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The recurrence equation
$ T(1) = 1$
$T(n) = 2T(n-1) + n, n \geq 2$

evaluates to

  1. $2^{n+1} - n - 2$
  2. $2^n - n$
  3. $2^{n+1} - 2n - 2$
  4. $2^n + n $
in Algorithms by Veteran (52.2k points)
edited by | 5.4k views
0
Initial equation should be T(0)=1 not T(1) =1.

7 Answers

+42 votes
Best answer
$T(n) = 2T(n-1) +n, n \geqslant 2 , T(1) = 1$

$  T(n)    = n + 2(n-1) + 2^2 (n-2) + \dots + 2^{(n-1)}(n -(n-1))$

           $=n( 1 + 2 + \dots + 2^{n-1}) - (1.2 + 2.2^{2} +3.2^{3}+\dots+ (n-1).2^{n-1})$

           $=n(2^n-1) -(n.2^n-2^{n+1}+2)$

           $=2^{n+1}-n -2$

Correct Answer: $A$
by Loyal (6.1k points)
edited by
+2
How?
+28

Here everything is straight forward, the only complex thing is evaluation of

1.2 + 2.22 +3.23+.....+ (n-1).2n-1 let us say Sn.

Sn  =  1.2 + 2.22 +3.23+.....+ (n-1).2n-1

2.Sn =          1.22+2.23+.......+(n-2).2n-1+(n-1).2n

_                  _      _                  _                _

.................................................................................................

-Sn  =  2 + 22 +23 + .................+2n-1 -(n-1).2n

Sn =  n.2n-2n+1+2

+2

I am stucked at this step ,plz clarify this 

T(n)=2kT(n-k)+2(k-1)T(n-(k-1))+2(k-2)T(n-(k-2))+...........+n

Now k=n-1 ,so 

T(n)=2(n-1)+2(n-2)(2)+2(n-3)(3)+...........+n

Now how to proceed from here ?

0
saw the above comment?
0

In above comments series formed is of the form 1.2 + 2.22 +3.23+.....+ (n-1).2n-1

While I am getting 2^n(1/2+2/4+3/8+.........)+n  ,so how is it same ?

–1

I was stuck at the same point , just multiply Sn (sumation)  with 2 and then subtract it with the original Sn.This will create a geometric series of 2s +(-n). The sumation of geometric series turns out to be 2^n+1 - 2. Hence final answer is A

0

T(n)=n+2(n−1)+22(n−2)+⋯+2(n−1)(n−(n−1))  in second line it should be T(n-(n-1))??????????????

0
an additional 2^(n) term should also come
+8

We can solve it using recurrence relation method that we used to do in discrete mathematics

The given recurrence is

an = 2an-1 +n when n$\geq$2   .........................(1)

1 when n=1

The solution of an is given as

an= an H (Solution to homogenous part) + an (Particular solution)

anH will be given  as

$\alpha$(2)

anP will be in form Cn +d

Substituting this in (1) we get

Cn+d = 2Cn+n+2d-2c

n(C+1) + (d-2C) =0

C=-1 and d=-2.

So our recurrence becomes

an = $\alpha$(2) -n-2

At n=1 an=1, so

we get $\alpha$ =2

so

an=2n+1-n-2

0
Why havent the leaf nodes been calculated into the total cost? They will incurr 2^n cost
0
Thank you
0
@Ayush Can you provide me tutorial links for that method ??
0
Kenneth Rosen
0

@Ayush Upadhyaya how to get value of A1, A0 ? In your equation they are C and d.

0

n(C+1) + (d-2C) =0

How did you calculate C and d after this ? @Ayush Upadhyaya

 might b silly but m stuck here. Plz help bro

+1

@tusharp-Equate coefficients of n and constant term to 0.

+1

One simple way to answer such question is:

put n=2 ( any small case and find its value and put in all the equations )

T(1) = 1

T(2) = 4

T(3) = 11

 

option a)  4

option b) 2

option c) 2

option d) 6

 

Now when we put n=2 in the equations we get the correct answer only in option a.

If multiple eq gives the correct answer then we have to take other case and check again on the remaining equations which were giving the correct answer.

 

Please correct me if this approach is wrong.

0
any one can tell me what is time complexity exist it  i.e. O(?)
+71 votes
T(1) = 1

T(2) = 4

T(3) = 11

T(4) = 26

T(5) = 57

T(6) = 120

T(7) = 247

So, $$T(n) = 2^{n+1} - n - 2$$
by Veteran (431k points)
+4
nice trick :)
+1
excellent trick bro
+1
Smart method
+1
is it required that we go till 7 value i put n=2 and get 4 then check ans which gave me 4 by putting n=2 only A option satisfied it..
+2

@Arjun

sir by applying muster theorem or subtract and conquer master theorem

ans comes out to be proportional to n*(2^n) .

Why it's far different from actual answer ?  Do muster theorem has any condition where it fails ?

Plz help.

+1

EXCELLENT !!

Does this technique works for every recurrence relation ?

+1
Yes, if options are given
0

@ same doubt do you got any answer?

+16 votes
Given that T(1) =1 //OMG here itself Option C & D fails now check A & B

T(2)=2T(1)+n=2+2=4 //Option B evaluates to 2 so it is wrong

Hence option A is ans.
by Boss (23.9k points)
+1
Nice reasoning for option C and D! Found useful!
0
Quickest way to solve this problem.
+9 votes

solution .

by (289 points)
+1
this is what i was looking for ,, thanx buddy :)
+1
@varun singh

you have written

T(n-k) = 2T(n-(k-1)) + (n-k)

so  for k=2

T(n-2) = 2T(n-(2-1)) + (n-2)

T(n-2) = 2T(n-1) + (n-2)

so it should be

T(n-k) = 2T(n-(k+1)) + (n-k)

please look into it and if i am wrong please do correct me.
0
Yes @pramodborana112 u are right ,  it should be T(n-k) = 2T(n-(k+1)) + (n-k), but everything else are correct.
0
I think  there would be change in arithmetric geometric expression  then as k =n-2 then in that case  how would we solve
+1 vote

One simple way to answer such question is:

put n=2 ( any small case and find its value and put in all the equations )

T(1) = 1

T(2) = 4

T(3) = 11

 

option a)  4

option b) 2

option c) 2

option d) 6

 

Now when we put n=2 in the equations we get the correct answer only in option a.

If multiple eq gives the correct answer then we have to take other case and check again on the remaining equations which were giving the correct answer.

by Junior (791 points)
+1 vote

Here, $T(1)=1$ and

$\begin{align} T(n)&=2T(n-1)+n\\&=2^2T(n-2)+2(n-1)+n;~[\text{Putting }T(n-1)=2T(n-2)+(n-1)]\\&=2^3T(n-3)+2^2(n-2)+2(n-1)+n;~[\text{Doing so}]\\&=\cdots\\&=2^{n-1}T(n-(n-1))+2^{n-2}\{n-(n-2)\}+2^{n-3}\{n-(n-3)\}+\cdots+2(n-1)+n\\&=2^{n-1}T(1)+2^{n-2}(2)+2^{n-3}(3)+\cdots+2^{1}(n-1)+2^{0}n\end{align}$

 

Now putting $T(1)=1$ as given,

$\begin{align}\therefore T(n)&=2^{n-1}(1)+2^{n-2}(2)+2^{n-3}(3)+\cdots+2^{1}(n-1)+2^{0}n \tag{i} \\ \Rightarrow 2T(n)&=2^n(1)+2^{n-1}(2)+2^{n-2}(3)+\cdots+2^2(n-1)+2^{1}n \tag{ii}\end{align} $

 

$\mathrm{no(ii)}-\mathrm{no(i)}\Rightarrow\\ \begin{align}T(n)&=2^n+2^{n-1}(2-1)+2^{n-2}(3-2)+\cdots+2^1(n-(n-1))-2^{0}n\\&=(2^{n}+2^{n-1}+2^{n-2}+\cdots+2)-n\\&=(2+2^2+2^3+\cdots+2^{n-1}+2^n)-n; ~[\text{Rearranging}]\\&=\frac{2(2^n-1)}{2-1}-n;~[\scriptsize\because a+ar+ar^2+\cdots+ar^n=\frac{a(r^{n+1}-1)}{r-1}\text{ as Geometric Series}]\\&=2^{n+1}-2-n \end{align}$

 

$\therefore T(n)=2^{n+1}-n-2$.

 

So the correct answer is A.

by Active (3.5k points)
edited by
+1
in eq 1 before (3) it will be $2^{n-3}$
0
Thanks, it was a typing mistake.
0 votes

T(2) = 4.

Option A.

Solved under 30 seconds.

by Loyal (6.3k points)
Answer:

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