Matrix Rotation

June 6, 2017

We have a two-part exercise today, based on a Microsoft interview question.

First, write a program to rotate an m × n matrix 90° to the right, as shown below; your solution should touch each matrix element only once:

    a b c
    d e f                 m j g d a
A = g h i        rot(A) = n k h e b
    j k l                 o l i f c
    m n o

Second, write a program to rotate a square matrix with n rows and columns in-place. where the source and target matrices are the same matrix and there is no intermediate matrix (be sure your solution works for both even and odd n):

    a b c d e                 u p k f a
    f g h i j                 v q l g b
B = k l m n o        rot(B) = w r m h c
    p q r s t                 x s n i d
    u v w x y                 y t o j e

Your task is to write the two programs that rotate matrices. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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7 Responses to “Matrix Rotation”

  1. Jussi Piitulainen said
    ;;; Implement a matrix as a vector in row-major order, with width.
    
    (define imp::matrix-contents cdr)
    
    (define matrix-width car)
    (define (matrix-height m)
      (/ (vector-length (imp::matrix-contents m))
         (matrix-width m)))
    (define (make-matrix h w . o)
      (let ((o (if (null? o) #f (car o))))
        (cons w (make-vector (* h w) o))))
    (define (matrix-ref m r k)
      (vector-ref (imp::matrix-contents m) (+ (* (matrix-width m) r) k)))
    (define (matrix-set! m r k o)
      (vector-set! (imp::matrix-contents m) (+ (* (matrix-width m) r) k) o))
    
    ;;; Some of the points appear to go from (r, k) to (k, h - r - 1), so
    ;;; guess they all do. By linearity or something.
    
    (define (rot m)
      (let* ((h (matrix-height m))
             (w (matrix-width m))
             (m1 (make-matrix w h)))
        (do ((r 0 (+ r 1))) ((= r h))
          (do ((k 0 (+ k 1))) ((= k w))
            (matrix-set! m1 k (- h r 1) (matrix-ref m r k))))
        m1))
    
    ;;; At each (r, k) apply the rotating transformation to the four
    ;;; corresponding points at once. But this rotates the matrix four
    ;;; times! So only do half of the rows and half of the columns to
    ;;; rotate only once. Rounding both ways seems to work already.
    
    (define (rot! m)
      (let ((hw (matrix-height m)))
        (do ((r 0 (+ r 1))) ((= r (floor (/ hw 2))))
          (do ((k 0 (+ k 1))) ((= k (ceiling (/ hw 2))))
            (let* ((nr r)  (nk k)           (n (matrix-ref m nr nk))
                   (er nk) (ek (- hw nr 1)) (e (matrix-ref m er ek))
                   (sr ek) (sk (- hw er 1)) (s (matrix-ref m sr sk))
                   (wr sk) (wk (- hw sr 1)) (w (matrix-ref m wr wk)))
              (matrix-set! m er ek n)
              (matrix-set! m sr sk e)
              (matrix-set! m wr wk s)
              (matrix-set! m nr nk w))))))    
    
    ;;; Nice display for ocular inspection.
    
    (define (lay m)
      (let* ((w (matrix-width m))
             (h (matrix-height m)))
        (do ((r 0 (+ r 1))) ((= r h))
          (do ((k 0 (+ k 1))) ((= k w) (newline))
            (write-char #\space)
            (write (matrix-ref m r k))))))
    
    ;;; Test some. Caught a few thinkoes and a typo.
    
    (define A (cons 3 '#(a b c  d e f  g h i  j k l  m n o)))
    (define B (cons 5 (apply vector '(a b c d e  f g h i j  k l m n o  p q r s t  u v w x y))))
    (define C (cons 4 (apply vector '(a b c d  e f g h  i j k l  m n o p))))
    
    (lay A) (newline)
    (lay (rot A)) (newline)
    
    (lay B) (newline)
    (lay (rot B)) (newline)
    (rot! B) (lay B) (newline)
    
    (lay C) (newline)
    (lay (rot C)) (newline)
    (rot! C) (lay C) (newline)
    
  2. Paul said

    The first part of the exercise in Python

    def rot_right(matrix):
        return list(zip(*reversed(matrix)))
    
  3. Zack said

    Both parts of the exercise in Julia. Just a single generic method covering all possibilities.

    function rotate{T <: Any}(X::Matrix{T})
    n = size(X, 1)
    n1 = n + 1
    Y = X'

    for i = 1:div(n, 2)
    j = n1 – i
    Y[:,i], Y[:,j] = Y[:,j], Y[:,i]
    end

    return Y
    end

  4. Paul said

    The second part in Python. The rotation is done by swapping two matrix elements at the time.

    def rot_right_in_place(matrix):
        n = len(matrix)
        c = n-1
        size = (n+1)//2
        def rot(row, col):
            col0, row0, col, row = col, row, row, c - col
            matrix[row0][col0], matrix[row][col] = matrix[row][col], matrix[row0][col0]  # swap
            return row, col
        for i in range(size):
            for j in range(size if n % 2 == 0 else size-1):
                rot(*rot(*rot(i, j)))
    
  5. Paul said

    A better version of the second part. This version sets the matrix elements only once.

    def rot_right_in_place(matrix):
        n = len(matrix)
        c = n-1
        size = (n+1)//2
        def rot(row, col):
            save = matrix[row][col]
            for _ in range(3):
                col0, row0, col, row = col, row, row, c - col
                matrix[row0][col0] = matrix[row][col]
            matrix[row][col] = save
        for i in range(size):
            for j in range(size if n % 2 == 0 else size-1):
                rot(i, j)
    
  6. Steve said

    Klong version

            :" assign array to a"
            a::[[:a :b :c] [:d :e :f] [:g :h :i] [:j :k :l] [:m :n :o]]
    [[:a :b :c] [:d :e :f] [:g :h :i] [:j :k :l] [:m :n :o]]
    
            |'+a   :"transpose the array and reverse each row"
    [[:m :j :g :d :a] [:n :k :h :e :b] [:o :l :i :f :c]]
            
            :" assign array to b"
            b::[[:a :b :c :d :e] [:f :g :h :i :j] [:k :l :m :n :o] [:p :q :r :s :t] [:u :v :w :x :y]]
    [[:a :b :c :d :e] [:f :g :h :i :j] [:k :l :m :n :o] [:p :q :r :s :t] [:u :v :w :x :y]]
    
            |'+b  :"transpose the array and reverse each row"
    [[:u :p :k :f :a] [:v :q :l :g :b] [:w :r :m :h :c] [:x :s :n :i :d] [:y :t :o :j :e]]
    
  7. Mike said

    Python 3.6.

    rot() is basically the same as Paul’s.

    rot_in_place() moves four corresponding elements at a time. ‘i’ can be thought of as the layer or ring, 0 being the outermost ring.
    ‘i’, ‘ir’, ‘j’, and ‘jr’ are the row or column indices of the elements being moved.

    rot = lambda m:[*zip(*m[::-1])]

    def rot_in_place(a):
    n = len(a)
    for i in range((n+1) // 2):
    ir = -i-1
    for j in range(i, n-i-1):
    jr = -j-1
    (a[i][j], a[j][ir],
    a[jr][i],a[ir][jr]) = (a[jr][i], a[i][j],
    a[ir][jr],a[j][ir])

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