### Table 3. Largest-volume empty tetrahedra in three-dimensional quasi-random point sets. ~ q stands for the sequence of bases (q1; q2; : : : ). The Hammersley and Sobol sequences seem to be best, and they beat the Halton

"... In PAGE 14: ... The largest empty triangles were usually unique, except for a few cases with small numbers of points. Table3 presents the analogous results for three dimensions. We computed the largest empty tetrahedron with vertices from the given point set, for the same reasons as in the planar case.... ..."

### Table 1: Comparison of Monte Carlo and quasi-Monte Carlo methods used to value a coupon bond

1998

"... In PAGE 21: ... For random Monte Carlo, the constant c is the standard deviation, and = :5. Table1 summarizes the results. For each method, the estimated size of the error at N = 10000 (based on the linear t), the convergence rate , and the approximate computation time for one run with this N are given.... In PAGE 25: ... Figure 2 displays these results in terms of the estimated computation time. In Table1 it can be seen that there is in fact a computational advantage to using quasi-random sequences over random for this problem. This is due to the time required for sequence generation.... ..."

Cited by 15

### Table 1: Convergence results for fractional absorptivity method using the Halton se- quence.

1994

"... In PAGE 15: ... A second consequence is that the improvement gained by using a quasi-random sequence is somewhat less pronounced. Table1 summarizes the results of the calculations using the fractional absorptivity method for several values of absorptivity, as well as for a mixed absorptivity case in which the walls are highly re ective (A = 0:1), while the wafer is highly absorbing (A = 0:7). The mixed case is more typical of an actual reactor.... ..."

Cited by 9

### Table 1. Largest-area empty triangles in quasi-random point sets We now discuss the results of table 1. We see that for the Hammersley sequence the area of the largest triangle decreases by a factor of q when N increases from an odd power of q to the subsequent even power (except for small N). In table 1, the corresponding entries are equal. On the other hand, when N increases from an even power of q to the subsequent odd power, the area of the largest empty triangle remains almost unchanged. (The corresponding entries in the table are approximately multiplied by q.) We conjecture that this is true in general. The data seem to support the hypothesis that the area of the largest empty triangle for the Hammersley sequence with N = 3n and N = 2n points is asymptotically equal to (4=9) 3?bn=2c and (3=8) 2?bn=2c, respectively. This would be in accordance with Theorem 3.2.

### Table 1. Average generalization ability and flnal training error (with standard de- viations between parenthesis) obtained for the network architectures constructed to compute all Boolean quasi-random functions with N=8 inputs.

### Table 2: Convergence results for discrete absorptivity method using the Halton sequence.

1994

"... In PAGE 15: ... Again, the advantage is more prominent for the directly visible surface 1. Figure 3 and Table2 provide a similar comparison of the Halton and pseudo-random sequences for the discrete absorption method. As mentioned above, the integrand being evaluated in this method has a larger variance than the integrand associated with the fractional absorption method, and therefore considerably more rays are need to obtain the same degree of accuracy.... In PAGE 15: ...4, 99 per cent of all rays have been absorbed after nine re ections. Nevertheless, this dimensional e ect on the quasi-random approach can be seen in Figure 3 and Table2 . The factor of improvement for Halton over random is noticeably smaller than for the fractional case, although signi cant gains are still made by using Halton.... ..."

Cited by 9

### Table 1 L1-error Comparisons

"... In PAGE 3: ... The number of the quasi-random numbers is M = 2000. From Table1 , we can see that the two schemes do not have significant differences. Table 1 L1-error Comparisons ... ..."

### Table 5 are superior to those of Table 3.

"... In PAGE 4: ... The loss amounts are then assigned to an in- dividual trial to produce the distribution of loss amounts summarized in Table 5. Table5 : Distribution of Loss Amounts using Quasi- random Numbers to Generate the Severity of Loss Percentile Point 0 $ 0 10 0 25 67,749 50 227,986 75 467,540 90 753,417 100 2,969,518 Because the quasi-random numbers were superior in our previous comparisons, we suspect that the results of Table 5 are superior to those of Table 3. ... ..."

### Table 1: Irradiance matrix of a patch for D = 3

"... In PAGE 6: ... Now each patch is represented by a triangle matrix I, where the (i; j) element stores the sum of those i-bounce irradiances where the last direction is !j. Table1 shows an example for D = 3. The complete combined algorithm is shown below: for m = 1 to M do Generate (!(m) 1 ;!(m) 2 ;:::;!(m) D ) for d = 0 to D ? 1 do // quasi-random walk I[1][D ? d] = A(!0 D?d) Le(!0 D?d) for b = 2 to d + 1 do I[b][D ? d] = 0 for pd = b ? 1 to d ? 1 do I[b][D ? d] += 4 A(!(m) D?d) F(!(m) D?pd;!(m) D?d) I[b ? 1][D ? pd] endfor endfor endfor Divide the b-bounce estimates (I[b][d]) by ?Db Calculate the image estimate from the I[b][d] estimates Divide the estimate by M and add to the Image endfor Display Image Furthermore, when the radiance is transferred to a direc- tion, the required information to transfer the radiance to the opposite direction is also available, that is when computing geometry matrix A(!0) for some direction, the matrix for the reverse direction A(?!0) is usually also known paying very little or no additional effort.... ..."

### Table 1: Irradiance matrix of a patch for D = 3

"... In PAGE 6: ... Now each patch is represented by a triangle matrix I, where the (i; j) element stores the sum of those i-bounce irradiances where the last direction is !j. Table1 shows an example for D = 3. The complete combined algorithm is shown below: for m = 1 to M do Generate (!(m) 1 ;!(m) 2 ;:::;!(m) D ) for d = 0 to D ? 1 do // quasi-random walk I[1][D ? d] = A(!0 D?d) Le(!0 D?d) for b = 2 to d + 1 do I[b][D ? d] = 0 for pd = b ? 1 to d ? 1 do I[b][D ? d] += 4 A(!(m) D?d) F(!(m) D?pd;!(m) D?d) I[b ? 1][D ? pd] endfor endfor endfor Divide the b-bounce estimates (I[b][d]) by ?Db Calculate the image estimate from the I[b][d] estimates Divide the estimate by M and add to the Image endfor Display Image Furthermore, when the radiance is transferred to a direc- tion, the required information to transfer the radiance to the opposite direction is also available, that is when computing geometry matrix A(!0) for some direction, the matrix for the reverse direction A(?!0) is usually also known paying very little or no additional effort.... ..."