WanderingPhilospher
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Shooters Shoot...
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September 07, 2024, 12:56:56 AM Last edit: September 07, 2024, 03:38:06 AM by WanderingPhilospher |
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The main difference between the Gaudry-Schost algorithm and the kangaroo algorithm is that when a distinguished point is hit, Gaudry and Schost restart the walk from a random starting point in a certain range, whereas the kangaroos keep on running. The theoretical analysis is different too: Gaudry and Schost use a variant of the birthday paradox whereas Pollard and van Oorschot and Wiener use a different probabilistic argument Do you agree with the above statement? (You first mentioned them, in your original post, thought you were comparing them, or maybe just the run time and the pain it is to program?) You forgot the fact that the "fruitless" one started from a known position that (if it is not taken randomly) Yes, I am saying the starting points are random. I thought you were too, I guess not. In your Kangaroo "variant" using 100s of GPUs, are you placing each thread on an exact / known starting point? If so, how are you doing this, on the fly? It can be done, I am curious if you are doing it? Are you using pseudorandom walks or something else as well? using random start points result in an improved lower runtime Where did I state this? Again, I am asking YOU questions about your new lower run time variant that you stumbled upon. Ignore everything else. How or what does the variant do, to lower average runtime? Is only having 3 starting points, feasible for a 130 bit range? Or would you break it up into sub-ranges? EDIT: I do agree with you about the fruitless, if starting point is known. These algos and these challenges are always fun and fascinating. It may take reading 10 different papers or posts, but eventually, one learns something new of sees something from a different perspective.
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COBRAS
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September 07, 2024, 01:11:11 AM |
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The main difference between the Gaudry-Schost algorithm and the kangaroo algorithm is that when a distinguished point is hit, Gaudry and Schost restart the walk from a random starting point in a certain range, whereas the kangaroos keep on running. The theoretical analysis is different too: Gaudry and Schost use a variant of the birthday paradox whereas Pollard and van Oorschot and Wiener use a different probabilistic argument Do you agree with the above statement? (You first mentioned them, in your original post, thought you were comparing them, or maybe just the run time and the pain it is to program?) You forgot the fact that the "fruitless" one started from a known position that (if it is not taken randomly) Yes, I am saying the starting points are random. I thought you were too, I guess not. In your Kangaroo "variant" using 100s of GPUs, are you placing each thread on an exact / known starting point? If so, how are you doing this, on the fly? It can be done, I am curious if you are doing it? Are you using pseudorandom walks or something else as well? using random start points result in an improved lower runtime Where did I state this? Again, I am asking YOU questions about your new lower run time variant that you stumbled upon. Ignore everything else. How or what does the variant do, to lower average runtime? Is only having 3 starting points, feasible for a 130 bit range? Or would you break it up into sub-ranges? Is starting point not random, did you take less operation then with random ? I think will be equal, or not ? then no method, not random operations will be equal to random.
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kTimesG
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September 07, 2024, 08:12:20 AM |
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Do you agree with the above statement? (You first mentioned them, in your original post, thought you were comparing them, or maybe just the run time and the pain it is to program?)
Of course, since it's correct. I never referred to GS as a kangaroo variant. Read the three/four kangaroo paper by the same authors & Pollard to understand better the challenges of implementing this correctly (mainly, restarts are not counted as group operations, but on EC this is a very heavy operation compared to the simple additive jumps). In your Kangaroo "variant" using 100s of GPUs, are you placing each thread on an exact / known starting point? If so, how are you doing this, on the fly? It can be done, I am curious if you are doing it?
Well, my kangaroos are kept on a central database, keeping track of the next starting point and how many times they jumped. The ones that jumped the less are scheduled to go in the next round when requested. I only extract XY after a lot of jumps (hundreds of thousands of jumps per thread, and a thread having several kangaroos) and verify they are correct (they should match the delta distance traveled, which is also counted during the jumps). This ensures nothing sketchy was done in the GPU kernel, like a bad field operation or other invisible bugs that can compromise the entire computation chain. Also for DPs the same thing, only needed piece of info from GPU is the distance it's at (relative to where computation started at - less bits needed to write and transfer) and the kang that found it, no need to ever store X coord and absolute distance like in JLP. I haven't discovered anything breaking new, to be honest. If we simply combine secp256k1 curve properties to a specific variant of using 3 types of walks, the complexity is lower than the expected, that is all. This simply happens because kangaroo is a generic algorithm, and there was no assumption in the paper of presentation, that an element and its inverse might only differ by a single bit (in our case, the Y sign). To be more clear than ever: the best known kangaroo method uses 4 kangaroos, and an expected number of 1.71 sqrt(b) operations on a generic group. In a generic group, an element and its inverse might be two totally different values with nothing in common. using random start points result in an improved lower runtime Where did I state this? You didn't, and it doesn't, random start positions and adding more kangaroos without serious thinking of the consequences just results in an increased number of operations and greater runtime. You can verify this over long number of runs on small intervals. Number of kang directly influences the choice of the optimal jump table length. If this is messed with after the algo started, cannot expect same results, or better results, only worse results
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stamun
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September 08, 2024, 07:57:42 PM |
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Which one will be most performant software to attack the #130 challenge public key? The CPU version of keyhunt is giving me around ~35 Pkeys/sec. Is there a more optimal solution for GPUs?
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COBRAS
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September 08, 2024, 08:13:37 PM |
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Which one will be most performant software to attack the #130 challenge public key? The CPU version of keyhunt is giving me around ~35 Pkeys/sec. Is there a more optimal solution for GPUs?
keyhant with 256 - 512 coree maybe
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nomachine
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September 09, 2024, 05:08:43 AM |
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512 coree maybe
1×10^20 cores
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pbies
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September 09, 2024, 02:57:29 PM |
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Has anyone public keys for known puzzle private keys?
I want to look at the public keys and maybe I will sort sth out from them...
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BTC: bc1qmrexlspd24kevspp42uvjg7sjwm8xcf9w86h5k
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mcdouglasx
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New ideas will be criticized and then admired.
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September 09, 2024, 04:48:59 PM |
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Has anyone public keys for known puzzle private keys?
I want to look at the public keys and maybe I will sort sth out from them...
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I'm not dead, long story... BTC bc1qxs47ttydl8tmdv8vtygp7dy76lvayz3r6rdahu
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citb0in
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September 09, 2024, 04:49:33 PM |
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Has anyone public keys for known puzzle private keys?
I want to look at the public keys and maybe I will sort sth out from them...
The private keys of the known puzzles are ... well, known! The rest is simple --> https://learnmeabitcoin.com/beginners/guide/public-keys/
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_______. ______ __ ______ ______ __ ___ .______ ______ ______ __ ______ .______ _______ / | / __ \ | | / __ \ / || |/ / | _ \ / __ \ / __ \ | | / __ \ | _ \ / _____| | (----`| | | | | | | | | | | ,----'| ' / | |_) | | | | | | | | | | | | | | | | |_) | | | __ \ \ | | | | | | | | | | | | | < | ___/ | | | | | | | | | | | | | | | / | | |_ | .----) | | `--' | | `----.| `--' | __| `----.| . \ | | | `--' | | `--' | | `----.__| `--' | | |\ \----.| |__| | |_______/ \______/ |_______| \______/ (__)\______||__|\__\ | _| \______/ \______/ |_______(__)\______/ | _| `._____| \______| | 2% fee anonymous solo bitcoin mining for all at https://solo.CKpool.org | No registration required, no payment schemes, no pool op wallets, no frills, no fuss. |
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mcdouglasx
Member
Offline
Activity: 258
Merit: 67
New ideas will be criticized and then admired.
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September 09, 2024, 09:07:32 PM |
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I'm not dead, long story... BTC bc1qxs47ttydl8tmdv8vtygp7dy76lvayz3r6rdahu
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kTimesG
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September 10, 2024, 12:18:04 AM |
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I'd like to take credit for the below informal conjectures. Solving ECDLP on an interval in 1.0 sqrt(b) group operations with 1.0 sqrt(b) stored items.This beats BSGS by a factor of 2x since BSGS requires 2 sqrt(b) operations (worse, I think they are actually scalar multiplications). This also beats any known kangaroo or other algorithm when DP = 0. T1 -> T2 -> -N/2. -4N/5 | 4N/5. N/5
Shift interval to [-N/2, N/2) (there's a twist to this though, 0 is not needed**) Start two tames at -4N/5 and 4N/5 Start two wilds at Q=kG and -Q Jump alternatively T1, W1, T2 and W2, use Y as jump function. Before each jump, check if the min(y, p - y) coordinate is mapped to a previous visit (type, distance). If so: if type = T1 and visit.type = T1: continue walk, kangaroo jumped from left to right through opposite points if type = T1 or T2 and visit.type = W1 or W2: DLP solved if type = W1 and visit.type = W1: DLP solved* (it sounds absurd, but we have (-k + d1) == -(-k + d2) if type = W2 and visit.type = W2: DLP solved* since (k + d1) == -(k + d2) if type = W1 and visit.type = W2: DLP solved*, since -k + d1 = k + d2 if type = W2 and visit.type = W1: DLP solved* since k + d1 = -k + d2 if dlp not solved, move one of the two kangaroos forward *: only if deltas don't cancel each other out (rarely but can happen) **: can get rid of 0 because 8045a5248dae8d1dea3c4eab7caac0b35142851038eea7d1f2556bf08e47ccf0What happens? Forward jumps by any kangaroo are equivalent to backward jumps of the same distance. Even if those points are not actually visited and stored. T1 moving forward == T2 moving backward W1 is moving forward while also mirroring a W2 walking in the opposite direction. Double win. If W1 or W2 catch each other, or catch some arbitrary point AND its opposite after passing through the middle, DLP is solved. That is right, we can solve DLP using a single wild kangaroo which sounds insane but is true, when k < 0 or -k < 0 and the walk passes through any -P & P points. When these details are factored into the usual analysis of catching kangaroos, the probabilities of collision are multiplied because now the chance of hitting a visited point is actually quadrupled or so at each jump of any kangaroo: - points that are symmetric are "the same point" basically = miss chance is halved - kangs go in a sort of "meet in the middle" walk = more "visited" spots per average jump size - kangs also jump in their usual forward walk ofcourse in the same time - classic analysis So now there are 4 walks instead of two, but because of the much higher density of "already visited" probabilities, number of group operations to solve ECDLP is on average < 1.03 sqrt(b) and number of scalar multiplications is basically zero (except initial setup) in 99.999% of the cases. So now the big question: what happens if we want to decrease storage and use DP? Well, I don't have any actual proof, but some parts of the above still hold, and depending on the ratio between the interval size and the chosen 2**DP there will still be an increased larger chance of collisions between DPs that are found on both halfs by misc. pairs of walks, and as such decreasing the total number of group operations that are expected, anywhere between 1.5 to 1.7 sqrt(b) - it is likely that as the interval increases, and the DP increases, complexity still stays around the same level or vey slightly increases.
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WanderingPhilospher
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Activity: 1148
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Shooters Shoot...
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September 10, 2024, 04:43:33 PM |
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I'd like to take credit for the below informal conjectures. Solving ECDLP on an interval in 1.0 sqrt(b) group operations with 1.0 sqrt(b) stored items.This beats BSGS by a factor of 2x since BSGS requires 2 sqrt(b) operations (worse, I think they are actually scalar multiplications). This also beats any known kangaroo or other algorithm when DP = 0. T1 -> T2 -> -N/2. -4N/5 | 4N/5. N/5
Shift interval to [-N/2, N/2) (there's a twist to this though, 0 is not needed**) Start two tames at -4N/5 and 4N/5 Start two wilds at Q=kG and -Q Jump alternatively T1, W1, T2 and W2, use Y as jump function. Before each jump, check if the min(y, p - y) coordinate is mapped to a previous visit (type, distance). If so: if type = T1 and visit.type = T1: continue walk, kangaroo jumped from left to right through opposite points if type = T1 or T2 and visit.type = W1 or W2: DLP solved if type = W1 and visit.type = W1: DLP solved* (it sounds absurd, but we have (-k + d1) == -(-k + d2) if type = W2 and visit.type = W2: DLP solved* since (k + d1) == -(k + d2) if type = W1 and visit.type = W2: DLP solved*, since -k + d1 = k + d2 if type = W2 and visit.type = W1: DLP solved* since k + d1 = -k + d2 if dlp not solved, move one of the two kangaroos forward *: only if deltas don't cancel each other out (rarely but can happen) **: can get rid of 0 because 8045a5248dae8d1dea3c4eab7caac0b35142851038eea7d1f2556bf08e47ccf0What happens? Forward jumps by any kangaroo are equivalent to backward jumps of the same distance. Even if those points are not actually visited and stored. T1 moving forward == T2 moving backward W1 is moving forward while also mirroring a W2 walking in the opposite direction. Double win. If W1 or W2 catch each other, or catch some arbitrary point AND its opposite after passing through the middle, DLP is solved. That is right, we can solve DLP using a single wild kangaroo which sounds insane but is true, when k < 0 or -k < 0 and the walk passes through any -P & P points. When these details are factored into the usual analysis of catching kangaroos, the probabilities of collision are multiplied because now the chance of hitting a visited point is actually quadrupled or so at each jump of any kangaroo: - points that are symmetric are "the same point" basically = miss chance is halved - kangs go in a sort of "meet in the middle" walk = more "visited" spots per average jump size - kangs also jump in their usual forward walk ofcourse in the same time - classic analysis So now there are 4 walks instead of two, but because of the much higher density of "already visited" probabilities, number of group operations to solve ECDLP is on average < 1.03 sqrt(b) and number of scalar multiplications is basically zero (except initial setup) in 99.999% of the cases. So now the big question: what happens if we want to decrease storage and use DP? Well, I don't have any actual proof, but some parts of the above still hold, and depending on the ratio between the interval size and the chosen 2**DP there will still be an increased larger chance of collisions between DPs that are found on both halfs by misc. pairs of walks, and as such decreasing the total number of group operations that are expected, anywhere between 1.5 to 1.7 sqrt(b) - it is likely that as the interval increases, and the DP increases, complexity still stays around the same level or vey slightly increases. So your 2 tames and 2 wilds are starting on the exact same points, mirrored of each other? Meaning same x, but different y. And then they both jump forward, positive jumps, of the same size?
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kTimesG
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September 10, 2024, 04:51:14 PM |
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So your 2 tames and 2 wilds are starting on the exact same points, mirrored of each other? Meaning same x, but different y. And then they both jump forward, positive jumps, of the same size?
Yes, all forward, same jump rules for all. But jump function is function of Y, so they don't jump forward with same size even if they have the same X. So the ones on the left are going towards the ones to the right, but the two don't go forward with equal jumps. This makes the covering more uniform.
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masken
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September 11, 2024, 03:45:33 AM |
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In its current state, yes, it seems impractical. However, it can be made usable by customizing and narrowing the range. Compared to regular vanity, it can find RIPEMD-160s with the same Hamming distance using less energy. just my opinion:)
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WanderingPhilospher
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Shooters Shoot...
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September 11, 2024, 04:00:49 AM |
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In its current state, yes, it seems impractical. However, it can be made usable by customizing and narrowing the range. Compared to regular vanity, it can find RIPEMD-160s with the same Hamming distance using less energy. just my opinion:)
What does this mean? You can use a variant of Vanity Search, to search in exact bit range or a specified range, with GPUs and/or CPUs, with prefix and suffix. How is this Rust, CPU only, program, going to be better or faster?
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masken
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September 11, 2024, 04:23:40 AM |
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Prefix: 13zb1hQb Hash: 20d45a6a76
effort spent to achieve this didn't seem the same to me.
Prefix: 13zb Suffix: h5so
Which target is closer to the hash? I'm genuinely asking to find out. The method seemed logical and useful to me, but I could be wrong. Please correct me if I'm mistaken.
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WanderingPhilospher
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September 11, 2024, 01:29:06 PM Last edit: September 11, 2024, 02:39:53 PM by WanderingPhilospher |
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Prefix: 13zb1hQb Hash: 20d45a6a76
effort spent to achieve this didn't seem the same to me.
Prefix: 13zb Suffix: h5so
Which target is closer to the hash? I'm genuinely asking to find out. The method seemed logical and useful to me, but I could be wrong. Please correct me if I'm mistaken.
It really does not matter what I tell you. People will argue for both sides so I would suggest you do your own key hunting and find out. I will tell you this, you could have a hash that out of 40 characters, 1 is different. An example: 20D45A6A762535700CE9E0B216E31994335DB8A5 20D45A6A762535700CE9E0B216E31994335DB8A6 and the private key of those two could be very far apart, one in a 66 bit range, one in a 255 bit range. Meaning the H160 doesn't signify where the private key is located (bit range), or having two almost identical H160s, does not mean they are close together, private key / bit range wise. The original point I was making, is you can use a variant of vanitysearch, which can use GPUs, and do the same type of search, but process many more keys per second. I haven't used it in quite a while but I believe if you use this: 13zb*h5so; in vanitysearch it's the same as Prefix: 13zb Suffix: h5so in the program you are referencing. Yes, all forward, same jump rules for all. But jump function is function of Y, so they don't jump forward with same size even if they have the same X. So the ones on the left are going towards the ones to the right, but the two don't go forward with equal jumps. This makes the covering more uniform. Can you give an example of what you mean by the jump function is function of Y, with an example? Thanks.
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kTimesG
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September 11, 2024, 05:09:26 PM |
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Yes, all forward, same jump rules for all. But jump function is function of Y, so they don't jump forward with same size even if they have the same X. So the ones on the left are going towards the ones to the right, but the two don't go forward with equal jumps. This makes the covering more uniform. Can you give an example of what you mean by the jump function is function of Y, with an example? Thanks. jump_index = current_element.y % jump_table_length Let's say we have two points with the same X, they are opposite points. P = (xp, +yp) -P = (xp, -yp) If they use xp as jump function, they jump forward with the same distance, hence a less random walk. Since yp = ± sqrt(xp**3 + 7) jumping by Y spreads the randomness. Another way to view this: in an [-N, N] interval we have N unique X values, and 2N unique Y values. Larger pool = better pseudo-randomness in an interval that's half of the length. But even jumping by X alone has very good results, it was just that jumping by Y had even better ones.
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