I don't know how to continue in this recurrence cause I don't see any pattern, any help??
T(n) = 2n + T(n/2)
= 3n + T(n/4)
= 7n/2 + T(n/8)
= 15n/4 + T(n/16)
and so on...
As I understand it's just simple reccurence.
T(n) = 2n + T(n/2)
Your notation can makes someone think different. For me it should be:
T(n) = 2n + T(n/2) ....(1)
T(n/2) = 2(n/2) + T(n/2/2) = n + T(n/4)
T(n) = 2n + n + T(n/4) = 3n + T(n/4) ....(2)
T(n/4) = 2(n/4) + T(n/4/2) = n/2 + T(n/8)
T(n) = 2n + n + n/2 + T(n/8) = 7n/2 + T(n/8) ....(3)
T(n/8) = 2(n/8) + T(n/8/2) = n/4 + T(n/16)
T(n) = 2n + n + n/2 + n/4 + T(n/16) = 15n/4 + T(n/16) ....(4)
T(n/16) = 2(n/16) + T(n/16/2) = n/8 + T(n/32)
T(n) = 15n/4 + n/8 + T(n/32) = 31n/4 + T(n/32) ....(5)
and so on...
This is a usual recurrence relation - if you are a CS student, you will soon know the result by heart.
If you want to find the result by hand, make a geometric sum appears from the recurrence:
T(n) = 2n + n + n/2 + ... + n/2^(k+1) + T(0)
= 2n(1 + 1/2+ ... + 1/2^(k+2)) + T(0)
Where k = INT(log2(n))
You can see a geometric sum of general term 1/2 appears
1 + 1/2 + ... + 1/2^(k+2) = (1 - 1/2^(k+3)) / (1 - 1/2)
Observing that 2^(k+2) = 8 * 2^(log2(n)) = 8n and simplifying
T(n) = 4n + T(0) - 1/2 = Theta(4n)
In addition to the expanding series till T(0) way shown by Alexandre Dupriez, You can also apply master theorem to solve it
For the recurrence equation
T(n) = 2n + T(n/2)
Master Theorem:
For recurrences of form,
T(n) = a T(n/b) + f(n)
where a >= 1 and b > 1
If f(n) is O(nc) then
If c < logba, then T(n) = O(n * logba)
If c = logba, then T(n) = O(nc * log n)
if c > logba, then T(n) = O(nc)
we have a = 1, b = 2, c = 1 and c > logba (case 3)[as c = 1 and log 1 (any base) = 0]
Therefore, T(n) = O (n1)
T(n) = O (n)
Related
what would be the big-o for
T(n)= T(n/2) + cn
I know the mergesort case T(n) = 2T(n/2)+cn i.e. linearithmic
and i was able to solve T(n) = 2T(n/2)+c to get linear but am confused in the first one...
The first one should be pretty simple:
T(n) = T(n/2) + cn = T(n/4) + cn/2 + cn = T(n/8) + cn/4 + cn/2 + cn
= T(1) + c(n/2^m + ... + n/4 + n/2 + n)
<= T(1) + c(n + n/2 + n/4 + n/8 + ...) = 2cn + T(1)
Where m = log(n).
Hence in terms of big-o notation T(n) ~ O(n).
BTW, not hard to prove this is actually theta of n,
T(n) = T(n/2) + cn = T(n/4) + cn/2 + cn = T(n/8) + cn/4 + cn/2 + cn
= T(1) + c(n/2^m + ... + n/4 + n/2 + n)
>= T(1) + c(n/2 + n/4 + n/8 + ...) = cn + T(1)
and therefore T(n) is actually theta of n, since both big-o and big-omega of n.
I got stuck in this recurrence:
T(n) = T(n − 1) + lg(1 + 1/n), T(1) = 1?
for a while and it seems the master method cannot be applied on this one.
We have:
lg(1 + 1/n) = lg((n + 1) / n) = lg(n+1) - lg(n)
Hence:
T(n) - T(n - 1) = lg(n + 1) - lg(n)
T(n-1) - T(n - 2) = lg(n) - lg(n - 1)
...
T(3) - T(2) = lg(3) - lg(2)
T(2) - T(1) = lg(2) - lg(1)
Adding and eliminating, we get:
T(n) - T(1) = lg(n + 1) - lg(1) = lg(n + 1)
or T(n) = 1 + lg(n + 1)
Hence T(n) = O(lg(n))
Same answer as the other correct answer here, just proved differently.
All the following equations are created from the given recurrence:
T(n) = T(n-1) + Log((n+1)/n)
T(n-1) = T(n-2) + Log(n/(n-1))
.
.
.
T(2) = T(1) + Log(3/2)
Summing all RHS and LHS in the above equations results in:
T(n) = T(1) + Log(3/2) + Log(4/3) + ... + Log((n+1)/n)
Since Log(a) + Log(b) = Log(ab),
T(n) = 1 + Log((n+1)/2)
T(n) = Log(10) + Log((n+1)/2) = Log(5n + 5) assuming that base was 10 and using 1 = Log1010
Therefore T(n) = O(Log(5n + 5)) = O(Log(n))
It is not linear as some people claim. It is O(log(n)). Here is mathematical analysis:
If you will start unrolling recursion you will get:
if you will do this till the end you will have
or in a short form:
Once you approximate the sum with an integral, you will get:
Finally if you will take a limit x -> infinity:
You will see that the first part is
Which gives you a final solution log(x + 1) which is O(log(n))
Given..
T(0) = 3 for n <= 1
T(n) = 3T(n/3) + n/3 for n > 1
So the answer's suppose to be O(nlogn).. Here's how I did it and it's not giving me the right answer:
T(n) = 3T(n/3) + n/3
T(n/3) = 3T(n/3^2) + n/3^2
Subbing this into T(n) gives..
T(n) = 3(3T(n/3^2) + n/3^2) + n/3
T(n/3^2) = 3(3(3T(n/3^3) + n/3^3) + n/3^2) + n/3
Eventually it'll look like..
T(n) = 3^k (T(n/3^k)) + cn/3^k
Setting k = lgn..
T(n) = 3^lgn * (T(n/3^lgn)) + cn/3^lgn
T(n) = n * T(0) + c
T(n) = 3n + c
The answer's O(n) though..What is wrong with my steps?
T(n) = 3T(n/3) + n/3
T(n/3) = 3T(n/9) + n/9
T(n) = 3(3T(n/9) + n/9) + n/3
= 9T(n/9) + 2*n/3 //statement 1
T(n/9)= 3T(n/27) + n/27
T(n) = 9 (3T(n/27)+n/27) + 2*n/3 // replacing T(n/9) in statement 1
= 27 T (n/27) + 3*(n/3)
T(n) = 3^k* T(n/3^k) + k* (n/3) // eventually
replace k with log n to the base 3.
T(n) = n T(1) + (log n) (n/3);
// T(1) = 3
T(n) = 3*n + (log n) (n/3);
Hence , O (n* logn)
Eventually it'll look like.. T(n) = 3^k (T(n/3^k)) + cn/3^k
No. Eventually it'll look like
T(n) = 3^k * T(n/3^k) + k*n/3
You've opened the parenthesis inaccurately.
These types of problems are easily solved using the masters theorem. In your case a = b = 3, c = log3(3) = 1 and because n^c grows with the same rate as your f(n) = n/3, you fall in the second case.
Here you have your k=1 and therefore the answer is O(n log(n))
This question can be solved by Master Theorem:
In a recursion form :
where a>=1, b>1, k >=0 and p is a real number, then:
if a > bk, then
if a = bk
a.) if p >-1, then
b.) if p = -1, then
c.) if p < -1, then
3. if a < bk
a.) if p >=0, then
b.) if p<0, then T(n) = O(nk)
So, the above equation
T(n) = 3T(n/3) + n/3
a = 3, b = 3, k =1, p =0
so it fall into 2.a case, where a = bk
So answer will be
O(n⋅log(n))
Need some help on solving this runtime recurrence, using Big-Oh:
T(n) = T(n/2) + T(n/2 - 1) + n/2 + 2
I don't quite get how to use the Master Theorem here
For n big enough you can assume T(n/2 - 1) == T(n/2), so you can change
T(n) = T(n/2) + T(n/2 - 1) + n/2 + 2
into
T(n) = 2*T(n/2) + n/2 + 2
And use Master Theorem (http://en.wikipedia.org/wiki/Master_theorem) for
T(n) = a*T(n/b) + f(n)
a = 2
b = 2
f(n) = n/2 + 2
c = 1
k = 0
log(a, b) = 1 = c
and so you have (case 2, since log(a, b) = c)
T(n) = O(n**c * log(n)**(k + 1))
T(n) = O(n * log(n))
Can someone please help me with this ?
Use iteration method to solve it. T(n) = T(n-1) +n
Explanation of steps would be greatly appreciated.
T(n) = T(n-1) + n
T(n-1) = T(n-2) + n-1
T(n-2) = T(n-3) + n-2
and so on you can substitute the value of T(n-1) and T(n-2) in T(n) to get a general idea of the pattern.
T(n) = T(n-2) + n-1 + n
T(n) = T(n-3) + n-2 + n-1 + n
.
.
.
T(n) = T(n-k) + kn - k(k-1)/2 ...(1)
For base case:
n - k = 1 so we can get T(1)
=> k = n - 1
substitute in (1)
T(n) = T(1) + (n-1)n - (n-1)(n-2)/2
Which you can see is of Order n2 => O(n2).
Expand it!
T(n) = T(n-1) + n = T(n-2) + (n-1) + n = T(n-3) + (n-2) + (n-1) + n
and so on, until
T(n) = 1 + 2 + ... + n = n(n+1)/2 [= O(n^2)]
provided that T(1) = 1
In pseudo code using iteration:
function T(n) {
int result = 0;
for (i in 1 ... n) {
result = result + i;
}
return result;
}
Another solution:
T(n) = T(n-1) + n
= T(n-2) + n-1 + n
= T(n-3) + n-2 + n-1 + n
// we can now generalize to k
= T(n-k) + n-k+1 + n-k+2 + ... + n-1 + n
// since n-k = 1 so T(1) = 1
= 1 + 2 + ... + n //Here
= n(n-1)/2
= n^2/2 - n/2
// we take the dominating term which is n^2*1/2 therefor 1/2 = big O
= big O(n^2)
Easy Method:
T (n) = T (n - 1) + (n )-----------(1)
//now submit T(n-1)=t(n)
T(n-1)=T((n-1)-1)+((n-1))
T(n-1)=T(n-2)+n-1---------------(2)
now submit (2) in (1) you will get
i.e T(n)=[T(n-2)+n-1]+(n)
T(n)=T(n-2)+2n-1 //simplified--------------(3)
now, T(n-2)=t(n)
T(n-2)=T((n-2)-2)+[2(n-2)-1]
T(n-2)=T(n-4)+2n-5---------------(4)
now submit (4) in (2) you will get
i.e T(n)=[T(n-4)+2n-5]+(2n-1)
T(n)=T(n-4)+4n-6 //simplified
............
T(n)=T(n-k)+kn-6
**Based on General form T(n)=T(n-k)+k, **
now, assume n-k=1 we know T(1)=1
k=n-1
T(n)=T(n-(n-1))+(n-1)n-6
T(n)=T(1)+n^2-n-10
According to the complexity 6 is constant
So , Finally O(n^2)