Category Archives: Mathematics

Generate random weighted variables

In this post I describe the common problem how to generate random weighted variables. For example you have a die with 6 faces that is fixed, so the probability of an 1 is not the same with the probability of a 2 etc.

The table below shows an example:

Face Probability
1  25
2  25
3  20
4  20
5  5
6  5

One easy solution is to reduce the problem to another one with a die with 100 faces that is not fixed.

You generate a random value from 0-99 that follows the uniform distribution. Do not use the modulo function to do that. See this post instead.

Once you have that value you perform the following steps:

You check the space within which that value relies.

  1. between 0 and 24, you generate 1 as the weighted variable
  2. between 25 and 49, you generate 2 as the weighted variable
  3. between 50 and 69, you generate 3 as the weighted variable
  4. between 70 and 89, you generate 4 as the weighted variable
  5. between 90 and 94, you generate 5 as the weighted variable
  6. between 95 and 99, you generate 6 as the weighted variable

 

 

 

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Generate uniformly distributed random variables

Formulas shown below generate values that follow uniform distribution.

Using simple modulo methods is not considered as a good solution (although used a lot by students) as the module function does not generate all the variables with the same frequency.

Random integer between [ 0, 1 ]

int r = (int) (rand()/(RAND_MAX + 0.0));

Random integer between [ 0, 1 )

int r = (int) (rand()/(RAND_MAX + 1.0));

Random integer between [ 0, N ]

int r = (int) (N * (rand() / (RAND_MAX + 0.0)));

Random integer between [ 0, N )

int r = (int) (N * (rand()/(RAND_MAX + 1.0)));

Random integer between [ M, N ]

int r = (int) (Μ + (rand()/(RAND_MAX + 0.0))*(N-M+1));

Random integer between [ M, N )

int r = (int) (Μ + (rand()/(RAND_MAX + 1.0))*(N-M+1));

Summary

The general formula is:

int r = (int) (A + (rand()/(RAND_MAX + C))*B);

Values of M, N, K are shown in the following table:

Range A B  C
 [ 0, 1 ]  0  1  0.0
 [ 0, N ]  0  N  0.0
 [ M, N ]  M  N – M + 1  0.0
 [ 0, 1 )  0  1  1.0
 [ 0, N )  0  N  1.0
 [ M, N )  M  N – M + 1  1.0

Generating values of normal distribution – Marsaglia method

Assuming you can already produce variables of uniform distribution, you can produce variables of normal distribution using various formulas. Two of the most important are:

  1. Box–Muller method
  2. Marsaglia polar method

Most methods are based on Box-Muller method.

The Marsaglia method is my favorite since it does not require using sin() or cos() functions and the steps are very easily to implement.

Here is a sequence of the steps:

  1. Generate a value that follows uniform distribution in any space you want (eg: [ 0 , 1 ))
  2. Map that value to space (-1, +1) and assign it to variable U
  3. Repeat steps 1 and 2 and assign the result to variable V
  4. Calculate S = U*U + V*V
  5. if S = 0 or S >= 1 then free all variables if needed and restart from the beginning
  6. The following two variables will be independent and standard normally distributed (mean = 0, variance = 1):

marsaglia

Optionally you can add m to the quantities above to change the mean value of the distribution.

 

Distance between point and rectangular box without case logic

Assuming you have a point (x,y) and a rectangle parallel to x and y axis defined by its upper left and lower right corner. How can we calculate its distance without using case logic?

P = (x,y)
A = { (x1,y1), (x2,y2) }

The answer is:

dx = max(A.x1 - p.x, 0, p.x - A.x2)
dy = max(A.y1 - p.y, 0, p.y - A.y2)
distance = sqrt(dx*dx + dy*dy)

Its very tricky.

Here is a great analysis here by MultiRRomero:

http://stackoverflow.com/questions/5254838/calculating-distance-between-a-point-and-a-rectangular-box-nearest-point

Determine if two rectangles intersect without case logic

Assuming you have two rectangles A and B that exist parallel to the x and y axis and are defined by their upper left (x1,x2) and lower right corners (x2,y2). How can you calculate if A and B intersect without using case logic?

A = { (x1,y1), (x2,y2) }
B = { (x1,y1), (x2,y2) }

 

intersect

 

The answer is:

if (A.x1 < B.x2 && 
    A.x2 > B.x1 &&  
    A.y1 < B.y2 && 
    A.y2 > B.y1) {
  then A and B intersect
}

There is a great analysis here:

http://stackoverflow.com/questions/306316/determine-if-two-rectangles-overlap-each-other

And an awesome interactive visualization by Matthew Crumley here:

https://silentmatt.com/rectangle-intersection/

Frequently used Unicode math operators

This is a selection of the operators i use frequently while taking notes. You can find a detailed table on wikipedia.

You can find ways to type the symbols here and a complete PDF for math operators here

Predicate logic
existential quantification U+2203 U+2204
universal quantification U+2200
logical conjunction
logical disjunction
U+2227
U+2228

material implication
implication
U+21D2
U+2192

material equivalence U+21D4
 negation  U+00AC  ¬
entails
provable from
U+22A8
U+22A2

sets

proper subset U+2282
U+2283

U+2284
U+2285

subset U+2286
U+2287

U+2288
U+2289

element of U+2208
U+220B

U+2209
U+220C

 empty set U+2205  ∅
intersection
union
U+22C2
U+22C3

cross product U+2a2f

Here are some Copy-Q tabs to use:

  1. With set operators here
  2. With logic operators here