Some Classes of Invertible Matrices in GF(2)

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1 Some Classes of Invertible Matrices in GF() James S. Plank Adam L. Buchsbaum Technical Report UT-CS Department of Electrical Engineering and Computer Science University of Tennessee August 16, 007 The home for this paper is plank/plank/papers/cs html. Please visit that link for up-to-date information about the publication status of this and related papers. Abstract Invertible matrices in GF () are important for constructing MDS erasure codes. This paper proves that certain classes of matrices in GF () are invertible, plus some additional properties about invertible matrices. 1 Introduction We are concerned ith the question of hether certain matrices in GF() are invertible. This question is important hen designing erasure codes for storage applications. If an erasure code is composed solely of exclusive-or operations [1,, 3, 4, 5, 8, 9], then it may be represented as a matrix-vector product in GF(). The act of decoding transforms an original distribution matrix into a square decoding matrix that must be inverted. The process is described for general GF ( ) by Plank [7] and is first used in GF() by Blomer et al. []. As such, a fundamental part of defining MDS erasure codes is to construct distribution matrices that result in invertible decoding matrices. This paper does not delve into erasures codes, but instead proves that certain classes of matrices in GF() are invertible. It also proves some properties of invertible matrices. Nomenclature In GF(), each element is either 0 or 1; addition is the binary exclusive-or operator (denoted ), and multiplication is the binary and operator. When e refer to a matrix M, that means that M is a square matrix in GF() ith ros and columns. Other information about the matrix is included in the subscripts. We refer to the element in ro r and column c of M as M [r, c]. These are zero-indexed, so the top-left element of M is M [0, 0], and the bottom-right element of M is M [ 1, 1]. We perform arithmetic of ro and column indices in M over the commutative ring Z/Z. We denote the quantity x modulo by x. In particular, because x + = x, e have 1 = 1. When context disambiguates, e drop the extra notation; e.g., 1 = 1. Department of Electrical Engineering and Computer Science, University of Tennessee, Knoxville, TN 37996, plank@cs.utk.edu. AT&T Labs-Research, Shannon Laboratory, 180 Park Avenue, Florham Park, NJ 0793, alb@research.att.com. 1

2 .1 Invertibility One ay to test hether a square matrix M is invertible is to perform Gaussian Elimination on it until it is in upper triangular form. Then M is invertible if and only if the result is unit upper triangular. (Basic facts about invertibility of matrices under simple operations are available in many textbooks, e.g., Lancaster and Tismenetsky [6].) We define steps of Gaussian Elimination as follos. Let c be the leftmost column ith at least to 1 s in some M ; let r be the topmost ro such that M [r, c] = 1 and M [r, c ] = 0 for 0 c < c. Then one step of Gaussian Elimination or Elimination Step replaces every ro r r such that M [r, c] = 1 ith the sum of ros r and r. An example is in Figure 1. The first step of Gaussian Elimination for the matrix in Figure 1(a) replaces ro ith the sum of ros 0 and, and ro 3 ith the sum of ros 0 and 3. The resulting matrix is in Figure 1(b). (a) (b) (c) Figure 1: One step of Gaussian Elimination, and deleting ros and columns that are upper-triangular. When the leftmost l columns of a matrix M have zeros belo the main diagonal i.e., M[i, j] = 0 for 0 i < l and i < j < e say the leftmost l columns are in upper triangular form or are upper triangular; if in addition M [i, i] = 1 for 0 i < l, e say the leftmost l columns are in unit upper triangular form or are unit upper triangular. Assume the leftmost l columns of M are unit upper triangular, and construct matrix M, here = l, by deleting the leftmost l columns and top l ros of M. Then M is invertible iff M is invertible. For example, since the leftmost to columns of the matrix in Figure 1(b) are in unit upper triangular form, e may delete the leftmost to columns and the top to ros to produce the matrix in Figure 1(c). This matrix is not invertible; therefore, the matrices in Figures 1(a) and 1(b) are also not invertible. There are other simple operations that preserve invertibility. The first are hat e call ro shifting and column shifting. There are four variants. Each takes an original matrix M and constructs a ne matrix M as follos: Shifting up by r ros: M [i, j] = M [i + r, j], for 0 i, j <. Shifting don by r ros: M [i, j] = M [i r, j], for 0 i, j <. Shifting left by c columns: M [i, j] = M [i, j + c ], for 0 i, j <. Shifting right by c columns: M [i, j] = M [i, j c ], for 0 i, j <. Obviously, shifting M up by r ros is equivalent to shifting it don by r ros, and shifting M left by r columns is equivalent to shifting it right by r columns. Sapping ros and columns preserves invertibility, and substituting any ro ith the sum of it and another ro also preserves invertibility. Examples are in Figure. We denote by I (rsp., I c ) the identity matrix (rsp., shifted c columns to the right) and by 0 the matrix of all zeros. Finally, e say a matrix class M is invertible iff all matrices in M are invertible. 3 The Matrix Classes D d,s and S d,s We no define to classes of matrices: Dd,s and S d,s. In both: >, 0 < d <, and 0 < s <. The letters are short for different and same. We define D d,s,0 to be the base element of D d,s. We construct D d,s,0 as follos: Start ith Dd,s,0 = I + I d.

3 (a) (b) (c) (d) (e) Figure : Operations that preserve invertibility. (a) is the original matrix. (b) shifts (a) up by three ros, or don by four ros. (c) shifts (a) left by four ros, or right by three ros. (d) saps ros 3 and 6. (e) replaces ro 6 ith the sum of ros 3 and 6. Set Dd,s,0 [0, 1] = D d,s,0 [0, 1] 1. Set D d,s,0 [s, d + s 1 ] = D d,s,0 [s, d + s 1 ] 1. There are elements of Dd,s, denoted D d,s,0,..., D d,s, 1. D d,s,i is equal to D d,s,0 shifted i ros don and i columns to the right. Therefore, all elements of D d,s have the same invertibility. Figure 3 gives various examples. The intuition is that elements of Dd,s are composed of to diagonals that differ by d columns. There are to extra bits flipped in the matrix, hich are s ros apart and adjacent to different diagonals. D 7 3,,0 D 7 3,, D 7 3,,6 D 7 1,3,0 D 7 1,3,3 Figure 3: Various examples of matrices in D d,s. The definition of Sd,s is similar, except the to extra bits that are flipped are adjacent to the same diagonal. As ith Dd,s, e define a base element S d,s,0 as follos: Start ith S d,s,0 = I + I d. Set Sd,s,0 [0, 1] = S d,s,0 [0, 1] 1. Set S d,s,0 [s, s 1 ] = S d,s,0 [s, s 1 ] 1. As ith Dd,s, there are elements of S d,s, denoted S d,s,0,..., S d,s, 1. S d,s,i is equal to S d,s,0 shifted i ros don and i columns to the right. Note that hen is even, there are only / distinct elements of S d,s, because S d,s,i is equal to S We give examples of S d,s,i+ d,s. in Figure 4. 4 Simple Relationships on D d,s and S d,s that Preserve Invertibility We use the folloing relationships on D d,s and S d,s. Lemma 1 Dd,s is invertible iff D d, s is invertible. 3

4 S 7 3,,0 S 7 3,, S 7 1,3,0 S 7 6,3,0 S 6,3,0 = S 6,3,3 Figure 4: Various examples of matrices in S d,s. D 11 3,4,0 D 11 8,7,0 Figure 5: D8,7,0 11 is constructed from D11 3,4,0 by shifting it four ros up and (4 + 3) ros to the left. Proof: D d, s,0 can be derived by shifting D d,s,0 s ros up and s + d columns left. Figure 5 demonstrates Lemma 1. Lemma Sd,s is invertible iff S d, s is invertible. Proof: S d,s, s is identical to S d, s,0. Lemma 3 For s > 1, Dd,s is invertible iff S d,s is invertible. Proof: Sd,s,0 can be constructed from D d,s,0 by substituting ro s ith ro s plus ro (s 1). Figure 6 demonstrates Lemma 3. D 11 3,4,0 S 11 3,4,0 Figure 6: S 11 3,4,0 is constructed from D 11 3,4,0 by substituting ro 4 ith ro 4 plus ro 3. Lemma 4 For 0 < s < 1, Sd,s is invertible iff S d,s is invertible. Proof: S d,s,0 can be constructed from S d,s,0 by first substituting ro s ith ro s plus ro (s 1) and ro 0 ith ro 0 plus ro 1, and then shifting the result d columns to the left. 4

5 (a) (b) (c) (d) S3,4,0 11 S8,4,0 11 Figure 7: (b) is created by substituting ro 4 ith ro 4 plus ro 3. (c) is created by substituting ro 0 ith ro 0 plus ro 10. (d) is created by shifting left three columns. Lemma 4 is demonstrated by Figure 7, here each step of converting S 11 3,4,0 to S 11 8,4,0 is shon. The constraints on s in Lemmas 3 and 4 are due to the folloing. When s = 1, adding ros 0 and 1 does not have the desired effect of moving ro s s one from one diagonal to the other, because ro 0 has three ones. When s = 1, adding ros 0 and 1 has the same problem. For convenience in the sequel, e consider invertibility to be an equivalence relation, so to matrices or matrix classes are equivalent iff they are both invertible or both not invertible. 5 Our Target Class of Matrices, L, and the Grand Liberation Theorem We define the class L to be the union of all Dd,s such that: > 1 is odd. GCD(d, ) = 1. If d is even, s = d. If d is odd, s = d. Theorem 5 (The Grand Liberation Theorem) All matrices in L are invertible. The rest of this paper proves the theorem. After demonstrating a fe special cases, hich include D 3 1,1, the proof proceeds as follos: 1. We prove by induction that D, 1 is invertible for all odd.. For d >, e first sho that for any odd d there exists some even d such that D d, d Hence e restrict our attention to even d >. is equivalent to D. d, d 3. We sho that for any even d >, D d, d is equivalent to S. d, d 4. We sho that the derived S d, d is equivalent to some S d, ith < <, even, and GCD(, d ) = We sho that any Sd, <. ith even > and GCD(, d) = 1 is equivalent to some D d,s L such that 6. A second inductive argument completes the proof, as e can iterate Steps 5 until = 3 or d = in Step 5. 5

6 5.1 Step 1: Base Cases for the Global Induction First, there are only to Dd,s 3 L: D3 1,1 and D,. 3 Their base elements are shon in Figure 8(a) and (b). It is easy to verify that they are invertible. Additionally, Figure 8(c) shos D,4,4 5, hich ill be used belo. It is also easy to verify that it is invertible. (a) (b) (c) (d) (e) D1,1,0 3 D,,0 3 D,4,4 5 D,10,10 11 D,10,10 11 after to steps of Gaussian Elimination. Figure 8: Base cases for the inductive proof. We no prove that D, 1 is invertible for all odd. We have already shon in Figure 8 that this is true for = 3 and = 5. Let > 5 be odd, and assume by induction that D, 1 is invertible for odd 1 < <. Consider D, 1, 1. An example is D,10,10, 11 depicted in Figure 8(d). This matrix has a very specific format: All elements of I and I are set to one, as are D, 1, 1 [ 1][ ] and D, 1, 1 [ ][ 1]. No, perform to steps of Gaussian Elimination. This ill set D, 1, 1 [, 0] and D, 1, 1 [ 1, 1] to zero, and D, 1, 1[, ] and D, 1, 1[ 1, 3] to one. Figure 8(e) demonstrates for = 11. The resulting matrix s first to columns are unit upper triangular, so the first to ros and columns may be deleted. This yields D, 3, 3, hich is of the form D, 1, 1 for some odd 1 < <. By induction, D, 1, 1 is invertible. Therefore, D, 1 is invertible for all odd > Steps 4: Reducing the Problem to S d, for Even, GCD(, d) = 1 No consider any D d, d L such that d is odd. By Lemma 1, this is equivalent to D. Since d d, d L. Therefore, every element Dd,s L for hich d is odd has a corresponding is an even number, D d, d element Dd,s L for hich d is even. Thus e need only prove that the elements D L ith even d are d, d invertible. We proved above that D, 1 is invertible, so e no prove that D is invertible for even d >. d, d Therefore, consider D such that d > is even, > 3 is odd, and GCD(, d) = 1. Since d >, it follos d, d that 1 = 1 d. Since > 3, the smallest value that d 5 1 may be is =. Therefore, by Lemma 3, D is equivalent to S, hich by Lemma is equivalent to S = d, d d, d d, ( d ) S. d, d So no consider S 17 d, d, d 1. An example is S6,3,13, depicted in Figure 9(a). Suppose > d. ( ill not equal d, because GCD(, d) = 1.) Perform d steps of Gaussian Elimination on S d, d, d 1. This moves the ones in ros d through 1 from columns 0 through d 1 to columns d through d 1. In our example of S 6,3,13, 17 six steps of Gaussian Elimination are shon in Figure 9(b). Therefore, hen e delete the first d ros and columns of the resulting matrix, e are left ith S d d, d, d d 1. Note: d is odd; d > d; and since GCD(, d) = 1, GCD( d, d) = 1. Our example continues in Figure 9(c), here e delete the first six ros and columns of Figure 9(b) to get S6,3,7 11. Iterate this process until it yields S for d < < d. We no perform ( d) steps of Gaussian d, d, d 1 Elimination. This moves the leftmost ones in ros ( d) through (( d) 1) over d columns to the right. When e delete the first d ros and columns, e are left ith S d = d ( d), d, d 1 Sd d, d, d 1. Since GCD(, d) = 1, GCD(d, d ) = 1 as ell. 6

7 (a) (b) (c) (d) (e) S6,3,13 17 Six Elimination Steps S6,3,7 11 Five Elimination Steps S1,3, 6 Figure 9: An example of converting S d, d to Sd x, d for = 17 and d = 6. Figure 9(d) shos 11 6 = 5 steps of Gaussian Elimination of S 11 6,3,7, and Figure 9(e) shos that S 6 1,3, results hen e delete the first five ros and columns from Figure 9(d). We have thus reduced the original problem to the folloing: Given S ith even > and GCD(, d ) = 1, d, invertible. We address this in the next section. determine hether S d, 5.3 Steps 5 6: Proving that S d, is Invertible for Even, GCD(, d) = 1 Since >, it follos that 1 < < 1. Therefore, by Lemma 4, S d, is equivalent to S d,, so e may assume that d >. We re going to break this proof into to cases. The first is hen d > + 1. Consider S d,, 1. An example of this is S11,8,7 16 displayed in Figure 10(a). We perform d steps of Gaussian Elimination on S d,, 1. Since d > + 1, e kno that d < 1, so the d steps of Gaussian Elimination simply move the leftmost ones in ros ( d) through (( d) 1) over d columns to the right. Deleting the first d ros and columns from the matrix, e are left ith Sd, d,d These steps are shon in Figures 10(b) and (c), as S 11,8,7 is converted into into S6,8, 11. (a) (b) (c) S11,8, Elimination Steps S6,8, 11 Figure 10: An example of converting S d, to Sd d, for = 16 and d = 11. As before, since GCD(, d) = 1, e kno that GCD(d, d) = 1. That is even implies that d is also even. Moreover, since d > + 1, e kno that d > 1. Therefore, by Lemma 3, S d, d is equivalent to Dd, d. Finally: d d = =. d d + 7

8 Therefore, Dd, d = D d, hich is an element of L. By induction, D d is invertible, implying that Sd, is invertible. d,d d d,d d The second case is for Sd, hen d = + 1 and GCD(, ) = 1. An example is S9,8,7 shon in Figure 11(a). Again, e ill perform d elimination steps. We ill do this in to parts, hoever. In the first part, e perform d 1 elimination steps. This moves the leftmost ones in ros ( d) through (( d) ) over d columns to the right. This is pictured in Figure 11(b). The last elimination step replaces to ros of the matrix, because ro ( d) has an extra one adjacent to the diagonal. Therefore, both ros ( d) and (( d) 1) move their leftmost ones into the column ( 1). This is pictured in Figure 11(c). (a) (b) (c) (d) S9,8, Elimination Steps 1 More S,8,0 9 Figure 11: An example of converting S d, to Sd, d 1 hen d = + 1 for = 16. When the first d ros and columns are deleted from the matrix, e are left ith S d d,,0. Since d = + 1, this is equal to S,d 1,0 d (shon as S9,8,0 in Figure 11(d)). That > is even implies that d = + 1 > is odd, so d 1 > 1; thus by Lemma 3, S,d 1 d is equivalent to Dd,d 1, hich e proved invertible in Section 5.1. Therefore Sd, is invertible. Q.E.D. 6 The Class O and the Little Liberation Theorem We no define a third class of matrices, Od such that > d 1. We define O d,0 to be the base element of O d and construct Od,0 as follos: Start ith O d = I + I d. Set Od,0 [0, 1] = O d,0 [0, 1] 1. Thus, Od,0 is similar to D d,s,0 and S d,s,0, except it only has one extra one in it, in the top-right corner. There are elements of Od, denoted O d,0,..., O d, 1. O d,i is equal to O d,0 shifted i ros don and i columns to the right. Therefore, all elements of Od are equivalent. We sho some examples of matrices in O d in Figure 1. (a) (b) (c) (d) O3,0 7 O3,6 7 O4,6 7 O1,1 Figure 1: Examples of matrices in O d. We start ith a simple lemma: 8

9 Lemma 6 Od is invertible iff O d is invertible. Proof: O d, 1 can be derived by replacing ro 1 of O d, 1 ith the sum of ros 1 and, and shifting the resulting matrix d columns to the left. An example is in Figure 1, here O 4,6 7 may be obtained by replacing ro 6 of O3,6 7 ith the sum of ros 5 and 6, and shifting the result three columns to the left. Define O to be the union of all O d such that GCD(, d) = 1. Theorem 7 (The Little Liberation Theorem) All matrices in O are invertible. Proof: This proof is far simpler than that of the Grand Liberation Theorem. It, too, is inductive. We start ith the base case O1, an element of hich is pictured in Figure 1(d). This matrix is already in unit upper triangular form and is therefore invertible. No, consider Od O and suppose by induction that Od O is invertible for all 1 d < <. By Lemma 6 and the hypothesis that GCD(, d) = 1, e may assume that d <, or else e consider O d in lieu of Od. Performing d elimination steps on O d, 1 moves the leftmost ones in ros d through 1 over d columns to the right. Since d <, these ones ill not be moved to the diagonal, nor ill the one at O d, 1 [ 1, ] be affected. Therefore deleting the leftmost d columns, hich are no unit upper triangular, and top d ros leaves O d d, d 1. Since GCD(, d) = 1, GCD( d, d) = 1, and therefore O d d, d 1 O. By induction, O d d is invertible; therefore Od is invertible. An example is depicted in Figure 13 here O 1 5,11 is converted to O 7 5,6 by five elimination steps. (a) (b) (c) O5,11 1 Five elimination steps O5,6 7 Figure 13: An example of converting O d, 1 to O d d, d 1. 7 Some Trivial Properties of Matrices in GF() The folloing to lemmas are likely folklore, but e include them for completeness. Lemma 8 If some matrix M has precisely ones, then M is invertible iff it is a permutation matrix. Proof: Any permutation matrix is invertible. Conversely, if M has precisely ones but is not a permutation matrix, then some ro or column contains all zeros, in hich case M is not invertible. Lemma 9 Let M 1 and M be permutation matrices. The sum M 1 + M is not invertible. Proof: Let M = M1 + M. Suppose there exist r and c such that M1 [r, c] = M [r, c] = 1. Then ro r of M contains all zeros, so M is not invertible. Thus, e assume there are no such r and c; in this case, M has precisely to ones in each ro and column. We prove by induction that such matrices are not invertible. The base case is shon in Figure 14(a), hich depicts the only M ith to ones in each ro and column. This matrix is clearly not invertible. 9

10 No, let matrix M for some > have exactly to ones in each ro and column. Let ros r 1 and r be the to ros that have ones in column zero, and let c 1, c > 0 be such that M [r 1, c 1 ] = M [r, c ] = 1. Sap ro r 1 ith ro 0, and perform one elimination step. This ill set M [r, 0] = 0 and M [r, c 1 ] = M [r, c 1 ] 1. If c 1 = c, then all of ro r s elements become zero, so M is not invertible. If c 1 c, then deleting the first ro and column leaves a matrix M 1 ith exactly to ones in each ro and column. By induction, this ne matrix in not invertible; therefore M is also not invertible. (a) (b) (c) (d) (e) (f) M. M 7, After one elimination M 7, After one M 6. c 1 = c = 3. step, Ro 4 is all zeros. c 1 = 3, c =. elimination step. Figure 14: Matrices ith to ones in each ro and column. We sho examples of the elimination step in Figure 14(b)-(f). In Figure 14(b), the elimination step, depicted in Figure 14(c) turns ro 4 into all zeros. In Figure 14(d), the elimination step of the matrix M 7 results in Figure 14(e), hich is equivalent to a matrix M 6 (Figure 14(f)). 8 A Final Theorem on the Invertibility of a Type of (k + ) k Matrix Let ro r of some matrix M contain precisely one one e call such a ro an identity ro and let the one be in column c. By cofactor expansion, deleting ro r and column c yields an equivalent matrix M 1. The remaining theorem concerns a (k + ) k block matrix A, structured as follos and pictured in Figure 15: Each block is. Block A[i, i] = I for 0 i < k. Blocks A[i, j] = A[j, i] = 0 for 0 i < j < k. Block A[k, j] = I for 0 j < k. Block A[k + 1, j] = X j for 0 j < k and some given X j. Consider the class A of ( ) k+ block matrices induced by deleting any to ros of blocks from A. Theorem 10 All matrices in A are invertible iff (1) every X i is invertible, and () for 0 i < j < k, X i + X j is invertible. Proof: Let A A. There are four cases. Case 1: A is composed of the first k ros of blocks of A, hich form I k. Case : A is composed of ro k and any k 1 of the first k ros of blocks of A. No, A has (k 1) identity ros. Deleting these ros and their associated columns yields I. Case 3: A is composed of ro k + 1 and any k 1 of the first k ros of blocks of A; let i be the omitted ro of blocks from the first k. Again, A has (k 1) identity ros, hich e can delete ith their associated columns to yield X i, so A is equivalent to X i. 10

11 Figure 15: The (k + ) k block matrix A of matrices over GF(). Figure 16: The matrix that results hen (k ) identity ros are deleted from A in Case 4. Case 4: A is composed of ros k, k + 1, and any k of the first k ros of blocks of A; let i and j be the omitted ros of blocks from the first k. No A has (k ) identity ros, hich e can delete ith their associated columns to yield the matrix pictured in Figure 16. No, perform elimination steps on this matrix. For each r and c such that X i [r, c] = 1, the elimination step for column c ill replace ro +r ith the sum of ros +r and c. This ill set X i [r, c] = 0 and X j [r, c] = X j [r, c] 1. After the elimination steps, the leftmost columns ill be upper triangular, and deleting them leaves X i + X j. Therefore, A is equivalent to X i + X j. 9 Acknoledgements This material is based upon ork supported by the National Science Foundation under grants CNS and CNS References [1] M. Blaum, J. Brady, J. Bruck, and J. Menon. EVENODD: An efficient scheme for tolerating double disk failures in RAID architectures. IEEE Transactions on Computing, 44():19 0, February [] J. Blomer, M. Kalfane, M. Karpinski, R. Karp, M. Luby, and D. Zuckerman. An XOR-based erasure-resilient coding scheme. Technical Report TR , International Computer Science Institute, August [3] P. Corbett, B. English, A. Goel, T. Grcanac, S. Kleiman, J. Leong, and S. Sankar. Ro diagonal parity for double disk failure correction. In 4th Usenix Conference on File and Storage Technologies, San Francisco, CA, March

12 [4] J. L. Hafner. WEAVER Codes: Highly fault tolerant erasure codes for storage systems. In FAST-005: 4th Usenix Conference on File and Storage Technologies, pages 11 4, San Francisco, December 005. [5] J. L. Hafner. HoVer erasure codes for disk arrays. In DSN-006: The International Conference on Dependable Systems and Netorks, Philadelphia, June 006. IEEE. [6] P. Lancaster and M. Tismenetsky. The Theory of Matrices. Computer Science and Applied Mathematics. Academic Press, San Diego, CA, second edition, [7] J. S. Plank. A tutorial on Reed-Solomon coding for fault-tolerance in RAID-like systems. Softare Practice & Experience, 7(9): , September [8] J. S. Plank and L. Xu. Optimizing Cauchy Reed-Solomon codes for fault-tolerant netork storage applications. In NCA-06: 5th IEEE International Symposium on Netork Computing Applications, Cambridge, MA, July 006. [9] J. J. Wylie and R. Saminathan. Determining fault tolerance of XOR-based erasure codes efficiently. In DSN- 007: The International Conference on Dependable Systems and Netorks, Edinburgh, Scotland, June 007. IEEE. 1

A New Minimum Density RAID-6 Code with a Word Size of Eight. James S. Plank

A New Minimum Density RAID-6 Code with a Word Size of Eight James S. Plank Department of Electrical Engineering and Computer Science University of Tennessee Knoxville, TN 37996 plank@cs.utk.edu Appearing