I have a question like, I should genearate 'k' random numbers lets say it is from 1 to 1000. But the generated numbers should have a mean value of 300. I used rand() function to generate random numbers. But I am stuck with the mean value. How can I do so that the numbers generated have a mean value.
I'd generate k-1 random numbers, and then set the K number to be (mean*k-[sum of all the numbers you generated so far]).
Unfortunately, the C standard does not guarantee that the random numbers are uniform (it doesn't specify any distribution, for that matter), so the only way to do it is to generate the 1000 numbers in advance, calculate the mean (M) and subtract M-300 from every element
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I want a simple (non-cryptographic) random number generation algorithm where I can freely choose the period.
One candidate would be a special instance of LCG:
X(n+1) = (aX(n)+c) mod m (m,c relatively prime; (a-1) divisible by all prime factors of m and also divisible by 4 if m is).
This has period m and does not restrict possible values of m.
I intend to use this RNG to create a permutation of an array by generating indices into it. I tried the LCG and it might be OK. However, it may not be "random enough" in that distances between adjacent outputs have very few possible values (i.e, plotting x(n) vs n gives a wrapped line). The arrays I want to index into have some structure that has to do with this distance and I want to avoid potential issues with this.
Of course, I could use any good PRNG to shuffle (using e.g. Fisher–Yates) an array [1,..., m]. But I don't want to have to store this array of indices. Is there some way to capture the permuted indices directly in an algorithm?
I don't really mind the method ending up biased w.r.t choice of RNG seed. Only the period matters and the permuted sequence (for a given seed) being reasonably random.
Encryption is a one-to-one operation. If you encrypt a range of numbers, you will get the same count of apparently random numbers back. In this case the period will be the size of the chosen range. So for a period of 20, encrypt the numbers 0..19.
If you want the output numbers to be in a specific range, then pick a block cipher with an appropriately sized block and use Format Preserving Encryption if needed, as #David Eisenstat suggests.
It is not difficult to set up a cipher with almost any reasonable block size, so long as it is an even number of bits, using the Feistel structure. If you don't require cryptographic security then four or six Feistel rounds should give you enough randomness.
Changing the encryption key will give you a different ordering of the numbers.
I'm trying to make this algorithm which inputs a lower and upper limit for two numbers (the two numbers may have different lower and upper limits) and outputs two random numbers within that range
The catch is however that when the two numbers are added, no "carry" should be there. This means the sum of the digits in each place should be no more than 9.
How can I make sure that the numbers are truly random and that no carrying occurs when adding the two numbers
Thanks a lot!
Edit: The ranges can vary, the widest range can be 0 to 999. Also, I'm using VBA (Excel)
An easy and distributionally correct way of doing this is to use Rejection Sampling, a.k.a. "Acceptance/Rejection". Generate the values independently, and if the carry constraint is violated repeat. In pseudocode
do {
generate x, y
} while (x + y > threshold)
The number of times the loop will iterate has a geometric distribution with an expected value of (proportion of sums below the threshold)-1. For example, if you're below the threshold 90% of the time then the long term number of iterations will average out to 10/9, 1.11... iterations per pair generated. For lower likelihoods of acceptance, it will take more attempts on average.
Assuming I can generate random bytes of data, how can I use that to choose an element out of an array of n elements?
If I have 256 elements I can generate 1 byte of entropy (8 bits), and then use that to pick my element simply be converting it to an integer.
If I have 2 elements I can generate 1 byte, discard 7 bits and use the remaining bit to select my element.
But what if I have 3 elements? 1 bit is too few and 2 is too many. How would I randomly select 1 of the 3 elements with equal probability?
Here is a survey of algorithms to generate uniform random integers from random bits.
J. Lumbroso's Fast Dice Roller in "Optimal Discrete Uniform Generation from Coin Flips, and Applications, 2013. See also the implementation at the end of this answer.
The Math Forum, 2004. See also "Bit Recycling for Scaling Random Number Generators".
D. Lemire, "A Fast Alternative to the Modulo Reduction".
M. O'Neill, "Efficiently Generating a Number in a Range".
Some of these algorithms are "constant-time", others are unbiased, and still others are "optimal" in terms of the number of random bits it uses on average. In the rest of this answer we will assume we have a "true" random generator that can produce unbiased and independent random bits.
For further discussion, see the following answer of mine:
How to generate a random integer in the range [0,n] from a stream of random bits without wasting bits?
You can generate the proper distribution by simply truncating into the necessary range. If you have N elements then simply generate ceiling(log(N))=K random bits. Doing so is inefficient, but still works as long as the K bits are generated randomly.
In your example where you have N=3, you need at least K=2 bits, you have the following outcomes [00, 01, 10, 11] of equal probability. To map this into the proper range, just ignore one of the outcomes, such as the last one. Think of this as creating a new joint probability distribution, p(x_1, x_2), over the two bits where p(x_1=1, x_2=1) = 0, while for each of the others it will be 1/3 due to renormalization (i.e., (1/4)/(3/4) = 1/3 ).
I was trying to solve this problem on SPOJ, in which I have to find how many numbers are there in a range whose digits sum up to a prime. This range can be very big, (upper bound of 10^8 is given). The naive solution timed out, I just looped over the entire range and checked the required condition. I cant seem find a pattern or a formula too. Could someone please give a direction to proceed in??
Thanks in advance...
Here are some tips:
try to write a function that finds how many numbers in a given range have a given sum of the digits. Easiest way to implement this is to write a function that returns the number of numbers with a given sum of digits up to a given value a(call this f(sum,a)) and then the number of such numbers in the range a to b will be f(sum,b) - f(sum, a - 1)
Pay attention that the sum of the digits itself will not be too high - up to 8 * 9 < 100 so the number of prime sums to check is really small
Hope this helps.
I (seriously) doubt whether this 'opposite' approach will be any faster than #izomorphius's suggestion, but it might prompt some thoughts about improving the performance of your program:
1) Get the list of primes in the range 2..71 (you can omit 1 and 72 from any consideration since neither is prime).
2) Enumerate the integer partitions of each of the prime numbers in the list. Here's some Python code. You'd want to modify this so as not to generate partitions which were invalid, such as those containing numbers larger than 9.
3) For each of those partitions, pad out with 0s to make a set of 8 digits, then enumerate all the permutations of the padded set.
Now you have the list of numbers you require.
Generate the primes using the sieve of Eratosthenes up to the maximum sum (9 + 9...). Put them in a Hash table. Then you could likely loop quickly through 10^8 numbers and add up their sums. There might be more efficient methods, but this should be quick enough.
I have to generate two random sets of matrices
Each containing 3 digit numbers ranging from 2 - 10
like that
matrix 1: 994,878,129,121
matrix 2: 272,794,378,212
the numbers in both matrices have to be greater then 100 and less then 999
BUT
the mean for both matrices has to be in the ratio of 1:2 or 2:3 what ever constraint the user inputs
my math skills are kind of limited so any ideas how do i make this happen?
In order to do this, you have to know how many numbers are in each list. I'm assuming from your example that there are four numbers in each.
Fill the first list with four random numbers.
Calculate the mean of the first list.
Multiply the mean by 2 or by 3/2, whichever the user input. This is the required mean of the second list.
Multiply by 4. This is the required total of the second list.
Generate 3 random numbers.
Subtract the total of the three numbers in step 5 from the total in step 4. This is the fourth number for the second list.
If the number in step 6 is not in the correct range, start over from step 5.
Note that the last number in the second list is not truly random, since it's based on the other values in the list.
You have a set of random numbers, s1.
s1= [ random.randint(100,999) for i in range(n) ]
For some other set, s2, to have a different mean it's simply got to have a different range. Either you select values randomly from a different range, or you filter random values to get a different range.
No matter how many random numbers you select from the range 100 to 999, the mean is always just about 550. The odds of being a different value are exactly the normal distribution probabilities on either side of the mean.
You can't have a radically different mean with values selected from the same range.