U.U.D.M. Project Report 2010:4
Examensarbete i matematik, 15 hp
Handledare och examinator: Sven Erick Alm April 2010
Department of Mathematics
Cover times of random walks on graphs
1. Introduction ... 3
2. Definitions and Notation ... 3
3. Exact results ... 4
The Complete graph ... 4
The Linear infinite graph ... 6
The Cyclic graph ... 7
4. Asymptotic results ... 8
5. The calculations ... 13
6. Results ... 16
References ... 21
In this thesis I study a simple random walk on a graph and am interested in the cover time, i.e. the time until all vertices of the graph have been visited at least once. The purpose of this thesis is to give a better picture of some already stated and proved results for cover times. The major part of the thesis focuses on the complete graph because of the close relation with the famous Dixie cup problem, and in the case of multiple covers with the double Dixie cup problem. There is no exact result on multiple covers for the complete graph and there will probably not be one. With some recursive equations I calculate the exact result for multiple covers of the complete graph for a given number of vertices. I am trying to answer questions like how does the cover time on the complete graph behave for big n and in particular for the multiple cover of the same graph. By comparison with the exact result given by the recursive equations I am analyzing the behavior of the multiple cover of the complete graph.
For a cover time on a given finite graph, it is always possible to calculate the expected cover time using the general method, that is solving a system of linear equations of what happens in each steps, conditioning on the previous step until we are done. But the method is quite bad because there will be lots of equations and unknowns to calculate. Blom and Sandell (1992) has quoted an example from an issue of the American Mathematical monthly (1989) where there is a random walk taking place on the eight corners of a cube, the expected number of steps to visit all the edges at least once will require the solution of an equation system with 387 unknowns. Here follows some definitions and notation for getting a better picture of cover times.
2. Definitions and Notation
Random walk on a graph: Assume we have a graph G with n vertices and e edges. Let d(v) be the number of neighbors to vertex v. If we start at vertex v at a given time point, in the next step we move to a neighbor vertex with probability 𝑑(𝑣)1 for each neighbor, and this continues for as many steps needed.
Random cover time: A random variable for the number of steps needed to cover a given graph G.
Cover time 𝑚𝑖,𝑛 : The expected number of steps it takes for the random walk to visit every vertex in a given graph, G. Here i is the starting vertex, and n is the number of vertices. Henceforth when I am talking about cover time I mean the expected cover time.
Extended cover time, 𝜇𝑖,𝑛 : The expected number of steps it takes the random walk to visit every vertex in a given graph and then return to the starting vertex. Here i is the starting vertex, and n is the number of vertices.
Extended partial cover time, 𝜇𝑖,𝑘 : The expected number of steps it takes the random walk to visit k vertices in a graph with n vertices and then return to the starting vertex. Here i is the starting vertex.
Multiple cover time : The expected number of steps it takes the random walk to visit all n vertices at least m times.
Eulers constant, 𝛾: Is the limit of the series 1 +1 2+
3+ ⋯ + 1
𝑛 − ln n , when 𝑛 → ∞. An approximate value is 0.5772.
3. Exact results
The Complete graph
Figure 1: Complete graph with 4 vertices.
A complete graph with n vertices is denoted 𝐾𝑛 and has 𝑛 𝑛 −1
2 edges . In the complete graph every pair of vertices is connected by an edge, as displayed in Figure 1. Regardless of which vertex the particle is located at, in the next step the particle can go to any of the other vertices.
The cover time for the complete graph with n edges is given by 𝑚𝑛 = 𝑛 − 1 1 +
2+ ⋯ + 1
𝑛−1 . (1)
Consider the coupon collector’s problem, which means that there exist 𝑛 bonus pictures, one with each package of food, and the question is how many packages you need to buy to own a complete set of pictures.
For the case of simple cover time the coupon collector’s problem is the same as the complete graph with loops.
Solution to the coupon collector’s problem (From Blom, Holst and Sandell (1994)). Write 𝑇 = 𝑌1+ 𝑌2+ 𝑌3+ ⋯ + 𝑌𝑛 ,
where 𝑌𝑖 is the number of packages a person has to buy to increase his collection from 𝑘 − 1 to 𝑘 different pictures.
𝐸 𝑇 = 𝐸 𝑌1+ 𝑌2+ 𝑌3 + ⋯ + 𝑌𝑛 = 𝐸 𝑌1 + 𝐸 𝑌2 + ⋯ + 𝐸(𝑌𝑛).
The probability of acquiring a new picture given that you already have 𝑖 − 1 pictures
Therefore, 𝑌𝑖 has a geometric distribution with
=𝑛 𝑛−𝑖+1 and hence 𝐸 𝑇 = 𝑛 𝑛−𝑖+1 𝑛 𝑖=1 = 𝑛 1 𝑛 + 1 𝑛−1+ ⋯ + 1 1 , which proves the coupon collector’s problem.
The cover time for the complete graph is similar, in the complete graph case you start with one picture (the starting vertex) . Therefore you only need 𝑛 − 1 pictures to have the whole set. The calculations are similar.
The partial cover time for a complete graph is given by 𝑚𝑘 = 𝑛 − 1 1 𝑛 −𝑘+1+ 1 𝑛−𝑘+2+ ⋯ + 1 𝑛 −1 , 𝑘 ≥ 2 , 𝑛 > 𝑘 . Proof: With similar calculations as in the previous situation with
=𝑛 −𝑖+1 𝑛 −1
=𝑛 −1 𝑛−𝑖+1
,𝐸 𝑇 = 𝑛−1 𝑛−𝑖+1 𝑘 𝑖=2 = 𝑛 − 1 1 𝑛−𝑘+1+ 1 𝑛−𝑘+2+ ⋯ + 1 𝑛−1 , which completes the proof.
The Linear infinite graph
Figure 2: Linear infinite graph.
A Linear infinite graph is for example the integer points. A random walk on the linear infinite graph is performed so that with equal probability, 1
2 , the particle either goes left or right. The walk continuous until k points have been visited.
The partial cover time is given by 𝑚0,𝑘 = 1 + 2 + 3 + ⋯ + 𝑘 − 1 =1
2𝑘 𝑘 − 1 = 𝑘
2 , (2)
where 0 is the starting point and k is the number of points. As seen from (2), the cover time is independent of the starting point.
Proof (From Blom & Sandell (1992)):
Consider the time when the random walk has visited 𝑘 different points, and call that time 𝑇𝑘. Then 𝑇𝑘 = 𝑌1+ 𝑌2+ ⋯ + 𝑌𝑘−1,
Let 𝑟 = 𝐸 𝑌𝑟 .
Then 𝑚0,𝑘 = 𝐸 𝑇𝑘 = 1+ 2 + ⋯ + 𝑘−1 .
To find 𝑟 , suppose the walk has visited exactly r points, when that happens for the first time, the particle is either at the leftmost or the rightmost point of one of the r-sets.
𝑅𝑗 = −𝑗, −𝑗 + 1, … ,0, … , 𝑖 − 1, 𝑖 ,
where 𝑖 + 𝑗 + 1 = 𝑟 and 𝑖 ∈ 0,1, … , 𝑟 − 1 . Assume we are in – 𝑗. Renumber the points by changing −𝑗 to 0, −𝑗 + 1 to 1 and so on. We use Fellers (1968) result that the expected time for a particle starting at the origin to be absorbed in 𝑥 = −𝑎 or 𝑥 = 𝑏 is 𝑎𝑏 .
By that result with 𝑎 = 1 𝑎𝑛𝑑 𝑏 = 𝑖 + 𝑗 + 1 we get 𝑟 = 𝑎𝑏 = 1 × 𝑖 + 𝑗 + 1 = 𝑟 ,
which completes the proof.
The Cyclic graph
Figure 3: Cyclic graph with 6 vertices.
The cyclic graph 𝐶𝑛 is a cyclic graph with n vertices. The first vertex is connected to the second vertex which is connected to the third vertex and so on to the final vertex, n which is connected to the first vertex.
The cover time is giver by: 𝑚𝑛 = 1 + 2 + ⋯ + 𝑛 − 1 =
Proof (From Wilf (1989)):
Consider the time point when the walk for the first time has visited exactly 𝑗 vertices, 1 ≤ 𝑗 ≤ 𝑛 − 1. When the walk has visited j vertices for the first time then the walk is at an end point, and these j vertices are contiguous. So the problem is what is the time to visit a new vertex? When we have visited j vertices there is only two vertices to increase the number of visited vertices from j to j+1 and that is the vertices at each end, of a (𝑗 + 2)-path. If we are at point r how long time does it take on average to reach an end point of a (𝑗 + 2)-path for the first time. If we call 𝑍𝑟 the time on average to reach an end point. Then 𝑍𝑟 = 1
2 𝑍𝑟−1+ 1 + 1
2 𝑍𝑟+1+ 1 2 ≤ 𝑟 ≤ 𝑗 + 1 , with end conditions 𝑍1 = 𝑍𝑗 +2 = 0,
and with the solution
𝑍𝑟 = 𝑗 + 3 𝑟 − 𝑟2− (𝑗 + 2) 𝑟 = 1,2, … , 𝑗 + 2 . In particular , 𝑍𝑗 +1 = 𝑗.
Therefore, for each 𝑗 = 1,2, … , 𝑛 − 1 , if we have, for the first time, visited j distinct points on 𝐶𝑛, it takes an average of 𝑗 steps before a new vertex is visited. This means the cover time of 𝐶𝑛 is
1 + 2 + ⋯ + 𝑛 − 1 = 12𝑛(𝑛 − 1),
and because of the symmetry of the graph it is independent of the starting point. This completes the proof.
Extended partial cover times are presented without proof. Theorem (Blom and Sandell (1992))
𝜇𝑘 =1 2𝑘 𝑘 − 1 + 1 6𝑘 3𝑛 − 2𝑘 + 1 = 1 6𝑘(3𝑛 + 𝑘 − 2).
4. Asymptotic results
There is a graph, the lollipop graph, with order of magnitude of 𝑛3 ,or more precisely close to 4
3. Feige (1995) proved that the expected time for a random walk to visit all n vertices of a connected graph is at most 274 𝑛3+ 𝑜 𝑛3 .
The lollipop graph is obtained by joining a complete graph to a path graph, see Figure 4.
Figure 4: Some lollipop graphs.
Figure 5: Lollipop graph with 4 vertices.
10 𝑚314 = 1 +1 2𝑚134, 𝑚213 = 1 +1 2𝑚312 + 1 2𝑚132, 𝑚132 = 1 + 1 3𝑚321 + 1 3𝑚213, 𝑚124 = 1 +1 3𝑚214 + 1 3𝑚412, 𝑚134 = 1 +1 3𝑚413 + 1 3𝑚314, 𝑚413 = 1 + 𝑚134.
Here the first index states the current location and the other indices points already visited, i.e. 𝑚312means that the random walk currently is in vertex 3 and have already visited 1 and 2.
This gives the solution 𝑚1 = 142
15 ~9.467 which is close to 4 274
The bound presented above is general for all graphs, but what is the correct order for the complete graph?
Figure 6: The exact result and the order of magnitude for single cover time of the complete graph.
The title above each plot states what is plotted. In the plot with all plots (bottom right), the bottom lines are the dashed and the dotted lines. Because of the small difference between the dashed and the dotted lines , at most 10.2103, it is hard to see that it is actually two lines. It is not much when the size of the end points is close to 9 × 104. According to the plot the difference between the solid line and the dashed line is increasing with n.
Figure 7: The exact result and the order of magnitude with constant for single cover time of the complete graph.
As seen in Figure 7, now the solid line and the dashed line coincides, and the difference between the end points is now 10.2875 with 10000 vertices and it is not increasing very much either. With 100000 vertices the difference between the exact result and 𝑛 × ln 𝑛 + 𝛾 is 12.5900.
How does the multiple cover time on the complete graph behave for big n? There is not much research done on this subject.
There is some research that is close to the problem. Consider the coupons collector’s problem, but now we want to collect m collections of n pictures. The problem is called the double Dixie cup problem, because the “pictures” appeared on the covers of Dixie cups. In 1960, Donald J. Newman , ,wrote an article, The double Dixie cup problem, and he proved two interesting results which will be presented without proofs.
𝐸𝑚 𝑛 = 𝑛 × 1 − 1 − 𝑚 −1𝑒−𝑡𝑗 !𝑡𝑗 𝑗 =0 𝑛 𝑑𝑡 ∞ 0 ,
The asymptotic result is 𝐸𝑚 𝑛 = 𝑛 × ln 𝑛 + 𝑚 − 1 × ln ln n + 𝐶𝑚 + 𝑜(1) ,
for fixed m as 𝑛 → ∞ . Here 𝐶𝑚 is some constant. In 1961, Erdős & Rényi showed that 𝐶𝑚 = 𝛾 − ln( 𝑚 − 1 !) where 𝛾 is Euler’s constant.
Covering the complete graph is similar to the Dixie cup problem, so there is a good chance that asymptotic results for multiple cover times on the complete graph should be similar to the Dixie cup result.
5. The calculations
To analyze the problem for multiple cover times on the complete graph, I have solved some systems of recursive equations displayed below. Some of the solutions are presented in the Appendix, Table 1.
The case m=2 :
15 𝑖, 𝑗, 𝑘, 𝑙, 𝑚 = 1 + 𝑖 𝑛 − 1𝑒 𝑖 − 1, 𝑗 + 1, 𝑘, 𝑙, 𝑚 + 𝑗 𝑛 − 1𝑓 𝑖, 𝑗 − 1, 𝑘 + 1, 𝑙, 𝑚 +𝑛−1𝑘 𝑔 𝑖, 𝑗, 𝑘 − 1, 𝑙 + 1, 𝑚 + 𝑙 𝑛−1 𝑖, 𝑗, 𝑘, 𝑙 − 1, 𝑚 + 1 +𝑚 −1 𝑛−1𝑑(𝑖, 𝑗, 𝑘, 𝑙, 𝑚 − 1) + 𝑛−𝑖−𝑗 −𝑘−𝑙−𝑚 𝑛−1 (𝑖, 𝑗, 𝑘, 𝑙, 𝑚) , 𝑑 𝑖, 𝑗, 𝑘, 𝑙, 𝑚 = 𝑛 − 1 𝑖 + 𝑗 + 𝑘 + 𝑙 + 𝑚+ 𝑖 𝑖 + 𝑗 + 𝑘 + 𝑙 + 𝑚𝑒 𝑖 − 1, 𝑗 + 1, 𝑘, 𝑙, 𝑚 + 𝑗 𝑖+𝑗 +𝑘+𝑙+𝑚𝑓 𝑖, 𝑗 − 1, 𝑘 + 1 + 𝑙, 𝑚 + 𝑘 𝑖+𝑗 +𝑘+𝑙+𝑚𝑔 𝑖, 𝑗, 𝑘 − 1, 𝑙 + 1, 𝑚 +𝑖+𝑗 +𝑘+𝑙+𝑚𝑙 𝑖, 𝑗, 𝑘, 𝑙 − 1, 𝑚 + 1 + 𝑚 𝑖+𝑗 +𝑘+𝑙+𝑚𝑑(𝑖, 𝑗, 𝑘, 𝑙, −1) .
A good start is to plot the results from the recursions with the asymptotic values, 𝑛 × ln 𝑛 + 𝑚 − 1 × ln ln n and (𝑛 − 1) × ln 𝑛 − 1 + 𝑚 − 1 × ln ln n − 1 to
see how the cover time looks for the exact result and the probable asymptotic values.
Figure 8: The case m=2 without constant for the complete graph.
There are actually three different lines in Figure 8. The solid line (top line), is the exact result from the recursion, the dashed line (middle line) is the asymptotic value with n and the dotted line (bottom line) is the asymptotic value with (n-1). It seems that the dashed and dotted line coincide, but there is actually a small difference between the lines. As seen in the graph, there is a big difference between the exact result and the asymptotic and it seems to increase with n. The difference between the end points for the exact result and the asymptotic value 𝑛 × ln 𝑛 + 𝑚 − 1 × ln 𝑙𝑛 𝑛 is 7.7760 × 103. If we use the same constant as in the double Dixie cup problem, 𝐶𝑚 = 𝛾 − ln( 𝑚 − 1 !) which means that in our case, m=2 , 𝐶𝑚 = 𝛾 the difference between the end points is 3.0480 × 103. With some experimentation, trying to find a good constant which makes the asymptotic value closer to the exact result with increasing n, it seems that the constants in the case with 𝑚 = 2 should be close to 1.
We use a constant which looks like the one in the double Dixie cup problem. Euler’s constant is approximate 0.5772, 2 × 𝛾 ≈ 1.1544 and 3 × 𝛾 ≈ 1.7316. ln 1 = 0, ln 2 ≈ 0.69315. Mixing them together in a similar way made me realize that 𝐶𝑚 = 𝑚 + 1 × 𝛾 − ln 𝑚 is a good approximation for the case 𝑚 = 2.
Figure 9:The case m=2 with constant for the complete graph.
As before there are three lines in Figure 9. The solid line (bottom) is the exact result given from the recursion, the dotted line (middle) is the asymptotic value with 𝑛 − 1 and the dashed line (top) is the asymptotic value with 𝑛. The constant 𝐶𝑚 = 𝑚 + 1 × 𝛾 − ln 𝑚 is in our case 𝐶𝑚 = 3 × 𝛾 − ln 2 ≈ 1.0385 and the maximum number of vertices in this case is 8191. The difference in the end points is 717.0296 between the dotted line and the solid line, and 730.3882 between the dashed line and the solid line, which is also the maximum difference. More than 700 steps seems like a big difference but the percentage of the exact result is as small as 0.72% . The minimum difference between the exact result and the asymptotic values are for the dotted line is when we have 720 vertices, and the difference is 0.0132 and for the dashed line is when we have 465 vertices and the difference at that point is 0.0124.
For 𝑚 = 3 we get the following picture.
The constant is now 𝐶𝑚 = 4 × 𝛾 − ln 3 ≈ 1.2103 and the maximum number of vertices in this case is 511.
Figure 10: The case m=3 with constant for the complete graph.
In Figure 10 we cannot distinguish the lines. The end points all lie in an interval of 13.8603, that is the difference between the solid line and the dashed line. In between lies the dotted line with a difference from the exact result of 1.4334. This means that the asymptotic result (𝑛 − 1) × ln 𝑛 − 1 + 𝑚 − 1 × ln 𝑙𝑛 𝑛 − 1 + 𝐶𝑚 is again closer to the exact result at the end point. The minimum difference between the exact result and the asymptotic values are for the dotted line when we have 500 vertices, and the difference is 0.0284 and for the dashed line is when we have 395 vertices and the difference at that point is 0.0443. The maximum difference are at 139 vertices for the dashed line, with value 16.4748 and at 151 vertices for the dotted line with value 27.268.
For 𝑚 = 4 we get the following picture.
The constant is now 𝐶𝑚 = 5 × 𝛾 − ln 4 ≈ 1.4998, and the maximum number of vertices in this case is 106.
Figure 11: The case m=4 with constant for the complete graph.
For m=5 we get the following picture.
The constant is now 𝐶𝑚 = 6 × 𝛾 − ln 5 ≈ 1.8539 , and the maximum number of vertices in this case is 39.
Figure 12: The case m=5 with constant for the complete graph.
Now we are able to distinguish the lines as we only have less than forty vertices for calculating the exact result of the multiple cover time with 𝑚 = 5 on the complete graph. The top line (solid) is above the dashed line which is above the dotted line. The difference between the dotted line and the solid line is increasing at first from 14.6897, which is the minimum difference, to 20.1676 which is the maximum difference, this happens when we have 20 vertices. After that the difference decreases to 17.1309 which is the difference between the end points. By the pattern of the other multiple covers the dotted line is getting closer to the exact result with increasing n. For the dashed line we have the maximum difference at 15 vertices, 8.861 and minimum at the end point, 4.3511.
 Aleliunas, R., Karp. R. M., Lipton, R., Lovasz, L. and Rackoff, C. (1979), Random walks, universal traversal sequences, and the complexity of maze problems. Proceedings of the 20th IEEE symposium on Foundations of computer science, 218-233.
 Blom, G., Holst, L. & Sandell, D. (1994), Problems and snapshots from the world
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 Blom, G and Sandell, D. (1992), Cover times for random walks on graphs. The mathematical scientist 17, 111-119.
 Erdős, P. & Rényi, A. (1961), On a classical problem of probability theory. Magy. Tud. Akad. Mat. Kutató Int. Közl. 6(1-2), 215-220.
 Feige, U. (1995), A tight upper bound on the cover time for random walks on graphs. Random Structures Algorithms 6, No.1, 51-54.
 Feller,W. (1968) An Introduction to probability Theory and its Applications, Vol. 1, 3rd edition. Wiley, New York.
 Holst, L. (1986), On Birthday, Collectors’, Occupancy and Other Classical Urn Problems . International Statistical Review, Vol.54, No. 1 , 15-27.
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Mathematical Monthly, Vol. 67 , No 1, 58-61.
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The multiple cover time cases and the asymptotic results have been calculated using Matlab, the code is presented here.
The cover time of the complete graph
% Program that compares the exact result for the covertime on the complete % graph to the order of magnitude.
antal = input('input number of vertices: '); gamma = 0.5772156649;
s = zeros(antal,1);
sn = zeros(antal,1); %vector to save the exact result in
slog1=zeros(antal,1);% %vector to save (n-1)ln(n-1) value
slog=zeros(antal,1); %vector to save nln(n) value in
for n=2:antal for i=1:(n-1) f=((n-1)/(n-i)); s(i)=f; end sn(n)=sum(s); slog1(n)=(n-1)*(reallog(n-1)+gamma); slog(n)=n*(reallog(n)+gamma); end x_axel=linspace(1,antal,antal); subplot(2,2,1); plot(x_axel,sn)
xlabel('number of vertices'); ylabel('number of steps'); subplot(2,2,2);
title('(n-1) x (ln(n-1) + gamma)'); xlabel('number of vertices');
ylabel('number of steps'); subplot(2,2,3);
title('n x (ln(n) + gamma)'); xlabel('number of vertices'); ylabel('number of steps'); subplot(2,2,4);
plot(x_axel,sn,x_axel,slog,'--',x_axel,slog1,':') title('All plots');
xlabel('number of vertices'); ylabel('number of steps');
The case m=2
23 [x,y]=textread('visits2res.txt'); gamma = 0.5772156649; z=size(x); s=zeros(z); t=zeros(z); for i=min(x):max(x) s(i-1)=(i-1)*(reallog(i-1)+reallog(reallog(i-1))+3*gamma-reallog(2)); t(i-1)=(i)*(reallog(i)+reallog(reallog(i))+3*gamma-reallog(2)); end plot(x,s,':') hold on plot(x,y) plot(x,t,'--') hold off
title('m=2 with Cm=(m+1)*gamma-ln(m)'); xlabel('number of vertices');
ylabel('number of steps');
The case m=3 clear all close all [x,y]=textread('visits3res.txt'); gamma = 0.5772156649; z=size(x); s=zeros(z); t=zeros(z); for i=min(x):max(x) s(i-1)=(i-1)*(reallog(i-1)+(3-1)*reallog(reallog(i-1))+4*gamma-reallog((3))); t(i-1)=i*(reallog(i)+(3-1)*reallog(reallog(i))+4*gamma-reallog((3))); end plot(x,s,':') hold on plot(x,y) plot(x,t,'--') hold off
title('m=3 with Cm=(m+1)*gamma-ln(m)'); xlabel('number of vertices');
ylabel('number of steps');
24 s(i-1)=(i-1)*(reallog(i-1)+(4-1)*reallog(reallog(i-1))+5*gamma-reallog((4))); % m=3 t(i-1)=i*(reallog(i)+(4-1)*reallog(reallog(i))+5*gamma-reallog((4))); end plot(x,s,':') hold on plot(x,y) plot(x,t,'--') hold off title('m=4 Cm=(m+1)*gamma-ln(m)'); xlabel('number of vertices'); ylabel('number of steps');
Table 1. Some of the data for multiple cover times on the complete graph.
n m=2 m=3 m=4 m=5
2 3.0000000e+000 5.0000000e+000 7.0000000e+000 9.0000000e+000 3 6.6250000e+000 1.0101563e+001 1.3500977e+001 1.6851593e+001 4 1.0792181e+001 1.5758362e+001 2.0563074e+001 2.5267170e+001 5 1.5353782e+001 2.1855841e+001 2.8104708e+001 3.4196519e+001 6 2.0212003e+001 2.8285346e+001 3.6008122e+001 4.3513832e+001 7 2.5307124e+001 3.4979700e+001 4.4199860e+001 5.3140122e+001 8 3.0599109e+001 4.1893772e+001 5.2630514e+001 6.3022221e+001 9 3.6059335e+001 4.8995303e+001 6.1264776e+001 7.3122143e+001 10 4.1666357e+001 5.6260143e+001 7.0076234e+001 8.3411488e+001 11 4.7403523e+001 6.3669564e+001 7.9044418e+001 9.3868258e+001 12 5.3257538e+001 7.1208629e+001 8.8153017e+001 1.0447494e+002 13 5.9217549e+001 7.8865158e+001 9.7388741e+001 1.1521725e+002 14 6.5274528e+001 8.6629033e+001 1.0674056e+002 1.2608337e+002 15 7.1420853e+001 9.4491718e+001 1.1619916e+002 1.3706333e+002
16 7.7650001e+001 1.0244592e+002 1.2575662e+002 1.4814860e+002 17 8.3956332e+001 1.1048533e+002 1.3540607e+002 1.5933184e+002 18 9.0334917e+001 1.1860445e+002 1.4514151e+002 1.7060664e+002 19 9.6781414e+001 1.2679844e+002 1.5495770e+002 1.8196737e+002 20 1.0329197e+002 1.3506299e+002 1.6484996e+002 1.9340903e+002 21 1.0986315e+002 1.4339429e+002 1.7481413e+002 2.0492718e+002 22 1.1649184e+002 1.5178887e+002 1.8484648e+002 2.1651780e+002 23 1.2317527e+002 1.6024364e+002 1.9494362e+002 2.2817731e+002 24 1.2991088e+002 1.6875577e+002 2.0510249e+002 2.3990242e+002 25 1.3669637e+002 1.7732269e+002 2.1532032e+002 2.5169016e+002 26 1.4352961e+002 1.8594204e+002 2.2559455e+002 2.6353780e+002 27 1.5040866e+002 1.9461166e+002 2.3592284e+002 2.7544286e+002 28 1.5733172e+002 2.0332957e+002 2.4630302e+002 2.8740301e+002 29 1.6429712e+002 2.1209392e+002 2.5673312e+002 2.9941614e+002 30 1.7130334e+002 2.2090300e+002 2.6721127e+002 3.1148027e+002 31 1.7834892e+002 2.2975521e+002 2.7773576e+002 3.2359357e+002 32 1.8543254e+002 2.3864909e+002 2.8830498e+002 3.3575433e+002 33 1.9255294e+002 2.4758325e+002 2.9891746e+002 3.4796096e+002