Can I decrease the factorization time with this formula ?
 

[B]Formula ( 3 Steps ) :[/B]

[B]Note :[/B] [TEX]\frac{1}{3}= 0.333[/TEX]

[B]Step 1 :[/B]

[TEX]prime number * prime number = product[/TEX]

[TEX]2 * \sqrt{0,333 * product} = N1[/TEX]

[B]Step 2 :[/B]

[TEX]\sqrt{product} = N2[/TEX]

[B]Step 3 :[/B]

[TEX]N1 + N2 = N3[/TEX]

[B]First Way[/B] [TEX]\frac{product}{N3} * 3,5 = N4[/TEX] [B]It work always (for prime factors similar value or distant)[/B]

[B]Second Way [/B] [TEX]\frac{product}{N3} = N4[/TEX] [B]It work only for prime factors distant value[/B]


[TEX]prime number < N4[/TEX]



[B]Example First Way :[/B]


[B]Step 1 :[/B]


[TEX]101 * 211 = 21311[/TEX]


[TEX]2 * \sqrt{0,333 * 21311} =168,... [/TEX] [TEX]N1[/TEX]

[B]Step 2 :[/B]

[TEX]\sqrt{21311} = 145,...[/TEX] [TEX]N2[/TEX]

[B]Step 3 :[/B]

[TEX]168 + 145 = 313 [/TEX] [TEX]N3[/TEX]

[B]First Way[/B] [TEX]\frac{21311}{313} * 3,5 = 238,..[/TEX] [TEX]N4[/TEX] [B]It work always (for prime factors similar value or distant)[/B]

[B]Second Way [/B] [TEX]\frac{21311}{313} = 68,..[/TEX] [TEX]N4[/TEX] [B]It work only for prime factors distant value[/B]


[TEX]prime number < N4[/TEX]


[TEX]211 < 238[/TEX]

--EDIT--


[B]Can I decrease the factorization time with this formula for [U]big numbers[/U]?[/B]



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What is the purpose of the "Second way" if the first one will "[B]It work always (for prime factors similar value or distant)[/B][B]"?

[/B]And by looking at it and simplifying it, [U]sqrt[B]([/B]product) [B]<[/B] N3 [B]<[/B] product[/U]. Always.
[ N3 = ( sqrt(3)+2 )/ sqrt(3) * sqrt(product) ] which is ~2,1547[SIZE=4][B] *[/B][/SIZE]sqrt(product)

N4 = product / (~2,15*sqrt(product) )* 3,5 <- which is like (sqrt(product)/~2,15 ) * 3,5 =
sqrt prod [SIZE=4]*[/SIZE] ~1,6 <- rough rounding
[I]
^- It might be wrong, and a mess, I'm little tired.[/I] [I]Also it's not scientific in any way.[/I]

Ok, so
product = 1852927
sqrt is 1361,222612213
N3 by using sqrt(3)s is 2933,027095
N4 by using ...*3,5 is 2211,109645
So what are the factors? (yafu: 0.0090 sec)

[QUOTE=thyw;452045]What is the purpose of the "Second way" if the first one will "[B]It work always (for prime factors similar value or distant)[/B][B]"?

[/B]And by looking at it and simplifying it, [U]sqrt[B]([/B]product) [B]<[/B] N3 [B]<[/B] product[/U]. Always.
[ N3 = ( sqrt(3)+2 )/ sqrt(3) * sqrt(product) ] which is ~2,1547[SIZE=4][B] *[/B][/SIZE]sqrt(product)

N4 = product / (~2,15*sqrt(product) )* 3,5 <- which is like (sqrt(product)/~2,15 ) * 3,5 =
sqrt prod [SIZE=4]*[/SIZE] ~1,6 <- rough rounding
[I]
^- It might be wrong, and a mess, I'm little tired.[/I] [I]Also it's not scientific in any way.[/I]

Ok, so
product = 1852927
sqrt is 1361,222612213
N3 by using sqrt(3)s is 2933,027095
N4 by using ...*3,5 is 2211,109645
So what are the factors? (yafu: 0.0090 sec)[/QUOTE]

Ok , [TEX]1277 * 1471 = 1852927 [/TEX]

[TEX]2 * \sqrt{0,333* 1852927} = 1571,...N1[/TEX]

[TEX]\sqrt{1852927} = 1361,...N2[/TEX]

[TEX]1361 + 1571 = 2932 [/TEX] [TEX]N3[/TEX]

[B]First Way[/B] [TEX]\frac{1852927}{2932} * 3.5 = 2211,...N4[/TEX]

[B]Second Way[/B] [TEX]\frac{1852927}{2932} = 631,...N4[/TEX]


Now trying First Way and Second Way , but only First Way is correct [TEX]1471 < 2211[/TEX] , than [TEX]\frac{1852927}{2211} [/TEX] .....[TEX]\frac{1852927}{2209} [/TEX] ...continue until [TEX]\frac{1852927}{1471} [/TEX]

[B]I think that for large numbers ( 300 digits ), could to be efficient ......[/B]


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[QUOTE=thyw;452045]What is the purpose of the "Second way" if the first one will "[B]It work always (for prime factors similar value or distant)[/B][B]"?


[/QUOTE]

997 * 3 = 2991

First way = N4 by using ...*3,5 is 89

Second Way = N4 is 25

[QUOTE=Godzilla;452030]Can I decrease the factorization time with this formula for [U]big numbers[/U]?[/QUOTE]

No, because you don't give any actionable suggestions/algorithms/etc. All you give are four numbers with no obvious connection to anything. But let's look at them anyway! :smile:

Let the product be N and its square root be S. Then

N1 = sqrt(3330)/50 * S = 1.154...S
N2 = S
N3 = (1 + sqrt(3330)/50)*S = 2.154...S
N4 = 35(3*sqrt(370) - 50)*S/166 = 1.62479...S

So basically you have four numbers, all of which are fixed algebraic multiples of the square root of the number you're trying to factor. How does this help?

[QUOTE=CRGreathouse;452077]No, because you don't give any actionable suggestions/algorithms/etc. All you give are four numbers with no obvious connection to anything. But let's look at them anyway! :smile:

Let the product be N and its square root be S. Then

N1 = sqrt(3330)/50 * S = 1.154...S
N2 = S
N3 = (1 + sqrt(3330)/50)*S = 2.154...S
N4 = 35(3*sqrt(370) - 50)*S/166 = 1.62479...S

So basically you have four numbers, all of which are fixed algebraic multiples of the square root of the number you're trying to factor. How does this help?[/QUOTE]

I think understand it ( but , I'm not sure ), ...for example (simple products)


[TEX]2*2 = 4[/TEX] and [B]First Way[/B] [TEX]N4 = 3,5[/TEX] and Second Way [TEX]N4 = 1[/TEX] [B]this is the minimum distance of the two prime factors for First Way[/B] ( [TEX]2 < 3,5[/TEX] )


[TEX]2*3 = 6[/TEX] and [B]First Way[/B] [TEX]N4 = 5,25[/TEX] and Second Way [TEX]N4 = 1,5[/TEX]

[TEX]2*5 = 10[/TEX] and [B]First Way[/B] [TEX]N4 = 5,8[/TEX] and Second Way [TEX]N4 = 1,6[/TEX]

[TEX]2*7 = 14[/TEX] and [B]First Way[/B] [TEX]N4 = 7[/TEX] and Second Way [TEX]N4 = 2[/TEX] [B]this is the maximum distance of the two prime factors for First Way[/B]


[TEX]2*11 = 4[/TEX] and First Way [TEX]N4 = 8,5[/TEX] and [B]Second Way[/B] [TEX]N4 = 2,4[/TEX] [B]Now Second Way is correct[/B]


[TEX]2*13 = 26[/TEX] and First Way [TEX]N4 = 9,1[/TEX] and [B]Second Way[/B] [TEX]N4 = 2,6[/TEX] [B]Now Second way is correct[/B]


You say it does not work in proportion also for large numbers?

P.S.

Should be tried with a program .

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Let's say I give you a number:

N = 67646988554084448110460372160440713890061829425106624449473216837898618103759093968747

Then I get (rounding down):

N1 = 9492406899940630172687271283182766199836725
N2 = 8224778936487256622736720394209992127637443
N3 = 17717185836427886795423991677392758327474168
N4 = 13363547807490383809719662986158762809329777

Knowing this, how do we factor the number N? Or what can be said about the factors of this number?

Note that N is not particularly large, and most of the people who post here could factor it without difficulty.

[QUOTE=CRGreathouse;452086]Let's say I give you a number:

N = 67646988554084448110460372160440713890061829425106624449473216837898618103759093968747

Then I get (rounding down):

N1 = 9492406899940630172687271283182766199836725
N2 = 8224778936487256622736720394209992127637443
N3 = 17717185836427886795423991677392758327474168
N4 = 13363547807490383809719662986158762809329777

Knowing this, how do we factor the number N? Or what can be said about the factors of this number?

Note that N is not particularly large, and most of the people who post here could factor it without difficulty.[/QUOTE]




But , I begin by "(N4)" N4-2 , N4-4 , N4-6...(First Way and Second Way) and not by"(N)" N-2 , N-4 , N-6.....! And I bet it will be faster.




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[QUOTE=Godzilla;452089]But , I begin by "(N4)" N4-2 , N4-4 , N4-6...(First Way and Second Way) and not by"(N)" N-2 , N-4 , N-6.....! And I bet it will be faster.[/QUOTE]

A 64-bit integer division takes 73 clocks, and a 286-bit division won't be faster than that. Assuming your clock speed is less than 100 GHz and you have no more than a million CPU cores it would take you more than 10^20 years to factor the number in this fashion. But it should only take a few minutes for SIQS.

[QUOTE=CRGreathouse;452093]A 64-bit integer division takes 73 clocks, and a 286-bit division won't be faster than that. Assuming your clock speed is less than 100 GHz and you have no more than a million CPU cores it would take you more than 10^20 years to factor the number in this fashion. But it should only take a few minutes for SIQS.[/QUOTE]



I was thinking in pencil, for example. I do not know how to work the clock.

[QUOTE=Godzilla;452095]I was thinking in pencil, for example. I do not know how to work the clock.[/QUOTE]

OK, let me rephrase. Your method, starting from N4 and working down through the odds, would take more than 10^40 divisions, and at one division per second would take 10^25 times the age of the universe to factor the number.