BurtW
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January 02, 2016, 07:11:19 PM 

We should form a cooperative and search the private key... Most dump idea ever. You made my day. You obviously do not know me very well. This idea is not even a mole on the ass of a gnat on the ass of the dumpest idea I have ever had. Thinking about is a bit further the cooperative method of finding the next prize can be simplified to just giving the entire prize, all 0.051 x $433.42 = $22.10 to whoever is lucky enough to be assigned the lucky section of private keys by the server. That way we would not have to figure out which charity, or how to prove work, etc. It would just be a lottery with a $22.10 prize for using up an immense amount of electricity.

Our family was terrorized by Homeland Security. Read all about it here: http://www.jmwagner.com/ and http://www.burtw.com/ Any donations to help us recover from the $300,000 in legal fees and forced donations to the Federal Asset Forfeiture slush fund are greatly appreciated!




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BG4
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PaperSafe


January 02, 2016, 09:53:54 PM 

After reading into this puzzle post ..I have a question maybe someone can answer 1 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreAnchuDf 2 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreAvUcVfH 3 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreB1FQ8BZ 4 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreB4AD8Yi 5 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBF8or94 6 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBKdE2NK 7 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBR6zCMU 8 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBbMaQX1 9 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBd7uGcN A 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBoNWTw6 B 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBquB4Rj C 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreC4p2u5o D 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreCAVyVnh E 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreCHK2Zzv F 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreCMUnXUo These are all valid private keys.... Just how many "NON valid" keys are in between each of the first 16 valid keys and is there a pattern.. I am a little rusty with my Base58 math skills and maybe I should have done it in WIF Compressed....




BurtW
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January 02, 2016, 09:58:29 PM 

After reading into this puzzle post ..I have a question maybe someone can answer 1 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreAnchuDf 2 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreAvUcVfH 3 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreB1FQ8BZ 4 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreB4AD8Yi 5 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBF8or94 6 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBKdE2NK 7 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBR6zCMU 8 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBbMaQX1 9 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBd7uGcN A 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBoNWTw6 B 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBquB4Rj C 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreC4p2u5o D 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreCAVyVnh E 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreCHK2Zzv F 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreCMUnXUo These are all valid private keys.... Just how many "NON valid" keys are in between each of the first 16 valid keys and is there a pattern.. I am a little rusty with my Base58 math skills and maybe I should have done it in WIF Compressed.... Read this post, just a few posts back: https://bitcointalk.org/index.php?topic=1306983.msg13424809#msg13424809If you want to know more please read the entire thread. It is only 7 pages long.

Our family was terrorized by Homeland Security. Read all about it here: http://www.jmwagner.com/ and http://www.burtw.com/ Any donations to help us recover from the $300,000 in legal fees and forced donations to the Federal Asset Forfeiture slush fund are greatly appreciated!



HugoTheSpider
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January 02, 2016, 11:05:50 PM 

After reading into this puzzle post ..I have a question maybe someone can answer 1 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreAnchuDf 2 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreAvUcVfH ... F 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreCMUnXUo These are all valid private keys.... Just how many "NON valid" keys are in between each of the first 16 valid keys and is there a pattern.. What you posted weren't the actual raw private keys but the base58 encoded and WIF formatted version of the private keys with the information stating it's uncompressed. The hex representation of a base58 encoded and WIF formatted uncompressed key is: 80 XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX YYYYYYYY The hex representation of a base58 encoded and WIF formatted compressed key is: 80 XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 01 YYYYYYYY where XX is the raw private key and YY the checksum. Your question regarding "how many invalid keys are in between" is ambiguous. You might also consider to open a new thread to reach appropriate audience.




Gleb Gamow
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January 02, 2016, 11:10:32 PM 

1 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreAnchuDf 2 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreAvUcVfH 3 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreB1FQ8BZ 4 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreB4AD8Yi 5 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBF8or94 6 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBKdE2NK 7 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBR6zCMU 8 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBbMaQX1 9 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBd7uGcN A 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBoNWTw6 B 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreBquB4Rj C 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreC4p2u5o D 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreCAVyVnh E 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreCHK2Zzv F 5HpHagT65TZzG1PH3CSu63k8DbpvD8s5ip4nEB3kEsreCMUnXUo NOW, that's interesting! I see a pattern that may now be solvable. I see somebody has a jumpstart: https://bitcointalk.org/index.php?topic=1306983.msg13424809#msg13424809 But why Log(2)?




shorena
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January 02, 2016, 11:25:38 PM 

Its an easy way to check whether the first bit is always 1 for a given step, which it is. Thus you can limit the search space. For step n you do not need to search for any possible solutions from step n1.




CatsLikeToStretch
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January 02, 2016, 11:51:33 PM 

Here are all keys from #1 to #40: Bits  Hex  Decimal  Binary  As raw text  Log(2)  1  0000000001  1  ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░█   0  2  0000000003  3  ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░██   1.5849625007212  3  0000000007  7  ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░███   2.8073549220576  4  0000000008  8  ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░█░░░   3  5  0000000015  21  ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░█░█░█   4.3923174227788  6  0000000031  49  ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░██░░░█  1  5.6147098441152  7  000000004C  76  ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░█░░██░░  L  6.2479275134436  8  00000000E0  224  ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░███░░░░░  à  7.8073549220576  9  00000001D3  467  ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░███░█░░██  Ó  8.8672787397097  10  0000000202  514  ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░█░░░░░░░█░   9.0056245491939  11  0000000483  1155  ░░░░░░░░░░░░░░░░░░░░░░░░░░░░░█░░█░░░░░██  ƒ  10.173677136303  12  0000000A7B  2683  ░░░░░░░░░░░░░░░░░░░░░░░░░░░░█░█░░████░██  {  11.389631339261  13  0000001460  5216  ░░░░░░░░░░░░░░░░░░░░░░░░░░░█░█░░░██░░░░░  `  12.348728154231  14  0000002930  10544  ░░░░░░░░░░░░░░░░░░░░░░░░░░█░█░░█░░██░░░░  )0  13.364134655008  15  00000068F3  26867  ░░░░░░░░░░░░░░░░░░░░░░░░░██░█░░░████░░██  hó  14.713547616913  16  000000C936  51510  ░░░░░░░░░░░░░░░░░░░░░░░░██░░█░░█░░██░██░  É6  15.652564919611  17  000001764F  95823  ░░░░░░░░░░░░░░░░░░░░░░░█░███░██░░█░░████  vO  16.548084361224  18  000003080D  198669  ░░░░░░░░░░░░░░░░░░░░░░██░░░░█░░░░░░░██░█   17.600007248708  19  000005749F  357535  ░░░░░░░░░░░░░░░░░░░░░█░█░███░█░░█░░█████  tŸ  18.447724952285  20  00000D2C55  863317  ░░░░░░░░░░░░░░░░░░░░██░█░░█░██░░░█░█░█░█  ,U  19.719530872026  21  00001BA534  1811764  ░░░░░░░░░░░░░░░░░░░██░███░█░░█░█░░██░█░░  ¥4  20.788963611792  22  00002DE40F  3007503  ░░░░░░░░░░░░░░░░░░█░██░████░░█░░░░░░████  ä  21.520134745822  23  0000556E52  5598802  ░░░░░░░░░░░░░░░░░█░█░█░█░██░███░░█░█░░█░  UnR  22.416686729788  24  0000DC2A04  14428676  ░░░░░░░░░░░░░░░░██░███░░░░█░█░█░░░░░░█░░  Ü*  23.782435585948  25  0001FA5EE5  33185509  ░░░░░░░░░░░░░░░██████░█░░█░████░███░░█░█  ú^å  24.984050066697  26  000340326E  54538862  ░░░░░░░░░░░░░░██░█░░░░░░░░██░░█░░██░███░  @2n  25.700781261713  27  0006AC3875  111949941  ░░░░░░░░░░░░░██░█░█░██░░░░███░░░░███░█░█  ¬8u  26.738278526959  28  000D916CE8  227634408  ░░░░░░░░░░░░██░██░░█░░░█░██░██░░███░█░░░  ‘lè  27.762143403295  29  0017E2551E  400708894  ░░░░░░░░░░░█░██████░░░█░░█░█░█░█░░░████░  âU  28.577979290797  30  003D94CD64  1033162084  ░░░░░░░░░░████░██░░█░█░░██░░██░█░██░░█░░  =”Íd  29.944419458082  31  007D4FE747  2102388551  ░░░░░░░░░█████░█░█░░███████░░███░█░░░███  }OçG  30.969382178281  32  00B862A62E  3093472814  ░░░░░░░░█░███░░░░██░░░█░█░█░░██░░░█░███░  ¸b¦.  31.526580209328  33  01A96CA8D8  7137437912  ░░░░░░░██░█░█░░█░██░██░░█░█░█░░░██░██░░░  ©l¨Ø  32.732759144628  34  034A65911D  14133072157  ░░░░░░██░█░░█░█░░██░░█░██░░█░░░█░░░███░█  Je‘  33.718356052473  35  04AED21170  20112871792  ░░░░░█░░█░█░███░██░█░░█░░░░█░░░█░███░░░░  ®Ò p  34.227400038686  36  09DE820A7C  42387769980  ░░░░█░░███░████░█░░░░░█░░░░░█░█░░█████░░  Þ‚   35.302929017097  37  1757756A93  100251560595  ░░░█░███░█░█░███░███░█░█░██░█░█░█░░█░░██  Wuj“  36.544833738747  38  22382FACD0  146971536592  ░░█░░░█░░░███░░░░░█░█████░█░██░░██░█░░░░  "8/¬Ð  37.096745824716  39  4B5F8303E9  323724968937  ░█░░█░██░█░██████░░░░░██░░░░░░█████░█░░█  K_ƒ é  38.235977688802  40  E9AE4933D6  1003651412950  ███░█░░██░█░███░░█░░█░░█░░██░░████░█░██░  é®I3Ö  39.868395419757 
Looks a lot like many of the Wolfram Automaton patterns, like this one: http://mathblog.com/wpcontent/uploads/2011/05/rule110.jpg




werty98
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January 04, 2016, 05:22:12 AM 

Your results are from CPU or GPU?
What's your performance?
I was involved in the challenge and earned some BTC. I speedhacked the code in the mealtime at work straight after I saw the puzzle back in January 2015 and started it on some server remotely. The first version could do about 100,000 keys/s using a CPU. After some improvements I boosted the rate to 700,000 keys/s on a single computer. Later I ran it on many computers and checked million of keys/s. Luckly I could recycle the code later in the August 2015 puzzle at https://bitcointalk.org/index.php?topic=1144807.0I had no time to learn how to code on a GPU. I've tried my first version using openssl and got about 25,000 keys/s (singlethread). You seem to use something other than openssl. Unlike hash operations, elliptic curve operations have unpredictable machine cycle count. Thus, speedup on SIMD processors (e.g. GPUs) won't be great. BTW, whoever created this challenge will be able to manipulate BTC price, won't she/he?




BurtW
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January 04, 2016, 09:03:28 AM 

Unlike hash operations, elliptic curve operations have unpredictable machine cycle count.
I would expect a single point addition operation (P _{3} = P _{1} + P _{2}) to have a very predictable machine cycle count. It should be something like: x _{3} = λ ^{2}+a _{1}λ−a _{2}−x _{1}−x _{2}y _{3} = −a _{1}x _{3}−a _{3}−λx _{3}+λx _{1}−y _{1}λ = (y _{2}−y _{1}) / (x _{2}−x _{1}) From: https://crypto.stanford.edu/pbc/notes/elliptic/explicit.html

Our family was terrorized by Homeland Security. Read all about it here: http://www.jmwagner.com/ and http://www.burtw.com/ Any donations to help us recover from the $300,000 in legal fees and forced donations to the Federal Asset Forfeiture slush fund are greatly appreciated!



werty98
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January 04, 2016, 09:24:18 AM 

Unlike hash operations, elliptic curve operations have unpredictable machine cycle count.
I would expect a single point addition operation (P _{3} = P _{1} + P _{2}) to have a very predictable machine cycle count. It should be something like: x _{3} = λ ^{2}+a _{1}λ−a _{2}−x _{1}−x _{2}y _{3} = −a _{1}x _{3}−a _{3}−λx _{3}+λx _{1}−y _{1}λ = (y _{2}−y _{1}) / (x _{2}−x _{1}) From: https://crypto.stanford.edu/pbc/notes/elliptic/explicit.htmlI did use that EC point increment method, otherwise the program would be 20x slower. Those multiplications/divisions are modulus but not conventional integer operators; they don't have predictable machine cycle count.




dalek


January 04, 2016, 10:23:45 AM 

I've tried my first version using openssl and got about 25,000 keys/s (singlethread). You seem to use something other than openssl.
Unlike hash operations, elliptic curve operations have unpredictable machine cycle count. Thus, speedup on SIMD processors (e.g. GPUs) won't be great.
I get the same performance as you, but EC_POINT_add isn't the slowest step, it's the call to EC_POINT_point2oct.




Bulista
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January 04, 2016, 11:09:31 AM Last edit: January 04, 2016, 11:28:56 AM by Bulista 

Guys with brute force we won't go anywhere. Let us assume we would be able somehow to generate a brute force of 5 TH/s (The max ASIC Miner speed today that I found out there). That's like 5000000000000 Hashes per second. Of course with CPU/GPU we won't get anywhere near. But let us assume for a second that this performance would be possible somehow, we would need ~303150594339750000000000000000000000000000000000000000000 years to break until the address 256 already counting with starting each address generation from 2^X1. (EDIT: Not even counting with all the addresses before ) So it's kind of useless to brute force this. Unlike hash operations, elliptic curve operations have unpredictable machine cycle count.
I would expect a single point addition operation (P _{3} = P _{1} + P _{2}) to have a very predictable machine cycle count. It should be something like: x _{3} = λ ^{2}+a _{1}λ−a _{2}−x _{1}−x _{2}y _{3} = −a _{1}x _{3}−a _{3}−λx _{3}+λx _{1}−y _{1}λ = (y _{2}−y _{1}) / (x _{2}−x _{1}) From: https://crypto.stanford.edu/pbc/notes/elliptic/explicit.htmlHow precise is this formula?




BurtW
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January 04, 2016, 12:06:45 PM Last edit: January 04, 2016, 12:18:47 PM by BurtW 

It is impossible to get all the rewards.
It is possible to get the next reward (0.051 BTC) in the sequence.
There are only 1,125,899,906,842,620 possible keys for the next unclaimed reward of 0.051 BTC.
At your rate of 5,000,000,000,000 trial per second it would only take a bit over 225 seconds to try all of them.
There is no hope of getting all the remaining rewards but it should be totally possible to get a few more small rewards from the sequence.

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werty98
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January 04, 2016, 12:14:04 PM 

I get the same performance as you, but EC_POINT_add isn't the slowest step, it's the call to EC_POINT_point2oct.
Do you know why EC_POINT_point2oct is slow? I called EC_POINT_get_affine_coordinates_GFp instead, but they seem to do the same thing. If EC_POINT_add stores the point in Cartesian coordination (XY), will extracting the public key be a trivial task? Guys with brute force we won't go anywhere.
How precise is this formula?
I think all of us know that brute force is not viable for sure, but it is fun to estimate how long it will take to break the 51st, 52nd, ... addresses. That division isn't the conventional division. It has perfect precision ( http://www.johannesbauer.com/compsci/ecc/#anchor07).




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January 04, 2016, 12:22:38 PM 

This is a basic description of the algorithm which should yield the fastest results: Initialization:
Set BitcoinAddresses[256] = the list of bitcoin addresses from the transaction, binary form without the checksum Set BitcoinAddressIndex = 0; Set PrivateKey = 1; Set PublicKey = G;
Loop Until BitcoinAddressIndex == 256: // == "forever"
Call Convert PublicKey to BitcoinAddress [but just to the binary form, do not calculate the checksum or encode to ASCII]
If BitcoinAddress == BitcoinAddresses[BitcoinAddressIndex] Then
Log BitcoinAddressIndex, PrivateKey, PublicKey, BitcoinAddress
Create transaction and claim Bitcoins if any available at BitcoinAddress
Endif
++PrivateKey;
Call Increment PublicKey by G // Highly optimized, very specialized function to just compute PublicKey = PublicKey + G
EndLoop Note on the PublicKey to BitcoinAddress conversion function: You only need to do the first 3 of the 9 steps in this process. 1  Take the PublicKey and format it properly (add the 1 byte of 0x04, change to compressed form if needed) 2  Perform SHA256 hashing on the result 3  Perform RIPEMD160 hashing on the result of SHA256 This result can be compared directly to the BitcoinAddresses[] array assuming you have stored the 256 Bitcoin addresses in the proper binary form. To get the proper values for this array simply undo the last 6 steps of the PublicKey to BitcoinAddress function for each of the 256 Bitcoin addresses in the transaction: 1  Decode the base58 string to a binary byte array 2  Strip off the 4 checksum bytes from the tail 3  Strip off the version byte (0x00) from the front 4  Store the result in the array Which step above is using the slow EC_POINT_point2oct function?

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January 04, 2016, 12:32:35 PM Last edit: January 04, 2016, 12:56:27 PM by BurtW 

Unlike hash operations, elliptic curve operations have unpredictable machine cycle count.
I would expect a single point addition operation (P _{3} = P _{1} + P _{2}) to have a very predictable machine cycle count. It should be something like: x _{3} = λ ^{2}+a _{1}λ−a _{2}−x _{1}−x _{2}y _{3} = −a _{1}x _{3}−a _{3}−λx _{3}+λx _{1}−y _{1}λ = (y _{2}−y _{1}) / (x _{2}−x _{1}) From: https://crypto.stanford.edu/pbc/notes/elliptic/explicit.htmlHow precise is this formula? That is the mathematics behind the point addition. The actual implementation of point addition would be optimized and very different. I was just trying to make the point that a single point addition operation should take the same amount of time (same number of machine instructions) no matter what the actual point values are. This is in contrast to the scalar multiplication function P = p*G which will take different amounts of time (different numbers of machine instructions) depending on the value of p. But I see now that the division operation (or equivalently calculating the inverse of the denominator) could take varying amounts of time. So, basically I was wrong because we have to calculate λ to add the two points.

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January 04, 2016, 12:55:27 PM 

BTW, whoever created this challenge will be able to manipulate BTC price, won't she/he?
I am not sure exactly what you mean. I do see a scenario where the creator of the challenge could possibly cause a panic sell off. Is that what you are talking about? The creator still has the private keys so they can spend the rewards at any time. So, they could claim a bunch of the rewards thus simulating a weakness in the Bitcoin crypto? This could possibly cause a panic sell off if the market believed it?

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Bulista
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January 04, 2016, 01:00:40 PM 

I get the same performance as you, but EC_POINT_add isn't the slowest step, it's the call to EC_POINT_point2oct.
Do you know why EC_POINT_point2oct is slow? I called EC_POINT_get_affine_coordinates_GFp instead, but they seem to do the same thing. If EC_POINT_add stores the point in Cartesian coordination (XY), will extracting the public key be a trivial task? Guys with brute force we won't go anywhere.
How precise is this formula?
I think all of us know that brute force is not viable for sure, but it is fun to estimate how long it will take to break the 51st, 52nd, ... addresses. That division isn't the conventional division. It has perfect precision ( http://www.johannesbauer.com/compsci/ecc/#anchor07). Well we wont get 5 TH/s or anything close for sure, but if we were able to generate 100 million keys per second we would break the #51 in about 3.5 months... And I believe it will be very hard to get to 100 million / s even with GPU. Yes of course we can try it for the fun, but it wont be much fun when you get home and your GPU is burning plus your electricity bill But of course if your formula brings down the range of keys to generate then we can have some chance. So, basically I was wrong because we have to calculate λ to add the two points.
If you find the right formula can you post the results for each address so we know exactly what ranges we can count with?




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January 04, 2016, 01:10:45 PM 

If you find the right formula can you post the results for each address so we know exactly what ranges we can count with?
So far there is nothing to indicate that there is a "right formula" to predict the next private key given all the found private keys. I believe the underlying sequence of private keys before masking to produce the shortened values was probably a secure RNG.

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werty98
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January 04, 2016, 01:25:03 PM 

BTW, whoever created this challenge will be able to manipulate BTC price, won't she/he?
I am not sure exactly what you mean. I do see a scenario where the creator of the challenge could possibly cause a panic sell off. Is that what you are talking about? The creator still has the private keys so they can spend the rewards at any time. So, they could claim a bunch of the rewards thus simulating a weakness in the Bitcoin crypto? This could possibly cause a panic sell off if the market believed it? Yeah, that's what I meant Anyway, as Einstein said, investors' behavior is the only thing in this universe that does not follow any physics law.




