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# Complex numbers

### #1

Posted 26 February 2014 - 05:26 PM

### #2

Posted 27 February 2014 - 11:06 AM

I can do it with 3 multiplications and 4 additions. Still looking.

*Vidi vici veni.*

### #3

Posted 04 March 2014 - 12:17 PM

*Vidi vici veni.*

### #4

Posted 11 March 2014 - 10:02 PM

As nobody found the solution, can you post it?

### #5

Posted 13 March 2014 - 12:48 PM

I = (a+b)(c+d) - R =ac+bc+ad+bd - (ac - bd) ... = ... +2*bd ....

**Edited by harey, 13 March 2014 - 12:50 PM.**

### #6

Posted 13 March 2014 - 04:14 PM

@bonanova - sorry, but I believe you have a sign error with the factor bd:

I = (a+b)(c+d) - R =ac+bc+ad+bd - (ac - bd) ... = ... +2*bd ....

Right. It's (a+b)(c+d) - ac - bd.

*Vidi vici veni.*

### #7

Posted 13 March 2014 - 04:46 PM

### #8

Posted 15 March 2014 - 03:27 PM

If our hardware uses fewer clock cycles to perform three additions than a single multiplication, we may well gain overall processing speed by using Eq. (2-4) and Eq. (2-5) instead of Eq. (1-2) for complex multiplication

If it worked, it would be great for fractal pictures. The bad news is that multiplications are in the processor heavily optimized, a fair guess 4-5 additions. Googling:

The latency is 1 cycle for an integer addition and 3 cycles for an integer multiplication. You can find the latencies and thoughput in Appendix C of the "Intel 64 and IA-32 Architectures Optimization Reference Manual", which is located on http://www.intel.com...cessor/manuals/.

The big question is how the 3 temporary variables were treated by the compiler. If we have to access the RAM...

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