Searches related to rsa formula filetype:pdf PDF

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

RSA: Period of Performance for Formula Grant Awards FAQs (3.21

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 ...

RSA-PD-05-01: Liquidation of Obligations Under RSAs Formula

RSA-IM-01-06 is being replaced by this Policy Directive because the extension of a liquidation period under RSA formula grant programs must.

Dear OSERS Grantee Letter: Prior approval applies to State formula

Oct 29 2019 OSEP and RSA Formula Grants. 1. What action is the Office of Special Education and Rehabilitative Services (OSERS) taking?

EVENT NAME

Jan 1 2018 RSA Free Friday Tech Huddle ... Support wants the information from the Calculation ... Validate the field in the Formula Builder window.

Understanding Your Retirement Plan and Options 2019

Benefit calculated using a formula. • Retirement benefits NOT dependent upon the 8.5% of earnable compensation. RSA participation is mandatory ...

RSA-IM-94-14: FY 1994 Reallotment Schedule for Formula Grants

To maximize the use of appropriated funds under the formula grant programs RSA establishes the following. FY 1994 reallotment schedules for the Basic Support (

RSA-1 - Carry Over Requirements for Formula Programs

Section 19(a)(1) of the Rehabilitation Act of 1973 as amended (Rehabilitation Act)

Member Handbook - TIER 1

The Retirement Systems of Alabama (RSA) is pleased to provide you with the Retirement Formula for Maximum Monthly Benefit .

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)

RSA Formula Grant Programs: Federal Reports and Deadlines

(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.

The RSA Cryptosystem

Dan Boneh

Stanford University

Page 2

The RSA cryptosystem

ØFirst published:

•Scientific American, Aug. 1977. (after some censorship entanglements) ØCurrently the "work horse" of Internet security: •Most Public Key Infrastructure (PKI) products. •SSL/TLS: Certificates and key-exchange. •Secure e-mail: PGP, Outlook, ...

Page 3

The RSA trapdoor permutation

ØParameters:N=pq. N »1024 bits. p,q »512 bits. e -encryption exponent. gcd(e, j(N) ) = 1 . ØPermutation:RSA(M) = Me(mod N) where MÎZNØTrapdoor:d-decryption exponent.

Where e×d= 1 (mod j(N) )

ØInversion:RSA(M)d= M(mod N)

Ø"Assumption":

no efficient alg. can invert RSA without trapdoor.

Page 4

Textbook RSA is insecure

ØTextbook RSA encryption:

•public key: (N,e)Encrypt: C = Me(mod N) •private key: dDecrypt: Cd= M(mod N) (M ÎZN)

ØCompletely insecure cryptosystem:

•Does not satisfy basic definitions of security. •Many attacks exist. ØThe RSA trapdoor permutation is not a cryptosystem !

Page 5

A simple attack on textbook RSA

ØSession-key K is 64 bits. View K Î{0,...,264}

Eavesdropper sees: C = Ke(mod N).

ØSuppose K = K1×K2where K1, K2< 234. (prob. »20%)

Then: C/K1e= K2e(mod N)

ØBuild table: C/1e, C/2e, C/3e, ..., C/234e. time: 234 For K2= 0,..., 234test if K2eis in table. time: 234×34

ØAttack time: »240 << 264Web

BrowserWeb

ServerCLIENT HELLO

SERVER HELLO (e,N)d

C=RSA(K)Random

session- key K

Page 6

Common RSA encryption

ØNever use textbook RSA.

ØRSA in practice:

ØMain question:

•How should the preprocessing be done? •Can we argue about security of resulting system?msgPreprocessingciphertextRSA

Page 7

PKCS1 V1.5

ØPKCS1 mode 2:(encryption)

ØResulting value is RSA encrypted.

ØWidely deployed in web servers and browsers.

ØNo security analysis !!02random padFFmsg

1024 bits

16 bits

Page 8

Attack on PKCS1

ØBleichenbacher98. Chosen-ciphertext attack.

ØPKCS1 used in SSL:

Þattacker can test if 16 MSBsof plaintext = '02'.

ØAttack: to decrypt a given ciphertext C do:

•Pick random r ÎZN. Compute C'= re×C = (rM)e. •Send C'to web server and use response.AttackerWeb

ServerdIs this

PKCS1?ciphertext

C=C

Yes: continue

No: error02

Page 9

Chosen ciphertext security (CCS)

ØNo efficient attacker can win the following game: (with non-negligible advantage)AttackerChallengerM

0, M1b'Î{0,1}

Attacker wins if b=b'C=E(M

b) bÎR{0,1}

ChallengeDecryption

oracle ¹C

Page 10

Chosen-ciphertext secure RSA

ØAre there CCS cryptosystems based on RSA?

•RSA-PKCS1 is not CCS !

ØAnswer: Yes!Dolev-Dwork-Naor(DDN). 1991.

•Problem: inefficient. ØOpen problem: efficient CCS system based on RSA.

ØWhat to do? Cheat!

•Build RSA system that is CCS in imaginary world. •"Assume" our-world = imaginary-world.

Page 11

PKCS1 V2.0 -OAEP

ØNew preprocessing function: OAEP (BR94).

ØThm: "trap-door permutation F ÞF-OAEP is CCS when H,G are "random oracles".

ØIn practice: use SHA-1 or MD5 for H and G.H+

G+

Plaintext to encryptwith RSArand.M0100..0

Check pad

on decryption.

Reject CT if invalid.

Î{0,1}n-1

Page 12

An incorrect proof

ØShoup2000: The OAEP thmcannot be correct !!

ØCounter ex: f(x)-xormalleable trapdoor permutation f(x), DÞf(xÅD) Define: h(x,y) = [ x, f(y) ](also trapdoor perm)

ØAttack on h-OAEP:AttackerChallengerM

0, M1C = h(OAEP(Mb)) = [x,f(y)]Rand D= r||01000

y'= yÅG(x)ÅG(xÅD)

C'= [ xÅD, f(y') ]Decrypt C' (¹C)M

b ÅDM b

Page 13

Consequences

ØOAEP is standardized due to an incorrect thm.

ØFortunately: Fujisaki-Okamoto-Pointcheval-Stern •RSA-OAEP is Chosen CiphertextSecure !! -Proof uses special properties of RSA.

ÞNo immediate need to change standards.

•Security proof less efficientthan original "proof". uMain proof idea [FOPS]: •For Shoup'sattack: given challenge C = RSA(x ||y) attacker must "know" x •RSA(x ||y) Þx then RSA is not one-way.

Page 14

OAEP Replacements

ØOAEP+: (Shoup'01)

"trap-door permutation F

F-OAEP+ is CCS when

H,G,W are "random oracles".

ØSAEP+: (B'01)

RSA trap-door perm Þ

RSA-SAEP+ is CCS when

H,W are "random oracle".R

H+

G+MW(M,R)

H+MW(M,R)

Page 15

Subtleties in implementing OAEP

OAEP-decrypt(C) {

error = 0; if (RSA-1(C) > 2n-1) { error =1; gotoexit; } if (pad(OAEP-1(RSA-1(C))) != "01000" ) { error = 1; gotoexit; }} ØProblem:timing information leaks type of error.

ÞAttacker can decrypt any ciphertextC.

ØLesson: Don't implement RSA-OAEPyourself ...

Part II:

Is RSA a One-Way Permutation?

Page 17

Is RSA a one-way permutation?

ØTo invert the RSA one-way function (without d) attacker must compute:

M from C = Me(mod N).

ØHow hard is computing e'throots modulo N??

ØBest known algorithm:

•Step 1: factor N. (hard) •Step 2: Find e'throots modulo p and q. (easy)

Page 18

Shortcuts?

ØMust one factor N in order to compute e'throots? Exists shortcut for breaking RSA without factoring?

ØTo prove no shortcut exists show a reduction:

•Efficient algorithm for e'throots mod N

Þefficient algorithm for factoring N.

•Oldest problem in public key cryptography.

ØEvidence no reduction exists:(BV'98)

•"Algebraic" reduction Þfactoring is easy. •Unlike Diffie-Hellman(Maurer'94).

Page 19

Improving RSA'sperformance

ØTo speed up RSA decryption use

small private key d.Cd= M (mod N) •Wiener87:if d < N0.25then RSA is insecure. •BD'98:if d < N0.292then RSA is insecure (open: d < N 0.5 ) •Insecure:priv. key d can be found from (N,e). •Small d should neverbe used.

Page 20

Wiener's attack

ØRecall:e×d = 1 (mod j(N) )

Þ$kÎZ : e×d = k×j(N) + 1

j(N) = N-p-q+1 Þ|N-j(N)| £p+q £3ÖN d £N0.25/3 Þ

Continued fraction expansion of e/N gives k/d.

e×d = 1 (mod k) Þgcd(d,k)=1e j(N) k d -£ 1 dj(N) e N k d -£ 1 2d2

Page 21

RSA With Low public exponent

ØTo speed up RSA encryption (and sig. verify)

use a small e.C = Me(mod N)

ØMinimal value: e=3( gcd(e, j(N) ) = 1)

ØRecommended value: e=65537=216+1

Encryption: 17 mod. multiplies.

ØSeveral weak attacks. Non known on RSA-OAEP.

ØAsymmetry of RSA: fast enc. / slow dec.

•ElGamal: approx. same time for both.

Page 22

Implementation attacks

ØAttack the implementation of RSA.

ØTiming attack: (Kocher97)

The time it takes to compute Cd(mod N)

can expose d.

ØPower attack: (Kocher99)

The power consumption of a smartcard while

it is computing Cd(mod N) can expose d.

ØFaults attack: (BDL 97)

A computer error during Cd(mod N)