Time Complexity Problem [closed]?

If the second machine (T_2) was 32 times as fast then T_1(n) = 32*T_2(m). So you can make the equation n^2 = 32*m^2 so m = sqrt( (n^2)/32 ) I think the key point of confusion for some is that for any given data size, the machine T_1 is 32 times slower than T_2. Once you have this, you then plug in the complexity equations for the two machines and get a resulting equation which you can easily re-arrange to solve.

Discussion moved to chat. – Robert Harvey? Sep 16 at 1:34.

Rewording would make this one easier to understand: t = 2n^2. T = 2m^2 / 32. What is m in terms of n?

Remember, t = T(n), which probably includes some constant regarding the machine. However, we know that the second machine is exactly 32 times faster, so we have no unknowns any more. The rewording is just that, a rewording.

We have some time, t, that is given. In this time, we can process n items. We also know that t = T(n) = 2n^2.In the same time, t, we can process another number of items, m.

We also know that the second machine is 32 times faster, so the time it takes is T(n)/32, or 1/32th of the time it takes the first machine. Rewriting these gives the first two equations.

Oh! I got it... thanks! – Nayefc Sep 15 at 1:58 So, substitute equation 1 into equation 2, and solve for m?

:) – Merlyn Morgan-Graham Sep 15 at 2:02 Yeah thats what it is – Nayefc Sep 15 at 2:05 @Merlyn Morgan-Graham what coefficient? 2's cancel out. – Nayefc Sep 15 at 2:22.

Complexity is a measure of the number of operations it takes to complete a task. Execution time is the ration of algorithm complexity and "processor speed" t = O / cpu So if the original run time is t, we can write two equations and solve for x: Original statement: t seconds = 2n^2 / cpu speed 32x faster with the same time: t seconds = 2*(x*n)^2 / (32*cpu speed).

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