Quote from: albert0bsd on Today at 12:44:15 PM

Quote from: kTimesG on October 16, 2024, 02:34:36 PM

Together with the target address I will also provide the range of the search interval.

This will be in the form (just an example):

Code:

minKey = 0xf2e542b46066c4e6f91abc80000000000000000000185e689447431d74c5b133
maxKey = 0xf2e542b46066c4e6f91abcbfffffffffffffffffffd85e689447431d74c5b133

The Hamming length of the range will therefore be 80 contiguous bits, but they may start anywhere.

Nice Challenge

Well If you provide the EXACT range it can be solve subtracting the bits that remain same in both start an end range example for some key as your example:

Code:

StarRange: f2e542b46066c4e6f91abc80000000000000000000185e689447431d74c5b133
ENDRange: f2e542b46066c4e6f91abcbfffffffffffffffffff185e689447431d74c5b133

Key: f2e542b46066c4e6f91abcc206b831c41b15d6bf49185e689447431d74c5b133
Publickey: 0259f57acc739162a68785c08baf201da1b2797d90ded6cd5dde5c3f26b6df24ae
MSB : f2e542b46066c4e6f91abc000000000000000000000000000000000000000000

Subtracting Most significan bits:

Code:

./keymath 0259f57acc739162a68785c08baf201da1b2797d90ded6cd5dde5c3f26b6df24ae - 0xf2e542b46066c4e6f91abc000000000000000000000000000000000000000000
Result: 03dc89a446662604456da1f1fdfba0e578fb69e9750751081b3e070cb747280bfc

With this step the original key may look like: 0xc206b831c41b15d6bf49185e689447431d74c5b133

Subtracting Less significan bits:

LSB: 000000000000000000000000000000000000000000185e689447431d74c5b133

Code:

./keymath 03dc89a446662604456da1f1fdfba0e578fb69e9750751081b3e070cb747280bfc - 0x185e689447431d74c5b133
Result: 029659ffde564a27b7e6ea76936dcc6c33ef969d1a089a05214151ebc9ef4f9db7

With this step the original key may look like: 0xc206b831c41b15d6bf490000000000000000000000

Now we need to remove zeros on the right:

0xc206b831c41b15d6bf490000000000000000000000
0x10000000000000000000000

Code:

./keymath 029659ffde564a27b7e6ea76936dcc6c33ef969d1a089a05214151ebc9ef4f9db7 / 0x10000000000000000000000
Result: 027fb98c7edb7aef934a25eaf02384dbb42a9b8a32a85b451a5e13e5ebdd613f85

With this step the original key may look like: 0xc206b831c41b15d6bf49
So once that you solve the public key: 027fb98c7edb7aef934a25eaf02384dbb42a9b8a32a85b451a5e13e5ebdd613f85

You need to do do the same steps backwards to get the original key.

So this is just a simple 80 bit challenge that can be solve with kangaroo:

Now

Quote from: kTimesG on September 24, 2024, 05:35:04 PM

Currently I have a 80-bits solver DP database that can solve any key in under 6 seconds on a GPU, and under a minute if using the CPU.

If you can solve it under 6 seconds on GPU or under a minute un CPU then this is not any challenge if you already have the right tools and scripts to the operations on the public key.

Since time ago you have such DP dataset the I bet nobody is going to beat you

Quote from: kTimesG on October 17, 2024, 10:12:02 AM

Well, that's it, we're screwed. My 3-kangaroo C program also runs on Raspberry Pi. I also have it an Android app with native JNI bindings to it. So every time I wake it up it runs a few batches to mine DPs. Next step is to implement kang on mobile GPUs using Vulkan.

hello! I'm sorry, it seemed to me or you explained everything somehow difficult. to solve the problem, you need to change the generation of kangaroos or bsgs. so that she leaves the beginning and the end and goes through the middle (the 80-bit segment selected by him in a 256-bit key). the public key, he will give. I'm sorry again...