In this session, we are going to introduce sets as another fundamental collection type. We are then going to combine sets and four expressions in a classical combinatorial search problem namely the m-queens problem. So far all the collection that we've seen were sequences of some sort or another. But actually also two other fundamental class of collections, one is sets and the other is map, and in this session we're going to take a closer look at sets. Sets are actually very close to sequences. You form sets just like sequences, so you could write fruit equals set of apple, banana, pears, let's say, or s equals 1 to 6 .toSet. So, toSet is an operator that takes a sequence and converts a sequence to a set. Most operations on sequences are also available on sets, so you could try s map and this is called +2 anytime you would expect to see a set that goes from 3, 4, 5, 6, 7, 8. Or you could write fruit filter where the fruit must start with "app," so that would give us a set of just apples. Or you could ask whether a set is nonEmpty, say all these operations are shared between sequences and sets. To find a complete set of operations, you could look in the class Iterable. The common super class of second sequence and find all the operations that are supported there, or you go and look up the operation same class set and you'll find, in addition, the operation specific to sets. The principle differences between sets and sequences are three. The first one is that sets are unordered. That means the element of a set do not have a predefined order in which they appear in the set. The second difference is that sets do not have duplicate elements. If I would map the set going from 1 to 6 with the function that divides each element by 2, I would obtain the set that contains elements 0, 1, 2, 3. So the set of six elements has just shrunk to four elements because duplicates were removed. And the third operation is where in a sequence the fundamental operations would be head and tail for lists or indexing for vectors. For set, the fundamental operation is contains so, the principal operation that you can do with the set is asking whether there is an element in the set. So let's now use that data structure and what we've seen so far in a somewhat more involved problem that highlights very well the techniques that we could apply to combinatorial search. The problem is a well known one. So you start with a chessboard. 8 by 8 rows normally, I will draw a smaller version with just 4 by 4. And the problem is to place queens into every row of the chessboard, such that none of the queens is threatened by another. So one way to do that here, would be to place a queen here, and here, and here, and here. So now we want to develop a solution for chessboards of any size, not just 4 or 8. One way to solve the problem would be to place a queen in each row. So we start with the first row, place a queen there, then place a queen in the second row, and so on. Once we've placed a number of queens, we must check for the next queen in the column that it does not threaten any of the other queens, so that it sits in its own column and it doesn't threaten the other queens by following a diagonal. That's lead to an algorithm for solving the problem. The algorithm is recursive. It says suppose that we've already generated all the solutions consisting of placing k-1 queens on a board of size n. First question is how do we represent each solution? So the idea there is we would just represent it by a list of length k-1. Obtaining the numbers of columns between 0 and n-1. So let's see in this example how that would work. So I number rows and columns from 0. Something like that. And then the solution of my first three queens would be the list of saying the last queen that I've placed that was the queen in the row two. That column was 0. The one before that I placed was the queen in column 3, and the first queen I've placed in the 0 row at column 1. So that would be a partial solution of the three queens. I can then complete it to a full solution by adding another queen in this column 2 that I've seen here. So that solution would evolve to the solution 2, 0, 3, 1 that you've seen on the board but, of course, in general, where there can be more solutions or none at all so, we're dealing not with the single solution here but, with sets of solution. So let's put this into an actual Scala program. I've opened a new work sheet, call it nqueens, and I've given you already the signature of these queens method that finds all solution to place nqueens on a chessboard of n rows. So the input to queens would be the number of rows of our chess board. And the output would be a set of solutions and each solution is a list of int, the one we've seen. So to implement the queens method, we use a recursive algorithm with an auxiliary method, call it place queens, which places a number k of queens on the board and produces the set of solutions. So the initial call would be placeQueens(n), which means we place all n and want to place all n queens. So now we've reduced the problem to how to implement the placeQueens. Well, let's deal with the boundary case first, if k=0, so we don't need to place any queen, what do we return? Well, you might be tempted to say return the empty set of solutions, but that's actually not quite right because if we're not asked to do anything, that there is a solution, namely, don't do anything. So that means we return just the empty list as our solution here. Okay, so that was the case where k=zero, so now, in the case where k is greater than 0, we have to do some real work. So, what do we need to do? So in general we have to place k queens, we first have to solve the sub problem of placing k-1 queen. So we call this for queens coming from place queens k-1. So that was the set of our partial solutions returned by placeQueens(k-1). And we let queens range over it. Now the second thing we would have to check for is we have to put our k queen into a certain column. And we'll simply try all the possible columns. So we would say the column range is from 0 until n. But, we can't place the queen in any column we place. Because, we still have to check that it doesn't threaten and the other queen. So, we put a test in there at filter which says that the column for the queen is safe which respect to the previous queens. And if it is then when we can yield a new solution what would that solution be? Well, it would be our partial solution here augmented by the queen in the new column. So it would be column followed by queens. But that gives the outline of our solution. There's still one thing to do namely define this method is safe. So I would like you to do that as an exercise. So the exercise for you is to write a function isSafe that takes a column for a new queen and an existing solution call it queens that returns a Boolean indicating whether it's safe to put the new queen in the given column. It's assumed, of course, that the new queen is place in the next available row after all the other place queen. So let's see how we would solve this. I've already given you the signature of the safe in the worksheet, the thing to do is work on its implementation. The first thing I want to do is I want to add rows to all the queens that we look at here. So the row of the queen to be placed, let's call it row. So that would be, simply queens.length because the other queens are in row 0 to row minus 1. The next thing that I want to do is I want to also add a row to each of these queens. So transforming the list of ints into a list of pairs of row and columns. So what I want to do is, let's say, if I have a solution. List of 0, 3, 1 partial solution. Then, I want to transform that into a solution that adds the rows. So, the first element was actually the last row to be placed, so it was row 2. So, I would get 2,0 1,3 and 0,1. So I want to go from here, to here. How could we do that? So the idea here is that we use a zip with a range, so the range that we want to apply here would be the range that goes from row-1. To 0 by minus 1 steps. And that sequence we zip with just the file queens. And we call that WithRow. So that would be now the partial solution of queens represented with rows. So what we can do now is we can simply check whether the queen at row row and column column here is in check with any of these here. So that would be for all. For all of these queens it must be that the new queen is not in check. So when we go through this partial solution their all pairs. So let's immediately take the row and the column out of the pair and now comes our check so what do we need to check here. Well, the first of this of course that the current column is not the same as any of the columns of the previous screen so that would be call different from z and the other thing to check is that the queen is also not in check over any of the diagonals. What that means is that the absolute difference between the two columns, we shall call that math.abs(col- c) must not be the same as the absolute difference between the two rows. So that's row minus r because in this case we know that row is always bigger than r. So in that case if that predicate is true then when we know that the queen is not in check over any of the diagonals with the queen in rc. So that's our definition of a safe. Let's try to test this with a simple example, let's just run queens of 4, and see what we get. So what we get is a list of two solutions, 1,3,0,2 and 2,0,3,1. And if you look at it, then it seems that the two solutions are both correct. But it's better to actually visualize this. So as a last step, I would like to add a component that shows all solutions as actual chessboards, so let's try to do that. I've put up a function here called show that takes a solution and will print it as a chessboard. We can see that immediately how it would work, let's call queens of 4 at now map with show, so for each of the solutions we want to see the check chessboard so that's what we get, we get a set of solutions. And each solution is a four line string, so here in this rectangle here you see the first one and in this rectangle you see the second one. And as you see, yes, it's true that none of the x's which represent the queens is in check with each of the other so their all on their own column in their own row. And none of them are the same, diagonal. So let's have a quick look at how we printed the queens here. So, what I did is I had a list of lines and that's just essentially I took the solution but I reversed it because we that in the solution later queens come first so we let the columns range over queens.reverse. And then for each of these rows I had to produce the string that consists of asterisks and the X where the queen is. So the first thing I did was I produced a vector with n elements that each read star and a space. And then I used the updated call at the column to replace the star and the space with an X and a space. And then I converted the whole vector of string elements to a string, there's a hand the utility functions also that make string will take any collection and simply print out all the elements in the collection one after the other. So, that was how we converted the single line to a string. To actually show a whole chessboard the ideas that we have to show each line and we have to separate them by new line characters. So, that's another usage of this make string which is a function that can also take a second argument. In that case it will print each lines separated by the second operand here, so in that case that you seen here lines would be separated by new lines, and I proceeded by that additional new line character, so that's what you saw here, we can even make this. Somewhat nicer by saying, instead of the scaffolding with the set here, I just want to convert it directly into a mkString. And I want to convert sets maybe by another new run character like that. So let's see what that would give. So now I have the solutions in a nicer way each nicely separated by blank lines. Let's play with it a little bit, if you have queens of eight then we will probably get too many solutions, yeah, that's a whole lot of solutions that we get here. So what we do instead is let's take the first three and there you would have all the solutions of size eight. And inspecting each of them you see that, yes, indeed it is a solution
