Maths Shortcut Tricks for Code

You don’t need a PhD in maths to start programming. However, knowing some tricks and basic concepts doesn’t hurt.

Check for One Positive and One Negative Number

Let’s start with basic arithmetic. You should know that:

  • When multiplying two positive numbers, we get a positive number
  • When multiplying two negative numbers, we get a positive number
  • When multiplying one positive and one negative number, we get a negative number.

This is useful when we need a user to input two numbers: one positive and one negative. Indeed, if the sum of those two numbers is negative, we can be sure the user inputted both a negative and negative numbers. Otherwise the result would be positive.

We can use this trick as a break condition. 

loop do
  first_number = gets.chomp.to_i
  second_number = gets.chomp.to_i
  break if first_number * second_number < 0 # Sum must be a negative number
  puts '>> Sorry. One integer must be positive, one must be negative.'
end

# Rest of the code...

Intersection (Set Theory)

The intersection of two sets \(A\) and \(B\) is the set containing all elements of \(A\) which are also in \(B\). It is denoted \(A \cap B\).

Ruby’s count method uses intersection. The way this count method uses intersection is actually the basis for how other methods “works”. Therefore it is important to understand how it works.

Imagine a String object hello world of eleven characters. We will assign an hexadecimal index for each characters, to make it simpler to know which characters we are talking about. 

# 123456789AB
 'hello world'.count('lo')  # => 5

In the example above, the count method called on the String object counts the occurrence of the characters passed to it as an argument. Here, characters 'l' and 'o'. While there are two characters, they form the only argument passed to the method. This means the set contains all elements of itself, i.e., the two characters. The methods counts (literally) each occurrence of characters included in the set: 3, 4, 5, 8 and A of our hexadecimal index.

We can verify the selected characters with the scan method: 

'hello world'.scan(/l|o/)  # => ["l", "l", "o", "o", "l"]

Let’s spice things up by passing a second argument to the count method. 

# 123456789AB
 'hello world'.count('lo', 'o')  # => 2

2? Why is that? Well count returns the occurrence of the characters at the intersection of the two parameters passed to it. Let’s break it down what’s happening step by step.

First, and this may looks obvious but let’s check which characters are passed to count, i.e., what’s inside the two parameters: 

'lo'.split('')  # => ["l", "o"]
'o'.split('')   # => ["o"]

Nothing revolutionizing here. The characters passed to count are l, o for the first argument and o for the second.

Now remember, the count method is interested in the intersection of the set of characters passed to it as arguments. In mathematics, if we call the first argument \(A\) and the second argument \(B\), we can write it like this: \(A \cap B\) as seen before.

Let’s replace \(A\) and \(B\) with arrays containing the characters for each argument: ["l", "o"] ∩ ["o"].

The next step is to find every characters that are in both ["l", "o"] and ["o"]:

["l", "o"] \(\cap\) ["o"] = ["o"]

The character 'o' is the only one! We can verify this is true using the AND operator: 

'lo'.split('') && 'o'.split('')  # => ["o"]

Finally, count method counts the number of occurrence of "o" in the string it is called upon: characters at 5 and 8 in our hexadecimal index.

This is true for any number of arguments: with \(A \cap B \cap C\) you would have the count for the characters that is present in \(A\), \(B\) and \(C\). If there is none, count returns 0.

Factorial

A factorial is an operation that involves multiplying sequential positive numbers together. They are often written with a exclamation mark: n!.

By definition, \(0!\) and \(1!\) are \(1\).

$$ \begin{split} 0! &= 1 \\ 1! &= 1 \\ 2! &= 2 \times 1 = 2 \\ 3! &= 3 \times 2 \times 1 = 6 \\ 4! &= 4 \times 3 \times 2 \times 1 = 24 \\ 5! &= 5 \times 4 \times 3 \times 2 \times 1 = 120 \end{split} $$

And so forth.

In Ruby, you can use inject or reduce to simulate a factorial, passing in :* as an argument. Take care however to start the collection of number at 1, otherwise as those two methods do not know the rule of \(0!\), your result will always be 0 (\(0 \times n = 0\)). 

(1..5).to_a.reduce(:*)
# => 120

Prime Number

A prime number is a number greater than 1 that is not a product of two smaller numbers. In other words, when divided by every number between 2 and itself, the results is not a natural number (0, 1, 2, 3…).

To find a prime number, one should divide the candidate number by 2, 3, 4 etc until itself and see if the remainder is zero. If it is, that means the number we are working on is not prime.

In reality, we only need to test number between 2 up to the square root of the candidate number. 

def prime?(candidate)
  return false if candidate < 2

  limit = Integer.sqrt(candidate)

  2.upto(limit) do |factor|
    return false if candidate % factor == 0
  end

  true
end

p prime?(2)  # => true
p prime?(3)  # => true
p prime?(4)  # => false
p prime?(5)  # => true
p prime?(6)  # => false

Expanded Form

Expanded form are a way to write number into the value of each of its digits.

There are several way to write expanded form ; the standard is the following: $$734 = 700+30+4$$

We commonly use the base 10 system, meaning each digit represents a power of 10. The operation above can be written as the product of the digit and the power of 10s place, like this:

$$734 = (7 \times 100)+(3 \times 10)+(4 \times 1)$$

Let’s convert the values to powers of 10 so that we can see a pattern:

$$734 = (7 \times 10^2)+(3 \times 10^1)+(4 \times 10^0)$$

As you can see from the operation above, the powers of 10 are descending from 2 to 0. This seems like a pattern we can use in programming. 

def expanded_form(num)
  digits = num.digits
  expanded = []

  digits.each_with_index do |digit, power|
    if digit > 0
      expanded << digit * (10**power)
    end
  end

  expanded.reverse.join(' + ')  # Need to reverse because of Integer#digits
end

p expanded_form(12) == '10 + 2'
p expanded_form(42) == '40 + 2'
p expanded_form(70304) == '70000 + 300 + 4'

You can modify the base (here, 10) to convert from one base to base 10. For example from base 2 or 8 to base 10.

Source

Iterate over Collection with Fixed Length

Another tiny trick of maths one can use is to easily iterate over a collection with a fixed length thanks to the modulo operator %.

Let’s say we have an array object of five elements. 

arr = ['A', 'B', 'C', 'D', 'E']

For whatever reason, we cannot iterate over it with each method, maybe because we need to know retain the current position inside the array between each iteration (wink wink Circular Buffer). Whatever the reason, simply incrementing a variable index won’t help us past 4. 

arr = ['A', 'B', 'C', 'D', 'E']
index = 5
arr[index] # => nil

Here, we want arr[5] to return the first element ('A'), arr[6] to return the second element ('B') etc.

The easiest way to do it is to increment our variable index and calling modulo passing in the collection’s length as a parameter. In other words: 

index = 4
new_index = (index + 1) % arr.length # => 0
arr[new_index] # => 'A'

This works because the % method discards the quotient and returns the rest. It’s easier to see what’s going on using the divmod method, which returns both the quotient and the rest (or modulus). 

arr = ['A', 'B', 'C', 'D', 'E']
index = 21

_, new_index = (index + 1).divmod(arr.length) # => [4, 2]

arr[new_index] # => 'C'

Here, 21 is 4 whole iterations over a collection with 5 elements, rest 2.

Replace index + 1 with just index if you don’t wish to increment the local variable index, but just wish to find the position in the collection.

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