commenting the if-line results in 2.6 seconds, that helped. However I don't see any GPU utilization. I tried raising num to 50 million, but then the process died before it ended. No result, GPU util = 0%

any clues ?

you still don't understand the meaning sense of the hash function? "more light" > what is random? you take a coin and flip it 1 time, it may come up heads or tails. what happens if you flip a coin 10 times in a row, according to the theory of probability, either 10 heads or 10 tails can fall out. if you want to get 5 heads and 5 tails (in any order) you will need to flip a coin in series of 10, 2^10 = 1024 (1024 by 10). to drop all possible combinations from 10 heads in a row to 10 tails in a row, for all tosses in a row 1024 by 10, you need to flip a coin, 1024^1024=... when we start looking at the numbers generated when creating a bitcoin address (any or the address itself or its private key or its 160 hash in any form, hex, dec, bin) we get some parts of this huge number 2^10 = 1024 (1024 by 10) <> 1024^1024=... if you take 3 bits, 2^3=8, 8^8=16777216, then we start looking from 1 to 2^160 what we have in rmd160 hash in hex or dec

someting blablabla
...

1111101110010000010011001101000011011011000101101111100110000011100011011110010 1101011011011101111100000100100110000001010101110011001000000111000100000111101
1111101110010000010011011101010001011110010001010101111001100100101001001001000 1010001010111001000001111010110110000110010101010100001100011111100000110111111
1111101110010000010000101000101001011101000101100110011001100110111110110010000 0011000111101001111110011101100110100010110011110100110011111010000101000000010
1111101110010000010011110100011011001001011101100011111101001110111010000111011 0001001100010000010011000110111010111010110010011001001111010010100000010100001
1111101110010000010011111001001101010000001111001001110001101011110000111011011 0111000010010000111110000100000111010010001000110010111010110111000110000011111
1111101110010000010011000111010011111010010100001000011000001100110010010110001 1010100111011011111111001101001111100000101000101000000110011010110110110111000
1111101110010000010000000110100101101111101110001100000111111001100110111011111 0010000010110100011110101010101010101000100110101100010001111000000000111110001
1111101110010000010000011010010110100011111111100110001101001111001000110101011 0010100010010101100111100010000000110101000001101101111001011101010100101010101

the principle is preserved that vertically and horizontally > 2^3=8, 8^8=16777216


11111011
11111011
11111011
111
111
111
111
111

how many such sections will fit into 2:160 > 1461501637330902918203684832716283019655932542976 16777216 = 87112285931760246646623899502532662132736

we will split our 160 bits by 3 bits into sections


111 110 111 001...
111 110 111 001...
111 110 111 001...
111 110 111 001...
111 110 111 001...
111 110 111 001...
111 110 111 001...
111 110 111 001...



11111011 10010000 01001100...
11111011 10010000 01001101...
11111011 10010000 01000010...

11111011 10010000 01001111...
11111011 10010000 01001111...
11111011 10010000 01001100...



of course we can't look further from where they fall from for each row > 16777216^16777216=...

with a horizontal representation, we get 160hesh/8bits = 20 parts.

8 bit 2^8 = 256, 256/20 = 12,8

2^1
1461501637330902918203684832716283019655932542976/12,8 = 114179815416476790484662877555959610910619729920 steps
2^160

256^256=... huge number 3231700607131100730071487668866995196044410266971548403213034542752465513886789 0893197201411522913463688717960921898019494119559150490921095088152386448283120 6308773673009960917501977503896521067960576383840675682767922186426197561618380 9433847617047058164585203630504288757589154106580860755239912393038552191433338 9668342420684974786564569494856176035326322058077805659331026192708460314150258 5928641771167259436037184618573575983511523016459044036976132332872312271256847 1082020972515710172693132346967854258065669793504599726835299863821552516638943 7335543602135433229604645318478604952148193555853611059596230656


16777216/12,8 =  1310720 steps, 2^1 to 2^160, 1461501637330902918203684832716283019655932542976 1310720 = 1115037259926531157076785913632418075299020,8

256^256=... huge number 1115037259926531157076785913632418075299020,8

3231700607131100730071487668866995196044410266971548403213034542752465513886789 0893197201411522913463688717960921898019494119559150490921095088152386448283120 6308773673009960917501977503896521067960576383840675682767922186426197561618380 9433847617047058164585203630504288757589154106580860755239912393038552191433338 9668342420684974786564569494856176035326322058077805659331026192708460314150258 5928641771167259436037184618573575983511523016459044036976132332872312271256847 1082020972515710172693132346967854258065669793504599726835299863821552516638943 7335543602135433229604645318478604952148193555853611059596230656 1115037259926531157076785913632418075299020,8 =~

2898289342675449871993098867672270812704240074863894775739976204579701715524344 5980591244385169006487474859731523184059403721470360895667790293862650383583527 8052308371150131655537851476057700318587873331791124614866020470257907019747447 1073288393330068706379548507509145858817458460347497468230852750261100385960260 0438313585964807278885206604007966494048122967982378718514987992376064111499830 2154324000491294951584421374564447449664912755949011836586456519390866498333142 8684237600062725568520340756025511278116285953620110095472786524359663371614992 7614893228693106196480

in general our 2^3=8, 8^8=16777216 fall into their position from the general 256^256=... huge number...

in the first column, theirs fall out all over, in the second one, etc. when presented vertically and similarly when recalculating and converting in horizontal...

instead of constantly flipping a coin, we initiate with Mersenne twister and hash functions (256,160hashs) by initiating the creation of a bitcoin address using a number, we get a fixed result instead of randomly generating a constantly new one (but it randomly takes it from a large number), but this number itself 2^160 or 2^256 is still fixed and falls out of the total huge number that goes into infinity 2^3=8, 8^8=16777216, 16777216^16777216... etc

in general, everyone is picking something and looking for their own ways))

https://youtu.be/AYWoDqQmm1o?t=128

I consider base58 to be one of the weakest methods for scanning, it slows down the speed and so on. As for the code, I was interested in this setting, it adds one letter to the base58 private key, such a pyramid, I used to deal with it with binary, but I forgot about it

---

here is the full speed version of the old code, don't forget this code will never crack 120 ddd why because it works like this it gives it some full number and the last 6 nulls it scans sequentially what gives you at 4m keys per second 3 full keys per second simply put it only scans the last numbers of the private keys

.hex() this slowed down the code because it had to convert the entire chain of 1,000,000 bytes public keys to hex

Please use scientific notation for large numbers, it's really hard navigating the page reading your post.
Btw, what's your point? I see no conclusion.

I have something for you, want to see if you can figure something out of it.
The following is a real private key:

Could you figure out what are the other possibilities regarding private keys with the huge number of keys that exist?

A fun fact about bitcoin is that private key is private but is not a secret, because they are all known, out there sitting for anyone to see them, however this is not the case about addresses, they are actually the secret element which nobody seems to realize, this fact is hiding in plain sight.
Now after reading some of your posts, can you find any relation between addresses and their private keys, such as in binary/hex/decimal/wif?
Let me give you a hint, there is a relation but it's spread throughout the entire key range, we can actually map out all the addresses just like there is a map of all private keys, but trying to map the addresses will take decades with 50 super computers, though if we had a QC and 10,000,000 TB of storage, we could create that map in less than 6 month.

I wanna know what are the all possibilities regarding private keys, for example, what else could we find on private keys, what sentences could go into a valid pk?
Since this challenge has somewhat a more educational approach towards blockchain infrastructure, I'm asking this here.

In the process of generating bitcoin addresses, what are the needed factors, a decimal string, public key string, hashing, adding checksum, calculating the private key and generating the address. What can we set/change manually? And how to know if our input will result in a valid address?

Hi mysterious entity, I am here to express my thoughts , and I will base my assumptions on you being the coy and famous Satoshi, the one and only BTCitcoin inventor.
First things first, Thank you for the work and efforts on making this new open source technology regardless of it's flaws and limitations! it's benefits and utilities outweighs the minor flaws. The outcome of your invention is paramount and the mark you have left will remain recognized and relevant for thousands of years in the future, you will be remembered as the founding father of the decentralized digital technology.

This kind of behavior shows a little bit of your character, it seems that you are reckless (maybe very busy having too much on your mind?), apparent if one goes back in history reading your posts, seeing your first few years of Bitcoin development leaving the scene for others to take control> irresponsibility in regards to the community you were so fond of that made you come up with the Bitcoin solution.(perhaps there was no need to continue publicly working on Bitcoin since you left it in good hands?) (which so far has proven to work as you intended, at least that's what I think). The irresponsible part is about the fact that you never showed up from time to time to share your thoughts with us, that just hurts our feelings because we all love you and respect you as an esteemed founder of Bitcoin.

Please, you are anything but stupid, I do realize at current price or the price back when you created the puzzle the amounts were not much, but giving it a few more years, people would sell their houses to buy hardware in order to solve any of the remaining puzzles.

Currently I'm working on a few ideas trying to formulate a new algorithm focused on breaking the [1]ECDSA and [2]RIPEMD-160 into smaller pieces, reconstructing the [3]SHA-256 to make a new hash function that displays the entire hashing operations on the screen or saves it on a file. These are the small steps I'm working on when I have spare time. my issue is funding, third party sponsors would require information and collected results/data, my work can NOT be made public if I achieve what I'm looking for. unfortunately further explanation is not wise.
I have an old laptop and need a strong computer with decent GPUs, that would cost at least 20,000 USD, Ca not afford it. If you are intrigued to see some new attack vectors on Bitcoin(2 years at least is needed) I will send you a PM to convey my request.

##This puzzle at this time with these baby tools available is wasting time, first quantify the numbers you need to search through to find the remaining ones, then open the calculator to enter the values, decimal/speed/ s,60 for one minute/60 for one hour, etc.

[1]= [2]= [3]= old, outdated.

You are correct, I was under the impression that whoever designed this challenge was after finding weaknesses and would sponsor any study done to accelerate the process, hence the mention of _"funding". I just couldn't help it to neglect mentioning the BIG elephant in the room, when I said using current technologies are a waste of time. Thank you for pointing out some facts which I missed myself. I'm sure someone will be able to sneak a peek over the event horizon of elliptic curve sooner rather than later.

Mister T. Hello, are you still here? I found your main account to be inactive for years and this one was active in 2019. How can we talk? I really need your help. By the way, what are you doing with puzzles? You are a genius yourself, no need to involve the community to solve the ECC problem.

Sent you an email from yoshimaka@grr.la to the email satoshin@gmx.com please read.

In this case, the number you are using as the starting value of the range() function is very large, and the range() function is trying to allocate memory to represent all the integers in that range. However, the integer is too large to be represented in memory, hence the OverflowError.

I know the key, but I won't tell you  

EDIT: Oh damn! I always thought it must be 1234567890123456789012345678901 but I was wrong.