posible prevompute part for ex from 1 to 99
then you start 100 use precomputed fo 101 to 199. 101 = 1+ 01..99(from file)
Also I think posible generated previously agresses for nex step , if calculate 1 to 99 use in 1xx after use what you generated in 1xx to calc 2xxx etc
What part of "walks that use different rules hit different DPs" was unclear? Besides that, only 50% of the higher-degree interval is covered by the previous interval. Adding DPs from lower intervals to higher intervals is just a waste of resources and slows down the lookup, if the jump rules are different, those older DPs will almost certainly NEVER be hit by any walk of any kangaroo. This is just the result of the probabilities that are involved, which at the end of the day produce the lambdas when the paths merge together.
It's like playing lottery tickets and then you go into a casino and play the tickets at the roulette, or the same if you want to play blackjack by rolling dices. There's nothing in common.
I don know anout DP, for good DP generation I think take a look for puzle privkeys are normal random,so if your generator generates no privkeys what have 50/50 0 and 1 they waste time....
1000 not walid for brute, valid 1010 1100, 0011, not 0100.... lenght of fake and good is same ,agree ? ))
another stoopid question, WanderPhilosepher tell , what for divide privkey to 2^30 need take 2^30 keys, but, why for duvide 2^130 to 2^90 and take only 2^30 in valu range needs 2^90 publick keys ,? How 2^90 was 'trsansormed' to 2^30 ? )))
Unequal bits scrypt what calculate no normal disrtributed numbers
Number of numbers with unequal bits: 431910
Maximum deviation from 50/50: 20
[Program finished]
def count_unequal_bits(start, end):
# Initialize count of numbers with unequal bits and maximum deviation
unequal_count = 0
max_deviation = 0
# Iterate over all numbers in the range
for i in range(start, end):
# Convert number to binary string and remove '0b' prefix
bin_str = bin(i)[2:]
# Count number of ones and zeros in binary string
ones = bin_str.count('1')
zeros = bin_str.count('0')
# Check if number of ones and zeros is not equal
if ones != zeros:
# Increment count of numbers with unequal bits
unequal_count += 1
# Calculate deviation from 50/50
deviation = abs(ones - zeros)
# Update maximum deviation if current deviation is greater
max_deviation = max(max_deviation, deviation)
# Return count of numbers with unequal bits and maximum deviation
return unequal_count, max_deviation
# Define range of numbers to check
start = 2**19
end = 2**20
# Run the function and print results
result, max_deviation = count_unequal_bits(start, end)
print(f"Number of numbers with unequal bits: {result}")
print(f"Maximum deviation from 50/50: {max_deviation}")
I modify scrypt, so, not normal in 2^20 buts is mach many:
Number of numbers with unequal bits: 431910
Number of numbers with equal bits: 92378
Maximum deviation from 50/50: 20
[Program finished]
def count_bits(start, end):
# Initialize counts of numbers with unequal and equal bits
unequal_count = 0
equal_count = 0
max_deviation = 0
# Iterate over all numbers in the range
for i in range(start, end):
# Convert number to binary string and remove '0b' prefix
bin_str = bin(i)[2:]
# Count number of ones and zeros in binary string
ones = bin_str.count('1')
zeros = bin_str.count('0')
# Check if number of ones and zeros is not equal
if ones != zeros:
# Increment count of numbers with unequal bits
unequal_count += 1
# Calculate deviation from 50/50
deviation = abs(ones - zeros)
# Update maximum deviation if current deviation is greater
max_deviation = max(max_deviation, deviation)
else:
# Increment count of numbers with equal bits
equal_count += 1
# Return counts of numbers with unequal and equal bits, and maximum deviation
return unequal_count, equal_count, max_deviation
# Define range of numbers to check
start = 2**19
end = 2**20
# Run the function and print results
unequal_count, equal_count, max_deviation = count_bits(start, end)
print(f"Number of numbers with unequal bits: {unequal_count}")
print(f"Number of numbers with equal bits: {equal_count}")
print(f"Maximum deviation from 50/50: {max_deviation}")
puzzle transactions especially in 50/50, but some of them 49
my scryot for you, darlings: ))
Number of numbers with unequal bits: 431910
Number of numbers with equal bits: 92378
Number of numbers differing by 1 bit: 0
Number of numbers differing by 2 bits: 167960
Number of numbers differing by 3 bits: 0
Number of numbers differing by 4 bits: 125970
Number of numbers differing by 5 bits: 0
Maximum deviation from 50/50: 20
[Program finished]
so in your search ranges not 2^20-2^19 = 524288, but 92378 calculations )) 5 times smaller, yes ?))
priv of puz 2^19-2^20 is 11010010110001010101 so, my darlings
, you are soend 75 percent of yor time for free then you brute all range
def count_bits(start, end):
# Initialize counts of numbers with unequal, equal, and differing by 1, 2, 3, 4, and 5 bits
unequal_count = 0
equal_count = 0
diff_1_count = 0
diff_2_count = 0
diff_3_count = 0
diff_4_count = 0
diff_5_count = 0
max_deviation = 0
# Iterate over all numbers in the range
for i in range(start, end):
# Convert number to binary string and remove '0b' prefix
bin_str = bin(i)[2:]
# Count number of ones and zeros in binary string
ones = bin_str.count('1')
zeros = bin_str.count('0')
# Check if number of ones and zeros is not equal
if ones != zeros:
# Increment count of numbers with unequal bits
unequal_count += 1
# Calculate deviation from 50/50
deviation = abs(ones - zeros)
# Update maximum deviation if current deviation is greater
max_deviation = max(max_deviation, deviation)
# Check if difference is 1, 2, 3, 4, or 5 bits
if deviation == 1:
diff_1_count += 1
elif deviation == 2:
diff_2_count += 1
elif deviation == 3:
diff_3_count += 1
elif deviation == 4:
diff_4_count += 1
elif deviation == 5:
diff_5_count += 1
else:
# Increment count of numbers with equal bits
equal_count += 1
# Return counts of numbers with unequal, equal, and differing by 1, 2, 3, 4, and 5 bits, and maximum deviation
return unequal_count, equal_count, diff_1_count, diff_2_count, diff_3_count, diff_4_count, diff_5_count, max_deviation
# Define range of numbers to check
start = 2**19
end = 2**20
# Run the function and print results
unequal_count, equal_count, diff_1_count, diff_2_count, diff_3_count, diff_4_count, diff_5_count, max_deviation = count_bits(start, end)
print(f"Number of numbers with unequal bits: {unequal_count}")
print(f"Number of numbers with equal bits: {equal_count}")
print(f"Number of numbers differing by 1 bit: {diff_1_count}")
print(f"Number of numbers differing by 2 bits: {diff_2_count}")
print(f"Number of numbers differing by 3 bits: {diff_3_count}")
print(f"Number of numbers differing by 4 bits: {diff_4_count}")
print(f"Number of numbers differing by 5 bits: {diff_5_count}")
print(f"Maximum deviation from 50/50: {max_deviation}")
