Pollard's kangaroo ECDLP solver - page 94.

Etar

sr. member

Activity: 616

Merit: 312

What i have done using proposed arulbero method.. I solve pazzle 59 bit and save work. Then i tame all wild DPs with previous private key in file.
And multiply distance of each DP by 32
After that i compile kangaroo.exe where G*inv(32)= (a3c9d9de2ba89d61c63af260be9759d752b8bfef56ee41b2dab2b99871af38a8,b639cb2d3e0f1940d5e5333061d489159d1783ba821eaef15f78802d002f63fb)
I can`t say if new wild DP collide old tame from workfile or it is collsion with both new DPs but here is some test with pazzle 64bit

Code:

DPs JUST MOVED TO NEW INTERVAL WITHOUT SHIFTING PUB
P=P/32
Start:10000000000000000
Stop :1FFFFFFFFFFFFFFFF
Range width: 2^64
Expected operations: 2^33.08
DP size: 15 [0xFFFE000000000000]
[15.25 MK/s][GPU 0.00 MK/s][Count 2^32.37][Dead 0][06:56 (Avg 09:53)][9.2/30.2MB]
Key# 0 [1S]Pub:  0x0299A240D5FD67F0F9DBC558AF072E26B285B5DF1FACD1E9509A168A9523F53958
       Priv: 0x1A838B13505B26867
Done: Total time 06:56

REPEAT FOR MORE TEST
DPs JUST MOVED TO NEW INTERVAL WITHOUT SHIFTING PUB
P=P/32
Start:10000000000000000
Stop :1FFFFFFFFFFFFFFFF
Keys :1
Range width: 2^64
DP size: 15 [0xFFFE000000000000]
[15.27 MK/s][GPU 0.00 MK/s][Count 2^32.43][Dead 1][07:18 (Avg 09:52)][9.5/30.9MB]
Key# 0 [1S]Pub:  0x0299A240D5FD67F0F9DBC558AF072E26B285B5DF1FACD1E9509A168A9523F53958
       Priv: 0x1A838B13505B26867
Done: Total time 07:20

PUB SHIFTED AND DPs MOVED TO NEW INTERVAL
P=(P-0x10000000000000000*G)/32
Start:0
Stop :FFFFFFFFFFFFFFFF
Range width: 2^64
Expected operations: 2^33.08
DP size: 15 [0xFFFE000000000000]
[15.13 MK/s][GPU 0.00 MK/s][Count 2^33.46][Dead 1][14:35 (Avg 09:58)][15.1/42.9MB]
Key# 0 [1S]Pub:  0x02B0090246C76FA061A11AF1165F652344FDF191FBC74B4295E7DF2FEDB388B5C7
       Priv: 0xA838B13505B26867+0x10000000000000000=0x1A838B13505B26867
Done: Total time 14:37

REPEAT FOR MORE TEST
PUB SHIFTED AND DPs MOVED TO NEW INTERVAL
P=(P-0x10000000000000000*G)/32
Start:0
Stop :FFFFFFFFFFFFFFFF
Range width: 2^64
DP size: 15 [0xFFFE000000000000]
[15.21 MK/s][GPU 0.00 MK/s][Count 2^32.29][Dead 0][06:32 (Avg 09:55)][9.0/29.6MB]
Key# 0 [1S]Pub:  0x02B0090246C76FA061A11AF1165F652344FDF191FBC74B4295E7DF2FEDB388B5C7
       Priv: 0xA838B13505B26867+0x10000000000000000=0x1A838B13505B26867
Done: Total time 06:32

REPEAT FOR MORE TEST
PUB SHIFTED AND DPs MOVED TO NEW INTERVAL
P=(P-0x10000000000000000*G)/32
Start:0
Stop :FFFFFFFFFFFFFFFF
Range width: 2^64
Expected operations: 2^33.08
DP size: 15 [0xFFFE000000000000]
[15.16 MK/s][GPU 0.00 MK/s][Count 2^32.61][Dead 0][08:01 (Avg 09:57)][10.2/32.6MB]
Key# 0 [1S]Pub:  0x02B0090246C76FA061A11AF1165F652344FDF191FBC74B4295E7DF2FEDB388B5C7
       Priv: 0xA838B13505B26867+0x10000000000000000=0x1A838B13505B26867
Done: Total time 08:02

REPEAT FOR MORE TEST
PUB SHIFTED AND DPs MOVED TO NEW INTERVAL
P=(P-0x10000000000000000*G)/32
Start:0
Stop :FFFFFFFFFFFFFFFF
Range width: 2^64
Expected operations: 2^33.08
DP size: 15 [0xFFFE000000000000]
[14.86 MK/s][GPU 0.00 MK/s][Count 2^33.54][Dead 1][15:45 (Avg 10:09)][15.7/43.9MB]
Key# 0 [1S]Pub:  0x02B0090246C76FA061A11AF1165F652344FDF191FBC74B4295E7DF2FEDB388B5C7
       Priv: 0xA838B13505B26867+0x10000000000000000=0x1A838B13505B26867
Done: Total time 15:46


NORAML SOLVING WITH RIGHT G
Start:10000000000000000
Stop :1FFFFFFFFFFFFFFFF
Range width: 2^64
DP size: 15 [0xFFFE000000000000]
[20.33 MK/s][GPU 0.00 MK/s][Count 2^33.38][Dead 1][10:02 (Avg 07:25)][12.4/37.8MB]
Key# 0 [1S]Pub:  0x0230210C23B1A047BC9BDBB13448E67DEDDC108946DE6DE639BCC75D47C0216B1B
       Priv: 0x1A838B13505B26867
Done: Total time 10:02

j2002ba2

full member

Activity: 204

Merit: 437

Quote from: arulbero on June 17, 2020, 11:46:40 AM

Quote from: j2002ba2 on June 17, 2020, 11:26:50 AM

Quote from: arulbero on June 17, 2020, 11:23:34 AM

Quote from: j2002ba2 on June 17, 2020, 11:20:20 AM

If you change the jumps old DPs are useless.

And I moved the point P in this new interval.

How?

I moved P in the interval C in this way: P -> P' = inv(32)*P

This would work properly only when the private key of P is zero mod 32. P=k*G, k=0 (mod 32).

Otherwise the point is guaranteed to not be in the interval.

The probability of being in the new interval is 1/32, which is about 3%. In all other 97% the algorithm would fail to find a point in any reasonable time.

If you somehow can deterministically move points from bigger interval to smaller, then no kangaroos are needed, ECDLP is solved. In the 1/32 case - in just 8 steps (!!!).

brainless

member

Activity: 313

Merit: 34

Quote from: Jean_Luc on June 17, 2020, 12:05:07 PM

It is necessary to use same jumps otherwise path are incompatible.
And the mean has also to be controlled.
Implementing changing of G is rather an heavy task so I would like to have at least an estimation of the loss due to the fact that all paths of new DP will have all points multiple of 32.
We will probably also use DP28 or DP27 for #120

finally you agree, to go back at normal work for 120 as worked for 115, no implements, no changes, no new developments, Huh

arulbero

legendary

Activity: 1914

Merit: 2071

Quote from: Jean_Luc on June 17, 2020, 12:05:07 PM

It is necessary to use same jumps otherwise path are incompatible.
And the mean has also to be controlled.
Implementing changing of G is rather an heavy task so I would like to have at least an estimation of the loss due to the fact that all paths of new DP will have all points multiple of 32.

I will do some tests tomorrow, but I fear that it is not worth it. For now don't work on it.

The mean is correct, the jumps are the same but the main point is:

each path must have the possibility to reach any point in the interval to maximize the probability, but in this case each path could reach only 1/32 of the points.

For example, only the paths starting from a multiple of 32 has the chance to collide with a old DPs, all the others haven't.

Jean_Luc

sr. member

Activity: 462

Merit: 696

It is necessary to use same jumps otherwise path are incompatible.
And the mean has also to be controlled.
Implementing changing of G is rather an heavy task so I would like to have at least an estimation of the loss due to the fact that all paths of new DP will have all points multiple of 32.
We will probably also use DP28 or DP27 for #120

arulbero

legendary

Activity: 1914

Merit: 2071

Quote from: j2002ba2 on June 17, 2020, 11:26:50 AM

Quote from: arulbero on June 17, 2020, 11:23:34 AM

Quote from: j2002ba2 on June 17, 2020, 11:20:20 AM

If you change the jumps old DPs are useless.

And I moved the point P in this new interval.

How?

I moved P in the interval C in this way: P -> P' = inv(32)*P

To be more precise: the wide of the interval is the same (2^119*G' = 2^114*G) as the old interval, and we use the same jumps (that look like they were x32 bigger, but they are not) and we perform the same number of steps for each path (2^25), then we shouldn't go out of the interval.

The points in this interval are closer (the distance between 2 consecutive points is (1/32)*G instead of G)

But you are right on a point: if we use the old jumps, from the point of view of G' they are all multiples of 32, each single path could visit only 1/32 of the points; in this case even if we used many kangaroos in parallel we would have to do more steps! Roll Eyes

j2002ba2

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Activity: 204

Merit: 437

Quote from: arulbero on June 17, 2020, 11:23:34 AM

Quote from: j2002ba2 on June 17, 2020, 11:20:20 AM

If you change the jumps old DPs are useless.

And I moved the point P in this new interval.

How?

arulbero

legendary

Activity: 1914

Merit: 2071

Quote from: j2002ba2 on June 17, 2020, 11:20:20 AM

If you change the jumps old DPs are useless.

I didn't change the jumps, I changed only the interval. In this interval the points are more by a factor of 32. And I moved the point P in this new interval.

j2002ba2

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Activity: 204

Merit: 437

Quote from: arulbero on June 17, 2020, 11:17:21 AM

Quote from: j2002ba2 on June 17, 2020, 11:10:48 AM

Quote from: arulbero on June 17, 2020, 09:26:46 AM

Jumps (the points) are the same, only their private keys are different;

instead of using, for example, 2543*G as jump, you use (2543*32)*G', where G' = inv(32)*G

It sounds quite useless to me.

To get the benefits of kangaroo algorithm, you need to use a specific mean step.

G' makes all the steps multiple of 32, visiting only 1/32 of the points. This is instant sqrt(32) times slowdown.

Why you are visiting only 1/32 of the points? You are working in a wider interval, then it is normal to use larger jumps.

The optimal would be increase the length of the jumps by sqrt(32) instead of 32, but you have to increase it.

If you change the jumps old DPs are useless.

arulbero

legendary

Activity: 1914

Merit: 2071

Quote from: j2002ba2 on June 17, 2020, 11:10:48 AM

Quote from: arulbero on June 17, 2020, 09:26:46 AM

Jumps (the points) are the same, only their private keys are different;

instead of using, for example, 2543*G as jump, you use (2543*32)*G', where G' = inv(32)*G

It sounds quite useless to me.

To get the benefits of kangaroo algorithm, you need to use a specific mean step.

G' makes all the steps multiple of 32, visiting only 1/32 of the points. This is instant sqrt(32) times slowdown.

Why you are visiting only 1/32 of the points? You are working in a interval with more points, then it is normal to use larger jumps.

The optimal would be increase the length of the jumps by sqrt(32) instead of 32, but you have to increase it.

Quote from: j2002ba2 on June 17, 2020, 11:10:48 AM

If you do P' = inv(32)*P, then you have 1/32 chance that it lays in the target interval, so 1/32 chance to solve it, 31/32 chance that algorithm takes forever.

I didn't understand that.

j2002ba2

full member

Activity: 204

Merit: 437

Quote from: arulbero on June 17, 2020, 09:26:46 AM

Jumps (the points) are the same, only their private keys are different;

instead of using, for example, 2543*G as jump, you use (2543*32)*G', where G' = inv(32)*G

It sounds quite useless to me.

To get the benefits of kangaroo algorithm, you need to use a specific mean step.

G' makes all the steps multiple of 32, visiting only 1/32 of the points. This is instant sqrt(32) times slowdown.

The other thing is transforming P.

If it stays the same or is translated, you get 32 times bigger interval, and sqrt(32) times slowdown.

If you do P' = inv(32)*P, then you have 1/32 chance that it lays in the target interval, so 1/32 chance to solve it, 31/32 chance that algorithm takes forever.

If you do 32 parallel P' kangaroo searches, so at least one fits in the target interval, you loose sqrt advantage. Instead of sqrt(32) slowdown of #120 compared to #115, you get 32 times slowdown.

arulbero

legendary

Activity: 1914

Merit: 2071

Quote from: Jean_Luc on June 17, 2020, 10:56:18 AM

OK I see, I will try to implement this.
Thanks Wink

Ok.

Remember: first shift P to P - 2^119*G and then inv(32)*(P - 2^119**G) = inv(32)*P - 2^(119-5)*G

inv(32)*(original P) is not correct.

You could let all the old wild DPs as they are, and computing at the end the private key only for the collision, if you store the old P (point #114) too

Jean_Luc

sr. member

Activity: 462

Merit: 696

OK I see, I will try to implement this.
Thanks Wink

brainless

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Activity: 313

Merit: 34

Quote from: brainless on June 17, 2020, 10:44:41 AM

Quote from: Jean_Luc on June 17, 2020, 10:36:41 AM

Quote from: arulbero on June 17, 2020, 09:26:46 AM

Jumps (the points) are the same, only their private keys are different;

So in practical, I let the old #115 DP as they are and for the new search #120 I use inv(32)*G instead of G ?

yes above is 1 point, and you need 2.5 points (25dp) in search
so if i generate 80 (2.5 per point) coresponding pubkeys, that could save more 80% time saving
those 80 keys only need to search in old 115bit save.work file
but i know u r hesitate to multipubkeys
so better use inv(32)*G instead of G with 1 pubeky

more over
for 25dp, you can go more downbit to 25bit down, final bit range would be 95bit, but with 2.5m corresponding pubkeys
can you imagine, where you can stand in finding solution ( time and resources )

arulbero

legendary

Activity: 1914

Merit: 2071

Quote from: Jean_Luc on June 17, 2020, 10:36:41 AM

Quote from: arulbero on June 17, 2020, 09:26:46 AM

Jumps (the points) are the same, only their private keys are different;

So in practical, I let the old #115 DP as they are and for the new search #120 I use inv(32)*G instead of G ?

Exactly.

You have to change P too, from P to P' = inv(32)*G.

The points are the same, not the distances, you have to multiply all the 32 distances of the jumps by 32.

And when you find a collision and you retrieve the private key, if the collision has happened with a old DP T = oldprivatekey*G:

P' + distance*G' = T (old DP point)

P' = T - distance*G'

P' = oldprivatekey*G - distance*G'

k*G' = oldprivatekey*32*G' - distance*G'

k = (oldprivatekey*32 - distance) mod n

I suggest do not check if a DP is 'old' or not, when you have a collision you can try both way:

k = privatekey of T - distance mod n

and

k = (privatekey of T)*32 - distance mod n

brainless

member

Activity: 313

Merit: 34

Quote from: Jean_Luc on June 17, 2020, 10:36:41 AM

Quote from: arulbero on June 17, 2020, 09:26:46 AM

Jumps (the points) are the same, only their private keys are different;

So in practical, I let the old #115 DP as they are and for the new search #120 I use inv(32)*G instead of G ?

yes above is 1 point, and you need 2.5 points (25dp) in search
so if i generate 80 (2.5 per point) coresponding pubkeys, that could save more 80% time saving
those 80 keys only need to search in old 115bit save.work file
but i know u r hesitate to multipubkeys
so better use inv(32)*G instead of G with 1 pubeky

Jean_Luc

sr. member

Activity: 462

Merit: 696

Quote from: arulbero on June 17, 2020, 09:26:46 AM

Jumps (the points) are the same, only their private keys are different;

So in practical, I let the old #115 DP as they are and for the new search #120 I use inv(32)*G instead of G ?

brainless

member

Activity: 313

Merit: 34

Quote from: arulbero on June 17, 2020, 09:26:46 AM

Quote from: Jean_Luc on June 17, 2020, 09:19:58 AM

Hi,
I don't understand well how this can work ?
If you change G, the path will also differ as jumps are a function of the X value.

Jumps (the points) are the same, only their private keys are different;

instead of using, for example, 2543*G as jump, you use (2543*32)*G', where G' = inv(32)*G

in this way the points are the same and the paths are the same. The same interval

A = [1G, 2G...., (2^114 - 1)*G]

can be viewed as

B = [32*G', 2*32*G'...., 32*(2^114 - 1)*G']

and this interval is a subset of the interval

C = [1*G', 2*G', 3*G',...., (2^119 - 1)*G']

Note that C has the same number of elements of

D = [1*G, 2*G, ...., (2^119 - 1)*G]

but C != D.

To recap:

A = B, A subset C
A subset D

What is the difference? Why using C instead of D?

The advantage is that the old DPs, that lie in A = B, lie in C too, but now they are uniformly spread out in C, while the same points are not uniformly spread out in D.

Now, P lies in the interval D, not in C, but P' = inv(32)*P does, and the private key of P' respect to G' is the same as the private key of P respect to G.

-----------------------------------------------------
Note:

C is not the original interval D = [0*G, 2*G, ...., (2^119 - 1)*G], C is a 'contraction' of D, on other hand the very original interval (where the real public key P lies) is E = [2^119 *G, ..., (2^120 - 1)*G].

Usually you shift E to D and P to P' = P - 2^119*G; note that it is like you changed G to G' = G - 2^119G.

Now I suggest to add another move:

move D to C and P' to P'' = inv(32)*P' ; I call this move a 'contraction', because I divide G by 32.

Actually you r saying upword in multiple mode
And i said long time ago downside multiple mode.

Upside *32 mean u still stand 1 part of 32 part
And down side u r stand for 32/1
So i know this stage of multiple pub keys you have 90% work n time savings.
Hope u will not understand. Other jean luc will leave everything and rush to develop multipub key search ver .
Bc after long junps in thinking he will come to this stage in last
Example i said
2 x 100 = 200
But he listening
2+2+2 till 200
2+3 till 200
2x2 x2 till 200

Hope u all moving slowly to multiple keys or multiple bits
But dont try to understand my words. As its from brainless

WhyMe

sr. member

Activity: 661

Merit: 250

Quote from: Jean_Luc on June 16, 2020, 09:10:13 AM

Yes thanks for this tip

I'm building a map to compute with accuracy the DP overhead as I didn't yet managed to get the correct analytical expression but only the 2 asymptote (0 and infinity)
Tomorrow I will also investigate how long would it take with the VanitySearch engine optimized for V100 to solve #64.
Then we will make our choice on attacking #64 or #120.

Can you please let the #64 for normal guys not owning/renting tons of gpu ?

arulbero

legendary

Activity: 1914

Merit: 2071

Quote from: Jean_Luc on June 17, 2020, 09:19:58 AM

Hi,
I don't understand well how this can work ?
If you change G, the path will also differ as jumps are a function of the X value.

Jumps (the points) are the same, only their private keys are different;

instead of using, for example, 2543*G as jump, you use (2543*32)*G', where G' = inv(32)*G

in this way the DP points are the same, the jumps are the same and all the old paths are the same too.

The same interval

A = [1G, 2G...., (2^114 - 1)*G]

can be viewed as

B = [32*G', 2*32*G'...., 32*(2^114 - 1)*G']

and this interval is a subset of the interval

C = [1*G', 2*G', 3*G',...., (2^119 - 1)*G']

Note that C has the same number of elements of

D = [1*G, 2*G, ...., (2^119 - 1)*G]

but C != D.

To recap:

A = B, A subset C
A subset D

What is the difference? Why to work in C instead of D, since the old points in A are already in D?

The advantage is that the old DPs, that lie in A, are uniformly spread out in C, while the same points are not uniformly spread out in D.

Now, P lies in the interval D, not in C, but P' = inv(32)*P does, and the private key of P' respect to G' is the same as the private key of P respect to G.

-----------------------------------------------------
Note:

C is not the original interval D = [0*G, 2*G, ...., (2^119 - 1)*G], C is a 'contraction' of D, on other hand the very original interval (where the real public key P lies) is E = [2^119 *G, ..., (2^120 - 1)*G].

Usually you shift E to D and P to P' = P - 2^119*G.

Now I suggest to add another move:

move D to C and P' to P'' = inv(32)*P' ; I call this move a 'contraction', because I divide G by 32.

Topic: Pollard's kangaroo ECDLP solver - page 94. (Read 55599 times)