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Von Neumann Extractor - Which bit is retained?


Is just a bit of random, “fully random”?Estimating random number entropy for input into 256 bit hashCSPRNG that cannot be used as random extractorReusing same source for single-source randomness extractorSecure entropy extractor for thermal noise collected from camera input?Fuzzy Extractor: Order of elements in setFuzzy Extractor constructor for sequence reconciliationFuzzy Extractor for Binary SequenceEntropy Rate of Bit StringProving von Neumann extractor correct













3












$begingroup$


Which bit is retained in the Von Neumann debiasing algorithm? 00 and 11 are discarded and 10, 01 are retained but is the first or the second bit retained or does it matter?



In other words:



first: 10 -> 1 , 01 -> 0



second: 10 -> 0 , 01 -> 1



Original Paper (appears to be first but could be interpreted either way):



https://dornsifecms.usc.edu/assets/sites/520/docs/VonNeumann-ams12p36-38.pdf



Examples (first digit accept):



https://en.wikipedia.org/wiki/Hardware_random_number_generator#Software_whitening



https://en.wikipedia.org/wiki/Randomness_extractor



http://pit-claudel.fr/clement/blog/generating-uniformly-random-data-from-skewed-input-biased-coins-loaded-dice-skew-correction-and-the-von-neumann-extractor/#more-410



https://people.seas.harvard.edu/~salil/pseudorandomness/extractors.pdf



Examples (second digit accept):



https://www.researchgate.net/publication/38359648_Iterating_Von_Neumann's_Procedure_for_Extracting_Random_Bits



https://www.esat.kuleuven.be/cosic/publications/article-2628.pdf



Examples of the second are present in papers regarding an Iterated Von Neumann algorithm.










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  • 2




    $begingroup$
    Does it matter? Hint: Write down the assumptions, and compute from the assumptions the probability distributions of the outcomes of both alternatives.
    $endgroup$
    – Squeamish Ossifrage
    2 hours ago












  • $begingroup$
    Why pick one when you can alternate?
    $endgroup$
    – Nat
    26 mins ago


















3












$begingroup$


Which bit is retained in the Von Neumann debiasing algorithm? 00 and 11 are discarded and 10, 01 are retained but is the first or the second bit retained or does it matter?



In other words:



first: 10 -> 1 , 01 -> 0



second: 10 -> 0 , 01 -> 1



Original Paper (appears to be first but could be interpreted either way):



https://dornsifecms.usc.edu/assets/sites/520/docs/VonNeumann-ams12p36-38.pdf



Examples (first digit accept):



https://en.wikipedia.org/wiki/Hardware_random_number_generator#Software_whitening



https://en.wikipedia.org/wiki/Randomness_extractor



http://pit-claudel.fr/clement/blog/generating-uniformly-random-data-from-skewed-input-biased-coins-loaded-dice-skew-correction-and-the-von-neumann-extractor/#more-410



https://people.seas.harvard.edu/~salil/pseudorandomness/extractors.pdf



Examples (second digit accept):



https://www.researchgate.net/publication/38359648_Iterating_Von_Neumann's_Procedure_for_Extracting_Random_Bits



https://www.esat.kuleuven.be/cosic/publications/article-2628.pdf



Examples of the second are present in papers regarding an Iterated Von Neumann algorithm.










share|improve this question







New contributor




bitblt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 2




    $begingroup$
    Does it matter? Hint: Write down the assumptions, and compute from the assumptions the probability distributions of the outcomes of both alternatives.
    $endgroup$
    – Squeamish Ossifrage
    2 hours ago












  • $begingroup$
    Why pick one when you can alternate?
    $endgroup$
    – Nat
    26 mins ago
















3












3








3





$begingroup$


Which bit is retained in the Von Neumann debiasing algorithm? 00 and 11 are discarded and 10, 01 are retained but is the first or the second bit retained or does it matter?



In other words:



first: 10 -> 1 , 01 -> 0



second: 10 -> 0 , 01 -> 1



Original Paper (appears to be first but could be interpreted either way):



https://dornsifecms.usc.edu/assets/sites/520/docs/VonNeumann-ams12p36-38.pdf



Examples (first digit accept):



https://en.wikipedia.org/wiki/Hardware_random_number_generator#Software_whitening



https://en.wikipedia.org/wiki/Randomness_extractor



http://pit-claudel.fr/clement/blog/generating-uniformly-random-data-from-skewed-input-biased-coins-loaded-dice-skew-correction-and-the-von-neumann-extractor/#more-410



https://people.seas.harvard.edu/~salil/pseudorandomness/extractors.pdf



Examples (second digit accept):



https://www.researchgate.net/publication/38359648_Iterating_Von_Neumann's_Procedure_for_Extracting_Random_Bits



https://www.esat.kuleuven.be/cosic/publications/article-2628.pdf



Examples of the second are present in papers regarding an Iterated Von Neumann algorithm.










share|improve this question







New contributor




bitblt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Which bit is retained in the Von Neumann debiasing algorithm? 00 and 11 are discarded and 10, 01 are retained but is the first or the second bit retained or does it matter?



In other words:



first: 10 -> 1 , 01 -> 0



second: 10 -> 0 , 01 -> 1



Original Paper (appears to be first but could be interpreted either way):



https://dornsifecms.usc.edu/assets/sites/520/docs/VonNeumann-ams12p36-38.pdf



Examples (first digit accept):



https://en.wikipedia.org/wiki/Hardware_random_number_generator#Software_whitening



https://en.wikipedia.org/wiki/Randomness_extractor



http://pit-claudel.fr/clement/blog/generating-uniformly-random-data-from-skewed-input-biased-coins-loaded-dice-skew-correction-and-the-von-neumann-extractor/#more-410



https://people.seas.harvard.edu/~salil/pseudorandomness/extractors.pdf



Examples (second digit accept):



https://www.researchgate.net/publication/38359648_Iterating_Von_Neumann's_Procedure_for_Extracting_Random_Bits



https://www.esat.kuleuven.be/cosic/publications/article-2628.pdf



Examples of the second are present in papers regarding an Iterated Von Neumann algorithm.







randomness entropy






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Check out our Code of Conduct.











share|improve this question







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asked 3 hours ago









bitbltbitblt

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bitblt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 2




    $begingroup$
    Does it matter? Hint: Write down the assumptions, and compute from the assumptions the probability distributions of the outcomes of both alternatives.
    $endgroup$
    – Squeamish Ossifrage
    2 hours ago












  • $begingroup$
    Why pick one when you can alternate?
    $endgroup$
    – Nat
    26 mins ago
















  • 2




    $begingroup$
    Does it matter? Hint: Write down the assumptions, and compute from the assumptions the probability distributions of the outcomes of both alternatives.
    $endgroup$
    – Squeamish Ossifrage
    2 hours ago












  • $begingroup$
    Why pick one when you can alternate?
    $endgroup$
    – Nat
    26 mins ago










2




2




$begingroup$
Does it matter? Hint: Write down the assumptions, and compute from the assumptions the probability distributions of the outcomes of both alternatives.
$endgroup$
– Squeamish Ossifrage
2 hours ago






$begingroup$
Does it matter? Hint: Write down the assumptions, and compute from the assumptions the probability distributions of the outcomes of both alternatives.
$endgroup$
– Squeamish Ossifrage
2 hours ago














$begingroup$
Why pick one when you can alternate?
$endgroup$
– Nat
26 mins ago






$begingroup$
Why pick one when you can alternate?
$endgroup$
– Nat
26 mins ago












1 Answer
1






active

oldest

votes


















3












$begingroup$

It doesn't matter at all, since both $01$ and $10$ have the same probability $p(1-p).$






share|improve this answer









$endgroup$









  • 1




    $begingroup$
    Furthermore, using the second bit instead of the first (or vice versa) just has the effect of inverting the output. And a uniformly random bitstream is still uniform even if inverted.
    $endgroup$
    – Ilmari Karonen
    2 hours ago












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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3












$begingroup$

It doesn't matter at all, since both $01$ and $10$ have the same probability $p(1-p).$






share|improve this answer









$endgroup$









  • 1




    $begingroup$
    Furthermore, using the second bit instead of the first (or vice versa) just has the effect of inverting the output. And a uniformly random bitstream is still uniform even if inverted.
    $endgroup$
    – Ilmari Karonen
    2 hours ago
















3












$begingroup$

It doesn't matter at all, since both $01$ and $10$ have the same probability $p(1-p).$






share|improve this answer









$endgroup$









  • 1




    $begingroup$
    Furthermore, using the second bit instead of the first (or vice versa) just has the effect of inverting the output. And a uniformly random bitstream is still uniform even if inverted.
    $endgroup$
    – Ilmari Karonen
    2 hours ago














3












3








3





$begingroup$

It doesn't matter at all, since both $01$ and $10$ have the same probability $p(1-p).$






share|improve this answer









$endgroup$



It doesn't matter at all, since both $01$ and $10$ have the same probability $p(1-p).$







share|improve this answer












share|improve this answer



share|improve this answer










answered 2 hours ago









kodlukodlu

9,45111332




9,45111332








  • 1




    $begingroup$
    Furthermore, using the second bit instead of the first (or vice versa) just has the effect of inverting the output. And a uniformly random bitstream is still uniform even if inverted.
    $endgroup$
    – Ilmari Karonen
    2 hours ago














  • 1




    $begingroup$
    Furthermore, using the second bit instead of the first (or vice versa) just has the effect of inverting the output. And a uniformly random bitstream is still uniform even if inverted.
    $endgroup$
    – Ilmari Karonen
    2 hours ago








1




1




$begingroup$
Furthermore, using the second bit instead of the first (or vice versa) just has the effect of inverting the output. And a uniformly random bitstream is still uniform even if inverted.
$endgroup$
– Ilmari Karonen
2 hours ago




$begingroup$
Furthermore, using the second bit instead of the first (or vice versa) just has the effect of inverting the output. And a uniformly random bitstream is still uniform even if inverted.
$endgroup$
– Ilmari Karonen
2 hours ago










bitblt is a new contributor. Be nice, and check out our Code of Conduct.










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