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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

ok.

next?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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But, between us I agree with you and TheDude and will stick to the notation you mentioned.

If two processes have a running time of 1111 log(n) and (1 / 111 )n^3 which one is slower? For what n will they be equal? ( here we are using the computer notation for log(n) )

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

hi bobbym

1111log(n) is much better.n^3/111 is slower

i don't know when they are equal.

*Last edited by anonimnystefy (2012-01-25 09:51:45)*

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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You have no idea when they are equal?

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

nope.i mean i know i have to solve an equation,but i'm not sure it can be done.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Want to try it?

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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yup.how?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Enter this into alpha, just as you see it.

Plot[{1111Log[n,2],n^3/111},{n,0,30}]

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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hi bobbym

i used "solve" instead.i got a real solution- 29.3544.now i'm not sure if i should take 19 or 30,but i'm more on the side of 29.

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**bobbym****Administrator**- From: Bumpkinland
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The plot would have shown you a little more. For n< that value you are better off using the slower process!

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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hi bobbym

i know that.but the big oh gives us results for large enough numbers.i was trying to find the exact value at which they are equal.

btw i looked it up first,but i couldn't see the exact result.

*Last edited by anonimnystefy (2012-01-25 10:15:24)*

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**bobbym****Administrator**- From: Bumpkinland
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You can not have n, other than a whole number so 29 and 30 are all you need..

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

next?

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**bobbym****Administrator**- From: Bumpkinland
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One more thing first. Did you see the reason for solving graphically or otherwise?

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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hi bobbym

well not really.i know when the latter will be better and when the first will be better.i just need to know when they are equal.

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**bobbym****Administrator**- From: Bumpkinland
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Yes, but the exact point tells you which one is best for particular n's, while Big(O) only tells you for gazillions! If those were two different type sorts and you only had 20 objects to sort the n^3 / 111 is the right one to use.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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hi

well,duh.i'm saying that using the mixed you can do better than using just one.

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**bobbym****Administrator**- From: Bumpkinland
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f(n) = 100n^2, g(n) = n^4

f(10) = 10000, g(10) = 10000

f(50) = 250000, g(50) = 6250000

f(100) = 1,000,000, g(100)= 100,000,000

The above table and a graph show that f(n) grows more slowly than g(n). Would you agree that f(n) = O(g(n))?

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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hi

it should be g(50)=6250000.

yes.i would say that f(n) is O(g(n))

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**bobbym****Administrator**- From: Bumpkinland
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Correct! I have fixed the above question.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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hi bobbym

4 zeros!

next?

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**bobbym****Administrator**- From: Bumpkinland
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I see that my cyber pen ran out of cyber ink.

Order these common functions according to growth rate:

1, n^n, 2^n, log(n), n log(n), n^2, n, n^2

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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hi

by what,complexity?

1,log(n),n,n log(n),n^2 (twice?),2^n,n^n

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**bobbym****Administrator**- From: Bumpkinland
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Kayrect a mundo! You certainly know your Big(O).

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,658

well it's not mine.

next?and this should be the last one for today.

*Last edited by anonimnystefy (2012-01-25 10:55:53)*

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