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CS100A, Fall 1997

CS100A, Fall 1997. Lecture, 6 November: More on two-dimensional arrays Declare and allocate two-dimensional array: int [ ] [ ] b; b= new int [3] [2]; Note: b.length is 3 and b[0].length is 2. Assigning to array elements could yield: b[0][0] 5

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CS100A, Fall 1997

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  1. CS100A, Fall 1997 Lecture, 6 November: More on two-dimensional arrays Declare and allocate two-dimensional array: int [ ] [ ] b; b= new int [3] [2]; Note: b.length is 3 and b[0].length is 2. Assigning to array elements could yield: b[0][0] 5 b[0] b[0][1] 3 b[1][0] 8 b[1] b[1][1] -1 b[2][0] 7 b[2] b[2][1] -5 CS100A, Fall 1997. Lecture, 6 Nov.

  2. Declare and allocate two-dimensional array: OthelloSquare [ ] [ ] b; b= new OthelloSquare [3] [2]; Note that each array element contains null. b[0][0]= new OthelloSquare(5,6); b[0][0] b[0] b[0][1] null b[1][0] null b[1] b[1][1] null b[2][0] null b[2] b[2][1] null CS100A, Fall 1997. Lecture, 6 Nov.

  3. Example of dealing with a two-dimensional array: Magic Squares 17 24 1 8 15 each row sums to 65 23 5 7 14 16 each col sums to 65 4 6 13 20 22 each diagonal sums 10 12 19 21 3 to 65 11 18 25 2 9 16 3 2 13 In Albert Duerer’s 5 10 11 8 engraving 9 6 7 12 “Melancolia”, done 4 15 14 1 in 1514! CS100A, Fall 1997. Lecture, 6 Nov.

  4. // Assign 0 to all elements of b staticvoid zeroOut(int b[][]) { for (int r= 0; r != b.length; r= r+1) for (int c= 0; c != b[r].length; c= c+1) b[r][c]= 0; } The for loop for (i= 0; i != n; i= i+1) S; equivalent to: i= 0; while (i != n) { S; i= i+1; } CS100A, Fall 1997. Lecture, 6 Nov.

  5. Two-dimensional array: array of rows or array of columns? Entirely up to the interpretation. Java doesn’t care. Othelloboard [ ] [ ] board; board[0] is a column, board[1] is a column, etc. Because that is how the GUI interprets things. In mathematics, int [ ] [ ] b = {{1,3,5}, {2,5,4}}; usually a “matrix” of rows and columns: 1 3 5 2 5 4 Use index names r,c to help the reader: board[c][r], b[r][c] CS100A, Fall 1997. Lecture, 6 Nov.

  6. 17 24 1 8 15 each row sums to 65 23 5 7 14 16 each col sums to 65 4 6 13 20 22 each diagonal sums 10 12 19 21 3 to 65 11 18 25 2 9 Encyclopedia Britannica: The smallest possible square of odd order has, of course, side-length 3! To construct n x n magic square n x n ( n odd): 0. Put 1 in b[0][n%2] 1. If i is in b[r][c], then i+1 goes in b[r’][c’], where r’,c’ determined as follows: (a) try r’= r-1, if -1, then use r’=n-1. (b) try c’= c+1, if n, then use c’= 0. (c) if b[r’][c’] is already filled, use r’=r+1, c’= c CS100A, Fall 1997. Lecture, 6 Nov.

  7. // Store a magic square in b staticpublicvoid magicSquare(int b[][]) { int r= 0; int c= 0; int i; int size= b.length; zeroOut(b); // Make up the magic square r= 0; c= size/2; i= 1; // invariant: Array elements have already been filled with 1, ..., i-1, and b[r][c] is supposed to get i. while (b[r][c] == 0) { b[r][c]= i; i= i+1; if (b[mod(r-1, size)][mod(c+1, size)] == 0) { r= mod(r-1, size); c= mod(c+1, size); } else r= mod(r+1,size); } } CS100A, Fall 1997. Lecture, 6 Nov.

  8. Pascal’s Triangle 1 1 1 1 2 1 1 3 3 1 1 46 4 1 1 5 10 10 5 1 1 6 1520 15 6 1 int [] [] p; p= new int[7][]; // p.length is 7 p[0]= new int[1]; // p[0].length is 1 p[1]= new int[2]; // p[1].length is 2 p[2]= new int[3]; // p]2].length is 3 Elements of an array can be arrays of different lengths! CS100A, Fall 1997. Lecture, 6 Nov.

  9. Blaise Pascal (1623-1662). • Age 20, constructed arithmetical calculator to help his father in his calculations. • Elements of Geometry. • One of creators of probability and statistics. • Deeply religious (Catholic) Expected value of an event (value if the event happens) * (probability that it will happen) Lottery: $1,000,000 * 10**(-9) = $.001 Get to heaven (liberation, self-realization, moksha) because of leading spiritual life: (infinity * ? ) = CS100A, Fall 1997. Lecture, 6 Nov.

  10. // Yield Pascal's triangle with size rows staticpublicint[][] calculatePascal(int size) { int[][]p= new int[size][]; //the triangle // Invariant: rows 0..r-1 have been // allocated and calculated for (int r= 0; r != size; r= r+1) { // Allocate row i of triangle --its r+1 values p[r]= new int[r + 1]; // Calculate row r of Pascal's triangle p[r][0]= 1; for (int c= 1; c < r; c= c+1) p[r][c]= p[r-1][c-1] + p[r-1][c]; p[r][r]= 1; } return p; } CS100A, Fall 1997. Lecture, 6 Nov.

  11. // Print Pascal's triangle p, assuming that its values are less than 1000 staticpublicvoid printPascal(int p[][]) { int size= pascal.length; for (int r= 0; r != size; r= r+1) { // Add ((size-r)/2)4 + ((r+1) % 2)*2 // blanks to s String s= ""; int num; if (size%2 == 1) num= (size-r)/2; else num= (size-r-1)/2; for (int c= 0; c< num; c= c+1) s= s + " "; if (r%2 == 0) s= s + " "; // Add the numbers in row r to s for (int c= 0; c <= r; c= c+1) s= s + printbrc(p, r,c,1000); System.out.println(s); } } CS100A, Fall 1997. Lecture, 6 Nov.

  12. // Return element b[r][c] of array b as a String, // with a blank before it. // Use as many columns as is required to print x. // So, if x = 325, use 4 characters in total (one // for the initial blank). // Precondition, x < 1000 staticpublic String printbrc( int[][] b, int r, int c, int x) The program used to demonstrate in this lecture will be placed on the CS100A web page. CS100A, Fall 1997. Lecture, 6 Nov.

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