How Large Can N Be for F(n) = Log N to Run in 1 Second?

[nuttynibbles]

Hi,

I'm doing this algorithm questions and i need to find the largest size n of a problem that can be solved in time t, assuming that the algorithm to solve the problem takes f(n) microseconds.

For example:

f(n) = log n
t = 1 seconds

how do i get the largest size of n in t time??

what i did is assume log is base 10, then:

lg x = y
x= 10^y
since:
f(n) = 1,000,000 microseconds
n = 10^1,000,000

am i right??

you can refer to the Q here: http://problems.datastructures.net/database/problems/compare-running-times/compare-running-times.pdf

 

[Char. Limit]

Hi,

I'm doing this algorithm questions and i need to find the largest size n of a problem that can be solved in time t, assuming that the algorithm to solve the problem takes f(n) microseconds.

For example:

f(n) = log n
t = 1 seconds

how do i get the largest size of n in t time??

what i did is assume log is base 10, then:

lg x = y
x= 10^y
since:
f(n) = 1,000,000 microseconds
n = 10^1,000,000

am i right??

you can refer to the Q here: http://problems.datastructures.net/database/problems/compare-running-times/compare-running-times.pdf

Well, first things first, you need to read the question a bit more carefully. The opening states clearly:

Note that log n means the logarithm in base 2 of n.

Other than that, you should be on the right track.

 

[nuttynibbles]

oh my.. tks

if it's base 2, then:

lg2 x = y
x= 2^y
since:
f(n) = 1,000,000 microseconds
n = 2^1,000,000

but why is the ans 10^3000000??

 

[Mentallic]

oh my.. tks

if it's base 2, then:

lg2 x = y
x= 2^y
since:
f(n) = 1,000,000 microseconds
n = 2^1,000,000

but why is the ans 10^3000000??

Do you mean 10^300,000 as opposed to 10^3,000,000? If

2^{10^6}=10^x

find x.

 

[nuttynibbles]

sorry i meant 300,000

since:
2^(10^6) = 10^x
That means:
since x = 10^6, therefore 10^x = 10^(10^6). but how do i get the value 10^300000??

 

[Char. Limit]

sorry i meant 300,000

since:
2^(10^6) = 10^x
That means:
10^(2^(10^6)) correct. but how do i calculate the value 10^300000?? i'll get error on my calculator

Not... quite. You have to take the log base 10 to isolate x, and then you get 10^6 log_10(2), or about 1000000 * log_10(2). That's the x you're looking for, and it's easy to prove that it's about 300000. Then you get that 2^(10^6) is about equal to 10^(300000).

 

[nuttynibbles]

hi char.limit, may i know which log rule define this??

 

[Mentallic]

hi char.limit, may i know which log rule define this??

If

a^b=c

then

b=\log_ac

So can you apply this to your example?

 

[nuttynibbles]

sorry but I am actually still confuse.. i don't really understand when char.limit said i can isolate x by taking log_10 of x and i get 10⁶ . log_10 (2).

i change x->y to avoid confusion. since the rule says:
y = log_10 x, then x=10^y

if we continue from
2¹⁰⁶ =10^y
then
y = log_10(2⁽¹⁰⁶⁾) ?
but how did char.limit get log_10(2).10⁶??

 

[nuttynibbles]

Hi,

i finally understood. the rule that i was looking for is
log_a(X^y) = y.log_a(X)

therefore,
2¹⁰⁶ =10^y
y = 10⁶ . log₁₀(2)

However, i got another Q. if the Q was asked to use log₂, must we give the final ans in log₁₀??

 

[Mentallic]

Hi,

i finally understood. the rule that i was looking for is
log_a(X^y) = y.log_a(X)

therefore,
2¹⁰⁶ =10^y
y = 10⁶ . log₁₀(2)

Yes, that's right

However, i got another Q. if the Q was asked to use log₂, must we give the final ans in log₁₀??

No, it's not necessary. I'd usually leave the answer as it is unless I was specifically asked to convert it into that form.

 

[nuttynibbles]

hmm okay tks a lot peeps