Jones Mike Alex Packard
SlidingBlocks Design Document
1. Our block object will contain a "current position" field denoted by x and y coordinates. Each position in the 2D array representing our tray object will either be null, or point to the block object that occupies that space. Each block object will have its position determined by its coordinates. So even if the array indices points to the block object the coordinates, only apply to the block objects coordinates of the top left position.
2. Inside of our tray object will be three (but not only three) fields, one of which will be "current position", possible paths, and current path. For all positions that we find that are possible, we will check if they are the solution, and if yes, stop. If they are not the solution, have we seen them before? If no, add them to the stack, if yes, ignore them.
3. Starting with our initial position, we will call our getPossibilities() method on the table, which will return multiple new possible positions. Starting with the last one reported, they will be added to a stack, and then will be popped and recursively have get possibilities called on them. We will run this algorithm depth first, meaning that we will run a single of path possibilities until we encounter either the solution or a repeated configuration. Once a known solution is encountered we know to stop searching and print out the solution.
4. Block moves on more than one space should not be considered. By considering these moves there may be solutions that are ignored unintentionally.
Mike Jones Alex Packard
5. As we recurse down the tree, we will add the configurations to a stack. When a configuration is determined to not lead to a solution, we will unmake the moves by recursing back up the tree, popping configurations off the stack as we go.
6. Tentatively, we plan to use a 2D array to store our tray and the configuration on the tray at the current time. The hash function we intend to use to calculate the hash for the current position entails:
[psuedocode]
String bob = "" //temporary string for all blocks{ bob+= "" + xco + "" + yco + ""+ height + "" + width; } bob.hash();
[end psuedocode]
If this position has been seen before (it is already inside the hash table), and if the position is not the solution position, than end this branch of recursion, if it hasn't been seen before than we just add it to the table and move on.