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 2011-03-28, 23:12 #1 Christenson     Dec 2010 Monticello 5×359 Posts TF Credit oddities Hi: I am running manual tests on a low-end GPU (mfaktc 0.16p1 on Galaxy GEForce210, with CUDA capability 1.2), and one core of a standard AMD 6-core CPU on a completely different machine doing TF work. Primenet seems to be granting the CPU much more credit for easier work than the GPU, as follows from my Primenet results details report: CPU Name Exponent Result-Type Received age-days Result GHz-Days Manual testing 82303141 NF 2011-03-28 13:27 6.6 no factor for M82303141 from 2^68 to 2^69 [mfaktc 0.16p1-Win 71bit_mul24] 0.7264 Eric-AMD-6-core 54008443 NF 2011-03-28 18:53 2.6 no factor from 2^68 to 2^69 1.1069 Why is this?
 2011-03-29, 00:55 #2 ATH Einyen     Dec 2003 Denmark 65728 Posts You are testing factors of the form 2*k*p, so for p=82303141 it is less work to factor to 2^69 than for p=54008443: 82303141: k = 2^69 / 2*82303141 = 3,586,107,426,682 54008443: k = 2^69 / 2*54008443 = 5,464,847,508,738 0.7264 credits * 82303141/54008443 = 1,10696 credits. Last fiddled with by ATH on 2011-03-29 at 00:57
2011-03-29, 02:11   #3

"Richard B. Woods"
Aug 2002
Wisconsin USA

22·3·641 Posts

Quote:
 Originally Posted by Christenson Primenet seems to be granting the CPU much more credit for easier work than the GPU, as follows from my Primenet results details report: CPU Name Exponent Result-Type Received age-days Result GHz-Days Manual testing 82303141 NF 2011-03-28 13:27 6.6 no factor for M82303141 from 2^68 to 2^69 [mfaktc 0.16p1-Win 71bit_mul24] 0.7264 Eric-AMD-6-core 54008443 NF 2011-03-28 18:53 2.6 no factor from 2^68 to 2^69 1.1069 Why is this?
As ATH's example illustrates, for TF (unlike other factoring methods, seemingly, at first glance) the work gets "easier" as exponents go higher.

Between any two given powers-of-2, the number of candidate factors for a higher exponent is less than the number of candidate factors for a lower exponent. Thus, if the power-of-2 limits are held constant, TF has fewer and fewer candidates to test as exponents go higher.

There is an O(log exponent) factor in the time needed for each candidate test as exponents go higher, but it is swamped by the 1/n decrease in number of candidates during that progression.

Last fiddled with by cheesehead on 2011-03-29 at 02:33

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