 mersenneforum.org 14° Primality test and factorization of Lepore ( conjecture )
 Register FAQ Search Today's Posts Mark Forums Read  2017-12-26, 21:31   #34
CRGreathouse

Aug 2006

3·1,993 Posts Quote:
 Originally Posted by Alberico Lepore I'm sorry I was wrong. the time taken is 2 ^ x where x is the depth of the tree. x depends on how the number is composed and not on the size of the digits.
How common is it for 2^x to be smaller than, say, N^(1/4) where N is the number to be factored?   2017-12-26, 22:00   #35
Alberico Lepore

May 2017
ITALY

2×32×29 Posts Quote:
 Originally Posted by CRGreathouse How common is it for 2^x to be smaller than, say, N^(1/4) where N is the number to be factored?

it does not depend on the number of digits
but by the composition of the number.

[149016617-2*a^2]=-a*(a-n)=-Z , Z-2*(a-n)^2=a-n , a^2+n*a=149016617 Time 2^1

67586227 Time 2^11

1500000021500000077 Time 2^2
[1500000021500000077-2*a^2]=-Z , Z=a*(a-n), Z-2*(a-n)^2=-(a-n)*[(a-n)-(n)]=-Q , Q-2*[(a-n)-(n)]^2=(a-n)-(n)-(n) , a^2+n*a=1500000021500000077

Last fiddled with by Alberico Lepore on 2017-12-26 at 22:31 Reason: add 3°example   2017-12-26, 22:59   #36
CRGreathouse

Aug 2006

175B16 Posts Quote:
 Originally Posted by Alberico Lepore it does not depend on the number of digits but by the composition of the number.
Well, what compositions are typical in numbers with (say) 60-400 decimal digits, and how long could we expect such factorizations to take?   2017-12-26, 23:18   #37
Alberico Lepore

May 2017
ITALY

2×32×29 Posts Quote:
 Originally Posted by CRGreathouse Well, what compositions are typical in numbers with (say) 60-400 decimal digits, and how long could we expect such factorizations to take?
it can range from 2 ^ 1 to never factoring it

[2286236812985825512574302734392158213081458619476870141764565-2*a^2]=-a*(a-n)=-Z , Z-2*(a-n)^2=a-n , a^2+n*a=2286236812985825512574302734392158213081458619476870141764565   2017-12-26, 23:31   #38
CRGreathouse

Aug 2006

135338 Posts Quote:
 Originally Posted by Alberico Lepore it can range from 2 ^ 1 to never factoring it
Much like the algorithm "guess numbers at random until one divides your number". Do you have any reasons to think your semi-algorithm is superior?   2017-12-26, 23:35   #39
Alberico Lepore

May 2017
ITALY

10128 Posts Quote:
 Originally Posted by CRGreathouse Much like the algorithm "guess numbers at random until one divides your number". Do you have any reasons to think your semi-algorithm is superior?
In my opinion it is the opposite,
that is, you are unlucky if you catch a low number
however it should be implemented and tested
are you willing to do it?   2017-12-27, 03:50   #40
CRGreathouse

Aug 2006

175B16 Posts Quote:
 Originally Posted by Alberico Lepore In my opinion it is the opposite, that is, you are unlucky if you catch a low number
I'd just like to have more to go on than an opinion.

Quote:
 Originally Posted by Alberico Lepore however it should be implemented and tested are you willing to do it?
I don't see any likelihood of it succeeding, and my finances preclude me from taking on non-paying freelance work at this time. But if you manage, please let us know.   2017-12-27, 09:44 #41 Alberico Lepore   May 2017 ITALY 2×32×29 Posts ENG I have come to a conclusion (conjecture): when the difference in the product of the child nodes is equal to the difference of the product of the parent node then it means that a> (a-n) when the difference in the product of the child nodes is different from the difference in the product of the parent node then it means that a <(a-n) ITA Sono giunto ad una conclusione (congettura) : quando la differenza del prodotto dei nodi figli è uguale alla differenza del prodotto del nodo padre allora significa che a>(a-n) quando la differenza del prodotto dei nodi figli è diversa dalla differenza del prodotto del nodo padre allora significa che a<(a-n) Example 9967*6781=67586227 9967*6781=67586227 | 9967*6781=3186 6781*3595=24377695 | 6781-3595=3186 3595*409=1470355 | 3595-409=3186 2777*409=1135793 | 2777-409=2368 | 14287665=2077*5145 | 5145-2077=3068 1959*409=801231 | ...... 1141*409=466669 | ...... 409*323=132107 | |409-323=86 | 202455=409*495 | 495-409=86 323*237=76551 | 323-237=86 237*151=35787 | 237-151=86 151*65=9815 | 151-65=86 65*21=483 | 65-21=44 23*21=1365 FINE 23-21=2 what do you think about it? Last fiddled with by Alberico Lepore on 2017-12-27 at 10:05 Reason: Format   2017-12-27, 09:59 #42 Alberico Lepore   May 2017 ITALY 2×32×29 Posts ENG If the conjecture was true then to face the problem of the small number (in this case the 409) we can make a log_2 at each step so a disadvantage would become an advantage ITA Se la congettura fosse vera allora per affrontare il problema del numero piccolo (in questo caso il 409) possiamo fare un log_2 ad ogni step così uno svantaggio diventerebbe un vantaggio what do you think about it?   2017-12-27, 14:41 #43 Alberico Lepore   May 2017 ITALY 2·32·29 Posts If the conjecture was true the computational complexity is k * log_2 (N) where k is a constant that depends on the number that is, it is the depth of the tree without repetitions   2017-12-28, 00:10   #44
CRGreathouse

Aug 2006

3×1,993 Posts Quote:
 Originally Posted by Alberico Lepore If the conjecture was true the computational complexity is k * log_2 (N) where k is a constant that depends on the number that is, it is the depth of the tree without repetitions
If it depends on the number, it's not a constant. What kind of dependency does it have?   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post carpetpool Miscellaneous Math 5 2018-02-05 05:20 Alberico Lepore Alberico Lepore 43 2018-01-17 15:55 Alberico Lepore Alberico Lepore 2 2018-01-01 21:31 Alberico Lepore Alberico Lepore 26 2017-12-17 18:44 Alberico Lepore Alberico Lepore 61 2017-09-23 21:52

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