CMCDragonkai

<?php

// O(n) time complexity with constant space complexity
function fibonacci_dynamic($n){

    if($n == 0){
        return 0;
    }elseif($n == 1 OR $n == 2){
        return 1;
    }

CMCDragonkai

let
    fib = number: fib' number 1 1;
    fib' = number: first: second:
        if number == 0
        then first
        else fib' (number - 1) second (first + second);
        # the second becomes the new first, the (first + second) becomes the new second
in
    fib 5

CMCDragonkai

# Note that you can also do things like for << b::2 <- <<1::1, 1::1, 1::1, 0::1>> >>, do: b
# It's pretty cool, it makes sure the b variable is composed of 2 bits. So it will segment the bitstring to represent and have create a b for every 2 bits.

# Binary Comprehension over a bitstring of size 1 bits
list = for << b::1 <- <<1::1, 1::1, 1::1, 0::1>> >>, do: b

# The Goal is the turn <<1::1, 1::1, 1::1, 0::1>> into a base 2 integer, which would be 14

# POWER is like 2^2 which means 2*2. So if x^y. Then there y recursions of x * x.
defmodule Math do

CMCDragonkai

%% this is a linked list append, which is O(n), because it needs to break up the list until the end, then add the list to be appended at the very end, then build up the list again by consing them!
%% unlike loops, each step is a recursion, and I guess there might some sort of tail call optimisation that makes sure that a stack doesn't grow forever in functional languages

:- pred append(list(T), list(T), list(T)).
:- mode append(in, in, out) is det.
:- mode append(out, out, in) is multi.

append(Xs, Ys, Zs) :-
    (
        Xs = [], Zs = Ys %% conjunction

CMCDragonkai

def depth_first_search(neighbour_list, root_node):
  visited_nodes = set()
  stack = [root_node]
  while stack:
    node = stack.pop()
    if node not in visited_nodes:
      visited_nodes.add(node)
      stack.extend(set(neighbour_list[node]) - visited_nodes)
  return list(visited_nodes)

CMCDragonkai

<?php

/**
 * Equal needle segregation
 * Given an zero-indexed array of values
 * Find the index where there is a equal number of X on the left hand side
 * To an equal number of Y on the right hand side.
 * X is the an arbitrary search needle.
 * Y is anything that is not the arbitrary search needle.
 * Example: [1, 2, 2, 3, 4];  X = 2

CMCDragonkai

<?php

/**
 * Logo Turtle Challenge
 * A turtle starts at (0,0).
 * You have an array of movements, where each element represents the distance of a single move.
 * Directions are cycled every four elements going through north, east, south, west.
 * Find the move where the turtle crosses through a point that it has passed through before.
 * This solution is somewhat space inefficient, and could become time inefficient if the distance range for a particular move is huge!
 * There is probably a solution that allows you to persist only the corner coordinates and then using line equations to figure out if they intersect.

CMCDragonkai

// our FizzBuzz language
var code = 
"for every number from 1 to 100\
    if the number is a multiple of 3 and it is a multiple of 5 then\
        print \"FizzBuzz\"\
    else if it is a multiple of 3 then\
        print \"Fizz\"\
    else if it is a multiple of 5 then\
        print \"Buzz\"\
    else\

CMCDragonkai

ometa List {
    listOf :p = apply(p):head (',' apply(p))*:tail -> [head].concat(tail),
    listOfDigits = listOf('digit'):l               -> l,
    END
}

var result = List.matchAll(
    '1,2,3,4,5',
    'listOfDigits'
);

CMCDragonkai

ometa Test {
    repeat :n :p = ?(n == 0) -> ('') // base case
                 | apply(p):l repeat(n - 1, p):r -> ((r.length > 0) ? [l].concat(r) : [l]), // concat the recursive matches
    tom = "TOM",
    repeatTOM = repeat(3, 'tom')
}

Test.matchAll(
    'TOMTOMTOM',
    'repeatTOM'