Timeline for Entropy of the key
Current License: CC BY-SA 3.0
15 events
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| Jun 17, 2020 at 8:17 | history | edited | CommunityBot |
Commonmark migration
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| Nov 8, 2016 at 10:39 | comment | added | CodesInChaos | @RAW You misread the question. In scenario 1, there are 995 uniformly random bits, so the entropy is exactly 995 bits. In scenario 2 all 1000 bits are random but biased. There is no scenario where there are 995 biased bits. | |
| Nov 8, 2016 at 5:18 | comment | added | RAW | Since the first 5 bits are fixed their entropy is 0 For the remaining 995 bits you first calculate the entropy for ONE bit: P(0)=0.54 and P(1)=0.46 −∑x∈XP(x)⋅log2(P(x))= −(0.54⋅log2(0.54)+0.46⋅log2(0.46))≈0.9954 bits of entropy PER bit The entropy for the 995 bits = 995⋅0.9954 = 990.423 bits of entropy (it's obvious the entropy for the 995 bits must be < 995) | |
| Oct 7, 2014 at 9:10 | vote | accept | Idonknow | ||
| Sep 4, 2014 at 1:33 | vote | accept | Idonknow | ||
| Sep 4, 2014 at 2:03 | |||||
| Sep 3, 2014 at 12:41 | history | edited | CodesInChaos | CC BY-SA 3.0 |
added 7 characters in body
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| Sep 3, 2014 at 12:40 | history | edited | e-sushi | CC BY-SA 3.0 |
edited body
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| Sep 3, 2014 at 11:46 | history | edited | Maarten Bodewes♦ | CC BY-SA 3.0 |
cleared things up according to comments
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| Sep 3, 2014 at 10:05 | comment | added | CodesInChaos | 1) Just search for a proof for additivity of Shannon entropy. This is such a basic property that you should find plenty. 2) 0.9954 is the entropy per bit. The entropy of the whole is obviously 1000 times that value. | |
| Sep 3, 2014 at 9:53 | comment | added | Idonknow | For part $(b)$, shouldn't we need to multiple $0.9954$ with $1000$ since the length of the key is $1000$? | |
| Sep 3, 2014 at 9:46 | comment | added | Idonknow | Actually for part$(b)$, I am required to prove that entropy of the key is the sum of the entropy of individual key. | |
| Sep 3, 2014 at 9:43 | history | edited | CodesInChaos | CC BY-SA 3.0 |
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| Sep 3, 2014 at 9:42 | comment | added | CodesInChaos | For the first 5 bits it's $P(0)=1$ and $P(1)=0$, for the other 995 bits it's $P(0)=0.5$ and $P(1)=0.5$. | |
| Sep 3, 2014 at 9:40 | comment | added | Idonknow | For part $(1)$, if I want to use the formula, what should be my $P(x)$? | |
| Sep 3, 2014 at 9:37 | history | answered | CodesInChaos | CC BY-SA 3.0 |