Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 161. (Read 215479 times)

From the moment you compare the puzzles with each other and try to derive and predict future events from existing values, you are talking about a pattern. On the subject of coincidence you can find enough worth reading on the Internet. What bothers you about the fact that Puzzle 66 consists of 44 zeros and 22 ones? Give a good reason why this could not occur.

Sometimes one can't see what one is not looking at and you can't find what you're not searching for but something you barely imagined can surface in your eyes beyond one's imaginations as to the fact that we didn't imagine what we are actually working with at the moment and a 66 bit key could be reduced to 60 bits if your guesses for the numbers of 1s are right, now imagine scanning randomly and scanning sequentially without missing a single key whereas randomness can almost iterate over the key without actually finding it because it moves randomly. only if there is a pool where your already scanned random key is stored for the scanned key not be scanned again will it be possible for one to actually find the key randomly, except otherwise stated, the search for the 66 bit random key will take 100 times more for the key to be gotten randomly.

If we have something that has potentials towards the search, it's better to give it a try but if you're hoping on luck or randomness to ascertain the 66 bit key. I have wasted over $23k renting the GPUs looking for the private key in question but all to no avail because I was trying to randomly search for the key in question and unfortunately I wasn't lucky because if I had tried to search sequentially I would barely have successfully search from 20-21 with that amount so I resorted to randomly scanning and nothing came out of the investment until I was down to the last $0. so if we have something with potentials I think its best to give it a try rather than trying to discourage ourselves. because right here, the sky is the limit

Indeed, the patterns compared between the keys 30-60, 32-64, and so on, might not truly exist. However, the project is not about patterns; it explores the property of randomness. If you generate a random key with 66 bits, how many 1s and 0s will it have? Most of the time, the proportion will align with what has been observed here. As the creator mentioned it was random, we have further evidence that a truly random key will exhibit an acceptable proportion of 1s and 0s. This isn't about patterns; it's about mathematics and probabilities.

I find it really interesting from the point of view of psychology how some people try to create a pattern that is not there. Especially when its existence was explicitly denied from the ground up by the creator. Seems to be a social phenomenon. Would not like to stop you from the project, of course, go ahead, use your calories energetically.

I am looking for an approach that is similar to the BSGS method, even though.

when will you publish the GPU version of the code for benchmarking and fully operational testing that way we can be sure which one is the time taken to get the key

This calculation is incorrect, all possible combinations are larger when the ratio of 1s and 0s is 50%, if 1 or 0 is smaller, the total decreases.

7,007,092,303,604,022,272/50,000,000,000,000/86,400=1.62 days with just 50 Trillion keys/s
We go faster than that on keyhunt BSGS if you have sufficient amount of RAM and a good number of CPU in your machine you can even go as fast as 500 Trillion

import math

n = 66
k = 32

combination = math.factorial(n) / (math.factorial(k) * math.factorial(n - k))

print(combination)


7007092303604022272  combination   2^62.6.....

Maybe James should publish the GPU code for the 1s and 0s searching and also make it possible for us to be able to arrange the positioning of the second 1s where we don't need to start all over again when the program stops working due to power loss or something and also make it in such a way that it program can continue from where it stopped so it doesn't have to keep doing the same thing over and over again. the code has so much potentials but we are going to give it a try too

zahid888 before you posted that my guess was the same :
my guess:
       #63       0=27   correct     1=36
       #64       0=34   correct     1=30
       #65       0=36   correct     1=29
       #66       0=34   guess       1=32
       #67       0=40   guess       1=27
       #68       0=33   guess       1=35
       #69       0=32   guess       1=37
       #70       0=36   correct     1=34

you skipped this part

Puzzle: 31       Zeros: 10       Ones: 21        Percent 0: 32.26%       Percent 1: 67.74%       Decimal: 2102388551 | Binary: 1111101010011111110011101000111
Puzzle: 62       Zeros: 28       Ones: 34        Percent 0: 45.16%       Percent 1: 54.84%       Decimal: 3908372542507822062 | Binary: 11011000111101010101000001111010110110000100011010101111101110

Puzzle: 30       Zeros: 14       Ones: 16        Percent 0: 46.67%       Percent 1: 53.33%       Decimal: 1033162084 | Binary: 111101100101001100110101100100
Puzzle: 60       Zeros: 28       Ones: 32        Percent 0: 46.67%       Percent 1: 53.33%       Decimal: 1135041350219496382 | Binary: 111111000000011110100001100000100101001101100111101110111110

Puzzle: 32       Zeros: 17       Ones: 15        Percent 0: 53.12%       Percent 1: 46.88%       Decimal: 3093472814 | Binary: 10111000011000101010011000101110
Puzzle: 64       Zeros: 34       Ones: 30        Percent 0: 53.12%       Percent 1: 46.88%       Decimal: 17799667357578236628 | Binary: 1111011100000101000111110010011110110000100100010001001011010100

Puzzle: 33       Zeros: 17       Ones: 16        Percent 0: 51.52%       Percent 1: 48.48%       Decimal: 7137437912 | Binary: 110101001011011001010100011011000
Puzzle: 66       Zeros: 34       Ones: 32        Percent 0: 51.52%       Percent 1: 48.48%       Decimal: ? ? ? ? ? ? ? | Binary: ? ? ? ? ? ? ?
       ☝☝             ☝☝           ☝☝                  ☝☝☝☝                  ☝☝☝☝       Little Prediction according to previous binaries 

If the private key of a point is known, you can divide the other point, for this you take the inverse of the private key of the known point and multiply by the point, multiplying with the inverse is equal to dividing the point.

Not: My native language is not english i translate from google sorry for the translation errors

work with me 🤣🤣

Your result is not correct because you should do the math mod n to have the correct result.

Anyways, point by point multiplication and division without knowing at least one point's private key is impossible, why else they call it crypto-graphy for?
If it was possible directly, we all could break ECC easily.

Ps, I'm not a mathematician, but I'll find a way to divide point by point or die trying! 🤣

for n = fffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141


my result is 2aaaaaaaaaaaaa00000000000000000000000000000000000000000000000000

 

but what I meant is whether there is a way to divide a point on the curve x, y by another point on the curve x, y?

If you mean numbers 66 and 99, they are mirrors of each other, also converting 66 and 99 to hex twice you will get 2a for 66 and 3f for 99. Converting 66 to decimal is 102, and 99 is 153.

Note that 6 and 9 are represented by each other in hex format, since there are only 6 alphabet letters, the remaining 9 digits have no alphabet representatives. So here is how it works :
0= f
1= e
2= d
3= c
4= b
5= a
6= 9
7= 8
8= 7
9= 6
It's a world of wonders this hexadecimal world!

---

Are you asking or telling? Of course there is a way, look at my personal text, e.g. dividing n by 6 will give you this :


You can multiply any point by above key, and it will divide your key by 6 if it's divisible by 6 you will get a correct answer.

Now if you divide a point by the key above, you will actually be multiplying your point by 6.

Now if you divide a point by 4 and multiply the result by 2 you should naturally get half of your original point, if your point is divisible by 4 then your results are correct, otherwise you will have a much much bigger point than your original point.

Essentially dividing a point by a number other than 2 will get you no where close if your key is not divisible by that number, you can try by dividing a number by 3, 4, 5, etc and if the fraction is something other than .5, then the result of point division will not be any where close, however as long as the fraction is .5, then you can always add n/your divisor to the result to get the correct result.


For example, divide 23 by 2, you will get -n/2-12 as a result, and if you add n/2 to your result, you'll have -12.

101 on how to break cryptography by ~dig.😉

what makes you think that it's automated and not manually entered as a transaction?
what do you mean by 'small transactions with numbers' in detail?