Comments on: Half a number, half it again and add it to the first half. http://www.parolski.com/2008/12/03/half-a-number-half-it-again-and-add-it-to-the-first-half/ Faith, Solaris, and Chicken Korma, by Anton Parol Mon, 20 Jun 2011 20:37:35 +0000 hourly 1 http://wordpress.org/?v=3.3 By: Anton Parol http://www.parolski.com/2008/12/03/half-a-number-half-it-again-and-add-it-to-the-first-half/comment-page-1/#comment-445 Anton Parol Fri, 05 Dec 2008 10:14:27 +0000 http://www.parolski.com/?p=136#comment-445 Liam. I've got an interesting reply coming up, once I've the chance to write it up!! Anton Liam. I’ve got an interesting reply coming up, once I’ve the chance to write it up!!

Anton

]]> By: Liam McBrien http://www.parolski.com/2008/12/03/half-a-number-half-it-again-and-add-it-to-the-first-half/comment-page-1/#comment-443 Liam McBrien Thu, 04 Dec 2008 14:04:56 +0000 http://www.parolski.com/?p=136#comment-443 Of course, you could just do away with the loop and use a formula: (starting number) - (starting number / (2 ^ (number of iterations)) Since that's effectively what you're doing. Adding half of a half to a half is the same as subtracting a quarter - which is a half of a half. So if we start with 10, iterations 1 to 5 would be represented by: 10 - (10 / 2^1) => 10 - (10 / 2) => 10 - 5 => 5 10 - (10 / 2^2) => 10 - (10 / 4) => 10 - 2.5 => 7.5 10 - (10 / 2^3) => 10 - (10 / 8) => 10 - 1.25 => 8.75 10 - (10 / 2^4) => 10 - (10 / 16) => 10 - 0.625 => 9.375 10 - (10 / 2^5) => 10 - (10 / 32) => 10 - 0.3125 => 9.6875 This also definitively answers the question of whether you ever reach ten - you won't, since you're always subtracting the result of a division by a power of two (which becomes infinitesimal for large powers, but will never equal zero). It's also a good lesson in optimisation. I haven't tried the Java implementation, but I'll bet it runs quicker than a loop! Right enough, a significant amount of the time spent in your program is probably all the System.out.println() calls - these actually take up a suprising amount of execution time when you're doing stuff like this (it is I/O, after all)! Of course, you could just do away with the loop and use a formula:

(starting number) – (starting number / (2 ^ (number of iterations))

Since that’s effectively what you’re doing. Adding half of a half to a half is the same as subtracting a quarter – which is a half of a half.

So if we start with 10, iterations 1 to 5 would be represented by:

10 – (10 / 2^1) => 10 – (10 / 2) => 10 – 5 => 5
10 – (10 / 2^2) => 10 – (10 / 4) => 10 – 2.5 => 7.5
10 – (10 / 2^3) => 10 – (10 / 8) => 10 – 1.25 => 8.75
10 – (10 / 2^4) => 10 – (10 / 16) => 10 – 0.625 => 9.375
10 – (10 / 2^5) => 10 – (10 / 32) => 10 – 0.3125 => 9.6875

This also definitively answers the question of whether you ever reach ten – you won’t, since you’re always subtracting the result of a division by a power of two (which becomes infinitesimal for large powers, but will never equal zero).

It’s also a good lesson in optimisation. I haven’t tried the Java implementation, but I’ll bet it runs quicker than a loop! Right enough, a significant amount of the time spent in your program is probably all the System.out.println() calls – these actually take up a suprising amount of execution time when you’re doing stuff like this (it is I/O, after all)!

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