculate the lowest possible open cell if the piece is placed on the board.
 * Used to later reduce the amount of time searching for open cells in the
 * solve function.
 */
func first_empty_cell(cell []int8, minimum int8) int8 {
	first_empty := minimum
	for first_empty == cell[0] || first_empty == cell[1] ||
		first_empty == cell[2] || first_empty == cell[3] ||
		first_empty == cell[4] {
		first_empty++
	}
	return first_empty
}

/* Generate the unsigned long long int that will later be anded with the
 * board to determine if it fits.
 */
func bitmask_from_cells(cell []int8) uint64 {
	var piece_mask uint64
	for i := 0; i < 5; i++ {
		piece_mask |= 1 << uint(cell[i])
	}
	return piece_mask
}

/* Record the piece and other important information in arrays that will
 * later be used by the solve function.
 */
func record_piece(piece int, minimum int8, first_empty int8, piece_mask uint64) {
	pieces[piece][minimum][piece_counts[piece][minimum]] = piece_mask
	next_cell[piece][minimum][piece_counts[piece][minimum]] = first_empty
	piece_counts[piece][minimum]++
}


/* Fill the entire board going cell by cell.  If any cells are "trapped"
 * they will be left alone.
 */
func fill_contiguous_space(board []int8, index int8) {
	if board[index] == 1 {
		return
	}
	board[index] = 1
	if !out_of_bounds(index, E) {
		fill_contiguous_space(board, shift(index, E))
	}
	if !out_of_bounds(index, SE) {
		fill_contiguous_space(board, shift(index, SE))
	}
	if !out_of_bounds(index, SW) {
		fill_contiguous_space(board, shift(index, SW))
	}
	if !out_of_bounds(index, W) {
		fill_contiguous_space(board, shift(index, W))
	}
	if !out_of_bounds(index, NW) {
		fill_contiguous_space(board, shift(index, NW))
	}
	if !out_of_bounds(index, NE) {
		fill_contiguous_space(board, shift(index, NE))
	}
}


/* To thin the number of pieces, I calculate if any of them trap any empty
 * cells at the edges.  There are only a handful of exceptions where the
 * the board can be solved with the trapped cells.  For example:  piece 8 can
 * trap 5 cells in the corner, but piece 3 can fit in those cells, or piece 0
 * can split the board in half where both halves are viable.
 */
func has_island(cell []int8, piece int) bool {
	temp_board := make([]int8, 50)
	var i int
	for i = 0; i < 5; i++ {
		temp_board[cell[i]] = 1
	}
	i = 49
	for temp_board[i] == 1 {
		i--
	}
	fill_contiguous_space(temp_board, int8(i))
	c := 0
	for i = 0; i < 50; i++ {
		if temp_board[i] == 0 {
			c++
		}
	}
	if c == 0 || (c == 5 && piece == 8) || (c == 40 && piece == 8) ||
		(c%5 == 0 && piece == 0) {
		return false
	}
	return true
}


/* Calculate all six rotations of the specified piece at the specified index.
 * We calculate only half of piece 3's rotations.  This is because any solution
 * found has an identical solution rotated 180 degrees.  Thus we can reduce the
 * number of attempted pieces in the solve algorithm by not including the 180-
 * degree-rotated pieces of ONE of the pieces.  I chose piece 3 because it gave
 * me the best time ;)
 */
func calc_six_rotations(piece, index int) {
	cell := make([]int8, 5)
	for rotation := 0; rotation < 6; rotation++ {
		if piece != 3 || rotation < 3 {
			calc_cell_indices(cell, piece, int8(index))
			if cells_fit_on_board(cell, piece) && !has_island(cell, piece) {
				minimum := minimum_of_cells(cell)
				first_empty := first_empty_cell(cell, minimum)
				piece_mask := bitmask_from_cells(cell)
				record_piece(piece, minimum, first_empty, piece_mask)
			}
		}
		rotate_piece(piece)
	}
}

/* Calculate every legal rotation for each piece at each board location. */
func calc_pieces() {
	for piece := 0; piece < 10; piece++ {
		for index := 0; index < 50; index++ {
			calc_six_rotations(piece, index)
			flip_piece(piece)
			calc_six_rotations(piece, index)
		}
	}
}


/* Calculate all 32 possible states for a 5-bit row and all rows that will
 * create islands that follow any of the 32 possible rows.  These pre-
 * calculated 5-bit rows will be used to find islands in a partially solved
 * board in the solve function.
 */
const (
	ROW_MASK    = 0x1F
	TRIPLE_MASK = 0x7FFF
)

var (
	all_rows = [32]int8{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16,
		17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31,
	}
	bad_even_rows   [32][32]int8
	bad_odd_rows    [32][32]int8
	bad_even_triple [32768]int8
	bad_odd_triple  [32768]int8
)

func rows_bad(row1, row2 int8, even bool) int8 {
	/* even is referring to row1 */
	var row2_shift int8
	/* Test for blockages at same index and shifted index */
	if even {
		row2_shift = ((row2 << 1) & ROW_MASK) | 0x01
	} else {
		row2_shift = (row2 >> 1) | 0x10
	}
	block := ((row1 ^ row2) & row2) & ((row1 ^ row2_shift) & row2_shift)
	/* Test for groups of 0's */
	in_zeroes := false
	group_okay := false
	for i := uint8(0); i < 5; i++ {
		if row1&(1<<i) != 0 {
			if in_zeroes {
				if !group_okay {
					return 1
				}
				in_zeroes = false
				group_okay = false
			}
		} else {
			if !in_zeroes {
				in_zeroes = true
			}
			if (block & (1 << i)) == 0 {
				group_okay = true
			}
		}
	}
	if in_zeroes {
		return boolInt(!group_okay)
	}
	return 0
}

/* Check for cases where three rows checked sequentially cause a false
 * positive.  One scenario is when 5 cells may be surrounded where piece 5
 * or 7 can fit.  The other scenario is when piece 2 creates a hook shape.
 */
func triple_is_okay(row1, row2, row3 int, even bool) bool {
	if even {
		/* There are four cases:
		 * row1: 00011  00001  11001  10101
		 * row2: 01011  00101  10001  10001
		 * row3: 011??  00110  ?????  ?????
		 */
		return ((row1 == 0x03) && (row2 == 0x0B) && ((row3 & 0x1C) == 0x0C)) ||
			((row1 == 0x01) && (row2 == 0x05) && (row3 == 0x06)) ||
			((row1 == 0x19) && (row2 == 0x11)) ||
			((row1 == 0x15) && (row2 == 0x11))
	}
	/* There are two cases:
	 * row1: 10011  10101
	 * row2: 10001  10001
	 * row3: ?????  ?????
	 */
	return ((row1 == 0x13) && (row2 == 0x11)) ||
		((row1 == 0x15) && (row2 == 0x11))
}

func calc_rows() {
	for row1 := int8(0); row1 < 32; row1++ {
		for row2 := int8(0); row2 < 32; row2++ {
			bad_even_rows[row1][row2] = rows_bad(row1, row2, true)
			bad_odd_rows[row1][row2] = rows_bad(row1, row2, false)
		}
	}
	for row1 := 0; row1 < 32; row1++ {
		for row2 := 0; row2 < 32; row2++ {
			for row3 := 0; row3 < 32; row3++ {
				result1 := bad_even_rows[row1][row2]
				result2 := bad_odd_rows[row2][row3]
				if result1 == 0 && result2 != 0 && triple_is_okay(row1, row2, row3, true) {
					bad_even_triple[row1+(row2*32)+(row3*1024)] = 0
				} else {
					bad_even_triple[row1+(row2*32)+(row3*1024)] = boolInt(result1 != 0 || result2 != 0)
				}

				result1 = bad_odd_rows[row1][row2]
				result2 = bad_even_rows[row2][row3]
				if result1 == 0 && result2 != 0 && triple_is_okay(row1, row2, row3, false) {
					bad_odd_triple[row1+(row2*32)+(row3*1024)] = 0
				} else {
					bad_odd_triple[row1+(row2*32)+(row3*1024)] = boolInt(result1 != 0 || result2 != 0)
				}
			}
		}
	}
}


/* Calculate islands while solving the board.
 */
func boardHasIslands(cell int8) int8 {
	/* Too low on board, don't bother checking */
	if cell >= 40 {
		return 0
	}
	current_triple := (board >> uint((cell/5)*5)) & TRIPLE_MASK
	if (cell/5)%2 != 0 {
		return bad_odd_triple[current_triple]
	}
	return bad_even_triple[current_triple]
}


/* The recursive solve algorithm.  Try to place each permutation in the upper-
 * leftmost empty cell.  Mark off available pieces as it goes along.
 * Because the board is a bit mask, the piece number and bit mask must be saved
 * at each successful piece placement.  This data is used to create a 50 char
 * array if a solution is found.
 */
var (
	avail          uint16 = 0x03FF
	sol_nums       [10]int8
	sol_masks      [10]uint64
	solutions      [2100][50]int8
	solution_count = 0
)

func record_solution() {
	for sol_no := 0; sol_no < 10; sol_no++ {
		sol_mask := sol_masks[sol_no]
		for index := 0; index < 50; index++ {
			if sol_mask&1 == 1 {
				solutions[solution_count][index] = sol_nums[sol_no]
				/* Board rotated 180 degrees is a solution too! */
				solutions[solution_count+1][49-index] = sol_nums[sol_no]
			}
			sol_mask = sol_mask >> 1
		}
	}
	solution_count += 2
}

func solve(depth, cell int8) {
	if solution_count >= *max_solutions {
		return
	}

	for board&(1<<uint(cell)) != 0 {
		cell++
	}

	for piece := int8(0); piece < 10; piece++ {
		var piece_no_mask uint16 = 1 << uint(piece)
		if avail&piece_no_mask == 0 {
			continue
		}
		avail ^= piece_no_mask
		max_rots := piece_counts[piece][cell]
		piece_mask := pieces[piece][cell]
		for rotation := 0; rotation < max_rots; rotation++ {
			if board&piece_mask[rotation] == 0 {
				sol_nums[depth] = piece
				sol_masks[depth] = piece_mask[rotation]
				if depth == 9 {
					/* Solution found!!!!!11!!ONE! */
					record_solution()
					avail ^= piece_no_mask
					return
				}
				board |= piece_mask[rotation]
				if boardHasIslands(next_cell[piece][cell][rotation]) == 0 {
					solve(depth+1, next_cell[piece][cell][rotation])
				}
				board ^= piece_mask[rotation]
			}
		}
		avail ^= piece_no_mask
	}
}

/* pretty print a board in the specified hexagonal format */
func pretty(b *[50]int8) {
	for i := 0; i < 50; i += 10 {
		fmt.Printf("%c %c %c %c %c \n %c %c %c %c %c \n", b[i]+'0', b[i+1]+'0',
			b[i+2]+'0', b[i+3]+'0', b[i+4]+'0', b[i+5]+'0', b[i+6]+'0',
			b[i+7]+'0', b[i+8]+'0', b[i+9]+'0')
	}
	fmt.Printf("\n")
}

/* Find smallest and largest solutions */
func smallest_largest() (smallest, largest *[50]int8) {
	smallest = &solutions[0]
	largest = &solutions[0]
	for i := 1; i < solution_count; i++ {
		candidate := &solutions[i]
		for j, s := range *smallest {
			c := candidate[j]
			if c == s {
				continue
			}
			if c < s {
				smallest = candidate
			}
			break
		}
		for j, s := range *largest {
			c := candidate[j]
			if c == s {
				continue
			}
			if c > s {
				largest = candidate
			}
			break
		}
	}
	return
}

func main() {
	flag.Parse()
	calc_pieces()
	calc_rows()
	solve(0, 0)
	fmt.Printf("%d solutions found\n\n", solution_count)
	smallest, largest := smallest_largest()
	pretty(smallest)
	pretty(largest)
}