Revisions to Ruby-ize for loop - counting all the n-digit numbers that contain the digit 5 anywhere

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edited Apr 13, 2017 at 12:19

1

So it turns out there is a better algorithm for this. The algorithm is \$10^n - 9^n\$ and you can find an explanation of it over on Mathematics Exchange Mathematics Exchange. The improved algorithm completely removed my need for a loop, so I separated the test (printing) logic from the actual function and "ruby-ized" that loop instead.

def count_of_numbers_containing_5(power)
  (10**power)-(9**power)
end 

def test_it(pwr)
  (1..pwr).each {|i| puts count_of_numbers_containing_5(i)}
end

test_it 20

So it turns out there is a better algorithm for this. The algorithm is \$10^n - 9^n\$ and you can find an explanation of it over on Mathematics Exchange. The improved algorithm completely removed my need for a loop, so I separated the test (printing) logic from the actual function and "ruby-ized" that loop instead.

def count_of_numbers_containing_5(power)
  (10**power)-(9**power)
end 

def test_it(pwr)
  (1..pwr).each {|i| puts count_of_numbers_containing_5(i)}
end

test_it 20

So it turns out there is a better algorithm for this. The algorithm is \$10^n - 9^n\$ and you can find an explanation of it over on Mathematics Exchange. The improved algorithm completely removed my need for a loop, so I separated the test (printing) logic from the actual function and "ruby-ized" that loop instead.

def count_of_numbers_containing_5(power)
  (10**power)-(9**power)
end 

def test_it(pwr)
  (1..pwr).each {|i| puts count_of_numbers_containing_5(i)}
end

test_it 20

deleted 21 characters in body

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edited Jun 6, 2014 at 15:34

RubberDuck

30.9k
6
71
174

So it turns out there is a better algorithm for this. The algorithm is \$10^n - 9^n\$ and you can find an explanation of it over on Mathematics Exchange. The improved algorithm completely removed my need for a loop, so I separated the test (printing) logic from the actual function and "ruby-ized" that loop instead.

def count_of_numbers_containing_5(power_of_tenpower)
  (10**power_of_ten10**power)-(9**power_of_ten9**power)
end 

def test_it(pwr)
  (1..pwr).each {|i| puts count_of_numbers_containing_5(i)}
end

test_it 20

So it turns out there is a better algorithm for this. The algorithm is \$10^n - 9^n\$ and you can find an explanation of it over on Mathematics Exchange. The improved algorithm completely removed my need for a loop, so I separated the test (printing) logic from the actual function and "ruby-ized" that loop instead.

def count_of_numbers_containing_5(power_of_ten)
  (10**power_of_ten)-(9**power_of_ten)
end 

def test_it(pwr)
  (1..pwr).each {|i| puts count_of_numbers_containing_5(i)}
end

test_it 20

So it turns out there is a better algorithm for this. The algorithm is \$10^n - 9^n\$ and you can find an explanation of it over on Mathematics Exchange. The improved algorithm completely removed my need for a loop, so I separated the test (printing) logic from the actual function and "ruby-ized" that loop instead.

def count_of_numbers_containing_5(power)
  (10**power)-(9**power)
end 

def test_it(pwr)
  (1..pwr).each {|i| puts count_of_numbers_containing_5(i)}
end

test_it 20

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created Jun 4, 2014 at 17:03

RubberDuck

30.9k
6
71
174

So it turns out there is a better algorithm for this. The algorithm is \$10^n - 9^n\$ and you can find an explanation of it over on Mathematics Exchange. The improved algorithm completely removed my need for a loop, so I separated the test (printing) logic from the actual function and "ruby-ized" that loop instead.

def count_of_numbers_containing_5(power_of_ten)
  (10**power_of_ten)-(9**power_of_ten)
end 

def test_it(pwr)
  (1..pwr).each {|i| puts count_of_numbers_containing_5(i)}
end

test_it 20

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