Magic Number

Fun with Coding

Finding primes between 2 to 50.

Ken Iverson’s APL solution on Wikipedia

(2 = +⌿ 0 = (⍳N) ∘.| ⍳N) / ⍳N

Using the fact that for numbers 1 to N, a prime only has 2 factors — itself and 1.

In our case, for numbers 2 to N, a prime has only one factor.

The ‘multiplication table’ APL solution.

(~ m ∊ m ∘.× m) / m ← 1↓⍳50

It’s short and sweet. Definitely not fast.

Spaces are used to separate items in the list.
Optionally we can use comma or semicolon for the sake of clarity.

Usually ⧻ counts how many elements in a list.
Here ⧻₁ counts how many elements satisfied the condition.
And you can use ⧻₀ to count what’s false.

Alternatively we can use the infix form of the maximum operator:

It’s easier to read in 2 lines.

Let’s use ⧻₁ to show our intention better.

We can use R instead of a normal variable.
R stands for result from a previous statement.
So if you have ‘1 + 2 ; R + 6’, you get 9 as R = 1 + 2 = 3

R allows us to break down a complex expression into smaller simpler chunks.

Using R to construct the solution:

Introducing ·❘·
While ÷ means division, ·❘· means divisible.
So 6 ·❘· 3 = 1 (true) and 6 ·❘· 4 = 0 (false).

Similar expression but factoring out 3 and 5.

Moment of truth — let’s find all the multiples under 1000:

Making use of the weakness that nested list is ignored, that is
[ [ 1 2 3 ] ] evaluates to [ 1 2 3 ] …
Let’s tidy up the parentheses: