the thing I wanted to know is that. Example: I want range of number from 1 to 10. and how to call a random number between the range from 1 to 10 automatically. Not using switches cases or if else statements. Any Idea ?
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I'm new to using BigInteger, so I'm trying my best reading through the documentation for it. I'm still confused however. I need to generate 500 random integers between the range of 2^70 and 2^80 and I don't know how to set a range for BigIntegers.
I'm getting the possible duplicate messages, so I guess I should add that I've already looked at the solutions in this one and still don't understand how to solve my issue: How to generate a random BigInteger value in Java?
There is a built-in method to generate a random BigInteger between 0 and 2^n - 1.
You can use that in a loop to generate numbers up to 2^80.
The chance that a number will fall below 2^70 is very small ( ~ 0.1%). If that happens (and with 500 iterations it might very well), just draw another one.
This will give you a uniform distribution of random numbers between 2^70 and 2^80.
The chance that a number will repeat is almost nonexistent. If you feel that it cannot be ignored, check your previous numbers for duplicates, and draw again.
First of all, I did a search on this topic but I could not found anything similar to what I'm trying to accomplished, so this could be a duplicate question.
I would like to have a function that returns a number (1 or 2) with a probability of 0.8% for the number one and 0.2% for the number two.
How can I do it?
Generate a random number between 0 and 1. If the number is between 0 and 0.8, return 1. else, return 2:
return rnd.nextFloat() < 0.8 ? 1 : 2;
You just make intervals, and if your random number fits in the interval, you have a hit. In this most simple example, you make intervals 1-4 and 5. So you fetch random number from 1 to 5. if it's a 1-4 (80% probability) you act as it's 1, and if it's 5 (20% probability) you act as it's 2.
For example, make an array of 100 numbers. 20 numbers with number two, and 80 numbers with number one.
Use the Random class to generate a number between 0 and 100. Then you use that number to get the result from the array.
This is just an example, you can use another values in the array.
I have three categories of input , each with a impact range.
Cat 1 : 20 - 16
Cat 2 : 15 - 5
Cat 3 : 4 -1
I have a file with say N randomly generated categories.
I am trying to take a sum of impact for all the 100 entries through a logic that looks something like this :
// calculate sum of impacts
getSum(){
Generate a random class with seed as current system execution time
for(as many entries in file){
switch(category)
case 1 : i = random input between 20 - 16
case 2 : i = random input between 15 - 5
case 3 : i = random input between 4 - 1
some default case here
sum = sum + i
}
return sum
}
.
.
// loop until you get a desired sum
while(true){
if(Call to getSum() returns value within a desired range){
display some statistics;
break;
}
}
However , i see that the program generally runs infinitely , as the random generation and subsequent summation is giving result beyond the desired range. So , to get things in range , I have to manually tune the max-min ranges for each execution.
Can someone suggest an algorithm that will automatically vary the max min ranges for each category , by learning the trend of obtained sum as the program is running , so as to quickly give a solution ?
Edit : i have just read about the 0/1 knapsack algorithm.. and it seems promising , but unsure if that is the algorithm for this case. Any help would be great.
A couple band-aids:
1) use a long int instead of a regular int t give you a longer range.
2) use an unsigned long, since all of your relevant numbers are positive (then be careful of underflow errors when you subtract.
A couple possible strategies:
1) (This contradicts your question, but it is how things are usually done.) Determine the static maximum for each category, and design to it, using long unsigned int if that is large enough, or some larger data structure as necessary.
2) (This is exactly what you are asking.) Use, and build if necessary, a data structure which expands when an overflow occurs.
Solution for strategy 2:
I will get back to you on this. :)
I am working on a project that needs to generate two random numbers from a given range (both of them at the same time, one after another) and check if they are equal to each other - if they are, proceed executing other code; if they aren't - generate the numbers again. Now my question is, if we have a range [0;10], and the first randomly generated number turned out to be 5, is the probability of the second number also being 5 as good as any other number? Specifically, does Math.random() have any "defense" against generating same number if it is called twice consecutively? or it "tries" to not generate the same number?
Generating the same number in the range [0,10] twice in succession is a perfectly valid occurrence for any random number generator. If it took any steps to prevent that it wouldn't be random.
On any invocation, the chances of any individual number being chosen should be 1:11, and each choice should be independent of previous choices, so the chances that in a pair the second number matches the first is 1 in 11.
As to how random Math.random() is, it's pseudo-random, meaning it uses an algorithm to generate a series of evenly distributed numbers starting with a "seed" value. It's not suitable for cryptography but quite good for simulations and other non-cryptographic uses.
first time here at Stackoverflow. I hope someone can help me with my search of an algorithm.
I need to generate N random numbers in given Ranges that sum up to a given sum!
For example: Generatare 3 Numbers that sum up to 11.
Ranges:
Value between 1 and 3.
Value between 5 and 8.
value between 3 and 7.
The Generated numbers for this examle could be: 2, 5, 4.
I already searched alot and couldnt find the solution i need.
It is possible to generate like N Numbers of a constant sum unsing modulo like this:
generate random numbers of which the sum is constant
But i couldnt get that done with ranges.
Or by generating N random values, sum them up and then divide the constant sum by the random sum and afterwards multiplying each random number with that quotient as proposed here.
Main Problem, why i cant adopt those solution is that every of my random values has different ranges and i need the values to be uniformly distributed withing the ranges (no frequency occurances at min/max for example, which happens if i cut off the values which are less/greater than min/max).
I also thought of an soultion, taking a random number (in that Example, Value 1,2 or 3), generate the value within the range (either between min/max or min and the rest of the sum, depending on which is smaller), substracting that number of my given sum, and keep that going until everything is distributed. But that would be horrible inefficiant. I could really use a way where the runtime of the algorithm is fixed.
I'm trying to get that running in Java. But that Info is not that importend, except if someone already has a solution ready. All i need is a description or and idea of an algorithm.
First, note that the problem is equivalent to:
Generate k numbers that sums to a number y, such that x_1, ..., x_k -
each has a limit.
The second can be achieved by simply reducing the lower bound from the number - so in your example, it is equivalent to:
Generate 3 numbers such that x1 <= 2; x2 <= 3; x3 <= 4; x1+x2+x3 = 2
Note that the 2nd problem can be solved in various ways, one of them is:
Generate a list with h_i repeats per element - where h_i is the limit for element i - shuffle the list, and pick the first elements.
In your example, the list is:[x1,x1,x2,x2,x2,x3,x3,x3,x3] - shuffle it and choose first two elements.
(*) Note that shuffling the list can be done using fisher-yates algorithm. (you can abort the algorithm in the middle after you passed the desired limit).
Add up the minimum values. In this case 1 + 5 + 3 = 9
11 - 9 = 2, so you have to distribute 2 between the three numbers (eg: +2,+0,+0 or +0,+1,+1).
I leave the rest for you, it's relatively easy to create a uniform distribution after this transformation.
This problem is equivalent to randomly distributing an excess of 2 over the minimum of 9 on 3 positions.
So you start with the minima (1/5/3) and then cycle 2 times, generating a (pseudo-)random value of [0-2] (3 positions) and increment the indexed value.
e.g.
Start 1/5/3
1st random=1 ... increment index 1 ... 1/6/3
2nd random=0 ... increment index 0 ... 2/6/3
2+6+3=11
Edit
Reading this a second time, I understand, this is exactly what #KarolyHorvath mentioned.