**Decrease key** refers to **reducing** the key value of any node.If we want to decrease a key in Binomial heap,we will replace the key with reduced value and will repeatedly **exchange** the reduced key with the parent in order to restore **min-heap** property.The** running time** to perform this operation is **O(log n)**.

**Example**

Say we want to decrease **50 to 4** in binomial heap given in Fig 1.The steps further followed are given in Fig 2,3,4,5 respectively.

4 and 21 are swapped in Fig 3 and 4 and 8 are swapped in Fig 4** **in order to retain min-heap property.

In **Extract-Min **operation node with **minimum key** is **deleted** from the **binomial heap h**.The running time to extract minimum value is **O(log n)**.The steps followed are :

- Find the
**root (say x)**with**minimum key**. **Delete**the root.**Break**the binomial heap into**h**and**h’**.- Perform the
**union**operation to**h**and**h’**.

Given the binomial heap h in **Fig 1**.The first step is to find the **root** with **minimum key**.Node with value **1** is smallest so it is** deleted** and **h** is broken down into** h** and **h’**(Fig 2).

Perform the **union** operation on **h** and **h’.**

After the **union** operation on **h** and **h’** the structure seems like as given in Fig 4

Different cases are performed according to the situation.**x** points to **11**,**next-x** points to **2**,**sibling[next-x]** points to **3** as shown in Fig 5

**Case 1** is applied in Fig 6.**x** now points to **2**,**next-x** points to **3** and **sibling[next-x]** points to **7**.

**Case 3** is applied to Fig 7.Trees with **root 2** and **3** are of **same** order i.e.** 1** and **2** is less than **3** so **3** becomes the child of **2**.**next-x** will now point to** 7** and **sibling[next-x]** will point to **4**.

**Case 3** is applied again in Fig 8.**2** is less than** 7** so **7** becomes the child of **2**.**next-x** will now point to **4**.

**Case 3** is applied in Fig 9 as **2<4** .So 4 becomes the child of 2.x now points to 2 and **next-x** points to **null**.Thus the process stops here.

The union of two heaps is the **merging** root lists by **root degree**.But if we simply merge two heaps then a **problem** can arise.There may be a chance that we get two or more trees of **same root degree**.This **violates** the property of binomial heap.To deal with this problem we have four cases and solution to these cases respectively. Referring to all figures we choose arbitrary names like :

**x**is the second node of merged heaps.**prev-x**points to previous node of x.**next-x**points to next node of x.**sibling[next-x]**points to next node of next-x.**B**_{k }and B_{1 }are**Binomial Heaps**of order**k**and**1**respectively**.**- We apply different cases according to the situation until next-x becomes nil.

**Case 1**

If we merge heaps and have a case like **x** points to binomial heap of **order k** and **next-x** points to binomial heap of **order 1** then ,then we need to shift our pointers towards right.**x** will now point to binomial heap with **order 1.next-k** will point to heap with root **node d** and **prev-x** will point to heap with root **node b** and **order k**.

**Case 2**

If in union operation the heaps with **root b,c,d** are of same **order k** and **x point**s to heap of **order k** and root **node b**,**next-x** points to next heap of** x** with **same order** and root **node c** and **sibling[next-x]** points to next heap of **next-x** again with **same order** and root **node d** then pointers shift by one.** x** now points to heap with root **node c,next-x** points heap with **root d** and **prev-x** points to heap with **root b**.

**Case 3**

After the union operation,if ** b,c** are of** same order** and** key[x]** is less than **key[next[x]]** then **c** becomes the **left child of b** and **x** remain on pointing to **b** but** next-z** now points to points to **d** which was earlier pointed by **sibling[next-x]**.

**Case 4**

The pointer points to same binomial heaps as in case 3 except the difference that** key of x** is greater than **key[next[x]]**.In this case,**b** becomes the **child of c**.**c** is pointed by **x** and** d** is pointed by **next-x**.

**Example**

Consider two heaps **h1** and **h2** as given in Fig 5.The union operation has to be performed on these two heaps**.h1** has binomial trees of order** 0,1** and **2**.h2 has binomial trees of order **0,1 and 3**.At the point of union operation both the heaps are combined together but the resultant structure does not follows **heap property** i.e the resultant structure has binomial trees of **same order**.In Fig 5,two trees with root 2 and 5 are of order 0 and two trees with roots **1** and **3** are of **order 1** violating the property of heaps.In order to make a structure satisfying heap properties we follow case according to the situation.The union of h1 and h2 is shown in Fig 6.

In Fig 7,**2** is less than **5** and both the trees are of same order i.e **0** so case** 3** is followed in which root pointed by **next-x(5)** becomes the child of** x(2)** and pointer **next-x** shifts to right i.e it shifts to** 1** and **sibling[next-x]** also shifts pointer to right i.e to **3**.

In Fig 8,all** three binomial trees** pointed by** x,next-x and sibling[next-x]** are of **same order** i.e **1** so** case 2** is followed in which all three pointers shift to right by one i.e **x** will point to** 1**,**next-x** will point to **3** and **sibling[next-x]** will point to **7**.

In Fig 9,**Case 3** is followed as **root 1** is less than **root 3** and their trees are of **same order**.Tree with **root 3** becomes child of tree with **root 1**.

In Fig 10,**Case 3** is followed again as **root 1** is less than **root 7** and their trees are of **same order** i.e. **2**.Tree with **root 1** becomes the **child** of tree with **root 7**.

In Fig 11,**Case 3** is followed as **1** is less than **4** and their trees are of **same order** i.e.**3**.Tree with **root 1** becomes the child of tree with **root 4**. The next-x becomes nil and so the process of applying cases stop here.We get two binomial trees of different orders.They are of order **1** and **4** respectively.Union of heaps is done satisfying all the properties of binomial heap.

**Binomial Heaps** are similar to **binary heaps** with additional feature of implementing binomial series as sequence of trees.The heap(Fig 1) is represented using **left child right sibling** pointers.It has three links per node **(parent,left,right)** and the roots of tree are connected using single linked list.The degrees of tree decrease from **left to right.**

- Each Binomial tree in the heap obey
**min-heap**property. - There can be only
**0 or 1 binomial trees**for each**order**.

The** root value** of every** binomial tree** is smaller than its children and there will be at most** log n +1** binomial trees of binomial heap with** n nodes.**

Each **binomial tree** corresponds to** one digit** in** binary representation** of a number n.For e.g.the binary representation of** 13**(Fig 2) is **1101**(2^{3}+2^{2}+2^{0}=8+4+1).The binomial heap will have **13 nodes** with binomial trees of order **3,2 and 0** respectively.The orders came from the powers of 2. Either there will be no tree or there will one tree of each order i.e we cannot have any two trees of same order in the binomial heap.Similarly for a number like 15,the binary representation is **1111**(2^{3}+2^{2 }+2^{1 }+2^{0} =8+4+2+1).The binomial heap will have **15 nodes**(Fig 4)with binomial trees of order **3,2,1,0** respectively.

The nodes in the binomial heap can be represented by a structure with parent,key,degree,child and sibling attributes ( Fig 5).

**Example**

Consider the binomial heap as given in Fig 6.Say we want to represent 9(Say) then the structure would be like as given in Fig 7 or if we want represent 31 the structure would be like as given in Fig 8

The various operations that can be performed on Binomial heaps are:

- Union
- Delete Min
- Decrease Key
- Delete
- Insert

The **Maxheap** sorts the elements in **increasing** order whereas **Minheap** sorts the element in **decreasing order**.To sort the heap three steps have to be followed

**Heapify**-The process picks the**largest child key**and compare it to the**parent key**. If parent key is larger than key then heapify**quit**s, otherwise it swaps the parent key with the largest child key. So that the parent now becomes larger than its children.**Build Heap**– Use the procedure ‘Heapify’ in a bottom-up fashion to convert an array into a**hea**p. The process starts from last nodes and not the leaves.**Heap Sort**– Sorting is done by removing the**max element**and storing in array .The storage starts from last position and goes till first index of array.

- Build Heap Algorithm will run in
**O(n)**time - There are n-1 calls to Heapify each call requires
**O(log n)**time - Heap sort program combine Build Heap program and Heapify, therefore it has the running time of
**O(n log n)**time - Total time complexity:
**O(n log n)**

**Example**

Given an array in Fig 1 .We need to convert it to Maxheap and sort it.

Array is converted to heap in **Fig 2**

The further steps are illustrated through following figures

Heaps of **n** **keys** can be represented by array of **length n+1**(Fig 2).For a node at **rank i** ,the **left child** is at **rank 2i +1** and the **right child** is at rank **2i +2**.The links between nodes are not stored explicitly.

- In Fig 1,heap has
**7**keys i.e**n =7.** - Array will be of
**length 8**(n+1=7+1).

If two heaps are to be **merged** and a **key(k)** has to be inserted then we follow these steps :

- A
**new heap**is created with**root node**storing**k**and two given heaps as**subtrees**. - Perform
**downheap**to restore**relational**property of heap.

**Example**

Given are the two heaps h1 and h2 in Fig 3.We need to merge them and add 5 to it.

A new heap** h3 ** is created with root node storing **5** and** h1,h2 as subtrees**(Fig 4).

Perform downheap operation to restore **heap property**(Fig 5).

A Heap is a** Binary tree**(Refer http://letslearncs.com/binary-trees/) that stores** keys** at it’s** internal nodes** and satisfies two additional properties which are :

- Relational Property
- Structural Property

It follows either **Min-Heap** or **Max-Heap **property.The **Max-Heap **property says that **keys** of each and every **child ** will be **less** than the** key ** of **parent**(Fig 2)whereas** ****Min-Heap ** property says that ** keys** of each and every child are either **greater** than or **equal** to the** key** of **parent**(Fig 1).

All the levels are **full** except the **last level** which is** left justified**.Left justified means that the parents of terminal nodes will have left child and may/may not have right child(Fig 4).Tree in Fig 3 fails to satisfy structural property.

The height of heap is **O(log n).**

- Let h be the
**height**of**heap**with**n keys**. - The maximum no. of keys in heap can be
**2**^{n.} - Since there 2
^{i }at depth i =0,1,2,3,….h-1 and at least one key at depth h,we have**1 +2 +4 + 2**^{h-1 }**+1**. - Thus n=
**2**and^{h }**height= O( log n).** **h=O(log n).**

The insertion in **Heap** follows three steps :

- Find the
**node**(Say n) where**key**(k) has to be inserted. - Store
**key(k)**at**node(n)**. - After the insertion of new key,heap property can get violated.
- Use
**Upheap**technique to restore heap property. - The upheap technique swaps the key along with the
**upward**path from the insertion node. - This method is followed until the key(k) reaches a node whose parent has a smaller/equal value to k.

**Example**

- This method removes the
**root**i.e the minimum value from the heap. - The
**root**is deleted and the**last key**takes the place at root node - Downheap technique is followed till the heap property is restored.
- Downheap swaps key along the
**downward**path. - Terminate downheap method leaf/terminal level is reached and key of parent is less than key of child.

**Example**