How to brute in float part of privkey ?

[COBRAS]

The you div 7 to 3 you get 2.5 = int2 float 0.5

7/3= 2.5

Then you sub 1 from 7 and div to 3 you get 2.

(7-1) / 3 = 2

So fo finding good range from 1 to 3, you need get EXACT integer without  float.


How to brute force in range from 0.1 to range 0.1111 ?

Br

But, why the I div
0xea1a5c66dcc11b5ad180 to 2**64 or 2**128 I always get some float result 0xea1a5c66dcc11b5ad180

[garlonicon]

How to brute in float part of privkey ?

There is no "float part of privkey". If you have 0.5, then you have 0x7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1. As you can see, that longer number has no "float part". It is like a clock, if you have a clock with modulo 24, then you don't have "25", you have "1". You can use smaller modulo than N=0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141, use N=24 and see that there is no "float part", there are only unsigned integers.

[COBRAS]

How to brute in float part of privkey ?

There is no "float part of privkey". If you have 0.5, then you have 0x7fffffffffffffffffffffffffffffff5d576e7357a4501ddfe92f46681b20a1. As you can see, that longer number has no "float part". It is like a clock, if you have a clock with modulo 24, then you don't have "25", you have "1". You can use smaller modulo than N=0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141, use N=24 and see that there is no "float part", there are only unsigned integers.

Then I div 0xea1a5c66dcc11b5ad180 to 2072576 result:


division   0x35ae38de0879fba remainder 0x9180

or enother "language 0x35ae38de0879fba,9180

And 0xea1a5c66dcc11b5ad180
 / 2072576 =
a158081848da8faf0d277663297c755f53593e29dcfc7de774e6193b35a70913

What is a range of this publick key 03B32D572B3A73B9E1212068030FDDB35522FEAF8520B829A4BE0CED43DAFD6901 (priv =a158081848da8faf0d277663297c755f53593e29dcfc7de774e6193b35a70913) ?

Original privkey not divided is a 2^80, but divided privkey mach more How to understand how to brute this ??

Big thank you.

[garlonicon]

How to understand how to brute this ??

Nobody knows how to break public keys in a fast way. Also, any private keys are what they are only under this base point G. If you switch to (1/2)*G, then you can multiply your private keys by two, and you will reach the same points. So, technically there is nothing like "parity" of the private key, because it is relative to the base point G. And there is no way to detect, if you reach any "fractions" or not, because there are no fractions.

In ECDSA, everything is always in modulo, rotating like in a clock, that's what makes the whole system secure, so you cannot look at it in the same way as you look at typical space of infinite numbers. Here, you can reach square root of some number that will behave in the same way as some integer! By understanding the clock analogy, you should understand that many numbers are connected with the same points, that's what makes some algorithms useless, because you cannot "just find out if that public key is in that range or not". You also cannot "just know if some public key has even or odd private key". Everything is relative to the base point, you should think about it as a pair of points, connected with some equation, like "A=x*B" or "A=x+B" (or even "A=x*B+c", in case of signatures).

Original privkey not divided is a 2^80, but divided privkey mach more

Of course it is. If you have some prime modulo, and you choose something small, for example 13, then your private keys are in range [1;12]. If you have some key from range [1;4], you can multiply that by 3, then it will be in range [1;12], but you will have different "granularity", that will change from 1 to 3. The same with dividing, which is really the same as multiplication by modulo inverse.

Maybe looking at those numbers modulo 13 will help, they are all equal in our "ECDSA world of modulo numbers":

Code:

1=(-12)=(-1/12)
2=(-11)=(-1/6)
3=(-10)=(-1/4)
4=(-9)=(-1/3)
5=(-8)=(-3/2)=(2/3)
6=(-7)=(-1/2)
7=(-6)=(1/2)
8=(-5)=(3/2)=(-2/3)
9=(-4)=(1/3)
10=(-3)=(1/4)
11=(-2)=(1/6)
12=(-1)=(1/12)

As you can see, in modulo 13, there are no fractions, because they are equal to integers. Also, there are no negative numbers, there are no square roots, nothing like that. There are only numbers that behave in the same way. And if you look at the real, huge, 256-bit modulo used in Bitcoin, it works in the same way as in this modulo 13 example. Just remember that there are no fractions.

[COBRAS]

How to understand how to brute this ??

Nobody knows how to break public keys in a fast way. Also, any private keys are what they are only under this base point G. If you switch to (1/2)*G, then you can multiply your private keys by two, and you will reach the same points. So, technically there is nothing like "parity" of the private key, because it is relative to the base point G. And there is no way to detect, if you reach any "fractions" or not, because there are no fractions.

In ECDSA, everything is always in modulo, rotating like in a clock, that's what makes the whole system secure, so you cannot look at it in the same way as you look at typical space of infinite numbers. Here, you can reach square root of some number that will behave in the same way as some integer! By understanding the clock analogy, you should understand that many numbers are connected with the same points, that's what makes some algorithms useless, because you cannot "just find out if that public key is in that range or not". You also cannot "just know if some public key has even or odd private key". Everything is relative to the base point, you should think about it as a pair of points, connected with some equation, like "A=x*B" or "A=x+B" (or even "A=x*B+c", in case of signatures).

Original privkey not divided is a 2^80, but divided privkey mach more

Of course it is. If you have some prime modulo, and you choose something small, for example 13, then your private keys are in range [1;12]. If you have some key from range [1;4], you can multiply that by 3, then it will be in range [1;12], but you will have different "granularity", that will change from 1 to 3. The same with dividing, which is really the same as multiplication by modulo inverse.

Maybe looking at those numbers modulo 13 will help, they are all equal in our "ECDSA world of modulo numbers":

Code:

1=(-12)=(-1/12)
2=(-11)=(-1/6)
3=(-10)=(-1/4)
4=(-9)=(-1/3)
5=(-8)=(-3/2)=(2/3)
6=(-7)=(-1/2)
7=(-6)=(1/2)
8=(-5)=(3/2)=(-2/3)
9=(-4)=(1/3)
10=(-3)=(1/4)
11=(-2)=(1/6)
12=(-1)=(1/12)

As you can see, in modulo 13, there are no fractions, because they are equal to integers. Also, there are no negative numbers, there are no square roots, nothing like that. There are only numbers that behave in the same way. And if you look at the real, huge, 256-bit modulo used in Bitcoin, it works in the same way as in this modulo 13 example. Just remember that there are no fractions.

Bros, thank you for your answers to me! I'll be honest, I'm a little drunk and very disappointed that it's so difficult to hack bitcoin, but I know a minisum with one person who hacked parrll. to a wallet with 50 bitcoins, and one more thing, who has 200 videlcards and he cracks passwords to bitcoin wallets, and that all wallets with wallet-dat.com almost all of them are already broken, and I still can't fucking. And I'm from Russia, we Russians treat with respect those who speak to us in essence and talk to us in a serious context, not thinking at us that we are joking when we are not joking at all.

Bratva, I can't figure out how when I point



7=(-6)=(1/2) but not x = 7 % 13 = 7 !! And not 1/2

[n0nce]

7=(-6)=(1/2) but not x = 7 % 13 = 7 !! And not 1/2

It's relatively simple:
1 ≡ 1 (mod 13)
1 ≡ 14 (mod 13)
1 ≡ 2*7 (mod 13)
1/2 ≡ 7 (mod 13)

Another example:
1 ≡ 1 (mod 13)
1 ≡ 27 (mod 13)
1 ≡ 3*9 (mod 13)
1/3 ≡ 9 (mod 13)

The other way round, if you have something like the number 1/6 (mod 13), you can go like this:
1/6 ≡ x (mod 13)
1 ≡ x*6 (mod 13)
1 ≡ 11*6 (mod 13)
Therefore, 1/6 ≡ 11 (mod 13) (first equation).

[garlonicon]

Also, when it comes to square roots, the situation is even more interesting in modulo 13:

Code:

1=sqrt(1^2)=sqrt(1)=sqrt(144)=sqrt(12^2)
2=sqrt(2^2)=sqrt(4)=sqrt(121)=sqrt(11^2)
3=sqrt(3^2)=sqrt(9)=sqrt(100)=sqrt(10^2)
4=sqrt(4^2)=sqrt(16)=sqrt(3)=sqrt(81)=sqrt(9^2)
5=sqrt(5^2)=sqrt(25)=sqrt(12)=sqrt(64)=sqrt(8^2)
6=sqrt(6^2)=sqrt(36)=sqrt(10)=sqrt(49)=sqrt(7^2)

See? In modulo 13, you only have private keys from 1 to 12. In the same way, if you have huge, 256-bit N, then you only have private keys from 1 to N-1. There are no more numbers, every other number is just in the same place as some unsigned integer. Here, in modulo 13, when you have sqrt(3), you don't have some infinite series of numbers after decimal point. You just have 4. Why? Because (sqrt(3))^2=4^2=16=3 (because we are in modulo 13). See? In modulo 13, (sqrt(3))^2=3, everything works! Instead of sqrt(3)=1.73205080757... we have just 4. It works in the same way in Bitcoin, for that huge 256-bit N, there are no fractions, there are even no infinite fractions, just some numbers behave like them.