Pollard's kangaroo ECDLP solver - page 62.

NotATether

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Activity: 1568

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Quote from: ?? on ??

Receive input when playing the Kangaroo executable
Please modify the program so that the kangaroo can run it.

The reason is that Bitcoin 1% is an integer
99% was done with the decimal point key.

0.0573619 =0x822557f3ae9c29d0940e305bbcc30babe59d36ccfad2ebc972a323a8acefefc8, 0x1d5954c43bc29bd412c887d67759a0465aca7b7ce5195212b671e2ba21094dd1
0.0573619 * 10,000,000 = 573619
priv(573619) = 0x3ec274bef2990b86b8c0740faf90c11e15b8b1891c110063e3d15f05ea9bd23, 0x273f184f12a2a029c0643628561f670de859013a7dd0ef0382d86463f4b75435

I guess if you want to scale all calculations down by 1/10 or 1/100, you can replace the point in the SetBase16 code above with the one for 0.1G or 0.01G.

So, do you want to change the program to look for 0.1G, 1G, 10G, 100G points like this? (I can not read CJK languages so it is hard for me to understand the last screenshot, can you explain what's going on in there?)

To my understanding you have these values [for example]:

0.0573619 <-- this is a private key "A"
10,000,000 <-- this is some kind of multiplier I think
1505297...the value in green <--- a private key
0x822557f3ae9c29d0940e305bbcc30babe59d36ccfad2ebc972a323a8acefefc8, 0x1d5954c43bc29bd412c887d67759a0465aca7b7ce5195212b671e2ba21094dd1
^-- this is the public key for 1505297...? Huh

Or for 0.0573619?

So is this 1505297... private key multiplied by 10 million to get the 573619 private key? Or more likely you're multiplying their pubkeys to get that privkey, am I understanding you correctly?

What is the relation between these private keys?
573619
0.0573619
1505297...
0x15cf751db (I saw this in your second screenshot, not sure what this is I'm assuming a hex version of a privkey)?

Quote from: ?? on ??

Friends will hear this story for the first time.
However, I did a lot of tests and found these things.

My goal is not to race the quiz.
Many people are sad because they have lost their private keys.
I think you can solve this problem with a newly developed calculator

Well, I can understand that we are multiplying public keys A and B to get C, but what's special about the relation between their private keys (private key A and C in particular, why use a large number to get 573619?)

Quote from: pacnload on April 26, 2021, 12:40:59 AM

is there any way to jump over some numbers for example never use numer 3 or number 1?

Skipping particular kangaroo values? I think it will be hard to exclude specific values because they are all very large and random, but I guess you could find some group of numbers that match a pattern and sequence and do another hop for those, without making them a distinguished point.

pacnload

newbie

Activity: 28

Merit: 1

is there any way to jump over some numbers for example never use numer 3 or number 1?

fxsniper

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Activity: 406

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Quote from: NotATether on April 25, 2021, 01:30:48 PM

Is this bolded number and point corresponding to G? Because this will make the problem easier to understand. I'm not really sure what you're trying to do when you say "oscillator".

I'm assuming you took the modular inverse G*(10G)^-1 to get the 0.1 result? (And not just did straight up division of the 10 by the 1

)

I don't really see the point of using G/10 as the generator point for Kangaroo but it's technically possible if you edit the jump table.

I think lostbitcoin need to be show python modify code to can easy compare with C++ code, what want to do

fxsniper

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Activity: 406

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Quote from: jacky19790729 on April 25, 2021, 08:51:32 AM

I don't understand "kangaroos"

try read description on JeanLucPons Kangaroo github

explain about how kangaroos works with images illustration figure

https://github.com/JeanLucPons/Kangaroo

NotATether

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Quote from: ?? on ??

I want to do a test.
Part of the oscillator
0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798, 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8

I'd like to know that this is the most basic so it shouldn't be modified
For testing purposes, I want to change this public key value and run the kangaroo.

1 = 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798, 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8

10 = 0xa0434d9e47f3c86235477c7b1ae6ae5d3442d49b1943c2b752a68e2a47e247c7, 0x893aba425419bc27a3b6c7e693a24c696f794c2ed877a1593cbee53b037368d7
1/10 = 0.1

0.1= 0x8c6154856794fcd5ee79ba8c96b7dd980e289a6b93a52d50dbdb9388a4c75604, 0x8fa03512509c5caa2238ef16720c9b787961e695e5401a6f5df6817ad0032d5c
priv = (81054462466121336796499689506081535496986294995352433067823614199062713046036)

Is this bolded number and point corresponding to G? Because this will make the problem easier to understand. I'm not really sure what you're trying to do when you say "oscillator".

I'm assuming you took the modular inverse G*(10G)^-1 to get the 0.1 result? (And not just did straight up division of the 10 by the 1

)

I don't really see the point of using G/10 as the generator point for Kangaroo but it's technically possible if you edit the jump table.

jacky19790729

jr. member

Activity: 82

Merit: 8

I don't understand "kangaroos"
but, if you want to printf P.x, P.y , pk value
you can use GetBase16()

Public Key (compressed) = (02 or 03) + P.x
Public Key (uncompressed) = 04 + P.x + P.y
Target Private key = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141 - pk

Code:

  Point P = secp->ComputePublicKey(&pk);

  printf("\nTestPrint1   %064s  %064s \nPK: %064s\n"
	  , P.x.GetBase16().c_str() , P.y.GetBase16().c_str()
	  , pk.GetBase16().c_str() );

-----my result output

TestPrint1   9FFC74F6136E60E9229A58544C0E9CFFA5A9D955D949E75E4BCCBE1C06B81C34  
FD2B45B2609A46E7C893F3409073557BF59D9E306FF0BEC04AD8644CBC0DEFE3
PK: FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A027244DA849B5C9F250

====  2021-04-25 21:38:34 =====
找到了 Key# 0 [1S]Pub:  0x0290E6900A58D33393BC1097B5AED31F2E4E7CBD3E5466AF958665BC0121248483
       Priv: 0x349B84B6431A6C4EF1

This Find time 06:35

test in #70

fxsniper

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Activity: 406

Merit: 47

Quote from: NotATether on April 25, 2021, 06:00:01 AM

Wait those spaces in between shouldn't be there, I meant for it to be zero-padded, can you change the code to this?

Code:

::printf("Point X: %016X%016X%016X%016X \n", P.x.bits64[3], P.x.bits64[2], P.x.bits64[1], P.x.bits64[0]);
::printf("Point Y: %016X%016X%016X%016X \n", P.y.bits64[3], P.y.bits64[2], P.y.bits64[1], P.y.bits64[0]);

Thank you very much

last code compare with old one, am I do correct?
if correct I will remove all old code that wrong

fxsniper

member

Activity: 406

Merit: 47

Quote from: ?? on ??

I'm using Kangaroo 2.2
Windows version

Kangaroo 2.2 using C++ language

Visual Studio 2019 Community Edition (Free)
https://visualstudio.microsoft.com/downloads/

compile using project file Kangaroo.sln on folder VC_CUDA102
require to install CUDA version 10.2

I am not programmer it is hard to modify it a lot
Kangaroo 2.2 only can work with 120 bit puzzle

python it easy to modify but it very very slow and limited with 50 bit puzzle

NotATether

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Activity: 1568

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Wait those spaces in between shouldn't be there, I meant for it to be zero-padded, can you change the code to this?

Code:

::printf("Point X: %016X%016X%016X%016X \n", P.x.bits64[3], P.x.bits64[2], P.x.bits64[1], P.x.bits64[0]);
::printf("Point Y: %016X%016X%016X%016X \n", P.y.bits64[3], P.y.bits64[2], P.y.bits64[1], P.y.bits64[0]);

fxsniper

member

Activity: 406

Merit: 47

Quote from: NotATether on April 25, 2021, 02:19:16 AM

Don't do it like that, print it like this using 16-character padding for the hex numbers (they are 64-bit):

Code:

::printf("Point X: %16X%16X%16X%16X \n", P.x.bits64[3], P.x.bits64[2], P.x.bits64[1], P.x.bits64[0]);
::printf("Point Y: %16X%16X%16X%16X \n", P.y.bits64[3], P.y.bits64[2], P.y.bits64[1], P.y.bits64[0]);

Thank you very much

code it is works

compare with my test print directly

this is 64 bit number right , how can decode to decimal

NotATether

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Quote from: bigvito19 on April 24, 2021, 02:20:30 PM

NotATether, Did you ever work on the 128 + bits DP mask?

Yes I finished the 128+ bit range quite some time ago, you can get it from https://github.com/ZenulAbidin/Kangaroo-256

As for dpmask, what I discovered is that the runtime goes UP the bigger the dpmask you use. Search time becomes too long (in hundreds of years) when your dpmask is at 128 so I scrapped the addition.

Quote from: WanderingPhilospher on April 24, 2021, 02:30:54 PM

That was a joke...it would take you 2^128 jumps to find a 128 DP mask. So let's say your CPU (1 thread) can do 4,000 jumps per second. Now divide 2^128 / 4000 jumps per second = 85,070,591,730,234,615,865,843,651,857,942,052,864 seconds to find one 128 DP.

~~I get that it wasn't a practical change to make but it's nice to be able to run it without it chopping the range while you're unaware of it.~~ My bad, I thought you were talking about range

Quote from: gsciservices on April 24, 2021, 05:54:12 PM

Dear Sir,
I have many time tried to download for this file; it was unable to download yet. Now it is required a access to the permission. Please given to me permission for download: https://drive.google.com/file/d/1wQWLCRsYY2s4DH2OZHmyTMMxIPn8kdsz

Regards,
Md. Ashraful Alam

I'm not sure where you got that Google Drive link from but I don't recall it ever being posted in this thread.

Quote from: fxsniper on April 24, 2021, 08:37:54 PM

yes, I print directly

I don't know C++ command to convert to binary (like 1010101)
or convert to hex

I try only
::printf("Point: %02X \n", P);
(it worsk)

Don't do it like that, print it like this using 16-character padding for the hex numbers (they are 64-bit):

Code:

::printf("Point X: %16X%16X%16X%16X \n", P.x.bits64[3], P.x.bits64[2], P.x.bits64[1], P.x.bits64[0]);
::printf("Point Y: %16X%16X%16X%16X \n", P.y.bits64[3], P.y.bits64[2], P.y.bits64[1], P.y.bits64[0]);

fxsniper

member

Activity: 406

Merit: 47

Quote from: WanderingPhilospher on April 24, 2021, 10:06:54 AM

@fx
Also:

Quote

if can get collision point from save.work file by kangaroo.exe may be help to easy to set DP to match with
example

puzzle #40
ec9d5049bd4509aa41e1f5c312de1900d3eac4adba0692a41286b2497e520000 1110317004752
ec9d5049bd4509aa41e1f5c312de1900d3eac4adba0692a41286b2497e520000 106665591802

The Kangaroo.exe and the kangaroo python script you have been running will not match. Notice in your python script the zeros are at the end, they are "trailing". Kangaroo.exe uses leading zeros. Plus, if you use a range, Kangaroo.exe subtracts start range and reduces search by 1 bit. The two will never match unless you do some deep programming modification in one of the programs.

EDIT: and the Kangaroo.exe only saves the least (last) bits of the points, where your python script saves the entire point.

old python script I use
http://bitchain.pl/100btc/pollard_kangaroo.txt
and
https://github.com/Telariust/pollard-kangaroo/blob/master/pollard-kangaroo-multi.py

for modify python just code on output display but for process and algorithm I am not modify keep work from original

Can you help to modify save function to save full data purpose for read and learning not for use continue works?
How can I do make it save full point

fxsniper

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Activity: 406

Merit: 47

Quote from: NotATether on April 24, 2021, 11:25:23 AM

Quote from: fxsniper on April 24, 2021, 07:49:15 AM

image I try to print P value out but fould like this may be is binary file need to be convert to hex or decimal

You are printing the P value directly, this will make binary output since it's a class, you have to print each of P.x and P.y separately, and each of those is split into 4 64-bit parts called P.x.bits64[0 1 2 3] which all need to be printed with a separate printf() format character.

yes, I print directly

I don't know C++ command to convert to binary (like 1010101)
or convert to hex

I try only
::printf("Point: %02X \n", P);
(it worsk)

look like every value is store and works in binary
pk
P
keyToSearchNeg

all is binary (when print directly will get Alien language)

gsciservices

newbie

Activity: 8

Merit: 0

Dear Sir,
I have many time tried to download for this file; it was unable to download yet. Now it is required a access to the permission. Please given to me permission for download: https://drive.google.com/file/d/1wQWLCRsYY2s4DH2OZHmyTMMxIPn8kdsz

Regards,
Md. Ashraful Alam

WanderingPhilospher

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Activity: 1232

Merit: 242

Shooters Shoot...

Quote from: bigvito19 on April 24, 2021, 02:20:30 PM

Quote from: NotATether on March 13, 2021, 11:32:48 PM

Quote from: WanderingPhilospher on March 12, 2021, 10:48:54 PM

This is just not true. The RAM consumed/used is not dependent on the bits being search...it comes down to your DP setting and how often you save the work file. I don't see your settings/flags when starting the program but you are using a very low -d setting or that plus using a long save to file time.

I can see now that searching a 256-bit interval will become feasible if we use 128+ bits of DP mask

Quote from: NotATether on March 23, 2021, 01:25:10 PM

Quote from: bigvito19 on March 15, 2021, 02:16:11 PM

How feasible are you talking? So you are working on searching 256-bit interval using 128 + bits of DP mask too?

I forgot about this thread - sorry about that.

Yeah I am, so far I have the actual program running on GPU, it's running at ~260MKeys/s with the expanded dpmask on my T4 though but it's making quite a large number of same herd collisions and dead kangaroos. I had actually expected the speed to be much faster, like around ~1500MKeys/s given that I saw someone's V100 do ~1100MKeys/s.

I doubt checking three more uint64's for equality within the main loop is what's causing this speed drop but it's a good opportunity to peek into the CUDA accelerated Int class and see what else can be sped up.

NotATether, Did you ever work on the 128 + bits DP mask?

That was a joke...it would take you 2^128 jumps to find a 128 DP mask. So let's say your CPU (1 thread) can do 4,000 jumps per second. Now divide 2^128 / 4000 jumps per second = 85,070,591,730,234,615,865,843,651,857,942,052,864 seconds to find one 128 DP.

bigvito19

full member

Activity: 706

Merit: 111

Quote from: NotATether on March 13, 2021, 11:32:48 PM

Quote from: WanderingPhilospher on March 12, 2021, 10:48:54 PM

This is just not true. The RAM consumed/used is not dependent on the bits being search...it comes down to your DP setting and how often you save the work file. I don't see your settings/flags when starting the program but you are using a very low -d setting or that plus using a long save to file time.

I can see now that searching a 256-bit interval will become feasible if we use 128+ bits of DP mask

Quote from: NotATether on March 23, 2021, 01:25:10 PM

Quote from: bigvito19 on March 15, 2021, 02:16:11 PM

How feasible are you talking? So you are working on searching 256-bit interval using 128 + bits of DP mask too?

I forgot about this thread - sorry about that.

Yeah I am, so far I have the actual program running on GPU, it's running at ~260MKeys/s with the expanded dpmask on my T4 though but it's making quite a large number of same herd collisions and dead kangaroos. I had actually expected the speed to be much faster, like around ~1500MKeys/s given that I saw someone's V100 do ~1100MKeys/s.

I doubt checking three more uint64's for equality within the main loop is what's causing this speed drop but it's a good opportunity to peek into the CUDA accelerated Int class and see what else can be sped up.

NotATether, Did you ever work on the 128 + bits DP mask?

NotATether

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Quote from: fxsniper on April 24, 2021, 07:49:15 AM

image I try to print P value out but fould like this may be is binary file need to be convert to hex or decimal

You are printing the P value directly, this will make binary output since it's a class, you have to print each of P.x and P.y separately, and each of those is split into 4 64-bit parts called P.x.bits64[0 1 2 3] which all need to be printed with a separate printf() format character.

WanderingPhilospher

full member

Activity: 1232

Merit: 242

Shooters Shoot...

@fx
Also:

Quote

if can get collision point from save.work file by kangaroo.exe may be help to easy to set DP to match with
example

puzzle #40
ec9d5049bd4509aa41e1f5c312de1900d3eac4adba0692a41286b2497e520000 1110317004752
ec9d5049bd4509aa41e1f5c312de1900d3eac4adba0692a41286b2497e520000 106665591802

The Kangaroo.exe and the kangaroo python script you have been running will not match. Notice in your python script the zeros are at the end, they are "trailing". Kangaroo.exe uses leading zeros. Plus, if you use a range, Kangaroo.exe subtracts start range and reduces search by 1 bit. The two will never match unless you do some deep programming modification in one of the programs.

EDIT: and the Kangaroo.exe only saves the least (last) bits of the points, where your python script saves the entire point.

WanderingPhilospher

full member

Activity: 1232

Merit: 242

Shooters Shoot...

Quote from: fxsniper on April 24, 2021, 07:49:15 AM

Quote from: NotATether on April 24, 2021, 05:38:39 AM

To my understanding I believe it contains records with each entry (they're in binary if I remember correctly) having the positions of the tame and wild kangaroos, and possibly the number of kangaroos present. I'm not entirely sure, I have to check again.

your correct may be is binary

I try to print P Value after check pk and found

  Point P = secp->ComputePublicKey(&pk);

  if(P.equals(keyToSearch)) {
  if(P.equals(keyToSearchNeg)) {

P is point of collision point right

image I try to print P value out but fould like this may be is binary file need to be convert to hex or decimal

@fx, again, what are you trying to see or accomplish? You have seen the tame and wild files from the kangaroo python script you have ran. There is nothing different with the kangaroo.exe files. They contain point and distances. I do not understand your obsession with wanting to see the collision points and distances. Seeing them will not help you figure out a DP or help you in any way.

JP and a lot of other smart people have thought of pretty much everything to make kangaroo.exe the best kangaroo program out there. the DP is auto calculated based on bit size you are searching and the number of kangaroos you have (based on your GPUs/CPUs) to limit the amount of DP overhead.

The work/save files contain headers and then the tame and wild points and distances.

I've told you to use the -wsplit option if you want to see collision points. After key is solved, extract points and distances via the python script I showed you. THEN you can see the collision points.

Code:

TAME
01f57df72a208ecb0dff27e1836f190850728 0000000000000000c6b916f827ecc764
01f7fe7bfd3dc4c604b3e708c2fb4bdd2ed46 0000000000000000f862dc916dfc4479
01f84404e1bbaf74e5e86ddf86fa2e96f165a 0000000000000000d190f29c6c0596f6
34f64327b6e6129a2af1cc6a106f2cec5fcf0 00000000000000001f9e0d582f7942fd
34f6889b8abd66be32f13ebe2cf24d5ab6b62 00000000000000009b0984d32bf64396
34f6fdd77d4d6a532866be4fdfa5923a09ead 0000000000000000bc3de9e206c157db
--
WILD
34f44f024c8997abff6b5f1ee364c414c3016 0000000000000000344f3351ce019232
34f6889b8abd66be32f13ebe2cf24d5ab6b62 -00000000000000000d2f2c61d9bc24d1
34f738fed0a483f198a7a44d04e073c45b962 -0000000000000000602f88bbfaa6b336
01f7a2ae1656a659818a81f9a87204c29f1e8 -000000000000000048ea5f42dfb5bcc4
01f7fe7bfd3dc4c604b3e708c2fb4bdd2ed46 0000000000000000502a2b5c6849dc12
01f8143c41d86963bf9df718f9d0ba2e80cd7 00000000000000000a350d95bb55cb79

Code:

TAME: 01f7fe7bfd3dc4c604b3e708c2fb4bdd2ed46 0000000000000000f862dc916dfc4479
WILD: 01f7fe7bfd3dc4c604b3e708c2fb4bdd2ed46 0000000000000000502a2b5c6849dc12

SECRET_PRIV_KEY = TAME_DIST - WILD_DIST + START
0xf862dc916dfc4479 - 0x502a2b5c6849dc12 + 2**64 = 0x1a838b13505b26867

fxsniper

member

Activity: 406

Merit: 47

Quote from: NotATether on April 24, 2021, 05:38:39 AM

To my understanding I believe it contains records with each entry (they're in binary if I remember correctly) having the positions of the tame and wild kangaroos, and possibly the number of kangaroos present. I'm not entirely sure, I have to check again.

your correct may be is binary

I try to print P Value after check pk and found

Point P = secp->ComputePublicKey(&pk);

if(P.equals(keyToSearch)) {
if(P.equals(keyToSearchNeg)) {

P is point of collision point right

image I try to print P value out but fould like this may be is binary file need to be convert to hex or decimal

Topic: Pollard's kangaroo ECDLP solver - page 62. (Read 60698 times)