1 / 30

CS252 Graduate Computer Architecture Lecture 17 ECC (continued), CRC

CS252 Graduate Computer Architecture Lecture 17 ECC (continued), CRC. John Kubiatowicz Electrical Engineering and Computer Sciences University of California, Berkeley http://www.eecs.berkeley.edu/~kubitron/cs252. Review: Code Vector Space. Code Space.

nathan-mueller
Télécharger la présentation

CS252 Graduate Computer Architecture Lecture 17 ECC (continued), CRC

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CS252Graduate Computer ArchitectureLecture 17ECC (continued), CRC John Kubiatowicz Electrical Engineering and Computer Sciences University of California, Berkeley http://www.eecs.berkeley.edu/~kubitron/cs252

  2. Review: Code Vector Space Code Space • Not every vector in the code space is valid • Hamming Distance (d): • Minimum number of bit flips to turn one code word into another • Number of errors that we can detect: (d-1) • Number of errors that we can fix: ½(d-1) C0=f(v0) Code Distance (Hamming Distance) v0 cs252-S09, Lecture 17

  3. G must be an nk matrix Review: How to Generate code words? • Consider a linear code. Need a Generator Matrix. • Let vi be the data value (k bits), Ci be resulting code (n bits): • Are there 2k unique code values? • Only if the k columns of G are linearly independent! • Of course, need some way of decoding as well. • Is this linear??? Why or why not? • A code is systematic if the data is directly encoded within the code words. • Means Generator has form: • Can always turn non-systematiccode into a systematic one (row ops) cs252-S09, Lecture 17

  4. Error vector Implicitly Defining Codes by Check Matrix • But – what is the distance of the code? Not obvious • Instead, consider a parity-check matrix H (n[n-k]) • Compute the following syndrome Si given code element Ci: • Define valid code words Ci as those that give Si=0 (null space of H) • Size of null space? (n-rank H)=k if (n-k) linearly independent columns in H • Suppose you transmit code word C, and there is an error. Model this as vector E which flips selected bits of C to get R (received): • Consider what happens when we multiply by H: • What is distance of code? • Code has distance d if no sum of d-1 or less columns yields 0 • I.e. No error vectors, E, of weight < d have zero syndromes • Code design: Design H matrix with these properties cs252-S09, Lecture 17

  5. P is (n-k)k, I is (n-k)(n-k) Result: H is (n-k)k P is (n-k)k, I is kk Result: G is nk How to relate G and H (Binary Codes) • Defining H makes it easy to understand distance of code, but hard to generate code (H defines code implicitly!) • However, let H be of following form: • Then, G can be of following form (maximal code size): • Notice: G generates values in null-space of H cs252-S09, Lecture 17

  6. Parity code (8-bits): Note: Complexity of logic depends on number of 1s in row! c8 v7v6v5v4v3v2v1v0 C8 C7C6 C5 C4 C3 C2 C1 C0 + + s0 Simple example (Parity, D=2) cs252-S09, Lecture 17

  7. Repetition code (1-bit): Positives: simple Negatives: Expensive: only 33% of code word is data Not packed in Hamming-bound sense (only D=3). Could get much more efficient coding by encoding multiple bits at a time C0 v0 C1 C2 C0 Error C1 C2 Simple example: Repetition (voting, D=3) cs252-S09, Lecture 17

  8. p1 p2 p3 Note: number bits from left to right. Simple Example: Hamming Code (d=3) • Example: (7,4) code: • Protect 4 data bits with 3 parity bits 1 2 3 4 5 6 7 p1 p2 d1 p3 d2 d3 d4 • Bit position number 001 = 110 011 = 310 101 = 510 111 = 710 010 = 210 011 = 310 110 = 610 111 = 710 100 = 410 101 = 510 110 = 610 111 = 710 cs252-S09, Lecture 17

  9. How to correct errors? • But – what is the distance of the code? Not obvious • Instead, consider a parity-check matrix H (n[n-k]) • Compute the following syndrome Si given code element Ci: • Suppose that two correctable error vectors E1 and E2 produce same syndrome: • But, since both E1 and E2 have  (d-1)/2 bits, E1 + E2  d-1 bits set this cannot be true! • So, syndrome is unique indicator of correctable error vectors cs252-S09, Lecture 17

  10. Example, d=4 code (SEC-DED) • Design H with: • All columns non-zero, odd-weight, distinct • Note that odd-weight refers to Hamming Weight, i.e. number of zeros • Why does this generate d=4? • Any single bit error will generate a distinct, non-zero value • Any double error will generate a distinct, non-zero value • Why? Add together two distinct columns, get distinct result • Any triple error will generate a non-zero value • Why? Add together three odd-weight values, get an odd-weight value • So: need four errors before indistinguishable from code word • Because d=4: • Can correct 1 error (Single Error Correction, i.e. SEC) • Can detect 2 errors (Double Error Detection, i.e. DED) • Example: • Note: log size of nullspace will be (columns – rank) = 4, so: • Rank = 4, since rows independent, 4 cols indpt • Clearly, 8 bits in code word • Thus: (8,4) code cs252-S09, Lecture 17

  11. Tweeks: • No reason cannot make code shorter than required • Suppose n-k=8 bits of parity. What is max code size (n) for d=4? • Maximum number of unique, odd-weight columns: 27 = 128 • So, n = 128. But, then k = n – (n – k) = 120. Weird! • Just throw out columns of high weight and make 72, 64 code! • But – shortened codes like this might have d > 4 in some special directions • Example: Kaneda paper, catches failures of groups of 4 bits • Good for catching chip failures when DRAM has groups of 4 bits • What about EVENODD code? • Can be used to handle two erasures • What about two dead DRAMs? Yes, if you can really know they are dead cs252-S09, Lecture 17

  12. cs252-S09, Lecture 17

  13. Galois Field • Definition: Field: a complete group of elements with: • Addition, subtraction, multiplication, division • Completely closed under these operations • Every element has an additive inverse • Every element except zero has a multiplicative inverse • Examples: • Real numbers • Binary, called GF(2)  Galois Field with base 2 • Values 0, 1. Addition/subtraction: use xor. Multiplicative inverse of 1 is 1 • Prime field, GF(p)  Galois Field with base p • Values 0 … p-1 • Addition/subtraction/multiplication: modulo p • Multiplicative Inverse: every value except 0 has inverse • Example: GF(5): 11  1 mod 5, 23  1mod 5, 44  1 mod 5 • General Galois Field: GF(pm)  base p (prime!), dimension m • Values are vectors of elements of GF(p) of dimension m • Add/subtract: vector addition/subtraction • Multiply/divide: more complex • Just like read numbers but finite! • Common for computer algorithms: GF(2m) cs252-S09, Lecture 17

  14. Consider polynomials whose coefficients come from GF(2). Each term of the form xnis either present or absent. Examples:0, 1, x, x2, and x7 + x6 + 1 = 1·x7 + 1· x6 + 0 · x5 + 0 · x4 + 0 · x3 + 0 · x2 + 0 · x1 + 1· x0 With addition and multiplication these form a field: “Add”: XOR each element individually with no carry: x4 + x3 + + x + 1 + x4 + + x2 + x x3 + x2 + 1 “Multiply”: multiplying by x is like shifting to the left. x2 + x + 1 x + 1 x2 + x + 1 x3 + x2 + x x3 + 1 Specific Example: Galois Fields GF(2n) cs252-S09, Lecture 17

  15. x4 + x3 So what about division (mod) x4 + x2 = x3 + x with remainder 0 x x4 + x2 + 1 = x3 + x2 with remainder 1 X + 1 x3 + x2 + 0x + 0 x4 + 0x3 + x2 + 0x + 1 X + 1 x3 + x2 x3 + x2 0x2 + 0x 0x + 1 Remainder 1 cs252-S09, Lecture 17

  16. These polynomials form a Galois (finite) field if we take the results of this multiplication modulo a prime polynomial p(x). A prime polynomial is one that cannot be written as the product of two non-trivial polynomials q(x)r(x) Perform modulo operation by subtracting a (polynomial) multiple of p(x) from the result. If the multiple is 1, this corresponds to XOR-ing the result with p(x). For any degree, there exists at least one prime polynomial. With it we can form GF(2n) Additionally, … Every Galois field has a primitive element, , such that all non-zero elements of the field can be expressed as a power of . By raising  to powers (modulo p(x)), all non-zero field elements can be formed. Certain choices of p(x) make the simple polynomial x the primitive element. These polynomials are called primitive, and one exists for every degree. For example, x4 + x + 1 is primitive. So  = x is a primitive element and successive powers of will generate all non-zero elements of GF(16). Example on next slide. Producing Galois Fields cs252-S09, Lecture 17

  17. 0 = 1 1 = x 2 = x2 3 = x3 4 = x + 1 5 = x2 + x 6 = x3 + x2 7 = x3 + x + 1 8 = x2 + 1 9 = x3 + x 10 = x2 + x + 1 11 = x3 + x2 + x 12 = x3 + x2 + x + 1 13 = x3 + x2 + 1 14 = x3 + 1 15 = 1 Note this pattern of coefficients matches the bits from our 4-bit LFSR example. In general finding primitive polynomials is difficult. Most people just look them up in a table, such as: Galois Fields – Primitives 4 = x4 mod x4 + x + 1 = x4 xor x4 + x + 1 = x + 1 cs252-S09, Lecture 17

  18. x2 + x +1 x3 + x +1 x4 + x +1 x5 + x2 +1 x6 + x +1 x7 + x3 +1 x8 + x4 + x3 + x2 +1 x9 + x4 +1 x10 + x3 +1 x11 + x2 +1 Primitive Polynomials x12 + x6 + x4 + x +1 x13 + x4 + x3 + x +1 x14 + x10 + x6 + x +1 x15 + x +1 x16 + x12 + x3 + x +1 x17 + x3 + 1 x18 + x7 + 1 x19 + x5 + x2 + x+ 1 x20 + x3 + 1 x21 + x2 + 1 x22 + x +1 x23 + x5 +1 x24 + x7 + x2 + x +1 x25 + x3 +1 x26 + x6 + x2 + x +1 x27 + x5 + x2 + x +1 x28 + x3 + 1 x29 + x +1 x30 + x6 + x4 + x +1 x31 + x3 + 1 x32 + x7 + x6 + x2 +1 Galois Field Hardware Multiplication by x shift left Taking the result mod p(x) XOR-ing with the coefficients of p(x) when the most significant coefficient is 1. Obtaining all 2n-1 non-zero Shifting and XOR-ing 2n-1 times. elements by evaluating xk for k = 1, …, 2n-1 cs252-S09, Lecture 17

  19. For k-bit LFSR number the flip-flops with FF1 on the right. The feedback path comes from the Q output of the leftmost FF. Find the primitive polynomial of the form xk + … + 1. The x0 = 1 term corresponds to connecting the feedback directly to the D input of FF 1. Each term of the form xn corresponds to connecting an xor between FF n and n+1. 4-bit example, uses x4 + x + 1 x4 FF4’s Q output x xor between FF1 and FF2 1 FF1’s D input To build an 8-bit LFSR, use the primitive polynomial x8 + x4 + x3 + x2 + 1 and connect xors between FF2 and FF3, FF3 and FF4, and FF4 and FF5. Building an LFSR from a Primitive Poly(Cycle through all non-zero values) cs252-S09, Lecture 17

  20. Reed-Solomon Codes • Galois field codes: code words consist of symbols • Rather than bits • Reed-Solomon codes: • Based on polynomials in GF(2k) (I.e. k-bit symbols) • Data as coefficients, code space as values of polynomial: • P(x)=a0+a1x1+… ak-1xk-1 • Coded: P(0),P(1),P(2)….,P(n-1) • Can recover polynomial as long as get any k of n • Properties: can choose number of check symbols • Reed-Solomon codes are “maximum distance separable” (MDS) • Can add d symbols for distance d+1 code • Often used in “erasure code” mode: as long as no more than n-k coded symbols erased, can recover data • Side note: Multiplication by constant in GF(2k) can be represented by kk matrix: ax • Decompose unknown vector into k bits: x=x0+2x1+…+2k-1xk-1 • Each column is result of multiplying a by 2i cs252-S09, Lecture 17

  21. Reed-Solomon Codes (con’t) • Reed-solomon codes (Non-systematic): • Data as coefficients, code space as values of polynomial: • P(x)=a0+a1x1+… a6x6 • Coded: P(0),P(1),P(2)….,P(6) • Called Vandermonde Matrix: maximum rank • Different representation(This H’ and G not related) • Clear that all combinations oftwo or less columns independent  d=3 • Very easy to pick whatever d you happen to want: add more rows • Fast, Systematic version of Reed-Solomon: • Cauchy Reed-Solomon, others cs252-S09, Lecture 17

  22. Another Example: Redundant Check • Send a message M and a “check” word C • Simple function on <M,C> to determine if both received correctly (with high probability) • Example: XOR all the bytes in M and append the “checksum” byte, C, at the end • Receiver XORs <M,C> • What should result be? • What errors are caught? *** bit i is XOR of ith bit of each byte cs252-S09, Lecture 17

  23. Example: TCP Checksum TCP Packet Format Application (HTTP,FTP, DNS) 7 Transport (TCP, UDP) 4 Network (IP) 3 Data Link (Ethernet, 802.11b) 2 TCP Checksum a 16-bit checksum, consisting of the one's complement of the one's complement sum of the contents of the TCP segment header and data, is computed by a sender, and included in a segment transmission. (note end-around carry) Summing all the words, including the checksum word, should yield zero Physical 1 cs252-S09, Lecture 17

  24. Example: Ethernet CRC-32 Application (HTTP,FTP, DNS) 7 Transport (TCP, UDP) 4 Network (IP) 3 Data Link (Ethernet, 802.11b) 2 Physical 1 cs252-S09, Lecture 17

  25. n bits of zero at the end tack on n bits of remainder Instead of the zeros CRC concept • I have a msg polynomial M(x) of degree m • We both have a generator poly G(x) of degree n • Let r(x) = remainder of M(x) xn / G(x) • M(x) xn = G(x)p(x) + r(x) • r(x) is of degree n • What is (M(x) xn – r(x)) / G(x) ? • So I send you M(x) xn – r(x) • m+n degree polynomial • You divide by G(x) to check • M(x) is just the m most signficant coefficients, r(x) the lower n • x-bit Message is viewed as coefficients of x-degree polynomial over binary numbers cs252-S09, Lecture 17

  26. Polynomial division • When MSB is zero, just shift left, bringing in next bit • When MSB is 1, XOR with divisor and shift eft 0 0 0 0 1 0 1 1 0 0 1 1 1 0 1 1 0 0 1 0 0 0 0 1 0 0 1 1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 1 0 0 cs252-S09, Lecture 17

  27. CRC encoding 1 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 0 0 1 0 0 0 0 0 0 1 0 1 1 0 0 1 0 0 0 0 0 1 0 1 1 0 0 1 0 0 0 0 1 0 1 1 0 0 1 0 0 0 0 0 1 01 0 1 0 0 0 0 1 01 0 1 0 0 0 0 01 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 1 1 0 0 0 1 0 1 0 1 0 1 0 Message sent: 1 0 1 1 0 0 1 1 0 1 0 cs252-S09, Lecture 17

  28. CRC decoding 1 0 1 1 0 0 1 1 0 1 0 0 0 0 0 0 0 0 1 0 1 1 0 0 1 1 0 1 0 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 1 0 1 1 0 0 1 1 0 1 0 0 1 01 0 1 1 0 1 0 1 01 0 1 1 0 1 0 01 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 cs252-S09, Lecture 17

  29. Generating Polynomials • CRC-16: G(x) = x16 + x15 + x2 + 1 • detects single and double bit errors • All errors with an odd number of bits • Burst errors of length 16 or less • Most errors for longer bursts • CRC-32: G(x) = x32 + x26 + x23 + x22 + x16 + x12 + x11 + x10 + x8 + x7 + x5 + x4 + x2 + x + 1 • Used in ethernet • Also 32 bits of 1 added on front of the message • Initialize the LFSR to all 1s cs252-S09, Lecture 17

  30. Conclusion • ECC: add redundancy to correct for errors • (n,k,d)  n code bits, k data bits, distance d • Linear codes: code vectors computed by linear transformation • Erasure code: after identifying “erasures”, can correct • Reed-Solomon codes • Based on GF(pn), often GF(2n) • Easy to get distance d+1 code with d extra symbols • Often used in erasure mode • Redundancy useful to gain reliability • Redundant disks+controllers+etc (RAID) • Geographical scale systems (OceanStore) • Disk technology: • Two innovations: GMR, Vertical recording • Disk Latency = Queuing Time + Seek Time + Rotation Time + Xfer Time + Ctrl Time cs252-S09, Lecture 17

More Related