Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 203. (Read 229790 times)

What is this mess? This doesn't prove anything XD

That is easy if you have the privatekey (X = X % 2 ? X/2 : (X-1)/2), but we can't do that if we don't have the privatekey, because there is no way that know if the publickey belongs to an even or odd privatekey.

But proof your self and show us.. here is a Publickey in the 110 bit range:

0301040c4e33b6ab297ba2a9c2858d15be5f6844febfc003ef7646796608f7f819

Private Key                                                                    Public key

1090246098153987172547740458951748   # puzzle 110   0309976ba5570966bf889196b7fdf5a0f9a1e9ab340556ec29f8bb60599616167d
545123049076993586273870229475874
272561524538496793136935114737937
136280762269248396568467557368968
68140381134624198284233778684484
34070190567312099142116889342242
17035095283656049571058444671121
8517547641828024785529222335560
4258773820914012392764611167780
2129386910457006196382305583890
1064693455228503098191152791945
532346727614251549095576395972
266173363807125774547788197986
133086681903562887273894098993
66543340951781443636947049496
33271670475890721818473524748
16635835237945360909236762374
8317917618972680454618381187
4158958809486340227309190593
2079479404743170113654595296
1039739702371585056827297648
519869851185792528413648824
259934925592896264206824412
129967462796448132103412206
64983731398224066051706103
32491865699112033025853051
16245932849556016512926525
8122966424778008256463262
4061483212389004128231631
2030741606194502064115815
1015370803097251032057907
507685401548625516028953
253842700774312758014476
126921350387156379007238
63460675193578189503619
31730337596789094751809
15865168798394547375904
7932584399197273687952
3966292199598636843976
1983146099799318421988
991573049899659210994
495786524949829605497
247893262474914802748
123946631237457401374
61973315618728700687
30986657809364350343
15493328904682175171
7746664452341087585
3873332226170543792
1936666113085271896
968333056542635948
484166528271317974
242083264135658987
121041632067829493
60520816033914746
30260408016957373
15130204008478686
7565102004239343
3782551002119671
1891275501059835
945637750529917
472818875264958
236409437632479
118204718816239
59102359408119
29551179704059
14775589852029
7387794926014
3693897463007
1846948731503
923474365751
461737182875
230868591437
115434295718
57717147859
28858573929
14429286964
7214643482
3607321741
1803660870
901830435
450915217
225457608
112728804
56364402
28182201
14091100
7045550
3522775
1761387
880693
440346
220173
110086
55043
27521
13760
6880
3440
1720
860
430
215
107
53
26
13
6
3
1  0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798

Don't believe in that, it is fake.


Agree.

Why from 115 to 28 bits? in that case why not up to 1 bit?

Guys please don't believe in that, in any case we only need to know one bit on the right side to break ECDSA

fake

Hello.
Could you share your algorithm?
I wonder how you managed this: "Puzzle 115 pubkey and reduced it to 2**28 bits in 2 seconds".

Sincerely,
be healthy.

I think that there isn't any way to find a solution for this puzzle; the only thing to solve this is the strength of community. It requires many and many power of hash/sec to be solved and with maths you can't do many things.

It would be cool to use something with a speed of an ASIC, in this way you'll be able to discover keys from 66 to 76 in very easy way !

If your algorithm is working for one key, it should work for all other keys, unless it's not working, if you can solve a known key and not any other key, you are doing things in a wrong way. Manage to find a key which you don't know it's private key, then you can claim success.

Hope you feel better.

My algorithm is working, it will be smoother with a few minor adjustments. I tried with Puzzle 115 pubkey and reduced it to 2**28 bits in 2 seconds. It took 1 second for the collider application to find the pubkey. Now writing the code to reverse the Algorithm .

I don't like working, my brain is almost numb from the blood pressure pills
i can't think

I'm sorry my English is bad, I'm writing from translate, but I hope you understand what I mean.

Look who is talking! Why would I offer something which could help breaking ECC? I can only hint at them, but there are already resourceful people around here explaining everything in details.

---

I already know that, we don't want +n key to be odd, halving it would halve the entire -n key which is a number near 2^256, and when we reach the middle range the halving reverses. Complex stuff! 😉

---

Mirror, quarter, third, one digit less, I'm not interested in those, I'm interested to find new ways and equations to solve a puzzle.

I live in this jungle, and I rather not pursuit shadows, they take you to dark parts of the woods.

Good luck to you too, may the help of God be with you all brothers! We can do this. 🤣

digaran
If you are surprised by the doubling of the original key, then you are still at the very beginning of this "jungle". numbers
I will share with you a little bit what the future holds for you if you have the strength and inspiration to explore the Bitcoin curve.
There are amazing things that can make 100 hours of your work useless and make you happy in the most unexpected place and time of working with a curve.
There are special numbers on the curve that can be used to get exactly 1 digit less.
A third of the number you are looking for.
A quarter of the sought number.
You can get the mirror part of the number you are looking for, which can also be compared with the original.
And this is not even a 10th part of what is there.
You can get a lot.



But by carrying out various manipulations with the numbers on the curve, as vjudeu said, you can get into an even more difficult situation and get a .....................labyrinth.

I wish you success.

You can always halve a key. However, if your key will be odd, then after halving it, you will end up in a worse situation.
See? You can reach a key with less bits, only if you know if it is even or odd. So, if you want to go from 125-bit key to 40-bit key, then guess what: you have to know the last 85 bits.

Digaran @
You can't offer anything useful. An endless stream of insanity.!

Do people really think Satoshi is reading all the posts here every day?
I don't know if anyone here is a giver, unfortunately we are here to take the loots.

Welcome to the jungle, and happy hunting.😉

---

Wow guys! I just discovered something, everytime I double a key I can get double the distance with my original base key!

I wonder, how many halving do I need to reduce a 125 bit key down to 40 bit?  A good thing about mod function in EC, you could keep halving for eternity without reaching destination, as we are moving towards bitcoin halving, we should start key halving to solve some puzzles before the halving event! 😉

It doesn't matter that much. If you have 123456 as your solution, and you know it is between 100000 and 200000, then it doesn't matter if you will try to solve 23456 (by subtracting lower range) or 76544 (by subtracting higher range). In the worst case, you have 100000 values to check. Also, people usually use subranges, and scan for example keys from 111000 to 112000. Sometimes they are lucky, scan subrange from 123000 to 124000, and reach the right answer, but still, in a typical scenario, your initial subtraction doesn't change that much.

Hey, I don't know who you are but don't give up yet!

Here I will give you some clues to keep going.

Imagine puzzle 125 is  number 25.

Now instead of subtracting a smaller key, we'd subtract it from a bigger key, we'd subtract it from 30, aka 2^125 = 0000000000000000000000000000000020000000000000000000000000000000
And the result =  0286936a275e6d53bb2b2718c93d8a5aa44f371f6e0300abb73b89dd851d2fbe88
Now the result is around 1/5 of our target, give or take. Now forget about puzzle #125 and focus on this new key.😉

Of course it is in binary. Each key is two times harder to solve than the previous one. You have 2^125 in the next puzzle for a reason. It is always 2^N for each puzzle. And how 2^125 looks in binary? Of course it is just one with 125 zeroes.

But even if you really want to calculate everything in hexadecimal, then still, there is the same pattern, just it is harder to see. If you check the first digit in the private key, it is always in the same range:

And then, it is simply repeated every four keys, just because if you have 16 hexadecimal digits, then they take 4 bits. It is that simple.

If you really want to see those leading ones in every single key, you can just multiply known keys, and see:
Now, can you see that every key has a leading one in binary? If you use hexadecimal, then you can just multiply it by 8,4,2,1, and that pattern repeats for every key in the puzzle. So, if you can see some key that has some leading one in hexadecimal, it doesn't mean it is easier. It is exactly as hard as other keys, because for every N-th output, you can just subtract 2^N. It always works.

For 2^125, you can simply see the leading one in your range, just because 125/4=31+1/4. That means, you have 31 hexadecimal numbers, and the leading one. But still, if you want to see the leading one in 115-bit puzzle, you can multiply it by 4, and you will see it (because 115/4=28+3/4, but 117/4=29+1/4, so you can move it by two bits to the left by multiplying it by 4).

Lol, I corrected my mistake. As for showing examples, if I had a working example, I would have used it already to solve the puzzle!
But here how it works in my mind, we have 2 and 4 as our base known points, and two unknown keys, we pick 15 and 20  for simplicity, we start by subtracting 2 from 15 and 4 from 15.

Now we have two new mutual keys, 11 and 13, if we subtract 11 from 20 we will have 9 which by subtracting 2*4 = 8 from it, we will have half of our first base point "2" ( aka 1).

Second step, we subtract 13 from 20 to have 7, now if we subtract 2+4 = 6 from it, we still have the half of our first base point "2" ( aka 1).

Therefore we have a triangle 2, 4 and 1, our known keys, and then we have 7, 9, 11, 13, 15, 20! Wait that turned into a threesome of triangles, 🤣 anyways the quantity of triangles is insignificant  compared to the quality of our equation.

The important part, if we do 2+4 = 6 *4 = 24, we now have 20 + 4 = 24, because we know 4, by adding it to x= 20(unknown key) we can observe whether any of our results are mutually related or not.

More complicated equation, we do the following, 2*4 = 8*2 = 16, now adding 4 to it, we will observe our x= 20 again.

Trick is picking your base points carefully, otherwise you will be doing things blindly. Find mutual keys that's all.😉

Yeah, if you know that K lies between A and B, you can subtract A from K, and thereby "reduce" the key, but any other subtractions is just a blind guessing.
As for the any number manipulations, secp256k1 operations obey all the basic math laws, i.e. A - 2*A = -A; A - -A = A+A; A+B+C = A + (B+C) = (A+B) + C; A*(B+C) = A*B + A*C and so on.

Subtracting -70 from +50 is resulting in +120.
I am still not getting trangular thing, can you explain with a simple example pl