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 November 2nd, 2018, 12:55 PM #1 Newbie   Joined: Oct 2017 From: Rumba Posts: 29 Thanks: 0 Algorithms question help Can someone give me guidance how to do this question: http://prntscr.com/ldp4l8
 November 2nd, 2018, 02:15 PM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,192 Thanks: 489 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. Thanks from sita Last edited by skipjack; November 3rd, 2018 at 10:49 AM.
November 3rd, 2018, 08:37 AM   #3
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Quote:
 Originally Posted by JeffM1 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.

November 3rd, 2018, 10:36 AM   #4
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Quote:
 Originally Posted by sita If I did for $n_2$ $\displaystyle \frac{1}{2} * (6 * 10^7)^2 \approx what?$ Would that be right?
Yes.

$\displaystyle 60 * 10^6 = 6 * 10^1 * 10^6 = 6 * 10^{(1 + 6)} = 6 * 10^7 \implies \\ \displaystyle \frac{1}{2} * (60 * 10^6) = \frac{1}{2} * (6 * 10^7) = 3 * 10^7.$

Last edited by skipjack; November 3rd, 2018 at 10:56 AM.

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