2

Ive been able to come up with one that is a completely solid star triangle but I'm having trouble creating a "hollow one" and a number one that both increments by whatever the number the user enters. Any help?

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6
  • Can you post some code showing you what you've attempted so far? Commented Jan 26, 2019 at 1:08
  • 1
    Okay sure Ive tried a couple of things so far. Commented Jan 26, 2019 at 1:10
  • The first and last lines of the triangle are special because the first only contains one * and the last is a full line. The rest is just a star plus (n-2) times a space plus another star, where n is the line number starting at 1. Your last try almost achieves what you want. Commented Jan 26, 2019 at 1:26
  • Okay thank you Ill give it a shot! Commented Jan 26, 2019 at 1:32
  • n = 1 while n >= 1 puts "* " * n n = n - 2 * " " + "*" end Commented Jan 26, 2019 at 1:43

5 Answers 5

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1

1st. iterate the lines

2nd. put "*" only in boundaries of the line: [0, i], filling " " inside:

3rd: exceptions for the 1st and last (n) case

n = 8

puts '*'

(n-2).times do |i|
  puts '*' + ' ' * (i) + '*'
end

puts '*' * n if n > 1
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2 Comments

That works! Could I try a similar method to produce the second triangle?
Sure, while iterating the lines you could iterate x <= [0, i] then puts x+1, you can refactor then after that.
1

This is my solution:

class RightTriangle
  class << self
    def draw_border(size:, char: '*')
      validate(size)

      1.upto(size) do |n|
        1.upto(n) do |o|
          break if n == size
          o == 1 || o == n ? print(char) : print(' ')
        end

        n.times { print char } if n == size

        puts if n > 0
      end
    end

    def draw_numbers(size:)
      validate(size)

      1.upto(size) do |n|
        1.upto(n) { |o| print o }
        puts
      end
    end

    private

    def validate(size)
      raise 'SizeError: `size` must be greater than 1' if size <= 1
    end
  end
end

# For the triangle border
RightTriangle.draw_border(size: 8) # character will be '*'
RightTriangle.draw_border(size: 8, char: 'a') # character will be 'a'

# For the numbers triangle
RightTriangle.draw_numbers(size: 8)
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3 Comments

What about the second triangle? Should it be nested?
@Void Oops, forgot that one. Fixed it.
draw_border if for the first triangle and draw_numbers is for the second triangle.
1 
def bt(n)
  1.upto(n) do |i|
    puts case i
    when 1
      '*'
    when n
      '*'*n
    else
      "*#{' '*(i-2)}*"
    end
  end
end

bt 8
*
**
* *
*  *
*   *
*    *
*     *
********


ROW = [*1..9, *'A'..'Z'].join
  #=> "123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ" 

def lt(n)
  1.upto(n) { |i| puts ROW[0,i] }
end

lt 8
1
12
123
1234
12345
123456
1234567
12345678


lt 22
1
12
123
1234
12345
123456
1234567
12345678
123456789
123456789A
123456789AB
123456789ABC
123456789ABCD
123456789ABCDE
123456789ABCDEF
123456789ABCDEFG
123456789ABCDEFGH
123456789ABCDEFGHI
123456789ABCDEFGHIJ
123456789ABCDEFGHIJK
123456789ABCDEFGHIJKL
123456789ABCDEFGHIJKLM


ROW = '😈'*10
lt 6
😈
😈😈
😈😈😈
😈😈😈😈
😈😈😈😈😈
😈😈😈😈😈😈
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0

You could create a simple method with two loops, for the second triangle:

def print_numbers_right_triangle(height)
  1.upto(height) do |row|
    1.upto(row) do |i|
      print i
    end
    puts
  end
end

print "Plese enter the height of the triangle:"
print_numbers_right_triangle(gets.strip.to_i)

Example Usage:

Plese enter the height of the triangle: 8
1
12
123
1234
12345
123456
1234567
12345678
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3 Comments

Are you sure? repl.it/repls/WelltodoGrowingOffice (Click run and look to the right of the screen)
I double checked everything and it works great! thank you.
@Void Updated answer with user input example
0

Other options, just for fun.

Hollow

n, cat, hyp = 4, "🎱", "🎱"

n.times do |n|
  a = Array.new(n+1){' '}
  a[0], a[-1] = cat, hyp
  puts a.join
end

It returns:

# 🎱
# 🎱🎱
# 🎱 🎱
# 🎱  🎱

Filled

n = 10
r = [*0..9, *'a'..'z']

n.times do |n|
  puts r[0..n].join
end

It returns

# 0
# 01
# 012
# 0123

If you shuffle the array r.shuffle[0..n].join, you could get random output:

# t
# 26
# ce0
# zqiw

Random emoji:

some_emotics = (0..9).each_with_object([]) { |n, o| o << "1F96#{n}".to_i(16) }.pack 'U*'
n.times do |n|
  puts some_emotics.split("").shuffle[0..n].join
end

to get:

# 🥡
# 🥦🥡
# 🥤🥥🥠
# 🥩🥣🥦🥥
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