How to generate random numbers with constraint?

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

Answer 1

[Torsten]

a = (1.2-0.05)*rand(25,1)+0.05;
a = sort(a,'descend')

[MIch]

Thank you very much.

I have another question

how to generate 25 random numbers in given range

complicated constraints 1>2>3>4>5>6>7>8>9>10>11>12

3>13>14>15>16>17

7>18>19>20>21>22

9>23>24>25

[Torsten]

Sort all 25 numbers in descending order, then all constraints you list above will be satisfied.

The question is only: Does the sorting destroy what you consider as "generate randomly" ?

I mean: Is it legitimate to sort the random numbers drawn afterwards ? Or is it necessary that the random numbers satisfying the constraints are drawn one after the other ?

[MIch]

I have to create solution of following problem.. 25 random generated numbers in given limits under mentioned constrtaints

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.

Is there any other way to sort these numbers because I can't aplly your sort solution because it doesn't suit my problem.

for example 13th number has to be greater than 3rd,2nd,1st, but 13th number doesnt necesaary have to be greater than 4th,5th,6th, it can also be <4,<5,<6... beacause these later numbers (13th, 14th,15th..) are not part of the first branch

Same applies for remaining two constraints.

[Walter Roberson]

@MIch

Are there upper or lower bounds on the numbers?

Is there a particular distribution that is required?

If you have a finite upper and lower bound and the values are all to be drawn from that distribution, then in practice the later numbers are going to be pretty "pinched"

[MIch]

upper bound is 1.2

lower bound is 0.05

no particular distribution

[Torsten]

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]

Answer 2

[Steven Lord]

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

[MIch]

Thank you

Answer 3

[Prakash S R]

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');    

[MIch]

Thank you

Answer 4

[Walter Roberson]

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

[Walter Roberson]

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.