Half of any bitcoin (crypto) public key - (public key half)

[EmptyZeroCoin]

To get half of any given bitcoin (crypt o) public key is possible, Technically it's possible, we made that.  The public key that's in the bitcoin network.

Public key x and y == Double(Half of the Public key x and y)

Example's:

Address

Code:

1HQ3Go3ggs8pFnXuHVHRytPCq5fGG8Hbhx

Public key in network

Code:

02545d2c25b98ec8827f2d9bee22b7a9fb98091b2008bc45b3b806d44624dc038c

Public key x and y

Code:

x = 545d2c25b98ec8827f2d9bee22b7a9fb98091b2008bc45b3b806d44624dc038c
y = f1e18224b09ed00841c5407e571829e41876d44522f97e05e405a91ef38d4f00

Half of the above public key

Code:

x = 8f870b1693cb408f96a3da9e3623b6d0315a403395d79f412f26044210bcddd2
y = 8a0a3e2399787a1f084d53500bc577ba1cc4e19ab7b2d9a2bd327e5e56e8afa0

Address

Code:

1JCe8z4jJVNXSjohjM4i9Hh813dLCNx2Sy

Public key in network

Code:

037f73b63c10ce53c1b73800751d859915badcf5423da1fd7fabfc8556a833f0bd

Public key x and y

Code:

x = 7f73b63c10ce53c1b73800751d859915badcf5423da1fd7fabfc8556a833f0bd
y = bb2402fa216c2eeafc8e5667c8290c8e28199cb7e48bef2fffbdee5dff0ad277

Half of the above public key

Code:

x = acbe1e6677d64cbbb501f3e1e43604bc478e57fcc473e78dd06a6cb3de223c6d
y = d3bf86be940c6a60352da18297d3c3c23a6db42e586fa74294c32a0d7c1b4a88

Half of the above public key

Code:

x = ae05a1d169edb67fe94fef428b6594ba218ce157bb04c8dac2889b805f6e8965
y = 18897d88bb3c44476d8c204bf29650f159c47f6138913ab04ba37fc7bf3081cb

Address

Code:

12ib7dApVFvg82TXKycWBNpN8kFyiAN1dr

Public key in network

Code:

04d6597d465408e6e11264c116dd98b539740e802dc756d7eb88741696e20dfe7d3588695d2e7ad23cbf0aa056d42afada63036d66a1d9b97070dd6bc0c87ceb0d

Public key x and y

Code:

x = d6597d465408e6e11264c116dd98b539740e802dc756d7eb88741696e20dfe7d
y = 3588695d2e7ad23cbf0aa056d42afada63036d66a1d9b97070dd6bc0c87ceb0d

half of the above public key

Code:

x = dc0f6e4ff971e22fffa77df43d24128046599f9b2299bc7a6fbc4e7e9fc4a0ff
y = 615f1884faa5197a00560ecb69f2b18ef223a3dfaee2407d4ea5a2286de2a41b

[1714951876]

1714951876

[1714951876]

1714951876

[1714951876]

1714951876

[1714951876]

1714951876

[1714951876]

1714951876

[1714951876]

1714951876

[Elliptic23]

Sure, just multiply the point by 2^-1 (mod n).

[keelperu]

Sure, just multiply the point by 2^-1 (mod n).

How its possible there is scalar addition, subtraction and multiplication only , can you explain more (multiply the point by 2^-1 (mod n))

[Elliptic23]

The goal is to "half" the point. You want to find the point Q where 2Q = P, your original point.

So you compute 2^-1 (mod n) where n is the order of the group (see the secp256k1 parameters: https://en.bitcoin.it/wiki/Secp256k1)

2^1 is the multiplicative inverse of 2 (mod n). 2 x 2^1 = 1 (mod n). When you do the scalar multiplication it cancels out a factor of 2.

2Q = P

(2^-1)2Q = (2^-1)P

Q = (2^-1)P

Hope that makes sense.

[keelperu]

The goal is to "half" the point. You want to find the point Q where 2Q = P, your original point.

So you compute 2^-1 (mod n) where n is the order of the group (see the secp256k1 parameters: https://en.bitcoin.it/wiki/Secp256k1)

2^1 is the multiplicative inverse of 2 (mod n). 2 x 2^1 = 1 (mod n). When you do the scalar multiplication it cancels out a factor of 2.

2Q = P

(2^-1)2Q = (2^-1)P

Q = (2^-1)P

Hope that makes sense.

here the another answer

https://crypto.stackexchange.com/questions/59972/half-of-any-bitcoin-crypto-public-key-public-key-half-is-possible


ECC is cryptography over an elliptic curve group.

Firstly you have an elliptic curve, e.g. Bitcoin uses a Koblitz curve secp256k1 y2=x3+7.

The group is defined over curve points over a finite field Fp (integer modular p). The group elements are points on the curve. A point in the affine form consists of two coordinates P=(x,y) where x,y∈Fp.

For group elements, you can do point addition P+Q, as well as scalar multiplication sP, where s is an integer in Zn where n is the order of the group (how many elements in the group).

A public key in bitcoin is a point P. To do P2, you multiply 12 to P where 12 is the multiplicative inverse of 2 in Zn. It is an integer that can be found using the extended Euclidean algorithm and is 57896044618658097711785492504343953926418782139537452191302581570759080747169 in the case of secp256k1.


but in crypt o ,

private key 3

x =  f9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9
y =  388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672

half of the above public key is 1 given below

x =  79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
y =  483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8

but the output is

x = c62c910e502cb615a27c58512b6cc2c94f5742f76cb3d12ec993400a3695d413
y = 17f3dadd767275ddd3b23f46723631778bf01dadaebb9a953cf068712457c010

what wrong in this how to correct this

3/2 = 1


  

[Elliptic23]

2^-1 (mod n) = 57896044618658097711785492504343953926418782139537452191302581570759080747169


2 * 57896044618658097711785492504343953926418782139537452191302581570759080747169 = 1 (mod n)

Multiplying a point by this number will "half" the point.

[maa@slm]

Hi,
How to Calculate x ,y in Addresses,can you explain more.. if any code in git-hub
can you explain step by  guide please..
I am New to bitcoin ....
please inbox me to learn something...

[pooya87]

How to Calculate x ,y in Addresses,can you explain more.. if any code in git-hub
can you explain step by  guide please..

you can't calculate x,y from a bitcoin address since addresses are basically the hash of a public key (ie. x,y coordinates) and hashes are not reversible.
what OP is explaining is where you already have the public key and want to compute (1/2)*pubkey which is a simple multiplication of the public key with modular multiplicative inverse of 2. any crypto library that has ECC also has a method for computing "ModInverse" and "point multiplication".

[athanred]

When you read about ECDSA used in Bitcoin, you probably learn about adding points and doubling it. But it is also possible to subtract any two points. You can also multiply or divide any point by any number, but you cannot multiply or divide two public keys without knowing at least one matching private key.

Halving a point is possible, but it does not break ECDSA security. It is a bit different than typical integer division when you can just divide everything by two and quickly come to zero and break everything. It is not the case when using ECDSA and doing calculations over finite field.

To better understand it, you can do some calculations on smaller finite fields, for example using modulo 67. You can start from 1 and multiply it by two. You will get 1,2,4,8,16,32,64 and then suddenly you will get 128-67=61. Here you can see that dividing 61 by two will give you 64.

[pooya87]

To better understand it, you can do some calculations on smaller finite fields, for example using modulo 67. You can start from 1 and multiply it by two. You will get 1,2,4,8,16,32,64 and then suddenly you will get 128-67=61. Here you can see that dividing 61 by two will give you 64.

your calculation is wrong, 61 divided by 2 in this group does give you 64. read my post above.
your mistake is that you are calculating 61/2=30.5 using simple arithmetic whereas all the calculations here are modular arithmetic meaning:

Code:

61/2 ≡ 61 * 34 ≡ 2047 ≡ 64 (mod 67)

where 34 is calculated by solving this:

Code:

2*x ≡ 1 (mod 67)

[ashraful1980]

Hi,
How to Calculate x ,y in Addresses,can you explain more.. if any code in git-hub
can you explain step by  guide please..
I am New to bitcoin ....
please inbox me to learn something...

It is simple XY coordinate python script:
import bitcoin as b
pubkey = b.decode_pubkey("02545d2c25b98ec8827f2d9bee22b7a9fb98091b2008bc45b3b806d44624dc038c")
print("X:", hex(pubkey[0]), "Y:", hex(pubkey[1]))

[BTCW]

This makes no sense at all.

First, the public address is a hashed public key, so unless there are major flaws in both SHA256 and RIPEMD-160 (hint: there aren't), you cannot take a public address and derive its public key (and its X and Y coordinates on the elliptic curve).

The public address and the public key are two different things.

Second, the math above is... a complete mess.

Third, even if you hypothetically could, what the heck would be the point of all this?

[archyone]

hello,

I read this post several, several, several times and i can't understand how it is possible to half a point like the first example.
Is it possible without the private key to divide a point by 2 ?

assume i have like the first example a point without the private key for it and divide it by 2

x = 545d2c25b98ec8827f2d9bee22b7a9fb98091b2008bc45b3b806d44624dc038c
y = f1e18224b09ed00841c5407e571829e41876d44522f97e05e405a91ef38d4f00

How get ?

x = 8f870b1693cb408f96a3da9e3623b6d0315a403395d79f412f26044210bcddd2
y = 8a0a3e2399787a1f084d53500bc577ba1cc4e19ab7b2d9a2bd327e5e56e8afa0

I clearly don't understand the concept of multiply by 2^-1 (mod n)
is it possible for someone to clarify this with an real example (not n, N P Q ETC... LOL ..I was the one sleeping in math classroom ^)

Thanks

[pooya87]

Is it possible without the private key to divide a point by 2 ?

why not. private key will tell you how many times to add G to itself. but after you are done and have the value for k*G as point P then you can do whatever you want with that point. for example you can add another G to that point and compute Q=P+G or R=P-G and similarly you can multiply that point with any number like computing S=3*P (the same way you would compute 3*G).
2^-1 is just another number that your point (P or G or Q,...) is multiplied by. you just have to first calculate what integer 2^-1 is in congruence with. to do that you compute its modular multiplicative inverse as i explained above. there is also an example in my second comment with small numbers.

[archyone]

Is it possible without the private key to divide a point by 2 ?

why not. private key will tell you how many times to add G to itself. but after you are done and have the value for k*G as point P then you can do whatever you want with that point. for example you can add another G to that point and compute Q=P+G or R=P-G and similarly you can multiply that point with any number like computing S=3*P (the same way you would compute 3*G).
2^-1 is just another number that your point (P or G or Q,...) is multiplied by. you just have to first calculate what integer 2^-1 is in congruence with. to do that you compute its modular multiplicative inverse as i explained above. there is also an example in my second comment with small numbers.

First thanks for the response.
Ok i begin to understand the concept but not ready to go yet ^^.
I know add or substract a point to another (different or equal) but i can't divide a point by any number if i only have the x and y coordinates.
If it's possible can you show me how divide this point by 2 for example ?

x = 72488970228380509287422715226575535698893157273063074627791787432852706183111
y = 62070622898698443831883535403436258712770888294397026493185421712108624767191

It is public key for private key: 10 , and we say here i haven't got this private key.

I expect to obtain the public key:

x = 21505829891763648114329055987619236494102133314575206970830385799158076338148
Y = 98003708678762621233683240503080860129026887322874138805529884920309963580118
(private key 5 )

Thanks in advance

[pooya87]

I know add or substract a point to another (different or equal) but i can't divide a point by any number if i only have the x and y coordinates.
If it's possible can you show me how divide this point by 2 for example ?

you shouldn't think in terms of "divide by 2" but in terms of "multiply by 2^-1" and 2^-1 is defined this way in modular arithmetic:
find x such that 2*x ≡ 1 (mod prime)

this is all because of how 2^-1 is defined and its value is 57896044618658097711785492504343953926418782139537452191302581570759080747169 because 2 * 57896044618658097711785492504343953926418782139537452191302581570759080747169 ≡ 1 (mod N)
it can be computed using Euclidean algorithm or the simplified 2^(prime-2) mode prime.
now multiplying that with the point we had returns the half point

x = 21505829891763648114329055987619236494102133314575206970830385799158076338148
Y = 98003708678762621233683240503080860129026887322874138805529884920309963580118

EDIT: I made the mistake of using P as the prime but it should have been N (curve order). now it is fixed.

[j2002ba2]

Is it possible without the private key to divide a point by 2 ?

why not. private key will tell you how many times to add G to itself. but after you are done and have the value for k*G as point P then you can do whatever you want with that point. for example you can add another G to that point and compute Q=P+G or R=P-G and similarly you can multiply that point with any number like computing S=3*P (the same way you would compute 3*G).
2^-1 is just another number that your point (P or G or Q,...) is multiplied by. you just have to first calculate what integer 2^-1 is in congruence with. to do that you compute its modular multiplicative inverse as i explained above. there is also an example in my second comment with small numbers.

First thanks for the response.
Ok i begin to understand the concept but not ready to go yet ^^.
I know add or substract a point to another (different or equal) but i can't divide a point by any number if i only have the x and y coordinates.
If it's possible can you show me how divide this point by 2 for example ?

x = 72488970228380509287422715226575535698893157273063074627791787432852706183111
y = 62070622898698443831883535403436258712770888294397026493185421712108624767191

It is public key for private key: 10 , and we say here i haven't got this private key.

I expect to obtain the public key:

x = 21505829891763648114329055987619236494102133314575206970830385799158076338148
Y = 98003708678762621233683240503080860129026887322874138805529884920309963580118
(private key 5 )

Thanks in advance

In secp256k1 there are two important primes - the prime p which is used for coordinates x and y (2^256 - 2^32 - 977), and the prime n, which is the group order (2^256 - 432420386565659656852420866394968145599).

The group is defined by two operations - addition of two points, and doubling a point.

From this we easily could multiply a point by scalar, this is series of additions and doublings.

Multiplying a point by scalar gives another point. Multiplying a point by n gives the point at infinity (0,0).

Dividing by 2 in a group with order n is equivalent to multiplying by the scalar 1/2 (mod n).

One can find the inverse of 2 modulo n by the Extended Euclidean Algorithm.
1/2 (mod n) = 57896044618658097711785492504343953926418782139537452191302581570759080747169

You'd have to multiply (x,y) by 1/2 (mod n), this gives
x = 21505829891763648114329055987619236494102133314575206970830385799158076338148
y = 98003708678762621233683240503080860129026887322874138805529884920309963580118

[archyone]

this is all because of how 2^-1 is defined and its value is 57896044618658097711785492504343953926418782139537452191302581570759080747169 because 2 * 57896044618658097711785492504343953926418782139537452191302581570759080747169 ≡ 1 (mod N)
it can be computed using Euclidean algorithm or the simplified 2^(prime-2) mode prime.
now multiplying that with the point we had returns the half point

x = 21505829891763648114329055987619236494102133314575206970830385799158076338148
Y = 98003708678762621233683240503080860129026887322874138805529884920309963580118

why not. private key will tell you how many times to add G to itself. but after you are done and have the value for k*G as point P then you can do whatever you want with that point. for example you can add another G to that point and compute Q=P+G or R=P-G and similarly you can multiply that point with any number like computing S=3*P (the same way you would compute 3*G).
2^-1 is just another number that your point (P or G or Q,...) is multiplied by. you just have to first calculate what integer 2^-1 is in congruence with. to do that you compute its modular multiplicative inverse as i explained above. there is also an example in my second comment with small numbers.

Hum! I'm clearly a noob and i can't get it working..
I'm ok with the concept of multiplicative inverse of 2 --> 57896044618658097711785492504343953926418782139537452191302581570759080747169
But how use this number to half a public key with x and y coordinates
I have a lot of difficulty doing research with my broken English, someone can clarify this for me, perhaps with a python code ---> how half public key (private key 10) to get public key (private key 5)

It's now that I realize that mathematics is a profession ^^ lol

Thanks in advance

[BASE16]

Half a private key of 10 is not private key 5.
That is something entirely different these are curve points.
It's 5 curve points away.

For example think of it this way,
You can plot the first 10 points from G then you have a ladder that you can climb up and down.
Then you start at 10 and move back 5 points you will arrive at another point on the curve
Maybe that will make it a lot easier for you to understand.

PrivateKey:  1

X:  0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
Y:  0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8

PrivateKey:  2

X:  0xc6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5
Y:  0x1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a

PrivateKey:  3

X:  0xf9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9
Y:  0x388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672

PrivateKey:  4

X:  0xe493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13
Y:  0x51ed993ea0d455b75642e2098ea51448d967ae33bfbdfe40cfe97bdc47739922

PrivateKey:  5

X:  0x2f8bde4d1a07209355b4a7250a5c5128e88b84bddc619ab7cba8d569b240efe4
Y:  0xd8ac222636e5e3d6d4dba9dda6c9c426f788271bab0d6840dca87d3aa6ac62d6

PrivateKey:  6

X:  0xfff97bd5755eeea420453a14355235d382f6472f8568a18b2f057a1460297556
Y:  0xae12777aacfbb620f3be96017f45c560de80f0f6518fe4a03c870c36b075f297

PrivateKey:  7

X:  0x5cbdf0646e5db4eaa398f365f2ea7a0e3d419b7e0330e39ce92bddedcac4f9bc
Y:  0x6aebca40ba255960a3178d6d861a54dba813d0b813fde7b5a5082628087264da

PrivateKey:  8

X:  0x2f01e5e15cca351daff3843fb70f3c2f0a1bdd05e5af888a67784ef3e10a2a01
Y:  0x5c4da8a741539949293d082a132d13b4c2e213d6ba5b7617b5da2cb76cbde904

PrivateKey:  9

X:  0xacd484e2f0c7f65309ad178a9f559abde09796974c57e714c35f110dfc27ccbe
Y:  0xcc338921b0a7d9fd64380971763b61e9add888a4375f8e0f05cc262ac64f9c37

PrivateKey:  10

X:  0xa0434d9e47f3c86235477c7b1ae6ae5d3442d49b1943c2b752a68e2a47e247c7
Y:  0x893aba425419bc27a3b6c7e693a24c696f794c2ed877a1593cbee53b037368d7


So for your info you also start with X and Y so that is already two coordinates on the curve.

[archyone]

Ok, this confirms what i thought. Even if it is possible to halve a point, it is useless ^^
On the other hand I still do not know how to do ^^ lol
Thanks for the explanations all.