How to generate random numbers with constraint?

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MIch
MIch 2022-4-27
编辑： Torsten 2022-4-28
I need ideas how to generate random numbers in given range in example under...constraint is--->1st number has to be greater than 2nd, 2nd greater than 3rd...24th greater than 25th
for n = 1 : 25
a(n) = (1.2-0.05)*rand(1)+0.05;
end

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采纳的回答

Torsten
Torsten 2022-4-27
a = (1.2-0.05)*rand(25,1)+0.05;
a = sort(a,'descend')
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显示 4更早的评论
MIch
MIch 2022-4-28
upper bound is 1.2
lower bound is 0.05
no particular distribution
Torsten
Torsten 2022-4-28
编辑：Torsten 2022-4-28
b = (1.2-0.05)*rand(12,1)+0.05;
b = sort(b,'descent');
c = (1.2-b(3))*rand(5,1)+b(3);
c = sort(c,'descent');
d = (1.2-b(7))*rand(5,1)+b(7);
d = sort(d,'descent');
e = (1.2-b(9))*rand(3,1)+b(9);
e = sort(e,'descent');
a = [b,c,d,e]

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更多回答（3 个）

Steven Lord
Steven Lord 2022-4-27
编辑：Steven Lord 2022-4-27
Ran in:
Generate the numbers then call sort on the array.
By the way, you don't need to use a for loop here. The rand function can generate a vector of values with a single call.
a = (1.2-0.05)*rand(1, 25)+0.05
a = 1×25
0.2278 0.6887 0.1720 0.1546 0.7384 0.5730 0.9304 0.4400 0.7577 0.7744 1.1934 1.1811 0.6472 0.8133 1.0554 0.4156 0.3485 0.0974 0.4555 0.3982 0.3111 1.0316 0.4099 0.5257 0.4588
b = sort(a, 'descend')
b = 1×25
1.1934 1.1811 1.0554 1.0316 0.9304 0.8133 0.7744 0.7577 0.7384 0.6887 0.6472 0.5730 0.5257 0.4588 0.4555 0.4400 0.4156 0.4099 0.3982 0.3485 0.3111 0.2278 0.1720 0.1546 0.0974
Prakash S R
Prakash S R 2022-4-27
If all you want is that the numbers are randomly drawn from the uniform distribution between 0.05 and 1.2, you could generate a as above, follwed by
a = sort(a, 'descend')
or simply
a = sort((1.2-0.05)*rand(1,25)+0.05, 'descend');
Walter Roberson
Walter Roberson 2022-4-28
Ran in:
First branch consists of following numbers 1>2>3>4>5>6>7>8>9>10>11>12, second brach starts at number 3 of first branch and consist folloving numbers 3>13>14>15>16>17, third branch starts at number 7 of first branch 7>18>19>20>21>22 and fourth branch starts at number 9 of first branch 9>23>24>25.
format long g
rmin = 0.05;
rmax = 1.2;
UB = { [], %1 < nothing
[1] %2 < 1
[1:2] %3 < 1,2
[1:3] %4 < 1,2,3
[1:4] %5 < 1,2,3,4
[1:5] %6 < 1,2,3,4,5
[1:6] %7 < 1,2,3,4,5,6
[1:7] %8 < 1,2,3,4,5,6,7
[1:8] %9 < 1,2,3,4,5,6,7,8
[1:9] %10 < 1,2,3,4,5,6,7,8,9
[1:10] %11 < 1,2,3,4,5,6,7,8,9,10
[1:11] %12 < 1,2,3,4,5,6,7,8,9,10,11
[3] %13 < 3
[3 13] %14 < 3,13
[3 13:14] %15 < 3,13,14
[3 13:15] %16 < 3,13,14,15
[3 13:16] %17 < 3,13,14,15,16
[7] %18 < 7
[7 18] %19 < 7,18
[7 18:19] %20 < 7,18,19
[7 18:20] %21 < 7,18,19,20
[7 18:21] %22 < 7,18,19,20,21
[9] %23 < 9
[9 23] %24 < 9,23
[9 23:24] %25 < 9,23,24
};
NV = numel(UB);
V = zeros(NV,1);
V(1) = RR(rmin,rmax);
for K = 2 : NV
least = min(V(UB{K}));
V(K) = RR(rmin,least);
end
[[1:11].', V(1:11)]
ans = 11×2
1 0.509337598303387 2 0.139316553971349 3 0.0584882780365813 4 0.0505019413423992 5 0.0504884664739595 6 0.0500210905816074 7 0.0500035847219492 8 0.0500022318950057 9 0.0500016859195006 10 0.0500009862376525
[[3,13:17].', V([3,13:17])]
ans = 6×2
3 0.0584882780365813 13 0.0563547623292359 14 0.0557441291649364 15 0.0512888627397752 16 0.0512332793800842 17 0.05071245049519
[[7,18:22].', V([7,18:22])]
ans = 6×2
7 0.0500035847219492 18 0.0500021857353128 19 0.0500006725624357 20 0.0500000578760855 21 0.0500000018122051 22 0.0500000010119062
[[9,23:25].', V([9,23:25])]
ans = 4×2
9 0.0500016859195006 23 0.0500001392274942 24 0.0500001168263988 25 0.0500000781080973
function x = RR(rmin, rmax)
x = rand() * (rmax - rmin) + rmin;
end
  1 个评论
Walter Roberson
Walter Roberson 2022-4-28
Notice how near the end, everything gets squashed into very close to the lower bound. This is to be expected for this kind of generating.

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