RSA Cryptosystem The RSA cryptosystem is a example of a “public
At the center of the RSA cryptosystem is the RSA modulus N. It is a positive and then Alice picks p and q so that equation (1) holds.
RSA Formula Grant Programs: Federal Reports and Deadlines
Quarterly. (PY). PD-19-03 · Joint PIRL. Data collected through the RSA-911 is used to assess the performance of the VR program through the calculation of
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Mar 21 2017 the unobligated balance of Federal funds to be carried over to the subsequent FFY (see FAQ 5 for additional information). For RSA formula awards ...
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RSA-IM-01-06 is being replaced by this Policy Directive because the extension of a liquidation period under RSA formula grant programs must.
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To maximize the use of appropriated funds under the formula grant programs RSA establishes the following. FY 1994 reallotment schedules for the Basic Support (
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N=pq N Ø (N) ) = 1 - Stanford University
Ø Permutation: RSA(M) = Me (mod N) where M?Z N Ø Trapdoor: d– decryption exponent Where e?d = 1 (mod ?(N) ) Ø Inversion: RSA(M) d = Me k??(N) +1 = M (mod N) Ø “Assumption”: no efficient alg can invert RSA without trapdoor Page 2 Textbook RSA is insecure Ø Textbook RSA encryption: • public key: (Ne) Encrypt: C = Me (mod N)
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(RSA-911) Quarterly (PY) PD-19-03 Joint PIRL Data collected through the RSA-911 is used to assess the performance of the VR program through the calculation of evaluation standards and performance indicators conduct annual reviews and periodic onsite monitoring of VR agencies and support disability research RSA-
Why is it important that phi(n) is kept a secret in RSA?
RSA modulus: N=pq So 55 = 5· 11 119 = 7· 17 and 10403 = 101· 103 could each be used as anRSA modulus although in practice one would use much larger numbers for bettersecurity to be explained below Also needed is an encoding exponente The only requirement oneis that gcd(e(p?1)(q?1)) = 1
The RSA Algorithm - University of Washington
The RSA Algorithm Evgeny Milanov 3 June 2009 In 1978 Ron Rivest Adi Shamir and Leonard Adleman introduced a cryptographic algorithm which was essentially to replace the less secure National Bureau of Standards (NBS) algorithm Most impor-tantly RSA implements a public-key cryptosystem as well as digital signatures RSA is motivated by
RSA and ASYMMETRIC (PUBLIC-KEY) ENCRYPTION
RSA: what to remember The RSA function f(x) = xe mod N is a trapdoor one way permutation: Easy forward: given N;e;x it is easy to compute f(x) Easy back with trapdoor: Given N;d and y = f(x) it is easy to compute x = f 1(y) = yd mod N Hard back without trapdoor: Given N;e and y = f(x) it is hard to compute x = f 1(y) Nadia Heninger UCSD 21
Searches related to rsa formula filetype:pdf
RSA With Low public exponent Ø To speed up RSA encryption (and sig verify) use a small e C = Me (mod N) Ø Minimal value: e=3 ( gcd(e ?(N) ) = 1) Ø Recommended value: e=65537=216+1 Encryption: 17 mod multiplies Ø Several weak attacks Non known on RSA-OAEP Ø Asymmetry of RSA: fast enc / slow dec • ElGamal: approx same time for both
How do you calculate RSA key?
- In fact that's just what happens during normal RSA key generation. You use that e ? d = 1 mod ?(n), and solve for d using the extended Euclidian algorithm. i.e., d is the multiplicative inverse of e mod ?(n). This is often computed using the extended Euclidean algorithm.
How do you factor n in RSA?
- Given ?(n) and n it's easy to factor n by solving the equations n = p ? q and ?(n) = (p ? 1) ? (q ? 1) for p and q. Remember that with RSA the number N is the product of two large secret primes. Let's call them P and Q. We will treat them as our unknowns: Now N is known, as part of the public key.
What is RSA and how does it work?
- Clifford Cocks, an English mathematician, had developed an equivalent system in 1973, but it was classified until 1997. A user of RSA creates and then publishes the product of two large prime numbers, along with an auxiliary value, as their public key.
RSA and ASYMMETRIC (PUBLIC-KEY) ENCRYPTION
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RSA generators
An RSA generator with security parameterkis an algorithmKrsathat returnsN;p;q;e;dsatisfying p;qare distinct odd primesN=pq, and is called the (RSA) modulus
jNj=k, meaning 2k1N2k e2Z'(N)is called the encryption exponent d2Z'(N)is called the decryption exponent edmod'(N) = 1Nadia Heninger UCSD 2
A formula for Phi
Fact:Supp oseN=pqfor distinct primespandq. Then
'(N) = (p1)(q1):Example:LetN= 15 = 35. Then the Fact says that
'(15) = (31)(51) = 8: As a check,Z15=f1;2;4;7;8;11;13;14gindeed has size 8.Nadia Heninger UCSD 3
Recall
Given'(N) ande2Z'(N), we can computed2Z'(N)satisfying edmod'(N) = 1 via d MOD-INV(e;'(N)): We have algorithms to eciently test whether a number is prime, and we know that a random number has a pretty good chance of being a prime.We use these facts to build RSA generators.
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Building RSA generators
Say we wish to havee= 3. (We will see that the smaller ise, the more ecient is encryption.) The generatorK3rsawith (even) security parameter kis as follows: repeat p;q$ f2k=21;:::;2k=21g;N pq;M (p1)(q1) untilN2k1andp;qare primeand gcd(e;M) = 1
d MOD-INV(e;M) returnN;p;q;e;dNadia Heninger UCSD 5
One-wayness of RSA
The following should be hard:
Given:N;e;ywherey=f(x) =xemodN
Find:x
Formalism picksxat random and generatesN;evia an RSA generator.Nadia Heninger UCSD 6
One-wayness of RSA, formally
LetKrsabe a RSA generator andIan adversary.
GameOWKrsa
procedure Initialize (N;p;q;e;d)$ Krsa x $ ZN;y xemodN returnN;e;yprocedure Finalize(x0) return(x=x0)The ow-advantage ofIis Adv owKrsa(I) = PrhOWIKrsa)truei
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Inverting RSA
Inverting RSA : givenN;e;yndxsuch thatxemodN=y
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Inverting RSA
Inverting RSA : givenN;e;yndxsuch thatxemodN=yEASY becausex=ydmodN KnowdNadia Heninger UCSD 9
Inverting RSA
Inverting RSA : givenN;e;yndxsuch thatxemodN=yEASY becausex=ydmodNKnowdEASY
becaused=MOD-INV(e;'(N))Know'(N)
Nadia Heninger UCSD 10
Inverting RSA
Inverting RSA : givenN;e;yndxsuch thatxemodN=yEASY becausex=ydmodNKnowdEASY
becaused=MOD-INV(e;'(N))Know'(N)EASY
because'(N) = (p1)(q1)Knowp;q
Nadia Heninger UCSD 11
Inverting RSA
Inverting RSA : givenN;e;yndxsuch thatxemodN=yEASY becausex=ydmodNKnowdEASY
becaused=MOD-INV(e;'(N))Know'(N)EASY
because'(N) = (p1)(q1)Knowp;q?
KnowNNadia Heninger UCSD 12
Factoring Problem
Given:NwhereN=pqandp;qare prime
Find:p;q
If we can factor we can invert RSA. We do not know whether the converse is true, meaning whether or not one can invert RSA without factoring.Nadia Heninger UCSD 13
A factoring algorithm
AlgFACTOR(N)//N=pqwherep;qare primes
fori= 2;:::;lpN m do ifNmodi= 0then p i;q N=i;returnp;qThis algorithm works but takes time
O(pN) =O(e0:5lnN)
which is prohibitive.Nadia Heninger UCSD 14
Factoring algorithms
AlgorithmTime taken to factorNNaiveO(e0:5lnN)Quadratic Sieve (QS)O(ec(lnN)1=2(lnlnN)1=2)Number Field Sieve (NFS)O(e1:92(lnN)1=3(lnlnN)2=3)Nadia Heninger UCSD 15
Factoring records
bit-length of numberWhen factoredAlgorithm used4001993QS
4281994QS
4311996NFS
4651999NFS
5151999NFS
5762003NFS
7682009NFS
7952019NFS
8292020NFS
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Moduli sizes
We estimate that a 1024-bit RSA modulus provides 80 bits of security, meaning factoring it takes 280time.
Factorization of a 1024-bit modulus hasn't been done yet in public, but is within reach of large organizations. Longer moduli, like 2048 bits, have been recommended since around 2010. Just because factoring some large numbers seems to be hard does not mean factoringalllarge numbers is hard. For example, a random integer has probability 1=2 of having 2 as a prime factor. This is why RSA uses moduliNdesigned to resist known factoring algorithms.Nadia Heninger UCSD 17
Choices of encryption exponent
Common choices aree= 3,e= 17 ande= 65;537. Why these?ebin(e)3111710001
65,53710000000000000001
Recall that the modular exponentiation algorithm computingx7! x emodNusesc(b) modular multiplications per bitb2 f0;1gin the binary expansion bin(e), wherec(0) = 1 andc(1) = 2. So the fewer the number of 1s in bin(e), the faster is the operation.Nadia Heninger UCSD 18
Textbook RSA is insecure
Choosing a good modulus and exponent is not enough to make an RSA implementation secure.For example:
Lete= 3,Nhave 1024 bits, and encrypty=xemodN.
To avoid this, typical RSA implementationspadmessages before encrypting. But even this is hard to get right: the most common RSA padding scheme in use is broken if it throws an error when it discovers incorrect padding.Nadia Heninger UCSD 19
RSA Video
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RSA: what to remember
The RSA functionf(x) =xemodNis a trapdoor one way permutation:Easy forward: givenN;e;xit is easy to computef(x)
Easy back with trapdoor: GivenN;dandy=f(x) it is easy to computex=f1(y) =ydmodN Hard back without trapdoor: GivenN;eandy=f(x) it is hard to computex=f1(y)Nadia Heninger UCSD 21
The quantum threat
On a quantum computer, Shor's algorithm can compute discrete logarithms and factor in polynomial time.Eorts to build quantum computers are underway.
Eorts are underway to standardize public-key cryptography based on computational problems like nding short vectors in lattices for which there are currently no known ecient quantum algorithms.Nadia Heninger UCSD 22
Two settings
Symmetric encryption:
Before Alice and Bob can communicate securely, they need to have a common secret keyKAB. If Alice wishes to also communicate with Charlie then she and Charlie must also have another common secret keyKAC. If Alice generatesKAB;KAC, they must be communicated to her partners over private and authenticated channels.Asymmetric (public-key) encryption:
Alice has a secret decryption keydkthat is shared with nobody, and an associated public encryption keyekthat is known to everybody. Anyone (Bob, Charlie,:::) can use Alice's encryption keyekto send her an encrypted message which only she can decrypt.Nadia Heninger UCSD 23
Syntax of a PKE scheme
A public-key (or asymmetric) encryption schemeAE= (K;E;D) consists of three algorithms that operate as follows: (ek;dk)$ K| generate an encryption keyekand matching decryption keydk C$ Eek(M) | encrypt messageMunder encryption keyekto get a ciphertextC. AlgorithmEmay be randomized. M0 Ddk(C) | decrypt ciphertextCunder decryption keydkto get an outputM02 f0;1g[ f?g.Nadia Heninger UCSD 24
Correct decryption requirement
LetAE= (K;E;D) be an asymmetric encryption scheme. The correct decryption requirement is thatPr[Ddk(Eek(M)) =M] = 1
for all (ek;dk) that may be output byKand all messagesMin the message spaceofAE. The probability is over the random choices ofE. This simply says that decryption correctly reverses encryption to recoverquotesdbs_dbs6.pdfusesText_11[PDF] rsa video maker
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