Recursión desmitificada

Para comprender la recursividad, primero debe comprender la recursividad.

Loco, ¿no?

Bueno, espero que al final de este artículo se sienta mucho más seguro sobre qué es la recursividad y, principalmente, cómo podemos encontrar una solución recursiva a un problema.

¿Qué es la recursividad?

¿Cómo se explica la recursividad a un niño de 4 años? Esta es una pregunta de entrevista bastante famosa y hay muchas respuestas disponibles en la web. No responderemos a esta pregunta porque es demasiado común.

Si eres tan inteligente como yo, le explicarías la recursividad a alguien un año menor que tú. Pídales que le expliquen la recursividad a alguien un año menor que ellos. Continúe hasta que tenga un niño de 5 años que le explique la recursividad a un niño de 4 años. Hecho. [Fuente: reddit].

En términos de programación, la recursividad es

Una función que se llama a sí misma.

La función anterior no hace un trabajo útil como tal, pero demuestra la recursividad. La relación recursiva anterior sería

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

Esto simplemente significa que la ejecución de la llamada a random_function(n)no puede continuar hasta que se complete la llamada a random_function(n-1)y así sucesivamente.

Esencialmente, retrasamos la ejecución del estado actual de la función hasta que se complete otra llamada a la misma función y devuelva su resultado.

El compilador sigue guardando el estado de la llamada de función ahora y luego pasa a la siguiente llamada de función y así sucesivamente. Por lo tanto, el compilador guarda los estados de las funciones en una pila y los usa para cálculos y retrocesos.

Esencialmente, si un problema se puede dividir en subproblemas similares que se pueden resolver individualmente y cuyas soluciones se pueden combinar para obtener la solución general, entonces decimos que podría existir una solución recursiva al problema.

En lugar de aferrarnos a esta definición aparentemente antigua de recursividad, veremos un montón de aplicaciones de recursividad. Entonces, con suerte, las cosas se aclararán.

Factorial de un número

Veamos cómo podemos encontrar el factorial de un número. Antes de eso, veamos qué representa el factorial de un número y cómo se calcula.

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

En pocas palabras, el factorial de un número es solo el producto de términos de 1 al número N multiplicado por otro.

Simplemente podemos tener un forbucle de 1 a N y multiplicar todos los términos iterativamente y tendremos el factorial del número dado.

Pero, si miras de cerca, existe una estructura recursiva inherente al factorial de un número.

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

Es como descargar el cálculo a otra llamada de función que opera en una versión más pequeña del problema original. Veamos cómo se desarrollaría esta relación para verificar si la solución aquí coincide con la proporcionada por el forciclo.

Entonces, está claro a partir de las dos figuras anteriores que la función recursiva que definimos anteriormente,

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

es de hecho correcto. Eche un vistazo al fragmento de código de Python utilizado para encontrar el factorial de una función, de forma recursiva.

Este ejemplo fue bastante simple. Consideremos un ejemplo un poco más grande pero estándar para demostrar el concepto de recursividad.

Secuencia Fibonacci

Ya debe estar familiarizado con la famosa secuencia de fibonacci. Para aquellos de ustedes que no han oído hablar de esta secuencia o no han visto un ejemplo antes, echemos un vistazo.

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

Veamos la fórmula para calcular el n ° número de Fibonacci.

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

Claramente, esta definición de la secuencia de fibonacci es de naturaleza recursiva, ya que el n º número de fibonacci depende de los dos números de fibonacci anteriores. Esto significa dividir el problema en subproblemas más pequeños y, por lo tanto, recursividad. Eche un vistazo al código para esto:

Todo problema recursivo debe tener dos cosas necesarias:

  1. La relación de recurrencia que define los estados del problema y cómo el problema principal se puede descomponer en subproblemas más pequeños. Esto también incluye el caso base para detener la recursividad.
  2. Un árbol de recursividad que muestra las primeras, si no todas, las llamadas a la función en consideración. Eche un vistazo al árbol de recursividad para la relación recursiva de las secuencias de fibonacci.

El árbol de recursividad nos muestra que los resultados obtenidos al procesar los dos subárboles de la raíz N se pueden utilizar para calcular el resultado del árbol con raíz en N. De manera similar para otros nodos.

Las hojas de este árbol de recursividad serían fibonacci(1)o fibonacci(2)ambas representan los casos base para esta recursividad.

Ahora que tenemos una comprensión muy básica de la recursividad, qué es una relación de recurrencia y el árbol de recursividad, pasemos a algo más interesante.

¡Ejemplos!

Creo firmemente en resolver innumerables ejemplos para cualquier tema dado en programación para convertirme en un maestro de ese tema. Los dos ejemplos que consideramos (factorial de un número y la secuencia de Fibonacci) tenían relaciones de recurrencia bien definidas. Veamos algunos ejemplos en los que la relación de recurrencia podría no ser tan obvia.

Altura de un árbol

Para simplificar las cosas en este ejemplo, solo consideraremos un árbol binario. Entonces, un árbol binario es una estructura de datos de árbol en la que cada nodo tiene como máximo dos hijos. Un nodo del árbol se designa como la raíz del árbol, por ejemplo:

Definamos lo que queremos decir con la altura del árbol binario.

La altura del árbol sería la longitud del camino más largo de la raíz a la hoja en el árbol.

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.

Veamos otro ejemplo que se puede resolver en líneas similares.

Número de nodos en un árbol

Aquí nuevamente, consideraremos un árbol binario por simplicidad, pero el algoritmo y el enfoque se pueden extender a cualquier tipo de árbol esencialmente.

El problema en sí mismo se explica por sí mismo. Dada la raíz de un árbol binario, necesitamos determinar el número total de nodos en el árbol. Esta pregunta y el enfoque que propondremos aquí son muy similares a la anterior. Solo tenemos que hacer cambios minúsculos y tendremos la cantidad de nodos en el árbol binario.

Eche un vistazo al siguiente diagrama.

El diagrama lo dice todo. Ya sabemos que un árbol se puede dividir en subárboles más pequeños. Aquí nuevamente, podemos preguntarnos,

¿Qué información podemos obtener de estos dos subárboles que nos ayudaría a encontrar el número de nodos en el árbol enraizados en 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.

If you’ve managed to read the article this far (not necessarily in one stretch ?), you’re awesome ?.

If you found this article helpful, share as much as possible and spread the ?. Cheers!