### Table 6: Records for the rst implementation GF(2155) GF(2196) GF(2300) GF(2400) GF(2500) GF(2601)

"... In PAGE 9: ... For these elds and for larger elds (the last one being the current record, as of February 1995), our implementation gave the following timings, for the curve: EX : y2 + xy = x3 + T16 + T14 + T13 + T9 + T8 + T7 + T6 + T5 + T4 + T3 (the coe cient was chosen as the binary expression of 91128 { our zip code { converted to a polynomial if GF(2n)). Table6 corresponds to the rst version of our implementation, using an... ..."

### Table 1: Computation times for GF (210) and GF (54). GF (210) GF (54)

1992

"... In PAGE 39: ... An extension GF (54) of GF (5) via the primitive and normal polynomial X4 + 4X3 + X + 2 over GF (5). Table1 shows the computation times for the three di erent representations of both elds. 12.... ..."

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### Table 8: Comparisons with GF(p) GF(2155 + 15) GF(2196 + 21) GF(2300 + 157)

"... In PAGE 10: ...lgorithms for doing series computations. The comparison for the case GF(2300) is striking. More details will be given in [27]. We also add Table8 which gives timings for elds GF(p) with p a large prime. It seems that the program for GF(2n) is slower than the program for large prime characteristic around n = 150.... ..."

### Table 4: Records for the second implementation GF(2155) GF(2196) GF(2300) GF(2701)

1995

"... In PAGE 31: ... But for all of these, we did not use equations (32), but rather the corresponding series. Table4 refers to the implementation that uses all features described in the present article. The comparison for the case GF(2300) is striking.... ..."

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### Table 3: Records for the rst implementation GF(2300) GF(2400) GF(2500) GF(2601)

1995

"... In PAGE 31: ...GF(265) GF(289) GF(2105) E = 2; C = 2; S = 0 E = 3; C = 4; S = 0 E = 3; C = 4; S = 24 min max avg min max avg min max avg `max 29 29 29 37 41 39 43 43 43 #U 1 4 2 1 6 3 4 6 5 #L 6 9 7 6 12 9 8 10 9 M 103 3:7105 5:8104 7:7102 2:8107 2:4106 1:5105 7:1108 6:6107 Xq 3:8 4:0 3:9 10:2 14:9 12:5 22:4 24:8 23:3 Xqr 1:3 3:2 2:2 3:8 12:0 8:1 11:5 18:3 14:7 Schoof 0:0 0:0 0:0 0:0 0:0 0:0 0:0 30:0 12:9 g 0:0 1:1 0:4 0:0 2:5 1:0 0:0 2:4 1:0 k 0:1 0:2 0:1 0:3 0:6 0:5 0:3 0:8 0:6 M-S 0:1 1:7 1:1 0:2 5:9 2:5 0:5 18:8 5:7 Total 6:1 8:8 7:7 17:9 32:2 24:6 43:0 73:9 58:1 the following timings, for the curve: EX : y2 + xy = x3 + T16 + T14 + T13 + T9 + T8 + T7 + T6 + T5 + T4 + T3 (the coe cient was chosen as the binary expression of 91128 { our zip code { converted to a polynomial if GF(2n)). Table3 corresponds to the rst version of our implementation, which used the backtrack approach. The cases n = 300 and n = 400 were done with a poor implementation of the treatment of Atkin primes; moreover no power of small primes were used.... ..."

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### Table 1: Logarithm tables for GF(16) For example, in GF(16):

1997

"... In PAGE 9: ... Then n = 3, and m = 3. We choose w to be four, since 2w gt; n + m, and since we can use the logarithm tables in Table1 to illustrate multiplication. Next, we set up gflog and gfilog to be as in Table 1.... In PAGE 9: ...egabyte. Then n = 3, and m = 3. We choose w to be four, since 2w gt; n + m, and since we can use the logarithm tables in Table 1 to illustrate multiplication. Next, we set up gflog and gfilog to be as in Table1 . We construct F to be a 3 3 Vandermonde matrix, de ned over GF(16): F = 2 6 6 4 10 20 30 11 21 32 12 22 32 3 7 7 5 = 2 6 6 4 1 1 1 1 2 3 1 4 5 3 7 7 5 Now, we can calculate each word of each checksum device using FD = C.... In PAGE 16: ... This is done in Table 2 below for GF(16), where q(x) = x4 + x + 1. It should be clear how this enumeration can be used to generate the gflog and gfilog arrays in Table1 . The C code in Figure 6 shows how to generate these arrays for w = 16:... ..."

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### Table 2: Logarithms for GF (256)

2003

"... In PAGE 6: ...able 1. ICONS data types versus management requirements................................................................................ 8 Table2 : Logarithms for GF (256).... ..."

### Table 1: Elements in GF(16).

"... In PAGE 5: ...n GF(16) such that all xkxi for 0 k 2 and 1 i 5 are distinct (Theorem 3.2). Clearly, we need at least 3 5 = 15 distinct nonzero numbers and GF(16) is (barely) adequate for the purposes of our construction. For completeness, the elements of GF(16) are shown in Table1 , including their power, polynomial vector and decimal representation. Choosing x1 = a, x2 = a2, x3 = a3, x4 = a4, x5 = a5 and x = a5 guarantees that the conditions of Theorem 3.... ..."

### Table 1: Elements in GF(16).

2004

"... In PAGE 5: ...n GF(16) such that all xkxi for 0 k 2 and 1 i 5 are distinct (Theorem 3.2). Clearly, we need at least 3 5 = 15 distinct nonzero numbers and GF(16) is (barely) adequate for the purposes of our construction. For completeness, the elements of GF(16) are shown in Table1 , including their power, polynomial vector and decimal representation. Choosing x1 = a, x2 = a2, x3 = a3, x4 = a4, x5 = a5 and x = a5 guarantees that the conditions of Theorem 3.... ..."