1.7k views

Consider the following recurrence relation

$T(1)=1$

$T(n+1) = T(n)+\lfloor \sqrt{n+1} \rfloor$ for all $n \geq 1$

The value of $T(m^2)$ for $m \geq 1$ is

1. $\frac{m}{6}\left(21m-39\right)+4$
2. $\frac{m}{6}\left(4m^2-3m+5\right)$
3. $\frac{m}{2}\left(3m^{2.5}-11m+20\right)-5$
4. $\frac{m}{6}\left(5m^3-34m^2+137m-104\right)+\frac{5}{6}$
hello..anybody idea about how to solve this recurrence ?
For quick answer use  option rejection method.

$T(m^2) = T(m^2-1) + \left\lfloor\sqrt{(m^2)} \right\rfloor$

$= T(m^2 - 2) + \left\lfloor\sqrt{(m^2 - 1)} \right\rfloor +\left\lfloor\sqrt{(m^2)} \right\rfloor$

$= T(m^2 - 3) + \left\lfloor\sqrt{(m^2 - 2)} \right\rfloor + \left\lfloor\sqrt{(m^2 - 1)} \right\rfloor +\left\lfloor\sqrt{(m^2)} \right\rfloor$

$\vdots$

$= T(1) + \left\lfloor\sqrt{(2)} \right\rfloor + \left\lfloor\sqrt{(3)} \right\rfloor + \ldots + \left\lfloor\sqrt{(m^2)} \right\rfloor$

$= 3 \times 1 + 5 \times 2 + \ldots + \left(2m - 1\right) \times (m-1) +m$

(We are taking floor of square root of numbers, and between successive square roots number of numbers are in the series $3,5,7 \dots$ like $3$ numbers from $1..4$, $5$ numbers from $5-9$ and so on).

We can try out options here or solve as shown at end:

Put $m = 5$, $T(25) = 3 \times 1 + 5 \times 2 + 7 \times 3 + 9 \times 4 + 5 = 75$

1. 59
2. 75
3. non-integer
4. 297.5

$T(m^2) = 3 \times 1 + 5 \times 2 + \dots + \left(2m - 1\right) \times (m-1) +m$
$= m + \sum_i=1^{m-1} (2i+1). (i)$
$= m + \sum_i=1^{m-1} 2i^2 + i$
$= m + \frac{(m-1) .m .(2m-1)}{3} + \frac{(m-1)m}{2}$
$= \frac{m}{6} \left(6 + 4m^2 -2m -4m + 2 + 3m - 3\right)$
$= \frac{m}{6} \left(4m^2 -3m + 5\right)$

[Sum of the first $n$ natural numbers $=\frac{n. (n+1)}{2}.$

Sum of the squares of first $n$ natural numbers $= \frac{n. (n+1). (2n+1)}{6}.$]

edited by
why 3*1+5*2...........?
Floor of squareroot(1), squareroot(2) and squareroot(3) is 1. Therefore, 3*1 and so on..
ok :)
Very good analysis. :) Thank You.
Edit it .... i am nt gd in latex ...

we can write this recurrence as T(n)=T(n-1)+√n

Now we can apply master theorem for subtract and conquer

here, a=1,b=1 f(n)=O(√n )

By case 2 of  subtract and conquer T(n)=O(n^1.5)

if we take n=m^2

T(m^2)=O(m^3),and which is tighter upper bound of  option B.

see below for subtract and conquer of master theorem.

https://www.eecis.udel.edu/~saunders/notes/recurrence-relations.pdf

Nice source of information!

T(1)=1

T(2)=T(1)+⌊√2⌋=1+1=2

T(3)=T(2)+⌊√3⌋=2+1=3

T(4)=T(3)+⌊√4⌋=3+2=5

....... so on

now find T(m2)

take m=2 so T(m2) =T(4)

now check options ,  only option B give value 5 which is equals to T(4)

so ans is B

@Gate Ranker18

Nopes , even option A will be also true , if u will do like this. So u can not say like this .
Yes exactly option A is also matching when m=2 and for m=3 also.
+1 vote

Using Integration bounds technique we can get approximate answer.

Integration bound method is explained in detail here(page 11, theorem 9.3.1).

$T(n) = T(n-1) + \left \lfloor \sqrt{n} \right \rfloor , n > 1$

Unrolling it gives,

$T_n = \sum_{k = 1}^{n} \left \lfloor \sqrt{k} \right \rfloor$

But, because $k - \left \lfloor k \right \rfloor < 1$

$\sum_{k = 1}^{n} \sqrt{k} - n \le T_n \le \sum_{k = 1}^{n} \sqrt{k}$

So, Let's find $S_n = \sum_{k = 1}^{n} \sqrt{k}$

Integration bound theorem: if $f$ is non decreasing function, $S = \sum_{k = 1}^{n} f(k)$ and $I = \int_{1}^{n} f(x) * dx$ then $I + f(1) \le S \le I + f(n)$.

$\frac{2}{3}*n^{\frac{3}{2}} - \frac{2}{3} + 1 \le S_n \le \frac{2}{3}*n^{\frac{3}{2}} - \frac{2}{3} + \sqrt{n}$

$\frac{2}{3}*n^{\frac{3}{2}} + \frac{1}{3} - n \le T_n \le \frac{2}{3}*n^{\frac{3}{2}} - \frac{2}{3} + \sqrt{n}$

Taking $n = m^2$

$\frac{2}{3}*m^3 - m^2 + \frac{1}{3} \le T_{m^2} \le \frac{2}{3}*m^3 - \frac{2}{3} + m$

Closet match is B.

edited