排序二维数组的二进制搜索

Question

I have a program that searches a 2d array using a binary search. In this case I am using the matrix below and searching for the integers 4,12,110,5,111. The program finds all of them except for 110 and 111 why is this?

       {1,3,7,8,8,9,12},
       {2,4,8,9,10,30,38},
       {4,5,10,20,29,50,60},
       {8,10,11,30,50,60,61},
       {11,12,40,80,90,100,111},
       {13,15,50,100,110,112,120},
       {22,27,61,112,119,138,153},

public static boolean searchMatrix(int[][] matrix, int p,int n) {  

int low = 0, high = n-1 ;  
while (low < high) {  
 int mid = (low + high) / 2;  
 if (p == matrix[mid][0])return true;  
 else if (p < matrix[mid][0]) high = mid - 1;  
 else if (p < matrix[mid+1][0]) { low = mid; break; }  
 else low = mid + 1;  
}  


int row = low;  
low = 0; high = matrix[row].length - 1;  
while (low <= high) {  
 int mid = (low + high) / 2;  
 if (p == matrix[row][mid]) return true;  
 else if (p < matrix[row][mid]) high = mid - 1;  
 else low = mid + 1;  
}  

return false;  
}  

Answer 1

I would rather say it's more or less a surprise that your algorithm does find 4, 5, and 12 in the first place. The reason for that is that 4 occurs in the first position of a row and 5 and 12 satisfy the condition that they are less than the first element in the next row. Only due to that fact are they discovered in the second half of your algorithm. The algorithm is a bit hard to read and I did not evaluate the +/-1 magic, but it seems the algorithm expects the 110 and 111 to occur actually in the last row (since both 110 and 111 are greater than 22) where they are not.

If I get you right, your approach is flawed in that it is actually impossible, by looking at a single number, to tell what row it will occur in, which is what your first tries to achieve. So any two-phase algorithm that first picks a row and then searches a column must fail.

With the few constraints you have on your matrix (each row and each column is sorted), it does not seem like binary search will work at all: Even if your bounds low and high would be 2D points it would not help a lot. Consider any element of the matrix that is greater than your search point. Then all you can say is that your search point is not below and right of that element (whereas what you were hoping to be able to conclude was that it was left and above, but that is not necessarily true - it can be above and right or left and below), so you are only cutting off only a too small part of the search space.

Answer 2

Your issue is that you are making the false assumption that you can first lock down the row of your search value, and then easily do binary search on that row. That's not the case at all.

For 110 and 111, the first element of each row is always less than your search value, and your algorithm comes to the false conclusion that that this means that your row must be the array with index 6 after the first loop. This is simply not true.

The reason it works for small numbers is because your algorithm is just lucky enough to lock down the right row in the first loop...

One correct algorithm to quickly search on a 2d matrix where every row and column is sorted in ascending order is as follows:

1) Start with top right element 2) Loop: compare this element e with x ….i) if they are equal then return its position …ii) e < x then move it to down (if out of bound of matrix then break return false) ..iii) e > x then move it to left (if out of bound of matrix then break return false) 3) repeat the i), ii) and iii) till you find element or returned false

I found this algorithm here: http://www.geeksforgeeks.org/search-in-row-wise-and-column-wise-sorted-matrix/

It's O(n) for an nxn matrix.