Question

Suppose you implement the disjoint-sets data structure usingunion-by-rank but not path compression. Give a sequence of$\mathbf{m}$ union and find operations on$\mathbf{n}$elements that take $\mathbf{\Omega }\left(m\mathrm{logn}\right)$time.

Short answer

The disjoint sets of data structure using union-by-rank a sequence of union and operations on elements by using forest implementation of disjoint sets that takes$\mathrm{\Omega }(\mathrm{m}\mathrm{logn})$ time.

Step-by-step solution

Forest Implementation

A forest implementation of disjoint sets with only the weighted union heuristics without path comparison. The disjoint sets of data structure using union-by-rank a sequence of m union and operations on n elements by using forest implementation of disjoint sets.

Create Sequence and Perform Operation

Let’s a forest implementation of disjoint sets with only the weighted union heuristics without path comparison. And it consists n distinct elements. For proving this given statement suppose n be the power of variable 2 and consider ${\mathrm{x}}_1,{\mathrm{x}}_2.........{\mathrm{x}}_{{\mathrm{n}}_{}}$ be the variables.

Let the sequence of unions. And there are ${\log }_2\mathrm{n}$levels of union operations. And $\frac{\mathrm{n}}{2}$ unions which is equals to the $\left(x_1,x_2\right),\left(x_3,x_4\right),...,\left(x_{n-1},x_n\right)$.

$\frac{n}{2}=\left({\mathrm{x}}_1,{\mathrm{x}}_2\right),\left({\mathrm{x}}_3,{\mathrm{x}}_4\right),...,\left({\mathrm{x}}_{\mathrm{n}-1},{\mathrm{x}}_{\mathrm{n}}\right)$

And after this, there are some disjoint sets.

Let ${\mathrm{T}}_1.............{\mathrm{T}}_{\mathrm{n}}$are some disjoint sets at the beginning of level $\mathcal{K}$.

Then the following unions are as follows:

$\left(root\left(T_1\right),\left(root\left(T_2\right)\right)....\right)$

Here the number of disjoint sets decreases by a factor of 2, and as there $\frac{\mathrm{n}}{2}$sets after level 1, after level k,

There are $\frac{\mathrm{n}}{2^{\mathrm{k}}}$sets.

This indicates after the level ${\log }_2\mathrm{n}-1$, there is one disjoint set T.

And the height of this set T is ${\log }_{2}n-1$. As the union the roots of the trees.

So far, n - 1 operations of unions are taking place.

And an extra union of $\left(x_1,x_2\right)$and the number of unions is exactly n, which does not change the structure of the given tree.

Now, after this sequence of unions, there is a node with height ${\log }_{2}n-1$in T.

After this step just find the operations on x. And without the path compression, it will not change the structure of the tree.

Since the depth of x, then the operation will take$\mathrm{\Omega }\left(\log \mathrm{n}\right)$time. So, the total time for finding the operation on x is $\mathrm{\Omega }\left(\mathrm{mlogn}\right)$.

The disjoint sets of data structure using union-by-rank a sequence of m union and operations on n elements by usingforest implementation of disjoint setsis given as

$\mathbf{const}\mathbf{}\mathbf{int}\mathbf{}\mathbf{N}\mathbf{=}\mathbf{10}\mathbf{;}$

Let the constant be ten, after that take integer,

int P[N+1]; //stores parent of each set, initially all elements set to -1 to indicate no parent.

int Min[N+1]; //stores min element of each set

$$\begin{gathered} \mathrm{int}\mathrm{Find}(\mathrm{int}\mathrm{u}) \\ \{ \\ \mathrm{return}\mathrm{P}[\mathrm{u}]<\mathrm{o}?\mathrm{uP}[\mathrm{u}]=\mathrm{Find}(\mathrm{P}\left[\mathrm{u}\right]; \\ \} \\ \mathrm{void}\mathrm{Union}(\mathrm{int}\mathrm{u},\mathrm{int}\mathrm{v}\left) \\ \right\{ \\ \mathrm{u}=\mathrm{Find}\left(\mathrm{u}\right); \\ \mathrm{v}=\mathrm{Find}\left(\mathrm{v}\right); \\ \mathrm{if}(\mathrm{u}==\mathrm{v})\mathrm{return}; \\ \mathrm{Min}\left[\mathrm{u}\right]=<\mathrm{Min}\left[\mathrm{v}\right]?\mathrm{Min}\left[\mathrm{u}\right]:\mathrm{Min}\left[\mathrm{v}\right]; \\ \mathrm{P}\left[\mathrm{v}\right]=\mathrm{u}; \\ \} \end{gathered}$$

$$\begin{gathered} \mathrm{int}\mathrm{FindMin}(\mathrm{int}\mathrm{u}) \\ \{ \\ \mathrm{return}\mathrm{Min}[\mathrm{Find}\left(\mathrm{u}\right)]; \\ \} \\ \mathrm{Therefore},\mathrm{the}\mathrm{total}\mathrm{time}\mathrm{complexity}\mathrm{is}\mathrm{\Omega }\left(\mathrm{mlogn}\right). \end{gathered}$$