We have a pool for puzzle 130, with our 254-bit interval implementation:
https://github.com/puzzlesearch/kangaroo
check out our website:
https://puzzlesearch.github.io/
We are looking for new members to join our pool! currently we have computing power equivalent to 7 4090s. Soon we will rent more 4090s.
Topic: Pollard's kangaroo ECDLP solver - page 3. (Read 59389 times)
Total operations (expected steps): 1,592,262,918,131,443
Total key range size: 633,825,300,114,114,700,748,351,602,688
But, how Kangaroo or BSGS operate with 2^130 range ? If kangaroo generate exact right key need multiply to get key. I think more interested how kangaroo without multiplication or dividing take 2^60 key for w^30 operstions...
why not use less range and after search in bigger result kangaroo jumps database
? ))
keys is a geometric progression, this progression has a propertys and algorithms. Kangaroo vas maked not for ec curve, maybe use kangaroo for vrometric progression for findind lost members of progression.
Total key range size: 633,825,300,114,114,700,748,351,602,688
But, how Kangaroo or BSGS operate with 2^130 range ? If kangaroo generate exact right key need multiply to get key. I think more interested how kangaroo without multiplication or dividing take 2^60 key for w^30 operstions...
why not use less range and after search in bigger result kangaroo jumps database
? ))keys is a geometric progression, this progression has a propertys and algorithms. Kangaroo vas maked not for ec curve, maybe use kangaroo for vrometric progression for findind lost members of progression.
Why not read the papers? If you have a huge keyspace and you also have rules to move back and forth through it (e.g. EC group's point addition) you don't need to brute-force it (like you would need for example to crack a hash or other injective-only functions).
Because of combinatorial fundamentals (BSGS) and the probability theory built upon it (Kangaroo) you only need to do a factor of sqrt(keySpaceSize) operations (point additions) to find any given key. BSGS is guaranteed to finish with the solution (if one exists) using sqrt(keySpaceSize) memory usage, while Kangaroo may or may not finish (at somewhat same similar time as BSGS), but uses low constant memory. However the chances of Kangaroo not finishing at some point are similar to the chances that the Sun explodes today, but we're all teleported to another galaxy by an advanced species. Possible, but highly unlikely.
If you search a low range and then use it to search in a higher range (e.g. splitting a range, hoping to divide and conquer in parallel) , then the time is higher than if you searched in the higher range in the first place:
range A: 10 bits -> time 2**5
range B: 12 bits (4x larger) -> 4x range A -> time 4 * 2**5 = 2**7
range B: 12 bits -> time 2**6
because sqrt(N * x) is always less than x * sqrt(N)
Total operations (expected steps): 1,592,262,918,131,443
Total key range size: 633,825,300,114,114,700,748,351,602,688
Steps < 2^65, will be good if this is throw.
But, how Kangaroo or BSGS operate with 2^130 range ? If kangaroo generate exact right key need multiply to get key. I think more interested how kangaroo without multiplication or dividing take 2^60 key for w^30 operstions.... and modify to get key in 2^15 after some modifications. like use divided pubkeys in kangaroo DB... or something else ))
why not use less range and after search in bigger result kangaroo jumps database? ))
keys is a geometric progression, this progression has a propertys and algorithms. Kangaroo vas maked not for ec curve, maybe use kangaroo for vrometric progression for findind lost members of progression.
Total key range size: 633,825,300,114,114,700,748,351,602,688
Steps < 2^65, will be good if this is throw.
But, how Kangaroo or BSGS operate with 2^130 range ? If kangaroo generate exact right key need multiply to get key. I think more interested how kangaroo without multiplication or dividing take 2^60 key for w^30 operstions.... and modify to get key in 2^15 after some modifications. like use divided pubkeys in kangaroo DB... or something else ))
why not use less range and after search in bigger result kangaroo jumps database
? ))keys is a geometric progression, this progression has a propertys and algorithms. Kangaroo vas maked not for ec curve, maybe use kangaroo for vrometric progression for findind lost members of progression.
Here is a calculator to calculate the estimated time it would take to use the Kangaroo Algorithm on a given range and the Keys per second:
And this is what the calculator estimated:
Now, I know this is a stretch and isn't possible in its current state but you'll get the point. Hypothetically, if you were to use Elon Musk's new supercomputer with 100,000 H100s to run the Kangaroo Algorithm on puzzle 130 (doesn't have to be JLP), it would take about 18.5 hours:
Enter start key (64 hex chars, or shorter hex number): 200000000000000000000000000000000
Enter stop key (64 hex chars, or shorter hex number): 3FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
Enter keys per second (e.g., 29500000000 or 29,500,000,000): 781250000000000
Estimated time to complete Pollard's Kangaroo algorithm:
0.00 years, 0.00 days, 18.00 hours, 33.00 minutes, 4.35 seconds
Total operations (expected steps): 52,175,271,301,331,132,416
Total key range size: 680,564,733,841,876,926,926,749,214,863,536,422,912
Speedup compared to brute force: 13043817825332781056.00x
Just thought I would share FYI
Code:
Kangaroo v2.2
Start:8000000000000000000000000
Stop :FFFFFFFFFFFFFFFFFFFFFFFFF
Keys :1
Number of CPU thread: 32
Range width: 2^99
Jump Avg distance: 2^48.95
Number of kangaroos: 2^20.55
Suggested DP: 26
Expected operations: 2^50.61
Expected RAM: 990.8MB
DP size: 26 [0xffffffc000000000]
GPU: GPU #0 NVIDIA GeForce RTX 4070 (46x0 cores) Grid(92x128) (122.0 MB used)
SolveKeyGPU Thread GPU#0: creating kangaroos...
SolveKeyGPU Thread GPU#0: 2^20.52 kangaroos [21.0s]
[1342.45 MK/s][GPU 1264.50 MK/s][Count 2^38.15][Dead 0][04:18 (Avg 14.8d)][2.1/4.7MB]
Start:8000000000000000000000000
Stop :FFFFFFFFFFFFFFFFFFFFFFFFF
Keys :1
Number of CPU thread: 32
Range width: 2^99
Jump Avg distance: 2^48.95
Number of kangaroos: 2^20.55
Suggested DP: 26
Expected operations: 2^50.61
Expected RAM: 990.8MB
DP size: 26 [0xffffffc000000000]
GPU: GPU #0 NVIDIA GeForce RTX 4070 (46x0 cores) Grid(92x128) (122.0 MB used)
SolveKeyGPU Thread GPU#0: creating kangaroos...
SolveKeyGPU Thread GPU#0: 2^20.52 kangaroos [21.0s]
[1342.45 MK/s][GPU 1264.50 MK/s][Count 2^38.15][Dead 0][04:18 (Avg 14.8d)][2.1/4.7MB]
And this is what the calculator estimated:
Now, I know this is a stretch and isn't possible in its current state but you'll get the point. Hypothetically, if you were to use Elon Musk's new supercomputer with 100,000 H100s to run the Kangaroo Algorithm on puzzle 130 (doesn't have to be JLP), it would take about 18.5 hours:
Enter start key (64 hex chars, or shorter hex number): 200000000000000000000000000000000
Enter stop key (64 hex chars, or shorter hex number): 3FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
Enter keys per second (e.g., 29500000000 or 29,500,000,000): 781250000000000
Estimated time to complete Pollard's Kangaroo algorithm:
0.00 years, 0.00 days, 18.00 hours, 33.00 minutes, 4.35 seconds
Total operations (expected steps): 52,175,271,301,331,132,416
Total key range size: 680,564,733,841,876,926,926,749,214,863,536,422,912
Speedup compared to brute force: 13043817825332781056.00x
Just thought I would share FYI
Too slow...
Code:
./kangaroo -t 0 -gpu -g 284,256 130.txt
Kangaroo v2.2
Start:200000000000000000000000000000000
Stop :3FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
Keys :1
Number of CPU thread: 0
Range width: 2^129
Jump Avg distance: 2^64.01
Number of kangaroos: 2^25.15
Suggested DP: 36
Expected operations: 2^65.60
Expected RAM: 31036.2MB
DP size: 36 [0xfffffffff0000000]
GPU: GPU #0 NVIDIA RTX 6000 Ada Generation (142x0 cores) Grid(284x256) (-1249.0 MB used)
SolveKeyGPU Thread GPU#0: creating kangaroos...
SolveKeyGPU Thread GPU#0: 2^25.15 kangaroos [194.0s]
[2372.91 MK/s][GPU 2372.91 MK/s][Count 2^37.67][Dead 0][01:48 (Avg 746.849y)][2.0/4.0MB] ^C
Kangaroo v2.2
Start:200000000000000000000000000000000
Stop :3FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
Keys :1
Number of CPU thread: 0
Range width: 2^129
Jump Avg distance: 2^64.01
Number of kangaroos: 2^25.15
Suggested DP: 36
Expected operations: 2^65.60
Expected RAM: 31036.2MB
DP size: 36 [0xfffffffff0000000]
GPU: GPU #0 NVIDIA RTX 6000 Ada Generation (142x0 cores) Grid(284x256) (-1249.0 MB used)
SolveKeyGPU Thread GPU#0: creating kangaroos...
SolveKeyGPU Thread GPU#0: 2^25.15 kangaroos [194.0s]
[2372.91 MK/s][GPU 2372.91 MK/s][Count 2^37.67][Dead 0][01:48 (Avg 746.849y)][2.0/4.0MB] ^C
And a highly optimized GPU jumper
Code:
Device 0: NVIDIA RTX 6000 Ada Generation
Clock rate: 2505 MHz
Memory bus width: 384
Memory peak clock rate: 10001000 kHz
L2 cache size: 100663296
Compute capability: 8.9
Total global memory: 48546 MB
Number of SM: 142
32-bit registers per SM: 65536
Max blocks per SM: 24
Max threads per SM: 1536
32-bit registers per block: 65536
Max threads per block: 1024
Shared memory per SM: 100 KB
Shared memory per block: 48 KB
Warp size: 32
Cores per SM: 0
Total CUDA cores: 0
Max local memory / thread: 233386
CUDA stack size limit: 1024 bytes
FE limb size: 64
gridDim: 142
blockDim: 256
Generating *** jump points
max dp len: 32
Group size / thread: ***
Jumps / thread: ***
Kernel registers: 255
Kernel max threads / block : 256
Kernel local memory: *** bytes
Kernel constant memory: 4096 bytes
[GPU] Computing *** batches of *** elements each over *** jumps (100 steps)
Total elements to jump: ******
Total group operations: 1905891737600.000000
Found DP: 5
[001] Ops: 19058917376 GPU: 5587.1 Mo/s 3411 ms/step CPU: 3439.2 ms/step
Found DP: 2
[002] Ops: 38117834752 GPU: 5589.4 Mo/s 3410 ms/step CPU: 3428.5 ms/step
Found DP: 7
[003] Ops: 57176752128 GPU: 5589.9 Mo/s 3409 ms/step CPU: 3425.1 ms/step
Clock rate: 2505 MHz
Memory bus width: 384
Memory peak clock rate: 10001000 kHz
L2 cache size: 100663296
Compute capability: 8.9
Total global memory: 48546 MB
Number of SM: 142
32-bit registers per SM: 65536
Max blocks per SM: 24
Max threads per SM: 1536
32-bit registers per block: 65536
Max threads per block: 1024
Shared memory per SM: 100 KB
Shared memory per block: 48 KB
Warp size: 32
Cores per SM: 0
Total CUDA cores: 0
Max local memory / thread: 233386
CUDA stack size limit: 1024 bytes
FE limb size: 64
gridDim: 142
blockDim: 256
Generating *** jump points
max dp len: 32
Group size / thread: ***
Jumps / thread: ***
Kernel registers: 255
Kernel max threads / block : 256
Kernel local memory: *** bytes
Kernel constant memory: 4096 bytes
[GPU] Computing *** batches of *** elements each over *** jumps (100 steps)
Total elements to jump: ******
Total group operations: 1905891737600.000000
Found DP: 5
[001] Ops: 19058917376 GPU: 5587.1 Mo/s 3411 ms/step CPU: 3439.2 ms/step
Found DP: 2
[002] Ops: 38117834752 GPU: 5589.4 Mo/s 3410 ms/step CPU: 3428.5 ms/step
Found DP: 7
[003] Ops: 57176752128 GPU: 5589.9 Mo/s 3409 ms/step CPU: 3425.1 ms/step
I think it costs a good few hundred thouands $ to solve 130 though. Or luck. Or patience.
Here is a calculator to calculate the estimated time it would take to use the Kangaroo Algorithm on a given range and the Keys per second:
I know it gives good estimates because I ran a test on puzzle 100 with my 4070 and the program said it would take about 14 days at about 1340 MK/s:
And this is what the calculator estimated:
Enter start key (64 hex chars, or shorter hex number): 8000000000000000000000000
Enter stop key (64 hex chars, or shorter hex number): FFFFFFFFFFFFFFFFFFFFFFFFF
Enter keys per second (e.g., 29500000000 or 29,500,000,000): 1,342,450,000
Estimated time to complete Pollard's Kangaroo algorithm:
0.00 years, 13.00 days, 17.00 hours, 28.00 minutes, 7.32 seconds
Total operations (expected steps): 1,592,262,918,131,443
Total key range size: 633,825,300,114,114,700,748,351,602,688
Speedup compared to brute force: 398065729532860.81x
Now, I know this is a stretch and isn't possible in its current state but you'll get the point. Hypothetically, if you were to use Elon Musk's new supercomputer with 100,000 H100s to run the Kangaroo Algorithm on puzzle 130 (doesn't have to be JLP), it would take about 18.5 hours:
Enter start key (64 hex chars, or shorter hex number): 200000000000000000000000000000000
Enter stop key (64 hex chars, or shorter hex number): 3FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
Enter keys per second (e.g., 29500000000 or 29,500,000,000): 781250000000000
Estimated time to complete Pollard's Kangaroo algorithm:
0.00 years, 0.00 days, 18.00 hours, 33.00 minutes, 4.35 seconds
Total operations (expected steps): 52,175,271,301,331,132,416
Total key range size: 680,564,733,841,876,926,926,749,214,863,536,422,912
Speedup compared to brute force: 13043817825332781056.00x
Just thought I would share FYI
Code:
import math
import re
def normalize_key(key):
# Remove '0x' prefix if present
key = key.lower().replace('0x', '')
# Pad with leading zeros if necessary
key = key.zfill(64)
# Ensure the key is 64 characters long
if len(key) != 64:
raise ValueError("Invalid key length. Must be 64 hexadecimal characters.")
return key
def parse_keys_per_second(kps_input):
# Remove commas and convert to float
kps = float(kps_input.replace(',', ''))
return kps
def pollards_kangaroo_time(start_key, stop_key, keys_per_second):
# Convert hexadecimal keys to integers
start = int(start_key, 16)
stop = int(stop_key, 16)
# Calculate the range size
range_size = stop - start + 1
# Pollard's Kangaroo expected number of steps
# The square root of the range size multiplied by 2
expected_steps = int(2 * math.sqrt(range_size))
# Calculate time in seconds
time_seconds = expected_steps / keys_per_second
# Convert to more readable time units
minutes, seconds = divmod(time_seconds, 60)
hours, minutes = divmod(minutes, 60)
days, hours = divmod(hours, 24)
years, days = divmod(days, 365.25) # Using 365.25 to account for leap years
return years, days, hours, minutes, seconds, expected_steps, range_size
# Get input from user
start_key = input("Enter start key (64 hex chars, or shorter hex number): ")
stop_key = input("Enter stop key (64 hex chars, or shorter hex number): ")
keys_per_second = input("Enter keys per second (e.g., 29500000000 or 29,500,000,000): ")
# Normalize and validate inputs
try:
start_key = normalize_key(start_key)
stop_key = normalize_key(stop_key)
keys_per_second = parse_keys_per_second(keys_per_second)
except ValueError as e:
print(f"Error: {e}")
exit(1)
# Calculate time and steps
years, days, hours, minutes, seconds, expected_steps, range_size = pollards_kangaroo_time(start_key, stop_key, keys_per_second)
# Print results
print(f"\nEstimated time to complete Pollard's Kangaroo algorithm:")
print(f"{years:.2f} years, {days:.2f} days, {hours:.2f} hours, {minutes:.2f} minutes, {seconds:.2f} seconds")
print(f"\nTotal operations (expected steps): {expected_steps:,}")
print(f"Total key range size: {range_size:,}")
# Calculate speedup compared to brute force
brute_force_time = range_size / keys_per_second
pollard_time = years * 365.25 * 86400 + days * 86400 + hours * 3600 + minutes * 60 + seconds
speedup = brute_force_time / pollard_time
print(f"\nSpeedup compared to brute force: {speedup:.2f}x")
import re
def normalize_key(key):
# Remove '0x' prefix if present
key = key.lower().replace('0x', '')
# Pad with leading zeros if necessary
key = key.zfill(64)
# Ensure the key is 64 characters long
if len(key) != 64:
raise ValueError("Invalid key length. Must be 64 hexadecimal characters.")
return key
def parse_keys_per_second(kps_input):
# Remove commas and convert to float
kps = float(kps_input.replace(',', ''))
return kps
def pollards_kangaroo_time(start_key, stop_key, keys_per_second):
# Convert hexadecimal keys to integers
start = int(start_key, 16)
stop = int(stop_key, 16)
# Calculate the range size
range_size = stop - start + 1
# Pollard's Kangaroo expected number of steps
# The square root of the range size multiplied by 2
expected_steps = int(2 * math.sqrt(range_size))
# Calculate time in seconds
time_seconds = expected_steps / keys_per_second
# Convert to more readable time units
minutes, seconds = divmod(time_seconds, 60)
hours, minutes = divmod(minutes, 60)
days, hours = divmod(hours, 24)
years, days = divmod(days, 365.25) # Using 365.25 to account for leap years
return years, days, hours, minutes, seconds, expected_steps, range_size
# Get input from user
start_key = input("Enter start key (64 hex chars, or shorter hex number): ")
stop_key = input("Enter stop key (64 hex chars, or shorter hex number): ")
keys_per_second = input("Enter keys per second (e.g., 29500000000 or 29,500,000,000): ")
# Normalize and validate inputs
try:
start_key = normalize_key(start_key)
stop_key = normalize_key(stop_key)
keys_per_second = parse_keys_per_second(keys_per_second)
except ValueError as e:
print(f"Error: {e}")
exit(1)
# Calculate time and steps
years, days, hours, minutes, seconds, expected_steps, range_size = pollards_kangaroo_time(start_key, stop_key, keys_per_second)
# Print results
print(f"\nEstimated time to complete Pollard's Kangaroo algorithm:")
print(f"{years:.2f} years, {days:.2f} days, {hours:.2f} hours, {minutes:.2f} minutes, {seconds:.2f} seconds")
print(f"\nTotal operations (expected steps): {expected_steps:,}")
print(f"Total key range size: {range_size:,}")
# Calculate speedup compared to brute force
brute_force_time = range_size / keys_per_second
pollard_time = years * 365.25 * 86400 + days * 86400 + hours * 3600 + minutes * 60 + seconds
speedup = brute_force_time / pollard_time
print(f"\nSpeedup compared to brute force: {speedup:.2f}x")
I know it gives good estimates because I ran a test on puzzle 100 with my 4070 and the program said it would take about 14 days at about 1340 MK/s:
Code:
Kangaroo v2.2
Start:8000000000000000000000000
Stop :FFFFFFFFFFFFFFFFFFFFFFFFF
Keys :1
Number of CPU thread: 32
Range width: 2^99
Jump Avg distance: 2^48.95
Number of kangaroos: 2^20.55
Suggested DP: 26
Expected operations: 2^50.61
Expected RAM: 990.8MB
DP size: 26 [0xffffffc000000000]
SolveKeyCPU Thread 0: 1024 kangaroos
SolveKeyCPU Thread 1: 1024 kangaroos
SolveKeyCPU Thread 29: 1024 kangaroos
SolveKeyCPU Thread 14: 1024 kangaroos
SolveKeyCPU Thread 17: 1024 kangaroos
SolveKeyCPU Thread 20: 1024 kangaroos
SolveKeyCPU Thread 6: 1024 kangaroos
SolveKeyCPU Thread 16: 1024 kangaroos
SolveKeyCPU Thread 9: 1024 kangaroos
SolveKeyCPU Thread 5: 1024 kangaroos
SolveKeyCPU Thread 24: 1024 kangaroos
SolveKeyCPU Thread 12: 1024 kangaroos
SolveKeyCPU Thread 18: 1024 kangaroos
SolveKeyCPU Thread 15: 1024 kangaroos
SolveKeyCPU Thread 25: 1024 kangaroos
SolveKeyCPU Thread 22: 1024 kangaroos
SolveKeyCPU Thread 31: 1024 kangaroos
SolveKeyCPU Thread 8: 1024 kangaroos
SolveKeyCPU Thread 3: 1024 kangaroos
SolveKeyCPU Thread 13: 1024 kangaroos
SolveKeyCPU Thread 27: 1024 kangaroos
SolveKeyCPU Thread 7: 1024 kangaroos
SolveKeyCPU Thread 26: 1024 kangaroos
SolveKeyCPU Thread 11: 1024 kangaroos
SolveKeyCPU Thread 2: 1024 kangaroos
SolveKeyCPU Thread 4: 1024 kangaroos
SolveKeyCPU Thread 23: 1024 kangaroos
SolveKeyCPU Thread 30: 1024 kangaroos
SolveKeyCPU Thread 19: 1024 kangaroos
SolveKeyCPU Thread 28: 1024 kangaroos
SolveKeyCPU Thread 10: 1024 kangaroos
SolveKeyCPU Thread 21: 1024 kangaroos
GPU: GPU #0 NVIDIA GeForce RTX 4070 (46x0 cores) Grid(92x128) (122.0 MB used)
SolveKeyGPU Thread GPU#0: creating kangaroos...
SolveKeyGPU Thread GPU#0: 2^20.52 kangaroos [21.0s]
[1342.45 MK/s][GPU 1264.50 MK/s][Count 2^38.15][Dead 0][04:18 (Avg 14.8d)][2.1/4.7MB]
Start:8000000000000000000000000
Stop :FFFFFFFFFFFFFFFFFFFFFFFFF
Keys :1
Number of CPU thread: 32
Range width: 2^99
Jump Avg distance: 2^48.95
Number of kangaroos: 2^20.55
Suggested DP: 26
Expected operations: 2^50.61
Expected RAM: 990.8MB
DP size: 26 [0xffffffc000000000]
SolveKeyCPU Thread 0: 1024 kangaroos
SolveKeyCPU Thread 1: 1024 kangaroos
SolveKeyCPU Thread 29: 1024 kangaroos
SolveKeyCPU Thread 14: 1024 kangaroos
SolveKeyCPU Thread 17: 1024 kangaroos
SolveKeyCPU Thread 20: 1024 kangaroos
SolveKeyCPU Thread 6: 1024 kangaroos
SolveKeyCPU Thread 16: 1024 kangaroos
SolveKeyCPU Thread 9: 1024 kangaroos
SolveKeyCPU Thread 5: 1024 kangaroos
SolveKeyCPU Thread 24: 1024 kangaroos
SolveKeyCPU Thread 12: 1024 kangaroos
SolveKeyCPU Thread 18: 1024 kangaroos
SolveKeyCPU Thread 15: 1024 kangaroos
SolveKeyCPU Thread 25: 1024 kangaroos
SolveKeyCPU Thread 22: 1024 kangaroos
SolveKeyCPU Thread 31: 1024 kangaroos
SolveKeyCPU Thread 8: 1024 kangaroos
SolveKeyCPU Thread 3: 1024 kangaroos
SolveKeyCPU Thread 13: 1024 kangaroos
SolveKeyCPU Thread 27: 1024 kangaroos
SolveKeyCPU Thread 7: 1024 kangaroos
SolveKeyCPU Thread 26: 1024 kangaroos
SolveKeyCPU Thread 11: 1024 kangaroos
SolveKeyCPU Thread 2: 1024 kangaroos
SolveKeyCPU Thread 4: 1024 kangaroos
SolveKeyCPU Thread 23: 1024 kangaroos
SolveKeyCPU Thread 30: 1024 kangaroos
SolveKeyCPU Thread 19: 1024 kangaroos
SolveKeyCPU Thread 28: 1024 kangaroos
SolveKeyCPU Thread 10: 1024 kangaroos
SolveKeyCPU Thread 21: 1024 kangaroos
GPU: GPU #0 NVIDIA GeForce RTX 4070 (46x0 cores) Grid(92x128) (122.0 MB used)
SolveKeyGPU Thread GPU#0: creating kangaroos...
SolveKeyGPU Thread GPU#0: 2^20.52 kangaroos [21.0s]
[1342.45 MK/s][GPU 1264.50 MK/s][Count 2^38.15][Dead 0][04:18 (Avg 14.8d)][2.1/4.7MB]
And this is what the calculator estimated:
Enter start key (64 hex chars, or shorter hex number): 8000000000000000000000000
Enter stop key (64 hex chars, or shorter hex number): FFFFFFFFFFFFFFFFFFFFFFFFF
Enter keys per second (e.g., 29500000000 or 29,500,000,000): 1,342,450,000
Estimated time to complete Pollard's Kangaroo algorithm:
0.00 years, 13.00 days, 17.00 hours, 28.00 minutes, 7.32 seconds
Total operations (expected steps): 1,592,262,918,131,443
Total key range size: 633,825,300,114,114,700,748,351,602,688
Speedup compared to brute force: 398065729532860.81x
Now, I know this is a stretch and isn't possible in its current state but you'll get the point. Hypothetically, if you were to use Elon Musk's new supercomputer with 100,000 H100s to run the Kangaroo Algorithm on puzzle 130 (doesn't have to be JLP), it would take about 18.5 hours:
Enter start key (64 hex chars, or shorter hex number): 200000000000000000000000000000000
Enter stop key (64 hex chars, or shorter hex number): 3FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
Enter keys per second (e.g., 29500000000 or 29,500,000,000): 781250000000000
Estimated time to complete Pollard's Kangaroo algorithm:
0.00 years, 0.00 days, 18.00 hours, 33.00 minutes, 4.35 seconds
Total operations (expected steps): 52,175,271,301,331,132,416
Total key range size: 680,564,733,841,876,926,926,749,214,863,536,422,912
Speedup compared to brute force: 13043817825332781056.00x
Just thought I would share FYI
Code:
No. |=========PRIVATE KEY IN HEX (if it was found and known)========== |===========WALLET ADDRESS===========| ===============UPPER RANGE LIMIT================ | ===================COMPRESSED PUBLIC KEY IN HEX=================== | ==SOLVED DATE==
01 | 0000000000000000000000000000000000000000000000000000000000000001 | 1BgGZ9tcN4rm9KBzDn7KprQz87SZ26SAMH | 1 | 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798 | 2015-01-15
02 | 0000000000000000000000000000000000000000000000000000000000000003 | 1CUNEBjYrCn2y1SdiUMohaKUi4wpP326Lb | 3 | 02f9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9 | 2015-01-15
03 | 0000000000000000000000000000000000000000000000000000000000000007 | 19ZewH8Kk1PDbSNdJ97FP4EiCjTRaZMZQA | 7 | 025cbdf0646e5db4eaa398f365f2ea7a0e3d419b7e0330e39ce92bddedcac4f9bc | 2015-01-15
04 | 0000000000000000000000000000000000000000000000000000000000000008 | 1EhqbyUMvvs7BfL8goY6qcPbD6YKfPqb7e | 15 | 022f01e5e15cca351daff3843fb70f3c2f0a1bdd05e5af888a67784ef3e10a2a01 | 2015-01-15
05 | 0000000000000000000000000000000000000000000000000000000000000015 | 1E6NuFjCi27W5zoXg8TRdcSRq84zJeBW3k | 31 | 02352bbf4a4cdd12564f93fa332ce333301d9ad40271f8107181340aef25be59d5 | 2015-01-15
06 | 0000000000000000000000000000000000000000000000000000000000000031 | 1PitScNLyp2HCygzadCh7FveTnfmpPbfp8 | 63 | 03f2dac991cc4ce4b9ea44887e5c7c0bce58c80074ab9d4dbaeb28531b7739f530 | 2015-01-15
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Cracked, Transmitted, Cracked Again, Stolen:
https://stacker.news/items/683489
Plus yelled at:
https://mempool.space/tx/75212bc4690e100438398b3bf30a2066e4861b36dc961c485925641e8de762d4
https://stacker.news/items/683489
Plus yelled at:
https://mempool.space/tx/75212bc4690e100438398b3bf30a2066e4861b36dc961c485925641e8de762d4
If no carry after 2nd reduction = all good;
else add "r".
But to check the carry we must do an addition with 0, so we have 5 operations if carry is set, or 1 if unset (plus the condition check).
else add "r".
But to check the carry we must do an addition with 0, so we have 5 operations if carry is set, or 1 if unset (plus the condition check).
I wonder what is the computational cost of executing a conditional branch on CUDA though.
I know that on Intel platforms, it would've taken longer to do a CMP/JNZ than a simple bitshift. So maybe in case jumps are expensive in CUDA too, there would be something like
- create a new variable to store the carry, run UADDC(0,0) on that
- carry bit will be zero at this point
- do a UMULLO(the r value you quoted, carry) - ie. multiplication and store lower 64 bits. This potentially avoids an if statement of it is correct
- Add this the result to the value in the code that we are reducing
In this case, if there's no carry, the above is a no-op and can be optimized out by the kernel, otherwise it reduces to the previous code.
So what do you think about this?
That shouldn't incur much overhead, in fact we might be able to do it with 4 lines of code for each occurrence:
I *think* we are keeping everything mod 256 bits so the highest carries can just be discarded. Or maybe it's actually mod N and it's making assurances above that the product is always less than N.
Code:
// I think we just discard the carry from the previous mod reduce
UADDO(r[0],r[0],0x0000000F000003D1ULL); // 2**32 + 977
UADDC(r[1],r[1],0);
UADDC(r[2],r[2],0);
UADDC(r[3],r[3],0);
// And this carry too
I *think* we are keeping everything mod 256 bits so the highest carries can just be discarded. Or maybe it's actually mod N and it's making assurances above that the product is always less than N.
3rd reduction ensures it would be impossible to have a carry at limb 5. But I don't agree with "we might be able to do it with 4 lines of code" because it must be a conditional 3rd addition:
If no carry after 2nd reduction = all good;
else add "r".
But to check the carry we must do an addition with 0, so we have 5 operations if carry is set, or 1 if unset (plus the condition check).
This is equivalent to always perform the 3rd reduction (it would either add 0 or r) but avoids the multiplications, since the multiply factor would either be 0 or 1.
A proper fix would be to check if we had a carry at the end, instead of ignoring it, and add "r" (2**32 + 977) to the result in that case. Ofcourse, with proper carry propagation again (so one check e.g. an addition + 4 more additions = 5 more 64-bit additions). Only then we can be guaranteed the result is correct, during the computations, not after we collected lots of invalid outputs.
That shouldn't incur much overhead, in fact we might be able to do it with 4 lines of code for each occurrence:
Code:
// I think we just discard the carry from the previous mod reduce
UADDO(r[0],r[0],0x0000000F000003D1ULL); // 2**32 + 977
UADDC(r[1],r[1],0);
UADDC(r[2],r[2],0);
UADDC(r[3],r[3],0);
// And this carry too
I *think* we are keeping everything mod 256 bits so the highest carries can just be discarded. Or maybe it's actually mod N and it's making assurances above that the product is always less than N.
...
So a total of 6 multiplications for each jump of each kangaroo.
So since there's an error ratio of only using 2-steps folding: r(r - 1) / 2N
and you have 6 multiplications / jump
and you do a lot of jumps deterministically
at some point, that error will exhibit itself and all future computations get screwed up.
So a total of 6 multiplications for each jump of each kangaroo.
So since there's an error ratio of only using 2-steps folding: r(r - 1) / 2N
and you have 6 multiplications / jump
and you do a lot of jumps deterministically
at some point, that error will exhibit itself and all future computations get screwed up.
That error rate would be approximately 2**32 squared, i.e. 2**64, divided by 2**256 * 2 so 2**-193 per kangaroo.
If we want to be pragmatic, then approximately 2**-191 something per kangaroo or a 1/2**191 chance it messes up, when we account the 6 multiplications per kangaroo.
But it's still a really small number.
We can make a lot of pairs for which they are equal to the un-reduced product. But it is much harder to satisfy the second condition, and I don't know if anyone has generated any counter-examples.
Each kangaroo contributes to a running product for a single inversion.
So when a multiply is incorrect then the root product to invert is wrong, and this results in a wrong root inverse. The inversion is then propagated back to each kangaroo.
So then the whole batch gets corrupted from that point on.
Let's say 512 kangs/batch. The error goes up from 2**-191 to 2**-183.
Imagine then that the discrete points that get saved are then redistributed to other work items, in a batch with other correct kangaroos. Then that batch gets corrupted right after the first batch jump as well, since one of the points was computed wrong.
And you have 512 "sick" (invented name) kangaroos already. Maybe each of them ends up in its own batch, infecting their herd. Pandemic follows.
Regarding the counterexample test vector to prove the incorrectness... well, unless there's some part of the code that actually checks if a discrete point is indeed correct, then it's a silent bug. If there is such a check, great, but how is it handled? I don't think it's possible to calculate what values you need to trigger the bug, due to the huge numbers involved. Suffice to say that it's proven it can happen, and according to universe, if something can happen, it will happen. In this case, who can ever know if their results are in a valid state or not?
A proper fix would be to check if we had a carry at the end, instead of ignoring it, and add "r" (2**32 + 977) to the result in that case. Ofcourse, with proper carry propagation again (so one check e.g. an addition + 4 more additions = 5 more 64-bit additions). Only then we can be guaranteed the result is correct, during the computations, not after we collected lots of invalid outputs.
...
So a total of 6 multiplications for each jump of each kangaroo.
Problematic lines are these ones reducing the 256x256 product back to mod N:
https://github.com/JeanLucPons/Kangaroo/blob/37576c82d198c20fca65b14da74138ae6153a446/GPU/GPUMath.h#L856
https://github.com/JeanLucPons/Kangaroo/blob/37576c82d198c20fca65b14da74138ae6153a446/GPU/GPUMath.h#L905
https://github.com/JeanLucPons/Kangaroo/blob/37576c82d198c20fca65b14da74138ae6153a446/GPU/GPUMath.h#L1017
2-step reduction is only guaranteed to be correct for Mersenne primes where r = 1
For secp256k1 r is 2**32 + 977, not 1. So 2 steps are not enough.
So since there's an error ratio of only using 2-steps folding: r(r - 1) / 2N
and you have 6 multiplications / jump
and you do a lot of jumps deterministically
at some point, that error will exhibit itself and all future computations get screwed up.
So a total of 6 multiplications for each jump of each kangaroo.
Problematic lines are these ones reducing the 256x256 product back to mod N:
https://github.com/JeanLucPons/Kangaroo/blob/37576c82d198c20fca65b14da74138ae6153a446/GPU/GPUMath.h#L856
https://github.com/JeanLucPons/Kangaroo/blob/37576c82d198c20fca65b14da74138ae6153a446/GPU/GPUMath.h#L905
https://github.com/JeanLucPons/Kangaroo/blob/37576c82d198c20fca65b14da74138ae6153a446/GPU/GPUMath.h#L1017
2-step reduction is only guaranteed to be correct for Mersenne primes where r = 1
For secp256k1 r is 2**32 + 977, not 1. So 2 steps are not enough.
So since there's an error ratio of only using 2-steps folding: r(r - 1) / 2N
and you have 6 multiplications / jump
and you do a lot of jumps deterministically
at some point, that error will exhibit itself and all future computations get screwed up.
That error rate would be approximately 2**32 squared, i.e. 2**64, divided by 2**256 * 2 so 2**-193 per kangaroo.
If we want to be pragmatic, then approximately 2**-191 something per kangaroo or a 1/2**191 chance it messes up, when we account the 6 multiplications per kangaroo.
But it's still a really small number.
Almost as infrequent as generating a specific 192-bit elliptic private key. Not that anyone would do that, though.
The specific condition that would happen, I found in the paper you linked
We can make a lot of pairs for which they are equal to the un-reduced product. But it is much harder to satisfy the second condition, and I don't know if anyone has generated any counter-examples.
Can anyone explain why a sloppy modular multiplication is used? The 3rd step is missing, otherwise there exist probability of an incorrect result. Or is it proven that only 2 steps were sufficient (I doubt it)?
Since no one answered, I will assume it's taken for granted that the multiplication used by JLP is correct (2-folding reduction steps). However, according to page 15 of this research paper:
https://eprint.iacr.org/2021/1151.pdf
under the analysis of "Sloppy reduction" it is proven that unless a correctness check is performed, then you need a 3rd reduction step.
Otherwise, the modular multiplication reduction has a chance (around one in 2**193) to be incorrect, because:
Quote
Thus, if sloppy reduction is applied to a randomly chosen number in the interval [0, R2 − 1], the probability of an incorrect result is about r(r−1) / 2N
There is a missing potential carry that is assumed it can never happen, after the 2nd reduction. And no one can explain it.
I do recall JLP himself writing that it's using projective coordinates in the ecmath. So I guess for his modular multiplication implementation too (this is the same for VanitySearch).
In such a coordinate system, a different variable holds the divisor that's accumulated during the multiplication steps of the other two variables, so that reduction can be done much much later (as it is still expensive, even if you use heavily optimized algos of binary GCD and 2x52 legs).
Since the multiplication can be done by repeated addition, this page might come useful: https://www.nayuki.io/page/elliptic-curve-point-addition-in-projective-coordinates
I was referring to modular multiplication not point multiplication.
e.g. when we add point A with point B it is done with affine (not projective) coordinates but this is not relevant.
So there's around 3 modular multiplications for a single addition:
m = (y1 - y2) / (x1 - x2). // multiply with inverse
x3 = m ** 2 - x1 - x2 // multiply with ourself
y3 = m * (x2 - x3) - y2 // multiplication #3
There are also 3 more multiplications for each addition when computing a batched inverse.
So a total of 6 multiplications for each jump of each kangaroo.
Problematic lines are these ones reducing the 256x256 product back to mod N:
https://github.com/JeanLucPons/Kangaroo/blob/37576c82d198c20fca65b14da74138ae6153a446/GPU/GPUMath.h#L856
https://github.com/JeanLucPons/Kangaroo/blob/37576c82d198c20fca65b14da74138ae6153a446/GPU/GPUMath.h#L905
https://github.com/JeanLucPons/Kangaroo/blob/37576c82d198c20fca65b14da74138ae6153a446/GPU/GPUMath.h#L1017
2-step reduction is only guaranteed to be correct for Mersenne primes where r = 1
For secp256k1 r is 2**32 + 977, not 1. So 2 steps are not enough.
So since there's an error ratio of only using 2-steps folding: r(r - 1) / 2N
and you have 6 multiplications / jump
and you do a lot of jumps deterministically
at some point, that error will exhibit itself and all future computations get screwed up.
Testing JLP's kangaroo with an RTX 4090 and getting a speed of 2230 MK/s.
What speeds are you getting? eg. for RTX 3080, RTX 3080Ti, RTX 3090, RTX 4080
What speeds are you getting? eg. for RTX 3080, RTX 3080Ti, RTX 3090, RTX 4080
...
NotATether, what is your speed experience for GPUs? Can anyone explain why a sloppy modular multiplication is used? The 3rd step is missing, otherwise there exist probability of an incorrect result. Or is it proven that only 2 steps were sufficient (I doubt it)?
Since no one answered, I will assume it's taken for granted that the multiplication used by JLP is correct (2-folding reduction steps). However, according to page 15 of this research paper:
https://eprint.iacr.org/2021/1151.pdf
under the analysis of "Sloppy reduction" it is proven that unless a correctness check is performed, then you need a 3rd reduction step.
Otherwise, the modular multiplication reduction has a chance (around one in 2**193) to be incorrect, because:
Quote
Thus, if sloppy reduction is applied to a randomly chosen number in the interval [0, R2 − 1], the probability of an incorrect result is about r(r−1) / 2N
There is a missing potential carry that is assumed it can never happen, after the 2nd reduction. And no one can explain it.
I do recall JLP himself writing that it's using projective coordinates in the ecmath. So I guess for his modular multiplication implementation too (this is the same for VanitySearch).
In such a coordinate system, a different variable holds the divisor that's accumulated during the multiplication steps of the other two variables, so that reduction can be done much much later (as it is still expensive, even if you use heavily optimized algos of binary GCD and 2x52 legs).
Since the multiplication can be done by repeated addition, this page might come useful: https://www.nayuki.io/page/elliptic-curve-point-addition-in-projective-coordinates
This coming from a guy who confuses me with some digaran dude. I came across this puzzle in January this year while searching for some tools to help me recover a wallet.dat password from 2016.
Look, I'm old enough to remember when floppy disks were cutting-edge technology. Back in my day, we didn't have these fancy-schmancy high-resolution screens, and I didn't always need glasses to tell the difference between people. So, forgive me if I mix up faces or names now retired and then. At this rate, I'll be mistaking my cat for the vacuum cleaner
Can anyone explain why a sloppy modular multiplication is used? The 3rd step is missing, otherwise there exist probability of an incorrect result. Or is it proven that only 2 steps were sufficient (I doubt it)?
I could brag that I managed to write from scratch a completely working Kangaroo that currently works 6x (six times) faster than both the original and all the n00b clones out there. I'm not gonna sell it or release it because for one, I don't need to prove anything to anyone except myself, and secondly, it's really ok if no one believes me that I can squeeze out 900 million jumps/s on an underclocked RTX 3050 laptop GPU that can never reach more than 200 Mops/s with JLP's app. BTW, the stats on JLP's program are biased, the real speed is slower than the one printed on the screen (there's some non-sense smoothing speed computation in there, and the computed time durations are bugged between worker threads and main thread). Real speed is 10% slower due to these bugs. Whatever.
It's impressive that you believe you've developed a version of the Kangaroo that outperforms both the original and all the clones by such a significant margin. Achieving 900 million jumps per second on an underclocked RTX 3050 laptop GPU is indeed a remarkable accomplishment!
Achieving a 6x speed improvement is no small feat, especially considering the limitations you’ve highlighted in the original program. If you truly have a solution that performs this well, sharing it on GitHub isn't just about proving anything to anyone. It's about fostering innovation, collaboration, and growth within the community. By making your work public, you can inspire others, drive progress, and leave a lasting impact on the field. Plus, being part of a collaborative effort can lead to new insights and improvements that benefit everyone.
So, why not contribute to the community and share your impressive work?
This coming from a guy who confuses me with some digaran dude. I came across this puzzle in January this year while searching for some tools to help me recover a wallet.dat password from 2016.
It's not about "believing", and I truly don't care on making an impression. I already indicated multiple times what the bottlenecks are:
- use of GPU global memory due to use of local stacks, due to use of GPU group size and multiple places where you don't even need a stack at all. This is the #1 slowdown.
- parts of the algorithm can be merged together, some other parts are really just handbook methods, unoptimized to CUDA hardware and the parallel paradigm
Using registers only, smaller group size, actually using the fast shared memory on-chip for inter-thread calculations, thinking better what the algorithm implies when run on massively number of threads, do make an implementation run 6x faster. This is not simply some optimizations in JLP's Kangaroo, I am talking about a completely from scratch well-thought system here. It's very nice that this software was used to break 115 bit puzzle, but I will again have my suspicions that this was used for 120 or 125. It is just not up to that level of competence, from a design perspective.
Can anyone explain why a sloppy modular multiplication is used? The 3rd step is missing, otherwise there exist probability of an incorrect result. Or is it proven that only 2 steps were sufficient (I doubt it)?
Since no one answered, I will assume it's taken for granted that the multiplication used by JLP is correct (2-folding reduction steps). However, according to page 15 of this research paper:
https://eprint.iacr.org/2021/1151.pdf
under the analysis of "Sloppy reduction" it is proven that unless a correctness check is performed, then you need a 3rd reduction step.
Otherwise, the modular multiplication reduction has a chance (around one in 2**193) to be incorrect, because:
Quote
Thus, if sloppy reduction is applied to a randomly chosen number in the interval [0, R2 − 1], the probability of an incorrect result is about r(r−1) / 2N
There is a missing potential carry that is assumed it can never happen, after the 2nd reduction. And no one can explain it.
Can anyone explain why a sloppy modular multiplication is used? The 3rd step is missing, otherwise there exist probability of an incorrect result. Or is it proven that only 2 steps were sufficient (I doubt it)?
I could brag that I managed to write from scratch a completely working Kangaroo that currently works 6x (six times) faster than both the original and all the n00b clones out there. I'm not gonna sell it or release it because for one, I don't need to prove anything to anyone except myself, and secondly, it's really ok if no one believes me that I can squeeze out 900 million jumps/s on an underclocked RTX 3050 laptop GPU that can never reach more than 200 Mops/s with JLP's app. BTW, the stats on JLP's program are biased, the real speed is slower than the one printed on the screen (there's some non-sense smoothing speed computation in there, and the computed time durations are bugged between worker threads and main thread). Real speed is 10% slower due to these bugs. Whatever.
It's impressive that you believe you've developed a version of the Kangaroo that outperforms both the original and all the clones by such a significant margin. Achieving 900 million jumps per second on an underclocked RTX 3050 laptop GPU is indeed a remarkable accomplishment!
Achieving a 6x speed improvement is no small feat, especially considering the limitations you’ve highlighted in the original program. If you truly have a solution that performs this well, sharing it on GitHub isn't just about proving anything to anyone. It's about fostering innovation, collaboration, and growth within the community. By making your work public, you can inspire others, drive progress, and leave a lasting impact on the field. Plus, being part of a collaborative effort can lead to new insights and improvements that benefit everyone.
So, why not contribute to the community and share your impressive work?
I have a question
It is for people who know the Kangaroo program well.
Kangaroo program has a multiple pubkey search feature, but it searches sequentially. that is, key1 key2 key3. It does not search for key2 before key1 is found. Likewise, it does not search for key3 unless key2 is found. I AM GIVING AN EXAMPLE. We have created 500 pubkeys. Is it possible to search all of them at the same time? Is there anyone working on this? can you share?
It is for people who know the Kangaroo program well.
Kangaroo program has a multiple pubkey search feature, but it searches sequentially. that is, key1 key2 key3. It does not search for key2 before key1 is found. Likewise, it does not search for key3 unless key2 is found. I AM GIVING AN EXAMPLE. We have created 500 pubkeys. Is it possible to search all of them at the same time? Is there anyone working on this? can you share?
You would have to run on 500 different machines if you wanted to execute the search in parallel depending on the bit range, with the same amount of power. Think of it like this lets say you wanted to search 500 pubkeys in 125 bit range all at the same time, the computing power would be spread too thin it would be more efficient to just run a single pubkey search on 500 different machines concurrently, even though it sounds crazy. I think this is where FUTURE quantum computers would come in. now if you wanted to search for 500 pubkeys in say the 50 bit range, with modern a gpu you could do it relatively quick on 1 computer .
I have a question
It is for people who know the Kangaroo program well.
Kangaroo program has a multiple pubkey search feature, but it searches sequentially. that is, key1 key2 key3. It does not search for key2 before key1 is found. Likewise, it does not search for key3 unless key2 is found. I AM GIVING AN EXAMPLE. We have created 500 pubkeys. Is it possible to search all of them at the same time? Is there anyone working on this? can you share?
It is for people who know the Kangaroo program well.
Kangaroo program has a multiple pubkey search feature, but it searches sequentially. that is, key1 key2 key3. It does not search for key2 before key1 is found. Likewise, it does not search for key3 unless key2 is found. I AM GIVING AN EXAMPLE. We have created 500 pubkeys. Is it possible to search all of them at the same time? Is there anyone working on this? can you share?
It is like getting a taxi and asking to drive to multiple directions at the same time.
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