PS: @digaran, you are awesome , 3 months ago im teaching you how to substract from public keys
Modified the quote. WP taught me subtraction/addition, I learned division from garlonicon's posts. Please don't take credit for things you haven't done. Thanks for kind words.! |
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Are you sure you know what bsgs is? because I see you confused
You mean Satoshi doesn't know what he is talking about? I'm 85% certain he is the man himself, very interested in brute force tools and very knowledgeable about math and cryptography.😉 |
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Take a deep breath man and relax ! There is no need to call somebody parasite even if it's clearly a scammer! Don't forget, this is a community, I don't see the point of being toxic in your comments!
Ok, lets bring a few virgins for that guy to make it up to him. Anyone here knows how to twerk? Or do I have to take another deep breath and do the twerking myself? You need to drink a glass of milk to wash the toxins away.😉 Why do you jump at people's throats so much? Do you have any frustration or do you simply think you are the owner of the absolute truth? Everyone has the right to use or recommend applications to others, without being offended by some like you. Maybe you should take your pills on time, so you don't go into fibrillation. ! Prove that PuzzleBTC contains viruses or trojans and then make claims ! Of course, I forgot my pills, now you mentioned it. Can you prove it doesn't have trojan/virus? No? Then kindly stay silent. See? I didn't jump at your throat, there have been many such newbie accounts posting a link for a closed source tool claiming their tool is better and faster. Firstly, if a program is better than the others, why don't they keep it for themselves to have an advantage in finding the keys? Secondly, if they want to help others by sharing something useful, why not posting a link to the source? You can't be good and helping, at the same time denying people the right to take a look at your code, unless there is something suspicious going on. Have you seen me physically stopping people from using these malwares? Of course not, they are still free to use them, but if nobody warns them about it, they'd assume "it's safe, because if it wasn't, surely someone with a brain would have warned us about it." If you post a script for example and say it does one thing that I want, but it also does something unwanted by me, if I'm a newbie and unsuspected, if nobody says anything, I'd assume it's perfectly safe and would run it unaware of the dangers and consequences. Next time come here defending such people, you will see what a real jump at throats mean!😉 |
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for compressed public keys import secp256k1 as ice
target_public_key = "032f3342152eff6aca5e7314db6d3301a28d6a90ddcfd189f96babadc2a053d392"
target = ice.pub2upub(target_public_key)
num = 100 # number of times.
sustract= 1 #amount to subtract each time.
sustract_pub= ice.scalar_multiplication(sustract)
res= ice.point_loop_subtraction(num, target, sustract_pub)
for t in range (num+1):
h= (res[t*65:t*65+65]).hex()
hc= ice.to_cpub(h)
data = open("data-base.txt","a")
data.write(str(hc)+"\n")
data.close() One last thing, how do I change the generator point in the script above? I even went to ice.secp256k1.py but didn't see x and y for G to change it, is there a way to include a new G in your script? Thank you so much for teaching us new things.👍 Ps, I have learned that we could use inverse keys to do multiplication on a script which only divides, lol.
Besides keyhunt and bitcrack.
What are you guys using here, I see a lot of different script some Python some C++ got confused!
are you guys collecting private keys with one script and checking them on another script?
Also there's no puzzle to think about! just a random guess in a range of private keys? right?
What do you know about elliptic curve math? If you are new, I suggest learning everything you can about *, +, -, / of public keys, without knowing that you'd be searching blindly using BSGS and kangaroo etc, and don't bother searching for #66, #67 etc, unless you know programming and can develop a new and faster brute force tool. |
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import secp256k1 as ice
target_public_key = "023d62d9d64a7164a2ae6f0561f7e8317e69b4a1ee61048fe768a1316b39b1d3a7"
target = ice.pub2upub(target_public_key)
num = 100 # number of times.
sustract= 1 #amount to subtract each time.
sustract_pub= ice.scalar_multiplication(sustract)
res= ice.point_loop_subtraction(num, target, sustract_pub)
for t in range (num+1):
h= res[t*65:t*65+65]
data = open("data-base.txt","a")
data.write(str(h.hex())+"\n")
data.close() Is it possible to make it possible to specify an unlimited number in num? If set num = 1000000000 That throws an error: Traceback (most recent call last): File "D :\PubSub\PubSub.py", line 8, in <module> res= ice.point_loop_subtraction(num, target, sustract_pub).hex() File "D :\PubSub\secp256k1.py", line 504, in point_loop_subtraction res = _point_loop_subtraction(num, pubkey1_bytes, pubkey2_bytes) File "D :\PubSub\secp256k1.py", line 497, in _point_loop_subtraction res = (b'\x00') * (65 * num) MemoryError More importantly, how to get compressed public keys as result? Lol I changed *65:*65+65 to *32:*32+33 but it messed up everything. |
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پازلهایی که به احتمال ۹۹ درصد خود ساتوشی طراحی کرده، چنین ادعایی باید بر اساس شواهد و مدارک باشه نه احساسات. یعنی مثلا ساتوشی باید سابقه پازلسازی داشته باشه یا مثلا به طور مستقیم گفته باشه که پازلی در بلاکچین گذاشته باشه. که تا جایی که میدونم چنین چیزهایی نبوده.
از نظر منطقی هم چنین چیزی برای سیستمی مثل بیتکوین امکانپذیر نیست، عقلانی هم نیست. چون چنین کاری باعث میشه اعتماد به بیتکوین از بین بره خصوصا اینکه از قبل هیچ اطلاعی داده نشده. اگه ناگهان یکی از کوینهای روزهای اول بخصوص اونهایی که منتصب به ساتوشی هستند خرج بشه (با فرض اینکه پازل بوده)، برای کل سیستم بده. چون تنها نتیجهگیری که میشه اینه که خود سیستم باگ داشته.
یعنی چی عقلانی نیست؟ انکار میکنی که یه نفر ۱۰۰۰ بیت کوین تو ۱۶۰ آدرس گذاشته تا ببینه با چه سرعتی میتونن پیداشون کنن؟ حالا یا خود ساتوشی یا هر ننه قمری، پیدا کردن و برداشتن اون بیت کوینهارو اسمش رو میذاری دزدی؟ حالا خوبه دوتا تاپیک چند صد صفحه ای در مورد این جریان وجود داره. و اینکه خود طراحش سال ۲۰۱۹ در موردش پست گذاشته و از ابزارهای بکارگرفته شده تعریف هم کرده و گفته که قصدش دقیقا اندازه گیری توان پیدا کردن اون بیت کوینها توسط اهالی کریپتو بوده. البته شاید یه فرد دیوانه که ۱۰۰۰ بیت کوین براش ارزشی نداشته این کار رو کرده، اسمش چیه و کی هست مهم نیست، مهم اینه که پیدا کردن اونا دزدی نیست. 😉 |
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Take a deep breath man and relax ! There is no need to call somebody parasite even if it's clearly a scammer! Don't forget, this is a community, I don't see the point of being toxic in your comments!
Ok, lets bring a few virgins for that guy to make it up to him. Anyone here knows how to twerk? Or do I have to take another deep breath and do the twerking myself? You need to drink a glass of milk to wash the toxins away.😉 |
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Yes there is, already taken care of, if you are the first one spending from that address, nobody else even you can not double spend it, RBF is turned off for all puzzle keys.
Where you get this info, If attacker send new transaction with 3BTC as fees then all pools will be so happy to replace the founder(who solve the puzzle) transaction that in best case have fraction of 1BTC as fees. Well, I have no useful info about RBF other than what I said, but you are right we can never stop double spending unless miners respect RBF signals from txs. But I'd say for puzzle 66 it would take at least 5 mins to find the key, but this should be automated where a script runs kangaroo after seeing the address has broadcasted the public key, and then after finding the key it needs to double spend it, so for our thief it takes around 5 mins. 66 solver is screwed.🤣 |
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IMO, if 2 or 3 DT members / merit sources vouch for an established member to become a source, there is no harm in trusting the already trustworthy DT / merit source members and accept the application, however if it's the matter of not wanting to add more merits in circulation, the new source could get his smerit source allocation from those who vouched for him.😉
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I recently started to approach the search for puzzle keys and I was wondering something.
how many chances are there that if i find for example the key of puzzle 66, which is between
0000000000000000000000000000000000000000000000020000000000000000
000000000000000000000000000000000000000000000003ffffffffffffffff
for example, find the key of the 66 puzzle, i send the btc from the address to my address thus exposing the public key. How long does it take for a person using kangaroo to find the private account, thus "stealing" my bitcoins, before my tx is confirmed? Are there any solutions to avoid all this? Tnx
Yes there is, already taken care of, if you are the first one spending from that address, nobody else even you can not double spend it, RBF is turned off for all puzzle keys. |
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دوست عزیز یک سوال؟ اگر کار شما درست بشه و ایدهای که بالاتر گفتی بگیره و چند میلیون دلار دربیاد، اون پول چند میلیون دلار از کجا قراره در بیاد؟ قراره برنامه کیفپول ملت رو خالی کنه دیگه. پس به نظرم بهتره پای خدا وسط نیاد . چون احتمالا با اوناس  د این مساله ی علمیه و مربوط به پیدا کردن راه حل برای حل کردن پازلهایی که به احتمال ۹۹ درصد خود ساتوشی طراحی کرده، و اینکه از عمد یک سری بیت کوین گذاشته که هرکسی تونست برداره برای خودش. . میدونی حالا که در مورد حرفتون فکر میکنم، با استدلال شما علم ساخت سلاحهای مرگبار، مواد مخدر خانمانسوز و خیلی چیزهای دیگه که مسلما از جانب خداوند بوده، دلیل بر اینه که خدا با جنایتکارها و آدمکشها بوده؟ این ماییم که در نهایت باید تصمیم بگیریم با علمی که خدا در اختیارمون
میذاره چیکار کنیم.
البته من از شما انتظاری ندارم، ظاهرا در مورد چالش ۱۰۰۰ بیت کوینی اطلاعی ندارید که تهمت دزدی به من میزنید. من توی نوشته هام معمولا یکم شوخ طبعی دارم. امیدوارم ناراحت نشده باشین دوست عزیز.
مثل اینکه سوتی دادم. نه خبری در مورد چالش ندارم. هرکسی باشه ناراحت میشه، اما سوء تفاهم پیش میاد و برطرف میشه و قابل گذشته، اما من در عجبم که استاد ماله کشی که در مورد چالش با خبره چرا باید با مریت دادن به پست شما بر این ادعای نادرست صحّه بذاره البته چنین مرامی از این قوم الظالمین دور از انتظار نیست 😅 |
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60f4d11574f5deee49961d9609ac6
strange_63_digest_PRIVATE
You could use a translator or simply use chat GPT, or ask from your native language board to assist you, anyways. This key 60f4d11574f5deee49961d9609ac6 is for puzzle #115, did you find this after 3 years? What is strange 63 digest private? Some sort of secret code?
Here i am, I got you, i have found that special magic number that meets exactly the description you described
I knew I could count on our crypto experts, could you give me the private key for the following point ( public key = 03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852 Solely for educational purposes.😅
Sorry if my idea sounds stupid, I'm a simpleton and can only think like one, easy and simple. |
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I'm guessing finding a way to implement a backdoor on a curve is extremely difficult, otherwise we could have seen such curves by now.
The only thing I could think of, is having a special number which when divided/multiplied by any point on curve mod some other special number resulting in the private key for that point.
I strongly believe there are such numbers/ values to just do that, but the question is how? Math+ECC expert could figure that out.😉
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It is a pity that such an operation will not work with an odd key))
Sure it will, if you add an odd key to another odd key ( our 2^256 key is odd ) you'll get an even key and all you have to do is to subtract the following key from the result : a2a8918ca85bafe22016d0b997e4df5f But why bother if you could add or subtract 1G from your odd key and divide it without having to do what I said. You can always divide an odd key, if by 2, you'd have to add n/2 to the result, if by 3, adding n/3 etc to get your actual and correct result. sorry my english is not good, i'm trying to understand by translating with translate, this gives me different sentences, so can you write the python code for the division you said. I'm not a coder but here is the idea. Translate the following. 11/3 = 3.6666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666667 Secp256k1 representation of the fraction above : 55555555555555555555555555555554e8e4f44ce51835693ff0ca2ef01215c0 11 decimal 0xb/3 = 55555555555555555555555555555554e8e4f44ce51835693ff0ca2ef01215c4 Subtract both above, result is 0x4.
Now the twisted division I talked about, simple : Target 7, is odd, we add 3 to get 10, divide 10 by 2 = 5, and if we subtract half of 3 = 1.5 from half of 7 = 3.5 we get 2, now we subtract 2 from 5 to get the real half of 7 in secp256k1. |
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It is a pity that such an operation will not work with an odd key))
Sure it will, if you add an odd key to another odd key ( our 2^256 key is odd ) you'll get an even key and all you have to do is to subtract the following key from the result : a2a8918ca85bafe22016d0b997e4df5f But why bother if you could add or subtract 1G from your odd key and divide it without having to do what I said. You can always divide an odd key, if by 2, you'd have to add n/2 to the result, if by 3, adding n/3 etc to get your actual and correct result. |
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you struggle in vain and miserably like many others to solve a puzzle. Find the mistake.
Don't you struggle like all other intelligent individuals hunting this puzzle? If not then that's understandable, no pressure. Around these woods men break under pressure, previous pages are my witness. You might break your nails by passing by. who broke SECP256K1
That's a promise, God willing I'd either deliver on that or die trying.😂 but doesn't want to share his genius findings.
Lol, what genius I use calculator to subtract 176 from 239, not lying. Anyways, do you have something useful that can help solving a puzzle? Stupid question, of course you don't.😘 |
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Some elliptic curve magic ahead! Public key for 2^256 in secp256k1 : 03dd3625faef5ba06074669716bbd3788d89bdde815959968092f76cc4eb9a9787 Private key : 000000000000000000000000000000014551231950b75fc4402da1732fc9bebf Public key divided by 2 : 02b23790a42be63e1b251ad6c94fdef07271ec0aada31db6c3e8bd32043f8be384 Private key : 2^255 8000000000000000000000000000000000000000000000000000000000000000 Now we add an even key to our 2^256 : Target Pub 034a4a6dc97ac7c8b8ad795dbebcb9dcff7290b68a5ef74e56ab5edde01bced775 Target Prv 0000000000000000000000000000000000000000000000000000000000008000 Result : pub 02c028224b2c45bd797143e8f32f025e24601ed85f27ef310d7d55020a192ddba5 Prv 000000000000000000000000000000014551231950b75fc4402da1732fca3ebf Divide by 2 , result : 0216ca7e1edb137684b4aa8a41fd0f8a89dcb773b9db807a9f3f864de2161735ff Now subtract pub 2^255 from above, result : 03111d6a45ac1fb90508907a7abcd6877649df662f3b3e2741302df6f78416824a Prv, also real half of our original target : 03111d6a45ac1fb90508907a7abcd6877649df662f3b3e2741302df6f78416824a , 0000000000000000000000000000000000000000000000000000000000004000 I just enjoy making a simple division difficult and twisted! 🤣, now chop chop start your brain's engine and do some calculation, large fractions could be solved by accounting for the above results, not telling you how. Dive deep and let your brain solve it.😉 |
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basically by changing Gx, Gy for example by pubkey of pk #1000
the sequence of the curve would be 1000,2000,3000,4000...
and you will wonder what happens with 1001,1002,1003...1999.
for that you subtract 1, 1000 times from the target and store it in the database.
If you search for 10924, and subtract 1000 times 1, the database would look like this: 10923,10922,10921...10000...9999,9998.
therefore you would find the pk 10000 in 10 steps, and you would know that your target is in the range 10000-10999
So the database is our actual target and some of it's offsets and we select a known key and add it as target so the script adds or subtracts from our known key until it lands on any one of the keys stored in database? Is data base file loaded into the RAM? How many keys per second can we generate and compare at the same time? This looks good as an idea, but in reality the performance is not what we need. Thanks for explanation.
it also feels like why am I sharing my hard work with those who have such negative thoughts about me,
Some say football some say soccer! Why is this hard work of yours in a downloadable file with closed source binaries? Share this hard work which seems to be magically correct and accurate by posting the ranges as a text file. Then we might discuss about you being a scammer or dishonest or both after that. |
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increase the search speed of the Pubkeys, dividing the search range. First of all, let me make it clear that this script is slow and basic, I did it like this in order to explain the range division method. If you want to apply it quickly, modify keyhunt or other c++ code before compiling to make it work with this method. The method consists in changing the parameters of the curve by x,y of a higher point (for the script I used pk#100). by default the curve uses pubkey #1 what works in sequence 1,2,3,4.... pubkey #100 which works in sequence 100,200,300,400.... if we want to find pk# 3922 in a range of 2000:6000 in the traditional way we would need to advance +1900 pubkeys In this method we will only need 19 pubkeys (1900/100). the database is done by subtracting 1 consecutively from the target pubkey (in my case I need to subtract( 1), 100 times because I chose pubkey pk# 100 as divisor. The trick is in your ability to use fast databases, the more pubkey you store in the database, the higher the divisor will be, therefore the search range is smaller. save as brute_div.py import os
import hashlib
import random
import binascii
#target pk decimal 3922
#Target_Upub "042f3342152eff6aca5e7314db6d3301a28d6a90ddcfd189f96babadc2a053d3929d0e4bef850141613e6bb3d3e2c480b3634af4332f125f4edfc65b88816f541d"
#Start_Range in decimal
Start_Range = 2000
#End_Range in decimal
End_Range = 6000
#divisor pk in decimal
boost = 100
#boost_upub "X=ed3bace23c5e17652e174c835fb72bf53ee306b3406a26890221b4cef7500f88, y=e57a6f571288ccffdcda5e8a7a1f87bf97bd17be084895d0fce17ad5e335286e"
class Point:
def __init__(self,
#change x and y for boost_upub
x=0xed3bace23c5e17652e174c835fb72bf53ee306b3406a26890221b4cef7500f88,
y=0xe57a6f571288ccffdcda5e8a7a1f87bf97bd17be084895d0fce17ad5e335286e,
p=2**256 - 2**32 - 2**9 - 2**8 - 2**7 - 2**6 - 2**4 - 1):
self.x = x
self.y = y
self.p = p
def __add__(self, other):
return self.__radd__(other)
def __mul__(self, other):
return self.__rmul__(other)
def __rmul__(self, other):
n = self
q = None
for i in range(256):
if other & (1 << i):
q = q + n
n = n + n
return q
def __radd__(self, other):
if other is None:
return self
x1 = other.x
y1 = other.y
x2 = self.x
y2 = self.y
p = self.p
if self == other:
l = pow(2 * y2 % p, p-2, p) * (3 * x2 * x2) % p
else:
l = pow(x1 - x2, p-2, p) * (y1 - y2) % p
newX = (l ** 2 - x2 - x1) % p
newY = (l * x2 - l * newX - y2) % p
return Point(newX, newY)
def toBytes(self):
x = self.x.to_bytes(32, "big")
y = self.y.to_bytes(32, "big")
return b"\x04" + x + y
def getPublicKey(privkey):
SECP256k1 = Point()
pk = int.from_bytes(privkey, "big")
publickey =(((SECP256k1 * pk).toBytes()).hex())
return publickey
if __name__ == "__main__":
#range Div
for i in range((Start_Range//boost), (End_Range//boost)):
print(i)
b = hex(i)[2:]
c = str.format(b)
pk= c.zfill(64)
Seq_Bytes = binascii.unhexlify(pk)
PublicKey = getPublicKey(Seq_Bytes)
#print(PublicKey)
Privatekey = (Seq_Bytes).hex()
#print(Privatekey)
data_base = 'data-base.txt'
with open(data_base, 'r') as file:
content= file.read()
#search publickey in data-base
if PublicKey in content:
print("Range Found")
print("Decimal:", int(Privatekey, base=16)*boost)
save as data-base.txt 040119630133edf190d14fdb00640f74cc4317e1ddb9dff1c9707eac3353ef9c21f16ab4e5e3a3ee6d743851876da07034caa4f5196106a0c803136d215420494c
04572dd1b78d9e4d185117860c71a224a05daee2b7bba651478ddbabadc71831c1d3fd83384c92bcaa326b75b10ef7cb33baf8b4d474cf8e5eac0e598a8a13200f
0462cc463c16ee8e8f16ea85175508278d4d5577e3556d3580ec8d90e3bbb97072cd294843e299c500846a6c91c913395b56b48efa2f9ab3fcb81863fe8a6a891e
040211625f596c1d65bd9f79fdd1fb4803f3165be3205f06e712aea0f191c08afa6ac2fd437b1bedadac240964a602346ea881d497f9a097a628f448006756c169
042144649d508cf5daa7b422f822e9125fd5a10965a9d32c6cc299bad46616771d1153178c0e33c445f03ef96b35067ceb76ddeacad9ddba1a74acd229d01ca76e
04627bbb09622ac837cef0922098ac6736db9599c855d32bb1d7f5d06d46dfb063fc5ee8575d03e1db328e6271f5e2c82eb7995ce60e80936976343faa67d329a7
04739fb43ab1b0715539c0b0ae43901bf2bc051a698ee8d50384fcc1f0ec37312ac12237043c8822262d4abe331de6ab4dff90d6b84c0b59e7a193e7636c6b7e28
04742771eed11bcc2c849c1ae6316eec9242dae6c581792b893802c01c2c6fb9b6cc3cc9dd51c8bcc37b93fd5b1c0d4026cecbf6b22637f4160229ddb0903505d7
0426ad559046e19dbf0e60953ed40a3958a357f6bc236422790cbcc35574fdab68af6fc5ef906721c0553c6a7460bc6894e5bb77e74dbc25e2184fb3f0fa795508
04abcf6e30599e5b9144892e59bbed1400f64fe0617d74b106fe7b61b250a84d165074e565f5908c234e798279d9ffff6394f8d94a2b7f0ee61e286b09a86b987a
0455ef868b6fc2923ee624fc0965f61852b3aeb38ce2c6b3bc54b86b3a60f0ab81ec198aeaab16b4cc5354d0adc66051fb76cff6df22dc1a0e72ce6e791622efab
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I currently don't have what it takes to code in C++ using the quick search methods, my resources are limited to the basics. I currently don't have an up-to-date pc to do the relevant tests (the disadvantages of the third world). However, later I will share faster methods, ECC tricks. Can you explain what this script does? I mean do we need to just provide it with a target upub, set a range and change G to whatever we want ( stride ) and then what about the database, where should we get our database keys from? |
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How do we know we are scanning a correct range, I mean what if someone scans all 588 ranges and finds nothing? You see, searching for addresses is a blind search no guarantee to land on the target unless you search the entire original range one by one. |
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