Skip to main content

Heap's Algorithm implementation with python

Heap's Algorithm is used to generate all the permutations of a given object. It was first proposed by B.R. Heap in the year 1963.

 In this post, we will write a program using Heap's Algorithm which will generate permutations for a given object. Remember that the object should be iterable. Otherwise, you will get an error.

If you don't know what permutations is all about then you can visit some of the following websites:

 Permutations - Wikipedia
Permutation - Wolfram math
Permutations and combinations - Math is fun  and
Permutations Khan Academy

 Here goes the program:
This function will take an iterable object as input and will generate all the permutations possible with the items in the object.
For example, if the input is:heap(["A", "B", "C"]) then the output will be a generator object. You can use a for loop to iterate over this object to see the output.
But remember that if you have any duplicates in the input(Ex: heap(["A", "B", "B"])) then you will also get duplicate permutations. (Ex output: [["A", "B", "B"], ["B", "A", "B"], ["B",  "A", "B"], ["A", "B", "B"], ["B", "B", "A"], ["B", "B", "A"]])

 You can eliminate duplicates by using python sets.

 One of the best ways to understand this algorithm is by looking at the pseudo-code on Wikipedia Heaps Algorithm page.

 We have used the nonrecursive method.

 If you are not sure of what yield does, then you can have a look at this answer on Stack overflow by e-satis for the question What does the "yield" keyword do

 If you use the above code and if you give a string as an input still you will get a list as output. To overcome this problem I have written a program, a slightly modified version of the above program so that we will get the output as a string.
The program is as follows:   
The above program will take a string as input and will give a generator object as output, which on further iteration will give you a list of strings.

 Line 14 is just swapping of the characters a[0] , a[i] in the string.

 Line 16 is again swapping of the characters a[i], a[c[i]] in the string. 

As we are doing string concatenation with more than two strings, the second program takes more time for execution than the first one.

If you have any doubt or didn't understand anything, then you can comment in the comment box below and I will be glad to help you.

 If you have any suggestions or if you have found any typo then please comment in the comment box. I will be very happy to see your comment.

 You can also contact me.

 Thank you. Have a nice day. 😃
Labels:

Popular posts from this blog

Project Euler Problem 67 Solution with Python

Maximum path sum II By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23. 3 7 4 2 4 6 8 5 9 3 That is, 3 + 7 + 4 + 9 = 23. Find the maximum total from top to bottom in triangle.txt (right click and 'Save Link/Target As...'), a 15K text file containing a triangle with one-hundred rows.
Read more

Problem 60 Project Euler Solution with python

Prime pair sets The primes 3, 7, 109, and 673, are quite remarkable. By taking any two primes and concatenating them in any order the result will always be prime. For example, taking 7 and 109, both 7109 and 1097 are prime. The sum of these four primes, 792, represents the lowest sum for a set of four primes with this property. Find the lowest sum for a set of five primes for which any two primes concatenate to produce another prime. This problem is j u st a brute force problem. If you have come here because you don't know the limit upto which you will h ave to gener ate the prime numbers t hen go ahe ad and t r y with 10,000 . When I first start ed solving the problem I chose 1 million(beca use most of the problem s on project E uler have this limit ), but it took very long for the computer to fin d the solution. After searching on the internet then I found many people choosing 10, 000 so I have changed my in put f rom 1 million to 10000 and the output was f ast. He...
Read more

Project Euler Problem 66 Solution with python

Diophantine equation ¶ Consider quadratic Diophantine equations of the form: $$ x^{2} – Dy^{2} = 1 $$ For example, when $D = 13$, the minimal solution in $ x $ is $ 649^{2} – 13 \times 180^{2} = 1 $ It can be assumed that there are no solutions in positive integers when $ D $ is square. By finding minimal solutions in $ x $ for $ D = {2, 3, 5, 6, 7} $, we obtain the following: $$ 3^{2} – 2×2^{2} = 1 $$ $$ 2^{2} – 3×1^{2} = 1 $$ $$ 9^{2} – 5×4^{2} = 1 $$ $$ 5^{2} – 6×2^{2} = 1 $$ $$ 8^{2} – 7×3^{2} = 1 $$ Hence, by considering minimal solutions in $ x $ for $ D ≤ 7 $, the largest $ x $ is obtained when $ D = 5 $. Find the value of $ D ≤ 1000 $ in minimal solutions of $ x $ for which the largest value of $ x $ is obtained.
Read more