This is Part 1 of a 4-part blog post on the mathematics of the 4x4 Rubik's Cube, its relation to algorithms, and some curious properties of Rubik's Cubes.

You are currently reading Part 1 of this blog post: Part 1: Representations

See Part 2 of this blog post here: Part 2: Permutations

See Part 3 of this blog post here: Part 3: Factoring Permutations

See Part 4 of this blog post here: Part 4: Sequence Order

Table of Contents

• Introduction: Why the Rubik's Cube

  • Why the 4x4 Rubik's Cube

• How the Rubik's Cube Works

  • The Pieces
  • Face Notation
  • Color Notation
  • Move Notation
    • Regular Face Rotations
    • Reverse Face Rotations
    • Double Face Rotations
    • Second Layer Face Rotations

• Computer Representation of a Rubik's Cube

  • Operations and Functionality
  • Face Numbering

• Tuple

  • Tuple Representation Requirements
  • Tuple Representation

• Preview of Part 2

• References

Introduction: Why The Rubik's Cube

In this series of four posts, we'll take a look at the 4x4 Rubik's Cube. The Rubik's Cube is an interesting puzzle that has some profound mathematical connections to group theory and combinatorics.

Group theory is a branch of mathematics that applies to any system that exhibits symmetry; combinatorics is the mathematics of counting things.

The Rubik's Cube gives us the opportunity to apply concepts in group theory and combinatorics to better understand how the cube works, and to learn how to apply these principles to real world systems.

Finally, we will wrap up by discussing some of the algorithms that are required to deal with a Rubik's Cube computationally.

In the next two posts, we'll talk more about the mathematical representation of permutations of the Rubik's Cube, and how to use this representation to understand some of the properties of move sequences when applied to the cube.

Why The 4x4 Rubik's Cube

The 4x4 Rubik's cube, also known as the Rubik's Revenge cube, is larger than the standard 3x3 Rubik's Cube - 96 faces, instead of the ususal 36. The 4x4 cube exhibits some particularly interesting properties as a result of having an even number of squares on each edge.

How the Rubik's Cube Works

Let's start with a discussion of cube mechanics, since this is important to coming up with an accurate mathematical model of the cube.

The Pieces

The 4x4 Rubik's Cube consists of six faces of sixteen squares each, for a total of 96 face squares. These face squares are not completely interchangeable, however - the 4x4 cube is actually composed of three types of pieces, called "cubies".

Figure 1: Corner pieces are green.

The first type of piece is a corner piece, which contains 3 faces. Note that it is impossible for the corner pieces to change their chirality (direction of rotation). There are 8 corner pieces, each of which can be oriented in 3 different ways.

Figure 2: Double edge pieces are blue.

The second type of piece is a double edge (dedge) piece. Each edge is composed of two double edges. There are 24 total double edge pieces, which can be further classified into 12 left-handed and 12 right-handed dedge pieces.

Figure 3: Center pieces are blue.

Lastly, there are 4 center pieces in the center of each face, for a total of 24 center pieces. Note that each of the center pieces of a given color are interchangeable, unlike the double edge pieces or corners.

Face Notation

To refer to particular faces on the cube, we use six letters to indicate different faces:

U - upper face (the top of the cube)

D - downward face (the bottom of the cube0

F - front face (the front of the cube)

B - back face (the back side of the cube)

L - left face of the cube (on the left side when facing the front F face)

R - right face of the cube

This will help refer to how we will rotate the cube.

Color Notation

In the solved state, each cube face has one of six colors. The orientation of these colors relative to one another is always fixed; the red and orange colored faces, for example, are never adjacent. This is due to the nature of the mechanical pieces that compose the Rubik's Cube.

The standard faces for each color on a solved cube are:

• U = White
• D = Yellow
• F = Green
• B = Blue (Back-Blue)
• L = Orange
• R = Red (Red-Right)

Note that on a 3x3 cube, we can always determine the final color a face will have, because the six center pieces on each side of a 3x3 cube always remain fixed.

On a 4x4 cube, however, all four center squares can rotate and move, meaning all 24 center squares are totally interchangeable, and there is no link between the center colors on a 4x4 cube and the final color that will be on that face when the cube is solved.

Move Notation

Using the face notation explained above, we can denote multiple types of moves on the 4x4 Rubik's Cube.

We have 36 total moves that we can make on the 4x4 Rubik's Cube, which can be grouped by the dozen:

L l r R
U u d D
B b f F

L' l' r' R'
U' u' d' D'
B' b' f' F'

2L 2L' 2R 2R'
2U 2U' 2D 2D'
2B 2B' 2F 2F'

Let's go through the details of the notation.

Regular Face Rotations

The regular face turns are denoted with capital letters: L R U D B F refer to a single clockwise rotation of the respective face. Here, "clockwise" means the direction that is clockwise when facing the given face head-on.

Reverse Face Rotations

The ' apostrophe following moves, as in L' R' U' D' B' F', indicates that the move shoud be a counter-clockwise rotation of the given face, instead of clockwise.

Double Face Rotations

Rotations that are indicated using a lowercase letter refer to two-layer rotations: l r u d b f.

Figure 4: Cube state after move u.

That is, the lowercase u refers to the clockwise rotation of the top two layers of the cube; the lowercase r refers to the clockwise rotation of the rightmost two layers of the cube; and so on.

The apostrophe also serves to indicate a counter-clockwise rotation: 'l r' u' d' b' f' indicate counter clockwise rotations of the two left, two right, two upper, two bottom, two back, and two front layers, respectively.

We have covered the first 24 moves - clockwise and counter-clockwise rotations of single and double layers.

Second Layer Face Rotations

Figure 5: Cube state after move 2U.

The 2 notation indicates a rotation of the second layer only. For example, 2U refers to the clockwise rotation of the second layer from the top. This is equivalent to the move sequence u U'.

Likewise, the apostrophe indicates a counterclockwise rotation.

Computer Representation of a Rubik's Cube

The computer representation we are using is the rubiks-cube-NxNxN-solver library by Github user @dwalton.

We have modified this library to provide additional functionality needed in the project; the fork used in this project is available at git.charlesreid1.com: rubiks-cube-nnn-solver

Using this library, here's how we create a 4x4 Rubik's Revenge cube:

In [1]: from rubikscubennnsolver.RubiksCube444 import RubiksCube444, solved_4x4x4

In [2]: order = 'URFDLB'

In [3]: cube = RubiksCube444(solved_4x4x4, order)

In [4]: cube.print_cube()
         U U U U
         U U U U
         U U U U
         U U U U

L L L L  F F F F  R R R R  B B B B
L L L L  F F F F  R R R R  B B B B
L L L L  F F F F  R R R R  B B B B
L L L L  F F F F  R R R R  B B B B

         D D D D
         D D D D
         D D D D
         D D D D

Operations and Functionality

Some important functionality:

• Obtaining each side
• Applying rotation
• Applying sequence of rotations
• Each side
• Side face numberings, centers, edges

To obtain each side, use the sides attribute:

In [8]: print(cube.sides)
OrderedDict([('U', <rubikscubennnsolver.RubiksSide.Side object at 0x11172d358>), 
             ('L', <rubikscubennnsolver.RubiksSide.Side object at 0x11172d240>), 
             ('F', <rubikscubennnsolver.RubiksSide.Side object at 0x11172d5c0>), 
             ('R', <rubikscubennnsolver.RubiksSide.Side object at 0x11172d5f8>), 
             ('B', <rubikscubennnsolver.RubiksSide.Side object at 0x11172d518>), 
             ('D', <rubikscubennnsolver.RubiksSide.Side object at 0x11172d390>)])

Each Side object has a long list of methods, including methods to obtain the index numbers of corner, edge, or center faces on a particular side.

To apply a rotation of a single face, use the rotate() method and pass the name of the face:

In [10]: cube.rotate("U")

In [11]: cube.print_cube()
         U U U U
         U U U U
         U U U U
         U U U U

F F F F  R R R R  B B B B  L L L L
L L L L  F F F F  R R R R  B B B B
L L L L  F F F F  R R R R  B B B B
L L L L  F F F F  R R R R  B B B B

         D D D D
         D D D D
         D D D D
         D D D D

Unfortunately, the rotate method does not take sequences of moves, but this is easily resolved:

In [12]: cube = RubiksCube444(solved_4x4x4, order)

In [13]: sequence = "U L U' L'"

In [14]: for move in sequence.split():
    ...:     cube.rotate(move)
    ...:

In [15]: cube.print_cube()
         L U U U
         U U U U
         U U U U
         U B B L

D F F F  R L L F  U R R R  B B B B
L L L L  F F F F  R R R R  B B B U
L L L L  F F F F  R R R R  B B B U
L L L L  F F F F  R R R R  B B B U

         D D D D
         D D D D
         D D D D
         B D D D

Face Numbering

Here is the numerical representation of the faces, which we will make extensive use of:

In [6]: cube.print_cube_layout()
             01 02 03 04
             05 06 07 08
             09 10 11 12
             13 14 15 16

17 18 19 20  33 34 35 36  49 50 51 52  65 66 67 68
21 22 23 24  37 38 39 40  53 54 55 56  69 70 71 72
25 26 27 28  41 42 43 44  57 58 59 60  73 74 75 76
29 30 31 32  45 46 47 48  61 62 63 64  77 78 79 80

             81 82 83 84
             85 86 87 88
             89 90 91 92
             93 94 95 96

Tuple Representation

We have a goal of finding a way of representing the state of the 4x4 Rubik's Revenge using a tuple, which is a mathematical object that will enable us to investigate properties of sequences, moves, and rotations.

It is important to note that the mechanics of the cube restrict some of the 96 total faces to only occur in particular configurations. By using a tuple of 96 integers, we are overspecifying the state of the cube, and we would be able to do much better if our goal were a minimal representation of the Rubik's Cube state.

However, our goal is not a minimal representation of the cube, but a unique representation of the cube. As we will see in a later post, the schema we use does not actually matter, so long as we can represent each unique state of the cube using a sequence of integers of arbitrary length.

Tuple Representation Requirements

The 4x4 cube, in the solved state, has a few characteristics that can be used to indicate a particular permutation or configuration:

• Face indciators UDFBLR
• Colors WYGBRO
• Integers 1-96 to number each face

Here is how the faces representation looks:

In [17]: cube.print_cube()
         U U U U
         U U U U
         U U U U
         U U U U

L L L L  F F F F  R R R R  B B B B
L L L L  F F F F  R R R R  B B B B
L L L L  F F F F  R R R R  B B B B
L L L L  F F F F  R R R R  B B B B

         D D D D
         D D D D
         D D D D
         D D D D

The equivalent color representation is:

         W W W W
         W W W W
         W W W W
         W W W W

O O O O  G G G G  R R R R  B B B B
O O O O  G G G G  R R R R  B B B B
O O O O  G G G G  R R R R  B B B B
O O O O  G G G G  R R R R  B B B B

         Y Y Y Y
         Y Y Y Y
         Y Y Y Y
         Y Y Y Y

However, the tuple representation cannot use colors to represent the state of the cube. This is because a tuple representation using "R" to represent each red face would give us no way of distinguishing between the (non-interchangeable) red faces on the cube. For example, if the red-green double edge piece were replaced with a red-blue double edge piece, oriented with the red face at the same location, the n-tuple needs to reflect that this face has a different value than it did the prior move.

For this reason, we must use an integer to index each distinct face:

             01 02 03 04
             05 06 07 08
             09 10 11 12
             13 14 15 16

17 18 19 20  33 34 35 36  49 50 51 52  65 66 67 68
21 22 23 24  37 38 39 40  53 54 55 56  69 70 71 72
25 26 27 28  41 42 43 44  57 58 59 60  73 74 75 76
29 30 31 32  45 46 47 48  61 62 63 64  77 78 79 80

             81 82 83 84
             85 86 87 88
             89 90 91 92
             93 94 95 96

We can rearrange this into a 96-tuple:

(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 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96)

If we apply a rotation, for example U R, we will end up with a different cube:

             13 09 05 01
             14 10 06 02
             15 11 07 03
             48 44 40 36

33 34 35 84  61 57 53 49  16 66 67 68  17 18 19 20
21 22 23 24  37 38 39 88  62 58 54 50  12 70 71 72
25 26 27 28  41 42 43 92  63 59 55 51  08 74 75 76
29 30 31 32  45 46 47 96  64 60 56 52  04 78 79 80

             81 82 83 77
             85 86 87 73
             89 90 91 69
             93 94 95 65

This particular sequence of moves results in a cube state uniquely represented by the following 96-tuple:

(13 9 5 1 14 10 6 2 15 11 7 3 48 44 40 36 33 34 35 84 21 22 23 24 25 26 27 28 29 30 31 32 61 57 53 49 37 38 39 88 41 42 43 92 45 46 47 96 16 66 67 68 62 58 54 50 63 59 55 51 64 60 56 52 17 18 19 20 12 70 71 72 8 74 75 76 4 78 79 80 81 82 83 77 85 86 87 73 89 90 91 69 93 94 95 65)

Tuple Representation

Now, we have managed to find a unique representation for any given cube state by labeling each individual face 1-96.

But we aren't quite done yet. It turns out that our statement, that our representation should treat each face as unique, is not strictly true for all 96 faces.

The square pieces are completely interchangeable, due to the fact that they are not connected to any other faces (and therefore have no orientation or way of differentiating them from one another).

If we are doing anything that involves counting configurations, it is important to account for this fact, by treating the following groups of face indices as interchangeable:

(6, 7, 10, 11)
(22, 23, 26, 27)
(38, 39, 42, 43)
(54, 55, 58, 59)
(70, 71, 74, 75)
(86, 87, 90, 91)

In Part 2 and Part 3 of this series, we will encounter these concepts again, and it will become more clear what these caveats and notes mean through example.

Following is a preview of Part 2 of this 3-part blog post.

Preview of Part 2

In Part 2 of this series, we will utilize the n-tuple representation of the 4x4 Rubik's Cube in order to write permutations of the cube corresponding to specific states, and turn a sequence of moves on the cube into permutations.

We will also create a map for each type of move, telling us where each face index will end up.

In Part 3 we will use these to predict properties of rotations applied to the 4x4 Rubik's Cube.

References

1. "Rubik's Cube". Charlesreid1.com wiki, Charles Reid. Edited 11 January 2017. Accessed 11 January 2017. <https://charlesreid1.com/wiki/Rubiks_Cube>

2. "Rubik's Revenge". Charlesreid1.com wiki, Charles Reid. Edited 11 January 2017. Accessed 11 January 2017. <https://charlesreid1.com/wiki/Rubiks_Revenge>

3. "Rubik's Cube/Tuple". Charlesreid1.com wiki, Charles Reid. Edited 11 January 2017. Accessed 11 January 2017. <https://charlesreid1.com/wiki/Rubiks_Cube/Tuple>

4. "Rubik's Cube/Permutations". Charlesreid1.com wiki, Charles Reid. Edited 11 January 2017. Accessed 11 January 2017. <https://charlesreid1.com/wiki/Rubiks_Cube/Permutations>

5. "Github - dwalton76/rubiks-cube-NxNxN-solver". dwalton76, Github Repository, Github Inc. Accessed 11 January 2017. <https://github.com/dwalton76/rubiks-cube-NxNxN-solver>

6. "Rubik's Cube NxNxN Solver". Git repository, git.charlesreid1.com. Charles Reid. Updated 11 January 2017. <https://git.charlesreid1.com/charlesreid1/rubiks-cube-nnn-solver>

7. "Rubiks Cube Cycles". Git repository, git.charlesreid1.com. Charles Reid. Updated 11 January 2017. <https://git.charlesreid1.com/charlesreid1/rubiks-cube-cycles>

Generating Permutations

In today's post we're going to discuss the generation of permutations.

Often, in combinatorics problems, we are interested in how many different instances or configurations of a particular thing we can have (what we'll call "enumeration" or "counting"). However, that is different from wanting to actually see all of those configurations. Indeed, if we are counting something with an astronomical number of configurations, we don't want to try to list all of them.

However, as usual, Donald Knuth, who covers the topic of permutation generation in Volume 4A of his classic work, The Art of Computer Programming, uncovers a problem that is much more complicated and subtle than it initially appears.

Background: Radix Method for Generating Permutations

In Volume 4A of his classic work, The Art of Computer Programming, Donald Knuth turns to the question of generating permutations for a given combinatoric system. The book opens with Knuth jumping immediately into the problem of generating permutations.

Algorith M is the first algorithm Knuth presents to generate all unique tuples. To pose the problem a little more clearly, consider a combinatoric system that has \(n\) independent parts, each part having a number of possible states. We can completely specify the state of the combinatoric system by specifying an \(n\)-tuple:

where each independent variable takes on one of its possible values \(0 \leq a_i \leq m_i\).

Knuth's Algorithm M starts by setting all a's to 0, and incrementing the right-most entry of the tuple (carrying if necessary).

This is equivalent to counting in binary from 0 to \(N-1\), or to labeling every possible outcome with a number between 0 and \(N-1\).

This becomes more clear if we consider the \(n\)-tuple to be a number in a variable-radix system:

The number is \(\left[ a_1, a_2, \dots, a_n \right]\)

The radix is \(\left[ m_1, m_2, \dots, m_n \right]\)

By repeatedly adding 1 to the number \(\left[ a_1, a_2, \dots, a_n \right]\), we iterate through every possible tuple \((a_1, a_2, \dots, a_n)\) and therefore through every possible combination of the independent variables.

Knuth's Algorithm M

Algorithm M (Mixed-radix generation). This algorithm visits all \(n\)-tuples that satisfy the number/radix expressions above, by repeatedly adding 1 to the mixed-radix number until overflow occurs. (Aux. variables \(a_0\) and \(m_0\) introduced for convenience only.)

M1. [Initialize.] Set \(a_j \rightarrow 0\) for \(0 \leq j \leq n\), set \(m_0 \rightarrow 2\).

M2. [Visit.] Visit the \(n\)-tuple \((a_1, \dots, a_n)\). (The program that wants to examine all \(n\)-tulpes now does its thing.)

M3. [Prepare to add one.] Set \(j \rightarrow n\).

M4. [Carry if necessary.] If \(a_j = m_j - 1\), set \(a_j \rightarrow 0, j \rightarrow j-1\) and repeat this step.

M5. [Increase, unless done.] If \(j=0\), terminate algorithm. Otherwise set \(a_j = a_j + 1\) and go back to step \(M2\).

Implementing Algorithm M

Unfortunately, this pseudocode takes some work to translate. Fortunately, that's already done in the method below.

The method below implements Algorithm M in Python to generate random sequences on a 4x4 Rubik's Cube (called the Rubik's Revenge, or RR for short). The RR cube has six faces that can each be rotated clockwise or counterclockwise by a single layer, denoted by the uppercase letters U, D, B, F, L, R (up, down, back, front, left, right, respectively) for clockwise rotations and U', D', B', F', L', R' for counterclockwise rotations, for 12 total moves.

However, on a 4x4 cube, we can also rotate two layers at a time. (That's the limit; moving three layers at a time is equivalent to a reversed rotation of the remaining single layer.) This type of rotation is denoted with a "w".

Thus, rotating the top two layers of each of the six faces clockwise is denoted Uw, Dw, Bw, Fw, Lw, Rw, and counterclockwise rotations are denoted Uw', Dw', Bw', Fw', Lw', Rw', for 12 additional moves.

We have one more type of move, which is where the second layer only is removed. This type of move is denoted with 2, and the face whose second layer is being rotated, for six moves: 2U, 2D, 2B, 2F, 2L, 2R. The prime notation denotes again a counterclockwise rotation, for an additional six moves: 2U', 2D', 2B', 2F', 2L', 2R'. This yields another 12 moves.

There are 36 total moves that can be executed on the 4x4 Rubik's Revenge cube.

def algorithm_m(n):
    """
    Knuth's Algorithm M for permutation generation,
    via AOCP Volume 4 Fascile 2.
    This is a generator that returns permtuations 
    generated using the variable-radix method. 

    This generates ALL permutations.
    Many of these are rotations of one another,
    so use the get_rotations() function
    (defined later) to eliminate redundant sequences.
    """
    moves = ['A','B','C','D']

    # M1 - Initialize
    a = [0,]*n
    m = [len(moves),]*n

    j = n-1

    nvisits = 0
    while True:

        # M2 - visit
        move_sequence = " ".join([ moves[int(aj)] for aj in a])
        yield move_sequence 

        nvisits += 1

        # M3 - prepare to +1
        j = n-1

        # M4 - carry
        while( a[j] == m[j]-1):
            a[j] = 0
            j = j-1

        # M5 - increase unless done
        if(j<0):
            break
        else:
            a[j] = a[j] + 1

Test Drive

Let's take a look at how Algorithm M looks when it is applied. No surprises here: Algorithm M generates each of the possible permutations in sequence.

from pprint import pprint

# (Algorithm M goes here) 

if __name__=="__main__":
    pprint(list(algorithm_m(3)))

and the result:

    ['A A A',
     'A A B',
     'A A C',
     'A A D',
     'A B A',
     'A B B',
     'A B C',
     'A B D',
     'A C A',
     'A C B',
     'A C C',
     'A C D',
     'A D A',
     'A D B',
     'A D C',
     'A D D',
     'B A A',
     'B A B',
     'B A C',
     'B A D',
     'B B A',
     'B B B',
     'B B C',
     'B B D',
     'B C A',
     'B C B',
     'B C C',
     'B C D',
     'B D A',
     'B D B',
     'B D C',
     'B D D',
     'C A A',
     'C A B',
     'C A C',
     'C A D',
     'C B A',
     'C B B',
     'C B C',
     'C B D',
     'C C A',
     'C C B',
     'C C C',
     'C C D',
     'C D A',
     'C D B',
     'C D C',
     'C D D',
     'D A A',
     'D A B',
     'D A C',
     'D A D',
     'D B A',
     'D B B',
     'D B C',
     'D B D',
     'D C A',
     'D C B',
     'D C C',
     'D C D',
     'D D A',
     'D D B',
     'D D C',
     'D D D']

What Other Ways Are There?

All of this may seem obvious or uninteresting, if you don't realize there are other ways of generating all possible \(n\)-tuples.

It's a bit easier to think about for binary numbers. Imagine you're trying to generate every possible 10-digit binary number. This means generating all binary numbers between \(0\) and \(2^{10}-1\).

Algorithm M, as we saw above, just counts from 0 to \(2^{10}-1\). But this can involve changing a large number of bits (for example, adding 1 to 001111111 results in 010000000, changing 8 digits. Knuth presents an alternative algorithm that only requires changing one bit to generate the next permutation, making the algorithm much faster.

More on that algorithm in a future blog post...

An Integral from Hardy

Every once in a while you run across an integral that looks straightforward at first glance, but the more you stare at it, the less obvious it becomes how to proceed. This is one of those integrals. It's the sort of problem that G. H. Hardy - the great English mathematician who taught at Cambridge for decades and famously collaborated with Ramanujan - liked to pose to students as a way to sharpen their instincts about integration.

Here's the integral:

At a glance, this doesn't fit any of the standard forms you memorize as an undergraduate. There's no obvious \(u\)-substitution. Integration by parts doesn't lead anywhere useful. And trying to attack it head-on with a trigonometric substitution leads to a mess. It's the kind of integral that makes you wonder whether it even has a closed-form solution at all.

It does. And the trick is a delightful one.

The Setup: Why This Is Hard

The difficulty here is entirely due to the shape of the denominator. If we had just \(\sqrt{x+a}\) in the denominator, this would be a trivial power-rule integral:

Similarly for \(\sqrt{x+b}\). But the sum of two square roots in the denominator is what gums up the works. You can't distribute the reciprocal across a sum. You can't easily substitute for the whole denominator, because its derivative involves both roots in a form that doesn't cleanly cancel anything in the numerator. The two square roots are entangled in a way that resists most of the standard tools.

The way out isn't a substitution at all. It's an algebraic manipulation you learned in high school, applied in a place you weren't expecting to see it.

The Trick: Multiply by the Conjugate

The move is to multiply the numerator and denominator by the conjugate of the denominator - that is, by \(\sqrt{x+a} - \sqrt{x+b}\):

This is the same trick you use to rationalize a denominator in algebra class. The reason it works is the difference-of-squares identity:

When \(u\) and \(v\) are square roots, squaring them strips them away entirely. Applying that identity with \(u = \sqrt{x+a}\) and \(v = \sqrt{x+b}\), the denominator becomes:

The two \(x\)'s cancel exactly, and we're left with a constant in the denominator. That's the whole game. The nasty entangled denominator collapses into \(a - b\), and the two square roots that were causing all the trouble have moved up into the numerator, where they can be integrated independently.

Putting It Together

After the conjugate multiplication, the integral becomes:

Now each piece is a straightforward power-rule integral. Using \(\int (x+c)^{1/2} \, dx = \tfrac{2}{3}(x+c)^{3/2}\):

Putting it all together:

That's the answer. Clean, symmetric, and once you see it, obvious in retrospect - which is the hallmark of a good trick.

Sanity Check by Differentiation

It's always worth verifying an integral by differentiating and checking that we recover the original integrand. Let

Differentiating term by term with the chain rule (the derivative of \(x+a\) is just 1, so the chain rule contributes nothing here):

Now we need to show that this equals \(\dfrac{1}{\sqrt{x+a} + \sqrt{x+b}}\). Multiply the top and bottom of our derivative by \(\sqrt{x+a} + \sqrt{x+b}\):

The numerator simplifies to \(a - b\), which cancels with the \(a - b\) in the denominator, leaving:

Confirmed.

What About \(a = b\)?

You may have noticed the answer contains \(a - b\) in the denominator, which looks alarming if \(a = b\). But if \(a = b\), the original integral degenerates: the denominator becomes \(2\sqrt{x+a}\), and the integral is just

So the interesting case - the one worth Hardy's attention - is when \(a \neq b\), and the formula is well-defined there. The apparent singularity at \(a = b\) is removable: if you take the limit of the closed-form answer as \(b \to a\) using L'Hôpital's rule or a Taylor expansion of \((x+b)^{3/2}\) around \(b = a\), you recover exactly \(\sqrt{x+a} + C\). The answer is consistent with the degenerate case, it just has to be teased out with a limit.

Why This Matters

The pattern here - multiply by the conjugate to rationalize away a sum of square roots - is worth stashing away in your bag of tricks. It shows up more often than you'd expect, and not just in integration. Any time you have an expression of the form \(\sqrt{P} + \sqrt{Q}\) that's obstructing an algebraic manipulation, ask yourself whether the conjugate \(\sqrt{P} - \sqrt{Q}\) would give you a cleaner form via the difference-of-squares identity.

That's the beauty of a problem like this one. It's not that the mathematics is deep, or that the answer is exotic. It's that the technique is transferable. Hardy's problems tend to have this quality: they force you to recognize a pattern that then generalizes far beyond the specific problem in front of you. You learn the pattern once, and afterwards you see it everywhere.

References

• G. H. Hardy, A Course of Pure Mathematics, Cambridge University Press. Available on Project Gutenberg.
• Wikipedia, Rationalisation (mathematics).