?Thx
Topic: Pollard's kangaroo ECDLP solver - page 102. (Read 60654 times)
Help someone. Kangaroo not found dead kangaroo and privkeys. Were is a problem ? What can help to fined privkey ? In 10 pubkeys I have one with dead kangaroo ? Pubkey with dead kandagoo moree good then pubkey without kangaroo ? What to do ?
Thx
?Thx
-snip-
mean 789.789 is value can u div pubkey by above value ?
Can`t and nobody can`t. devider must be integer not float. all pubkeys Q=k*G, where k - integer.mean 789.789 is value can u div pubkey by above value ?
Sorry for continuing an off topic discussion, but:
Of course you can divide by 789.789 if you want.
because 789.789=789789/1000 , which doesn't have decimals in it AND you can divide and multiply by scalar numbers.
Don't know why would anyone want to divide by 789.789, but it s possible
Sorry. Move talk about calculator to enother thread !!!
-snip-
mean 789.789 is value can u div pubkey by above value ?
Can`t and nobody can`t. devider must be integer not float. all pubkeys Q=k*G, where k - integer.mean 789.789 is value can u div pubkey by above value ?
Sorry for continuing an off topic discussion, but:
Of course you can divide by 789.789 if you want.
because 789.789=789789/1000 , which doesn't have decimals in it AND you can divide and multiply by scalar numbers.
Don't know why would anyone want to divide by 789.789, but it s possible
@mrxtraf
In the private key group (mod n) we can add, negate, and invert - this allows for multiplication and division.
In the public key group (elliptic curve mod p of size n) we can add, negate, and double only. This leads to multiplication by a scalar.
One public key corresponds to exactly one private key, and vice versa. The proof is very easy. Let G is the generator of secp256k1. Let P=k*G is a point on the curve. Let also P=k'*G. Then (k-k')*G=O => (k-k') divides n. But n is prime, hence k=k' (mod n).
That is, you can’t divide the public key by 10?In the private key group (mod n) we can add, negate, and invert - this allows for multiplication and division.
In the public key group (elliptic curve mod p of size n) we can add, negate, and double only. This leads to multiplication by a scalar.
One public key corresponds to exactly one private key, and vice versa. The proof is very easy. Let G is the generator of secp256k1. Let P=k*G is a point on the curve. Let also P=k'*G. Then (k-k')*G=O => (k-k') divides n. But n is prime, hence k=k' (mod n).
Give me any public key from which you know the private key, I will divide it by 10. And I will give in return the result in the form of a public key. And you yourself divide the private key by 10, get the public key from it and compare.
The multiplication (and division, which is multiplication with the inverse) is by scalar only. You cannot multiply two public keys without solving ECDLP first. And if you somehow can, then all coins are belong to you.
mrxtraf is saying they can "divide" the public key by 10 by multiplying the public key by the multiplicative inverse of 10 mod n.
apply your algo and show me pubkey div by 789 for
03D041CF467F485A96AB21EC0E1E1E26A344B28A12244320C4BDE48C123653D88F
Sure.
The inverse of 789 mod n = 90549707299650307447836879278023370272751301850683416988678562304583910826370
Multiply your public key by that and you get:
035776B3684B6A5E9A6307AA53C3D484AABB90244E6371405C114CF8910A9A3BD0
Multiplying that point by 789 will result in the original public key. In otherwords, 789 * 90549707299650307447836879278023370272751301850683416988678562304583910826370 = 1 (mod n).
Are you guys still discussing the off-topic things?
It was said several times, that there is no division for public points. Our brains want to visualize the division process and we only imagine integers. But group field works under different principles.
The only thing we can do - is to multiply by the inverse scalar. That's it. But we still just multiplying a public point by the integer number, and divide it.
For every integer x we can find inverse y which is such a number where x*y%order = 1, where order if the order of the group.
No inverse exist for 0, and for order itself.
Easy explanation for the group of 7 elements (let's say week days from Monday to Sunday):
If you multiply Wednesday by 4, you will receive Friday (3*4%7 = 12%7 = 5)
For division, we also make multiplication by the inverse scalar. Here is the list of inverse scalars for every number:
inverse(1) = 1
inverse(2) = 4
inverse(3) = 5
inverse(4) = 2
inverse(5) = 3
inverse(6) = 6
inverse(7) does not exist
Now if you want to "divide" Friday by 4, you should multiply it by inverse(4) = 2. So, we have 5*2%7 = 10%7 = 3
You can see that equation is correct as it was in the 1st example (where we multiplied Wednesday by 4).
You should never divide 5 by 4 and imagine number 1.25, and so on. It is not correct.
All the elements in the group is a wheel, and repeating after reaching the order.
So, if we divide some number x by k within the group with finite order, we are searching such a number y, which will give us x if we multiply it by k: y * k = x
In order to find it, we calculate inverse k'=(1/k), and make the multiplication: y = x * (1/k)
It was said several times, that there is no division for public points. Our brains want to visualize the division process and we only imagine integers. But group field works under different principles.
The only thing we can do - is to multiply by the inverse scalar. That's it. But we still just multiplying a public point by the integer number, and divide it.
For every integer x we can find inverse y which is such a number where x*y%order = 1, where order if the order of the group.
No inverse exist for 0, and for order itself.
Easy explanation for the group of 7 elements (let's say week days from Monday to Sunday):
If you multiply Wednesday by 4, you will receive Friday (3*4%7 = 12%7 = 5)
For division, we also make multiplication by the inverse scalar. Here is the list of inverse scalars for every number:
inverse(1) = 1
inverse(2) = 4
inverse(3) = 5
inverse(4) = 2
inverse(5) = 3
inverse(6) = 6
inverse(7) does not exist
Now if you want to "divide" Friday by 4, you should multiply it by inverse(4) = 2. So, we have 5*2%7 = 10%7 = 3
You can see that equation is correct as it was in the 1st example (where we multiplied Wednesday by 4).
You should never divide 5 by 4 and imagine number 1.25, and so on. It is not correct.
All the elements in the group is a wheel, and repeating after reaching the order.
So, if we divide some number x by k within the group with finite order, we are searching such a number y, which will give us x if we multiply it by k: y * k = x
In order to find it, we calculate inverse k'=(1/k), and make the multiplication: y = x * (1/k)
-snip-
so each and every digit have value, and i can div them in integar and float too actual low then 1 is called is fractional point,and i can play with each and every point
each representation K like positive or negative or fraction is allways positive integer from 1 to n.so each and every digit have value, and i can div them in integar and float too actual low then 1 is called is fractional point,and i can play with each and every point
if you say me positive integer representation of 789.789 i can easy devide any pub key by this valus. Here no problem to do this.
Like two fingers on the asphalt.
K * 1000 = NK
f(NK / 1000) = MK
-snip-
so each and every digit have value, and i can div them in integar and float too actual low then 1 is called is fractional point,and i can play with each and every point
each representation K like positive or negative or fraction is allways positive integer from 1 to n.so each and every digit have value, and i can div them in integar and float too actual low then 1 is called is fractional point,and i can play with each and every point
if you say me positive integer representation of 789.789 i can easy devide any pub key by this valus. Here no problem to do this.
-snip-
mean 789.789 is value can u div pubkey by above value ?
devider must be integer not float. all pubkeys Q=k*G, where k - integer.mean 789.789 is value can u div pubkey by above value ?
highest value in ecc is 1
1 = 115792089237316195423570985008687907852837564279074904382605163141518161494337. 0
and above to 0 is
0.11579208923731619542357098500868790785283756427907490438260516314151816149433 7
so each and every digit have value, and i can div them in integar and float too actual low then 1 is called is fractional point,and i can play with each and every point
Good day. Do you know how get wright range for kangaroo search ? And how maybe manipulate pubkey for so range ?
Thx
-snip-
mean 789.789 is value can u div pubkey by above value ?
devider must be integer not float. all pubkeys Q=k*G, where k - integer.mean 789.789 is value can u div pubkey by above value ?
highest value in ecc is 1
1 = 115792089237316195423570985008687907852837564279074904382605163141518161494337. 0
and above to 0 is
0.11579208923731619542357098500868790785283756427907490438260516314151816149433 7
so each and every digit have value, and i can div them in integar and float too actual low then 1 is called is fractional point,and i can play with each and every point
-snip-
mean 789.789 is value can u div pubkey by above value ?
Can`t and nobody can`t. devider must be integer not float. all pubkeys Q=k*G, where k - integer. mean 789.789 is value can u div pubkey by above value ?
-snip-
if need div 789.789 then ?
what you mean? you can devide point only by integer.if need div 789.789 then ?
-snip-
if need div 789.789 then ?
what you mean? you can devide point only by integer. if need div 789.789 then ?
-snip-
BitCrack
apply your algo and show me pubkey div by 789 for
03D041CF467F485A96AB21EC0E1E1E26A344B28A12244320C4BDE48C123653D88F
035776b3684b6a5e9a6307aa53c3d484aabb90244e6371405c114cf8910a9a3bd0BitCrack
apply your algo and show me pubkey div by 789 for
03D041CF467F485A96AB21EC0E1E1E26A344B28A12244320C4BDE48C123653D88F
-snip-
BitCrack
apply your algo and show me pubkey div by 789 for
03D041CF467F485A96AB21EC0E1E1E26A344B28A12244320C4BDE48C123653D88F
035776b3684b6a5e9a6307aa53c3d484aabb90244e6371405c114cf8910a9a3bd0BitCrack
apply your algo and show me pubkey div by 789 for
03D041CF467F485A96AB21EC0E1E1E26A344B28A12244320C4BDE48C123653D88F
And what is problem in division pub by 789?
@mrxtraf
In the private key group (mod n) we can add, negate, and invert - this allows for multiplication and division.
In the public key group (elliptic curve mod p of size n) we can add, negate, and double only. This leads to multiplication by a scalar.
One public key corresponds to exactly one private key, and vice versa. The proof is very easy. Let G is the generator of secp256k1. Let P=k*G is a point on the curve. Let also P=k'*G. Then (k-k')*G=O => (k-k') divides n. But n is prime, hence k=k' (mod n).
That is, you can’t divide the public key by 10?In the private key group (mod n) we can add, negate, and invert - this allows for multiplication and division.
In the public key group (elliptic curve mod p of size n) we can add, negate, and double only. This leads to multiplication by a scalar.
One public key corresponds to exactly one private key, and vice versa. The proof is very easy. Let G is the generator of secp256k1. Let P=k*G is a point on the curve. Let also P=k'*G. Then (k-k')*G=O => (k-k') divides n. But n is prime, hence k=k' (mod n).
Give me any public key from which you know the private key, I will divide it by 10. And I will give in return the result in the form of a public key. And you yourself divide the private key by 10, get the public key from it and compare.
The multiplication (and division, which is multiplication with the inverse) is by scalar only. You cannot multiply two public keys without solving ECDLP first. And if you somehow can, then all coins are belong to you.
mrxtraf is saying they can "divide" the public key by 10 by multiplying the public key by the multiplicative inverse of 10 mod n.
apply your algo and show me pubkey div by 789 for
03D041CF467F485A96AB21EC0E1E1E26A344B28A12244320C4BDE48C123653D88F
@mrxtraf
In the private key group (mod n) we can add, negate, and invert - this allows for multiplication and division.
In the public key group (elliptic curve mod p of size n) we can add, negate, and double only. This leads to multiplication by a scalar.
One public key corresponds to exactly one private key, and vice versa. The proof is very easy. Let G is the generator of secp256k1. Let P=k*G is a point on the curve. Let also P=k'*G. Then (k-k')*G=O => (k-k') divides n. But n is prime, hence k=k' (mod n).
That is, you can’t divide the public key by 10?In the private key group (mod n) we can add, negate, and invert - this allows for multiplication and division.
In the public key group (elliptic curve mod p of size n) we can add, negate, and double only. This leads to multiplication by a scalar.
One public key corresponds to exactly one private key, and vice versa. The proof is very easy. Let G is the generator of secp256k1. Let P=k*G is a point on the curve. Let also P=k'*G. Then (k-k')*G=O => (k-k') divides n. But n is prime, hence k=k' (mod n).
Give me any public key from which you know the private key, I will divide it by 10. And I will give in return the result in the form of a public key. And you yourself divide the private key by 10, get the public key from it and compare.
The multiplication (and division, which is multiplication with the inverse) is by scalar only. You cannot multiply two public keys without solving ECDLP first. And if you somehow can, then all coins are belong to you.
mrxtraf is saying they can "divide" the public key by 10 by multiplying the public key by the multiplicative inverse of 10 mod n.
@mrxtraf
In the private key group (mod n) we can add, negate, and invert - this allows for multiplication and division.
In the public key group (elliptic curve mod p of size n) we can add, negate, and double only. This leads to multiplication by a scalar.
One public key corresponds to exactly one private key, and vice versa. The proof is very easy. Let G is the generator of secp256k1. Let P=k*G is a point on the curve. Let also P=k'*G. Then (k-k')*G=O => (k-k') divides n. But n is prime, hence k=k' (mod n).
That is, you can’t divide the public key by 10?In the private key group (mod n) we can add, negate, and invert - this allows for multiplication and division.
In the public key group (elliptic curve mod p of size n) we can add, negate, and double only. This leads to multiplication by a scalar.
One public key corresponds to exactly one private key, and vice versa. The proof is very easy. Let G is the generator of secp256k1. Let P=k*G is a point on the curve. Let also P=k'*G. Then (k-k')*G=O => (k-k') divides n. But n is prime, hence k=k' (mod n).
Give me any public key from which you know the private key, I will divide it by 10. And I will give in return the result in the form of a public key. And you yourself divide the private key by 10, get the public key from it and compare.
The multiplication (and division, which is multiplication with the inverse) is by scalar only. You cannot multiply two public keys without solving ECDLP first. And if you somehow can, then all coins are belong to you.
@mrxtraf
In the private key group (mod n) we can add, negate, and invert - this allows for multiplication and division.
In the public key group (elliptic curve mod p of size n) we can add, negate, and double only. This leads to multiplication by a scalar.
One public key corresponds to exactly one private key, and vice versa. The proof is very easy. Let G is the generator of secp256k1. Let P=k*G is a point on the curve. Let also P=k'*G. Then (k-k')*G=O => (k-k') divides n. But n is prime, hence k=k' (mod n).
That is, you can’t divide the public key by 10?In the private key group (mod n) we can add, negate, and invert - this allows for multiplication and division.
In the public key group (elliptic curve mod p of size n) we can add, negate, and double only. This leads to multiplication by a scalar.
One public key corresponds to exactly one private key, and vice versa. The proof is very easy. Let G is the generator of secp256k1. Let P=k*G is a point on the curve. Let also P=k'*G. Then (k-k')*G=O => (k-k') divides n. But n is prime, hence k=k' (mod n).
Give me any public key from which you know the private key, I will divide it by 10. And I will give in return the result in the form of a public key. And you yourself divide the private key by 10, get the public key from it and compare.
@mrxtraf
In the private key group (mod n) we can add, negate, and invert - this allows for multiplication and division.
In the public key group (elliptic curve mod p of size n) we can add, negate, and double only. This leads to multiplication by a scalar.
One public key corresponds to exactly one private key, and vice versa. The proof is very easy. Let G is the generator of secp256k1. Let P=k*G is a point on the curve. Let also P=k'*G. Then (k-k')*G=O => (k-k') divides n. But n is prime, hence k=k' (mod n).
In the private key group (mod n) we can add, negate, and invert - this allows for multiplication and division.
In the public key group (elliptic curve mod p of size n) we can add, negate, and double only. This leads to multiplication by a scalar.
One public key corresponds to exactly one private key, and vice versa. The proof is very easy. Let G is the generator of secp256k1. Let P=k*G is a point on the curve. Let also P=k'*G. Then (k-k')*G=O => (k-k') divides n. But n is prime, hence k=k' (mod n).
Jean_Luc
Can you insert this function in the code ?
Please
Big thank you.
Can you insert this function in the code ?
Code:
And one more trick, if the program showed not only just dead kangaroos, but also their type.
Please
Big thank you.
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