How to test your cell phone warranty

[starrynte]

Here is a version not yet implementing the "fast modulus", but it now uses text output which seems to be faster, plus a few minor optimizations.

[henryzz]

Originally Posted by starrynte

Here is a version not yet implementing the "fast modulus", but it now uses text output which seems to be faster, plus a few minor optimizations.

i will test on my slowcoach phone
hopefully now it is text based i will be able to run overnight tests without it stopping(you cant measure cpu time can you?)

[henryzz]

Originally Posted by henryzz

i will test on my slowcoach phone
hopefully now it is text based i will be able to run overnight tests without it stopping(you cant measure cpu time can you?)

i left my phone at school but on my mums(a W580i which was almost twice as fast as mine) it still stops after a while

[Damian]

I tested starrynte's new version (I think you should number your versions) and compared it to the previous version.

On my f305 it seems the new version is slower than the previous one, in fact, the older version was somewhat 1.014 times faster than the new.

#    Mersenne  old midlet        new midlet         ratio
m13 | M521   | 0.38 sec        | 0.39 sec         | 1.026
m14 | M607   | 0.54 sec        | 0.56 sec         | 1.037
m15 | M1279  | 4.29 sec        | 4.4 sec          | 1.026 
m16 | M2203  | 20.58 sec       | 20.9 sec         | 1.015
m17 | M2281  | 23.11 sec       | 23.46 sec        | 1.015
m18 | M3217  | 1 min 2.57 sec  | 1 min 3.43 sec   | 1.014

Does it run faster on other phones?

[starrynte]

Interesting. It was (much) slower on my phone too, but faster on the emulator...
From what I read from a quick glance over Google, you can't detect CPU usage on mobile applications.
I'll try a precomputed table for the percentages (which won't count as part of the time needed for the test)
Also, I'll try "starting" the midlet when pauseApp() is called, assuming that when it "stops" it's only "paused"

[henryzz]

Originally Posted by starrynte

Also reposting questions:
1) Do all LLR tests perform <exponent - 2> iterations?
2) Can you do (mod k * 2^n - 1) after each iteration for
i) only where k=1?
ii) all LLR tests?
iii) something else?

read this http://isthe.com/chongo/src/calc/lucas-calc
here are my answers based on it
1) yes
2) ii

it also explains selecting all starting values(his program wont select absolutely all as he relies on tables for the variables D, a, and b)
also look at llr's source code for that

[henryzz]

1802 seconds/30 minutes to test 2^23209-1(M26) on one core of a Samsung Galaxy S II. Anyone ever tested higher?

[LaurV]

Originally Posted by henryzz

Samsung Galaxy S II

My daughter is owning one since 2 weeks (I am still Motorola C116 guy, hehe). Willing to try what the 2-cores MPU can do. Any link for the code/application you used?

[henryzz]

Originally Posted by LaurV

My daughter is owning one since 2 weeks (I am still Motorola C116 guy, hehe). Willing to try what the 2-cores MPU can do. Any link for the code/application you used?

I used the latest .jar above converted to apk with http://www.netmite.com/android/

[LaurV]

Sorry, did not see the thread has 2 pages , just now found the jar file. I will give it a try.

Originally Posted by starrynte

ok
expect a much smaller and faster version some time this week

(I found a faster way to do modulus,
a % 2^n - 1 = (a + 2^n - 1 * (a % 2^n)) / 2^n
and of course a % 2^n = a & (2^n - 1), and dividing by 2^n = shift right n bits
)

This still sounds complicate for me. To do x modulo m (m=2^n-1) you just have to add the higher n bits of x with the lower n bits of x (x<m^2 if it comes from squaring) and if the result is m, then make it 0, else you got the right result. If you do LL, you will never get a 0 "on the way", just the last one can be 0, so you don't really need the test.

This works because if y=x mod m, then y=m*k+x, so y=k*(2^n-1)+x that you can write like x=(y+k)-k*(2^n).

So your "mod 2^n-1" operation is just a mask on n bits, a shift on n bits, and an addition on n bits (maximum).
For example, 13*13 (mod 31) = (binarry) 1101*1101=101 01001 (note the space) and you have to add 101+01001=01110=14(decimal) that is 169 mod 31.

[science_man_88]

Originally Posted by LaurV

Sorry, did not see the thread has 2 pages , just now found the jar file. I will give it a try.




This still sounds complicate for me. To do x modulo m (m=2^n-1) you just have to add the higher n bits of x with the lower n bits of x (x<m^2 if it comes from squaring) and if the result is m, then make it 0, else you got the right result. If you do LL, you will never get a 0 "on the way", just the last one can be 0, so you don't really need the test.

This works because if y=x mod m, then y=m*k+x, so y=k*(2^n-1)+x that you can write like x=(y+k)-k*(2^n).

So your "mod 2^n-1" operation is just a mask on n bits, a shift on n bits, and an addition on n bits (maximum).
For example, 13*13 (mod 31) = (binarry) 1101*1101=101 01001 (note the space) and you have to add 101+01001=01110=14(decimal) that is 169 mod 31.

flaw:

(13:13)>binary(128)
%15 = [1, 0, 0, 0, 0, 0, 0, 0]
(13:18)>[1,0,0]+[0,0,0]
%16 = [1, 0, 0]
(13:18)>128%7
%17 = 2
(13:19)>2^3-1
%18 = 7

also I don't see how you get the last equation. according to you to do 128 mod (7=2^3-1) I take 100 + 000 = 100 and solve it as 4 but 128 = (2*49) + (4*7) +2 so the correct result is 2.