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edited Jun 4, 2014 at 17:49

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Ruby-ize for loop - findingcounting all the n-digit numbers less than 10000 that contain the digit 5 anywhere

I wrote this a while back when my fiance was taking a Number Theory class. I wrote about it here and it recently came back to my attention. Anytime I write a for loop in Ruby I feel kind of dirty. I would like to "Ruby-ize" this routine if there is a more Ruby way to do it. Also, is this the most efficient algorithm for this?

Original Problem:

Count all theHow many numbers less than 10000 that contain the digit 5 anywhere.?

(Even thought the original problem was for < 10000, I've run this much farther out)

The algorithm:

\$f(x) = 9y + \dfrac{x}{10}\$

Where \$y\$ is the previous result and \$x\$ is the power of 10 we're checking against.

i.e. \$y = 1\$ and \$x = 100\$, or \$y = 19\$ and \$x = 1000\$

or, for the more mathematically inclined

\$f(10^n)=9f(10^{n-1})+10^{n-1}\$

The code:

def numbersContaining5(pwr)
# Prints on screen the count of numbers containing a "5"
#    for each power of ten up and including the one passed in.
# pwr = the power of ten you wish to calculate to
 
    prev = 0
    for i in 1..pwr
        prev = (prev*9) + (10**i)/10 
        puts prev
    end
end
 
#call function
numbersContaining5(4) # 10^4 = 10000

Ruby-ize for loop - finding all the numbers less than 10000 that contain the digit 5 anywhere

I wrote this a while back when my fiance was taking a Number Theory class. I wrote about it here and it recently came back to my attention. Anytime I write a for loop in Ruby I feel kind of dirty. I would like to "Ruby-ize" this routine if there is a more Ruby way to do it. Also, is this the most efficient algorithm for this?

Original Problem:

Count all the numbers less than 10000 that contain the digit 5 anywhere.

(Even thought the original problem was for < 10000, I've run this much farther out)

The algorithm:

\$f(x) = 9y + \dfrac{x}{10}\$

Where \$y\$ is the previous result and \$x\$ is the power of 10 we're checking against.

i.e. \$y = 1\$ and \$x = 100\$, or \$y = 19\$ and \$x = 1000\$

or, for the more mathematically inclined

\$f(10^n)=9f(10^{n-1})+10^{n-1}\$

The code:

def numbersContaining5(pwr)
# Prints on screen the count of numbers containing a "5"
#    for each power of ten up and including the one passed in.
# pwr = the power of ten you wish to calculate to
 
    prev = 0
    for i in 1..pwr
        prev = (prev*9) + (10**i)/10 
        puts prev
    end
end
 
#call function
numbersContaining5(4) # 10^4 = 10000

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

I wrote this a while back when my fiance was taking a Number Theory class. I wrote about it here and it recently came back to my attention. Anytime I write a for loop in Ruby I feel kind of dirty. I would like to "Ruby-ize" this routine if there is a more Ruby way to do it. Also, is this the most efficient algorithm for this?

Original Problem:

How many numbers less than 10000 that contain the digit 5 anywhere?

(Even thought the original problem was for < 10000, I've run this much farther out)

The algorithm:

\$f(x) = 9y + \dfrac{x}{10}\$

Where \$y\$ is the previous result and \$x\$ is the power of 10 we're checking against.

i.e. \$y = 1\$ and \$x = 100\$, or \$y = 19\$ and \$x = 1000\$

or, for the more mathematically inclined

\$f(10^n)=9f(10^{n-1})+10^{n-1}\$

The code:

def numbersContaining5(pwr)
# Prints on screen the count of numbers containing a "5"
#    for each power of ten up and including the one passed in.
# pwr = the power of ten you wish to calculate to
 
    prev = 0
    for i in 1..pwr
        prev = (prev*9) + (10**i)/10 
        puts prev
    end
end
 
#call function
numbersContaining5(4) # 10^4 = 10000

added 1 character in body

Source Link

edited Jun 4, 2014 at 17:48

RubberDuck

30.9k
6
71
174

I wrote this a while back when my fiance was taking a Number Theory class. I wrote about it here and it recently came back to my attention. Anytime I write a for loop in Ruby I feel kind of dirty. I would like to "Ruby-ize" this routine if there is a more Ruby way to do it. Also, is this the most efficient algorithm for this?

Original Problem:

FindCount all the numbers less than 10000 that contain the digit 5 anywhere.

(Even thought the original problem was for < 10000, I've run this much farther out)

The algorithm:

\$f(x) = 9y + \dfrac{x}{10}\$

Where \$y\$ is the previous result and \$x\$ is the power of 10 we're checking against.

i.e. \$y = 1\$ and \$x = 100\$, or \$y = 19\$ and \$x = 1000\$

or, for the more mathematically inclined

\$f(10^n)=9f(10^{n-1})+10^{n-1}\$

The code:

def numbersContaining5(pwr)
# Prints on screen the count of numbers containing a "5"
#    for each power of ten up and including the one passed in.
# pwr = the power of ten you wish to calculate to
 
    prev = 0
    for i in 1..pwr
        prev = (prev*9) + (10**i)/10 
        puts prev
    end
end
 
#call function
numbersContaining5(4) # 10^4 = 10000

I wrote this a while back when my fiance was taking a Number Theory class. I wrote about it here and it recently came back to my attention. Anytime I write a for loop in Ruby I feel kind of dirty. I would like to "Ruby-ize" this routine if there is a more Ruby way to do it. Also, is this the most efficient algorithm for this?

Original Problem:

Find all the numbers less than 10000 that contain the digit 5 anywhere.

(Even thought the original problem was for < 10000, I've run this much farther out)

The algorithm:

\$f(x) = 9y + \dfrac{x}{10}\$

Where \$y\$ is the previous result and \$x\$ is the power of 10 we're checking against.

i.e. \$y = 1\$ and \$x = 100\$, or \$y = 19\$ and \$x = 1000\$

or, for the more mathematically inclined

\$f(10^n)=9f(10^{n-1})+10^{n-1}\$

The code:

def numbersContaining5(pwr)
# Prints on screen the count of numbers containing a "5"
#    for each power of ten up and including the one passed in.
# pwr = the power of ten you wish to calculate to
 
    prev = 0
    for i in 1..pwr
        prev = (prev*9) + (10**i)/10 
        puts prev
    end
end
 
#call function
numbersContaining5(4) # 10^4 = 10000

I wrote this a while back when my fiance was taking a Number Theory class. I wrote about it here and it recently came back to my attention. Anytime I write a for loop in Ruby I feel kind of dirty. I would like to "Ruby-ize" this routine if there is a more Ruby way to do it. Also, is this the most efficient algorithm for this?

Original Problem:

Count all the numbers less than 10000 that contain the digit 5 anywhere.

(Even thought the original problem was for < 10000, I've run this much farther out)

The algorithm:

\$f(x) = 9y + \dfrac{x}{10}\$

Where \$y\$ is the previous result and \$x\$ is the power of 10 we're checking against.

i.e. \$y = 1\$ and \$x = 100\$, or \$y = 19\$ and \$x = 1000\$

or, for the more mathematically inclined

\$f(10^n)=9f(10^{n-1})+10^{n-1}\$

The code:

def numbersContaining5(pwr)
# Prints on screen the count of numbers containing a "5"
#    for each power of ten up and including the one passed in.
# pwr = the power of ten you wish to calculate to
 
    prev = 0
    for i in 1..pwr
        prev = (prev*9) + (10**i)/10 
        puts prev
    end
end
 
#call function
numbersContaining5(4) # 10^4 = 10000

added 81 characters in body

Source Link

edited Jun 4, 2014 at 15:28

RubberDuck

30.9k
6
71
174

I wrote this a while back when my fiance was taking a Number Theory class. I wrote about it here and it recently came back to my attention. Anytime I write a for loop in Ruby I feel kind of dirty. I would like to "Ruby-ize" this routine if there is a more Ruby way to do it. Also, is this the most efficient algorithm for this?

Original Problem:

Find all the numbers less than 10000 that contain the digit 5 anywhere.

(Even thought the original problem was for < 10000, I've run this much farther out)

The algorithm:

\$f(x) = 9y + \dfrac{x}{10}\$

Where \$y\$ is the previous result and \$x\$ is the power of 10 we're checking against.

i.e. \$y = 1\$ and \$x = 100\$, or \$y = 19\$ and \$x = 1000\$

or, for the more mathematically inclined

\$f(10^n)=9f(10^{n-1})+10^{n-1}\$

The code:

def numbersContaining5(pwr)
# Prints on screen the count of numbers containing a "5"
#    for each power of ten up and including the one passed in.
# pwr = the power of ten you wish to calculate to
 
    prev = 0
    for i in 1..pwr
        prev = (prev*9) + (10**i)/10 
        puts prev
    end
end
 
#call function
numbersContaining5(4) # 10^4 = 10000

I wrote this a while back when my fiance was taking a Number Theory class. I wrote about it here and it recently came back to my attention. Anytime I write a for loop in Ruby I feel kind of dirty. I would like to "Ruby-ize" this routine if there is a more Ruby way to do it. Also, is this the most efficient algorithm for this?

Original Problem:

Find all the numbers less than 10000 that contain the digit 5 anywhere.

(Even thought the original problem was for < 10000, I've run this much farther out)

The algorithm:

\$f(x) = 9y + \dfrac{x}{10}\$

Where \$y\$ is the previous result and \$x\$ is the power of 10 we're checking against.

i.e. \$y = 1\$ and \$x = 100\$, or \$y = 19\$ and \$x = 1000\$

The code:

def numbersContaining5(pwr)
# Prints on screen the count of numbers containing a "5"
#    for each power of ten up and including the one passed in.
# pwr = the power of ten you wish to calculate to
 
    prev = 0
    for i in 1..pwr
        prev = (prev*9) + (10**i)/10 
        puts prev
    end
end
 
#call function
numbersContaining5(4) # 10^4 = 10000

I wrote this a while back when my fiance was taking a Number Theory class. I wrote about it here and it recently came back to my attention. Anytime I write a for loop in Ruby I feel kind of dirty. I would like to "Ruby-ize" this routine if there is a more Ruby way to do it. Also, is this the most efficient algorithm for this?

Original Problem:

Find all the numbers less than 10000 that contain the digit 5 anywhere.

(Even thought the original problem was for < 10000, I've run this much farther out)

The algorithm:

\$f(x) = 9y + \dfrac{x}{10}\$

Where \$y\$ is the previous result and \$x\$ is the power of 10 we're checking against.

i.e. \$y = 1\$ and \$x = 100\$, or \$y = 19\$ and \$x = 1000\$

or, for the more mathematically inclined

\$f(10^n)=9f(10^{n-1})+10^{n-1}\$

The code:

def numbersContaining5(pwr)
# Prints on screen the count of numbers containing a "5"
#    for each power of ten up and including the one passed in.
# pwr = the power of ten you wish to calculate to
 
    prev = 0
    for i in 1..pwr
        prev = (prev*9) + (10**i)/10 
        puts prev
    end
end
 
#call function
numbersContaining5(4) # 10^4 = 10000