![]() They help you count things: for example, different combinations of numbers in a list or different permutations of a string. ![]() Putting these two pieces together, we end up with combinatoric iterators. A related concept is an iterator, which is an object that returns the next element of an iterable. A string itself is an iterable since you can loop through all the characters in it. These iterables are commonly filled with integer values, floats, or strings. Some common iterables you may be familiar with are lists, tuples, sets, and arrays, which you can iterate through using a for loop. In Python, iterable objects, more commonly referred to as iterables, are groups of data. ![]() For this to work, we need a collection of objects to act on – something to iterate through. It can help us count the number of permutations of something (how many possible arrangements of a deck of cards) or the number of combinations (how many unique arrangements of different colored balls). In this article, I would like to provide a brief introduction to combinatoric iterators in Python.Ĭombinatorics in the mathematical sense is about counting things. This introduction shows you some of the most useful ones in Python. So, let's make a permutation function to do this.Combinatoric iterators are tools that provide building blocks to make code more efficient. So, recursion seems to be the most generic way to solve the problem. Thus, we are recurring to make permutations here. It can be easily noticed that for the number 1234, we are first making permutations of 234 first and for 234, permutations of 34 and so on. So, you have understood the logic of making the permutations. Similarly, after having the permutation of last three digits, we will replace the first digit and will again get all the permutations of the last three digits. Now, we made the permutation of these digits and got 13. For example, After making all the permutations of 34 (34 and 43) and getting the numbers 12, we replaced 2 with 3 (2 was the last fixed digit in the number). After having all the permutations of the smaller number, we are just replacing one of the digits of this new number with the last digit which was fixed and again making permutations of the newer number. You can see that we are breaking the problem into smaller problems and then making the permutations of these smaller ones. Similarly, we will keep all other digits at the first position and get the corresponding permutations. Now, we have all the numbers which can be made by keeping 1 at the first position. Thus the numbers obtained are:Īgain, keeping 4 fixed out of 2, 3 and 4. So, we will make the permutations of 2, 3 and 4 by keeping 2 fixed. And we have to make all the permutations of the digits 2, 3 and 4. Thus, we are left with the digits 2, 3 and 4. We will start by keeping 1 at the first position. So, let's use this logic to make the permutations of the digits 1, 2, 3 and 4. And of course, making permutations of only 3 digits is quite easy. So, now we have all our permutations which can be made by the digits 1, 2 and 3. Similarly, keeping 3 at the first position, the numbers are: So, let's keep 2 at the first position this time and make the permutations. Thus the numbers obtained by keeping 1 fixed are: 1, fixed, and will make the permutations of the other numbers. One way I am going to make the permutation is: I will start by keeping the first number, i.e. So before going into the coding part, let's first understand the logic of making the permutations in reality and then we will code that logic. Writing the code for a problem is not a big deal if you know how to solve the problem practically or understand the logic of solving the problem in reality. This is also a very common question of computer programming. This post is about printing all the permutations of an array with the use of recursion.
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