cpe 481 - minesweeper hafeez jaffer. introduction -rules of the game -history -originally created by...

CPE 481 - Minesweeper

Hafeez Jaffer

Introduction- Rules of the game- History

- originally created by Robert Donner and Curt Johnson

New Features

- open and save minefield

- more freedom in choosing minefield dimensions

New Features Cont.

-Turn Rete engine on/off-Display Rete engine facts-Try all combinations (beta)

Goal of the Assignment Create a JESS agent that will find

mines in a given Minefield minesweeper.clp

Found in the Jess directory, each rule that is written in this file will be loaded and executed by the MineSweeper program to determine mine locations

Grading Criteria Logically deduce and flag mine

locations in a given solvable minefield (one that doesn't require guessing)

Solve all the test cases in the "testCases" directory

Solve expert minefields in an efficient and fast manner

Point breakdown 5 points for implementing the

deffunctions "flagAllSurroundingUnknowns" and "revealAllSurroundingUnknowns"

5 points for performance 10 points for the test runs 5 points for the README file and

documentation of the source code

Submission "minesweeper.clp" file

containing the JESS rules to solve a given minefield

A README file documenting how your agent deduces mine locations and any known limitations

Test Cases This presentation contains 6 basic

test cases These may not be the ones I use to

test your program

Test Case1

Test Case1 Solution

A

B

D

C

A B C – 1A B – 1

B C D – 1 C D – 1

1 – 1 = 0.0 means all that’s left are not minesC is not a mine.1 – 1 = 0.0 means all that’s left are not minesB is not a mine.

Test Case2

Test Case2 Solution

A

B DC E

D E – 1D E C – 2

A B C D – 1E C D – 2

C is left after DE is taken out. C must be a mine. (2 – 1 = 1)

E is left after CD is taken out. C must be a mine.(2 – 1 = 1)

Jess functions used: Intersection, complement, and test

2 – 1 = 1, the number that’s left is equal to this, therefore mine

Test Case3

Test Case3 Solution

A B C F H I – 3 B C D J I – 5

B C D I J – 5 C D E J G K – 3

E

GF

H

A CB D

I J K

BCI was taken out,5 – 3 = 2. That’s the number remaining. If equal, all are mines. DJ are mines

CDJ was taken out,5 – 3 = 2. That’s the number remaining. If equal, all are mines. BI are mines

Jess functions used: Intersection, complement, and test

Test Case4

Test Case4 Solution

E

G

F

H

A CB D

I J K

C D F – 1C D F I J K – 3

B C H I J – 4A B C H I E G – 3

BCHI removed, 4 – 3 = 1, That’s the number remaining. So J is a mine. For the bottom, AEG are left. 3 – 4 = -1. This is less than or equal to 0. AEG are not mines.

Jess functions used: Intersection, complement, and test

Test Case5

Test Case5 Solution

A BC D

B can be revealed because there is only one mine left.B must be 1 so A and D can be revealed

Test Case6

Test Case6 Solution-Use the “Try all combinations” option from the menu

Implementation Details - Deftemplates

(MineSquare (slot id) - integer (slot row) - integer (slot col) - integer (slot status ?status) - integer (multislot surroundingUnknowns) - integer list (slot numSurroundingFlags) - integer )

(MinefieldVariables (slot numberMines) - integer (slot mineHeight) - integer (slot mineWidth) - integer )

Implementation Details - Defglobals

EMPTY_SQUARE (=0) FLAGGED_MINE (=-2) UNKNOWN (=-4)

Implementation Details – GUI hooks

(reveal-square ?id) (flag-square ?id)

Implementation Details – Jess commands

http://www.cs.vu.nl/~ksprac/2002/doc/Jess60/functions.html

length$ Returns the number of fields in a multifield value.

Subsetp Returns TRUE if the first argument is a subset of the second (i.e.,

all the elements of the first multifield appear in the second multifield); otherwise, returns FALSE.

complement$ Returns a new multifield consisting of all elements of the second

multifield not appearing in the first multifield.

intersection$ Returns the intersection of two multifields. Returns a multifield

consisting of the elements the two argument multifields have in common.

Test Returns true if the argument is true, false otherwise.

Conclusion Any questions?