# Heap’s Algorithm

In this post, I describe an alternative to the Narayana Pandita’s algorithm, an algorithm used to generate permutations of a set of elements which I have already written about.

# Comparison with the alternative

This is one of the most efficient ways to generate the permutations of a list. It is more efficient than the Narayana Pandita’s algorithm if both are properly implemented.

This algorithm does not generate permutations in lexicographic order, which is required in some scenarios.

This algorithm does not require any comparison between elements of the list it is sorting, making it more appealing for sequences without defined order or whose elements are difficult to compare.

# Python implementations

Based on implementations found on the Internet, I wrote two PEP 8 compliant Python 3 versions of Heap’s algorithm. The first one, which is also the most elegant one in my opinion, is a recursive implementation. As one may assume, the cost of the several levels of recursion make it substantially more expensive than it needs to be.

## Recursive implementation

def swap(elements, i, j):
elements[i], elements[j] = elements[j], elements[i]

def generate_permutations(elements, n):
if n == 0:
yield elements
else:
for i in range(n - 1):
# Generate permutations with the last element fixed.
yield from generate_permutations(elements, n - 1)
# Swap the last element.
if i % 2 == 0:
swap(elements, i, n - 1)
else:
swap(elements, 0, n - 1)
# Generate the last permutations after the final swap.
yield from generate_permutations(elements, n - 1)

def permutations(elements):
yield from generate_permutations(elements, len(elements))



This code uses the yield from syntax, added in Python 3.3. You can read more about it here.

The algorithm is not trivially understood. Essentially, what is happening is a locking of the rightmost element and the recursive permutation of all other elements, then an intelligently chosen swap involving the rightmost element and the repetition of the process until all elements have been in the rightmost position.

## Non-recursive implementation

def swap(elements, i, j):
elements[i], elements[j] = elements[j], elements[i]

def generate_permutations(elements, n):
# As by Robert Sedgewick in Permutation Generation Methods
c =  * n
yield elements
i = 0
while i < n:
if c[i] < i:
if i % 2 == 0:
swap(elements, 0, i)
else:
swap(elements, c[i], i)
yield elements
c[i] += 1
i = 0
else:
c[i] = 0
i += 1

def permutations(elements):
return generate_permutations(elements, len(elements))



## Benchmarking

I have benchmarked the algorithm for lexicographic permutations and both implementations above using a sequence of 10 integers and a sequence of 10 strings of 100 characters which only differed by the last character. Note that although an input of length 10 may seem small for an unaware reader, there are 10! = 3,628,800 results generated even for such small input.

The benchmarking results follow.

Lexicographic with 10 integers took 10.99 seconds.
Recursive Heap's with 10 integers took 14.10 seconds.
Non-recursive Heap's with 10 integers took 4.29 seconds.

Lexicographic with 10 strings took 11.66 seconds.
Recursive Heap's with 10 strings took 14.43 seconds.
Non-recursive Heap's with 10 strings took 4.31 seconds.


As you can see, the cost of recursion makes the more elegant implementation of Heap’s algorithm even slower than the lexicographic permutation generator. However, the non-recursive implementation is substantially more efficient. It is also visible from these numbers that the performance hit from the use of big strings (which are more expensive to compare than small integers) was bigger on the lexicographic permutation generator than on the generators that do not compare elements.

# Balancing Chemical Equations

On this post, I describe an efficient way to balance chemical equations using linear algebra.

# Procedure

From any chemical equation, we can derive a vector equation that describes the number of atoms of each element present in the reaction. Row reducing this vector matrix leads to the general solution, from which we can easily derive a solution with integer coefficients.

# Example

Find the smallest integer coefficients that balance the following chemical equation.

MnS

• As2Cr10O35
• H2SO4 → HMnO4
• AsH3
• CrS3O12
• H2O

## Solution

From the chemical equation above, we can derive the following matrix. Notice that the first row is for Mn, the second is for S, and so on. However, how you choose to order does not matter, as long as all columns respect the order.

1    0    0   -1    0    0    0
1    0    1    0    0   -3    0
0    2    0    0   -1    0    0
0   10    0    0    0   -1    0
0   35    4   -4    0  -12   -1
0    0    2   -1   -3    0   -2


The last four columns all have negative atom counts because they have been subtracted from both sides of the original vector equation.

After row reducing the matrix with Octave, we get

1.00000   0.00000   0.00000   0.00000   0.00000   0.00000  -0.04893
0.00000   1.00000   0.00000   0.00000   0.00000   0.00000  -0.03976
0.00000   0.00000   1.00000   0.00000   0.00000   0.00000  -1.14373
0.00000   0.00000   0.00000   1.00000   0.00000   0.00000  -0.04893
0.00000   0.00000   0.00000   0.00000   1.00000   0.00000  -0.07951
0.00000   0.00000   0.00000   0.00000   0.00000   1.00000  -0.39755


However, this is not of much help as we want to end up with integer coefficients. Therefore, we change the number display format to rational by using format rat. This makes the last column six rationals with the same denominator.

 -16/327
-13/327
-374/327
-16/327
-26/327
-130/327


After multiplying the last column by 327 and changing its sign, we get the coefficients for the balanced equation.

 16
13
374
16
26
130


Note that the free variable is the coefficient of H2O, which is 327 as we are interested on the smallest integer that makes all other coefficients integers.

Wrapping up, the balanced equation has coefficients 16, 13, 374, 16, 26, 130, 327, in this order. Not something easy to find by trial and error.

# Lexicographic Permutations

This post describes how to generate the lexicographic permutations of a sequence. The lexicographic order is a generalization of the way the alphabetical order of words is based on the alphabetical order of their component letters. This generalization consists primarily in defining a total order over the sequences of elements of a finite totally ordered set. You may understand that this is a way to establish ordering between sequences based on how their elements compare.

# Algorithm description

There exist several ways to generate all permutations of a given sequence. The simple algorithm which I will discuss here is based on finding the next permutation in lexicographic ordering, if it exists, or reversing the last permutation to get back to the minimal permutation. This method goes back to Narayana Pandita in 14th century India.

The algorithm is quite straightforward and may be memorized:

Find the biggest i such that a[i] < a[i + 1];
Find the biggest j greater than i such that a[j] > a[i];
Swap a[i] and a[j];
Reverse the elements from a[i + 1] to the last element.


If the first step fails (because such index does not exist) the current permutation is the last one.

Given any permutation of a list, this generates the next one. It should be noted, however, that when given the greatest lexicographic permutation, this algorithm returns this same permutation, so it should be checked to ensure that if the permutation at hand is the last one, we reverse the sequence to get back to the first permutation.

This algorithm is simple to implement correctly, computationally efficient, and it only generates each distinct permutation once, which is convenient when there are many repeated elements.

# Python implementation

Below is an in-place Python 3 implementation of the described algorithm. If you do not want mutability, wrap the algorithm calls with a function that copies the list. However, take into consideration that for very large lists this may use a significant amount of memory if you are interested in several permutations.

def swap(elements, i, j):
elements[i], elements[j] = elements[j], elements[i]

def reverse(elements, i, j):
for offset in range((j - i + 1) // 2):
swap(elements, i + offset, j - offset)

def next_permutation(elements):
last_index = len(elements) - 1
if last_index < 1:
return

i = last_index - 1
while i >= 0 and not elements[i] < elements[i + 1]:
i -= 1

# If there is no greater permutation, return to the first one.
if i < 0:
reverse(elements, 0, last_index)
else:
j = last_index
while j > i + 1 and not elements[j] > elements[i]:
j -= 1

swap(elements, i, j)
reverse(elements, i + 1, last_index)


Documentation was omitted for the sake of brevity. One could also consider this an example of code as documentation.

## Algorithm output

Generating eight permutations of the string ‘abc’ in ascending lexicographic order produces the following sequence:

abc, acb, bac, bca, cab, cba, abc, acb, ...


Notice that after the sixth permutation we get back to the first one, as there are only six distinct permutations of the string ‘abc’.

## Algorithm complexity

This algorithm uses constant additional space (as it is in-place) and has linear time complexity in the worst case but is amortized to constant time.

# Using the C++ standard library

If you can use the C++ STL, you have access to std::next_permutation.

## Example

std::string text = "abbc";
do {
cout << text << '\n';
} while (std::next_permutation(text.begin(), text.end()));

abbc
abcb
acbb
babc
bacb
bbac
bbca
bcab
bcba
cabb
cbab
cbba


As per the documentation, the return value of the function is true if the new permutation is lexicographically greater than the old, or false if the last permutation was reached and the range was reset to the first permutation.

# Concealed Squares

This is my solution to Project Euler problem number 206. Although a brute force solution can be easily derived, this approach makes use of mathematical analysis to find the answer in less time. The problem statement follows.

Find the unique positive integer whose square has the form 1_2_3_4_5_6_7_8_9_0, where each “_” is a single digit.

It can be shown that the answer lies in the range R.

R = [floor(sqrt(1020304...00)), ceiling(sqrt(192939..90))] = [1010101010, 1389026624]


R has 378,925,614 integers, which is a lot to test quickly. Fortunately, one can reduce it by noticing that the square of the integer is divisible by 10, which implies that the integer too is divisible by 10. This leaves us with the possible solution set P.

P = {1010101010, 1010101020, ..., 1389026620}


The largest possible number is 19293949596979899990, which requires 65 bits to be represented as an unsigned integer.

However, because we know that the answer is a multiple of 10, we also know that the square of the answer is a multiple of 100. And, therefore, that it ends with zeros. This allows for another optimization: divide all numbers in P by 10 and find the one whose square is of the form 1_2_3_4_5_6_7_8_9. These numbers fit into 64-bit words, so arithmetic with them will not be specially expensive.

P has 37,892,561 elements, but we do not need to check all of them. If x2 ends in 9, x ends in 3 or 7, which reduces by 80% the number of values we have to test.

Using Clang with the O3 flag, my C++ implementation of the algorithm finishes after 120 ms, which is a good enough solution for this problem.

# Solving Tic-Tac-Toe Using Racket

I decided to implement a perfect AI (using Minimax) for Tic-tac-toe. I had not decided though whether I’d use JavaScript or Racket, two languages I’ve been using quite a bit recently and about which I am quite interested. As the post title gives off, I’ve decided to write a functional programming solution using Racket. It is open-source and available on GitHub. See the repository. There is also a fairly decent GUI-based game that allows you to play against the AI. When writing this I decided to use vectors instead of lists, expecting it would give me better performance. However, it is quite slow. Even though the current implementation is good enough for real-time gameplay, it could be much more efficient for such a simple decision tree as the one used in Tic-tac-toe. Maybe using lists would have been a better choice as some parts of the application now have to convert vectors to lists. Additionally, slicing and merging vectors seem to generate a lot of work for the garbage collector. Finally, as the biggest vector holds only nine elements, the constant-time random access vectors provide is likely not making too much of a difference.

Being an automation enthusiast myself. This project also got its amount of continuous integration and delivery. Every push and pull request is built and tested by Travis CI. Here you can find the build history of the project. Travis also prepares and deploys a Linux-only standalone distribution to GitHub releases on tags. See the releases page. In the future I may make Windows and Mac OS X standalone distributions available. You should be able to play the standalone game on Linux even without having Racket installed.

There is also alpha-beta pruning left to be implemented, which should also significantly improve performance. Even though this is still a work in progress, the game works and the AI is indeed perfect. It also runs in any operating system as long as it has a Racket 6 distribution installed.