ruby - rotate Matrix anti-clockwise by n position - arrays

Question

Given a 2D matrix:
matrix = [
[ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ],
[ 9, 10, 11, 12 ],
[ 13, 14, 15, 16 ]
]
How can we rotate the matrix anti-clockwise so that values are pushed like this?
matrix = [
[ 2, 3, 4, 8 ]
[ 1, 7, 11, 12 ]
[ 5, 6, 10, 16 ]
[ 9, 13, 14, 15 ]
]
Note
This question is not a duplicate of this & this because what I'm trying to achieve is by rotating the values in anti-clockwise fashion.
My current implementation & Problem
My current implementation only prints out the values in anti-clockwise fashion, but it does not rotate the values.
layers = [_rows, _cols].min / 2
r1, r2, c3, c4 = 0, _rows, _cols, _cols
new_matrix = Array.new(_rows + 1) { Array.new(_cols + 1) }
(0..layers).each do |layer|
row_top_left, row_bottom_left, col_top_right, col_bottom_right = r1, r2, c3, c4
result = []
while row_top_left < row_bottom_left
result << matrix[row_top_left][layer]
row_top_left += 1
end
row_bottom_left = layer
while row_bottom_left < col_bottom_right
result << matrix[row_top_left][row_bottom_left]
row_bottom_left += 1
end
temp_col_bottom_right = col_bottom_right
temp_col_top_right = layer
while col_bottom_right > temp_col_top_right
result << matrix[col_bottom_right][temp_col_bottom_right]
col_bottom_right -= 1
end
# p row_top_left
tmp_row_top_left = layer
while col_top_right > tmp_row_top_left
result << matrix[tmp_row_top_left][col_top_right]
col_top_right -= 1
end
p result.cycle
r1 += 1
r2 -= 1
c3 -= 1
c4 -= 1
update v0.1
The key idea is that the matrix needs to be rotated in the correct way. For example, let's say our matrix requires 2 rotation. Therefore:
matrix_rotation(
matrix.length - 1, # rows
matrix[0].length - 1, # columns
2, # Nom. of rotation
matrix # The matrix
)
matrix = [
# Original Iter: 1 Iter: 2
[ 1, 2, 3, 4 ], # [ 2, 3, 4, 8 ] # [ 3, 4, 8, 12 ]
[ 5, 6, 7, 8 ], # [ 1, 7, 11, 12 ] # [ 2, 11, 10, 16 ]
[ 9, 10, 11, 12 ], # [ 5, 6, 10, 16 ] # [ 1, 7, 6, 15 ]
[ 13, 14, 15, 16 ] # [ 9, 13, 14, 15 ] # [ 5, 9, 13, 14 ]
]
Update v0.2
The dimension of the array is denoted: NxM where N and M can be any numbers, even or odd. For example 5x4, 4,4, 4x8 etc..
There is no such thing as "empty squares".

Answer 1

TL:DR
If you want to jump straight to the solution code, jump to the bottom section of this answer.
Explanation
You need to break down the problem and solve each one independently.
Problems
Get the number of layers
Loop in reverse spiral form to just get the expected values
Shift them based on the rotation parameter given
Let us walk through each point separately:
Get the number of layers
You need a way to get the number of layers. The below matrix has 2 layers. How?
given a matrix:
matrix layers
--------------------------------
| 1, 2, 3, 4 | 0 0 0 0 |
| 5, 6, 7, 8 | 0 1 1 0 |
| 9, 10, 11, 12 | 0 1 1 0 |
| 13, 14, 15, 16 | 0 0 0 0 |
--------------------------------
To find the number of layers, simply do:
[rows, cols].min / 2
Thus the first problem is done.
Loop in reverse spiral form to just get the expected values
This part requires a lot of thinking. Let us visualise:
matrix layers
--------------------------------
| 1, 2, 3, 4 | ↓ ← ← ↰ | 0 0 0 0 |
| 5, 6, 7, 8 | ↓ 1 1 ↑ | 0 ↓ ↰ 0 |
| 9, 10, 11, 12 | ↓ 1 1 ↑ | 0 ↳ → 0 |
| 13, 14, 15, 16 | ↳ → → → | 0 0 0 0 |
--------------------------------
This is achievable. We will have 4 for loops. Each loop will take care of:
left (top to bottom)
bottom (left to right)
right (bottom to top)
top (right to left)
Before I get into the loops, we need some container to store our values in spiral form.
Let us have a temp array to store the values:
# this array will get us the output of borders of the layer
row = []
For the sake of explanation, let us only work on the outer most layer. (i.e. 0th layer:
1st Loop (Left: top to bottom)
# this loop will output the top-left side of the matrix
# ==============================
# ↓ [ 1, 2, 3, 4 ]
# ↓ [ 5, 6, 7, 8 ]
# ↓ [ 9, 10, 11, 12 ]
# ↓ [ 13, 14, 15, 16 ]
# Output: [[1, 5, 9], [6] ]
# ==============================
(0...rows - 1 - layer).each do |i|
row << matrix[i][layer]
end
Note: 0 means the 0th layer.
2nd Loop (Bottom: Left to Right)
# this loop will output the bottom side of the matrix
# ==============================
# ↓ [ 1, 2, 3, 4 ]
# ↓ [ 5, 6, 7, 8 ]
# ↓ [ 9, 10, 11, 12 ]
# ↓ [ 13, 14, 15, 16 ]
# ↪ → → → →
# Output: [[1, 5, 9, 13, 14, 15], [6, 10]]
# ==============================
(0...cols - 1 - layer).each do |i|
row << matrix[rows - 1 - layer][i]
end
3rd Loop (Right: Bottom to Top)
# this loop will output the right side of the matrix
# ==============================
# ↓ [ 1, 2, 3, 4 ] ↑
# ↓ [ 5, 6, 7, 8 ] ↑
# ↓ [ 9, 10, 11, 12 ] ↑
# [ 13, 14, 15, 16 ] ↑
# ↪ → → → → ⤻
# Output: [[1, 5, 9, 13, 14, 15, 16, 12, 8], [6, 10, 11]]
# ==============================
(rows - 1 - layer).step(0 + 1, -1).each do |i|
row << matrix[i][cols - 1 - layer]
end
4th Loop (Top: Right to Left)
# this loop will output the top side of the matrix
# ==============================
# ← ← ← ← ↰
# ↓ [ 1, 2, 3, 4 ] ↑
# ↓ [ 5, 6, 7, 8 ] ↑
# ↓ [ 9, 10, 11, 12 ] ↑
# [ 13, 14, 15, 16 ] ↑
# ↪ → → → → ⤻
# Output: [[1, 5, 9, 13, 14, 15, 16, 12, 8, 4, 3, 2], [6, 10, 11, 7]]
# ==============================
(cols - 1 - layer).step(0 + 1, -1).each do |i|
row << matrix[layer][i]
end
Shift them based on the rotation parameter given
So now at this point, we have the values in the spiral form. But the most important aspect of this problem lies in this section. How does one shift the values? Funnily enough, we will use modulo.
The modulo will do the main thing here. It will allow us to shift values based on the rotate. But also give us the correct index in the array to start the shift. For example, if we want to rotate 2 times: 2 % 12 = 2 for the outer most layer.
# row = [1, 5, 9, 13, 14, 15, 16, 12, 8, 4, 3, 2]
shift = rotate % row.size
# if we negate shift variable, we can get correct index
# i.e. row[-2] = 3
idx = -shift
Before we shift values, let us create another matrix which will contain the correct values:
# let us create a new matrix
result = (1..( rows * cols )).each_slice(rows).to_a
We will loop again in the same manner, but get the values from the idx in row. For example:
(0...rows - 1 - 0).each do |i|
result[i][layer] = row[idx]
idx += 1
idx %= row.size
end
(0...cols - 1 - 0).each do |i|
result[rows - 1 - 0][i] = row[idx]
idx += 1
idx %= row.size
end
(rows - 1 - 0).step(0 + 1, -1).each do |i|
result[i][cols - 1 - 0] = row[idx]
idx += 1
idx %= row.size
end
(cols - 1 - 0).step(0 + 1, -1).each do |i|
result[0][i] = row[idx]
idx += 1
idx %= row.size
end
Note: 0 is the 0th layer (for the sake of explanation)
Solution
matrix_4_x_4 = (1..16).each_slice(4).to_a
matrix_8_x_8 = (1..64).each_slice(8).to_a
def matrix_rotation(*args)
# let us extract rows & cols from our matrix. We also need to know how
# times to rotate.
rows, cols, rotate, matrix = args
# to find out how many layers our matrix have, simply get the min of the two (rows, cols)
# and divide it
layers, str_cols = [rows, cols].min / 2, ""
# needed to beatify our console output in table format
cols.times do str_cols << "%5s " end
# we will work on a temporary array
temp_rows = []
# so the first task is to loop n times, where n is the number of layers
(0...layers).each do |layer|
# this array will get us the output of borders of the layer
row = []
# this loop will output the top-left side of the matrix
# ==============================
# ↓ [ 1, 2, 3, 4 ]
# ↓ [ 5, 6, 7, 8 ]
# ↓ [ 9, 10, 11, 12 ]
# ↓ [ 13, 14, 15, 16 ]
# Output: [[1, 5, 9], [6] ]
# ==============================
(layer...rows - 1 - layer).each do |i|
row << matrix[i][layer]
end
# this loop will output the bottom side of the matrix
# ==============================
# ↓ [ 1, 2, 3, 4 ]
# ↓ [ 5, 6, 7, 8 ]
# ↓ [ 9, 10, 11, 12 ]
# ↓ [ 13, 14, 15, 16 ]
# ↪ → → → →
# Output: [[1, 5, 9, 13, 14, 15], [6, 10]]
# ==============================
(layer...cols - 1 - layer).each do |i|
row << matrix[rows - 1 - layer][i]
end
# this loop will output the right side of the matrix
# ==============================
# ↓ [ 1, 2, 3, 4 ] ↑
# ↓ [ 5, 6, 7, 8 ] ↑
# ↓ [ 9, 10, 11, 12 ] ↑
# [ 13, 14, 15, 16 ] ↑
# ↪ → → → → ⤻
# Output: [[1, 5, 9, 13, 14, 15, 16, 12, 8], [6, 10, 11]]
# ==============================
(rows - 1 - layer).step(layer + 1, -1).each do |i|
row << matrix[i][cols - 1 - layer]
end
# this loop will output the top side of the matrix
# ==============================
# ← ← ← ← ↰
# ↓ [ 1, 2, 3, 4 ] ↑
# ↓ [ 5, 6, 7, 8 ] ↑
# ↓ [ 9, 10, 11, 12 ] ↑
# [ 13, 14, 15, 16 ] ↑
# ↪ → → → → ⤻
# Output: [[1, 5, 9, 13, 14, 15, 16, 12, 8, 4, 3, 2], [6, 10, 11, 7]]
# ==============================
(cols - 1 - layer).step(layer + 1, -1).each do |i|
row << matrix[layer][i]
end
temp_rows << row
end
# let us create a new matrix
result = (1..( rows * cols )).each_slice(rows).to_a
# we're going to loop in the same manner as before
(0...layers).each do |layer|
# based on current layer, get the values around that layer
row = temp_rows[layer]
# !important: the modulo will do the main thing here:
# It will allow us to shift values based on the rotate. But also
# gives us the correct index in the array to start the shift.
# For example, if we want to rotate 2 times: 2 % 12 = 2 for the outer most layer
shift = rotate % row.size
# when whe negate the shift value, we will get the correct index from the end of the array.
# row = [1, 5, 9, 13, 14, 15, 16, 12, 8, 4, 3, 2]
# So -2 in row[-2] for the outer layer is 3. We increment idx, then row[-1] is 2 etc..
idx = -shift
(layer...rows - 1 - layer).each do |i|
result[i][layer] = row[idx]
idx += 1
idx %= row.size
end
(layer...cols - 1 - layer).each do |i|
result[rows - 1 - layer][i] = row[idx]
idx += 1
idx %= row.size
end
(rows - 1 - layer).step(layer + 1, -1).each do |i|
result[i][cols - 1 - layer] = row[idx]
idx += 1
idx %= row.size
end
(cols - 1 - layer).step(layer + 1, -1).each do |i|
result[layer][i] = row[idx]
idx += 1
idx %= row.size
end
end
result.each do |row| printf("#{str_cols}\n", *row) end
end
matrix_rotation(
matrix_8_x_8.size,
matrix_8_x_8.first.size,
2,
matrix_8_x_8
)

Answer 2

Code
def nxt(rows, cols, row, col)
case row
when rows[:first]
col == cols[:last] ? [row+1, col] : [row, col+1]
when rows[:last]
col == cols[:first] ? [row-1, col] : [row, col-1]
else
col == cols[:last] ? [row+1, col] : [row-1, col]
end
end
def rotate_array_times(matrix, n)
arr = matrix.dup.map(&:dup)
nrows, ncols = arr.size, arr.first.size
0.upto([nrows, ncols].min/2-1) do |m|
rows = { first: m, last: nrows-m-1 }
cols = { first: m, last: ncols-m-1 }
rect_size = 2 * (nrows + ncols) - 8*m - 4
rotations = n % rect_size
row = col = rrow = rcol = m
rotations.times { rrow, rcol = nxt(rows, cols, rrow, rcol) }
rect_size.times do
arr[row][col] = matrix[rrow][rcol]
row, col = nxt(rows, cols, row, col)
rrow, rcol = nxt(rows, cols, rrow, rcol)
end
end
arr
end
Examples
matrix = [
[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12],
[13, 14, 15, 16]
]
(1..3).each { |n| p rotate_array_times(matrix, n) }
[[2, 3, 4, 8],
[1, 7, 11, 12],
[5, 6, 10, 16],
[9, 13, 14, 15]]
[[3, 4, 8, 12],
[2, 11, 10, 16],
[1, 7, 6, 15],
[5, 9, 13, 14]]
[[4, 8, 12, 16],
[3, 10, 6, 15],
[2, 11, 7, 14],
[1, 5, 9, 13]]
matrix = (1..24).each_slice(4).to_a
#=> [[ 1, 2, 3, 4],
# [ 5, 6, 7, 8],
# [ 9, 10, 11, 12],
# [13, 14, 15, 16],
# [17, 18, 19, 20],
# [21, 22, 23, 24]]
(1..3).each { |n| p rotate_array_times(matrix, n) }
#=> [[ 2, 3, 4, 8],
# [ 1, 7, 11, 12],
# [ 5, 6, 15, 16],
# [ 9, 10, 19, 20],
# [13, 14, 18, 24],
# [17, 21, 22, 23]]
# [[ 3, 4, 8, 12],
# [ 2, 11, 15, 16],
# [ 1, 7, 19, 20],
# [ 5, 6, 18, 24],
# [ 9, 10, 14, 23],
# [13, 17, 21, 22]]
# [[ 4, 8, 12, 16],
# [ 3, 15, 19, 20],
# [ 2, 11, 18, 24],
# [ 1, 7, 14, 23],
# [ 5, 6, 10, 22],
# [ 9, 13, 17, 21]]
matrix = (1..48).each_slice(8).to_a
#=> [[ 1, 2, 3, 4, 5, 6, 7, 8],
# [ 9, 10, 11, 12, 13, 14, 15, 16],
# [17, 18, 19, 20, 21, 22, 23, 24],
# [25, 26, 27, 28, 29, 30, 31, 32],
# [33, 34, 35, 36, 37, 38, 39, 40],
# [41, 42, 43, 44, 45, 46, 47, 48]]
(1..3).each { |n| p rotate_array_times(matrix, n) }
[[ 2, 3, 4, 5, 6, 7, 8, 16],
[ 1, 11, 12, 13, 14, 15, 23, 24],
[ 9, 10, 20, 21, 22, 30, 31, 32],
[17, 18, 19, 27, 28, 29, 39, 40],
[25, 26, 34, 35, 36, 37, 38, 48],
[33, 41, 42, 43, 44, 45, 46, 47]]
[[ 3, 4, 5, 6, 7, 8, 16, 24],
[ 2, 12, 13, 14, 15, 23, 31, 32],
[ 1, 11, 21, 22, 30, 29, 39, 40],
[ 9, 10, 20, 19, 27, 28, 38, 48],
[17, 18, 26, 34, 35, 36, 37, 47],
[25, 33, 41, 42, 43, 44, 45, 46]]
[[ 4, 5, 6, 7, 8, 16, 24, 32],
[ 3, 13, 14, 15, 23, 31, 39, 40],
[ 2, 12, 22, 30, 29, 28, 38, 48],
[ 1, 11, 21, 20, 19, 27, 37, 47],
[ 9, 10, 18, 26, 34, 35, 36, 46],
[17, 25, 33, 41, 42, 43, 44, 45]]
Explanation
nxt
Given row and column indices row and col, nxt(rows, cols, row, col) returns the indices [next_row, next_col] of the "next" element on the perimeter of a subarray that is to replace the element (also on the perimeter) at indices [row, col] in a single iteration. The subarray is given by the hashes rows and cols which each have keys :first and :last.
Let's consider an an array arr with 4 elements (rows), each element (row) having 6 values (columns). Then
nrows, ncols = arr.size, arr.first.size
#=> [4, 6]
If m = 0
rows = { first: m, last: nrows-m-1 }
#=> {:first=>0, :last=>3}
cols = { first: m, last: ncols-m-1 }
#=> {:first=>0, :last=>5}
It is seen that rows and cols describes the "perimeter" of he array matrix. We can see how nxt works as follows.
first_row, first_col = rows[:first], cols[:first]
row, col = first_row, first_col
print "[#{row}, #{col}]"
loop do
next_row, next_col = nxt(rows, cols, row, col)
print "->[#{next_row}, #{next_col}]"
row, col = next_row, next_col
(puts; break) if [row, col] == [first_row, first_col]
end
[0, 0]->[0, 1]->[0, 2]->[0, 3]->[0, 4]->[0, 5]->[1, 5]->[2, 5]->[3, 5]->
[3, 4]->[3, 3]->[3, 2]->[3, 1]->[3, 0]->[2, 0]->[1, 0]->[0, 0]
If m = 1, the above calculation yields
[1, 1]->[1, 2]->[1, 3]->[1, 4]->[2, 4]->[2, 3]->[2, 2]->[2, 1]->[1, 1]
rotate_array_times
This method constructs a deep copy of matrix, arrr, whose elements are rotated in the prescribed matter n times and then returns the resulting array.
To speed calculations, n is replaced by a modulus of itself. For a 4x4 array, for example, after 12 iterations the perimeter of the array would be back to its original value. Therefore, it is sufficient to perform n % 12 rotations.
matrix contains n = [matrix.size, matrix.first.size].min subarrays whose perimeters are to be rotated. The top-left corner of each subarray is given by the coordinate [m,m], where m = 0..n-1.
For the subarray specified by m the first step is to determine the location of the element of matrix that is to replace the element of arr at [m,m]. That is done in the line
rotations.times { rrow, rcol = nxt(rows, cols, rrow, rcol) }
("rrow" and "rcol" for "replacement row" and "replacement col", respectively). At this time the element of arr at location row #=> m, col #=> m is to be replaced the element of matrix at the location given by rrow and rcol. The following operations then performed as many times as there are elements in the perimeter of the subarray which are to be rotated:
arr[row][col] = matrix[rrow][rcol]
row, col = nxt(rows, cols, row, col)
rrow, rcol = nxt(rows, cols, rrow, rcol)
Tweaking efficiency
A modest improvement in efficiency could be achieved by replacing the line
rotations.times { rrow, rcol = nxt(rows, cols, rrow, rcol) }
with
rrow, rcol = first_replacement_loc(rows, cols, rotations)
and adding the following method.
def first_replacement_loc(rows, cols, rotations)
ncm1 = cols[:last]-cols[:first]
nrm1 = rows[:last]-rows[:first]
return [rows[:first], cols[:first]+rotations] if rotations <= ncm1
rotations -= ncm1
return [rows[:first]+rotations, cols[:last]] if rotations <= nrm1
rotations -= nrm1
return [rows[:last], cols[:last]-rotations] if rotations <= ncm1
rotations -= ncm1
[rows[:last]-rotations, cols[:first]]
end

Answer 3

This is another implementation (I didn't make a method, just the logic that needs to be improved).
array = (1..24).each_slice(6).to_a
array.each { |e| p e }
puts
n = 4 # sub matrix rows
m = 6 # sub matrix cols
x = 0 # x row origin (corner) of the rotation
y = 0 # y col origin (corner) of the rotation
rotations = 2 # negative is ccw, positive is cw
raise "Sub matrix too small, must be 2x2 at least" if m < 2 || n < 2
# to add: check if the submatrix is inside the matrix, given the origin x, y
y_size = array.size
x_size = array.size
idx_map = Array.new(n){ [] }
m.times.map { |mm| n.times.map { |nn| idx_map[nn][mm] = [nn + x, mm + y] } }
before = [(idx_map.map(&:shift)).concat(idx_map.pop).concat(idx_map.map(&:pop).reverse).concat(idx_map.shift.reverse)].flatten(1)
after = before.rotate(rotations)
tmp = array.map(&:dup)
before.size.times.map { |idx| array[before[idx][0]][before[idx][1]] = tmp[after[idx][0]][after[idx][1]]}
array.each { |e| p e }
#=> [1, 2, 3, 4, 5, 6]
#=> [7, 8, 9, 10, 11, 12]
#=> [13, 14, 15, 16, 17, 18]
#=> [19, 20, 21, 22, 23, 24]
#=>
#=> [13, 7, 1, 2, 3, 4]
#=> [19, 8, 9, 10, 11, 5]
#=> [20, 14, 15, 16, 17, 6]
#=> [21, 22, 23, 24, 18, 12]
You can also rotate a 3x3 sub-matrix starting in (1, 1), so, for example n = 3, m = 3, x = 1, y = 1 and rotations = -1:
#=> [1, 2, 3, 4, 5, 6]
#=> [7, 9, 10, 16, 11, 12]
#=> [13, 8, 15, 22, 17, 18]
#=> [19, 14, 20, 21, 23, 24]

Answer 4

I thought it would be interesting to benchmark my code against #Humbledore's. (#iGian: I can add your code to the benchmark if you can edit your answer to wrap it in a method with arguments matrix and nbr_rotations).
def nxt(rows, cols, row, col)
case row
when rows[:first]
col == cols[:last] ? [row+1, col] : [row, col+1]
when rows[:last]
col == cols[:first] ? [row-1, col] : [row, col-1]
else
col == cols[:last] ? [row+1, col] : [row-1, col]
end
end
def cary1(matrix, n)
arr = matrix.dup.map(&:dup)
nrows, ncols = arr.size, arr.first.size
0.upto([nrows, ncols].min/2-1) do |m|
rows = { first: m, last: nrows-m-1 }
cols = { first: m, last: ncols-m-1 }
rect_size = 2 * (nrows + ncols) - 8*m - 4
rotations = n % rect_size
row = col = rrow = rcol = m
rotations.times { rrow, rcol = nxt(rows, cols, rrow, rcol) }
rect_size.times do
arr[row][col] = matrix[rrow][rcol]
row, col = nxt(rows, cols, row, col)
rrow, rcol = nxt(rows, cols, rrow, rcol)
end
end
arr
end
def first_replacement_loc(rows, cols, rotations)
ncm1 = cols[:last]-cols[:first]
nrm1 = rows[:last]-rows[:first]
return [rows[:first], cols[:first]+rotations] if rotations <= ncm1
rotations -= ncm1
return [rows[:first]+rotations, cols[:last]] if rotations <= nrm1
rotations -= nrm1
return [rows[:last], cols[:last]-rotations] if rotations <= ncm1
rotations -= ncm1
[rows[:last]-rotations, cols[:first]]
end
def cary2(matrix, n)
arr = matrix.dup.map(&:dup)
nrows, ncols = arr.size, arr.first.size
0.upto([nrows, ncols].min/2-1) do |m|
rows = { first: m, last: nrows-m-1 }
cols = { first: m, last: ncols-m-1 }
rect_size = 2 * (nrows + ncols) - 8*m - 4
rotations = n % rect_size
row = col = m
rrow, rcol = first_replacement_loc(rows, cols, rotations)
rect_size.times do
arr[row][col] = matrix[rrow][rcol]
row, col = nxt(rows, cols, row, col)
rrow, rcol = nxt(rows, cols, rrow, rcol)
end
end
arr
end
def humbledore(matrix, rotate)
rows, cols = matrix.size, matrix.first.size
layers, str_cols = [rows, cols].min / 2, ""
# cols.times do str_cols << "%5s " end
temp_rows = []
(0...layers).each do |layer|
row = []
(layer...rows - 1 - layer).each do |i|
row << matrix[i][layer]
end
(layer...cols - 1 - layer).each do |i|
row << matrix[rows - 1 - layer][i]
end
(rows - 1 - layer).step(layer + 1, -1).each do |i|
row << matrix[i][cols - 1 - layer]
end
(cols - 1 - layer).step(layer + 1, -1).each do |i|
row << matrix[layer][i]
end
temp_rows << row
end
result = (1..( rows * cols )).each_slice(rows).to_a
(0...layers).each do |layer|
row = temp_rows[layer]
shift = rotate % row.size
idx = -shift
(layer...rows - 1 - layer).each do |i|
result[i][layer] = row[idx]
idx += 1
idx %= row.size
end
(layer...cols - 1 - layer).each do |i|
result[rows - 1 - layer][i] = row[idx]
idx += 1
idx %= row.size
end
(rows - 1 - layer).step(layer + 1, -1).each do |i|
result[i][cols - 1 - layer] = row[idx]
idx += 1
idx %= row.size
end
(cols - 1 - layer).step(layer + 1, -1).each do |i|
result[layer][i] = row[idx]
idx += 1
idx %= row.size
end
end
result
end
require 'benchmark'
def test(rows, cols, rotations)
puts "\nrows = #{rows}, cols = #{cols}, rotations = #{rotations}"
matrix = (1..rows*cols).each_slice(cols).to_a
Benchmark.bm do |x|
x.report("Cary1") { cary1(matrix, rotations) }
x.report("Cary2") { cary2(matrix, rotations) }
x.report("Humbledore") { humbledore(matrix, rotations) }
end
end
test 10,10,1
rows = 10, cols = 10, rotations = 1
user system total real
Cary1 0.000000 0.000000 0.000000 ( 0.000077)
Cary2 0.000000 0.000000 0.000000 ( 0.000074)
Humbledore 0.000000 0.000000 0.000000 ( 0.000051)
test 10,10,78
rows = 10, cols = 10, rotations = 78
user system total real
Cary1 0.000000 0.000000 0.000000 ( 0.000079)
Cary2 0.000000 0.000000 0.000000 ( 0.000061)
Humbledore 0.000000 0.000000 0.000000 ( 0.000053)
test 100,100,378
rows = 100, cols = 100, rotations = 378
user system total real
Cary1 0.000000 0.000000 0.000000 ( 0.007673)
Cary2 0.015625 0.000000 0.015625 ( 0.005168)
Humbledore 0.000000 0.000000 0.000000 ( 0.002919)
test 500,500,1950
rows = 500, cols = 500, rotations = 1950
user system total real
Cary1 0.171875 0.000000 0.171875 ( 0.166671)
Cary2 0.140625 0.000000 0.140625 ( 0.137141)
Humbledore 0.046875 0.000000 0.046875 ( 0.053705)
test 500,1000,2950
rows = 500, cols = 1000, rotations = 2950
user system total real
Cary1 0.296875 0.000000 0.296875 ( 0.292997)
Cary2 0.234375 0.000000 0.234375 ( 0.248384)
Humbledore 0.125000 0.000000 0.125000 ( 0.103964)
Benchmark reports execution times in seconds. The results are found to be quite consistent.
Notice that in all of the tests I performed the number of columns of the array is at least as large as the number of rows. That's because a NoMethodError (undefined method '[]=' for nil:NilClass) exception was raised in Humbledore's code whenever the number of rows exceeded the number of columns. (Try test 3,2,1, for example.) The error message occurred in the second line of the following block of code.
(layer...cols - 1 - layer).each do |i|
result[rows - 1 - layer][i] = row[idx]
idx += 1
idx %= row.size
end
I expect the problem is easily fixable.