== Bitcoin challenge transaction: ~1000 BTC total bounty to solvers! ==UPDATED== - page 18.

sssergy2705

copper member

Activity: 205

Merit: 1

Quote from: MrlostinBTC on June 22, 2023, 04:17:27 PM

Quote from: sssergy2705 on June 22, 2023, 03:43:11 PM

Interested here is a question. If the end of the key is known, but there is no possibility of reverse recovery. Is it possible to restore the full key in this case? I remember there used to be a wif-solver-cuda program for wif, but is there such a program for hex private key?

It doesn't matter that the keys are in HEX. You can convert them to WIF and then use the according base keys to generate valid address. I encountered the same problem. WIF validation may be correct however there are way too many possible keys. What is interesting is uncompressed format there is a few less bytes to deal with, which in turn could be converted back to compressed if needed. Ill leave a couple examples here.

EC 0000000000000000000000000000000000000000000000000000000000000001
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU73sVHnoWn
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreAnchuDf

EC 0000000000000000000000000000000000000000000000000000000000000003
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU74sHUHy8S
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreB1FQ8BZ

EC 00000000000000000000000000000000000000000000000000000022382FACD0
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9P3MahktLW5315v
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB4BW8dsj4c9a6g

EC 000000000000000000000000000000000000000000000001A838B13505B26867 - puzzle 65
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZM21gaY8WN2CdwnTG57
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ipCnYRNeQuRFKarWVVs

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU73sVHnoWn
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU74sHUHy8S
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9P3MahktLW5315v
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZM21gaY8WN2CdwnTG57

So we could say upcoming puzzles private key 66 will start with...

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3

Can you show an example, for example on puzzle 80, by removing the middle part of the key? And how long will this work take?
00000000000000000000000000000000000000000000ExxxxxxxxxC11B5AD180

MrlostinBTC

jr. member

Activity: 50

Merit: 30

Basically yes it is the same, But because there are a lot of invalid WIF formats you could print only the valid ones, and then check them for valid keys. I guess if you could output valid keys very quickly and dump invalids even quicker it might be faster then what we are currently doing. Based on hardware limitations that I know about software is much faster at ignoring a input. However with the uncompressed byte being less it may be at least 1 byte less to scan. I guess its just a different way of looking at EC operations from the other end. And BTC core can handle 20,000 plus keys at a time during a scan. I know because I lost my own genesis wallet, and have tested it. And software can ignore something much quicker then process it.

If this helps someone please don't forget the little guy still learning python. And sorry if I sound a little special.

racminer

member

Activity: 245

Merit: 17

Quote from: MrlostinBTC on June 22, 2023, 04:17:27 PM

Quote from: sssergy2705 on June 22, 2023, 03:43:11 PM

Interested here is a question. If the end of the key is known, but there is no possibility of reverse recovery. Is it possible to restore the full key in this case? I remember there used to be a wif-solver-cuda program for wif, but is there such a program for hex private key?

It doesn't matter that the keys are in HEX. You can convert them to WIF and then use the according base keys to generate valid address. I encountered the same problem. WIF validation may be correct however there are way too many possible keys. What is interesting is uncompressed format there is a few less bytes to deal with, which in turn could be converted back to compressed if needed. Ill leave a couple examples here.

EC 0000000000000000000000000000000000000000000000000000000000000001
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU73sVHnoWn
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreAnchuDf

EC 0000000000000000000000000000000000000000000000000000000000000003
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU74sHUHy8S
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreB1FQ8BZ

EC 00000000000000000000000000000000000000000000000000000022382FACD0
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9P3MahktLW5315v
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB4BW8dsj4c9a6g

EC 000000000000000000000000000000000000000000000001A838B13505B26867 - puzzle 65
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZM21gaY8WN2CdwnTG57
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ipCnYRNeQuRFKarWVVs

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU73sVHnoWn
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU74sHUHy8S
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9P3MahktLW5315v
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZM21gaY8WN2CdwnTG57

So we could say upcoming puzzles private key 66 will start with...

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3

True, but you are just saying that in hex representation key to puzzle 66 is
000000000000000000000000000000000000000000000002XXXXXXXXXXXXXXXX
or
000000000000000000000000000000000000000000000003XXXXXXXXXXXXXXXX

which we already know lol

Denis_Hitov

newbie

Activity: 49

Merit: 0

Quote from: sssergy2705 on June 22, 2023, 03:43:11 PM

Interested here is a question. If the end of the key is known, but there is no possibility of reverse recovery. Is it possible to restore the full key in this case? I remember there used to be a wif-solver-cuda program for wif, but is there such a program for hex private key?

https://github.com/Mizogg/Tkinter-Power-Mini

WIF HEX DEC Recovery Tools

https://user-images.githubusercontent.com/88630056/207971831-5d34f507-f19a-4659-b731-b27fee75ea35.png

MrlostinBTC

jr. member

Activity: 50

Merit: 30

Quote from: sssergy2705 on June 22, 2023, 03:43:11 PM

Interested here is a question. If the end of the key is known, but there is no possibility of reverse recovery. Is it possible to restore the full key in this case? I remember there used to be a wif-solver-cuda program for wif, but is there such a program for hex private key?

It doesn't matter that the keys are in HEX. You can convert them to WIF and then use the according base keys to generate valid address. I encountered the same problem. WIF validation may be correct however there are way too many possible keys. What is interesting is uncompressed format there is a few less bytes to deal with, which in turn could be converted back to compressed if needed. Ill leave a couple examples here.

EC 0000000000000000000000000000000000000000000000000000000000000001
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU73sVHnoWn
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreAnchuDf

EC 0000000000000000000000000000000000000000000000000000000000000003
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU74sHUHy8S
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreB1FQ8BZ

EC 00000000000000000000000000000000000000000000000000000022382FACD0
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9P3MahktLW5315v
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB4BW8dsj4c9a6g

EC 000000000000000000000000000000000000000000000001A838B13505B26867 - puzzle 65
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZM21gaY8WN2CdwnTG57
5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ipCnYRNeQuRFKarWVVs

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU73sVHnoWn
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU74sHUHy8S
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9P3MahktLW5315v
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZM21gaY8WN2CdwnTG57

So we could say upcoming puzzles private key 66 will start with...

KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3

sssergy2705

copper member

Activity: 205

Merit: 1

Interested here is a question. If the end of the key is known, but there is no possibility of reverse recovery. Is it possible to restore the full key in this case? I remember there used to be a wif-solver-cuda program for wif, but is there such a program for hex private key?

Denis_Hitov

newbie

Activity: 49

Merit: 0

Quote from: digaran on June 21, 2023, 01:05:39 PM

Quote from: WanderingPhilospher on June 20, 2023, 10:04:12 PM

But mine all end with DP of 29

Can you tell me what is dp of 29, you mean the keys ending with 29?

DP - number of trailing zeros distinguished point

digaran

copper member

Activity: 1330

Merit: 899

🖤😏

Quote from: WanderingPhilospher on June 20, 2023, 10:04:12 PM

But mine all end with DP of 29

Can you tell me what is dp of 29, you mean the keys ending with 29?

Etar

sr. member

Activity: 617

Merit: 312

Quote from: WanderingPhilospher on June 20, 2023, 10:00:11 PM

I can run either wild only or tame only. In any range. I sent you an email weeks, or maybe months ago asking if you had source code for your workfile reconfiguration program.

Sent you PM.

WanderingPhilospher

full member

Activity: 1162

Merit: 237

Shooters Shoot...

Quote from: digaran on June 20, 2023, 11:14:57 AM

Any one here looking for a specific private key related to #125? You might have a lot of unknown public keys but you never know, someone might know the private key for just one of them, and to solve the puzzle all we need is 3 keys, 2 known in order to find the third unknown one.

So if you have a few keys and would like to work together, post them here to see if we have what you need. I will start by myself.

If anyone knows the private key for any of the following keys, we will share the prize 50-50 😉.

Code:

0239ddd9a2a1a113c105175e17903c1f72326ff89b109efc8b976cc9916429c9c4

Code:

0364696e1656dd8b0cca211c9cac57bfaedd1e4bc5b0f85405c9f3a76592f28f17

Code:

0304c40326f5350a090a16e3e3b8fad1ebd433989e825780d7586cd77f19c2652f

Code:

032a6b135b5792e58f9e3cea8c64efdb677d3be09decdaea575320faf73a89a0a7

Code:

027371ddafc0da5c86cc3800ce98faeee13469cab4636e534ad5d6181af4e2924a

Code:

0283a7519d110b317f20675715767762c950e8e4e8690a3e7e126fafa5759c43d5

Code:

0299bcb3387f04c75b423c86099e9ac48e42d9166f4d6fb7634d5eb24074d9c59f

This is not #125 key, it's a twin.

Code:

02a81ea7b0f52f34e249a8308300ae9f2bc6d28db649137343e093e5c84ec4f00e

Another twin.

Code:

0329ec62f37968906686bfad34ad5fdd9008fe187d868c0c4f9de1055b3062f00e

This one I have the private key for

Code:

0399ae0cf361425cbea86fc0c1fecdff5f61ca8a4c4c28a1db891f3d651dc7f00e

Code:

0286936a275e6d53bb2b2718c93d8a5aa44f371f6e0300abb73b89dd851d2fbe88

Code:

03a64a0b3739ddccddece6d90407c925717c75467cc8ce46321d73ec2663320130

Might post more later, lets see what you got, I showed you mine, it's your turn!😂

I have billions, where do you want to start lol.

But mine all end with DP of 29

WanderingPhilospher

full member

Activity: 1162

Merit: 237

Shooters Shoot...

Quote from: Etar on June 20, 2023, 05:17:23 AM

If there is a need to look for alot of different public keys in the same search range using kangaroo,
then you can make your work faster.
For example, our search range is 74 bits 0x4000000000000000000:0x7ffffffffffffffffff
First we need to prepare a working file. To do this, you need to run a kangaroo in this range with -w, and some public key that is definitely not in this range.
This is necessary in order to get enough tame kangaroos.
For 74 bit expected operations: 2^38.47 with DP16 it is 2^22.47 DPs and we need compute at least 3 times more kangaroos around 2^24.1 DPs.
Then we need remove wild kangaroos from working file. Our file wil contain around 2^23.05 tame kangaroos that is randomly distributed and continue their path.
After working file ready we must change the public key in the this file to the one of interest.
Also we need a version of kangaroo that produce only wild kangaroo, we don't need new tame kangaroo anymore.
Don`t forget that we need change the public key in the working file every time when change interest.
So every time we start with the same working file that contain only tame kangaroos but with changed public key in file.
And kangaroo programm got job from this working file.

For testing i a use randomly generated public keys in range 2^74, some are very close to the beginning, some near the end.
Here is result:

Expected operations: 2^38.47

Code:

[728.59 MK/s][GPU 728.59 MK/s][Count 2^37.44][Dead 2][04:49 (Avg 08:41)][347.2/440.5MB]
Key# 0 [1S]Pub: 0x03B043758AC54072BB816BDE28ECF833560D85286E1D86A3CD365C76882A8C87C6
   Priv: 0x402E66057958F4E9E0E

[727.18 MK/s][GPU 727.18 MK/s][Count 2^37.41][Dead 0][04:43 (Avg 08:42)][345.2/438.0MB]
Key# 0 [1S]Pub: 0x02633305FFE6B6238982F24044C864479C8A7DCFF444D7B9446732617F888D788B
   Priv: 0x414C10C6D7D89F38873

[730.19 MK/s][GPU 730.19 MK/s][Count 2^36.35][Dead 0][02:15 (Avg 08:40)][300.8/382.5MB]
Key# 0 [1S]Pub: 0x026ACEBA7AD37487DFAF24CAFE4302B379C3688ACA7CF332DC0FFB9050EA72F27F
   Priv: 0x46D59B75FE1AD6FCBB3

[749.64 MK/s][GPU 749.64 MK/s][Count 2^34.70][Dead 0][42s (Avg 08:26)][273.1/347.9MB]
Key# 0 [1S]Pub: 0x036E152745E2CB09DDA8F0E8FE5E6939111F751895E5E222EF50E25361E4751702
   Priv: 0x49CA4F75FC4DD4A0E98

[736.65 MK/s][GPU 736.65 MK/s][Count 2^36.19][Dead 0][02:01 (Avg 08:35)][296.6/377.3MB]
Key# 0 [1S]Pub: 0x02AE1FDEF6FB06E68EAAC3E0342A65E37ADDE369AD82B2C5B41AAE488A6C30BB48
   Priv: 0x539F9F353128FA9E829

[730.97 MK/s][GPU 730.97 MK/s][Count 2^36.92][Dead 0][03:20 (Avg 08:39)][320.5/407.2MB]
Key# 0 [1S]Pub: 0x022E7A857A9891F872A4610396FCAE10AC638B0D89D588B127A3C1423DA41174EF
   Priv: 0x554AEDB740506B9EFD1

[729.24 MK/s][GPU 729.24 MK/s][Count 2^37.33][Dead 0][04:27 (Avg 08:40)][340.6/432.2MB]
Key# 0 [1S]Pub: 0x027BBD8C6F41B43046A3E128037326EEC9807EC767C20E9529CD51034B5361963E
   Priv: 0x6BDFFBDB6411F328BB9

[738.33 MK/s][GPU 738.33 MK/s][Count 2^34.87][Dead 0][48s (Avg 08:34)][274.7/349.9MB]
Key# 0 [1S]Pub: 0x020DFBA9B7C4D7A53A6EE985AB11ACC06BEACC47EBC53D0751B455323F21CBD1B6
   Priv: 0x6D9D613445EFF402945

[741.74 MK/s][GPU 741.74 MK/s][Count 2^36.69][Dead 1][02:50 (Avg 08:32)][311.9/396.4MB]
Key# 0 [1S]Pub: 0x022FCF70783B01AE1EDDE1CAD273926E089D65CDC4D1510A5C642D9E10A527743F
   Priv: 0x71C02F72D735ABC20C9

[733.28 MK/s][GPU 733.28 MK/s][Count 2^35.80][Dead 0][01:33 (Avg 08:38)][288.1/366.6MB]
Key# 0 [1S]Pub: 0x026893801A8A18773DEC6C166C8E9081B09FB6B724F8D3B9825E5C20EDEA9085D6
   Priv: 0x7A54720FF5235105605

[722.31 MK/s][GPU 722.31 MK/s][Count 2^36.94][Dead 0][03:26 (Avg 08:45)][321.7/408.6MB]
Key# 0 [1S]Pub: 0x03150058CFFA65835192FBC8076C40377355189A810D04150882578E55FEB50672
   Priv: 0x7DBA59C1B7ACF79825E

As you can see all time we do not reach even expected operations/2
We do not count tame kangaroos that were produced once at the beginning.

I can run either wild only or tame only. In any range. I sent you an email weeks, or maybe months ago asking if you had source code for your workfile reconfiguration program.

I think the key is to set better traps, that is what I am currently doing, trying to set better traps for the wilds.

EDIT: I can generate 2^23.05 DP 29 tames in about 32 hours, while only running 8 GPUs.

digaran

copper member

Activity: 1330

Merit: 899

🖤😏

Any one here looking for a specific private key related to #125? You might have a lot of unknown public keys but you never know, someone might know the private key for just one of them, and to solve the puzzle all we need is 3 keys, 2 known in order to find the third unknown one.

So if you have a few keys and would like to work together, post them here to see if we have what you need. I will start by myself.

If anyone knows the private key for any of the following keys, we will share the prize 50-50 😉.

Code:

0239ddd9a2a1a113c105175e17903c1f72326ff89b109efc8b976cc9916429c9c4

Code:

0364696e1656dd8b0cca211c9cac57bfaedd1e4bc5b0f85405c9f3a76592f28f17

Code:

0304c40326f5350a090a16e3e3b8fad1ebd433989e825780d7586cd77f19c2652f

Code:

032a6b135b5792e58f9e3cea8c64efdb677d3be09decdaea575320faf73a89a0a7

Code:

027371ddafc0da5c86cc3800ce98faeee13469cab4636e534ad5d6181af4e2924a

Code:

0283a7519d110b317f20675715767762c950e8e4e8690a3e7e126fafa5759c43d5

Code:

0299bcb3387f04c75b423c86099e9ac48e42d9166f4d6fb7634d5eb24074d9c59f

This is not #125 key, it's a twin.

Code:

02a81ea7b0f52f34e249a8308300ae9f2bc6d28db649137343e093e5c84ec4f00e

Another twin.

Code:

0329ec62f37968906686bfad34ad5fdd9008fe187d868c0c4f9de1055b3062f00e

This one I have the private key for

Code:

0399ae0cf361425cbea86fc0c1fecdff5f61ca8a4c4c28a1db891f3d651dc7f00e

Code:

0286936a275e6d53bb2b2718c93d8a5aa44f371f6e0300abb73b89dd851d2fbe88

Code:

03a64a0b3739ddccddece6d90407c925717c75467cc8ce46321d73ec2663320130

Might post more later, lets see what you got, I showed you mine, it's your turn!😂

sssergy2705

copper member

Activity: 205

Merit: 1

Quote from: Etar on June 20, 2023, 05:17:23 AM

If there is a need to look for alot of different public keys in the same search range using kangaroo,
then you can make your work faster.
For example, our search range is 74 bits 0x4000000000000000000:0x7ffffffffffffffffff
First we need to prepare a working file. To do this, you need to run a kangaroo in this range with -w, and some public key that is definitely not in this range.
This is necessary in order to get enough tame kangaroos.
For 74 bit expected operations: 2^38.47 with DP16 it is 2^22.47 DPs and we need compute at least 3 times more kangaroos around 2^24.1 DPs.
Then we need remove wild kangaroos from working file. Our file wil contain around 2^23.05 tame kangaroos that is randomly distributed and continue their path.
After working file ready we must change the public key in the this file to the one of interest.
Also we need a version of kangaroo that produce only wild kangaroo, we don't need new tame kangaroo anymore.
Don`t forget that we need change the public key in the working file every time when change interest.
So every time we start with the same working file that contain only tame kangaroos but with changed public key in file.
And kangaroo programm got job from this working file.

For testing i a use randomly generated public keys in range 2^74, some are very close to the beginning, some near the end.
Here is result:

Expected operations: 2^38.47

Code:

[728.59 MK/s][GPU 728.59 MK/s][Count 2^37.44][Dead 2][04:49 (Avg 08:41)][347.2/440.5MB]
Key# 0 [1S]Pub: 0x03B043758AC54072BB816BDE28ECF833560D85286E1D86A3CD365C76882A8C87C6
Priv: 0x402E66057958F4E9E0E

[727.18 MK/s][GPU 727.18 MK/s][Count 2^37.41][Dead 0][04:43 (Avg 08:42)][345.2/438.0MB]
Key# 0 [1S]Pub: 0x02633305FFE6B6238982F24044C864479C8A7DCFF444D7B9446732617F888D788B
Priv: 0x414C10C6D7D89F38873

[730.19 MK/s][GPU 730.19 MK/s][Count 2^36.35][Dead 0][02:15 (Avg 08:40)][300.8/382.5MB]
Key# 0 [1S]Pub: 0x026ACEBA7AD37487DFAF24CAFE4302B379C3688ACA7CF332DC0FFB9050EA72F27F
Priv: 0x46D59B75FE1AD6FCBB3

[749.64 MK/s][GPU 749.64 MK/s][Count 2^34.70][Dead 0][42s (Avg 08:26)][273.1/347.9MB]
Key# 0 [1S]Pub: 0x036E152745E2CB09DDA8F0E8FE5E6939111F751895E5E222EF50E25361E4751702
Priv: 0x49CA4F75FC4DD4A0E98

[736.65 MK/s][GPU 736.65 MK/s][Count 2^36.19][Dead 0][02:01 (Avg 08:35)][296.6/377.3MB]
Key# 0 [1S]Pub: 0x02AE1FDEF6FB06E68EAAC3E0342A65E37ADDE369AD82B2C5B41AAE488A6C30BB48
Priv: 0x539F9F353128FA9E829

[730.97 MK/s][GPU 730.97 MK/s][Count 2^36.92][Dead 0][03:20 (Avg 08:39)][320.5/407.2MB]
Key# 0 [1S]Pub: 0x022E7A857A9891F872A4610396FCAE10AC638B0D89D588B127A3C1423DA41174EF
Priv: 0x554AEDB740506B9EFD1

[729.24 MK/s][GPU 729.24 MK/s][Count 2^37.33][Dead 0][04:27 (Avg 08:40)][340.6/432.2MB]
Key# 0 [1S]Pub: 0x027BBD8C6F41B43046A3E128037326EEC9807EC767C20E9529CD51034B5361963E
Priv: 0x6BDFFBDB6411F328BB9

[738.33 MK/s][GPU 738.33 MK/s][Count 2^34.87][Dead 0][48s (Avg 08:34)][274.7/349.9MB]
Key# 0 [1S]Pub: 0x020DFBA9B7C4D7A53A6EE985AB11ACC06BEACC47EBC53D0751B455323F21CBD1B6
Priv: 0x6D9D613445EFF402945

[741.74 MK/s][GPU 741.74 MK/s][Count 2^36.69][Dead 1][02:50 (Avg 08:32)][311.9/396.4MB]
Key# 0 [1S]Pub: 0x022FCF70783B01AE1EDDE1CAD273926E089D65CDC4D1510A5C642D9E10A527743F
Priv: 0x71C02F72D735ABC20C9

[733.28 MK/s][GPU 733.28 MK/s][Count 2^35.80][Dead 0][01:33 (Avg 08:38)][288.1/366.6MB]
Key# 0 [1S]Pub: 0x026893801A8A18773DEC6C166C8E9081B09FB6B724F8D3B9825E5C20EDEA9085D6
Priv: 0x7A54720FF5235105605

[722.31 MK/s][GPU 722.31 MK/s][Count 2^36.94][Dead 0][03:26 (Avg 08:45)][321.7/408.6MB]
Key# 0 [1S]Pub: 0x03150058CFFA65835192FBC8076C40377355189A810D04150882578E55FEB50672
Priv: 0x7DBA59C1B7ACF79825E

As you can see all time we do not reach even expected operations/2
We do not count tame kangaroos that were produced once at the beginning.

Can you explain in more detail how to prepare a working file?
Than it can be opened for viewing and editing.
And of course, where to get a version of kangaroo that only creates wild kangaroos.

Etar

sr. member

Activity: 617

Merit: 312

If there is a need to look for alot of different public keys in the same search range using kangaroo,
then you can make your work faster.
For example, our search range is 74 bits 0x4000000000000000000:0x7ffffffffffffffffff
First we need to prepare a working file. To do this, you need to run a kangaroo in this range with -w, and some public key that is definitely not in this range.
This is necessary in order to get enough tame kangaroos.
For 74 bit expected operations: 2^38.47 with DP16 it is 2^22.47 DPs and we need compute at least 3 times more kangaroos around 2^24.1 DPs.
Then we need remove wild kangaroos from working file. Our file wil contain around 2^23.05 tame kangaroos that is randomly distributed and continue their path.
After working file ready we must change the public key in the this file to the one of interest.
Also we need a version of kangaroo that produce only wild kangaroo, we don't need new tame kangaroo anymore.
Don`t forget that we need change the public key in the working file every time when change interest.
So every time we start with the same working file that contain only tame kangaroos but with changed public key in file.
And kangaroo programm got job from this working file.

For testing i a use randomly generated public keys in range 2^74, some are very close to the beginning, some near the end.
Here is result:

Expected operations: 2^38.47

Code:

[728.59 MK/s][GPU 728.59 MK/s][Count 2^37.44][Dead 2][04:49 (Avg 08:41)][347.2/440.5MB]
Key# 0 [1S]Pub: 0x03B043758AC54072BB816BDE28ECF833560D85286E1D86A3CD365C76882A8C87C6
Priv: 0x402E66057958F4E9E0E

[727.18 MK/s][GPU 727.18 MK/s][Count 2^37.41][Dead 0][04:43 (Avg 08:42)][345.2/438.0MB]
Key# 0 [1S]Pub: 0x02633305FFE6B6238982F24044C864479C8A7DCFF444D7B9446732617F888D788B
Priv: 0x414C10C6D7D89F38873

[730.19 MK/s][GPU 730.19 MK/s][Count 2^36.35][Dead 0][02:15 (Avg 08:40)][300.8/382.5MB]
Key# 0 [1S]Pub: 0x026ACEBA7AD37487DFAF24CAFE4302B379C3688ACA7CF332DC0FFB9050EA72F27F
Priv: 0x46D59B75FE1AD6FCBB3

[749.64 MK/s][GPU 749.64 MK/s][Count 2^34.70][Dead 0][42s (Avg 08:26)][273.1/347.9MB]
Key# 0 [1S]Pub: 0x036E152745E2CB09DDA8F0E8FE5E6939111F751895E5E222EF50E25361E4751702
Priv: 0x49CA4F75FC4DD4A0E98

[736.65 MK/s][GPU 736.65 MK/s][Count 2^36.19][Dead 0][02:01 (Avg 08:35)][296.6/377.3MB]
Key# 0 [1S]Pub: 0x02AE1FDEF6FB06E68EAAC3E0342A65E37ADDE369AD82B2C5B41AAE488A6C30BB48
Priv: 0x539F9F353128FA9E829

[730.97 MK/s][GPU 730.97 MK/s][Count 2^36.92][Dead 0][03:20 (Avg 08:39)][320.5/407.2MB]
Key# 0 [1S]Pub: 0x022E7A857A9891F872A4610396FCAE10AC638B0D89D588B127A3C1423DA41174EF
Priv: 0x554AEDB740506B9EFD1

[729.24 MK/s][GPU 729.24 MK/s][Count 2^37.33][Dead 0][04:27 (Avg 08:40)][340.6/432.2MB]
Key# 0 [1S]Pub: 0x027BBD8C6F41B43046A3E128037326EEC9807EC767C20E9529CD51034B5361963E
Priv: 0x6BDFFBDB6411F328BB9

[738.33 MK/s][GPU 738.33 MK/s][Count 2^34.87][Dead 0][48s (Avg 08:34)][274.7/349.9MB]
Key# 0 [1S]Pub: 0x020DFBA9B7C4D7A53A6EE985AB11ACC06BEACC47EBC53D0751B455323F21CBD1B6
Priv: 0x6D9D613445EFF402945

[741.74 MK/s][GPU 741.74 MK/s][Count 2^36.69][Dead 1][02:50 (Avg 08:32)][311.9/396.4MB]
Key# 0 [1S]Pub: 0x022FCF70783B01AE1EDDE1CAD273926E089D65CDC4D1510A5C642D9E10A527743F
Priv: 0x71C02F72D735ABC20C9

[733.28 MK/s][GPU 733.28 MK/s][Count 2^35.80][Dead 0][01:33 (Avg 08:38)][288.1/366.6MB]
Key# 0 [1S]Pub: 0x026893801A8A18773DEC6C166C8E9081B09FB6B724F8D3B9825E5C20EDEA9085D6
Priv: 0x7A54720FF5235105605

[722.31 MK/s][GPU 722.31 MK/s][Count 2^36.94][Dead 0][03:26 (Avg 08:45)][321.7/408.6MB]
Key# 0 [1S]Pub: 0x03150058CFFA65835192FBC8076C40377355189A810D04150882578E55FEB50672
Priv: 0x7DBA59C1B7ACF79825E

As you can see all time we do not reach even expected operations/2
We do not count tame kangaroos that were produced once at the beginning.

digaran

copper member

Activity: 1330

Merit: 899

🖤😏

Quote from: Etar on June 17, 2023, 09:56:23 AM

you can try fraction-bsgs or fraction-kangaroo it do what you need.
Step1 substruct range begin from public key
Step2 devide G by 2^24, it will be fraction
Step3 devide public key by 2^24, it was first public key to serach in range 1.. 2^100
Step4 substuct fraction from first public key , it wil be second public key to search
Step5 substruct fraction from second public key.......

In simple numbers how it work. For ex. our desired key is 379 that lie in range 2^9
We want search in range 2^6=64, divisor is 2^3=8
Devide 379/8 = 47,375
fraction = 1/8 = 0,125

Stage 0 search 47,375 in range 2^6 => result negative
Stage 1 subtract from it 0,125 => 47.25, search in range 2^6 => result negative.
Stage 2 subtract from it 0,125 => 47.125, search in range 2^6 => result negative.
Stage 3 subtract from it 0,125 => 47, search in range 2^6 => result positive.
So now we need multiply 47 by divisor 2^3=8 = 376 and add stage number =>376 +3 = 379
Desired key 379 founded in 3 stage of 8

From myself that if you are not lucky, then this is a bad idea, very bad idea.
With kangaroo puzzle#125 need +/- 2^63 op.
With division you will have 2^24 public keys with search range 2^100 that need 2^51 op (maxstep 2) for every pub.
Total 2^24*2^51=2^75 op and this is not a guarantee that the key will be found due to maxstep.

It seems you know how to sing a lullaby, yet you can't sleep yourself. 😉
Only if it was so easy, though what could be faster than dividing or finding a divisor, is to figure out what comes after 1 for #125, I have worked on this for 2 months now, God willing if my calculations are correct, then I am 98% certain what comes next.

If finding a single character of a low range key is so hard, imagine higher bit range keys, so I was wondering, now that I have managed to drop 2 digits from 32 digits of #125, how long would it take for kangaroo/BSGS to search in this new range which is a few times greater than 2^120?

kalos15btc

jr. member

Activity: 50

Merit: 1

Quote from: GR Sasa on June 17, 2023, 07:50:46 AM

I have other question. I want for fun reduce puzzle key #125 to total of 100 bits. Can anyone tell me the steps? AND tell me if this is actually possible? lol..

I know i will land on tons new pub keys but is there a program that reduce bits ?

yes and with easy steps
https://github.com/WanderingPhilosopher/Windows-KeySubtractor

Etar

sr. member

Activity: 617

Merit: 312

Quote from: GR Sasa on June 17, 2023, 07:50:46 AM

I have other question. I want for fun reduce puzzle key #125 to total of 100 bits. Can anyone tell me the steps? AND tell me if this is actually possible? lol..

I know i will land on tons new pub keys but is there a program that reduce bits ?

you can try fraction-bsgs or fraction-kangaroo it do what you need.
Step1 substruct range begin from public key
Step2 devide G by 2^24, it will be fraction
Step3 devide public key by 2^24, it was first public key to serach in range 1.. 2^100
Step4 substuct fraction from first public key , it wil be second public key to search
Step5 substruct fraction from second public key.......

In simple numbers how it work. For ex. our desired key is 379 that lie in range 2^9
We want search in range 2^6=64, divisor is 2^3=8
Devide 379/8 = 47,375
fraction = 1/8 = 0,125

Stage 0 search 47,375 in range 2^6 => result negative
Stage 1 subtract from it 0,125 => 47.25, search in range 2^6 => result negative.
Stage 2 subtract from it 0,125 => 47.125, search in range 2^6 => result negative.
Stage 3 subtract from it 0,125 => 47, search in range 2^6 => result positive.
So now we need multiply 47 by divisor 2^3=8 = 376 and add stage number =>376 +3 = 379
Desired key 379 founded in 3 stage of 8

From myself that if you are not lucky, then this is a bad idea, very bad idea.
With kangaroo puzzle#125 need +/- 2^63 op.
With division you will have 2^24 public keys with search range 2^100 that need 2^51 op (maxstep 2) for every pub.
Total 2^24*2^51=2^75 op and this is not a guarantee that the key will be found due to maxstep.

GR Sasa

member

Activity: 194

Merit: 14

I have other question. I want for fun reduce puzzle key #125 to total of 100 bits. Can anyone tell me the steps? AND tell me if this is actually possible? lol..

I know i will land on tons new pub keys but is there a program that reduce bits ?

digaran

copper member

Activity: 1330

Merit: 899

🖤😏

I have a question, is it even possible to brute force a key like puzzle #121? If it's possible I wonder what would be the cost?
And if the cost is more than the prize, no use, if the cost is less than the prize, again no use or purpose for designing this puzzle.

Think about it, revealing the exact range of every puzzle in order to see if people could brute force them while either the cost is more than the prize or less, this doesn't make any sense and has no reasonable purpose honestly.

It wouldn't even prove anything, because it's obvious brute forcing keys especially 256 bit keys without known public key is impossible even if you know the estimated range as this challenge is the evidence.

So in my opinion working on any puzzle is a waste of time and effort if the prize is less than the cost, having a bigger prize than the cost of solving a key is also nonsense, because what? I have enough money and could use GPUs to solve a puzzle in order to earn something more than I have spent?
And if I have the money and calculate the cost to see it's not worth it, I would never attempt to solve a puzzle.

I think @Satoshi, or the actual designer of this puzzle should rethink and revise the purpose of this challenge.

Imagine even if the whole world wanted to brute force a 256 bit key, they'd have to guess the range blindly and keep searching forever.

It would be nice to have unsolved puzzles above #67 moved to higher bit ranges with their public keys exposed, because I don't think there is any shortcut to solve a key with no known public key other than brute forcing sequentially, though having a public key is at least something we could work with and improve on existing tools.
No public key = waste of time.

bestie1549

jr. member

Activity: 75

Merit: 5

By the end of the day, all that matters is the prize. Whether you use -R or mathematics. your target is the prize. Do whatever you have to do to get the prize. Please don't forget to drop your closest calculations to the range of the puzzle. I'm always available for us to share some funds.

Topic: == Bitcoin challenge transaction: ~1000 BTC total bounty to solvers! ==UPDATED== - page 18. (Read 54326 times)