I would like to set up the xcor and ycor using a random number inside a precise interval.
In particular, I would like my turtles to be spread out in the first half of the world and then make them move to the other half if certain conditions are satisfied.
So I was thinking that I could set xcor and ycor by letting netlogo choose a random number in the interval [0-16] which should coincide with the first half of my world. How can I do it ?
I was thinking something like
setxy random-integer [0-16]
but it clearly does not work. Can you help me ?
There are two reasons, why it isn't working.
setxy expects two inputs, one for the x and one for the y coordinate. (See setxy entrance in the netlogo-dictionary)
The reporter, that returns random integers is simply random and returns values stricly less than the given (positive) number. (See random entrance in the netlogo-dictionary)
The solution for your problem is:
setxy random 17 random 17
assuming that the world is rectangular.
Related
Let X be a random variable on probability space (Ω,P), Suppose X~U({1,2,3}), does this mean the space (Ω,P) is uniform space.
I tried to come up with counter example and did not work out but i still think this statement isn't right.
You are correct that the statement isn't right. Here is a concrete counterexample with a bit of R code to illustrate it.
A standard 6-sided die has 3 pairs: (1,6), (2,5), (3,4) where each number in the pair is on the opposite side of the other. Suppose that such a die is biased so that each pair is equally likely but that in a pair the larger of the two numbers is twice as likely as the smaller. For example, 6 is twice as likely as 1. This is easily seen to imply that the numbers 1,2,3 appear with probability 1/9 and the numbers 4,5,6 appear with probability 2/9.
You can simulate 1000 rolls like this:
rolls <- sample(1:6,1000,replace = TRUE, prob = c(1/9,1/9,1/9,2/9,2/9,2/9))
Here is a display created by making a barplot of the tabulation of the results:
Confirming the obvious fact that the distribution is not uniform.
We can define X on it as the function which indicates what pair a rolls is in (so that 1,6 are in the first pair, 2,5 in the second, 3,4 in the third):
X = function(x){min(x,7-x)}
and then:
barplot(table(sapply(rolls,X)))
leading to:
which confirms the obvious fact that X is uniform.
This is a purely theoretical question.
We all know that most, if not all, random-number generators actually only generate pseudo-random numbers.
Let's say I want a random number from 10 to 20. I can do this as follows (myRandomNumber being an integer-type variable):
myRandomNumber = rand(10, 20);
However, if I execute this statement:
myRandomNumber = rand(5, 10) + rand(5, 10);
Is this method more random?
No.
The randomness is not cumulative. The rand() function uses a uniform distribution between your two defined endpoints.
Adding two uniformly distributions invalidates the uniform distribution. It will make a strange looking pyramid, with the most probability tending toward the center. This is because of accumulation of the probability density function with increasing degrees of freedom.
I urge you to read this:
Uniform Distribution
and this:
Convolution
Pay special attention to what happens with the two uniform distributions on the top right of the screen.
You can prove this to yourself by writing to a file all the sums and then plotting in excel. Make sure you give yourself a large enough sample size. 25000 should be sufficient.
The best way to understand this is by considering the popular fair ground game "Lucky Seven".
If we roll a six sided die, we know that the probability of obtaining any of the six numbers is the same - 1/6.
What if we roll two dice and add the numbers that appear on the two ?
The sum can range from 2 ( both dice show 'one') uptil 12 (both dice show 'six')
The probabilities of obtaining different numbers from 2 to 12 are no longer uniform. The probability of obtaining a 'seven' is the highest. There can be a 1+6, a 6+1, a 2+5, a 5+2, a 3+4 and a 4+3. Six ways of obtaining a 'seven' out of 36 possibilities.
If we plot the distribution we get a pyramid. The probabilities would be 1,2,3,4,5,6,5,4,3,2,1 (of course each of these has to be divided by 36).
The pyramidal figure (and the probability distribution) of the sum can be obtained by 'convolution.
If we know the 'expected value' and standard deviation ('sigma') for the two random numbers, we can perform a quick a ready calculation of the expected value of the sum of the two random numbers.
The expected value is simply the addition of the two individual expected values.
The sigma is obtained by applying the "pythagoras theorem" on the two individual sigmas (square root of the sum of the square of each sigma).
Scenario:
Drawing a graph. Have data points which range from A to B, and want to decide on a granularity for drawing the axis scales. Eg, for 134 to 151 the scale might run from 130 to 155, to start and end on "round" numbers in the decimal system. But the numbers might run from 134.31 to 134.35, in which case a scale from 130 to 135 would (visually) compress out the "significance" in the data -- it would be better to draw the scale from 134 to 135, or maybe even 134.3 to 134.4. And the data values might instead run from 0.013431 to 0.013435, or from 1343100 to 1343500.
So I'm trying to figure out an elegant way to calculate the "granularity" to round the low bound down to and the upper bound up to, to produce a "pleasing" chart. One could just "hack" it somehow, but that produces little confidence that "odd" cases will be handled well.
Any ideas?
Just an idea:
Add about 10% to your range, tune this figure empirically
Divide size of range by number of tick marks you want to have
Take the base 10 logarithm of that number
Multiply the result by three, then round to the nearest integer
The remainder modulo 3 will tell you whether you want the least significant decimal to change in steps of 1, 2, or 5
The result of an integer division by 3 will tell you the power of ten to use
Take the (extended) range and compute the extremal tick points it contains, according to the tick frequencey just computed
Ensure that all data points actually lie within that range, add ticks if not
If needed, add minor ticks by decreasing the integer above by one
I found a very helpful calculation which is very similar to the axis scale of excel graphs:
It is written for excel but I used and transformed it into objective-c code for setting up my graph axis.
Is there an elegant method to create a number that does not exist in a given list of floating point numbers? It would be nice if this number were not close to the existing values in the array.
For example, in the list [-1.5, 1e+38, -1e38, 1e-12] it might be nice to pick a number like 20 that's "far" away from the existing numbers as opposed to 0.0 which is not in the list, but very close to 1e-12.
The only algorithm I've been able to come up with involves creating a random number and testing to see if it is not in the array. If so, regenerate. Is there a better deterministic approach?
Here's a way to select a random number not in the list, where the probability is higher the further away from an existing point you get.
Create a probability distribution function f as follows:
f(x) = <the absolute distance to the point closest to x>
such function gives a higher probability the further away from the a given point you are. (Note that it should be normalized so that the area below the function is 1.)
Create the primitive function F of f (i.e. the accumulated area below f up to a given point).
Generate a uniformly random number, x, between 0 and 1 (that's easy! :)
Get the final result by applying the inverse of F to that value: F-1(x).
Here's a picture describing a situation with 1.5, 2.2 and 2.9 given as existing numbers:
Here's the intuition of why it works:
The higher probability you have (the higher the blue line is) the steeper the red line is.
The steeper the red line is, the more probable it is that x hits the red line at that point.
For example: At the given points, the blue lines is 0, thus the red line is horizontal. If the red line is horizontal, probability that x hits that point is zero.
(If you want the full range of doubles, you could set min / max to -Double.MAX_VALUE and Double.MAX_VALUE respectively.)
If you have the constraint, that the new value must be somewhere in between [min, max] then you could sort your values and insert the mean value of the two adjacent values with the largest absolute difference.
In your sample case [-1e38, -1.5, 1e-12, 1e+38] is the ordered list. As you calculate the absolute differences, you'll find the maximum difference for the values (1e-12, 1e+38) so you calculate the new value to be ((n[i+1] - n[i]) / 2) + n[i] (simple mean value calculation).
Update:
Additionally you could also check if the FLOAT_MAX or FLOAT_MIN values will give good candidates. Simply check their distance to min and max and if the result values are larger than the maximum difference for two adjacent values, pick them.
If there is no upper bound, just sum up the absolute value of all the numbers, or subtract them all.
Another possible solution would be to get the smallest number and the greatest number in the list, and choose something outside their bounds (maybe double the greatest number).
Or probably the best way would be to compute the average, the smalelst and the biggest number, as long as the standard deviation. Then, with all this data, you know how the numbers are structured, and can choose accordingly (all clustered around a given negative value? Chosoe a positive one. All small numbers? Choose a big one. etc.)
Something along the lines of
number := 1
multiplier := random(1000)+1
if avg>0
number:= -number
if min < 1 and max > 1
multiplier:= 1 / (random(1000)+1)
if stdDev > 1000
number := avg+random(500)-250
multiplier:= multiplier / (random(1000)+1)
(just an example from the top of my head)
Or another Possibility would be to XOR all the numbers together. Should yield a good result.
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.