Suppose we have a solution to the n -Queens problem instance...

Suppose we have a solution to the $n$ -Queens problem instance in which $n=4 .$ Can we extend this solution to find a solution to the problem instance in which $n=5 ?$ Can we then use the solutions for $n=4$ and $n=5$ to construct a solution to the instance in which $n=6$ and continue this dynamic programming approach to find a solution to any instance in which $n>4 ?$ Justify your answer.

Key Concepts

n-Queens Problem

The n-Queens Problem is a classic example of a constraint satisfaction problem where the goal is to place n queens on an n×n chessboard such that no two queens threaten each other. This requires careful consideration of rows, columns, and both sets of diagonals, making the problem a challenging exercise in combinatorial optimization and search algorithms.

Dynamic Programming

Dynamic programming is a method for solving complex problems by breaking them down into simpler subproblems and storing the solutions to these subproblems to avoid redundant computations. It is most effective when the problem exhibits overlapping subproblems and an optimal substructure, meaning the optimal solution to the entire problem can be constructed efficiently from optimal solutions of its parts.

Subproblem Extension and Independence

In many constraint satisfaction and combinatorial problems, the overall structure does not lend itself to simple extension from a solution of a smaller instance to a larger one because the additional constraints introduced in the larger instance can invalidate or conflict with the smaller solution. Therefore, assuming that a valid solution for a smaller problem instance can be incrementally grown to solve a larger instance typically fails if the problem does not have the necessary property of optimal substructure across instance sizes.

Transcript

00:01 Okay, so here we're given a n is equal to five times a n minus 1 minus 6 times a .n minus 2 plus 2 to the power of n plus 3 times n.

00:12 So first we want the roots characteristic equation.

00:24 And so to do that, we're going to substitute where a .n is equal to r squared.

00:32 A n minus 1 will be equal to r and a n minus 2 is equal to 1 and all other functions of n will be equal to 0.

00:49 So that transforms a n then to r squared is equal to 5r minus 6.

01:00 Okay, and if we subtract 5r and add 6 to both sides, and then a factor, we get r minus 2 times r minus 3 is equal to 0.

01:17 So r is equal to 2 and 3.

01:22 So this means that our homogeneous portion of our solution, so we'll put a little h for homogeneous, is going to be equal to some constant alpha 1 times 2 to the power of n, plus alpha 2, another constant, times 3 to the power of n, which is each of our roots raised to the power of n.

01:49 I'm just going to have this in our pocket for one ready for it, and figure out the particular solution component.

02:00 So the particular solution, we have f of n being equal to 2n plus 3 times n.

02:14 Of the components we said equal to zero previously.

02:18 And this is equal to 2 to the power of n plus 3 times n times 1 to the power of n.

02:41 Okay, so 2 is a root of our characteristic equation with multiplicity 1, and 1 is not a root.

02:52 So we're looking for a particular solution.

02:57 Going to be in the form of q times n times 2 to the power of n plus p1 times n plus p .0.

03:26 So the thing is this particular solution needs to satisfy our recurrence relation.

03:34 Right? so what we're going to do then is put it into our recurrence relation...

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