Pollard's kangaroo ECDLP solver

[brainless]

compress to uncompress

Code:

import binascii

p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F

def decompress_pubkey(pk):
    x = int.from_bytes(pk[1:33], byteorder='big')
    y_sq = (pow(x, 3, p) + 7) % p
    y = pow(y_sq, (p + 1) // 4, p)
    if y % 2 != pk[0] % 2:
        y = p - y
    y = y.to_bytes(32, byteorder='big')
    return b'\x04' + pk[1:33] + y

with open('add.txt') as f:
  for line in f:
    line=line.strip()
    print(binascii.hexlify(decompress_pubkey(binascii.unhexlify(line))).decode(),file=open("uncomp.txt", "a"))

uncompress to compress

Code:

def cpub(x,y):
 prefix = '02' if y % 2 == 0 else '03'
 c = prefix+ hex(x)[2:].zfill(64)
 return c
with open('add.txt') as f:
  for line in f:
    line=line.strip()
    x = int(line[2:66], 16)
    y = int(line[66:], 16)
    pub04=cpub(x,y)

    print(pub04,file=open("comp.txt", "a"))

[1714469119]

1714469119

[1714469119]

1714469119

[ssxb]

compress to uncompress

Code:

import binascii

p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F

def decompress_pubkey(pk):
    x = int.from_bytes(pk[1:33], byteorder='big')
    y_sq = (pow(x, 3, p) + 7) % p
    y = pow(y_sq, (p + 1) // 4, p)
    if y % 2 != pk[0] % 2:
        y = p - y
    y = y.to_bytes(32, byteorder='big')
    return b'\x04' + pk[1:33] + y

with open('add.txt') as f:
  for line in f:
    line=line.strip()
    print(binascii.hexlify(decompress_pubkey(binascii.unhexlify(line))).decode(),file=open("uncomp.txt", "a"))

uncompress to compress

Code:

def cpub(x,y):
 prefix = '02' if y % 2 == 0 else '03'
 c = prefix+ hex(x)[2:].zfill(64)
 return c
with open('add.txt') as f:
  for line in f:
    line=line.strip()
    x = int(line[2:66], 16)
    y = int(line[66:], 16)
    pub04=cpub(x,y)

    print(pub04,file=open("comp.txt", "a"))

cool , will try it today

[Alpaste]

Hey guys can anyone help me? i'm noob to all this.


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
 / / / /
i got this small private key part (1A838B13505B26867) what is this exaclty? does this belongs to the public key i put in my work file?
another question also, does the private key starts with that 1A838B13505B26867. . . . or they mean it end at . . ..1A838B13505B26867 ?
Thank you guys for help.

[NotATether]

i got this small private key part (1A838B13505B26867) what is this exaclty? does this belongs to the public key i put in my work file?
another question also, does the private key starts with that 1A838B13505B26867. . . . or they mean it end at . . ..1A838B13505B26867 ?
Thank you guys for help.

Exactly where in the private key is the fragment "1A838B13505B26867" located in?

You can only input beginning parts of public (edit: after re-reading the README it is actually supposed to be private) keys inside the Kangaroo input file, there is no way to search based on private key parts.

[WanderingPhilospher]

Hey guys can anyone help me? i'm noob to all this.


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
 / / / /
i got this small private key part (1A838B13505B26867) what is this exaclty? does this belongs to the public key i put in my work file?
another question also, does the private key starts with that 1A838B13505B26867. . . . or they mean it end at . . ..1A838B13505B26867 ?
Thank you guys for help.

Ok.
The Priv: 0x1A838B13505B26867 is the actual entire private key. You could also look at it as:
000000000000000000000000000000000000000000000001A838B13505B26867

and it is the private key of Pub:  0x0230210C23B1A047BC9BDBB13448E67DEDDC108946DE6DE639BCC75D47C0216B1B

Look at the start and stop range in your search:
Start:10000000000000000
Stop :1FFFFFFFFFFFFFFFF
and now the priv key:
priv  :1A838B13505B26867

So everything matches up. You searched in a range and found a private key in that range, by using the private key's pub address.

[Alpaste]

Thank you for this great explanation!
How does this "Kangaroo program actually finds the private key from only the public key? isn't that impossible?

[COBRAS]

Please remind me of the link to the script that allows you to convert the saved files(.ws) so that you can see the kangaroo's  data in a readable form ?

Thank you very mach

[wedom]

Please remind me of the link to the script that allows you to convert the saved [Suspicious link removed]) so that you can see the kangaroo's  data in a readable form ?

Thank you very mach

https://gist.github.com/PatatasFritas/a0409df4306fb1bb81f9a53e70151ddc
This script?

[COBRAS]

Please remind me of the link to the script that allows you to convert the saved [Suspicious link removed]) so that you can see the kangaroo's  data in a readable form ?

Thank you very mach

https://gist.github.com/PatatasFritas/a0409df4306fb1bb81f9a53e70151ddc
This script?

Yes, Thank you very mach !!!

[ssxb]

guys lets say 32 divisors will return 32 keys , one of will be guaranteed for lower range but what about other all keys , i guess they all are also valid ~~ so my question is all other keys are random in 256 range or what?

[wedom]

guys lets say 32 divisors will return 32 keys , one of will be guaranteed for lower range but what about other all keys , i guess they all are also valid ~~ so my question is all other keys are random in 256 range or what?

all keys are valid but are in the other 31 ranges. The difficulty is that we do not know where which key is located, so we are looking for all 32 keys in our range.

[brainless]

guys lets say 32 divisors will return 32 keys , one of will be guaranteed for lower range but what about other all keys , i guess they all are also valid ~~ so my question is all other keys are random in 256 range or what?

No random
Pure calculated area in n order
Let me give you examples in details

[COBRAS]

guys lets say 32 divisors will return 32 keys , one of will be guaranteed for lower range but what about other all keys , i guess they all are also valid ~~ so my question is all other keys are random in 256 range or what?

No random
Pure calculated area in n order
Let me give you examples in details

Yes, Brainless provide examples pls ?

Brainless, can you find a private key for one of  "second"basepoint https://bitcointalk.org/index.php?topic=5357248.0 ? What you thin about a rage of "second basepoins" privateeys ?

Thx.

[brainless]

guys lets say 32 divisors will return 32 keys , one of will be guaranteed for lower range but what about other all keys , i guess they all are also valid ~~ so my question is all other keys are random in 256 range or what?

No random
Pure calculated area in n order
Let me give you examples in details

Yes, Brainless provide examples pls ?

Brainless, can you find a private key for one of  "second"basepoint https://bitcointalk.org/index.php?topic=5357248.0 ? What you thin about a rage of "second basepoins" privateeys ?

Thx.

if you divide 1 by 32, you will get 0.3125
first zone is 0000000000000000000000000000000000000000000000000000000000000001
2nd zone is for 0.03125
F7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFEC4D965FF79CE5B39E1D3CB9869B48F37 (bitcrack start range)
pubkey 03bb2228d3ea32cb3c1eb160cc824a4ba8115f9a7f415d18ddcaac8193defc2c47
same each step add into 0.03125 to 0.0625, till 31th key you will back to first zone

if you take pubkey and multiply by 32 you will get 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798

with this basic calc you can create all 32 zone for your each key to find in that range
hope these will help you to start

[brainless]

guys lets say 32 divisors will return 32 keys , one of will be guaranteed for lower range but what about other all keys , i guess they all are also valid ~~ so my question is all other keys are random in 256 range or what?

No random
Pure calculated area in n order
Let me give you examples in details

Yes, Brainless provide examples pls ?

Brainless, can you find a private key for one of  "second"basepoint https://bitcointalk.org/index.php?topic=5357248.0 ? What you thin about a rage of "second basepoins" privateeys ?

Thx.

if you divide 1 by 32, you will get 0.3125
first zone is 0000000000000000000000000000000000000000000000000000000000000001
2nd zone is for 0.03125
F7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFEC4D965FF79CE5B39E1D3CB9869B48F37 (bitcrack start range)
pubkey 03bb2228d3ea32cb3c1eb160cc824a4ba8115f9a7f415d18ddcaac8193defc2c47
same each step add into 0.03125 to 0.0625, till 31th key you will back to first zone

if you take pubkey and multiply by 32 you will get 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798

with this basic calc you can create all 32 zone for your each key to find in that range
hope these will help you to start

little bit more example
dec 33
hex 21
pubkey 021697ffa6fd9de627c077e3d2fe541084ce13300b0bec1146f95ae57f0d0bd6a5
use your program to get 32 keys of above pubkey

you will see 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 in your list
one step ahead pubkey is 1.03125 ( 03bb2228d3ea32cb3c1eb160cc824a4ba8115f9a7f415d18ddcaac8193defc2c47 + 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798  )

and in private key F7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFEC4D965FF79CE5B39E1D3CB9869B48F38
029eda0cebe3c594b59add6dccbff3347f06ad09e83e0b9279dd821cc94284c5d0

hope you will test each zone for your understand

[ssxb]

guys lets say 32 divisors will return 32 keys , one of will be guaranteed for lower range but what about other all keys , i guess they all are also valid ~~ so my question is all other keys are random in 256 range or what?

No random
Pure calculated area in n order
Let me give you examples in details

Yes, Brainless provide examples pls ?

Brainless, can you find a private key for one of  "second"basepoint https://bitcointalk.org/index.php?topic=5357248.0 ? What you thin about a rage of "second basepoins" privateeys ?

Thx.

if you divide 1 by 32, you will get 0.3125
first zone is 0000000000000000000000000000000000000000000000000000000000000001
2nd zone is for 0.03125
F7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFEC4D965FF79CE5B39E1D3CB9869B48F37 (bitcrack start range)
pubkey 03bb2228d3ea32cb3c1eb160cc824a4ba8115f9a7f415d18ddcaac8193defc2c47
same each step add into 0.03125 to 0.0625, till 31th key you will back to first zone

if you take pubkey and multiply by 32 you will get 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798

with this basic calc you can create all 32 zone for your each key to find in that range
hope these will help you to start

little bit more example
dec 33
hex 21
pubkey 021697ffa6fd9de627c077e3d2fe541084ce13300b0bec1146f95ae57f0d0bd6a5
use your program to get 32 keys of above pubkey

you will see 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 in your list
one step ahead pubkey is 1.03125 ( 03bb2228d3ea32cb3c1eb160cc824a4ba8115f9a7f415d18ddcaac8193defc2c47 + 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798  )

and in private key F7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFEC4D965FF79CE5B39E1D3CB9869B48F38
029eda0cebe3c594b59add6dccbff3347f06ad09e83e0b9279dd821cc94284c5d0

hope you will test each zone for your understand

even your saying is correct , like they got relationships with each other but fact is only 1 key will be in down range other all will be present somewhere on curve but not in down range.

lets say if i do divisor 32  and got some key but i know that key is not from down range , it is from 100 bit upper range , and if i know private key of that public address ,~ so how to reach back to target in this case

there must be some way as this key appear in 32 divisor , but we cant use  (privkeyfound*divisor+keyfoundinposition) in this situation i guess.

[ssxb]

Code:

is there any specific way to recognize which exactly is the divisor key which will lead us 5 bit down?  

i guess not  but brainless can you confirm it ;D

i mean if i do divisor this key ~ 0348e843dc5b1bd246e6309b4924b81543d02b16c8083df973a89ce2c7eb89a10d

will get this output

Code:
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where 0335f2e3d043050272c3185f50fbea2321a9b63954c19928ff05a879faf62847a5  is the only key leading to the target

so my question & my opinion it is not necessary to jump to the target from only 5 bit down specific key , i guess we can jump from any key listed above if we know the private key? @brainless dont use your brain less  , tell me this is possible or not? 

[brainless]

Code:

is there any specific way to recognize which exactly is the divisor key which will lead us 5 bit down?  

i guess not  but brainless can you confirm it ;D

i mean if i do divisor this key ~ 0348e843dc5b1bd246e6309b4924b81543d02b16c8083df973a89ce2c7eb89a10d

will get this output

Code:
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where 0335f2e3d043050272c3185f50fbea2321a9b63954c19928ff05a879faf62847a5  is the only key leading to the target

so my question & my opinion it is not necessary to jump to the target from only 5 bit down specific key , i guess we can jump from any key listed above if we know the private key? @brainless dont use your brain less  , tell me this is possible or not? 

Everything is possible

[ssxb]

today i successfully jump from 110 bit back toward target key range but i landed 5 bit down the range of target key , still figuring it out lets see ~~

math is beautiful and hell big  in range

[bigvito19]

today i successfully jump from 110 bit back toward target key range but i landed 5 bit down the range of target key , still figuring it out lets see ~~

math is beautiful and hell big  in range

How long did it take for you to do that?