Recursion

Introduction

Want to get done the basics for some of this.

Why

Breaks down large chunks into smaller chunks.

When

You use it when problems contain smaller instances of the same problem. E.g. finding how many combinations there are of n items.

Aspects of Recursion

They all have a

• base case
• recursive case

Base Case

This is where the smallest instance of the case can be solved easily. E.g. For number of combinations when the array is empty.

Recursive Case

This is an instance of the problem that shrink the size of the input toward the base case.

Tail Call Recursion

This is a strange topic and the example stranger to. Often feel I do not speak the same language.

const factorial_non_tail = (n) => {
    if(n === 1) return 1
    return n * factorial_non_tail(n - 1);
}

const factorial_tail = (n, acc = 1) => {
    if(n === 1) return acc
    return factorial_tail(n - 1, n * acc);
}

In the first example above the last thing it does is n * factorial_non_tail(n - 1). Which means 5 x something but in the second example the last thing it does is a recursion call. Felt quite chuffed when I found out why - always why. Shown below

Tail recursion is important because it can be optimized by compilers or interpreters through a technique called tail call optimization (TCO) or tail call elimination. In essence, when a function is tail-recursive, the compiler can transform the recursive call into a simple jump instruction, effectively reusing the current stack frame instead of creating a new one for each recursive call.

Examples

Example Factorial

<syntaxhightlight lang="js"> const factorial = (n) => {

   if(n === 1) return 1
   return n * factorial(n-1)

} </syntaxhightlight>

Example Palindrome

const palindrome = (str) => {
    console.log("calling with",str)
    if(str.length === 0) return true
    if(str.length === 1) return true
    if(str[0] !== str[str.length-1]) return false
    
    return palindrome(str.substr(1, str.length-2))

}

Example Sum

This was a bit of enlightenment for improving speed. Did the sum as you would normally do and you can see on line 3 we slice which means we copy.

const _sum = (numbers, idx) =>  {
    if(numbers.length == 0) return 0
    return n + sum(numbers.slice(1))
}

A better approach is to create a sub function which takes an index and therefore no more slices

const sum = (numbers) => {
    return _sum(numbers, 0)
}

const _sum = (numbers, idx) =>  {
    if(idx == numbers.length) return 0
    return numbers[idx] + _sum(numbers, idx + 1)
}

Example String Reversal

This is where I struggle with recursive. Separating the a) reduce the argument and b) forming the result. It seems very odd to me. Maybe my dyslexic brain is fried

const string_rev = (str) => {
    if(str.length === 0 || str.length === 1) return str

    const result = string_rev(str.slice(1))
    return  result + str[0]
}

console.log("Reverse:", string_rev("FRED"))

So this is really

string_rev("RED") + "F"
 string_rev("ED") + "R"
  string_rev("D") + "E"
   string_rev("") + "D"

Example dec2bin

Dividing a number by 2 and concatenating the rem making a binary number

const dec2bin2  = (number) => {
    if(number === 0 ) return 0
    
    const quotent = Math.floor(number / 2)
    const rem = (number % 2).toString()

    const result = dec2bin(quotent)
    return result + rem
}

Example Binary Search

find 10 in an array of [-1,0,1,2,3,4,7,9,10,20]. Key thing is in order. Took a while to solve this as I need floor and forgot to put brackets around the calculation - argggghhhhhhhhhh.

const binSearch = (val, numbers, left, right) => {
    if(left > right) return -1

    console.log("calling with left=%d, right=%d", left,right)

    // Calculate the mid-point
    const mid_point = Math.floor((left + right) / 2)
    // const  mid_point = (left+right) /2

    console.log("mid_point now:%d", mid_point)

    // Check if in middle
    if(val === numbers[mid_point]) return mid_point

    // Test is val is less than mid_point and search lefside
    if(val < numbers[mid_point]) {
        return binSearch(val, numbers, left, mid_point-1)
    }

    return binSearch(val, numbers, mid_point + 1, right)
}
console.log("dec2bin", binSearch(10,[-1,0,1,2,3,4,7,9,10,20], 0, 9))

Example Merge Sort

Kinda insane. This splits and array and for each half split down to two characters, sorts, unwinds so 4 characters, sorts until done

const merge = (array, start, mid, end, side) => {
  // console.log("Merging start=%d, mid=%d, end=%d", start, mid, end);
  console.log(">>--------------------------------")
  console.log(array.slice(start, end + 1));
  let i = start;
  let j = mid + 1;
  let k = 0;

  let temp = Array(end - start);
  while (i <= mid && j <= end) {
    if (array[i] <= array[j]) {
      temp[k++] = array[i++];
    } else {
      temp[k++] = array[j++];
    }
  }

  // Add the rest of the values from left sub-array into the result
  while (i <= mid) {
    temp[k] = array[i];
    k++;
    i++;
  }

  // Add the rest of the values from right sub-array into the result
  while (j <= end) {
    temp[k] = array[j];
    k++;
    j++;
  }

  for (i = start; i <= end; i++) {
    data[i] = temp[i - start];
  }

  console.log(array.slice(start, end + 1));
  console.log("<<--------------------------------");
};

const mergeSort = (array, start, end) => {
  if (array.length === 0) return [];
  if (array.length === 1) return array;

  if (start < end) {
    const mid = Math.floor((start + end) / 2);

    mergeSort(array, start, mid);
    mergeSort(array, mid + 1, end);
    merge(array, start, mid, end);
  }
};

const data = [4, 1, 3, 2, 5, 6];
mergeSort(data, 0, data.length - 1);
console.log("Sorted data", data);

The output from merge looks like this for

input data [4, 1, 3, 2, 5, 6];
>>left ---------------------------
[ 4, 1 ]
[ 1, 4 ]
<<--------------------------------
>>left -----------------------------
[ 1, 4, 3 ]
[ 1, 3, 4 ]
<<--------------------------------
>>right---------------------------
[ 2, 5 ]
[ 2, 5 ]
<<--------------------------------
>>right---------------------------
[ 2, 5, 6 ]
[ 2, 5, 6 ]
<<--------------------------------
>>all-----------------------------
[ 1, 3, 4, 2, 5, 6 ]
[ 1, 2, 3, 4, 5, 6 ]
<<--------------------------------
output [ 1, 2, 3, 4, 5, 6 ]

Example Link List

Yum link lists my favourite. This is another head bender for me. I struggled to understand how link_list.next.next is not null. But got there with some debug.

const linklist_reverse = (linked_list) => {
    // We dont process an empty linked list
    // Or a linked list not link to anything
    if(link_list === null) return link_list
    if(linked_list.next === null) {
        // We need a node with a next.next pointer
        console.log("Base case reached with")
        return linked_list
    }

    console.log("Calling with")
    prettyPrintLinkedList(linked_list.next);
    const result = linklist_reverse(linked_list.next)
    console.log("changing node-----------")
    prettyPrintLinkedList(linked_list.next);
    console.log("using node-----------")
    prettyPrintLinkedList(linked_list);
    linked_list.next.next = linked_list
    linked_list.next = null
    console.log("Returning with")
    prettyPrintLinkedList(result);
    console.log("------------------------")
    return result
}

And here is the debugging

Calling with B -> C -> D -> null
Calling with C -> D -> null
Calling with D -> null

// Base case as we have no linked_list.next.next.
4/                                              returning D -> null 
3/ changing node D -> null with  C -> D -> null returning D -> C -> null
2/ changing node C -> null with  B -> C -> null returning D -> C -> B -> null
1/ changing node B -> null with  A -> B -> null returning D -> C -> B -> A -> null

Example Binary Search Tree Insert

The great thing about a bst is you do not need to move nodes to do and insert. Only identify the node it belongs to. And now the robot will build and pretty print it.

• left value must be less than the parent
• right value must be greater than the parent
  

const binary_search_insert = (bst, value) => {
  if (bst === null) {
    bst = { value, left: null, right: null };
    return bst;
  }

  if (value < bst.value) {
    bst.left = binary_search_insert(bst.left, value);
    return bst;
  } else {
    bst.right = binary_search_insert(bst.right, value);
    return bst;
  }
};

Example DFS Depth First Search

This is where we traverse a graph going downwards first.

This particular diagram confused a dyslexic like me. So I redrew it

Now here is the structure

const newNodes = {
  a: ["b"],
  b: ["a", "c"],
  c: ["b", "d"],
  d: ["c", "e", "g"],
  e: ["d", "f"],
  f: ["e", "i", "k"],
  g: ["d", "h"],
  h: ["g"],
  i: ["f"],
  j: ["f"],
  k: ["f"],
};

Finally the dfs search.

const dfs_search = (nodes, current, goal, visited = new Set()) => {
  const adjacentNodes = nodes[current];
  console.log("Comparing %s with %s", current, goal);

  // Base Case: Not Search so false
  if (current === null) return false;
  // Base Case: Found the goal return true
  if (current === goal) return true;

  // Iterate over current values
  for (let thisNode of adjacentNodes) {

    // Already been to this node skip
    if (visited.has(thisNode)) continue;

    // Not visited, mark as visited
    visited.add(thisNode);

    const found = dfs_search(nodes, thisNode, goal, visited);
    if (found) return true;
  }

  return false;
};

More Fib

With Fib we use two calls to the same function. This is known as multi branch recursion.

const fib = (n) => {
    if(n < 0) return 0
    if(n <= 1) return 1
    if(n === 2) return 1

    return fib(n-1) + fib(n-2) 
}

This could be shown as a tree

Climbing Stairs (Recursion)

You are climbing a staircase. It takes n steps to reach the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
Still trying to connect pictures with questions. So here goes for climb stairs. First a picture for climb stairs where there are 4 stairs. The successful paths (leaf nodes) in green

In the yellow square, the two child nodes, the left +1 step away, and the right +2 steps away. When you see a binary tree, and the children are k+1 and k+2 then you can use Fibonacci f(n) = f(n-1) + f(n-2) to calculate the number of leaf nodes

Combinations

Common questions are

• Given N things, in how many ways can we arrange them
• In how many ways can we do X
• What is the shortest way to do Y

What is a combination

It is a collection of things where the order does not matter. Important is give [a,b] in a combination it is the same as [b,a]. Given a set of n things, there are 2^n possible combinations.

Example

Looking at [a,b,c] this gives the following

To at each level, the left is if you don't take the level value. So at level a on the left branch we take the parent ([]), on the right we take the level value (a).