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2. The Math of Asymptotics
Solve the two following problems:
• Algorithm A uses 8n primitive operations, while Algorithm B uses n2 primitive operations. Determine the value n0 such that A runs faster than B for all n ≥ n0.
• Algorithm A uses 10n primitive operations, while Algorithm B uses 2nlog n prim- itive operations. Determine the value n0 such that A runs faster than B for all n ≥ n0.
• Algorithm A uses 10nlog n primitive operations, while Algorithm B uses n2 prim- itive operations. Determine the value n0 such that A runs faster than B for all n ≥ n0.
Remark: log n stands for log2 n (logarithm in base 2 of n).

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1 Antwort

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8 n < n 2

<=> n 2 - 8 n > 0

<=> n 2 - 8 n + 16 > 16

<=> ( n - 4 ) 2 > 16

<=> | n - 4 | > 4

<=> n - 4 > 4 OR - n + 4 > 4

<=> n > 8 OR n < 0

Since n < 0 is meaningless concerning the given problem, the (only) solution is:

n0 > 8

 

10 n < 2 n log n

<=> 5 < log n

<=> 2 5 < n

<=> n > 2 5 =  32

Solution is: n0  > 32

 

10 n log n < n 2

<=> 10 log n < n 

<=> log n < n / 10

<=> n < 2 n / 10

<=> n 10 < 2 n

<=> ...

This can not be calculated in closed form.
Using a numerical algorithm (or just the method of trial and error) gives:

n0 ≥ 59

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