PEG SOLITAIRE PEGS LEFT HOW TO

If there is no solution, solve should return null The easiest way to build the solution string is to consider the return value from a given recursive call. Each move should be separated by a space (or newline) and should look like this: oldX oldYnewX newY where oldX and oldY are the x and y location of the peg before the jump occurred and newX and newY are the x and y location of where the peg lands.

Upon finding a successful solution you should return a String that indicates the sequence of moves from beginning to end.

PEG SOLITAIRE PEGS LEFT HOW TO

It is tricky to assemble the solution, so read the next paragraphs carefully to help you think about how to do this. Instead, we have to assemble the solution in a different way. Your solve method will be terminated and you will not receive credit for that example if it takes longer Notice that, unlike Knight's Tour, moves can revisit spaces in Peg Solitaire so we can't leave the moves that we made in the original array to indicate the solution. Note that in the first row, (0, 0), (1, 0), (5, 0), and (6, 0) are all invalid locations that can be skipped -you can consider those indexes if it simplifies your program, but never move a peg into those locations! The test program will only allow your solve method to run for up to 5 minutes for a given board. You must examine each valid position in the board for a peg to move. Unlike rather than jus ith ni r, there an iec moves. You should attempt to solve the game in the same manner as we solved Knight's Tour in class. The solution format is specified on the next page pegs are not allowed to land, 4 in each corner. solve should return a solution as a String if the board is solvable, and null if it is not solvable. A solution leaves a single peg in the center location (3, 3). Note that the array includes 16 locations where will be true if a peg is in the corresponding position and false if it is not. It could look like this: private static String solve(booleanI] pegs, int startX, int starty, int jumpX, int jumpY, int endx, int endY) When a peg is moved, it starts at a location, jumps over a location, and ends at a location (Notice that we don't necessarily know which peg to move first! You could do a special call to the helper method with, for example, negative numbers to start with.) solve works with a board which you may assume to be a 7x7 boolean array. Please implement the following method in the PegSolitaire class: public static String solve (boolean pegs) You will need to create a helper method which keeps track of the current move. Here are three example starting conditions that you will be expected to solve Standard Rhombus us You will need to be able to solve all three configurations shown above, some additional symmetrical configurations, one or more very simple single-move boards, as well as deal with unsolvable boards. The 'before' and 'after' pictures are separated with: Bottom peg jumps up, top peg jumps down, left peg jumps right, right peg jumps left. The four kinds of moves that can be made are depicted below. The player wins if a single peg remains in the center of the 7 x 7 grid We represent a peg with the symboland an empty spot with the symbol O.

Peg B is removed and play continues until no moves are possible. Peg A must land on a spot where there is no peg and within the specified constraints. Peg A lands in the adjacent spot in the same direction. A move consists of a peg A jumping over a peg B which is directly above, below, to the left, or to the right of peg A. A constrained grid is presented with locations where pegs may be present or absent. Transcribed image text: Consider the Peg Solitaire game demonstrated in class and found online.

Peg solitaire pegs left