## Description

1. (24 Points) isLegalPosition(board,n)

Write a function isLegalPosition(board,n) that takes a (possibly partial)

position and n as arguments and returns TRUE if and only if no two Queens

attack each other.

Here is a solution to the 8-Queens Problem:

(1 6 8 3 7 4 2 5)

QZ0Z0Z0Z Z0Z0ZQZ0

0Z0Z0Z0L Z0L0Z0Z0 0Z0Z0ZQZ Z0ZQZ0Z0

0L0Z0Z0Z Z0Z0L0Z0

Thus, isLegalPosition((1 6 8 3 7 4 2 5),8) should return TRUE.

1

Because we are implementing a backtracking algorithm, we will restrict ourselves to positions which fill from the top of the board. We will insist then

that the first k ≤ n positions be filled, i.e. non-zero, but the remaining n − k

positions may be zeroes. So the partial solution:

(1 6 8 3 7 0 0 0)

QZ0Z0Z0Z Z0Z0ZQZ0 0Z0Z0Z0L Z0L0Z0Z0

0Z0Z0ZQZ Z0Z0Z0Z0 0Z0Z0Z0Z

Z0Z0Z0Z0

should also have isLegalPosition((1 6 8 3 7 0 0 0),8) return TRUE, while

(1 6 8 3 5 0 0 0)

QZ0Z0Z0Z Z0Z0ZQZ0

0Z0Z0Z0L Z0L0Z0Z0 0Z0ZqZ0Z

Z0Z0Z0Z0 0Z0Z0Z0Z Z0Z0Z0Z0

should cause isLegalPosition((1 6 8 3 5 0 0 0),8) to return FALSE.

Why?

Do you see an elegant way to check that?

2. (18 Points) NextLegalPosition(board,n)

From any (possibly partial) position, we need to be able to find the next legal

position. There are, perhaps, three cases here. First, the next legal position

from an illegal partial position; second, the next legal position from a legal

partial position, and third, the next legal position after a full-fledged solution.

We will fill our board from the top down and from left to right, so the next

legal position after (illegal) partial position:

(1 6 8 3 5 0 0 0)

QZ0Z0Z0Z Z0Z0ZQZ0 0Z0Z0Z0L

Z0L0Z0Z0 0Z0ZqZ0Z Z0Z0Z0Z0 0Z0Z0Z0Z

Z0Z0Z0Z0

is

(1 6 8 3 7 0 0 0)

QZ0Z0Z0Z Z0Z0ZQZ0

0Z0Z0Z0L Z0L0Z0Z0 0Z0Z0ZQZ Z0Z0Z0Z0

0Z0Z0Z0Z Z0Z0Z0Z0

And the next legal position after (legal!) partial position:

(1 6 8 3 7 0 0 0)

QZ0Z0Z0Z Z0Z0ZQZ0 0Z0Z0Z0L

Z0L0Z0Z0 0Z0Z0ZQZ Z0Z0Z0Z0 0Z0Z0Z0Z

Z0Z0Z0Z0

is

(1 6 8 3 7 4 0 0)

QZ0Z0Z0Z Z0Z0ZQZ0

0Z0Z0Z0L Z0L0Z0Z0 0Z0Z0ZQZ Z0ZQZ0Z0

0Z0Z0Z0Z Z0Z0Z0Z0

Will the next legal position from a legal position always add a Queen to the

next rank?

Why? / Why not?

Lastly, the next legal position after our solution:

(1 6 8 3 7 4 2 5)

QZ0Z0Z0Z Z0Z0ZQZ0

0Z0Z0Z0L Z0L0Z0Z0 0Z0Z0ZQZ Z0ZQZ0Z0

0L0Z0Z0Z Z0Z0L0Z0

is

(1 6 8 5 0 0 0 0)

QZ0Z0Z0Z Z0Z0ZQZ0

0Z0Z0Z0L Z0Z0L0Z0 0Z0Z0Z0Z

Z0Z0Z0Z0 0Z0Z0Z0Z Z0Z0Z0Z0

(Understanding this is understanding the backtracking–and then the forwarding–

we are doing. This is the crux of the method.)

Write a function NextLegalPosition(board,n) that takes a (possibly partial) position and n as arguments and returns a board/position/array that represents the next legal position, or (01, 02, . . . , 0n) if no legal position succeeds

“board”.

Hint: It may be useful to write another function Successor(board, n) that

returns the next position to “board”, whether legal or not.

3. (12 Points) Find the “first” solution to the n-Queens Problem for n = 4…100.

With isLegalPosition(board,n) and NextLegalPosition(board,n) in

your hip pocket, write a program which solves the n-Queens problem for all

values between 4 and 100, inclusive.

Your output should give a single solution to each instance of the problem, and

it should be the first solution lexicographically.

We saw that the 4-Queens problem has solutions (2, 4, 1, 3) and (3, 1, 4, 2) as

its distinct solutions. Your output should be the first of these.

Is our solution to the 8-Queens problem the first one?

4. (6 Points) Find all solutions to the n-Queens Problem for a particular n.

With isLegalPosition(board,n) and NextLegalPosition(board,n) in

your hip pocket, write a program which finds all solutions to the n-Queens

problem for a single instance of the problem with 4 ≤ n ≤ 20, input by the

user. Your output should consist of a list of solutions in lexicographical order.

And keep a running count/index!

(What constitutes a “solution”?)

For parts 2-4, you can get some gains in efficiency by modifying isLegalPosition(board,n).

We will build positions from legal positions. This means that only the last Queen,

the last non-zero entry, can cause a position to be illegal. Do you see?

Why are you going to want increased efficiency?

You may find the Wikipedia entry on the “8-Queens puzzle” to be helpful. . . maybe

even interesting.