Here a=0.75 and the dependence of the deviation from the number of steps N is still linear.
ν= 0.50402
Wednesday, February 20, 2008
Week 4b(Valentines Day) - Modified Random Path - a=0.75
Here a=0.75 and the dependence of the deviation from the number of steps N is still linear.
ν= 0.50402
Week 4b(Valentines Day) - Modified Random Path - a=0.25
The ν= 0.498765 so the <Δx^2> depends almost linear from N for a=0.25
Week 4b(Valentines Day) - Modified Random Path
Below is the solution for the Modified Random Path simulation.
This solution for choosing the next step isn't unique but it is the simplest in my opinion.
I have used 100000 paths for each N.
The average should be zero because this simulation chooses each step depending from the very first step of the path and that is always 50% left or right so the average over many paths should be zero.
This solution for choosing the next step isn't unique but it is the simplest in my opinion.
I have used 100000 paths for each N.
The average should be zero because this simulation chooses each step depending from the very first step of the path and that is always 50% left or right so the average over many paths should be zero.
Week 3 - Brown motion in 2-D
Below is the solution for the Brown motion in 2-D problem using simple Monte-Carlo simulation.
The mean x^2 and y^2 have a linear dependence from the number of steps (N) and the displacement R has a non-linear (y~f(x^0.5)) dependence. The mean values are calculated for a 1000 paths.
Please notice that neither x^2 or y^2 can have a value larger than N^2. This should have been a good checking point for the validity of your simulation.
The mean x^2 and y^2 have a linear dependence from the number of steps (N) and the displacement R has a non-linear (y~f(x^0.5)) dependence. The mean values are calculated for a 1000 paths.
Please notice that neither x^2 or y^2 can have a value larger than N^2. This should have been a good checking point for the validity of your simulation.
Week 2 - Buffon
Below is the solution for the calculation of π with the Buffon method.
The graph that presents the results.
The graph that presents the results.
Tuesday, February 12, 2008
Week 2 - Ogkos Prismatos
Below is the program that calculates the volume of a prism in 2,3,4,5 and 10 dimensions.
The results are shown below. For 10 dimensions a calculation with length=1 is very difficult unless we use a lot of random points.
The results are shown below. For 10 dimensions a calculation with length=1 is very difficult unless we use a lot of random points.
Week 1 - Triades
Below is the program that produces points in 3D space using all three of the random functions.
We observe strong correlations in the random function with and without overflow correction while there isn't any correlation in the ran1 function from numerical recipes.
We observe strong correlations in the random function with and without overflow correction while there isn't any correlation in the ran1 function from numerical recipes.
Friday, February 8, 2008
Week 1 - Diades
The solution for the production of random points in two dimensions.
Option 1 uses the random function without overflow, option 2 uses the random function with overflow correction and option 3 the ran1 function (Numerical Recipes).
Results from the random function without overflow.
We can observe correlations between the points.
Results from the random function with overflow correction.
We observe the reduction correlations compared with the previous one.
Results from the ran1 function (Numerical Recipes)
We observe the lack of any correlation.
Option 1 uses the random function without overflow, option 2 uses the random function with overflow correction and option 3 the ran1 function (Numerical Recipes).
Results from the random function without overflow.We can observe correlations between the points.
Results from the random function with overflow correction.We observe the reduction correlations compared with the previous one.
Results from the ran1 function (Numerical Recipes)We observe the lack of any correlation.
Week 1 - Zari
These are the solutions for the Part 1, 2 and 5a. Please notice that for the proper operation of the two first random functions all of their variables should be integers, especially when the correction for the overflow is needed. Also in the function from the numerical recipes the seed must be a negative integer.
From the results we can observe that the best function is the first one (although this shouldn't be the case).
Zari with normal mod use without overflow
Zari using overflow correction
Zari using the ran1 function from Numerical Recipes
From the results we can observe that the best function is the first one (although this shouldn't be the case).
Zari with normal mod use without overflow
Zari using overflow correction
Zari using the ran1 function from Numerical Recipes
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