I have tried to give a more precise approximation to Sieve of Eratosthenes.
Elementary operations and weights that I used:
prime[p] -> 1 operation
m = p * p -> 2 operations
prime[m] = false -> 1 operation
m = m + p -> 2 operations
My proof:
Is my proof correct? I found in the literature that the complexity is O(nlog(log(n))) or O(nlog(log(n))/log(n)).
Yes, that's correct, O(nloglogn)==O(nloglog(sqrt(n))):
Related
I struggle to fill this table in even though I took calculus recently and good at math. It is only specified in the chapter how to deal with lim(n^k/c^n), but I have no idea how to compare other functions. I checked the solution manual and no info on that, only a table with answers which provides little insight.
When I solve these I don't really think about limits -- I lean on a couple facts and some well-known properties of big-O notation.
Fact 1: for all functions f and g and all exponents p > 0, we have f(n) = O(g(n)) if and only if f(n)p = O(g(n)p), and likewise with o, Ω, ω, and Θ respectively. This has a straightforward proof from the definition; you just have to raise the constant c to the power p as well.
Fact 2: for all exponents ε > 0, the function lg(n) is o(nε). This follows from l'Hôpital's rule for limits: lim lg(n)/nε = lim (lg(e)/n)/(ε nε−1) = (lg(e)/ε) lim n−ε = 0.
Fact 3:
If f(n) ≤ g(n) + O(1), then 2f(n) = O(2g(n)).
If f(n) ≤ g(n) − ω(1), then 2f(n) = o(2g(n)).
If f(n) ≥ g(n) − O(1), then 2f(n) = Ω(2g(n)).
If f(n) ≥ g(n) + ω(1), then 2f(n) = ω(2g(n)).
Fact 4: lg(n!) = Θ(n lg(n)). The proof uses Stirling's approximation.
To solve (a), use Fact 1 to raise both sides to the power of 1/k and apply Fact 2.
To solve (b), rewrite nk = 2lg(n)k and cn = 2lg(c)n, prove that lg(c) n − lg(n) k = ω(1), and apply Fact 3.
(c) is special. nsin(n) ends up anywhere between 0 and n. Since 0 is o(√n) and n is ω(√n), that's a solid row of NO.
To solve (d), observe that n ≥ n/2 + ω(1) and apply Fact 3.
To solve (e), rewrite nlg(c) = 2lg(n)lg(c) = 2lg(c)lg(n) = clg(n).
To solve (f), use Fact 4 and find that lg(n!) = Θ(n lg(n)) = lg(nn).
Exp(n)
If n = 0
Return 1
End If
If n%2==0
temp = Exp(n/2)
Return temp × temp
Else //n is odd
temp = Exp((n−1)/2)
Return temp × temp × 2
End if
how can i prove by strong induction in n that for all n ≥ 1, the number of multiplications made by
Exp (n) is ≤ 2 log2 n.
ps: Exp(n) = 2^n
A simple way is to use strong induction.
First, prove that Exp(0) terminates and returns 2^0.
Let N be some arbitrary even nonnegative number.
Assume the function Exp correctly calculates and returns 2^n for every n in [0, N].
Under this assumption, prove that Exp(N+1) and Exp(N+2) both terminate and correctly return 2^(N+1) and 2^(N+2).
You're done! By induction it follows that for any nonnegative N, Exp(N) correctly returns 2^N.
PS: Note that in this post, 2^N means "two to the power of N" and not "bitwise xor of the binary representations of 2 and N".
The program exactly applies the following recurrence:
P[0] = 1
n even -> P[n] -> P[n/2]²
n odd -> P[n] -> P[(n-1)/2]².2
the program always terminates, because for n>0, n/2 and (n-1)/2 < n and the argument of the recursive calls always decreases.
P[n] = 2^n is the solution of the recurrence. Indeed,
n = 0 -> 2^0 = 1
n = 2m -> 2^n = (2^m)²
n = 2m+1 -> 2^n = 2.(2^n)²
and this covers all cases.
As every call decreases the number of significant bits of n by one and performs one or two multiplications, the total number does not exceed two times the number of significant bits.
I am trying to prove that for any constant,k, log^k N = o(N) (little O of N)
All that I know for little o is that it follows the form T(n) = o(p(n)) where T(n) grows at a rate slower than p(n). Also I can't really do a limit and use L'hopital rule because I do not know what f(n) or g(n) is. Can someone please help getting me started!
You need to show that
lim (log^k N)/N = 0
N -> infinity
Write N = e^x, and that becomes
lim (x^k)/(e^x) = 0
Now, use the Power series expansion of the exponential function,
e^x = ∑ (x^n)/n!
to get an estimate for that quotient.
Spoiler: Picking the term with n = k+1 gives e^x > x^(k+1)/(k+1)! and from that easily follows (x^k)/(e^x) < (k+1)!/x.
I am trying to learn number theory for RSA cryptography by reading the CLR algorithms book. I was looking at exercise 31.2-5 which claims a bound of 1 + logΦ(b / gcd(a,b)).
The full question is:
If a > b >= 0, show that the invocation EUCLID(a,b) makes at most 1 + logΦb recursive calls. Improve this bound to 1 + logΦ(b / gcd(a,b)).
Does anyone know how to show this? There are already several other questions and answers to Euclid's algorithm on this site already but none of them seem to have this exact precise answer.
Refer to the analysis of Euclid's algorithm by Donald Knuth, in TAOCP Vol.2 p.356
This is how I solved the first part. I am still working on the second
We know that the runtime of Euclid algorithm is a function of the number of steps involved (pg of the book)
Let k be the number of recursive steps needed.
Hence b >= F k+1 >= φ k -1
b >= φ k -1
Taking log to find k we have
Log φ b >= k-1
1 + Log φ b > = k
Hence the run time is O (1 + Log φ b)
Show that if Euclid(a,b) takes more than N steps, then
a>=F(n+1) and b>=F(n), where F(i) is the ith Fibonacci
number. This can easily be done by Induction.
Show that F(n) ≥ φn-1, again by
Induction.
Using results of Step 1 and 2, we have b ≥ F(n) ≥
φn-1 Taking logarithm on both sides,
logφb ≥ n-1. Hence proved, n ≤ 1 +
logφb
The second part trivially follows.
No. of recursive calls in EUCLID(ka,kb) is the same as in EUCLID(a,b), where k is some integer.
Hence, the bound is improved to 1 + logφ( b / gcd(a,b) ).
f(n)=(log(n))^log(n)
g(n)= n/log(n)
f = O(g(n))?
Take the log of both sides:
log(f(n)) = log(log n) * log n
log(g(n)) = log(n) - log(log(n)) = log(n)(1 - log(log(n))/log(n))
Clearly log(log(n)) dominates (1 - log(log(n))/log(n)), so g is O(f). f is not O(g). Since it's homework, you may need to fill in the details.
It's also fairly easily to get an idea what the answer should be just by trying it with a large number. 1024 is 2^10, so taking n=1024:
f(n) = 10^10
g(n) = 1024/10.
Obviously that's not a proof, but I think we can see who's winning this race.
f(n) grows faster than g(n) if and only if f(en) also grows faster than g(en) since exp is strictly increasing to infinity (prove it yourself).
Now f(en) = nn and g(en) = en / n, and you can quote the known results.
If Limit[f[x] / g[x], x -> Infinity] = Infinity, then f[x] grows faster than g[x].
Limit[Log[x] ^ Log[x] / (x / Log[x]), x -> Infinity] = + Infinity
So, Log[x] ^ Log[x] grows faster than x / Log[x]
Mathematica gives the limit of f(n) / g(n) as n tends towards infinity as infinity, which means that f grows faster. This means that g(n) belongs to (=) O(f(n)).
You can use this for example if you don't have Mathematica.
f is vastly bigger. By n^loglog(n) -1 . log n