Bitcoin puzzle transaction ~32 BTC prize to who solves it

[optioncmdPR]

Puzzle 71 will be solved within next week, stay tuned.I am confident that I have found mathematical magic.

 curious , if your mathematical magic always seems to take a dive about halfway through the integer. I ask because this happens to me, regardless of approach angle.

🎯 REVISED P71 POSITIONING:
✅ Position:  77.3%  ( Success Probability 88.0% )
📍 Target Range: 0x7174 area
🔬 Improved Calibration: Enhanced mathematical modeling
📊 Confidence: Higher precision positioning
🎯 Search Strategy: COVERAGE deployment recommended
Reason for sharing this update:

"The mathematical positioning keeps evolving as I refine the φ-based calculations. This has higher success probability."

So anyone want too hash this ? 😅🤔

🚀 ADAPTIVE AI RANGE DEPLOYMENT:
─────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
📦 PRECISION: 717BF1E8E60C65E698 → 717D955714BE2A1968
              Width:  0.01% (118,059,162,071,741,136 keys)
              Surgical precision + AI

📦  BALANCED: 717B2D02DED7C1B1E0 → 717E5A3D1BF2CE4E20
              Width:  0.02% (228,903,190,186,990,656 keys)
              Optimal balance + AI

📦    SAFETY: 7178CB173103783CC0 → 7180BC28C9C717C340
              Width:  0.05% (572,257,975,467,476,608 keys)
              Enhanced safety + AI

📦  COVERAGE: 7174D28E64A1A87980 → 7184B4B19628E78680
              Width:  0.10% (1,144,515,950,934,953,216 keys)
              Maximum coverage + AI

Seems pi theory not so bad hmmmm. I predict also near here.. let see 😅

           I took the first 65536 decimals of pi and with it made a 256 x 256 grid, then rotated the grid 90 degrees, therebye transposing the digits that were verticle into horizonal rows, 256 per line. When you take this matrix and resize its  containment box, very very interesting results appear at the edges. Then it will piss you off because this method, just like any others its so close to matching known values but not enough to accurately predict. As usual.

[fecell]

Rhetorical question:
Why use sha256 multiple times after secp256k1?

Code:

[+] [18,067] 0.00% S: 331895.0 p/s, E: 18:19:32, TE: 16d 11:59:41, ETA: 1623936135880879259797946368y 5m 23d 17:21:04
255215748667055164621267256750573
---
[+] [18,068] 0.00% S: 331860.1 p/s, E: 18:20:48, TE: 16d 12:00:57, ETA: 1623933104747034416043261952y 3m 24d 20:39:28
143991517358151100293789707953175
---
[+] [18,069] 0.00% S: 331802.2 p/s, E: 18:22:09, TE: 16d 12:02:18, ETA: 1623935310097885652527874048y 8m 28d 09:18:56
32767286049247035966312159155777
---
[+] [18,070] 0.00% S: 331762.4 p/s, E: 18:23:27, TE: 16d 12:03:35, ETA: 1623933431784606125021724672y 5m 8d 16:40:32
246061608398769734450787649898602
---
[+] [18,071] 0.00% S: 331735.1 p/s, E: 18:24:41, TE: 16d 12:04:50, ETA: 1623928731610614548238696448y 9m 18d 04:11:44
134837377089865652108911591619220
---
[+] [18,072] 0.00% S: 331691.1 p/s, E: 18:25:59, TE: 16d 12:06:08, ETA: 1623927852853047085750026240y 6d 07:38:40
23613145780961605795832552303806
---
[+] [18,073] 0.00% S: 331644.7 p/s, E: 18:27:18, TE: 16d 12:07:27, ETA: 1623927559057088759526850560y 4m 20d 16:21:20
236907468130484250237112514600679
---
[+] [18,074] 0.00% S: 331586.9 p/s, E: 18:28:39, TE: 16d 12:08:48, ETA: 1623929873722913526030270464y 6m 12d 00:23:28
125683236821580203924033475285265
---
[+] [18,075] 0.00% S: 331558.6 p/s, E: 18:29:54, TE: 16d 12:10:03, ETA: 1623925481386162456780341248y 4m 20d 07:36:32
14459005512676139596555926487867

is a statistic well?

current state:

Code:

[+] [19,979] 0.00% S: 324569.5 p/s, E: 06:26:12, TE: 18d 04:45:24, ETA: 1619688167311456692992999424y 9m 16d 13:30:40
154138132311997239066806131546382

it was 2 days and 4237314074705763787341824 years reduce.

it's one thread test python script.
I'm shocked.

Code:

---
[+] [23,071] 0.00% S: 284020.6 p/s, E: 08:11:16, TE: 20d 20:46:44, ETA: 1608218584649767732054589440y 10m 8d 07:21:36
238481803112972095737066543953970
---
[+] [23,072] 0.00% S: 283910.9 p/s, E: 08:12:49, TE: 20d 20:48:16, ETA: 1608231297785976453800656896y 9m 18d 20:30:56
127257571804068049423987504638556
---
[+] [23,073] 0.00% S: 283839.8 p/s, E: 08:14:17, TE: 20d 20:49:45, ETA: 1608240481529862446628470784y 5m 20d 11:48:16
16033340495163967082111446359174
---
[+] [23,074] 0.00% S: 283801.5 p/s, E: 08:15:42, TE: 20d 20:51:10, ETA: 1608246644233578118603866112y 6m 11d 15:32:16
229327662844686647552188427620015
---
[+] [23,075] 0.00% S: 283790.8 p/s, E: 08:17:04, TE: 20d 20:52:32, ETA: 1608250225446504813796786176y 9m 2d 01:57:20
118103431535782583224710878822617
---
[+] [23,076] 0.00% S: 283786.7 p/s, E: 08:18:26, TE: 20d 20:53:53, ETA: 1608253203113222630635732992y 6m 29d 02:40:00
6879200226878518897233330025219
---
[+] [23,077] 0.00% S: 283800.3 p/s, E: 08:19:45, TE: 20d 20:55:13, ETA: 1608254515208766918852870144y 1m 6d 04:09:36
220173522576401181352911801804076
---
[+] [23,078] 0.00% S: 283814.7 p/s, E: 08:21:05, TE: 20d 20:56:32, ETA: 1608255732800132037635735552y 5m 24d 07:00:16
108949291267497117025434253006678
---
[+] [23,079] 0.00% S: 283798.3 p/s, E: 08:22:28, TE: 20d 20:57:55, ETA: 1608259876753818796962086912y 5m 3d 18:42:08
322243613617019779481112724785535
---

reduce 15669300082259363815751680 years within 3 days.



seems best DLP solve algo. but hidden for anyone exept me, as U know.
'cuze me. must complete research B4 share all iИf0

[teguh54321]

Code:

[+] [18,067] 0.00% S: 331895.0 p/s, E: 18:19:32, TE: 16d 11:59:41, ETA: 1623936135880879259797946368y 5m 23d 17:21:04
255215748667055164621267256750573
---
[+] [18,068] 0.00% S: 331860.1 p/s, E: 18:20:48, TE: 16d 12:00:57, ETA: 1623933104747034416043261952y 3m 24d 20:39:28
143991517358151100293789707953175
---
[+] [18,069] 0.00% S: 331802.2 p/s, E: 18:22:09, TE: 16d 12:02:18, ETA: 1623935310097885652527874048y 8m 28d 09:18:56
32767286049247035966312159155777
---
[+] [18,070] 0.00% S: 331762.4 p/s, E: 18:23:27, TE: 16d 12:03:35, ETA: 1623933431784606125021724672y 5m 8d 16:40:32
246061608398769734450787649898602
---
[+] [18,071] 0.00% S: 331735.1 p/s, E: 18:24:41, TE: 16d 12:04:50, ETA: 1623928731610614548238696448y 9m 18d 04:11:44
134837377089865652108911591619220
---
[+] [18,072] 0.00% S: 331691.1 p/s, E: 18:25:59, TE: 16d 12:06:08, ETA: 1623927852853047085750026240y 6d 07:38:40
23613145780961605795832552303806
---
[+] [18,073] 0.00% S: 331644.7 p/s, E: 18:27:18, TE: 16d 12:07:27, ETA: 1623927559057088759526850560y 4m 20d 16:21:20
236907468130484250237112514600679
---
[+] [18,074] 0.00% S: 331586.9 p/s, E: 18:28:39, TE: 16d 12:08:48, ETA: 1623929873722913526030270464y 6m 12d 00:23:28
125683236821580203924033475285265
---
[+] [18,075] 0.00% S: 331558.6 p/s, E: 18:29:54, TE: 16d 12:10:03, ETA: 1623925481386162456780341248y 4m 20d 07:36:32
14459005512676139596555926487867

is a statistic well?

current state:

Code:

[+] [19,979] 0.00% S: 324569.5 p/s, E: 06:26:12, TE: 18d 04:45:24, ETA: 1619688167311456692992999424y 9m 16d 13:30:40
154138132311997239066806131546382

it was 2 days and 4237314074705763787341824 years reduce.

it's one thread test python script.
I'm shocked.

Code:

---
[+] [23,071] 0.00% S: 284020.6 p/s, E: 08:11:16, TE: 20d 20:46:44, ETA: 1608218584649767732054589440y 10m 8d 07:21:36
238481803112972095737066543953970
---
[+] [23,072] 0.00% S: 283910.9 p/s, E: 08:12:49, TE: 20d 20:48:16, ETA: 1608231297785976453800656896y 9m 18d 20:30:56
127257571804068049423987504638556
---
[+] [23,073] 0.00% S: 283839.8 p/s, E: 08:14:17, TE: 20d 20:49:45, ETA: 1608240481529862446628470784y 5m 20d 11:48:16
16033340495163967082111446359174
---
[+] [23,074] 0.00% S: 283801.5 p/s, E: 08:15:42, TE: 20d 20:51:10, ETA: 1608246644233578118603866112y 6m 11d 15:32:16
229327662844686647552188427620015
---
[+] [23,075] 0.00% S: 283790.8 p/s, E: 08:17:04, TE: 20d 20:52:32, ETA: 1608250225446504813796786176y 9m 2d 01:57:20
118103431535782583224710878822617
---
[+] [23,076] 0.00% S: 283786.7 p/s, E: 08:18:26, TE: 20d 20:53:53, ETA: 1608253203113222630635732992y 6m 29d 02:40:00
6879200226878518897233330025219
---
[+] [23,077] 0.00% S: 283800.3 p/s, E: 08:19:45, TE: 20d 20:55:13, ETA: 1608254515208766918852870144y 1m 6d 04:09:36
220173522576401181352911801804076
---
[+] [23,078] 0.00% S: 283814.7 p/s, E: 08:21:05, TE: 20d 20:56:32, ETA: 1608255732800132037635735552y 5m 24d 07:00:16
108949291267497117025434253006678
---
[+] [23,079] 0.00% S: 283798.3 p/s, E: 08:22:28, TE: 20d 20:57:55, ETA: 1608259876753818796962086912y 5m 3d 18:42:08
322243613617019779481112724785535
---

reduce 15669300082259363815751680 years within 3 days.



seems best DLP solve algo. but hidden for anyone exept me, as U know.
'cuze me. must complete research B4 share all iИf0

Need comment from KtimesG 😅

[kTimesG]

reduce 15669300082259363815751680 years within 3 days.

seems best DLP solve algo. but hidden for anyone exept me, as U know.
'cuze me. must complete research B4 share all iИf0

Need comment from KtimesG 😅

The only comment I have is that if someone gives me 10 BTC, puzzle 135 is solved within 30 days, not quadrillions of eons.

[optioncmdPR]

A solution, sort of a hint.
Have any of you studied G point throughly? It has some interesting characteristics, it was generated by someone we don't know anything about, other than that, N also is interesting and you should research both N and G.

One other thing is the concept of adding and multiplying G by k, which obviously is not what I thought, I always assumed that if G is 5, and k is 20, we'd just multiply 5*20=100 =p. Well that was a misconception from my part.

Now instead of wasting your time doing useless stuff, start doing some research and experiment on elliptic curve.

Worth mentioning that almost 99% of you are unaware that bitcoin elliptic curve is a mirror. Now that you know, you should study the mirror verse to see what cool stuff are lurking there. Good luck and happy hunting.


I just wanted to shock Satoshi for a second. Are you shocked?🤣 .

This is true.


First private key
000000000000000000000000000000000000000000000000000000000000000000000000000001
EC Point
             x:79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
             y:483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
Last private key
115792089237316195423570985008687907852837564279074904382605163141518161494336
EC point
             x:79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
             y:b7c52588d95c3b9aa25b0403f1eef75702e84bb7597aabe663b82f6f04ef2777

[brainless]

A solution, sort of a hint.
Have any of you studied G point throughly? It has some interesting characteristics, it was generated by someone we don't know anything about, other than that, N also is interesting and you should research both N and G.

One other thing is the concept of adding and multiplying G by k, which obviously is not what I thought, I always assumed that if G is 5, and k is 20, we'd just multiply 5*20=100 =p. Well that was a misconception from my part.

Now instead of wasting your time doing useless stuff, start doing some research and experiment on elliptic curve.

Worth mentioning that almost 99% of you are unaware that bitcoin elliptic curve is a mirror. Now that you know, you should study the mirror verse to see what cool stuff are lurking there. Good luck and happy hunting.


I just wanted to shock Satoshi for a second. Are you shocked?🤣 .

This is true.


First private key
000000000000000000000000000000000000000000000000000000000000000000000000000001
EC Point
             x:79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
             y:483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
Last private key
115792089237316195423570985008687907852837564279074904382605163141518161494336
EC point
             x:79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
             y:b7c52588d95c3b9aa25b0403f1eef75702e84bb7597aabe663b82f6f04ef2777

You learned this after 15 years....?
These are basics,

[TheMissingNTLDR]

A solution, sort of a hint.
Have any of you studied G point throughly? It has some interesting characteristics, it was generated by someone we don't know anything about, other than that, N also is interesting and you should research both N and G.

One other thing is the concept of adding and multiplying G by k, which obviously is not what I thought, I always assumed that if G is 5, and k is 20, we'd just multiply 5*20=100 =p. Well that was a misconception from my part.

Now instead of wasting your time doing useless stuff, start doing some research and experiment on elliptic curve.

Worth mentioning that almost 99% of you are unaware that bitcoin elliptic curve is a mirror. Now that you know, you should study the mirror verse to see what cool stuff are lurking there. Good luck and happy hunting.


I just wanted to shock Satoshi for a second. Are you shocked?🤣 .

This is true.


First private key
000000000000000000000000000000000000000000000000000000000000000000000000000001
EC Point
             x:79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
             y:483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
Last private key
115792089237316195423570985008687907852837564279074904382605163141518161494336
EC point
             x:79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
             y:b7c52588d95c3b9aa25b0403f1eef75702e84bb7597aabe663b82f6f04ef2777

If you go to Learnmeabitcoin website, Insert the value for x in integer format, type any integer e.g. 111 in for y value, then an error will be displayed and the correct expected 2 values of Y will also be displayed to the user.

[eggsylacer]

A solution, sort of a hint.
Have any of you studied G point throughly? It has some interesting characteristics, it was generated by someone we don't know anything about, other than that, N also is interesting and you should research both N and G.

One other thing is the concept of adding and multiplying G by k, which obviously is not what I thought, I always assumed that if G is 5, and k is 20, we'd just multiply 5*20=100 =p. Well that was a misconception from my part.

Now instead of wasting your time doing useless stuff, start doing some research and experiment on elliptic curve.

Worth mentioning that almost 99% of you are unaware that bitcoin elliptic curve is a mirror. Now that you know, you should study the mirror verse to see what cool stuff are lurking there. Good luck and happy hunting.


I just wanted to shock Satoshi for a second. Are you shocked?🤣 .

This is true.


First private key
000000000000000000000000000000000000000000000000000000000000000000000000000001
EC Point
             x:79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
             y:483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
Last private key
115792089237316195423570985008687907852837564279074904382605163141518161494336
EC point
             x:79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
             y:b7c52588d95c3b9aa25b0403f1eef75702e84bb7597aabe663b82f6f04ef2777

My idea is to create local symmetry within a symmetric elliptical curve.
For example, if we take points S1 = G and S2 = -G, start adding to S1 + G and subtracting from S2 + (-G), and take Y(S1) mod 2, Y(S2) mod 2, and construct sequences from the remainders, we will obtain sequences 1111101100 and 0000010011, which are symmetrical to each other.
Let's assume we can define “symmetry” for small ranges, for example, there is a range 2^10 - 2^11, the center of the range is (2^10+2^11)/2, the center is 0 and 1, let's start counting from the center +G and -G and take Y(P) mod 2 from these points and build a sequence, we will get sequences that are not symmetrical to each other, but they themselves are scalars of points symmetrical relative to the center of the range.
If we can determine from these two sequences which one belongs to the left or right side relative to the center, then we can do the same with a random point in the range. That is, there is a point A = G*k, where k = [2^10, 2^11-1], then point B = G* (2^10+2^11) - A, we construct sequences from Y(A) mod 2 and Y(B) mod 2, and determine which point from A, B lies on the “left” or “right” side.

[farou9]

My idea is to create local symmetry within a symmetric elliptical curve.
For example, if we take points S1 = G and S2 = -G, start adding to S1 + G and subtracting from S2 + (-G), and take Y(S1) mod 2, Y(S2) mod 2, and construct sequences from the remainders, we will obtain sequences 1111101100 and 0000010011, which are symmetrical to each other.
Let's assume we can define “symmetry” for small ranges, for example, there is a range 2^10 - 2^11, the center of the range is (2^10+2^11)/2, the center is 0 and 1, let's start counting from the center +G and -G and take Y(P) mod 2 from these points and build a sequence, we will get sequences that are not symmetrical to each other, but they themselves are scalars of points symmetrical relative to the center of the range.
If we can determine from these two sequences which one belongs to the left or right side relative to the center, then we can do the same with a random point in the range. That is, there is a point A = G*k, where k = [2^10, 2^11-1], then point B = G* (2^10+2^11) - A, we construct sequences from Y(A) mod 2 and Y(B) mod 2, and determine which point from A, B lies on the “left” or “right” side.

where did you bring those binary sequences , and what are relying on that y somehow gives you a symmetry relation if you make it mod a number

[Xal0lex]

-snip-

where did you bring those binary sequences , and what are relying on that y somehow gives you a symmetry relation if you make it mod a number

This account has been inactive for almost a year. Plus, it's been banned. So it won't be able to respond to you. Before commenting on such old posts, I recommend checking the account profile for activity.

[kTimesG]

If we can determine from these two sequences which one belongs to the left or right side relative to the center

All the points are a center of some range

All the points can be used as generators relative to one another.

There's no "left" and "right" or "center", just the distinct set of unique points that create a cycle once you pick one of them as a generator.

So, the only thing you'll get is statistical uniformity. Unless of course, the curve's broken.

[farou9]

-snip-

where did you bring those binary sequences , and what are relying on that y somehow gives you a symmetry relation if you make it mod a number

This account has been inactive for almost a year. Plus, it's been banned. So it won't be able to respond to you. Before commenting on such old posts, I recommend checking the account profile for activity.

i mistakenly deleted the wrong post owner in the quote section when i was writing my response the post i replied to is from
eggsylacer

[eggsylacer]

My idea is to create local symmetry within a symmetric elliptical curve.
For example, if we take points S1 = G and S2 = -G, start adding to S1 + G and subtracting from S2 + (-G), and take Y(S1) mod 2, Y(S2) mod 2, and construct sequences from the remainders, we will obtain sequences 1111101100 and 0000010011, which are symmetrical to each other.
Let's assume we can define “symmetry” for small ranges, for example, there is a range 2^10 - 2^11, the center of the range is (2^10+2^11)/2, the center is 0 and 1, let's start counting from the center +G and -G and take Y(P) mod 2 from these points and build a sequence, we will get sequences that are not symmetrical to each other, but they themselves are scalars of points symmetrical relative to the center of the range.
If we can determine from these two sequences which one belongs to the left or right side relative to the center, then we can do the same with a random point in the range. That is, there is a point A = G*k, where k = [2^10, 2^11-1], then point B = G* (2^10+2^11) - A, we construct sequences from Y(A) mod 2 and Y(B) mod 2, and determine which point from A, B lies on the “left” or “right” side.

where did you bring those binary sequences , and what are relying on that y somehow gives you a symmetry relation if you make it mod a number

G - generator point.
The first sequence is constructed based on adding point to point G:

Code:

A = G
ABitSeq = ''
for i in range(10):
    ABitSeq += str(A.y % 2)
    A += G
ABitSeq -> 0000010011

The second sequence is constructed in exactly the same way, but starts from point -G and adds point -G instead of G:

Code:

_G = -G # (G.x, -G.y % P)
B = _G
BBitSeq = ''
for i in range(10):
    BBitSeq += str(B.y % 2)
    B += _G
BBitSeq -> 1111101100

The sequences will be symmetric because operations on Y values are performed modulo P, which means that "-Y mod P" and "Y mod P" are symmetric with respect to each other and their center.

I assume that this symmetry will be preserved in smaller ranges, but I apply it to scalar of points. And since we perform this operation in small ranges relative to the range N, the symmetry itself will not resemble the symmetry that is usually discussed. Most likely, it will be a symmetry with its own conditions.

[farou9]

My idea is to create local symmetry within a symmetric elliptical curve.
For example, if we take points S1 = G and S2 = -G, start adding to S1 + G and subtracting from S2 + (-G), and take Y(S1) mod 2, Y(S2) mod 2, and construct sequences from the remainders, we will obtain sequences 1111101100 and 0000010011, which are symmetrical to each other.
Let's assume we can define “symmetry” for small ranges, for example, there is a range 2^10 - 2^11, the center of the range is (2^10+2^11)/2, the center is 0 and 1, let's start counting from the center +G and -G and take Y(P) mod 2 from these points and build a sequence, we will get sequences that are not symmetrical to each other, but they themselves are scalars of points symmetrical relative to the center of the range.
If we can determine from these two sequences which one belongs to the left or right side relative to the center, then we can do the same with a random point in the range. That is, there is a point A = G*k, where k = [2^10, 2^11-1], then point B = G* (2^10+2^11) - A, we construct sequences from Y(A) mod 2 and Y(B) mod 2, and determine which point from A, B lies on the “left” or “right” side.

where did you bring those binary sequences , and what are relying on that y somehow gives you a symmetry relation if you make it mod a number

G - generator point.
The first sequence is constructed based on adding point to point G:

Code:

A = G
ABitSeq = ''
for i in range(10):
    ABitSeq += str(A.y % 2)
    A += G
ABitSeq -> 0000010011

The second sequence is constructed in exactly the same way, but starts from point -G and adds point -G instead of G:

Code:

_G = -G # (G.x, -G.y % P)
B = _G
BBitSeq = ''
for i in range(10):
    BBitSeq += str(B.y % 2)
    B += _G
BBitSeq -> 1111101100

The sequences will be symmetric because operations on Y values are performed modulo P, which means that "-Y mod P" and "Y mod P" are symmetric with respect to each other and their center.

I assume that this symmetry will be preserved in smaller ranges, but I apply it to scalar of points. And since we perform this operation in small ranges relative to the range N, the symmetry itself will not resemble the symmetry that is usually discussed. Most likely, it will be a symmetry with its own conditions.

the only thing that is symmetric is the x values not the y values , the connection between the y values are not symetric its just that they are two partsof the same number p where when you subtract one of y or -y from p you get the other , what is the symety of y you are talking about

[eggsylacer]

If we can determine from these two sequences which one belongs to the left or right side relative to the center

All the points are a center of some range

All the points can be used as generators relative to one another.

There's no "left" and "right" or "center", just the distinct set of unique points that create a cycle once you pick one of them as a generator.

So, the only thing you'll get is statistical uniformity. Unless of course, the curve's broken.

I rely not on the mathematics of elliptic curves, but on the general theory/rules/laws of numbers.
For example:

Code:

M1 = 2**10
M2 = 2**11
x = random.randint(M1+1, M2-1) # random number from M1+1 to M2-1
y = (M2+M1)-x # Y symmetrical with respect to X

Now we know with 100% probability that one number from x and y lies in the range from M1 to (M1+M2)/2, and the second number lies in the range from (M1+M2)/2 to M2.

Now let's apply this property to curves.

Code:

M1 = 2**10
M2 = 2**11
x = random.randint(M1+1, M2-1) # random number from M1+1 to M2-1
A = G*x # Suppose that point A lies in the range from M1 to M2.
B = G*(M1+M2) - A

Now, from the code above, we know with 100% probability that the scalars(in binary form) of points A and B are symmetric with respect to each other, symmetric with respect to their center((M1+M2)/2), and symmetric with respect to field M2+M1 (since G*(M1+M2) - A = B, and G*(M1+M2) - B = A)

My idea is to create local symmetry within a symmetric elliptical curve.
For example, if we take points S1 = G and S2 = -G, start adding to S1 + G and subtracting from S2 + (-G), and take Y(S1) mod 2, Y(S2) mod 2, and construct sequences from the remainders, we will obtain sequences 1111101100 and 0000010011, which are symmetrical to each other.
Let's assume we can define “symmetry” for small ranges, for example, there is a range 2^10 - 2^11, the center of the range is (2^10+2^11)/2, the center is 0 and 1, let's start counting from the center +G and -G and take Y(P) mod 2 from these points and build a sequence, we will get sequences that are not symmetrical to each other, but they themselves are scalars of points symmetrical relative to the center of the range.
If we can determine from these two sequences which one belongs to the left or right side relative to the center, then we can do the same with a random point in the range. That is, there is a point A = G*k, where k = [2^10, 2^11-1], then point B = G* (2^10+2^11) - A, we construct sequences from Y(A) mod 2 and Y(B) mod 2, and determine which point from A, B lies on the “left” or “right” side.

where did you bring those binary sequences , and what are relying on that y somehow gives you a symmetry relation if you make it mod a number

G - generator point.
The first sequence is constructed based on adding point to point G:

Code:

A = G
ABitSeq = ''
for i in range(10):
    ABitSeq += str(A.y % 2)
    A += G
ABitSeq -> 0000010011

The second sequence is constructed in exactly the same way, but starts from point -G and adds point -G instead of G:

Code:

_G = -G # (G.x, -G.y % P)
B = _G
BBitSeq = ''
for i in range(10):
    BBitSeq += str(B.y % 2)
    B += _G
BBitSeq -> 1111101100

The sequences will be symmetric because operations on Y values are performed modulo P, which means that "-Y mod P" and "Y mod P" are symmetric with respect to each other and their center.

I assume that this symmetry will be preserved in smaller ranges, but I apply it to scalar of points. And since we perform this operation in small ranges relative to the range N, the symmetry itself will not resemble the symmetry that is usually discussed. Most likely, it will be a symmetry with its own conditions.

the only thing that is symmetric is the x values not the y values , the connection between the y values are not symetric its just that they are two partsof the same number p where when you subtract one of y or -y from p you get the other , what is the symety of y you are talking about

I didn't just give an example of constructing a binary sequence for no reason. Take a closer look at these two sequences (they are symmetrical to each other).

If we follow your logic, then the values of x are not symmetrical; they are literally identical.

[SRG02289]

1PWo3JeB9SyEuPyjoqKm2Ur6xUNJigqkmF

1PWo3JeB9d5CYrv5dzc4P8cr59ibFRMZMf

1PWo3JeB9sYfCE3gqL4LADHDLuhtr3kFH3

1PWo3JeB9Xa4x6UjH5tqGTkb8WE6H9MJuv

1PWo3JeB9UnXe1CR3K7rEt9te92E6SBH5p

1PWo3JeB9HvP415UbBEyMW9VEkUj619KGD

1PWo3JeB9NmXAKPPnPLNnuEo5pLCKASzs9

1PWo3JeB9JykV7ixxyAr8QfTQWFRN7THDJ

1PWo3JeB9tWXkfLAyxnnq6ykiBXAv4g5nc

still nothing guys

1PWo3JeB9jakyr87nZvq34SGhWNMvDbay4

1PWo3JeB9jFakrTGTTdpsB24cxtmyf6Nu6   02E5B40D7EC2177AF5F182EC29D66123B028F8DACAE9620083562600056501A5CC

[Mafioso246]

1PWo3JeB9SyEuPyjoqKm2Ur6xUNJigqkmF

1PWo3JeB9d5CYrv5dzc4P8cr59ibFRMZMf

1PWo3JeB9sYfCE3gqL4LADHDLuhtr3kFH3

1PWo3JeB9Xa4x6UjH5tqGTkb8WE6H9MJuv

1PWo3JeB9UnXe1CR3K7rEt9te92E6SBH5p

1PWo3JeB9HvP415UbBEyMW9VEkUj619KGD

1PWo3JeB9NmXAKPPnPLNnuEo5pLCKASzs9

1PWo3JeB9JykV7ixxyAr8QfTQWFRN7THDJ

1PWo3JeB9tWXkfLAyxnnq6ykiBXAv4g5nc

still nothing guys

1PWo3JeB9jakyr87nZvq34SGhWNMvDbay4

1PWo3JeB9jFakrTGTTdpsB24cxtmyf6Nu6   02E5B40D7EC2177AF5F182EC29D66123B028F8DACAE9620083562600056501A5CC

Between 40x and 477x, there are two addresses with the prefix 1PWo3JeB9jr. If you fully scanned that range, I’m curious could you tell me one of them?

[kTimesG]

All the points are a center of some range

I rely not on the mathematics of elliptic curves, but on the general theory/rules/laws of numbers.

Yeah those rules won't help much at all, they're pretty much, by themselves, the basis of why they do the exact opposite of helping. Have you heard of the Discrete Logarithm Problem?

Here, let's simply find the symmetric point of some arbitrary point P which is in some known range:

Q = P - (minKey + rangeSize/2) * G      // move center from range middle to 0 / point at infinity
Q' = {Q.x, p - Q.y}       // math magic because of symmetry!

So now Q and Q' are both in a symmetric range of the exact size as the original.

No need to mess around with weird binary sequences, all the points are now symmetrical (everything on the left side is symmetric to everything on the right side).

Now, your question is simple: let's see where Q stands relative to the center (e.g. relative to [-G, G]): left or right? If it's left, then Q' is on he right. If it's right, then Q' is on the left.

Good luck solving that problem. It's been asked for around 50 years or so.

[eggsylacer]

All the points are a center of some range

I rely not on the mathematics of elliptic curves, but on the general theory/rules/laws of numbers.

Yeah those rules won't help much at all, they're pretty much, by themselves, the basis of why they do the exact opposite of helping. Have you heard of the Discrete Logarithm Problem?

Here, let's simply find the symmetric point of some arbitrary point P which is in some known range:

Q = P - (minKey + rangeSize/2) * G      // move center from range middle to 0 / point at infinity
Q' = {Q.x, p - Q.y}       // math magic because of symmetry!

So now Q and Q' are both in a symmetric range of the exact size as the original.

No need to mess around with weird binary sequences, all the points are now symmetrical (everything on the left side is symmetric to everything on the right side).

Now, your question is simple: let's see where Q stands relative to the center (e.g. relative to [-G, G]): left or right? If it's left, then Q' is on he right. If it's right, then Q' is on the left.

Good luck solving that problem. It's been asked for around 50 years or so.

Yeah, this will give us a point whose scalar is the distance from the center of the range to the point. After performing the operation module -Y mod P on this point, we obtain a symmetric point relative to a group of points of size N. But I am talking about symmetry relative to the scalar of the point in the specified range, not the entire range from 1 to N. In any case, if we are looking for a solution, we need to look outside the system of elliptical curves.

Also, I have another idea. Example:

Code:

StartP = G
referenceBitSeq = bitarray()

for _ in range(2**15):
    referenceBitSeq.append(StartP.y % 2)
    StartP += G

Thus, we obtain a bit sequence where 1 bit of memory is assigned to each point.
Now, if we take a random scalar point from the range 1 to 2^15 and construct a bit sequence in the same way and compare it with the reference sequence (whether this sequence is in the reference), we can determine whether this random point is in the given range.

Code:

randScalar = random.randint(1, 2**15)
findP = randScalar*G

findBitSeq = bitarray()
for i in range(40): # The more bits (the longer the bit sequence), the more accurate the verification will be
    findBitSeq.append(findP.y % 2)
    findP += G

state = next(referenceBitSeq.search(findBitSeq), None) is not None
print(state)

As far as I understand, the complexity will be O(M*N). M - reference sequence length; N - length of the sequence being checked.
Of course, we must be sure that the scalar of the point is between 1+N and 2^15-N.
In essence, this can be improved by implementing a more efficient mechanism for searching for a bit string in a string. However, the issue remains in generating a massive reference bit string, which will take up a significant amount of memory, and the search will take considerably longer.

[kTimesG]

Yeah, this will give us a point whose scalar is the distance from the center of the range to the point. After performing the operation module -Y mod P on this point, we obtain a symmetric point relative to a group of points of size N. But I am talking about symmetry relative to the scalar of the point in the specified range, not the entire range from 1 to N.

The range has the same size in both cases. However, the only ever symmetric-ensuring range center you will ever find is for inexistent scalar 0 (which is, the neutral element, which doesn't exist on the curve).

In essence, this can be improved by implementing a more efficient mechanism for searching for a bit string in a string. However, the issue remains in generating a massive reference bit string, which will take up a significant amount of memory, and the search will take considerably longer.

The computational complexity skyrockets when "storing" a single point using a single bit, and trying to do those matches requires multiplying the total number of group operations by a factor, which invalidates any computational savings whatsoever. This has been discussed already.