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Can anyone help me to understand that Is 2^(n^2 )=Θ(2^(n^3 )) ? it will be great if also provide the proof for this. As per my view this does not need to be equal.
The given assumption is not true:
First of all, 2^(n^2) is a function and Theta(2^(n^3)) is a set of functions, so it would be correct to say that 2^(n^2) ∈ Theta(2^(n^3)). The = is just a common abuse of notation, but it actually means ∈. To find out whether that statement is true, solve the following limit:
lim (n->infinity) of (2^(n^2)) / (2^(n^3))
If the result is 0 or infinite then the function does not belong to that particular Theta class. If it is some other value, it does belong to that class.
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I try to increase number in while-loop as in c\c++, but its not correct working, whats wrong?
summ=0
i=0
while summ<p do
summ=p0+aug
i=i+1
end
There are several things wrong with the provided code.
p is nil which will cause an error for comparing a nil value with a number value.
p0 and aug are both nil which will cause an error for performing arithmetic operations on a nil value
Assuming that p, p0 and aug weren't nil they would not change value within the while loop. Hence p0+aug is a constant. So depending on p0+aug>p your loop would either never start, stop after the first run or run forever.
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This may have been asked before but none of the answers I saw worked for me. I tried Lucas Theorem,Fermat's theorem but none of them worked. Is there an efficient way to find the value of:
nCr mod 10^9+7 where n<=10^9 and r<=1000
Any help will be very useful
n is large while r is small, you are better off compute nCr by n(n-1)...(n-r+1)/(1*2*...*r)
You may need to find multiplicate inverse of 1, 2, ... r mod 10^9+7
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An electric current, I, in amps, is given by
I=cos(wt)+√(8)sin(wt),
where w≠0 is a constant. What are the maximum and minimum values of I?
I have tried finding the derivative, but after that, I do not know how to solve for 0 because of the constant w.
Well David,you can convert this function into one trigonometric function by multiplying and dividing it by
√(1^2 + 8) i.e, 3. So your function becomes like this
I = 3*(1/3 cos(wt) + √8/3 sin(wt))
= 3* sin(wt + atan(1/√8))
Now, you can easily say its maximum value is
I = 3 amp
and minimum value is
I = 0 amp.
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I need to prove that for every k, there's a DFA M with k+2 states, so in every automat M' who accepts the language reverse(L(M)) there are at least 2^k states.
Help would be really appreciated.
Thanks :)
Assuming that the alphabet set contains at least two elements, let it be {0,1}.
Next, let M be the automata accepting the language L defined as:
All the strings which k-th position is 1
defined as:
M = {Q,{0,1},q0,{qk+1},δ}, where
Q={q0,q1,...,qk,qF}
δ(qi,a) = qi+1, for a in {0,1} and i=0,1,...,k-2
δ(qk-1,0) = qF,
δ(qk-1,1) = qk,
δ(qF,a) = qF, for a in {0,1}
Note that M has exactly k+2 states, and that it accepts the language L.
Now, note that the language reverse(L(M)) can be translated as:
All the strings which k-th position from the end is 1
To recognize that language, note that we need to remember the last k symbols, because we don't know when the string will end. We know that there are at least 2k possible strings of length k (since the alphabet size is at least 2).
So using a DFA, we need at least 2k states, each to represent one possible string of length k. ▢
Author's note:
The idea of this proof is to find a language which is "easy" to be recognized in normal way, but "difficult" when is read backward. Through experience, I remember that fixing the k-th position from the beginning is "easy", while k-th position from the end is "difficult", hence my answer.
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If you evaluate the following in VBA, the output is True:
(2 And 3) = 2
Can someone explain this to me? Thanks!
You have:
2 (base 10) == 10 (base 2)
3 (base 10) == 11 (base 2)
Performing a bitwise AND gives
10 (base 2) == 2 (base 10)
0 = false, everything else is true in VB
I tried to find some link for MS that explained this but I could not.