Ruby AI Challenge

ai-challenge-prompt.md

You are an expert software engineer that understands all aspects of the Ruby programming language (version 3).

Consider the following tree structure:

.
├── 1
│   ├── 1.1
│   │   ├── 1.1.1
│   │   ├── 1.1.2
│   │   └── 1.1.3
│   ├── 1.2
│   │   ├── 1.2.1
│   │   ├── 1.2.2
│   │   └── 1.2.3
│   └── 1.3
│       ├── 1.3.1
│       ├── 1.3.2
│       └── 1.3.3
├── 2
│   ├── 2.1
│   │   ├── 2.1.1
│   │   ├── 2.1.2
│   │   └── 2.1.3
│   ├── 2.2
│   │   ├── 2.2.1
│   │   ├── 2.2.2
│   │   └── 2.2.3
│   └── 2.3
│       ├── 2.3.1
│       ├── 2.3.2
│       └── 2.3.3
└── 3
    ├── 3.1
    │   ├── 3.1.1
    │   ├── 3.1.2
    │   └── 3.1.3
    ├── 3.2
    │   ├── 3.2.1
    │   ├── 3.2.2
    │   └── 3.2.3
    └── 3.3
        ├── 3.3.1
        ├── 3.3.2
        └── 3.3.3

Each node within this tree is labed by the parent node plus it's position with the parent.

For example, 3.3.2 is the 2nd child of 3.3, 3.3 is the 3rd child of 3, and 3 is the 3rd child of the root (.)

Write a Ruby method that will take two properties:

• max_depth (Integer) indicates how many times nodes should be nested. The node 3.3.3 is at a depth of 2, meaning the are two parent nodes before the root node.

• items_per_depth (Integer) is the number of children each node should have.

Have the method simply print the node values. For example, print_tree(max_depth: 2, items_per_depth: 3) should print out the following:

1
1.1
1.1.1
1.1.2
1.1.3
1.2
1.2.1
1.2.2
1.2.3
1.3
1.3.1
1.3.2
1.3.3  
2
2.1
2.1.1
2.1.2
2.1.3
2.2
2.2.1
2.2.2
2.2.3
2.3
2.3.1
2.3.2
2.3.3  
3
3.1
3.1.1
3.1.2
3.1.3
3.2
3.2.1
3.2.2
3.2.3
3.3
3.3.1
3.3.2
3.3.3  

ai-challenge-results.md

ChatGPT (using MindMac)

Response

You can achieve this output using a recursive method to generate and print the node values. Here's a Ruby method that meets the requirements:

```ruby
def print_tree(max_depth:, items_per_depth:)
  (1..items_per_depth).each do |num|
    print_node(num.to_s, 1, max_depth, items_per_depth)
  end
end

def print_node(prefix, level, max_depth, items_per_depth)
  puts prefix
  return if level == max_depth

  (1..items_per_depth).each do |i|
    child_prefix = "#{prefix}.#{i}"
    print_node(child_prefix, level + 1, max_depth, items_per_depth)
  end
end

When you call print_tree(max_depth: 3, items_per_depth: 3), it will print the desired output as per the provided tree structure. This method uses recursion to generate and print the node values with the specified maximum depth and number of children per node.

Results

1
1.1
1.2
1.3
2
2.1
2.2
2.3
3
3.1
3.2
3.3

Google Bard

Response

Sure, here is the Ruby method that will print the node values based on the specified max_depth and items_per_depth:

def print_tree(current_depth, parent_value, items_per_depth)
  if current_depth == 0
    puts parent_value
    return
  end

  for child_position in 1..items_per_depth
    child_value = "#{parent_value}.#{child_position}"
    print_tree(current_depth - 1, child_value, items_per_depth)
  end
end

To use this method, you can call it with the desired max_depth and items_per_depth values. For example, to print the tree structure with a maximum depth of 2 and 3 items per depth, you would call the method like this:

print_tree(2, 1, 3)

This will print the following output:

1
1.1
1.1.1
1.1.2
1.1.3
1.2
1.2.1
1.2.2
1.2.3
1.3
1.3.1
1.3.2
1.3.3
2
2.1
2.1.1
2.1.2
2.1.3
2.2
2.2.1
2.2.2
2.2.3
2.3
2.3.1
2.3.2
2.3.3
3
3.1
3.1.1
3.1.2
3.1.3
3.2
3.2.1
3.2.2
3.2.3
3.3
3.3.1
3.3.2
3.3.3

Results (part 1)

Sure, here is the Ruby method that will print the node values based on the specified max_depth and items_per_depth:

def print_tree(current_depth, parent_value, items_per_depth)
  if current_depth == 0
    puts parent_value
    return
  end

  for child_position in 1..items_per_depth
    child_value = "#{parent_value}.#{child_position}"
    print_tree(current_depth - 1, child_value, items_per_depth)
  end
end

To use this method, you can call it with the desired max_depth and items_per_depth values. For example, to print the tree structure with a maximum depth of 2 and 3 items per depth, you would call the method like this:

print_tree(2, 1, 3)

This will print the following output:

1
1.1
1.1.1
1.1.2
1.1.3
1.2
1.2.1
1.2.2
1.2.3
1.3
1.3.1
1.3.2
1.3.3
2
2.1
2.1.1
2.1.2
2.1.3
2.2
2.2.1
2.2.2
2.2.3
2.3
2.3.1
2.3.2
2.3.3
3
3.1
3.1.1
3.1.2
3.1.3
3.2
3.2.1
3.2.2
3.2.3
3.3
3.3.1
3.3.2
3.3.3

I replied with the following:

Your solution did not follow the requirements. The method that is called should only have the two arguments and none others.

Results (part 2)

I apologize for the previous incorrect solution. Here's the corrected version of the print_tree method that only takes two arguments:

def print_tree(max_depth, items_per_depth)
  current_depth = 1
  parent_value = 1

  while current_depth <= max_depth
    for child_position in 1..items_per_depth
      child_value = "#{parent_value}.#{child_position}"
      puts child_value
      parent_value = child_value
    end

    parent_value = 1
    current_depth += 1
  end
end

To use this method, you can call it with the desired max_depth and items_per_depth values. For example, to print the tree structure with a maximum depth of 2 and 3 items per depth, you would call the method like this:

print_tree(2, 3)

This will print the following output:

1.1
1.2
1.3
2.1
2.2
2.3
3.1
3.2
3.3

RubyMine AI

Response

Sure! You need a recursive method to obtain this. Here's a code that accomplishes what you need:

def print_tree(prefix = '', depth = 0, max_depth:, items_per_depth:)
  return if depth > max_depth

  items_per_depth.times do |i|
    item = (i + 1).to_s
    puts "#{prefix}#{item}"
    print_tree("#{prefix}#{item}.", depth + 1, max_depth: max_depth, items_per_depth: items_per_depth)
  end
end

print_tree(max_depth: 2, items_per_depth: 3)

This simple recursive method keeps track of the current depth of the recursion and the current prefix of the item, adding to it as it dives deeper. It checks if the current depth has exceeded the maximum allowed depth before recursive dive for each item. If not, it recurses with an incremented depth and an updated prefix.

Results

1
1.1
1.1.1
1.1.2
1.1.3
1.2
1.2.1
1.2.2
1.2.3
1.3
1.3.1
1.3.2
1.3.3
2
2.1
2.1.1
2.1.2
2.1.3
2.2
2.2.1
2.2.2
2.2.3
2.3
2.3.1
2.3.2
2.3.3
3
3.1
3.1.1
3.1.2
3.1.3
3.2
3.2.1
3.2.2
3.2.3
3.3
3.3.1
3.3.2
3.3.3