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November 2nd, 2018, 12:55 PM   #1
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Algorithms question help

Can someone give me guidance how to do this question:
http://prntscr.com/ldp4l8
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November 2nd, 2018, 02:15 PM   #2
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This is basically just arithmetic.

You calculate

$\displaystyle \frac{1}{2} * (11 * 10^6)^2 \approx what?$

$\displaystyle \frac{1}{2} * (60 * 10^6)^2 \approx what?$

$(11 * 10^6) * \log_2(11 * 10^6) = (11 * 10^6)\{\log_2(11) + 6\log_2(10)\} = what?$

$(60 * 10^6) * \log_2(60 * 10^6) = (60 * 10^6)\{\log_2(60) + 6\log_2(10)\} = what?$

Then you fill out the rest of the table.
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Last edited by skipjack; November 3rd, 2018 at 10:49 AM.
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November 3rd, 2018, 08:37 AM   #3
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Quote:
Originally Posted by JeffM1 View Post
This is basically just arithmetic.

You calculate

$\displaystyle \frac{1}{2} * (11 * 10^6)^2 \approx what?$

$\displaystyle \frac{1}{2} * (60 * 10^6)^2 \approx what?$

$(11 * 10^6) * \log_2(11 * 10^6) = (11 * 10^6)\{\log_2(11) + 6\log_2(10)\} = what?$

$(60 * 10^6) * \log_2(60 * 10^6) = (60 * 10^6)\{\log_2(60) + 6\log_2(10)\} = what?$

Then you fill out the rest of the table.
If I did for $n_2$ $\displaystyle \frac{1}{2} * (6 * 10^7)^2 \approx what?$

Would that be right?

Last edited by skipjack; November 3rd, 2018 at 10:51 AM.
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November 3rd, 2018, 10:36 AM   #4
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Quote:
Originally Posted by sita View Post
If I did for $n_2$ $\displaystyle \frac{1}{2} * (6 * 10^7)^2 \approx what?$

Would that be right?
Yes.

$\displaystyle 60 * 10^6 = 6 * 10^1 * 10^6 = 6 * 10^{(1 + 6)} = 6 * 10^7 \implies \\

\displaystyle \frac{1}{2} * (60 * 10^6) = \frac{1}{2} * (6 * 10^7) = 3 * 10^7.$
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Last edited by skipjack; November 3rd, 2018 at 10:56 AM.
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