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Old 2017-12-26, 21:31   #34
CRGreathouse
 
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Quote:
Originally Posted by Alberico Lepore View Post
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?
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Old 2017-12-26, 22:00   #35
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Originally Posted by CRGreathouse View Post
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
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Old 2017-12-26, 22:59   #36
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Quote:
Originally Posted by Alberico Lepore View Post
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?
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Old 2017-12-26, 23:18   #37
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Quote:
Originally Posted by CRGreathouse View Post
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
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Old 2017-12-26, 23:31   #38
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Quote:
Originally Posted by Alberico Lepore View Post
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?
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Old 2017-12-26, 23:35   #39
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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?
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Old 2017-12-27, 03:50   #40
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Quote:
Originally Posted by Alberico Lepore View Post
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 View Post
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.
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Old 2017-12-27, 09:44   #41
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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
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Old 2017-12-27, 09:59   #42
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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?
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Old 2017-12-27, 14:41   #43
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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
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Old 2017-12-28, 00:10   #44
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Quote:
Originally Posted by Alberico Lepore View Post
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?
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