1. CS1020 SIT IN LAB 1 Pattern Matching

2. Problem Given a RxC letters matrix and a RpxCp pattern matrix,find how many ways to match the patterns and letters. Rp <= R <= 50 && Cp <= C <= 50 An asterisk “*” is used (in the pattern matrix) if any letter can be used for that cell.

3. Example Sample Input: 6 2 AAAXAA CCDXCX 2 2 AA C* Sample Output: 3

4. Algorithm Brute Force: Match the letters 1 character by 1 character. Letters: A A A X A A C C D X C X Patterns: A A C *

5. Algorithm Brute Force: Match the letters 1 character by 1 character. Letters: A A A X A A C C D X C X Patterns: A A C *

6. Algorithm Brute Force: Match the letters 1 character by 1 character. Letters: A A A X A A C C D X C X Patterns: A A C *

7. Algorithm Brute Force: Match the letters 1 character by 1 character. Letters: A A A X A A C C D X C X Patterns: A A C *

8. Algorithm Brute Force: Match the letters 1 character by 1 character. Letters: A A A X A A C C D X C X Patterns: A A C *

9. Sample Code (Pseudocode) While ( Checking each cell in letters matrix) { While (Matching each cell in patterns matrix) { if (Not all match) { break; } else { add one way; } } }

10. Java code (ANY_CHARACTER = ‘*’) int calculateNumberOfWays(int letterRow, int letterColumn, char[][] letterMatrix, int patternRow, int patternColumn, char[][] patternMatrix) { int ways = 0; for (int i = 0; i <= letterRow - patternRow; i++) { for (int j = 0; j <= letterColumn - patternColumn; j++) { boolean isMatched = true; for (int k = 0; k < patternRow; k++) { for (int l = 0; l < patternColumn; l++) { if (patternMatrix[k][l] == '*' || patternMatrix[k][l] == letterMatrix[i+k][j+l]) { continue; } else { isMatched = false; break; } } } if (isMatched) { ways++; } } } return ways; }