non-repetitive

I want to create a C program to generate numbers from 0 to 999999, keeping in mind that the number generated should not have any digits that are repetitive within it. For example, "123" is an acceptable value but not "121" as the '1' is repeated. I have sourced other program codes that check if an integer has repeated digits: Check if integer has repeating digits. No string methods or arrays What is the fastest way to check for duplicate digits of a number? However these do not really solve my problem and they are very inefficient solutions if I were to perform the check for 1,000,000

I want to generate random numbers within the range 1 to 4, 4 including. Here is my code: int num = r.nextInt(4) + 1; //r is instance of Random. However, I am running the above code in a loop and don't want repeating random number. What happens now is often I am getting: 1,1,1,2,3,1,4,2,2,1,4,2,4,4,2,1,4,3,3,1,4,2,4,1 as my output. Here, though the numbers are random within the range(1-4), but often repeated like the number "1"in the first 3 iterations. What I am looking for is a way to get non repeating random number within the loop. One simple way I know is of keeping the last random number

I'm looking for an efficient algorithm that produces random values within a range, without repetitions. In pseudo-code: (in class Rand) Rand(long from, long to) { this.from = from; this.to = to; // ... } long getNumber() { // returns a random number in the [from, to] range // which has never been returned before } Usage: Rand r = new Rand(1, 100000000); long x = r.getNumber(); long y = r.getNumber(); ... The numbers returned from r.getNumber() should always be different from the ones previously returned. Of course, if to - from + 1 numbers were returned, the algorithm should start again (or

My question is: How can I generate non-repetitive random numbers in numpy? list = np.random.random_integers(20,size=(10)) If you don't insist on using NumPy, you can use random.sample() from the standard library: print random.sample(range(20), 10) With NumPy, you will have to use numpy.random.shuffle() and slicing: a = numpy.arange(20) numpy.random.shuffle(a) print a[:10] strnam I think numpy.random.sample doesn't work right, now. This is my way: import numpy as np np.random.choice(range(20), 10, replace=False) You can get this by sorting as well: random_numbers = np.random.random([num_samples