million

For this case, two level factorial design would require 1048576 points and the CCD would require 1048617 points.

> X:={seq(ithprime(2*i),i=1..20)}; X := {3, 7, 13, 19, 29, 37, 43, 53, 61, 71, 79, 89, 101, 107, 113, 131, 139, 151, 163, 173} > N:=sum (X[ithprime(i)],i=1..8); N := 574 > count:=0: subsetsum(X,{},N);count; {3, 7, 13, 19, 29, 37, 43, 53, 61, 71, 107, 131} true 223 > N:=sum (X[i],i=1..8); N := 204 > count:=0: subsetsum(X,{},N);count; {3, 7, 13, 19, 29, 37, 43, 53} true 9 > N:=sum (X[i],i=11..18); N := 910 > count:=0: subsetsum(X,{},N);count; {3, 7, 13, 19, 29, 37, 43, 53, 61, 71, 107, 131, 163, 173} true 27803 Given that the number of subsets to be searched through using an exhaustive search is [22.sup.20] = 1048576, we see that backtracking can provide a better alternative.

FIGURE 3 The results of the penny-doubling problem A B 1 1 2 2 3 4 4 8 5 16 6 32 7 64 8 128 9 256 10 512 11 1024 12 2048 13 4096 14 8192 15 16384 16 32768 17 65535 18 131072 19 262144 20 524288 21 1048576 22 2097152 23 4194304 24 8388608 25 16777216 26 33554432 27 67108864 28 134217728 29 268435456 30 536870912 31 1073741824 FIGURE 4 The sum of the penny doubling A B C 1 1 1 2 2 3 3 4 7 4 8 15 5 16 31 6 32 63 7 64 127 8 128 255 9 256 511 10 512 1023 The students wanted to put "penny doubling" into a spreadsheet.

[n.sub.x] [beta] = 1.3 [beta] = 1.7 [beta] = 1.3 [beta] = 1.7 32768 6.0 (0.43) 7.0 (0.47) 6.8 (0.90) 6.0 (0.81) 65536 6.0 (0.96) 7.0 (0.97) 6.4 (1.93) 6.0 (1.75) 131072 6.0 (1.85) 7.0 (2.23) 5.9 (3.93) 5.9 (3.89) 262144 6.0 (7.10) 7.1 (8.04) 5.4 (12.78) 6.0 (13.52) 524288 6.0 (15.42) 7.8 (19.16) 5.1 (25.71) 6.0 (27.40) 1048576 6.0 (34.81) 8.0 (41.76) 4.9 (51.02) 6.3 (62.57) TABLE 5.3 The solver IKPIK is presented, for a variety of meshes and two different values of [beta] in both the constant and the variable coefficient cases.

(1) Long keystream data set test: it begins by generating [2.sup.20] bits for each of the 300 random CK (set IV to zero in all the cases) to obtain 300 vectors of 1048576 bits.