I have written two functions where in one of the the break if array.length == num statements works, while the other does not. Here are the methods and their tests, starting with the working:
def primes(n)
ret = []
return [] if n < 1
for num in 2..+1.0/0.0
ret << num if prime?(num)
break if ret.count == n
end
ret
end
def prime?(num)
(1..num).select {|x| num % x == 0}.count == 2
end
puts "\nPrimes:\n" + "*" * 15 + "\n"
puts primes(0) == []
puts primes(1) == [2]
puts primes(2) == [2,3]
puts primes(6) == [2,3,5,7,11,13]
Related
Now I have a ruby like this:
def fizzbuzz(numSize)
result = []
1.upto(numSize) do |num|
if num % 15 == 0
result << "FizzBuzz"
elsif num % 3 == 0
result << "Fizz"
elsif num % 5 == 0
result << "Buzz"
else
result << num.to_s
end
end
result
end
print fizzbuzz(10) {|item| "-#{i1tem}-"}
If I want to print the result like this:
["-1-", "-2-", "-Fizz-", "-4-", "-Buzz-", "-Fizz-", "-7-", "-8-", "-Fizz-", "-Buzz-"]
What can I modify my code in method fizzbuzz if I can not change the code:
print fizzbuzz(10) {|item| "-#{i1tem}-"}
Thanks
That block is being given to your method, but you're not making use of it. That's an easy fix:
def fizzbuzz(numSize, &block)
# ... (existing code) ...
result.map(&block)
end
Where that transforms the result value using map.
Note this requires fixing the typo in your print block which is i1tem not item.
It's also worth noting you should avoid this pattern:
x = [ ]
y.each do |v|
x << f(v)
end
x
That's just a long-winded version of this:
y.map do |v|
f(v)
end
Where when you're transforming on a 1:1 basis from the source just use map.
In your case that reduces the code to this more minimal form that has a lot less repetition:
def fizzbuzz(numSize, &block)
1.upto(numSize).map do |num|
if num % 15 == 0
"FizzBuzz"
elsif num % 3 == 0
"Fizz"
elsif num % 5 == 0
"Buzz"
else
num.to_s
end
end.map(&block)
end
I have written the logic for the program to perform FizzBuzz operations:
fizzbuzz
module FizzBuzz
class Operation
def input
puts 'Enter a number upto which Fizz/Buzz needs to be printed'
num = gets.chomp.to_i
fizzbuzz_function(num)
end
def fizzbuzz_function(num)
for i in 1..num
if i % 3 == 0 && i % 5 == 0
puts 'FizzBuzz'
elsif i % 3 == 0
puts 'Fizz'
elsif i % 5 == 0
puts 'Buzz'
else
puts i
end
end
end
end
res = Operation.new
res.input
end
But I am trying to print the output in form of a table.
Here is FizzBuzz in form of a table:
def fizzbuzz_gen(num)
Enumerator.new do |y|
(1..num).each do |i|
if i % 3 == 0 && i % 5 == 0
y << 'FizzBuzz'
elsif i % 3 == 0
y << 'Fizz'
elsif i % 5 == 0
y << 'Buzz'
else
y << i.to_s
end
end
end
end
def fill_to_width(width, e)
result = ""
future_length = -1
while result.length + future_length < width
result << e.next
result << " "
future_length = e.peek.length
end
result.center(width)
end
def format_table(num)
fb = fizzbuzz_gen(num)
begin
puts fill_to_width(75, fb)
puts fill_to_width(75, fb)
loop do
puts "%10s%s%31s%s" % ["", fill_to_width(12, fb), "", fill_to_width(12, fb)]
end
rescue StopIteration
end
end
format_table(100)
There may be less numbers output than specified, in order for one leg not to be shorter than another.
I'm working on project Euler #35. I am getting the wrong number returned and I can't find where I have done wrong!
def is_prime?(num)
(2..Math.sqrt(num)).each { |i| return false if num % i == 0}
true
end
def is_circular?(num)
len = num.to_s.length
return true if len == 1
(len - 1).times do
new_n = cycle(num)
break unless is_prime?(new_n)
end
end
def cycle(num)
ary = num.to_s.split("")
return ary.rotate!.join.to_i
end
def how_many
circulars = []
(2..1000000).each do |num|
if is_prime?(num) && is_circular?(num)
circulars << num
end
end
p circulars.count
end
how_many #=> 14426
The returned number is '14426'. I am only returning the circular primes, supposedly the correct answer is '55'
I have edited your code with few fixes in Ruby way. Your mistake was including corect set of [a, b, c] three times to count, instead of counting them as a one circular prime number. Your answer was correct, while 55 is the number of unique sets.
require 'prime'
class Euler35
def is_circular?(num)
circulars_for(num).all?{ |el| ::Prime.instance.prime?(el) }
end
def circulars_for(a)
a.to_s.split("").length.times.map{|el| a.to_s.split("").rotate(el).join.to_i }
end
def how_many
circulars = []
::Prime.each(1_000_000) do |num|
continue if circulars.include?(num)
if is_circular?(num)
circulars << circulars_for(num)
end
end
circulars.count
end
end
puts Euler35.new.how_many # => 55
I often find myself checking multiple conditions. How can I cut down on the number of lines used to achieve the same effect?
def super_fizzbuzz(array)
final = []
for num in array
if num % 15 == 0
final << 'FizzBuzz'
elsif num % 5 == 0
final << 'Buzz'
elsif num % 3 == 0
final << 'Fizz'
else
final << num
end
end
final
end
def super_fizzbuzz(array)
array.map do |num|
a = []
a << 'Fizz' if num % 3 == 0
a << 'Buzz' if num % 5 == 0
a.empty? ? num : a.join()
end
end
def super_fizzbuzz(array)
final = []
array.each do |num|
num % 15 == 0 ? final << 'FizzBuzz' : num % 5 == 0 ? final << 'Buzz' : num % 3 == 0 ? final << 'Fizz' : final << num
end
final
end
But your way is more readable.
def super_fizzbuzz(array)
array.map do |num|
case 0
when num % 15 then "FizzBuzz"
when num % 5 then "Buzz"
when num % 3 then "Fizz"
else num
end
end
end
This is slightly more complex, but reduces number of explicit coded conditionals to 2:
FIZZBUZZ = { 3 => 'Fizz', 5 => 'Buzz' }
def super_fizzbuzz(array)
array.map do |num|
fbs = FIZZBUZZ.select do |divisor,cat|
num % divisor == 0
end.values
fbs.empty? ? num : fbs.join
end
end
There is always the danger when coding for DRY that you take things too far. In this case, with only two overlapping categories, I think the above is a little unwieldy. However, add another category or two:
FIZZBUZZ = { 3 => 'Fizz', 5 => 'Buzz', 7 => 'Boom', 11 => 'Whizz' }
and it starts to look smarter.
Quote:
I think Fizz-Buzz is "hard" for some programmers because (#1) it doesn't fit into any of the patterns that were given to them in school assignments, and (#2) it isn't possible to directly and simply represent the necessary tests, without duplication, in just about any commonly-used modern programming language.
Source: c2.com Wiki
Another way:
def super_fizzbuzz(arr)
arr.map do |e|
s = ''
s << 'Fizz' if (e%3).zero?
s << 'Buzz' if (e%5).zero?
s = e if s.empty?
s
end
end
super_fizzbuzz [9, 25, 225, 31]
#=> ["Fizz", "Buzz", "FizzBuzz", 31]
Spoiler alert: I am a true novice. Tasked with figuring out fizz buzz in
ruby for a class and while I have found more than a few versions of code
that solve the problem, my understanding is so rudimentary that I cannot
figure out how these examples truly work.
First question(refer to spoiler alert if you laugh out loud at this):
How do i print out numbers one through 100 in Ruby?
Second question: can 'if else" be used to solve this? My failed code is
below(attachment has screen shot):
puts('Lets play fizzbuzz')
print('enter a number: ')
number = gets()
puts(number)
if number == % 3
puts ('fizz')
elsif number == % 5
puts ('buzz')
elsif number == %15
puts ('fizzbuzz')
end
Thanks,
Thats ok being a novice, we all have to start somewhere right? Ruby is lovely as it get us to use blocks all the time, so to count to 100 you can use several methods on fixnum, look at the docs for more. Here is one example which might help you;
1.upto 100 do |number|
puts number
end
For your second question maybe take a quick look at the small implementation i whipped up for you, it hopefully might help you understand this problem:
1.upto 100 do |i|
string = ""
string += "Fizz" if i % 3 == 0
string += "Buzz" if i % 5 == 0
puts "#{i} = #{string}"
end
First question: this problem has several solutions. For example,
10.times { |i| puts i+1 }
For true novice: https://github.com/bbatsov/ruby-style-guide
another method that can be helpful :
puts (1..100).map {|i|
f = i % 3 == 0 ? 'Fizz' : nil
b = i % 5 == 0 ? 'Buzz' : nil
f || b ? "#{ f }#{ b }" : i
}
As a one liner
(1..100).map { |i| (i % 15).zero? ? 'FizzBuzz' : (i % 3).zero? ? 'Fizz' : (i % 5).zero? ? 'Buzz' : i }
In Regards to your failed code, your conditional statements should be like this:
if number % 3 == 0
puts "Fizz"
end
if number % 5 == 0
puts "Buzz"
end
You don't want the last elsif statement because it will never get executed
(if a number is not divisible by 3 or divisible by 5, then it is certainly not divisible by 15)
Adjust for this by changing the second elsif to simply and if and if the number is divisble by 5 and not by 3, then Fizz will not be outputted but Buzz Will be
I'm just showing you how to correct your code, but as others have pointed out, there are far more elegant solutions in Ruby.
Not the most beautiful way to write it but good for beginners and for readability.
def fizzbuzz(n)
(1..n).each do |i|
if i % 3 == 0 && i % 5 == 0
puts 'fizzbuzz'
elsif i % 3 == 0
puts 'fizz'
elsif i % 5 == 0
puts 'buzz'
else
puts i
end
end
end
fizzbuzz(100)
1.upto(100).each do |x| # Question #1 The 'upto' method here takes is
# what you would use to count in a range.
if (x % 3 == 0) && (x % 5 == 0)
puts " Fizzbuzz"
elsif x % 3 == 0
puts " Fizz"
elsif x % 5 == 0
puts " Buzz"
else
puts x
end
end
Question #2 Yes you can but I would look for a more elegant way to write this as a part of a definition like
def fizzbuzz(last_number)
1.upto(last_number).each do |x|
if (x % 3 == 0) && (x % 5 == 0)
puts " Fizzbuzz"
elsif x % 3 == 0
puts " Fizz"
elsif x % 5 == 0
puts " Buzz"
else
puts x
end
end
end
This is the answer that helped me to understand that no variables are being created with the .each method. Sorry about my indenting. Still learning how to use Stackoverflow text editing.
As for a more complex solution, that's one way you could build
a simple DSL for quickly modifying the FizzBuzz programme (adding new divisors with their own keywords)
class FizzBuzzer
# #return [Hash{String, Symbol => Integer}]
attr_reader :keywords
# #param keywords [Hash{String, Symbol => Integer}]
def initialize(keywords)
#keywords = keywords
end
# #param range [Range]
# #return [void]
def call(range)
range.each do |num|
msg = ''
#keywords.each do |name, divisor|
msg << name.to_s if (num % divisor).zero?
end
msg = num if msg.empty?
puts msg
end
puts
end
end
# create a fizz buzzer with custom keywords for divisors
CLASSIC_FIZZ_BUZZER = FizzBuzzer.new Fizz: 3, Buzz: 5
# print for a particular range
CLASSIC_FIZZ_BUZZER.call(1..25)
# you can easily define an extended fizz buzzer
EXTENDED_FIZZ_BUZZER = FizzBuzzer.new Fizz: 3, Buzz: 5, Bazz: 7, Fuzz: 11 # print 'Fuzz' when divisible by 11
EXTENDED_FIZZ_BUZZER.call(1..25)
Here's a quite elegant solution.
(1..100).each do |num|
msg = ''
msg << 'Fizz' if (num % 3).zero?
msg << 'Buzz' if (num % 5).zero?
msg = num if msg.empty?
puts(msg)
end
It can be even more compact
(1..100).each do |num|
(msg ||= '') << 'Fizz' if (num % 3).zero?
(msg ||= '') << 'Buzz' if (num % 5).zero?
puts msg || num
end
FizzBuzz
(1..100).each do |num|
if num % 3 == 0 && num % 5 == 0
puts "#{num}. FIZZBUZZ!"
elsif num % 3 == 0
puts "#{num}. FIZZ!"
elsif num % 5 == 0
puts "#{num}. BUZZ!"
else
puts "#{num}."
end
end
First question:
for i in 1..100
puts i
end
Here is my most "idiomatic ruby" solution:
class FizzBuzz
def perform
iterate_to(100) do |num,out|
out += "Fizz" if num.divisable_by?(3)
out += "Buzz" if num.divisable_by?(5)
out || num
end
end
def iterate_to(max)
(1..max).each do |num|
puts yield num,nil
end
end
end
class Fixnum
def divisable_by?(num)
self % num == 0
end
end
class NilClass
def +(other)
other
end
end
FizzBuzz.new.perform
And it works:
https://gist.github.com/galori/47db94ecb822de2ac17c