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Efficient way of generating a random number of N (less than 64) bits with exactly M bits equal to one
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Efficient way of generating a random number of N (less than 64) bits with exactly M bits equal to one
Generating unsigned, bounded random value using signed bounded random valuesIs deniable error-correction possible?What is the most computationally efficient way of generating pseudo-random permutations?Has there been any research on entropy efficient information-theoretically secure PRNGs?Secure entropy extractor for thermal noise collected from camera input?Generating DH key with specific bit-lengthRSA-KEM: minimal number of random bitsRandomizing Prime Field Elements
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Would there be an efficient way to implement a function with the following signature:
unsigned long long int random_word(size_t n, size_t m)
that would generate a random machine word (64 bits here) such that exactly m
bits over the n
least significant ones at set to 1
. For example: random_word(10, 3)
would generate a 64-bit random number such that 3 bits over the 10 LSBs are set to 1
. For a given n
and m
every possible output should have equal probability (uniform distribution of possible permutations).
If assembly bit twiddling hacks to do that are known, great, if not, I am looking for references and research directions.
algorithm-design random-number-generator implementation randomness pseudo-random-permutation
New contributor
$endgroup$
|
show 3 more comments
$begingroup$
Would there be an efficient way to implement a function with the following signature:
unsigned long long int random_word(size_t n, size_t m)
that would generate a random machine word (64 bits here) such that exactly m
bits over the n
least significant ones at set to 1
. For example: random_word(10, 3)
would generate a 64-bit random number such that 3 bits over the 10 LSBs are set to 1
. For a given n
and m
every possible output should have equal probability (uniform distribution of possible permutations).
If assembly bit twiddling hacks to do that are known, great, if not, I am looking for references and research directions.
algorithm-design random-number-generator implementation randomness pseudo-random-permutation
New contributor
$endgroup$
$begingroup$
If the percentage of samples that match the requirement from the full range of numbers isn't too small, then rejection sampling should work (only needs to be done on those n bits, the prefix can be randomized once and then you rejection sample the n bits)
$endgroup$
– Natanael
8 hours ago
5
$begingroup$
Seems more like a programming challenge than crypto-related. Define "efficient":. Code size? Minimal number of uniformlly random bit used? Is that on average, for the first call, or..? Speed: for the first call, for a million calls with the same n,m..? Is the time to generate uniform random bits counted in the performance?
$endgroup$
– fgrieu
8 hours ago
$begingroup$
Why do you need this in cryptography?
$endgroup$
– Conrado
8 hours ago
1
$begingroup$
There's actually a second way that's likely more efficient (at least when m is far from n/2). Generate 64 minus n bits of random bits for the prefix, then simply generate a bitstring of m 1's and n-m 0's, and then you perform a randomized bitwise sort (with some random sort algorithms with a sufficiently small bias, using a unique random seed), and concatenate the two strings.
$endgroup$
– Natanael
7 hours ago
$begingroup$
I don't get your sort. But if you just randomize the positions of the ones or zeros in the n LSB bits, and let the rest consist of random bits, then that should not introduce any bias, right?
$endgroup$
– Maarten Bodewes♦
7 hours ago
|
show 3 more comments
$begingroup$
Would there be an efficient way to implement a function with the following signature:
unsigned long long int random_word(size_t n, size_t m)
that would generate a random machine word (64 bits here) such that exactly m
bits over the n
least significant ones at set to 1
. For example: random_word(10, 3)
would generate a 64-bit random number such that 3 bits over the 10 LSBs are set to 1
. For a given n
and m
every possible output should have equal probability (uniform distribution of possible permutations).
If assembly bit twiddling hacks to do that are known, great, if not, I am looking for references and research directions.
algorithm-design random-number-generator implementation randomness pseudo-random-permutation
New contributor
$endgroup$
Would there be an efficient way to implement a function with the following signature:
unsigned long long int random_word(size_t n, size_t m)
that would generate a random machine word (64 bits here) such that exactly m
bits over the n
least significant ones at set to 1
. For example: random_word(10, 3)
would generate a 64-bit random number such that 3 bits over the 10 LSBs are set to 1
. For a given n
and m
every possible output should have equal probability (uniform distribution of possible permutations).
If assembly bit twiddling hacks to do that are known, great, if not, I am looking for references and research directions.
algorithm-design random-number-generator implementation randomness pseudo-random-permutation
algorithm-design random-number-generator implementation randomness pseudo-random-permutation
New contributor
New contributor
New contributor
asked 8 hours ago
VincentVincent
1062 bronze badges
1062 bronze badges
New contributor
New contributor
$begingroup$
If the percentage of samples that match the requirement from the full range of numbers isn't too small, then rejection sampling should work (only needs to be done on those n bits, the prefix can be randomized once and then you rejection sample the n bits)
$endgroup$
– Natanael
8 hours ago
5
$begingroup$
Seems more like a programming challenge than crypto-related. Define "efficient":. Code size? Minimal number of uniformlly random bit used? Is that on average, for the first call, or..? Speed: for the first call, for a million calls with the same n,m..? Is the time to generate uniform random bits counted in the performance?
$endgroup$
– fgrieu
8 hours ago
$begingroup$
Why do you need this in cryptography?
$endgroup$
– Conrado
8 hours ago
1
$begingroup$
There's actually a second way that's likely more efficient (at least when m is far from n/2). Generate 64 minus n bits of random bits for the prefix, then simply generate a bitstring of m 1's and n-m 0's, and then you perform a randomized bitwise sort (with some random sort algorithms with a sufficiently small bias, using a unique random seed), and concatenate the two strings.
$endgroup$
– Natanael
7 hours ago
$begingroup$
I don't get your sort. But if you just randomize the positions of the ones or zeros in the n LSB bits, and let the rest consist of random bits, then that should not introduce any bias, right?
$endgroup$
– Maarten Bodewes♦
7 hours ago
|
show 3 more comments
$begingroup$
If the percentage of samples that match the requirement from the full range of numbers isn't too small, then rejection sampling should work (only needs to be done on those n bits, the prefix can be randomized once and then you rejection sample the n bits)
$endgroup$
– Natanael
8 hours ago
5
$begingroup$
Seems more like a programming challenge than crypto-related. Define "efficient":. Code size? Minimal number of uniformlly random bit used? Is that on average, for the first call, or..? Speed: for the first call, for a million calls with the same n,m..? Is the time to generate uniform random bits counted in the performance?
$endgroup$
– fgrieu
8 hours ago
$begingroup$
Why do you need this in cryptography?
$endgroup$
– Conrado
8 hours ago
1
$begingroup$
There's actually a second way that's likely more efficient (at least when m is far from n/2). Generate 64 minus n bits of random bits for the prefix, then simply generate a bitstring of m 1's and n-m 0's, and then you perform a randomized bitwise sort (with some random sort algorithms with a sufficiently small bias, using a unique random seed), and concatenate the two strings.
$endgroup$
– Natanael
7 hours ago
$begingroup$
I don't get your sort. But if you just randomize the positions of the ones or zeros in the n LSB bits, and let the rest consist of random bits, then that should not introduce any bias, right?
$endgroup$
– Maarten Bodewes♦
7 hours ago
$begingroup$
If the percentage of samples that match the requirement from the full range of numbers isn't too small, then rejection sampling should work (only needs to be done on those n bits, the prefix can be randomized once and then you rejection sample the n bits)
$endgroup$
– Natanael
8 hours ago
$begingroup$
If the percentage of samples that match the requirement from the full range of numbers isn't too small, then rejection sampling should work (only needs to be done on those n bits, the prefix can be randomized once and then you rejection sample the n bits)
$endgroup$
– Natanael
8 hours ago
5
5
$begingroup$
Seems more like a programming challenge than crypto-related. Define "efficient":. Code size? Minimal number of uniformlly random bit used? Is that on average, for the first call, or..? Speed: for the first call, for a million calls with the same n,m..? Is the time to generate uniform random bits counted in the performance?
$endgroup$
– fgrieu
8 hours ago
$begingroup$
Seems more like a programming challenge than crypto-related. Define "efficient":. Code size? Minimal number of uniformlly random bit used? Is that on average, for the first call, or..? Speed: for the first call, for a million calls with the same n,m..? Is the time to generate uniform random bits counted in the performance?
$endgroup$
– fgrieu
8 hours ago
$begingroup$
Why do you need this in cryptography?
$endgroup$
– Conrado
8 hours ago
$begingroup$
Why do you need this in cryptography?
$endgroup$
– Conrado
8 hours ago
1
1
$begingroup$
There's actually a second way that's likely more efficient (at least when m is far from n/2). Generate 64 minus n bits of random bits for the prefix, then simply generate a bitstring of m 1's and n-m 0's, and then you perform a randomized bitwise sort (with some random sort algorithms with a sufficiently small bias, using a unique random seed), and concatenate the two strings.
$endgroup$
– Natanael
7 hours ago
$begingroup$
There's actually a second way that's likely more efficient (at least when m is far from n/2). Generate 64 minus n bits of random bits for the prefix, then simply generate a bitstring of m 1's and n-m 0's, and then you perform a randomized bitwise sort (with some random sort algorithms with a sufficiently small bias, using a unique random seed), and concatenate the two strings.
$endgroup$
– Natanael
7 hours ago
$begingroup$
I don't get your sort. But if you just randomize the positions of the ones or zeros in the n LSB bits, and let the rest consist of random bits, then that should not introduce any bias, right?
$endgroup$
– Maarten Bodewes♦
7 hours ago
$begingroup$
I don't get your sort. But if you just randomize the positions of the ones or zeros in the n LSB bits, and let the rest consist of random bits, then that should not introduce any bias, right?
$endgroup$
– Maarten Bodewes♦
7 hours ago
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
I'd guess that you can simply split this into two problems:
- create 64 - n random bits, call this R
- shuffle n bits where m bits (at any location) are set to 1, call this P
Finally you can simply perform R | P (presuming big endian notation).
Shuffling lists of elements is an operation present in almost any language. If there is any inefficiency it would be in the shuffling algorithm (although Fisher-Yates is optimal, so you'd expect some form of that algorithm, possibly the inefficiency is getting values in a range...).
$endgroup$
$begingroup$
I'm thinking that you could also just generate a value x within of 0..n - i where i goes from 0 to m, where you set the x'th bit that is not set. That would be equivalent and easier to implement.
$endgroup$
– Maarten Bodewes♦
3 hours ago
add a comment |
$begingroup$
The problem for choosing $k$ bits from $64$ ultimately comes down to computing a uniformly random integer $r$ with $0 leq r < frac64!k!(64-k)!$ then decoding it to determine which bits. The $k!$ in the denominator is annoying, but we can ignore it, because we can just allow our algorithm to have $k!$ random numbers that map to the same output (setting bit 0 then bit 4 is the same as setting bit 4 then bit 0). Now we just have multiplying a decreasing sequence starting from $64$: with $k=4$ this equals $64 * 63 * 62 * 61$.
So for efficiency, you select a random number in $0 le r_0 < 64$, then another $0 le r_1 < 63$ ... through $0 le r_k-1 < 64-(k-1)$ each time using $r_n$ to select among the remaining unset bits.
I threw the following Python code together showing the idea, though it's not fast or anything:
# b = size of integer type
# n = number of set bits
# random_limited(x) is some function returning [0, x) sufficiently uniformly
def random_n_set_bits(b, n):
assert b > 0
assert n >= 0 and n <= b
result = 0
available = list(range(b))
for i in range(n):
index = random_limited(len(available))
bit = available[index]
available = available[:index] + available[index + 1:]
result |= (1 << bit)
return result
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'd guess that you can simply split this into two problems:
- create 64 - n random bits, call this R
- shuffle n bits where m bits (at any location) are set to 1, call this P
Finally you can simply perform R | P (presuming big endian notation).
Shuffling lists of elements is an operation present in almost any language. If there is any inefficiency it would be in the shuffling algorithm (although Fisher-Yates is optimal, so you'd expect some form of that algorithm, possibly the inefficiency is getting values in a range...).
$endgroup$
$begingroup$
I'm thinking that you could also just generate a value x within of 0..n - i where i goes from 0 to m, where you set the x'th bit that is not set. That would be equivalent and easier to implement.
$endgroup$
– Maarten Bodewes♦
3 hours ago
add a comment |
$begingroup$
I'd guess that you can simply split this into two problems:
- create 64 - n random bits, call this R
- shuffle n bits where m bits (at any location) are set to 1, call this P
Finally you can simply perform R | P (presuming big endian notation).
Shuffling lists of elements is an operation present in almost any language. If there is any inefficiency it would be in the shuffling algorithm (although Fisher-Yates is optimal, so you'd expect some form of that algorithm, possibly the inefficiency is getting values in a range...).
$endgroup$
$begingroup$
I'm thinking that you could also just generate a value x within of 0..n - i where i goes from 0 to m, where you set the x'th bit that is not set. That would be equivalent and easier to implement.
$endgroup$
– Maarten Bodewes♦
3 hours ago
add a comment |
$begingroup$
I'd guess that you can simply split this into two problems:
- create 64 - n random bits, call this R
- shuffle n bits where m bits (at any location) are set to 1, call this P
Finally you can simply perform R | P (presuming big endian notation).
Shuffling lists of elements is an operation present in almost any language. If there is any inefficiency it would be in the shuffling algorithm (although Fisher-Yates is optimal, so you'd expect some form of that algorithm, possibly the inefficiency is getting values in a range...).
$endgroup$
I'd guess that you can simply split this into two problems:
- create 64 - n random bits, call this R
- shuffle n bits where m bits (at any location) are set to 1, call this P
Finally you can simply perform R | P (presuming big endian notation).
Shuffling lists of elements is an operation present in almost any language. If there is any inefficiency it would be in the shuffling algorithm (although Fisher-Yates is optimal, so you'd expect some form of that algorithm, possibly the inefficiency is getting values in a range...).
edited 4 hours ago
answered 5 hours ago
Maarten Bodewes♦Maarten Bodewes
58.5k6 gold badges85 silver badges213 bronze badges
58.5k6 gold badges85 silver badges213 bronze badges
$begingroup$
I'm thinking that you could also just generate a value x within of 0..n - i where i goes from 0 to m, where you set the x'th bit that is not set. That would be equivalent and easier to implement.
$endgroup$
– Maarten Bodewes♦
3 hours ago
add a comment |
$begingroup$
I'm thinking that you could also just generate a value x within of 0..n - i where i goes from 0 to m, where you set the x'th bit that is not set. That would be equivalent and easier to implement.
$endgroup$
– Maarten Bodewes♦
3 hours ago
$begingroup$
I'm thinking that you could also just generate a value x within of 0..n - i where i goes from 0 to m, where you set the x'th bit that is not set. That would be equivalent and easier to implement.
$endgroup$
– Maarten Bodewes♦
3 hours ago
$begingroup$
I'm thinking that you could also just generate a value x within of 0..n - i where i goes from 0 to m, where you set the x'th bit that is not set. That would be equivalent and easier to implement.
$endgroup$
– Maarten Bodewes♦
3 hours ago
add a comment |
$begingroup$
The problem for choosing $k$ bits from $64$ ultimately comes down to computing a uniformly random integer $r$ with $0 leq r < frac64!k!(64-k)!$ then decoding it to determine which bits. The $k!$ in the denominator is annoying, but we can ignore it, because we can just allow our algorithm to have $k!$ random numbers that map to the same output (setting bit 0 then bit 4 is the same as setting bit 4 then bit 0). Now we just have multiplying a decreasing sequence starting from $64$: with $k=4$ this equals $64 * 63 * 62 * 61$.
So for efficiency, you select a random number in $0 le r_0 < 64$, then another $0 le r_1 < 63$ ... through $0 le r_k-1 < 64-(k-1)$ each time using $r_n$ to select among the remaining unset bits.
I threw the following Python code together showing the idea, though it's not fast or anything:
# b = size of integer type
# n = number of set bits
# random_limited(x) is some function returning [0, x) sufficiently uniformly
def random_n_set_bits(b, n):
assert b > 0
assert n >= 0 and n <= b
result = 0
available = list(range(b))
for i in range(n):
index = random_limited(len(available))
bit = available[index]
available = available[:index] + available[index + 1:]
result |= (1 << bit)
return result
$endgroup$
add a comment |
$begingroup$
The problem for choosing $k$ bits from $64$ ultimately comes down to computing a uniformly random integer $r$ with $0 leq r < frac64!k!(64-k)!$ then decoding it to determine which bits. The $k!$ in the denominator is annoying, but we can ignore it, because we can just allow our algorithm to have $k!$ random numbers that map to the same output (setting bit 0 then bit 4 is the same as setting bit 4 then bit 0). Now we just have multiplying a decreasing sequence starting from $64$: with $k=4$ this equals $64 * 63 * 62 * 61$.
So for efficiency, you select a random number in $0 le r_0 < 64$, then another $0 le r_1 < 63$ ... through $0 le r_k-1 < 64-(k-1)$ each time using $r_n$ to select among the remaining unset bits.
I threw the following Python code together showing the idea, though it's not fast or anything:
# b = size of integer type
# n = number of set bits
# random_limited(x) is some function returning [0, x) sufficiently uniformly
def random_n_set_bits(b, n):
assert b > 0
assert n >= 0 and n <= b
result = 0
available = list(range(b))
for i in range(n):
index = random_limited(len(available))
bit = available[index]
available = available[:index] + available[index + 1:]
result |= (1 << bit)
return result
$endgroup$
add a comment |
$begingroup$
The problem for choosing $k$ bits from $64$ ultimately comes down to computing a uniformly random integer $r$ with $0 leq r < frac64!k!(64-k)!$ then decoding it to determine which bits. The $k!$ in the denominator is annoying, but we can ignore it, because we can just allow our algorithm to have $k!$ random numbers that map to the same output (setting bit 0 then bit 4 is the same as setting bit 4 then bit 0). Now we just have multiplying a decreasing sequence starting from $64$: with $k=4$ this equals $64 * 63 * 62 * 61$.
So for efficiency, you select a random number in $0 le r_0 < 64$, then another $0 le r_1 < 63$ ... through $0 le r_k-1 < 64-(k-1)$ each time using $r_n$ to select among the remaining unset bits.
I threw the following Python code together showing the idea, though it's not fast or anything:
# b = size of integer type
# n = number of set bits
# random_limited(x) is some function returning [0, x) sufficiently uniformly
def random_n_set_bits(b, n):
assert b > 0
assert n >= 0 and n <= b
result = 0
available = list(range(b))
for i in range(n):
index = random_limited(len(available))
bit = available[index]
available = available[:index] + available[index + 1:]
result |= (1 << bit)
return result
$endgroup$
The problem for choosing $k$ bits from $64$ ultimately comes down to computing a uniformly random integer $r$ with $0 leq r < frac64!k!(64-k)!$ then decoding it to determine which bits. The $k!$ in the denominator is annoying, but we can ignore it, because we can just allow our algorithm to have $k!$ random numbers that map to the same output (setting bit 0 then bit 4 is the same as setting bit 4 then bit 0). Now we just have multiplying a decreasing sequence starting from $64$: with $k=4$ this equals $64 * 63 * 62 * 61$.
So for efficiency, you select a random number in $0 le r_0 < 64$, then another $0 le r_1 < 63$ ... through $0 le r_k-1 < 64-(k-1)$ each time using $r_n$ to select among the remaining unset bits.
I threw the following Python code together showing the idea, though it's not fast or anything:
# b = size of integer type
# n = number of set bits
# random_limited(x) is some function returning [0, x) sufficiently uniformly
def random_n_set_bits(b, n):
assert b > 0
assert n >= 0 and n <= b
result = 0
available = list(range(b))
for i in range(n):
index = random_limited(len(available))
bit = available[index]
available = available[:index] + available[index + 1:]
result |= (1 << bit)
return result
edited 5 hours ago
answered 5 hours ago
MyriaMyria
1,0265 silver badges14 bronze badges
1,0265 silver badges14 bronze badges
add a comment |
add a comment |
Vincent is a new contributor. Be nice, and check out our Code of Conduct.
Vincent is a new contributor. Be nice, and check out our Code of Conduct.
Vincent is a new contributor. Be nice, and check out our Code of Conduct.
Vincent is a new contributor. Be nice, and check out our Code of Conduct.
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If the percentage of samples that match the requirement from the full range of numbers isn't too small, then rejection sampling should work (only needs to be done on those n bits, the prefix can be randomized once and then you rejection sample the n bits)
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– Natanael
8 hours ago
5
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Seems more like a programming challenge than crypto-related. Define "efficient":. Code size? Minimal number of uniformlly random bit used? Is that on average, for the first call, or..? Speed: for the first call, for a million calls with the same n,m..? Is the time to generate uniform random bits counted in the performance?
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– fgrieu
8 hours ago
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Why do you need this in cryptography?
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– Conrado
8 hours ago
1
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There's actually a second way that's likely more efficient (at least when m is far from n/2). Generate 64 minus n bits of random bits for the prefix, then simply generate a bitstring of m 1's and n-m 0's, and then you perform a randomized bitwise sort (with some random sort algorithms with a sufficiently small bias, using a unique random seed), and concatenate the two strings.
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– Natanael
7 hours ago
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I don't get your sort. But if you just randomize the positions of the ones or zeros in the n LSB bits, and let the rest consist of random bits, then that should not introduce any bias, right?
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– Maarten Bodewes♦
7 hours ago