August 1, 2017 9:00 AM
The solution starts from the top-right corner and works down and left, one cell at a time. If the current cell is larger than the target, move left. If the current cell is smaller than the target, move down. If the current cell is equal to the target, report success. If you reach the left or bottom edge of the matrix without finding the target, report failure:
(define (target m x)
(let loop ((r 0) (c (- (matrix-cols m) 1)))
(cond ((or (negative? c) (= (matrix-rows m) r)) #f)
((< x (matrix-ref m r c)) (loop r (- c 1)))
((< (matrix-ref m r c) x) (loop (+ r 1) c))
(else (values r c)))))
We return the row and column coordinates of the target, or #f if it is not found. Here’s an example:
> (define m '#(
#( 1 12 43 87)
#( 9 25 47 88)
#(17 38 48 92)
#(45 49 74 95)))
> (target m 72)
#f
> (target m 38)
2
1
The algorithm is clearly O(m + n) for an m by n matrix. You can run the program at http://ideone.com/YR0tv0.
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Categories: Exercises
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In Python3. Starting inspecting upper right index for the target value. Due to the fact that the Matrix is already ordered, this means that a larger target value should be beneath current index. If it is smaller, move index one to the left and repeat.
By Rutger on August 1, 2017 at 2:49 PM
Recursive algorithm in Python 3.6.
The function ‘find’ returns a generator of indices. This should be efficient for testing whether the target is present or for finding the location of all occurrences.
def find(A, t): def rec(i, j, m, n): if m == 0 or n == 0 or t < A[i][j] or A[i+m-1][j+n-1] < t: return else: if m == 1 and n == 1: if A[i][j] == t: yield (i, j) else: return else: yield from rec(i, j, m//2, n//2) yield from rec(i+m//2, j, m-m//2, n//2) yield from rec(i, j+n//2, m//2, n-n//2) yield from rec(i+m//2, j+n//2, m-m//2, n-n//2) return rec(0, 0, len(A), len(A[0])) def printmat(A): print("[", ",\n ".join(map(str, A)), "]") example = """ 1 12 43 87 9 25 47 88 17 38 48 92 45 49 74 95 """ A = [ list(map(int, line.strip().split())) for line in example.strip().split("\n") ] printmat(A) for row in A: for el in row: #print(el) result = list(find(A, el)) #print(result) print([(el, A[i][j]) for (i, j) in result]) for el in [0, 26, 50]: print(el, list(find(A, el))) m, n = 7, 9 B = [ [ i**2 + j**2 for j in range(n) ] for i in range(m) ] printmat(B) print(list(find(B, 25)))By Jan Van lent on August 1, 2017 at 4:24 PM
Slightly more compact:
def find(A, t): def rec(i, j, m, n): if m > 0 and n > 0 and A[i][j] <= t <= A[i+m-1][j+n-1]: if m == 1 and n == 1 and A[i][j] == t: yield (i, j) else: yield from rec(i, j, m//2, n//2) yield from rec(i+m//2, j, m-m//2, n//2) yield from rec(i, j+n//2, m//2, n-n//2) yield from rec(i+m//2, j+n//2, m-m//2, n-n//2) return rec(0, 0, len(A), len(A[0]))By Jan Van lent on August 1, 2017 at 4:36 PM
Here’s an O(log(n) + log(m)) solution in Python.
It first uses binary search on the first row to find a candidate column. It then uses binary search on that column.
If that fails, it uses binary search on the first column to find a candidate row. It then uses binary search on that row.
Numpy’s searchsorted function was used for binary search.
import numpy as np def array_find(array, element): idx = np.searchsorted(array, element, side='right') - 1 return element == array[idx] def matrix_find(matrix, element): col_idx = np.searchsorted(matrix[0, :], element, side='right') - 1 if array_find(matrix[:, col_idx], element): return True row_idx = np.searchsorted(matrix[:, 0], element, side='right') - 1 if array_find(matrix[row_idx, :], element): return True return False M = np.array([ [ 1, 12, 43, 87], [ 9, 25, 47, 88], [17, 38, 48, 92], [45, 49, 74, 95] ]) for x in range(-100,100): if matrix_find(M, x): print xBy Daniel on August 1, 2017 at 6:16 PM
Here’s the output of the program I just posted.
By Daniel on August 1, 2017 at 6:16 PM
My previous solution does not work.
The following example doesn’t find the 48.
M = np.array([ [1, 12, 43, 87], [9, 25, 47, 88], [17, 48, 50, 92], [45, 49, 74, 95] ]) for x in range(-100, 100): if matrix_find(M, x): print xBy Daniel on August 2, 2017 at 3:30 AM
Here’s another binary-search solution in Python. This one is O(n log m), but unlike my last solution it works (I hope).
It iterates over the rows, performing binary search on each row.
import numpy as np def matrix_find(matrix, element): for row in matrix: idx = np.searchsorted(row, element, side='right') - 1 if idx > -1 and row[idx] == element: return True return False M = np.array([ [ 1, 12, 43, 87], [ 9, 25, 47, 88], [17, 38, 48, 92], [45, 49, 74, 95] ]) for x in range(-100, 100): if matrix_find(M, x): print xBy Daniel on August 2, 2017 at 3:49 AM
Two more solutions in Python. The second is a correct implementation of what Daniel tried to do with numpy in his first post, I think.
def find_target(m, target): ncols = len(m[0]) i, j = 0, ncols - 1 while i < ncols and j >=0: if m[i][j] == target: return True i, j = (i+1, j) if m[i][j] < target else (i, j-1) return False def matrix_find(matrix, target): nrows, ncols = matrix.shape row, col = 0, ncols-1 while row < nrows and col >= 0: elt = matrix[row, col] if elt == target: return True if elt < target: row += np.searchsorted(matrix[row:, col], target, side='left') else: col = np.searchsorted(matrix[row, :col], target, side='right') - 1 return FalseBy Paul on August 3, 2017 at 2:47 PM
I’ve never used this website before, and i’m also fairly new to Programing. But this is my solution. Nested for loop that moves through the array until the value is found.
//——————————————————
// Consider a two-dimensional matrix
//containing integer entries in which
//all rows and all columns are sorted in ascending order
//——————————————————
var Matrix = [
[1, 12, 43, 87],
[9, 25, 47, 88],
[17, 38, 48, 92],
[45, 49, 74, 95]
];
//——————————————————
//Your task is to write a program that takes a matrix as
//describe above, and a target integer, and determines
//if the target integer is present in the matrix.
//——————————————————
var Target = 1;
check: {
for (var i = 0; i < Matrix.length; i++) {
//console.log(Matrix.length[0]);
for (var j = 0; j < Matrix[i].length; j++) {
if (Matrix[i][j] == Target) {
console.log("The Target is present in the Matrix, and is located at Matrix[" + i + ", " + j + "]");
break check;
}
}
}
console.log("The Target is not present in the Matrix");
}
By Musicman on August 4, 2017 at 4:42 AM
@Musicman: Welcome! We’re glad to have you.
You might look at the menu bar for a description about how to format source code.
By programmingpraxis on August 4, 2017 at 6:14 PM