Can we generate random numbers using irrational numbers like π and e?
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Irrational numbers like $pi$, $e$ and $sqrt{2}$ have a unique and non-repeating sequence after the decimal point. If we extract the $n$-th digit from such numbers (where $n$ is the number of times the method is called) and make a number with the digits as it is, should we not get a perfect random number generator? For example, if we're using $sqrt{2}$, $e$ and $pi$, the first number is 123, second one is 471, the next one is 184 and so on.
randomized-algorithms randomness random-number-generator
New contributor
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add a comment |
$begingroup$
Irrational numbers like $pi$, $e$ and $sqrt{2}$ have a unique and non-repeating sequence after the decimal point. If we extract the $n$-th digit from such numbers (where $n$ is the number of times the method is called) and make a number with the digits as it is, should we not get a perfect random number generator? For example, if we're using $sqrt{2}$, $e$ and $pi$, the first number is 123, second one is 471, the next one is 184 and so on.
randomized-algorithms randomness random-number-generator
New contributor
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Related: mathoverflow.net/questions/26942/…
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– BurnsBA
10 hours ago
9
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You have a strange definition of "random" in your head. "Random" means "unpredictable". How is your sequence unpredictable? What definition of "random" do you have in mind? Perhaps what you are calling "random" has another name.
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– Eric Lippert
9 hours ago
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Essentially the same question was asked here: cs.stackexchange.com/q/65078/755. (Community votes requested -- should these be merged? Should one be closed as a dup of the other?)
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– D.W.♦
9 hours ago
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Note the spigot algorithm can be used to generate any hex digit in pi, without having to generate prior digits.
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– rcgldr
2 hours ago
add a comment |
$begingroup$
Irrational numbers like $pi$, $e$ and $sqrt{2}$ have a unique and non-repeating sequence after the decimal point. If we extract the $n$-th digit from such numbers (where $n$ is the number of times the method is called) and make a number with the digits as it is, should we not get a perfect random number generator? For example, if we're using $sqrt{2}$, $e$ and $pi$, the first number is 123, second one is 471, the next one is 184 and so on.
randomized-algorithms randomness random-number-generator
New contributor
$endgroup$
Irrational numbers like $pi$, $e$ and $sqrt{2}$ have a unique and non-repeating sequence after the decimal point. If we extract the $n$-th digit from such numbers (where $n$ is the number of times the method is called) and make a number with the digits as it is, should we not get a perfect random number generator? For example, if we're using $sqrt{2}$, $e$ and $pi$, the first number is 123, second one is 471, the next one is 184 and so on.
randomized-algorithms randomness random-number-generator
randomized-algorithms randomness random-number-generator
New contributor
New contributor
edited 13 hours ago
dkaeae
2,3421922
2,3421922
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asked 14 hours ago
Abhradeep SarkarAbhradeep Sarkar
312
312
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Related: mathoverflow.net/questions/26942/…
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– BurnsBA
10 hours ago
9
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You have a strange definition of "random" in your head. "Random" means "unpredictable". How is your sequence unpredictable? What definition of "random" do you have in mind? Perhaps what you are calling "random" has another name.
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– Eric Lippert
9 hours ago
$begingroup$
Essentially the same question was asked here: cs.stackexchange.com/q/65078/755. (Community votes requested -- should these be merged? Should one be closed as a dup of the other?)
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– D.W.♦
9 hours ago
$begingroup$
Note the spigot algorithm can be used to generate any hex digit in pi, without having to generate prior digits.
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– rcgldr
2 hours ago
add a comment |
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Related: mathoverflow.net/questions/26942/…
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– BurnsBA
10 hours ago
9
$begingroup$
You have a strange definition of "random" in your head. "Random" means "unpredictable". How is your sequence unpredictable? What definition of "random" do you have in mind? Perhaps what you are calling "random" has another name.
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– Eric Lippert
9 hours ago
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Essentially the same question was asked here: cs.stackexchange.com/q/65078/755. (Community votes requested -- should these be merged? Should one be closed as a dup of the other?)
$endgroup$
– D.W.♦
9 hours ago
$begingroup$
Note the spigot algorithm can be used to generate any hex digit in pi, without having to generate prior digits.
$endgroup$
– rcgldr
2 hours ago
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Related: mathoverflow.net/questions/26942/…
$endgroup$
– BurnsBA
10 hours ago
$begingroup$
Related: mathoverflow.net/questions/26942/…
$endgroup$
– BurnsBA
10 hours ago
9
9
$begingroup$
You have a strange definition of "random" in your head. "Random" means "unpredictable". How is your sequence unpredictable? What definition of "random" do you have in mind? Perhaps what you are calling "random" has another name.
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– Eric Lippert
9 hours ago
$begingroup$
You have a strange definition of "random" in your head. "Random" means "unpredictable". How is your sequence unpredictable? What definition of "random" do you have in mind? Perhaps what you are calling "random" has another name.
$endgroup$
– Eric Lippert
9 hours ago
$begingroup$
Essentially the same question was asked here: cs.stackexchange.com/q/65078/755. (Community votes requested -- should these be merged? Should one be closed as a dup of the other?)
$endgroup$
– D.W.♦
9 hours ago
$begingroup$
Essentially the same question was asked here: cs.stackexchange.com/q/65078/755. (Community votes requested -- should these be merged? Should one be closed as a dup of the other?)
$endgroup$
– D.W.♦
9 hours ago
$begingroup$
Note the spigot algorithm can be used to generate any hex digit in pi, without having to generate prior digits.
$endgroup$
– rcgldr
2 hours ago
$begingroup$
Note the spigot algorithm can be used to generate any hex digit in pi, without having to generate prior digits.
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– rcgldr
2 hours ago
add a comment |
4 Answers
4
active
oldest
votes
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It is cryptographically useless because an adversary can predict every single digit. It is also very time consuming.
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6
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OP never mentions cryptography...
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– AnoE
10 hours ago
3
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@AnoE So? That this process would be cryptographically useless is still relevant because crypto is an avid user of randomness. If you bring up the devices/dev/random
and/dev/urandom
someone will invariably bring up cryptography.
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– Greg Schmit
10 hours ago
2
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You would be amazed at how useless cryptographic security is in real time PRNG generation. irrational numbers are often used in GPU PRNGs. There are a lot of applications where how "secure" your PRNG is simply irrelevant. What matters in something like coherent noise generation is the quality of distribution and how often your period repeats, and correlation effects due to adjacent seeds (which would require avalanche mixers to fix). Quite honestly your answer is wrong, doesn't belong here, and should probably be deleted.
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– opa
9 hours ago
2
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This is anot an answer to the question. Note the OP of the linked question uses random numbers for seeding a monte carlo analysis. An update to address the question asked should be considered. mathoverflow.net/questions/26942/…
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– CramerTV
9 hours ago
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Mersenne Twister isn't cryptographically secure either, but it's one of the most common PRNGs out there, so I wouldn't say that's grounds to call it useless. I do agree that it would be unreasonably time consuming, though.
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– DarthFennec
7 hours ago
|
show 1 more comment
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For any reasonable definition of perfect, the mechanism you describe is not a perfect random number generator.
Non-repeating isn't enough. The decimal number $0.10100100010001dots$ is non-repeating but it's a terrible generator of random digits, since the answer is "always" zero, occasionally one, and never anything else.
We don't actually know if every digit occurs equally often in the decimal expansion of $pi$ or $mathrm{e}$ (though we suspect they do).
In many situations, we require random numbers to be unpredictable (indeed, if you asked a random person what "random" means, they'd probably say something about unpredictability). The digits of well-known constants are totally predictable.
We usually want to generate random numbers reasonably quickly, but generating successive digits of mathematical constants tends to be quite expensive.
It is, however, true that the digits of $pi$ and $mathrm{e}$ look statistically random, in the sense that every possible sequence of digits seems to occur about as often as it should. So, for example, each digit does occur very close to one time in ten; each two-digit sequence very close to one in a hundred, and so on.
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For the third point, there must be some sort of 'secret' input to your generation process for it to be unpredictable (the generation process itself should be deterministic if we don't want to rely on yet another random number generator.). This extra input is often called a seed.
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– Discrete lizard♦
13 hours ago
1
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@Discretelizard This is true but there's not much scope for seeding beyond "return successive digits starting with position $s$." By the time you've seen $2log s$ digits, that sequence occurs only a few times within the first $s^2$ digits of $pi$, so it's unique within the first $s$ digits with high probability and you know the seed.
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– David Richerby
10 hours ago
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I was not suggesting to try seeding with this approach. However, now that you mention it, would it help to not look at successive digits, but at the $scdot n$'th digit? I guess that would probably be too expensive to compute, but I don't directly see any other problems.
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– Discrete lizard♦
10 hours ago
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Computing digits of irrational numbers is expensive, but hasn't that work already been done? We can just download or access a copy of all the digits. A class time-space tradeoff.
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– Barmar
4 hours ago
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@Barmar: At that point you have to ask whether this technique is really more performant (and more space-efficient) than a "standard" PRNG would be.
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– Kevin
2 hours ago
add a comment |
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Duplicate, but nice question. It was already answered https://math.stackexchange.com/questions/51829/distribution-of-the-digits-of-pi:
...it is believed that $pi$ is a normal number (~uniform distribution of every digits sequence).
For digit distribution data, see e.g. http://www.eveandersson.com/pi/precalculated-frequencies or https://thestarman.pcministry.com/math/pi/RandPI.html (first 1000 digits):
At mathoverflow, there are also nice answers at:
- What is the state of our ignorance about the normality of pi?
- Does pi contain 1000 consecutive zeroes?
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1
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If you believe the question is a duplicate, then why are you answering it? You should simply flag it, not reinforce undesired posting behavior.
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– dkaeae
10 hours ago
3
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@dkaeae There is no support for duplicates of questions on other sites. Furthermore, the same question on different sites can get different answers. In this case, a site such as Mathematics might not give much consideration to security concerns. See also this answer. Do note that we discourage asking the same question on multiple sites at the same time, since this tends to lead to wasted efforts. But the same question by different persons at different times on different sites is usually ok.
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– Discrete lizard♦
10 hours ago
3
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Unfortunately, just because a number is normal doesn't mean that outputting its digits gives you a good RNG. The outputs of such a RNG are still entirely predictable. Whether that's acceptable might depend on the application. So, I don't think it's quite as simple as saying "pi is normal, case closed".
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– D.W.♦
9 hours ago
add a comment |
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In general, this approach does not work: "randomness" does not mean that you get a lot of different digits, but there are other aspects as well. For example, a classic test is to see if all two-digit, or three-digit etc. combinations occur with the same frequency. This would be a very simple test, which can rule out obvious non-random results, but is still by far too simplistic to check for really random behaviour.
See the Wikipedia page about Randomness Tests as a collection of links to primary sources regarding this. They do mention a good amount of quite complicated-sounding concepts; it is it not so important to go into deep detail about this - but it is clear that it is not intuitively possible to declare a specific number to be a good source for such digits.
On a positive note: For a specific irrational number, you are of course free to just try it; i.e., calculate the number to a sufficiantly large degree of digits, and run it through all known tests (there are tools for that, see above link). If the measure is good enough for your use case, and if you are aware that this is obviously useless for cryptographical applications, and always get the same numbers if you should start over, and that the quality might degrade if you get past the n
you picked for testing the randomness, you could use those numbers. But it will be far better to use a dedicated (pseudo-)random number generator; and nothing beats a good physical source of randomness.
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OK but $pi$ and $mathrm{e}$ have the property that all the 2, 3, 4, ... digit sequences do empirically turn up with the right frequency. Nobody's managed to prove it but it seem to be true.
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– David Richerby
10 hours ago
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Ayrat's answer links to other sites where mathematicians have done these tests. They believe, but haven't proved, that π meets the statistical tests.
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– Barmar
4 hours ago
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is cryptographically useless because an adversary can predict every single digit. It is also very time consuming.
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6
$begingroup$
OP never mentions cryptography...
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– AnoE
10 hours ago
3
$begingroup$
@AnoE So? That this process would be cryptographically useless is still relevant because crypto is an avid user of randomness. If you bring up the devices/dev/random
and/dev/urandom
someone will invariably bring up cryptography.
$endgroup$
– Greg Schmit
10 hours ago
2
$begingroup$
You would be amazed at how useless cryptographic security is in real time PRNG generation. irrational numbers are often used in GPU PRNGs. There are a lot of applications where how "secure" your PRNG is simply irrelevant. What matters in something like coherent noise generation is the quality of distribution and how often your period repeats, and correlation effects due to adjacent seeds (which would require avalanche mixers to fix). Quite honestly your answer is wrong, doesn't belong here, and should probably be deleted.
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– opa
9 hours ago
2
$begingroup$
This is anot an answer to the question. Note the OP of the linked question uses random numbers for seeding a monte carlo analysis. An update to address the question asked should be considered. mathoverflow.net/questions/26942/…
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– CramerTV
9 hours ago
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Mersenne Twister isn't cryptographically secure either, but it's one of the most common PRNGs out there, so I wouldn't say that's grounds to call it useless. I do agree that it would be unreasonably time consuming, though.
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– DarthFennec
7 hours ago
|
show 1 more comment
$begingroup$
It is cryptographically useless because an adversary can predict every single digit. It is also very time consuming.
$endgroup$
6
$begingroup$
OP never mentions cryptography...
$endgroup$
– AnoE
10 hours ago
3
$begingroup$
@AnoE So? That this process would be cryptographically useless is still relevant because crypto is an avid user of randomness. If you bring up the devices/dev/random
and/dev/urandom
someone will invariably bring up cryptography.
$endgroup$
– Greg Schmit
10 hours ago
2
$begingroup$
You would be amazed at how useless cryptographic security is in real time PRNG generation. irrational numbers are often used in GPU PRNGs. There are a lot of applications where how "secure" your PRNG is simply irrelevant. What matters in something like coherent noise generation is the quality of distribution and how often your period repeats, and correlation effects due to adjacent seeds (which would require avalanche mixers to fix). Quite honestly your answer is wrong, doesn't belong here, and should probably be deleted.
$endgroup$
– opa
9 hours ago
2
$begingroup$
This is anot an answer to the question. Note the OP of the linked question uses random numbers for seeding a monte carlo analysis. An update to address the question asked should be considered. mathoverflow.net/questions/26942/…
$endgroup$
– CramerTV
9 hours ago
$begingroup$
Mersenne Twister isn't cryptographically secure either, but it's one of the most common PRNGs out there, so I wouldn't say that's grounds to call it useless. I do agree that it would be unreasonably time consuming, though.
$endgroup$
– DarthFennec
7 hours ago
|
show 1 more comment
$begingroup$
It is cryptographically useless because an adversary can predict every single digit. It is also very time consuming.
$endgroup$
It is cryptographically useless because an adversary can predict every single digit. It is also very time consuming.
answered 14 hours ago
gnasher729gnasher729
11.8k1218
11.8k1218
6
$begingroup$
OP never mentions cryptography...
$endgroup$
– AnoE
10 hours ago
3
$begingroup$
@AnoE So? That this process would be cryptographically useless is still relevant because crypto is an avid user of randomness. If you bring up the devices/dev/random
and/dev/urandom
someone will invariably bring up cryptography.
$endgroup$
– Greg Schmit
10 hours ago
2
$begingroup$
You would be amazed at how useless cryptographic security is in real time PRNG generation. irrational numbers are often used in GPU PRNGs. There are a lot of applications where how "secure" your PRNG is simply irrelevant. What matters in something like coherent noise generation is the quality of distribution and how often your period repeats, and correlation effects due to adjacent seeds (which would require avalanche mixers to fix). Quite honestly your answer is wrong, doesn't belong here, and should probably be deleted.
$endgroup$
– opa
9 hours ago
2
$begingroup$
This is anot an answer to the question. Note the OP of the linked question uses random numbers for seeding a monte carlo analysis. An update to address the question asked should be considered. mathoverflow.net/questions/26942/…
$endgroup$
– CramerTV
9 hours ago
$begingroup$
Mersenne Twister isn't cryptographically secure either, but it's one of the most common PRNGs out there, so I wouldn't say that's grounds to call it useless. I do agree that it would be unreasonably time consuming, though.
$endgroup$
– DarthFennec
7 hours ago
|
show 1 more comment
6
$begingroup$
OP never mentions cryptography...
$endgroup$
– AnoE
10 hours ago
3
$begingroup$
@AnoE So? That this process would be cryptographically useless is still relevant because crypto is an avid user of randomness. If you bring up the devices/dev/random
and/dev/urandom
someone will invariably bring up cryptography.
$endgroup$
– Greg Schmit
10 hours ago
2
$begingroup$
You would be amazed at how useless cryptographic security is in real time PRNG generation. irrational numbers are often used in GPU PRNGs. There are a lot of applications where how "secure" your PRNG is simply irrelevant. What matters in something like coherent noise generation is the quality of distribution and how often your period repeats, and correlation effects due to adjacent seeds (which would require avalanche mixers to fix). Quite honestly your answer is wrong, doesn't belong here, and should probably be deleted.
$endgroup$
– opa
9 hours ago
2
$begingroup$
This is anot an answer to the question. Note the OP of the linked question uses random numbers for seeding a monte carlo analysis. An update to address the question asked should be considered. mathoverflow.net/questions/26942/…
$endgroup$
– CramerTV
9 hours ago
$begingroup$
Mersenne Twister isn't cryptographically secure either, but it's one of the most common PRNGs out there, so I wouldn't say that's grounds to call it useless. I do agree that it would be unreasonably time consuming, though.
$endgroup$
– DarthFennec
7 hours ago
6
6
$begingroup$
OP never mentions cryptography...
$endgroup$
– AnoE
10 hours ago
$begingroup$
OP never mentions cryptography...
$endgroup$
– AnoE
10 hours ago
3
3
$begingroup$
@AnoE So? That this process would be cryptographically useless is still relevant because crypto is an avid user of randomness. If you bring up the devices
/dev/random
and /dev/urandom
someone will invariably bring up cryptography.$endgroup$
– Greg Schmit
10 hours ago
$begingroup$
@AnoE So? That this process would be cryptographically useless is still relevant because crypto is an avid user of randomness. If you bring up the devices
/dev/random
and /dev/urandom
someone will invariably bring up cryptography.$endgroup$
– Greg Schmit
10 hours ago
2
2
$begingroup$
You would be amazed at how useless cryptographic security is in real time PRNG generation. irrational numbers are often used in GPU PRNGs. There are a lot of applications where how "secure" your PRNG is simply irrelevant. What matters in something like coherent noise generation is the quality of distribution and how often your period repeats, and correlation effects due to adjacent seeds (which would require avalanche mixers to fix). Quite honestly your answer is wrong, doesn't belong here, and should probably be deleted.
$endgroup$
– opa
9 hours ago
$begingroup$
You would be amazed at how useless cryptographic security is in real time PRNG generation. irrational numbers are often used in GPU PRNGs. There are a lot of applications where how "secure" your PRNG is simply irrelevant. What matters in something like coherent noise generation is the quality of distribution and how often your period repeats, and correlation effects due to adjacent seeds (which would require avalanche mixers to fix). Quite honestly your answer is wrong, doesn't belong here, and should probably be deleted.
$endgroup$
– opa
9 hours ago
2
2
$begingroup$
This is anot an answer to the question. Note the OP of the linked question uses random numbers for seeding a monte carlo analysis. An update to address the question asked should be considered. mathoverflow.net/questions/26942/…
$endgroup$
– CramerTV
9 hours ago
$begingroup$
This is anot an answer to the question. Note the OP of the linked question uses random numbers for seeding a monte carlo analysis. An update to address the question asked should be considered. mathoverflow.net/questions/26942/…
$endgroup$
– CramerTV
9 hours ago
$begingroup$
Mersenne Twister isn't cryptographically secure either, but it's one of the most common PRNGs out there, so I wouldn't say that's grounds to call it useless. I do agree that it would be unreasonably time consuming, though.
$endgroup$
– DarthFennec
7 hours ago
$begingroup$
Mersenne Twister isn't cryptographically secure either, but it's one of the most common PRNGs out there, so I wouldn't say that's grounds to call it useless. I do agree that it would be unreasonably time consuming, though.
$endgroup$
– DarthFennec
7 hours ago
|
show 1 more comment
$begingroup$
For any reasonable definition of perfect, the mechanism you describe is not a perfect random number generator.
Non-repeating isn't enough. The decimal number $0.10100100010001dots$ is non-repeating but it's a terrible generator of random digits, since the answer is "always" zero, occasionally one, and never anything else.
We don't actually know if every digit occurs equally often in the decimal expansion of $pi$ or $mathrm{e}$ (though we suspect they do).
In many situations, we require random numbers to be unpredictable (indeed, if you asked a random person what "random" means, they'd probably say something about unpredictability). The digits of well-known constants are totally predictable.
We usually want to generate random numbers reasonably quickly, but generating successive digits of mathematical constants tends to be quite expensive.
It is, however, true that the digits of $pi$ and $mathrm{e}$ look statistically random, in the sense that every possible sequence of digits seems to occur about as often as it should. So, for example, each digit does occur very close to one time in ten; each two-digit sequence very close to one in a hundred, and so on.
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2
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For the third point, there must be some sort of 'secret' input to your generation process for it to be unpredictable (the generation process itself should be deterministic if we don't want to rely on yet another random number generator.). This extra input is often called a seed.
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– Discrete lizard♦
13 hours ago
1
$begingroup$
@Discretelizard This is true but there's not much scope for seeding beyond "return successive digits starting with position $s$." By the time you've seen $2log s$ digits, that sequence occurs only a few times within the first $s^2$ digits of $pi$, so it's unique within the first $s$ digits with high probability and you know the seed.
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– David Richerby
10 hours ago
$begingroup$
I was not suggesting to try seeding with this approach. However, now that you mention it, would it help to not look at successive digits, but at the $scdot n$'th digit? I guess that would probably be too expensive to compute, but I don't directly see any other problems.
$endgroup$
– Discrete lizard♦
10 hours ago
$begingroup$
Computing digits of irrational numbers is expensive, but hasn't that work already been done? We can just download or access a copy of all the digits. A class time-space tradeoff.
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– Barmar
4 hours ago
$begingroup$
@Barmar: At that point you have to ask whether this technique is really more performant (and more space-efficient) than a "standard" PRNG would be.
$endgroup$
– Kevin
2 hours ago
add a comment |
$begingroup$
For any reasonable definition of perfect, the mechanism you describe is not a perfect random number generator.
Non-repeating isn't enough. The decimal number $0.10100100010001dots$ is non-repeating but it's a terrible generator of random digits, since the answer is "always" zero, occasionally one, and never anything else.
We don't actually know if every digit occurs equally often in the decimal expansion of $pi$ or $mathrm{e}$ (though we suspect they do).
In many situations, we require random numbers to be unpredictable (indeed, if you asked a random person what "random" means, they'd probably say something about unpredictability). The digits of well-known constants are totally predictable.
We usually want to generate random numbers reasonably quickly, but generating successive digits of mathematical constants tends to be quite expensive.
It is, however, true that the digits of $pi$ and $mathrm{e}$ look statistically random, in the sense that every possible sequence of digits seems to occur about as often as it should. So, for example, each digit does occur very close to one time in ten; each two-digit sequence very close to one in a hundred, and so on.
$endgroup$
2
$begingroup$
For the third point, there must be some sort of 'secret' input to your generation process for it to be unpredictable (the generation process itself should be deterministic if we don't want to rely on yet another random number generator.). This extra input is often called a seed.
$endgroup$
– Discrete lizard♦
13 hours ago
1
$begingroup$
@Discretelizard This is true but there's not much scope for seeding beyond "return successive digits starting with position $s$." By the time you've seen $2log s$ digits, that sequence occurs only a few times within the first $s^2$ digits of $pi$, so it's unique within the first $s$ digits with high probability and you know the seed.
$endgroup$
– David Richerby
10 hours ago
$begingroup$
I was not suggesting to try seeding with this approach. However, now that you mention it, would it help to not look at successive digits, but at the $scdot n$'th digit? I guess that would probably be too expensive to compute, but I don't directly see any other problems.
$endgroup$
– Discrete lizard♦
10 hours ago
$begingroup$
Computing digits of irrational numbers is expensive, but hasn't that work already been done? We can just download or access a copy of all the digits. A class time-space tradeoff.
$endgroup$
– Barmar
4 hours ago
$begingroup$
@Barmar: At that point you have to ask whether this technique is really more performant (and more space-efficient) than a "standard" PRNG would be.
$endgroup$
– Kevin
2 hours ago
add a comment |
$begingroup$
For any reasonable definition of perfect, the mechanism you describe is not a perfect random number generator.
Non-repeating isn't enough. The decimal number $0.10100100010001dots$ is non-repeating but it's a terrible generator of random digits, since the answer is "always" zero, occasionally one, and never anything else.
We don't actually know if every digit occurs equally often in the decimal expansion of $pi$ or $mathrm{e}$ (though we suspect they do).
In many situations, we require random numbers to be unpredictable (indeed, if you asked a random person what "random" means, they'd probably say something about unpredictability). The digits of well-known constants are totally predictable.
We usually want to generate random numbers reasonably quickly, but generating successive digits of mathematical constants tends to be quite expensive.
It is, however, true that the digits of $pi$ and $mathrm{e}$ look statistically random, in the sense that every possible sequence of digits seems to occur about as often as it should. So, for example, each digit does occur very close to one time in ten; each two-digit sequence very close to one in a hundred, and so on.
$endgroup$
For any reasonable definition of perfect, the mechanism you describe is not a perfect random number generator.
Non-repeating isn't enough. The decimal number $0.10100100010001dots$ is non-repeating but it's a terrible generator of random digits, since the answer is "always" zero, occasionally one, and never anything else.
We don't actually know if every digit occurs equally often in the decimal expansion of $pi$ or $mathrm{e}$ (though we suspect they do).
In many situations, we require random numbers to be unpredictable (indeed, if you asked a random person what "random" means, they'd probably say something about unpredictability). The digits of well-known constants are totally predictable.
We usually want to generate random numbers reasonably quickly, but generating successive digits of mathematical constants tends to be quite expensive.
It is, however, true that the digits of $pi$ and $mathrm{e}$ look statistically random, in the sense that every possible sequence of digits seems to occur about as often as it should. So, for example, each digit does occur very close to one time in ten; each two-digit sequence very close to one in a hundred, and so on.
answered 13 hours ago
David RicherbyDavid Richerby
70k15106196
70k15106196
2
$begingroup$
For the third point, there must be some sort of 'secret' input to your generation process for it to be unpredictable (the generation process itself should be deterministic if we don't want to rely on yet another random number generator.). This extra input is often called a seed.
$endgroup$
– Discrete lizard♦
13 hours ago
1
$begingroup$
@Discretelizard This is true but there's not much scope for seeding beyond "return successive digits starting with position $s$." By the time you've seen $2log s$ digits, that sequence occurs only a few times within the first $s^2$ digits of $pi$, so it's unique within the first $s$ digits with high probability and you know the seed.
$endgroup$
– David Richerby
10 hours ago
$begingroup$
I was not suggesting to try seeding with this approach. However, now that you mention it, would it help to not look at successive digits, but at the $scdot n$'th digit? I guess that would probably be too expensive to compute, but I don't directly see any other problems.
$endgroup$
– Discrete lizard♦
10 hours ago
$begingroup$
Computing digits of irrational numbers is expensive, but hasn't that work already been done? We can just download or access a copy of all the digits. A class time-space tradeoff.
$endgroup$
– Barmar
4 hours ago
$begingroup$
@Barmar: At that point you have to ask whether this technique is really more performant (and more space-efficient) than a "standard" PRNG would be.
$endgroup$
– Kevin
2 hours ago
add a comment |
2
$begingroup$
For the third point, there must be some sort of 'secret' input to your generation process for it to be unpredictable (the generation process itself should be deterministic if we don't want to rely on yet another random number generator.). This extra input is often called a seed.
$endgroup$
– Discrete lizard♦
13 hours ago
1
$begingroup$
@Discretelizard This is true but there's not much scope for seeding beyond "return successive digits starting with position $s$." By the time you've seen $2log s$ digits, that sequence occurs only a few times within the first $s^2$ digits of $pi$, so it's unique within the first $s$ digits with high probability and you know the seed.
$endgroup$
– David Richerby
10 hours ago
$begingroup$
I was not suggesting to try seeding with this approach. However, now that you mention it, would it help to not look at successive digits, but at the $scdot n$'th digit? I guess that would probably be too expensive to compute, but I don't directly see any other problems.
$endgroup$
– Discrete lizard♦
10 hours ago
$begingroup$
Computing digits of irrational numbers is expensive, but hasn't that work already been done? We can just download or access a copy of all the digits. A class time-space tradeoff.
$endgroup$
– Barmar
4 hours ago
$begingroup$
@Barmar: At that point you have to ask whether this technique is really more performant (and more space-efficient) than a "standard" PRNG would be.
$endgroup$
– Kevin
2 hours ago
2
2
$begingroup$
For the third point, there must be some sort of 'secret' input to your generation process for it to be unpredictable (the generation process itself should be deterministic if we don't want to rely on yet another random number generator.). This extra input is often called a seed.
$endgroup$
– Discrete lizard♦
13 hours ago
$begingroup$
For the third point, there must be some sort of 'secret' input to your generation process for it to be unpredictable (the generation process itself should be deterministic if we don't want to rely on yet another random number generator.). This extra input is often called a seed.
$endgroup$
– Discrete lizard♦
13 hours ago
1
1
$begingroup$
@Discretelizard This is true but there's not much scope for seeding beyond "return successive digits starting with position $s$." By the time you've seen $2log s$ digits, that sequence occurs only a few times within the first $s^2$ digits of $pi$, so it's unique within the first $s$ digits with high probability and you know the seed.
$endgroup$
– David Richerby
10 hours ago
$begingroup$
@Discretelizard This is true but there's not much scope for seeding beyond "return successive digits starting with position $s$." By the time you've seen $2log s$ digits, that sequence occurs only a few times within the first $s^2$ digits of $pi$, so it's unique within the first $s$ digits with high probability and you know the seed.
$endgroup$
– David Richerby
10 hours ago
$begingroup$
I was not suggesting to try seeding with this approach. However, now that you mention it, would it help to not look at successive digits, but at the $scdot n$'th digit? I guess that would probably be too expensive to compute, but I don't directly see any other problems.
$endgroup$
– Discrete lizard♦
10 hours ago
$begingroup$
I was not suggesting to try seeding with this approach. However, now that you mention it, would it help to not look at successive digits, but at the $scdot n$'th digit? I guess that would probably be too expensive to compute, but I don't directly see any other problems.
$endgroup$
– Discrete lizard♦
10 hours ago
$begingroup$
Computing digits of irrational numbers is expensive, but hasn't that work already been done? We can just download or access a copy of all the digits. A class time-space tradeoff.
$endgroup$
– Barmar
4 hours ago
$begingroup$
Computing digits of irrational numbers is expensive, but hasn't that work already been done? We can just download or access a copy of all the digits. A class time-space tradeoff.
$endgroup$
– Barmar
4 hours ago
$begingroup$
@Barmar: At that point you have to ask whether this technique is really more performant (and more space-efficient) than a "standard" PRNG would be.
$endgroup$
– Kevin
2 hours ago
$begingroup$
@Barmar: At that point you have to ask whether this technique is really more performant (and more space-efficient) than a "standard" PRNG would be.
$endgroup$
– Kevin
2 hours ago
add a comment |
$begingroup$
Duplicate, but nice question. It was already answered https://math.stackexchange.com/questions/51829/distribution-of-the-digits-of-pi:
...it is believed that $pi$ is a normal number (~uniform distribution of every digits sequence).
For digit distribution data, see e.g. http://www.eveandersson.com/pi/precalculated-frequencies or https://thestarman.pcministry.com/math/pi/RandPI.html (first 1000 digits):
At mathoverflow, there are also nice answers at:
- What is the state of our ignorance about the normality of pi?
- Does pi contain 1000 consecutive zeroes?
$endgroup$
1
$begingroup$
If you believe the question is a duplicate, then why are you answering it? You should simply flag it, not reinforce undesired posting behavior.
$endgroup$
– dkaeae
10 hours ago
3
$begingroup$
@dkaeae There is no support for duplicates of questions on other sites. Furthermore, the same question on different sites can get different answers. In this case, a site such as Mathematics might not give much consideration to security concerns. See also this answer. Do note that we discourage asking the same question on multiple sites at the same time, since this tends to lead to wasted efforts. But the same question by different persons at different times on different sites is usually ok.
$endgroup$
– Discrete lizard♦
10 hours ago
3
$begingroup$
Unfortunately, just because a number is normal doesn't mean that outputting its digits gives you a good RNG. The outputs of such a RNG are still entirely predictable. Whether that's acceptable might depend on the application. So, I don't think it's quite as simple as saying "pi is normal, case closed".
$endgroup$
– D.W.♦
9 hours ago
add a comment |
$begingroup$
Duplicate, but nice question. It was already answered https://math.stackexchange.com/questions/51829/distribution-of-the-digits-of-pi:
...it is believed that $pi$ is a normal number (~uniform distribution of every digits sequence).
For digit distribution data, see e.g. http://www.eveandersson.com/pi/precalculated-frequencies or https://thestarman.pcministry.com/math/pi/RandPI.html (first 1000 digits):
At mathoverflow, there are also nice answers at:
- What is the state of our ignorance about the normality of pi?
- Does pi contain 1000 consecutive zeroes?
$endgroup$
1
$begingroup$
If you believe the question is a duplicate, then why are you answering it? You should simply flag it, not reinforce undesired posting behavior.
$endgroup$
– dkaeae
10 hours ago
3
$begingroup$
@dkaeae There is no support for duplicates of questions on other sites. Furthermore, the same question on different sites can get different answers. In this case, a site such as Mathematics might not give much consideration to security concerns. See also this answer. Do note that we discourage asking the same question on multiple sites at the same time, since this tends to lead to wasted efforts. But the same question by different persons at different times on different sites is usually ok.
$endgroup$
– Discrete lizard♦
10 hours ago
3
$begingroup$
Unfortunately, just because a number is normal doesn't mean that outputting its digits gives you a good RNG. The outputs of such a RNG are still entirely predictable. Whether that's acceptable might depend on the application. So, I don't think it's quite as simple as saying "pi is normal, case closed".
$endgroup$
– D.W.♦
9 hours ago
add a comment |
$begingroup$
Duplicate, but nice question. It was already answered https://math.stackexchange.com/questions/51829/distribution-of-the-digits-of-pi:
...it is believed that $pi$ is a normal number (~uniform distribution of every digits sequence).
For digit distribution data, see e.g. http://www.eveandersson.com/pi/precalculated-frequencies or https://thestarman.pcministry.com/math/pi/RandPI.html (first 1000 digits):
At mathoverflow, there are also nice answers at:
- What is the state of our ignorance about the normality of pi?
- Does pi contain 1000 consecutive zeroes?
$endgroup$
Duplicate, but nice question. It was already answered https://math.stackexchange.com/questions/51829/distribution-of-the-digits-of-pi:
...it is believed that $pi$ is a normal number (~uniform distribution of every digits sequence).
For digit distribution data, see e.g. http://www.eveandersson.com/pi/precalculated-frequencies or https://thestarman.pcministry.com/math/pi/RandPI.html (first 1000 digits):
At mathoverflow, there are also nice answers at:
- What is the state of our ignorance about the normality of pi?
- Does pi contain 1000 consecutive zeroes?
edited 13 hours ago
Discrete lizard♦
4,45011538
4,45011538
answered 13 hours ago
AyratAyrat
4781419
4781419
1
$begingroup$
If you believe the question is a duplicate, then why are you answering it? You should simply flag it, not reinforce undesired posting behavior.
$endgroup$
– dkaeae
10 hours ago
3
$begingroup$
@dkaeae There is no support for duplicates of questions on other sites. Furthermore, the same question on different sites can get different answers. In this case, a site such as Mathematics might not give much consideration to security concerns. See also this answer. Do note that we discourage asking the same question on multiple sites at the same time, since this tends to lead to wasted efforts. But the same question by different persons at different times on different sites is usually ok.
$endgroup$
– Discrete lizard♦
10 hours ago
3
$begingroup$
Unfortunately, just because a number is normal doesn't mean that outputting its digits gives you a good RNG. The outputs of such a RNG are still entirely predictable. Whether that's acceptable might depend on the application. So, I don't think it's quite as simple as saying "pi is normal, case closed".
$endgroup$
– D.W.♦
9 hours ago
add a comment |
1
$begingroup$
If you believe the question is a duplicate, then why are you answering it? You should simply flag it, not reinforce undesired posting behavior.
$endgroup$
– dkaeae
10 hours ago
3
$begingroup$
@dkaeae There is no support for duplicates of questions on other sites. Furthermore, the same question on different sites can get different answers. In this case, a site such as Mathematics might not give much consideration to security concerns. See also this answer. Do note that we discourage asking the same question on multiple sites at the same time, since this tends to lead to wasted efforts. But the same question by different persons at different times on different sites is usually ok.
$endgroup$
– Discrete lizard♦
10 hours ago
3
$begingroup$
Unfortunately, just because a number is normal doesn't mean that outputting its digits gives you a good RNG. The outputs of such a RNG are still entirely predictable. Whether that's acceptable might depend on the application. So, I don't think it's quite as simple as saying "pi is normal, case closed".
$endgroup$
– D.W.♦
9 hours ago
1
1
$begingroup$
If you believe the question is a duplicate, then why are you answering it? You should simply flag it, not reinforce undesired posting behavior.
$endgroup$
– dkaeae
10 hours ago
$begingroup$
If you believe the question is a duplicate, then why are you answering it? You should simply flag it, not reinforce undesired posting behavior.
$endgroup$
– dkaeae
10 hours ago
3
3
$begingroup$
@dkaeae There is no support for duplicates of questions on other sites. Furthermore, the same question on different sites can get different answers. In this case, a site such as Mathematics might not give much consideration to security concerns. See also this answer. Do note that we discourage asking the same question on multiple sites at the same time, since this tends to lead to wasted efforts. But the same question by different persons at different times on different sites is usually ok.
$endgroup$
– Discrete lizard♦
10 hours ago
$begingroup$
@dkaeae There is no support for duplicates of questions on other sites. Furthermore, the same question on different sites can get different answers. In this case, a site such as Mathematics might not give much consideration to security concerns. See also this answer. Do note that we discourage asking the same question on multiple sites at the same time, since this tends to lead to wasted efforts. But the same question by different persons at different times on different sites is usually ok.
$endgroup$
– Discrete lizard♦
10 hours ago
3
3
$begingroup$
Unfortunately, just because a number is normal doesn't mean that outputting its digits gives you a good RNG. The outputs of such a RNG are still entirely predictable. Whether that's acceptable might depend on the application. So, I don't think it's quite as simple as saying "pi is normal, case closed".
$endgroup$
– D.W.♦
9 hours ago
$begingroup$
Unfortunately, just because a number is normal doesn't mean that outputting its digits gives you a good RNG. The outputs of such a RNG are still entirely predictable. Whether that's acceptable might depend on the application. So, I don't think it's quite as simple as saying "pi is normal, case closed".
$endgroup$
– D.W.♦
9 hours ago
add a comment |
$begingroup$
In general, this approach does not work: "randomness" does not mean that you get a lot of different digits, but there are other aspects as well. For example, a classic test is to see if all two-digit, or three-digit etc. combinations occur with the same frequency. This would be a very simple test, which can rule out obvious non-random results, but is still by far too simplistic to check for really random behaviour.
See the Wikipedia page about Randomness Tests as a collection of links to primary sources regarding this. They do mention a good amount of quite complicated-sounding concepts; it is it not so important to go into deep detail about this - but it is clear that it is not intuitively possible to declare a specific number to be a good source for such digits.
On a positive note: For a specific irrational number, you are of course free to just try it; i.e., calculate the number to a sufficiantly large degree of digits, and run it through all known tests (there are tools for that, see above link). If the measure is good enough for your use case, and if you are aware that this is obviously useless for cryptographical applications, and always get the same numbers if you should start over, and that the quality might degrade if you get past the n
you picked for testing the randomness, you could use those numbers. But it will be far better to use a dedicated (pseudo-)random number generator; and nothing beats a good physical source of randomness.
$endgroup$
1
$begingroup$
OK but $pi$ and $mathrm{e}$ have the property that all the 2, 3, 4, ... digit sequences do empirically turn up with the right frequency. Nobody's managed to prove it but it seem to be true.
$endgroup$
– David Richerby
10 hours ago
$begingroup$
Ayrat's answer links to other sites where mathematicians have done these tests. They believe, but haven't proved, that π meets the statistical tests.
$endgroup$
– Barmar
4 hours ago
add a comment |
$begingroup$
In general, this approach does not work: "randomness" does not mean that you get a lot of different digits, but there are other aspects as well. For example, a classic test is to see if all two-digit, or three-digit etc. combinations occur with the same frequency. This would be a very simple test, which can rule out obvious non-random results, but is still by far too simplistic to check for really random behaviour.
See the Wikipedia page about Randomness Tests as a collection of links to primary sources regarding this. They do mention a good amount of quite complicated-sounding concepts; it is it not so important to go into deep detail about this - but it is clear that it is not intuitively possible to declare a specific number to be a good source for such digits.
On a positive note: For a specific irrational number, you are of course free to just try it; i.e., calculate the number to a sufficiantly large degree of digits, and run it through all known tests (there are tools for that, see above link). If the measure is good enough for your use case, and if you are aware that this is obviously useless for cryptographical applications, and always get the same numbers if you should start over, and that the quality might degrade if you get past the n
you picked for testing the randomness, you could use those numbers. But it will be far better to use a dedicated (pseudo-)random number generator; and nothing beats a good physical source of randomness.
$endgroup$
1
$begingroup$
OK but $pi$ and $mathrm{e}$ have the property that all the 2, 3, 4, ... digit sequences do empirically turn up with the right frequency. Nobody's managed to prove it but it seem to be true.
$endgroup$
– David Richerby
10 hours ago
$begingroup$
Ayrat's answer links to other sites where mathematicians have done these tests. They believe, but haven't proved, that π meets the statistical tests.
$endgroup$
– Barmar
4 hours ago
add a comment |
$begingroup$
In general, this approach does not work: "randomness" does not mean that you get a lot of different digits, but there are other aspects as well. For example, a classic test is to see if all two-digit, or three-digit etc. combinations occur with the same frequency. This would be a very simple test, which can rule out obvious non-random results, but is still by far too simplistic to check for really random behaviour.
See the Wikipedia page about Randomness Tests as a collection of links to primary sources regarding this. They do mention a good amount of quite complicated-sounding concepts; it is it not so important to go into deep detail about this - but it is clear that it is not intuitively possible to declare a specific number to be a good source for such digits.
On a positive note: For a specific irrational number, you are of course free to just try it; i.e., calculate the number to a sufficiantly large degree of digits, and run it through all known tests (there are tools for that, see above link). If the measure is good enough for your use case, and if you are aware that this is obviously useless for cryptographical applications, and always get the same numbers if you should start over, and that the quality might degrade if you get past the n
you picked for testing the randomness, you could use those numbers. But it will be far better to use a dedicated (pseudo-)random number generator; and nothing beats a good physical source of randomness.
$endgroup$
In general, this approach does not work: "randomness" does not mean that you get a lot of different digits, but there are other aspects as well. For example, a classic test is to see if all two-digit, or three-digit etc. combinations occur with the same frequency. This would be a very simple test, which can rule out obvious non-random results, but is still by far too simplistic to check for really random behaviour.
See the Wikipedia page about Randomness Tests as a collection of links to primary sources regarding this. They do mention a good amount of quite complicated-sounding concepts; it is it not so important to go into deep detail about this - but it is clear that it is not intuitively possible to declare a specific number to be a good source for such digits.
On a positive note: For a specific irrational number, you are of course free to just try it; i.e., calculate the number to a sufficiantly large degree of digits, and run it through all known tests (there are tools for that, see above link). If the measure is good enough for your use case, and if you are aware that this is obviously useless for cryptographical applications, and always get the same numbers if you should start over, and that the quality might degrade if you get past the n
you picked for testing the randomness, you could use those numbers. But it will be far better to use a dedicated (pseudo-)random number generator; and nothing beats a good physical source of randomness.
answered 10 hours ago
AnoEAnoE
1,02749
1,02749
1
$begingroup$
OK but $pi$ and $mathrm{e}$ have the property that all the 2, 3, 4, ... digit sequences do empirically turn up with the right frequency. Nobody's managed to prove it but it seem to be true.
$endgroup$
– David Richerby
10 hours ago
$begingroup$
Ayrat's answer links to other sites where mathematicians have done these tests. They believe, but haven't proved, that π meets the statistical tests.
$endgroup$
– Barmar
4 hours ago
add a comment |
1
$begingroup$
OK but $pi$ and $mathrm{e}$ have the property that all the 2, 3, 4, ... digit sequences do empirically turn up with the right frequency. Nobody's managed to prove it but it seem to be true.
$endgroup$
– David Richerby
10 hours ago
$begingroup$
Ayrat's answer links to other sites where mathematicians have done these tests. They believe, but haven't proved, that π meets the statistical tests.
$endgroup$
– Barmar
4 hours ago
1
1
$begingroup$
OK but $pi$ and $mathrm{e}$ have the property that all the 2, 3, 4, ... digit sequences do empirically turn up with the right frequency. Nobody's managed to prove it but it seem to be true.
$endgroup$
– David Richerby
10 hours ago
$begingroup$
OK but $pi$ and $mathrm{e}$ have the property that all the 2, 3, 4, ... digit sequences do empirically turn up with the right frequency. Nobody's managed to prove it but it seem to be true.
$endgroup$
– David Richerby
10 hours ago
$begingroup$
Ayrat's answer links to other sites where mathematicians have done these tests. They believe, but haven't proved, that π meets the statistical tests.
$endgroup$
– Barmar
4 hours ago
$begingroup$
Ayrat's answer links to other sites where mathematicians have done these tests. They believe, but haven't proved, that π meets the statistical tests.
$endgroup$
– Barmar
4 hours ago
add a comment |
Abhradeep Sarkar is a new contributor. Be nice, and check out our Code of Conduct.
Abhradeep Sarkar is a new contributor. Be nice, and check out our Code of Conduct.
Abhradeep Sarkar is a new contributor. Be nice, and check out our Code of Conduct.
Abhradeep Sarkar is a new contributor. Be nice, and check out our Code of Conduct.
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Related: mathoverflow.net/questions/26942/…
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– BurnsBA
10 hours ago
9
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You have a strange definition of "random" in your head. "Random" means "unpredictable". How is your sequence unpredictable? What definition of "random" do you have in mind? Perhaps what you are calling "random" has another name.
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– Eric Lippert
9 hours ago
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Essentially the same question was asked here: cs.stackexchange.com/q/65078/755. (Community votes requested -- should these be merged? Should one be closed as a dup of the other?)
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– D.W.♦
9 hours ago
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Note the spigot algorithm can be used to generate any hex digit in pi, without having to generate prior digits.
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– rcgldr
2 hours ago