Can someone try brute force this pu in 80 bit range ?

[COBRAS]

This is a 120 bit puzzle with modified pubkey, try brute force please,if we success, we share pazzle:

pub for brute 02834a2aaf9441265a847282c488100b99d1bafcbae69f5445574674f8bd05c884

range 1:0xfffffffffffffffffffff

0xfffffffffffffffffffffffffffffffebaaedce6af38a03bbfd25e8cd0364142:0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141

[albert0bsd]

how do you can be 100% sure that the key is in the 80 bits range?

[COBRAS]

how do you can be 100% sure that the key is in the 80 bits range?

Because I substract 0x1ffffffffff00000..... from publick key this is not exact because shit happens, but if my substracted range not so big, all mast be ok...
.

[albert0bsd]

That subtraction DON'T guaranty that the result is under 80 bits.

[COBRAS]

That subtraction DON'T guaranty that the result is under 80 bits.

No Guaranty, but, if point go to negative side, I was provide range for this negative side.. 80 bit, not so big range...if someone try, and find something we share with him some cents...

[bigvito19]

Its 80 bits in the 120 bit range, not 80 bits in the 80 bit range. Its not like that at all, its a 80 bit difference on the 120 key.

[COBRAS]

Its 80 bits in the 120 bit range, not 80 bits in the 80 bit range. Its not like that at all, its a 80 bit difference on the 120 key.

0x1ffffffffff000000000000000000000 + 80 bit(0xfffffffffffffffffffff) = 125 bit

[albert0bsd]

No Guaranty, but, if point go to negative side, I was provide range for this negative side.. 80 bit

That is also NOT true, it can be in under the 2^120 Negative.

we share with him some cents...

Some cents??

[albert0bsd]

0x1ffffffffff000000000000000000000 + 80 bit(0xfffffffffffffffffffff) = 125 bit

That is also incorrect.

[COBRAS]

No Guaranty, but, if point go to negative side, I was provide range for this negative side.. 80 bit

That is also NOT true, it can be in under the 2^120 Negative.

we share with him some cents...

Some cents??

"Cents"...

Then substract from 100 bit key 110 range,range will be 10 bit in N-10:N

[COBRAS]

0x1ffffffffff000000000000000000000 + 80 bit(0xfffffffffffffffffffff) = 125 bit

That is also incorrect.

2^125 up range 0x1fffffffffffffffffffffffffffffff - 0x1ffffffffff000000000000000000000 = 80 bit....

[albert0bsd]

But that is ONLY valid if the key start those digits:

Random key in the 120 bit:

Code:

b125bbec43b33ceffae7e41e7ee68e

subtraction:

Code:

$ ./modmath 0x1ffffffffff000000000000000000000 - 0xb125bbec43b33ceffae7e41e7ee68e
Result: 1f4eda4413ac4cc31005181be1811972

now did you see the problem?

[garlonicon]

Code:

SHA-256("120-bit")=7914af6a8fe938aacca6c2e60ae391bc7da6ecd858d33169be8ea60f42ec9078
           mask120=0000000000000000000000000000000000ffffffffffffffffffffffffffffff
  samplePrivKey120=0000000000000000000000000000000000a6ecd858d33169be8ea60f42ec9078
            mask80=00000000000000000000000000000000000000000000ffffffffffffffffffff
   samplePrivKey80=000000000000000000000000000000000000000000003169be8ea60f42ec9078
 subtractionMask40=0000000000000000000000000000000000ffffffffff00000000000000000000
 subtractPrivKey40=0000000000000000000000000000000000a6ecd858d300000000000000000000

And you think that you can get from "samplePrivKey120" to "samplePrivKey80" without knowing "subtractPrivKey40". I have bad news: you cannot. Knowing mask or range alone is not enough, because bits can be flipped, and from 120-bit key you will also get another 120-bit key. It can be negative, it does not matter. Remember: it is like rotating in a clock: going to negative numbers will bring you no luck, because it will still be 120-bit negative number, just with leading ones, instead of zeros, but that's all.

[COBRAS]

But that is ONLY valid if the key start those digits:

Random key in the 120 bit:

Code:

b125bbec43b33ceffae7e41e7ee68e

subtraction:

Code:

$ ./modmath 0x1ffffffffff000000000000000000000 - 0xb125bbec43b33ceffae7e41e7ee68e
Result: 1f4eda4413ac4cc31005181be1811972

now did you see the problem?

No, not see problem Bro
Try 125 but key generate

And my priv start with 1

[albert0bsd]

randomkey in the 125 bit space: 143a1858c689bd58d0285a83b7d11f50

Subtraction:

Code:

./modmath 0x1ffffffffff000000000000000000000 - 0x143a1858c689bd58d0285a83b7d11f50
Result: bc5e7a7396642a72fd7a57c482ee0b0

Result is in the 124 bit space

[COBRAS]

randomkey in the 125 bit space: 143a1858c689bd58d0285a83b7d11f50

Subtraction:

Code:

./modmath 0x1ffffffffff000000000000000000000 - 0x143a1858c689bd58d0285a83b7d11f50
Result: bc5e7a7396642a72fd7a57c482ee0b0

Result is in the 124 bit space

So need substract 124.8 but ...

[albert0bsd]

You don't have idea of what are you doing.

[COBRAS]

Code:

SHA-256("120-bit")=7914af6a8fe938aacca6c2e60ae391bc7da6ecd858d33169be8ea60f42ec9078
           mask120=0000000000000000000000000000000000ffffffffffffffffffffffffffffff
  samplePrivKey120=0000000000000000000000000000000000a6ecd858d33169be8ea60f42ec9078
            mask80=00000000000000000000000000000000000000000000ffffffffffffffffffff
   samplePrivKey80=000000000000000000000000000000000000000000003169be8ea60f42ec9078
 subtractionMask40=0000000000000000000000000000000000ffffffffff00000000000000000000
 subtractPrivKey40=0000000000000000000000000000000000a6ecd858d300000000000000000000

And you think that you can get from "samplePrivKey120" to "samplePrivKey80" without knowing "subtractPrivKey40". I have bad news: you cannot. Knowing mask or range alone is not enough, because bits can be flipped, and from 120-bit key you will also get another 120-bit key. It can be negative, it does not matter. Remember: it is like rotating in a clock: going to negative numbers will bring you no luck, because it will still be 120-bit negative number, just with leading ones, instead of zeros, but that's all.

Then you substract from 120 -(80+40) you get around119.9 !!!