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### SICP Exercise 2.42: Eight queens puzzle

2021-10-29

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Exercise 2.42: The “eight-queens puzzle” asks how to place eight queens on a chessboard so that no queen is in check from any other (i.e., no two queens are in the same row, column, or diagonal).

One way to solve the puzzle is to work across the board, placing a queen in each column. Once we have placed `k - 1` queens, we must place the `k`th queen in a position where it does not check any of the queens already on the board. We can formulate this approach recursively: Assume that we have already generated the sequence of all possible ways to place `k - 1` queens in the first `k - 1` columns of the board. For each of these ways, generate an extended set of positions by placing a queen in each row of the `k`th column. Now filter these, keeping only the positions for which the queen in the `k`th column is safe with respect to the other queens. This produces the sequence of all ways to place `k` queens in the first `k` columns. By continuing this process, we will produce not only one solution, but all solutions to the puzzle.

We implement this solution as a procedure `queens`, which returns a sequence of all solutions to the problem of placing `n` queens on an `n×n` chessboard. Queens has an internal procedure `queen-cols` that returns the sequence of all ways to place queens in the first `k` columns of the board.

``` 1(define (queen board-size)
2    (define (queen-cols k)
3        (if (= k 0)
4            (list empty-board)
5            (filter
6                (lambda (positions) (safe? k positions))
7                (flatmap
8                    (lambda (rest-of-queens)
9                        (map (lambda (new-row)
11                             (enumerate-interval 1 board-size)))
12                    (queen-cols (- k 1))))))
13    (queen-cols board-size))
```

In this procedure `rest-of-queens` is a way to place `k - 1` queens in the first `k - 1` columns, and `new-row` is a proposed row in which to place the queen for the `k`th column. Complete the program by implementing the representation for sets of board positions, including the procedure `adjoin-position`, which adjoins a new row-column position to a set of positions, and `empty-board`, which represents an empty set of positions. You must also write the procedure `safe?`, which determines for a set of positions, whether the queen in the `k`th column is safe with respect to the others. (Note that we need only check whether the new queen is safe – the other queens are already guaranteed safe with respect to each other.)

I will try to implement the simpler procedures first.

`empty-board` represents an empty set of positions:

```1(define empty-board '())
```

`adjoin-position` adjoins a new row-column position to a set of positions:

```1(define (adjoin-position row col rest-of-queens)
2    (cons (list row col) rest-of-queens))
```

`attack?` checks if two queens are in the same row, column, or diagonal:

``` 1(define (attack? q1 q2)
2    (or (= (row q1) (row q2))
3        (= (abs (- (row q1) (row q2)))
4           (abs (- (col q1) (col q2))))))
5
6(define (row position)
7    (car position))
8
9(define (col position)
```

(No need to check if they are in the same column)

The most difficult procedure is `safe?`: which determines for a set of positions, whether the queen in the `k`th column is safe with respect to the others.

The steps should be something like this:

• get the queen in the `k`th column
• get the first element of the `rest-of-queens`
• check if two queens are under attack
• if yes, return `#f` (false)
• else: remove the first element from `rest-of-queens` and do the same with the remaining elements

Here’s the code:

``` 1(define (safe? k positions)
2    (if (= (length positions) 1)
3        #t
4        (if (attack? (k-queen positions) (first-queen positions))
5            #f
6            (safe? k (remove (first-queen positions) positions)))))
7
8(define (k-queen positions)
9    (car positions))
10
11(define (first-queen positions)
```

The `remove` procedure is similar to the one at the end of Nested Mappings section but uses `equals?` to compare list:

```1(trace-define (remove item sequence)
2    (filter (lambda (x) (not (equal? x item)))
3            sequence))
```

The completed program:

``` 1(trace-define (queen board-size)
2    (define (queen-cols k)
3        (if (= k 0)
4            (list empty-board)
5            (filter
6                (lambda (positions) (safe? k positions))
7                (flatmap
8                    (lambda (rest-of-queens)
9                        (map (lambda (new-row)
11                             (enumerate-interval 1 board-size)))
12                    (queen-cols (- k 1))))))
13    (trace queen-cols)
14    (queen-cols board-size))
15
16(define empty-board '())
17
19    (cons (list row col) rest-of-queens))
20
21(define (safe? k positions)
22    (if (= (length positions) 1)
23        #t
24        (if (attack? (k-queen positions) (first-queen positions))
25            #f
26            (safe? k (remove (first-queen positions) positions)))))
27
28(define (k-queen positions)
29    (car positions))
30
31(define (first-queen positions)
33
34(define (attack? q1 q2)
35    (or (= (row q1) (row q2))
36        (= (abs (- (row q1) (row q2)))
37           (abs (- (col q1) (col q2))))))
38
39(define (row position)
40    (car position))
41
42(define (col position)
44
45(define (remove item sequence)
46    (filter (lambda (x) (not (equal? x item)))
47            sequence))
```

Test:

``` 1\$ racket 2.42-eight-queens.rkt
2>(queen 4)
3>(queen-cols 4)
4> (queen-cols 3)
5> >(queen-cols 2)
6> > (queen-cols 1)
7> > >(queen-cols 0)
8< < <'(())
9< < '(((1 1)) ((2 1)) ((3 1)) ((4 1)))
10< <'(((3 2) (1 1))
11     ((4 2) (1 1))
12     ((4 2) (2 1))
13     ((1 2) (3 1))
14     ((1 2) (4 1))
15     ((2 2) (4 1)))
16< '(((2 3) (4 2) (1 1))
17    ((1 3) (4 2) (2 1))
18    ((4 3) (1 2) (3 1))
19    ((3 3) (1 2) (4 1)))
20<'(((3 4) (1 3) (4 2) (2 1)) ((2 4) (4 3) (1 2) (3 1)))
21'(((3 4) (1 3) (4 2) (2 1)) ((2 4) (4 3) (1 2) (3 1)))
```
```1|---|---|---|---|
2|   |   | X |   |
3|---|---|---|---|
4| X |   |   |   |
5|---|---|---|---|
6|   |   |   | X |
7|---|---|---|---|
8|   | X |   |   |
9|---|---|---|---|
```
```1|---|---|---|---|
2|   | X |   |   |
3|---|---|---|---|
4|   |   |   | X |
5|---|---|---|---|
6| X |   |   |   |
7|---|---|---|---|
8|   |   | X |   |
9|---|---|---|---|
```

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