Рекурсія Демістифікована

Для того, щоб зрозуміти рекурсію, спочатку потрібно зрозуміти рекурсію.

Божевільний, чи не так?

Ну, я сподіваюся, що до кінця цієї статті ви почуватиметесь набагато впевненіше в тому, що таке рекурсія, і головним чином, як ми можемо придумати рекурсивне рішення проблеми.

Що таке рекурсія?

Як ви пояснюєте рекурсію у 4-річного віку? Це досить відоме запитання для інтерв’ю, і в Інтернеті є безліч відповідей. Ми не будемо відповідати на це запитання, оскільки воно надто поширене.

Якщо ти такий розумний, як я ??, ти пояснив би рекурсію комусь на рік молодший за тебе. Нехай вони пояснять рекурсію комусь на рік молодший за них. Продовжуйте, поки у вас не буде 5-річного віку, що пояснює рекурсію до 4-річного. Готово. [Джерело: reddit].

З точки зору програмування, рекурсія є

Функція, що викликає себе.

Вищевказана функція не робить корисної роботи як такої, але демонструє рекурсію. Рекурсивне відношення вище було б

T(N) = T(N - 1) + O(1)

Це просто означає, що виконання для дзвінка не random_function(n)може тривати до завершення виклику random_function(n-1)тощо.

По суті, ми затримуємо виконання поточного стану функції до завершення чергового виклику тієї ж функції та повернення її результату.

Компілятор продовжує зберігати стан виклику функції зараз, а потім переходить до наступного виклику функції тощо. Отже, компілятор зберігає стани функцій у стек і використовує їх для обчислень та зворотного відстеження.

По суті, якщо проблему можна розбити на подібні підзадачі, які можна розв’язати індивідуально, і рішення яких можна об’єднати разом, щоб отримати загальне рішення, тоді ми говоримо, що може існувати рекурсивне рішення проблеми.

Замість того, щоб чіплятися за це, здавалося б, старе визначення рекурсії, ми розглянемо цілу купу застосувань рекурсії. Тоді, сподіваємось, все буде зрозуміло.

Факториал числа

Давайте подивимося, як ми можемо з’ясувати факторіал числа. До цього давайте подивимося, що представляє факторіал числа і як воно обчислюється.

factorial(N) = 1 * 2 * 3 * .... * N - 1 * N

Простіше кажучи, факторіал числа - це просто добуток доданків від 1 до числа N, помноженого один на одного.

Ми можемо просто мати forцикл від 1 до N і помножити всі доданки ітеративно, і ми отримаємо факторіал заданого числа.

Але, якщо придивитися уважніше, існує факторична рекурсивна структура, що відноситься до факторіалу числа.

factorial(N) = N * factorial(N - 1)

Це все одно, що розвантажити обчислення до іншого виклику функції, що працює на меншій версії вихідної проблеми. Давайте подивимося, як буде розвиватися це відношення, щоб перевірити, чи відповідає розв’язання тут рішенню, яке надає forцикл.

Отже, з двох малюнків вище видно, що рекурсивна функція, яку ми визначили раніше,

factorial(N) = N * factorial(N - 1)

це справді правильно. Погляньте на фрагмент коду Python, який використовується для пошуку факторіалу функції, рекурсивно.

Цей приклад був досить простим. Розглянемо трохи більший, але стандартний приклад, щоб продемонструвати концепцію рекурсії.

Послідовність Фібоначчі

Ви, мабуть, уже знайомі з відомою послідовністю Фібоначчі. Для тих з вас, хто раніше не чув про цю послідовність або не бачив приклад, давайте подивимось.

1 1 2 3 5 8 13 ..... 

Давайте розглянемо формулу для обчислення n ^ -го числа Фібоначчі.

F(n) = F(n - 1) + F(n - 2)where F(1) = F(2) = 1

Очевидно, що це визначення послідовності Фібоначчі має рекурсивний характер, оскільки n ^ -те число Фібоначчі залежить від попередніх двох чисел Фібоначчі. Це означає розподіл проблеми на менші підзадачі, а отже, і рекурсію. Погляньте на код цього:

Кожна рекурсивна проблема повинна мати дві необхідні речі:

  1. Відношення рецидивів, що визначає стан проблеми та спосіб, як основну проблему можна розбити на менші підзадачі. Сюди також входить базовий випадок зупинки рекурсії.
  2. Дерево рекурсії, яке демонструє перші кілька, якщо не всі виклики функції, що розглядається. Погляньте на дерево рекурсії для рекурсивного відношення послідовностей Фібоначчі.

Дерево рекурсії показує нам, що результати, отримані в результаті обробки двох піддерев кореня N, можуть бути використані для обчислення результату для дерева, корінного в N. Аналогічно для інших вузлів.

Листя цього дерева рекурсії будуть fibonacci(1)або fibonacci(2)обоє з них представляють базові випадки для цієї рекурсії.

Тепер, коли ми дуже просто розуміємо рекурсію, що таке відношення рекурсивності та дерево рекурсії, давайте перейдемо до чогось більш цікавого.

Приклади!

Я дуже впевнений у вирішенні безлічі прикладів для будь-якої заданої теми в програмуванні, щоб стати майстром цієї теми. Два розглянуті нами приклади (факторіал числа та послідовність Фібоначчі) мали чітко визначені рекурентні співвідношення. Давайте розглянемо кілька прикладів, коли відношення рецидивів може бути не таким очевидним.

Висота дерева

Для спрощення ситуацій у цьому прикладі ми розглянемо лише бінарне дерево. Отже, двійкове дерево - це деревоподібна структура даних, в якій кожен вузол має щонайбільше двох дітей. Один вузол дерева позначається як корінь дерева, наприклад:

Давайте визначимо, що ми маємо на увазі під висотою двійкового дерева.

Висота дерева буде довжиною найдовшого шляху від кореня до листя на дереві.

So, for the example diagram displayed above, considering that the node labelled as A as the root of the tree, the longest root to leaf path is A → C → E → G → I . Essentially, the height of this tree is 5 if we count the number of nodes and 4 if we just count the number of edges on the longest path.

Now, forget about the entire tree and just focus on the portions highlighted in the diagram below.

The above figure shows us that we can represent a tree in the form of its subtrees. Essentially, the structure to the left of node A and the structure to the right of A is also a binary tree in itself, just smaller and with different root nodes. But, they are binary trees nonetheless.

What information can we get from these two subtrees that would help us find the height of the main tree rooted at A ?

If we knew the height of the left subtree, say h1, and the height of the right subtree, say h2, then we can simply say that the maximum of the two + 1 for the node A would give us the height of our tree. Isn’t that right?

Formalizing this recursive relation,

height(root) = max(height(root.left), height(root.right)) + 1

So, that’s the recursive definition of the height of a binary tree. The focus is on binary here, because we used just two children of the node root represented by root.left and root.right. But, it is easy to extend this recursive relation to an n-ary tree. Let’s take a look at this in code.

The problem here was greatly simplified because we let recursion do all the heavy lifting for us. We simply used optimalanswers for our subproblems to find a solution to our original problem.

Давайте розглянемо ще один приклад, який можна розв’язати за подібними рядками.

Кількість вузлів у дереві

Тут ми знову розглянемо бінарне дерево для простоти, але алгоритм та підхід можуть бути поширені на будь-яке дерево по суті.

Сама проблема сама собою пояснюється. Враховуючи корінь двійкового дерева, нам потрібно визначити загальну кількість вузлів у дереві. Це питання та підхід, який ми тут придумаємо, дуже схожі на попереднє. Нам просто потрібно внести незначні зміни, і ми отримаємо кількість вузлів у двійковому дереві.

Погляньте на схему нижче.

Діаграма говорить про все. Ми вже знаємо, що дерево можна розбити на менші піддерева. Тут ми знову можемо запитати себе,

Яку інформацію ми можемо отримати з цих двох піддерев, що допомогло б нам знайти кількість вузлів у дереві, коріння якого лежить у A?

Well, if we knew the number of nodes in the left subtree and the number of nodes in the right subtree, we can simply add them up and add one for the root node and that would give us the total number of nodes.

Formalizing this we get,

number_of_nodes(root) = number_of_nodes(root.left) + number_of_nodes(right) + 1

If you look at this recursion and the previous one, you will find that they are extremely similar. The only thing that is varying is what we do with the information we obtained from our subproblems and how we combined them to get some answer.

Now that we have seen a couple of easy examples with a binary tree, let’s move onto something less trivial.

Merge Sort

Given an array of numbers like

4 2 8 9 1 5 2

we need to come up with a sorting technique that sorts them either in ascending or descending order. There are a lot of famous sorting techniques out there for this like Quick Sort, Heap Sort, Radix Sort and so on. But we are specifically going to look at a technique called the Merge Sort.

It’s possible that a lot of you are familiar with the Divide and Conquer paradigm, and this might feel redundant. But bear with me and read on!

The idea here is to break it down into subproblems.

That’s what the article is about right ? ?

What if we had two sorted halves of the original array. Can we use them somehow to sort the entire array?

That’s the main idea here. The task of sorting an array can be broken down into two smaller subtasks:

  • sorting two different halves of the array
  • then using those sorted halves to obtain the original sorted array

Now, the beauty about recursion is, you don’t need to worry about how we will get two sorted halves and what logic will go into that. Since this is recursion, the same method call to merge_sort would sort the two halves for us. All we need to do is focus on what we need to do once we have the sorted haves with us.

Let’s go through the code:

At this point, we trusted and relied on our good friend recursion and assumed that left_sorted_half and right_sorted_half would in fact contain the two sorted halves of the original array.

So, what next?

The question is how to combine them somehow to give the entire array.

The problem now simply boils down to merging two sorted arrays into one. This is a pretty standard problem and can be solved by what is known as the “two finger approach”.

Take a look at the pseudo code for better understanding.

let L and R be our two sorted halves. let ans be the combined, sorted array
l = 0 // The pointer for the left halfr = 0 // The pointer for the right halfa = 0 // The pointer for the array ans
while l < L.length and r < R.length { if L[l] < R[r] { ans[a] = L[l] l++ } else { ans[a] = R[r] r++ }}
copy remaining array portion of L or R, whichever was longer, into ans.

Here we have two pointers (fingers), and we position them at the start of the individual halves. We check which one is smaller (that is, which value pointed at by the finger is smaller), and we add that value to our sorted combined array. We then advance the respective pointer (finger) forward. In the end we copy the remaining portion of the longer array and add it to the back of the ans array.

So, the combined code for merge-sort is as follows:

We will do one final question using recursion and trust me, it’s a tough one and a pretty confusing one. But before moving onto that, I will iterate the steps I follow whenever I have to think of a recursive solution to a problem.

Steps to come up with a Recursive Solution

  1. Try and break down the problem into subproblems.

2. Once you have the subproblems figured out, think about what information from the call to the subproblems can you use to solve the task at hand. For example, the factorial of N — 1 to find the factorial of N , height of the left and right subtrees to find the height of the main tree, and so on.

3. Keep calm and trust recursion! Assume that your recursive calls to the subproblems will return the information you need in the most optimal fashion.

4. The final step in this process is actually using information we just got from the subproblems to find the solution to the main problem. Once you have that, you’re ready to code up your recursive solution.

Now that we have all the steps lined up, let’s move on to our final problem in this article. It’s called Sum of Distances in a Tree.

Sum of Distances in a Tree

Let’s look at what the question is asking us to do here. Consider the following tree.

In the example above, the sum of paths for the node A (the number of nodes on each path from A to every other vertex in the tree) is 9. The individual paths are mentioned in the diagram itself with their respective lengths.

Similarly, consider the sum of distances for the node C.

C --> A --> B (Length 2)C --> A (Length 1)C --> D (Length 1)C --> E (Length 1)C --> D --> F (Length 2)Sum of distances (C) = 2 + 1 + 1 + 1 + 2 = 7

This is known as the sum of distances as defined for just a single node A or C. We need to calculate these distances for each of the nodes in the tree.

Before actually solving this generic problem, let us consider a simplified version of the same problem. It says that we just need to calculate the sum of distances for a given node, but we will only consider the tree rooted at the given node for calculations.

So, for the node C, this simplified version of the problem would ask us to calculate:

C --> D (Length 1)C --> E (Length 1)C --> D --> F (Length 2)Simplified Sum of Distances (C) = 1 + 1 + 2 = 4

This is a much simpler problem to tackle recursively and would prove to be useful in solving the original problem.

Consider the following simple tree.

The nodes B and C are the children of the root (that is, A).

We are trying to see what information can we use from subproblems (the children) to compute the answer for the root A .

Note: here we simply want to calculate the sum of paths for a given node X to all its successors in its own subtree (the tree rooted at the node X).

There are no downwards going paths from the node B, and so the sum of paths is 0 for the node B in this tree. Let’s look at the node C . So this node has 3 different successors in F, D and E . The sum of distances are as follows:

C --> D (Path containing just 1 edge, hence sum of distances = 1)C --> D --> F (Path containing 2 edges, hence sum of distances = 2)C --> E (Path containing just 1 edge, hence sum of distances = 1)

The sum of all the paths from the node C to all of its decedents is 4, and number of such paths going down is 3.

Note the difference here. The sum_of_distances here counts the number of edges in each path — with each edge repeating multiple times, probably because of their occurrence on different paths — unlike number_of_paths , which counts, well, the number of paths ?.

If you look closely, you will realize that the number of paths going down is always going to be the number of nodes in the tree we are considering (except the root). So, for the tree rooted at C, we have 3 paths, one for the node D, one for E, and one for F. This means that the number of paths from a given node to the successor nodes is simply the total number of descendent nodes since this is a tree. So, no cycles or multiple edges.

Now, consider the node A. Let us look at all the new paths that are being introduced because of this node A. Forget the node B for now and just focus on the child node C corresponding to A. The new sets of paths that we have are:

A --> C (Path containing just 1 edge, hence sum of distances = 1)A --> (C --> D) (Path containing 2 edges, hence sum of distances = 2)A --> (C --> E) (Path containing 2 edges, hence sum of distances = 2)A --> (C --> D --> F) (Path containing 3 edges, hence sum of distances = 3)

Except for the first path A → C, all the others are the same as the ones for the node C, except that we have simply changed all of them and incorporated one extra node A.

If you look at the diagram above you will see a tuple of values next to each of the nodes A, B, and C.

(X, Y) where X is the number of paths originating at that node and going down to the decedents. Y is the sum of distances for the tree rooted at the given node. 

Since the node B doesn’t have any further children, the only path it is contributing to is the path A -->; B to A's tuple of (5, 9) above. So let’s talk about C.

C had three paths going to its successors. Those three paths (extended by one more node for A) also become three paths from A to its successors, among others.

N-Paths[A] = (N-Paths[C] + 1) + (N-Paths[B] + 1)

That is the exact relation we are looking for as far as the number of paths (= number of successor nodes in the tree) are concerned. The 1 is because of the new path from the root to it’s child, that is A -->; C in our case.

N-Paths[A] = 3 + 1 + 0 + 1 = 5

As far as the sum of distances is concerned, take a look at the diagram and the equations we just wrote. The following formula becomes very clear:

Sum-Dist[A] = (N-Paths[C] + 1 + Sum-Dist[C]) + (N-Paths[B] + 1 + Sum-Dist[B])
Sum-Dist[A] = (3 + 1 + 4 + 0 + 1 + 0) = 9

The main thing here is N-Paths[C] + Sum-Dist[C] . We sum these up because all of the paths from C to its descendants ultimately become the paths from A to its descendants — except that they originate at A and go through C, and so each of the path lengths are increased by 1. There are N-Paths[C] paths in all originating from C and their total length is given by Sum-Dist[C] .

Hence the tuple corresponding to A = (5, 9). The Python code for the algorithm we discussed above is as follows:

The Curious Case of the Visited Dictionary :/

If you look at the code above closely, you’ll see this:

# Prevents the recursion from going into a cycle. self.visited[vertex] = 1

The comment says that this visited dictionary is for preventing the recursion from entering a cycle.

If you’ve paid attention til now, you know that we are dealing with a tree here.

The definition of a tree data structure doesn’t allow cycles to exist. If a cycle exists in the structure, then it is no longer a tree, it becomes a graph. In a tree, there is exactly one path between any two pair of vertices. A cycle would mean there is more than one path between a pair of vertices. Look at the figures below.

The structure on the left is a tree. It has no cycles in it. There is a unique path between any two vertices.

The structure on the right is a graph, there exists a cycle in the graph and hence there are multiple paths between any pair of vertices. For this graph, it so happens that any pair of vertices have more than one path. This is not necessary for every graph.

Almost always, we are given the root node of the tree. We can use the root node to traverse the entire tree without having to worry about any cycles as such.

However, if you’ve read the problem statement clearly, it does not state anything about root of the tree.

That means that there is no designated root for the tree given in the question. This could mean that a given tree can be visualized and processed in so many different ways depending upon what we consider as the root. Have a look at multiple structures for the same tree but with different root nodes.

So many different interpretations and parent child relationships are possible for a given unrooted tree.

So, we start with the node 0 and do a DFS traversal of the given structure. In the process we fix the parent child relationships. Given the edges in the problem, we construct an undirected graph-like structure which we convert to the tree structure. Taking a look at the code should clear up some of your doubts:

Every node would have one parent. The root won’t have any parent, and the way this logic is, the node 0 would become the root of our tree. Note that we are not doing this process separately and then calculating the sum of distances downwards. Given a tree, we were trying to find, for every node, the simplified sum of distances for the tree rooted at that node.

So, the conversion from the graph to the tree happens in one single iteration along with finding out the sum of distances downwards for each and every node.

I posted the code again so that the visited dictionary makes much more sense now. So, one single recursion doing all that for us. Nice!

Bringing it all together

Now that we have our tree structure defined, and also the values of sum of distances going downward defined for us, we can use all of this information to solve the original problem of Sum of Distances in a Tree.

How do we do that? It’s best to explain this algorithm with the help of an example. So we will consider the tree below and we will dry run the algorithm for a single node. Let’s have a look at the tree we will be considering.

The node for which we want to find the sum of distances is 4. Now, if you remember the simpler problem we were trying to solve earlier, you know that we already have two values associated with each of the nodes:

  1. distances_down Which is the sum of distances for this node while only considering the tree beneath.
  2. number_of_paths_down which is the number of paths / nodes in the tree rooted at the node under consideration.

Let’s look at the annotated version of the above tree. The tree is annotated with tuples (distances_down, number_of_paths_down) .

Let’s call the value we want to compute for each node as sod which means sum of distances, which is what the question originally asks us to compute.

Let us assume that we have already computed the answer for the parent node of 4 in the diagram above. So, we now have the following information for the node labelled 2 (the parent node) available:

(sod, distances_down, number_of_paths_down) = (13, 4, 3)

Let’s rotate the given tree and visualize it in a way where 2 is the root of the tree essentially.

Now, we want to remove the contribution of the tree rooted at 4 from sod(2). Let us consider all of the paths from the parent node 2 to all other nodes except the ones in the tree rooted at 4 .

2 --> 5 (1 edge)2 --> 1 (1 edge)2 --> 1 -->7 (2 edges)2 --> 1 --> 7 --> 9 (3 edges)2 --> 1 --> 7 --> 10 (3 edges)
Number of nodes considered = 6Sum of paths remaining i.e. sod(2) rem = 1 + 1 + 2 + 3 + 3 = 10

Let’s see how we can use the values we already have calculated to get these updated values.

* N = 8 (Total number of nodes in the tree. This will remain the same for every node. )* sod(2) = 13
* distances_down[4] = 1* number_of_paths_down[4] = 1
* (distances_down[4] does not include the node 4 itself)N - 1 - distances_down[4] = 8 - 1 - 1 = 6
* sod(2) - 1 - distances_down[4] - number_of_paths_down[4] = 13 - 1 - 1 - 1 = 10

If you remember this from the function we defined earlier, you will notice that the contribution of a child node to the two values distances_down and number_of_paths_down is n_paths + 1 and n_paths + s_paths + 1 respectively. Naturally, that is what we subtract to obtain the remaining tree.

sod(4) represents the sum of edges on all the paths originating at the node 4 in the tree above. Let’s see how we can find this out using the information we have calculated till now.

distances_down[4] represents the answer for the tree rooted at the node 4 but it only considers paths going to its successors, that is all the nodes in the tree rooted at 4. For our example, the successor of 4 is the node 6. So, that will directly add to the final answer. Let’s call this value own_answer . Now, let’s account for all the other paths.

4 --> 2 (1 edge)4 --> 2 --> 5 (1 + 1 edge)4 --> 2 --> 1 (1 + 1 edge)4 --> 2 --> 1 -->7 (1 + 2 edges)4 --> 2 --> 1 --> 7 --> 9 (1 + 3 edges)4 --> 2 --> 1 --> 7 --> 10 (1 + 3 edges)own_answer = 1
sod(4) = 1 + 1 + 2 + 2 + 3 + 4 + 4 = 17
sod(4) = own_answer + (N - 1 - distances_down[4]) + (sod(2) - 1 - distances_down[4] - number_of_paths_down[4]) = 1 + 6 + 10 = 17

Before you go bonkers and start doing this, let’s look at the code and bring together all of the things we discussed in the example above.

The recursive relation for this portion is as follows:

Did I just see “MEMOIZATION” in the code?

Yes, indeed you did!

Consider the following example tree:

The question asks us to find the sum of distances for all the nodes in the given tree. So, we would do something like this:

for i in range(N): ans.append(find_distances(N))

But, if you look at the tree above, the recursive call for the node 5 would end up calculating the answers for all the nodes in the tree. So, we don’t need to recalculate the answers for the other nodes again and again.

Hence, we end up storing the already calculated values in a dictionary and use that in further calculations.

Essentially, the recursion is based on the parent of a node, and multiple nodes can have the same parent. So, the answer for the parent should only be calculated once and then be used again and again.

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