Algorithm to rotate a matrix using transpose

This is a horrible question but there is a neat trick.

To rotate 90 degrees just take transpose then reverse the rows. To rotate -90 (270) degrees take the transpose then reverse the columns.

val matrix = Array(Array(1,2), Array(3,4), Array(5,6))
print(matrix)

1 2
3 4
5 6

Transpose

def transpose(matrix: Array[Array[Int]]): Array[Array[Int]] = {
  matrix.head.indices.map(i => matrix.map(_(i))).toArray
}

print(transpose(matrix))

1 3 5
2 4 6

Rotate 90 degrees clockwise 

Transpose and reverse rows:

def rotate90(matrix: Array[Array[Int]]): Array[Array[Int]] = {
  transpose(matrix).map(_.reverse)
}

print(rotate90(matrix))

5 3 1
6 4 2

Voila.

Rotate 90 degrees anticlockwise

Transpose and reverse columns:

def rotateMinus90(matrix: Array[Array[Int]]): Array[Array[Int]] = {
  val t = transpose(matrix)
  t.head.indices.foreach { i =>
    (0 until (t.length/2)).foreach { j =>
      val temp = t(j)(i)
      t(j)(i) = t(t.length - j - 1)(i)
      t(t.length - j - 1)(i) = temp
    }
  }
  t
}

print(rotateMinus90(matrix))

2 4 6
1 3 5

Full code:

The 12 Days of Christmas

Question. 

On the first day of Christmas my true love gave to me: a partridge in a pear tree

On the second day of Christmas my true love gave to me: two turtle doves and a partridge in a pear tree

Etc.

On the 12th day of Christmas, of which type of bird/person do you have the greatest number? 

Answer:

Exhaustively, on day 12:

1 * 12 partridges
2 * 11 turtle doves
3 * 10 french hens
4 * 9 calling birds
5 * 8 gold rings
6 * 7 geese a-laying
7 * 6 swans a-swimming
8 * 5 maids a-milking
9 * 4 ladies dancing
10 * 3 lords a-leaping
11 * 2 pipers piping
12 * 1 drummers drumming

(yes i had to google what was what on each day)

So the answer is that you have 42 geese and 42 swans.

Intuitively we can say that the biggest value is when we are multiplying the numbers that are closest together, in this case 6*7. So the answer for the max is $floor(\frac{13}{2}) * (13-floor(\frac{13}{2}) )$

How to prove this mathematically:

Firstly, is there a general formula?

Yes, where $i$ is the day you first getting a particular type of animal then clearly on day twelve you have $i*(13 - i) $

To find the max we differentiate:

$$\frac{d}{di} i*(13 - i) $$

$$ = \frac{d}{di} 13i - i^2 $$

$$ = 13 - 2i $$

Set this to zero and solve

$$ 13 - 2i = 0 => i = \frac{13}{2}$$

Is this definitely maximum? Differentiating again = $-2$ which means this is maximal.

Obviously, our data is discrete, so we check the values on either side of 6.5, 6 and 7. In this case they give the same answer: 6 * 7 and 7*6 = 42.

(Fun fact when I was fourteen my friends and I made up our own version of this song. All I can remember is the following:

5 g strings
4 manly hugs
3 french kisses
2 bulging biceps
And a nice piece of ass for me

I've forgotten everything I ever learned in A level chemistry, but it's good to know that the important stuff sticks. )