Randi(imax,m,n)

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Moazam Ali
Moazam Ali 2020-6-17
How to increase the size of the argument imax in randi function. I used the code e=randi(phi,1,1) but there comes an error saying value of phi has to be less than 2^52. What to do to increase this range?
  2 个评论
Matt J
Matt J 2020-6-17
Increase it by how much?
Moazam Ali
Moazam Ali 2020-6-18
The phi I used here is of the order of 2^1024

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回答(2 个)

James Tursa
James Tursa 2020-6-17
编辑:James Tursa 2020-6-17
The point is that once the eps of the max value is greater than 1, you cannot represent contiguous sets of integer values in double precision. E.g.,
>> eps(2^52)
ans =
1
>> eps(2^53)
ans =
2
So double precision could represent 2^52 + 1 exactly, but you see it will not be able to represent 2^53 + 1 exactly because eps(2^53) is greater than 1. So then the question becomes what makes sense for randi to produce as output when it can't represent all of the integers in the desired range? It doesn't make sense, hence the restriction.
  1 个评论
Matt J
Matt J 2020-6-17
And that is why my answer involves a conversion to uint64.

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Matt J
Matt J 2020-6-17
Perhaps as follows,
num = uint64(randi(2^53-1,1))*2^8 +randi(2^8-1,1)
  2 个评论
Moazam Ali
Moazam Ali 2020-6-18
Uint 64 represents the variables as 64bit data but anything beyond that limit is stored as 2^64-1. Isn't it going to effect the random selection
James Tursa
James Tursa 2020-6-18
编辑:James Tursa 2020-6-18
Using uint64 extends the allowed range beyond double, but of course it has a limit as well as you note. If you really need more range then use more than one variable for each value you need and extend Matt's approach.

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