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The following advanced formulas are generally used when utilizing lists, arrays and matrices created directly into Math Expression custom variables or Google Sheet variable import.

You can create these formulas by adding new Math Expression custom variables or using several register nodes directly on the canvas.

As presented on our GDD integration presentation video, lists and matrices are the most useful tools in representing, referencing, and searching a given data set. Here are some common use cases for which arrays simply model possible inputs:

Gaming:
- Items stats – cost, damage, weight, rarity, defense, ability cost;
- Minion wave attacks – number of enemies, enemy type, XP rewards, difficulty, chance of defeat;
- Inventory management – inventory size, maximum capacity, lists of stored items;
- Crafting tables – item cost1, item cost2, item costx, crafting level requirement, generated item;
- Player stats based on level – HP, defense, basic & special attack, dexterity;

Other industries:
- E- commerce, shops & order management – number of purchased items, price, number of days for shipping, weight of the package, priority;
- Human resources management – name, age, salary, working hours, experience, department;
- Customer management – name, customer payment plan, customer age, activity time, engagement;
- Weather forecasting – day of the year, temperature, precipitation, wind speed, humidity;

With the modelling phase out of the way, we can now explore different operations, formulas, and structures to get specific results:

1. Generate random arrays of given length

🛑Problem: Create a one-dimensional or two-dimensional array of a given length [X] or [X,Y] filled with random values in given ranges between min and max.

❇️Solution: randomInt(size, min, max) where:

size = the length of an array or matrix to be filled with the random values;
min = minimum boundary for the random value, included;
max = maximum boundary for the random value, excluded;

Examples:
1. randomInt([3], 1, 5) – [rand1..5, rand1..5,rand1..5] – [2, 4, 1] – generate an array of length 3 with random integers between 1 and 5;

2. randomInt([4], 10) – [rand0..10, rand0..10,rand0..10, rand0..10] – [5, 9,1,9] – generate an array of length 4 with random integers between 0 and 10;

3. randomInt([3,3], 2, 9) – [[rand2..9, rand2..9,rand2..9],[rand2..9, rand2..9,rand2..9],[rand2..9, rand2..9,rand2..9]] – [[2,5,7],[3,3,8],[1,5,7]] – generate a matrix with 3 rows and 3 columns with random integers between 2 and 9;

2. Select a single random value from a list or a random row from a matrix

🛑 Problem: Get any singular random value from a one-dimensional array(list) or a 2-dimentional array (matrix). The element should be accessible on a different register, and the initial array is not modified upon selecting the random element.

❇️ List Solution: list[randomInt(1,squeeze(size(list)))] where:

list = the initial list of elements given as input (the pool of possible choices)
size(list) = returns the length of the initial list to get the maximum number of possible elements to choose from – returns the result as an array with a single element
squeeze() = a math.js function that removes the brackets [,] from matrices and lists when they contain a single element
randomInt(1, length_list) = returns a random number between 1 and the number of elements of the list. This value is used as an index value, meaning that the function will return the actual value from the randomly generated position in the list.

❇️ Matrix Solution: row(matrix, randomInt(1,subset(size(matrix), index(1)))) where:

matrix = the initial 2D array of elements given as input (the pool of possible choices)
size(matrix) = returns the length of the initial list to get the maximum number of possible elements to choose from – returns the result as an array with the number of rows and the number of columns;
subset(size, index(1)) = since the size() function returns both the number of rows and columns, we can use the subset() function and the first index to return only the number of rows
randomInt(1, number of rows) = returns a random number between 1 and the number of rows of the matrix.

Examples:
1. list = [1,2,3,4,5]
list[randomInt(1 ,squeeze(size(list)))] = 🎲a, where a= 1,2,3,4, or 5 randomly – returns only one random element from the given list;

2. list = [100,200,300]
list[randomInt(1 ,squeeze(size(list)))] = 🎲a, where a= 100, 200, or 300 randomly – returns only one random element from the given list;

3. matrix =[[1,1,1],[2,2,2],[3,3,3],[4,4,4]]
row(matrix, randomInt(1,subset(size(matrix), index(1)))) = 🎲[a,a,a], where a= 1,2,3, or 4 randomly – returns only one random row from the input matrix;

3. Get x random rows from a 2D array

🛑 Problem: Select a given number of rows (x) from a 2-dimensional array. The rows are selected randomly.

❇️ Solution: subset(array, index(randomInt([x],1,no_rows+1),range(1, no_columns))) where:

array = the entire inital 2D array given as an input;
x = the number of randomly selected rows to be returned;
no_rows = the number of rows included in the initial 2D array;
no_columns = the number of columns included in the initial 2D array;

Examples:
1. array = [[1, 1, 1], [2, 2, 2], [3, 3, 3]]
x = 1
subset(array, index(randomInt([x],1,4),range(1, 3))) = 🎲[a,a,a] – returns only one random list from the initial 2-dimensional array with 3 columns and 3 rows.

2. array = [[1,1],[2,2],[3,3],[4,4]]
x = 2
subset(array, index(randomInt([x],1,5),range(1, 2))) = [🎲[a,a,a],🎲[b,b,b]] – returns 2 (x) random lists creating a new matrix from the initial 2D array with 2 columns and 4 rows.

4. Traversing a list element by element during a simulation

🛑 Problem: During a simulation, using a register node, reference each element on a list by traversing it one element for each step. When the final element of the array is reached, the traversal should restart from the first element of the list.

❇️ Solution: list[index] and index.reset when index==size(list) where:

list = the entire initial list given as input;
index = the index of the traversed list; it is equivalent to the simulation step number;
index.reset = a Machinations construction using drains and state connections to reset the index to 1 when the last element of the array is reached;

🔧 Step-by-step guide to create the Machinations building block:

1. Create the index – an integer value which is incremented each simulation step, used to traverse the subsequent element of the list; Since “simulation step” == “index”, we can use an automatic source which increases the value of an index pool each step by 1.

2. Reference the elements from the list using the index: Since the index is now automatically updating each step, we can use the formula: list[index] in a register to store the traversed element from the list. Using the indexing syntax [ ], the mentioned register will generate the following values: step 1 = list[1]; step 2 = list[2], step x = list[x];

3. Reset the index back to 1 when the last element of the list is traversed: To complete this step, we must first determine the length of the list by using the formula squeeze(size(list)) on a register. Next, using a conditional register, we perform the check if index==size by referencing both values using state connections. Lastly, using an activator state connection, we will enable a drain node that removes all resources from the index pool when the index reset should be triggered.

The final building block can be observed below; in this example, the original array is set to [100,200,300,400], meaning that the values will loop once every 4 steps:

5. Find the index/position of a given element in a list

🛑 Problem: Given a list containing only distinct elements and a specific value X, return the position of the element X in the array. If the given value X is not found in the array, 0 should be returned instead.

❇️ Solution: (list==target)*(1:squeeze(size(list))) where:

list = the entire initial array given as input;
target = the value needed to be searched for in the list;
squeeze(size(list)) = returns the length of the initial list as an integer value;
1:size = range(1, size) = traverses the indexes of the list incrementally during a single simulation step;

Examples:
1. list = [10,20,30,40,50,60]
target = 40
(list==target)*(1:squeeze(size(list))) = 4 returns the position in the list where the target value is found

2. list= [323,6565,3434]
target = 6565
(list==target)*(1:squeeze(size(list))) = 2 returns the position in the list where the target value is found

3. list= [1,2,3,4,5]
target = 7
(list==target)*(1:squeeze(size(list))) = 0 returns 0 if the target element is not found in the initial list

6. Count the occurences of an element in an array

🛑 Problem: Given a constant value X, and a list with any number of elements (not distinct), return the number of occurrences of element X in the given list. If the target element is not found in the list, 0 is returned.

❇️ Solution: sum(equal(list,target)) where:

list = the entire initial list given as input;
target = the number you would want to search in the list;

Examples:

1. list = [10,20,10,20,10,20]
target = 20
sum(equal(list,target)) = 3 returns the number of occurrences where the target value is found

2. list= [20,20,40,88,88]
target = 20
sum(equal(list,target)) = 2 returns the number of occurrences where the target value is found

3. list= [1,2,3,4,5]
target = 7
sum(equal(list,target)) = 0 returns 0 if the target element is not found in the initial list

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