Rewrote the solved question below for a reference:
Original question:
T(n) = 2T(n - 1) - 1, if n > 0
1, otherwise
1st iteration
T(n) = 2T(n - 1) - 1
T(n - 1) = 2T(n - 1 - 1) - 1
T(n - 1) = 2T(n - 2) - 1
2nd iteration
T(n) = 2(2T(n - 2) - 1) - 1
T(n) = 4T(n - 2) - 2 - 1
T(n) = 4T(n - 2) - 3
T(n - 2) = 2T(n - 2 - 1) - 1
T(n - 2) = 2T(n - 3) - 1
3rd iteration
T(n) = 4(2T(n - 3) - 1) - 3
T(n) = 8T(n - 3) - 4 - 3
T(n) = 8T(n - 3) - 7
T(n - 3) = 2T(n - 3 - 1) - 1
T(n - 3) = 2T(n - 4) - 1
4th iteration
T(n) = 8T(n - 3) - 7
T(n) = 8(2T(n - 4) - 1) - 7
T(n) = 16T(n - 4) - 8 - 7
T(n) = 16T(n - 4) - 15
Final table
At k=1, T(n) = 2T(n - 1) - 1
At k=2, T(n) = 4T(n - 2) - 3
At k=3, T(n) = 8T(n - 3) - 7
At k=4, T(n) = 16T(n - 4) - 15
T(n) = 2ᴷT(n - k) - (2ᴷ - 1)
At k=n T(n) = 2ᴺT(n - n) - (2ᴺ - 1)
T(n) = 2ᴺT(0) - (2ᴺ - 1)
T(n) = 2ᴺ - (2ᴺ - 1)
T(n) = 2ᴺ - 2ᴺ + 1
T(n) = 1
The inductive proof that T(n) = 1 looks like this:
Induction start.
What is T(0)? 1.
Induction step.
Assume for any n >= 0 that T(n) = 1. Now let's see if the claim holds for T(n+1).
T(n+1) = 2*T(n) - 1 = 2*1 - 1 = 1
Conclusion.
T(n) = 1 for all natural n. And 1 is in O(1).
Your error.
The first error I found in your calculation is here:
T(n) = 2²(2T(n - 3) - 1) - 3
T(n) = 2³T(n - 3) - 2 - 3
^ this should be a 4
T(n) = 2T(n-1) -1
= 2(2T(n-2)-1) - 1 = 4T(n-2)-3 = (2^2)T(n-2)-(2^2-1)
= 2(4T(n-3)-3) - 1 = 8T(n-3)-7 = (2^3)T(n-3)-(2^3-1)
...
= (2^n)T(n-n)-(2^n-1) = (2^n)1 - 2^n + 1 = 1
Related
I have tried recursion-tree but failed. I also could not solve the problem using math skills...and the Master theorem also seemed unavailable.
If we only define T(1) then we always have T(2) = 1 regardless of what T(1) is. So the answer is T(n) = 1 it comes from the fact that if T(n-1) = 1 then T(n) = [(n-2) * 1 + 2]/n = 1.
So let us suppose T starts from 2 rather than 1. Consider multiple scenarios :
T(2) > 1 : Then for n>2 we have T(n) < T(n - 1), also T(n) > 1 so limit T(n) = 1 :
because if T(n-1) > 1 then T(n) = [(n-2) * T(n-1) + 2]/n we have : T(n-1) > 1 so [(n-2) * T(n-1) + 2]/n > n/n = 1. Also (n-2) * T(n) + 2 T(n) = (n-2) T(n-1) + 2 and we have T(n) > 1 so : (n-2) * (T(n) - T(n-1)) = 2 * (1 - T(n)) < 0 so T(n) - T(n-1) < 0 so T(n) < T(n-1).
You have to prove there are no boundaries other than 1.
T(2) < 1 (including negatives) : Using a similar approach to the one mentioned above(not exact) you could argue that for n>2 we have T(n) > T(n-1) and T(n) < 1. And again prove that limit T(n) = 1.
T(2) = 1 : As I mentioned above if T(2) = 1 then T(n) = 1.
PS: I didn't provide a complete solution because it seems to be homework. I hope you can figure out the rest.
I'm trying to find the big O of T(n) = 2T(n/2) + 1. I figured out that it is O(n) with the Master Theorem but I'm trying to solve it with recursion and getting stuck.
My solving process is
T(n) = 2T(n/2) + 1 <= c * n
(we know that T(n/2) <= c * n/2)
2T(n/2) + 1 <= 2 * c * n/2 +1 <= c * n
Now I get that 1 <= 0.
I thought about saying that
T(n/2) <= c/2 * n/2
But is it right to do so? I don't know if I've seen it before so I'm pretty stuck.
I'm not sure what you mean by "solve it with recursion". What you can do is to unroll the equation.
So first you can think of n as a power of 2: n = 2^k.
Then you can rewrite your recurrence equation as T(2^k) = 2T(2^(k-1)) + 1.
Now it is easy to unroll this:
T(2^k) = 2 T(2^(k-1)) + 1
= 2 (T(2^(k-2) + 1) + 1
= 2 (2 (T(2^(k-3) + 1) + 1) + 1
= 2 (2 (2 (T(2^(k-4) + 1) + 1) + 1) + 1
= ...
This goes down to k = 0 and hits the base case T(1) = a. In most cases one uses a = 0 or a = 1 for exercises, but it could be anything.
If we now get rid of the parentheses the equation looks like this:
T(n) = 2^k * a + 2^k + 2^(k-1) + 2^(k-2) + ... + 2^1 + 2^0
From the beginning we know that 2^k = n and we know 2^k + ... + 2 + 1 = 2^(k+1) -1. See this as a binary number consisting of only 1s e.g. 111 = 1000 - 1.
So this simplifies to
T(n) = 2^k * a + 2^(k+1) - 1
= 2^k * a + 2 * 2^k - 1
= n * a + 2 * n - 1
= n * (2 + a) - 1
And now one can see that T(n) is in O(n) as long as a is a constant.
Let's say I had the equation T = sum(A**n) for n from 1 to M.
Now let's say I knew M and T, but wanted A. How would I solve for A?
I want to do an exponential backoff in the event of an error, but I don't want the total time spent backing off to be greater than T, nor the maximum number of retries to exceed M. So I'd need to find A.
The closed-form solution for sum(A**n) for n from 1 to M is (A^(M+1) - 1) / (A - 1) - 1. To see this work, let M = 3 and A = 2. Then 2^1 + 2^2 + 2^3 = 14, and (2^4 - 1) / (2 - 1) - 1 = 15 / 1 - 1 = 14.
So, we have the closed form expression T = (A ^ (M + 1) - 1) / (A - 1) - 1. This is a transcendental equation and has no closed-form solution. However, because the RHS is monotonically increasing in A (bigger values of A always give bigger values of the expression) then we can do what amounts to binary search to find an answer to arbitrary precision:
L = 0
H = MAX(T, 2)
A = (L + H) / 2
while |(A ^ (M + 1) - 1) / (A - 1) - 1 - T| > precision
if (A ^ (M + 1) - 1) / (A - 1) - 1 > T then
H = A
else then
L = A
end if
A = (L + H) / 2
loop
Example: T = 14, M = 3, epsilon = 0.25
L = 0
H = MAX(15, 2) = 14
A = L + H / 2 = 7
|(A ^ (M + 1) - 1) / (A - 1) - 1 - T|
= 385 > 0.25
H = A = 7
A = (L + H) / 2 = 3.5
|(A ^ (M + 1) - 1) / (A - 1) - 1 - T|
= 44.625 > 0.25
H = A = 3.5
A = (L + H) / 2 = 1.75
|(A ^ (M + 1) - 1) / (A - 1) - 1 - T|
= 3.828125 > 0.25
L = A = 1.75
A = (L + H) / 2 = 2.625
|(A ^ (M + 1) - 1) / (A - 1) - 1 - T|
= 13.603515625 > 0.25
H = A = 2.625
A = (L + H) / 2 = 2.1875
|(A ^ (M + 1) - 1) / (A - 1) - 1 - T|
= 3.440185546875 > 0.25
H = A = 2.1875
A = (L + H) / 2 = 1.96875
|(A ^ (M + 1) - 1) / (A - 1) - 1 - T|
= 0.524444580078125 > 0.25
L = A = 1.96875
A = (L + H) / 2 = 2.078125
|(A ^ (M + 1) - 1) / (A - 1) - 1 - T|
= 1.371326446533203125 > 0.25
H = A = 2.078125
A = (L + H) / 2 = 2.0234375
|(A ^ (M + 1) - 1) / (A - 1) - 1 - T|
= 0.402295589447021484375 > 0.25
H = A = 2.0234375
A = (L + H) / 2 = 1.99609375
|(A ^ (M + 1) - 1) / (A - 1) - 1 - T|
= 0.066299498081207275390625 < 0.25
Solution: 1.99609375
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Consider the following recurrence relation.
T(n) = 5 if n <= 2
T(n-1) + n otherwise
Closed form solution for T(n) is
I got solution as n(n+1)/2 + 7 for all the values. But in my university exam they gave the solution n(n+1)/2 + 2. However this solution doesn't terminate at 5 for values n<2. Can some body please explain ?
Let's solve it; first let's expand in telescopic sums:
T(k) = T(k)
T(k + 1) = T(k) + k + 1
T(k + 2) = T(k + 1) + k + 2 = T(k) + k + 1 + k + 2
...
T(k + m) = T(k) + k + 1 + k + 2 + ... + k + m =
= T(k) + mk + 1 + 2 + ... + m =
= T(k) + mk + (1 + m) * m / 2
...
Now we have
T(k + m) = T(k) + mk + (1 + m) * m / 2
Let k = 2:
T(m + 2) = T(2) + 2m + (1 + m) * m / 2 = 5 + 2m + (1 + m) * m / 2
Finally, let m + 2 = n or m = n - 2:
T(n) = 5 + 2 * (n - 2) + (n - 1) * (n - 2) / 2 = n * (n + 1) / 2 + 2
We have
T(n) = n * (n + 1) / 2 + 2 when n > 2
T(n) = 5 when n <= 2
As a simple non-rigorous reasoning exercise, we known that T(n) = T(n-1) + n yields the sum of the first n numbers: S(n) = n * (n + 1) / 2
Now, when n=2, S(2) = 3, so the value of 5 is actually an increment by 5 - S(2) = 2. So we could say:
T(n) = S(n) + (5 - S(2)) for n >=2
or
T(n) = S(n) + 2 for n >= 2
T(n) = 5 for n <= 2
Note that at n=2, the two relations are identical.
Please verify my logic to see if what I'm attempting is valid or shady.
W(n) = W(n/2) + nlg(n)
W(1) = 1
n =2^k
By trying the pattern
line 1 : W (2^k) = W(2^k-1) + nlgn
line 2 : = W(2^k-2) + nlgn + nlgn
...
line i : = W(2^k-i) + i*nlgn
and then solve the rest for i.
I just want to make sure it's cool if I substitute in 2k in one place (on line 1) but not in the other for the n lg n.
I find by subbing in 2^k for 2^k lg(2^k) gets really greasy.
Any thoughts are welcome (specifically if I should be subbing in 2^k, and if I should then how would you suggest the solution)
It's fine to switch back and forth between n and 2k as needed because you're assuming that n = 2k. However, that doesn't mean that what you have above is correct. Remember that as n decreases, the value of n log n keeps decreasing as well, so it isn't the case that the statement
line i = W(2k-i) + i * n lg n
is true.
Let's use the iteration method one more time:
W(n) = W(n / 2) + n log n
= (W(n / 4) + (n/2) log (n/2)) + n log n
= W(n / 4) + (n/2) (log n - 1) + n log n
= W(n / 4) + n log n / 2 - n / 2 + n log n
= W(n / 4) + (1 + 1/2) n log n - n / 2
= (W(n / 8) + (n / 4) log(n/4)) + (1 + 1/2) n log n - n / 2
= W(n / 8) + (n / 4) (log n - 2) + (1 + 1/2) n log n - n / 2
= W(n / 8) + n log n / 4 - n / 2 + (1 + 1/2) log n - n / 2
= W(n / 8) + (1 + 1/2 + 1/4) n log n - n
= (W(n / 16) + (n / 8) log(n/8)) + (1 + 1/2 + 1/4) n log n - n
= W(n / 16) + (n / 8) (log n - 3)) + (1 + 1/2 + 1/4) n log n - n
= W(n / 16) + n log n / 8 - 3n / 8 + (1 + 1/2 + 1/4) n log n - n
= W(n / 16) + (1 + 1/2 + 1/4 + 1/8) n log n - n - 3/8n
We can look at this to see if we spot a pattern. First, notice that the n log n term has a coefficient of (1 + 1/2 + 1/4 + 1/8 + ...) as we keep expanding outward. This series converges to 2, so when the iteration stops that term will be between n log n and 2n log n. Next, look at the coefficient of the -n term. If you look closely, you'll notice that this coefficient is -1 times
(1 / 2 + 2 / 4 + 3 / 8 + 4 / 16 + 5 / 32 + ... )
This is the sum of i / 2i, which converges to 2. Therefore, if we iterate this process, we'll find at each step that the value is Θ(n log n) - Θ(n), so the overall recurrence solves to Θ(n log n).
Hope this helps!