Peculiar Behavior for a Newby to Undertstand


#1

I created a class for prime numbers as so:

class Primes
def initialize
end

def prime?(number)
    # Method returns true if number is prime.
    limit = Math.sqrt(number).ceil
    flag = true
    if number % 2 == 0
        flag = false
    else
        3.step(limit,2) {|i|
            if number % i == 0
                flag = false
                break
            end
        }
    end
    return flag
end

def show_primes(lower, upper)
    # Prints all primes between lower and upper
    # Lower is incremented 1 if it is even.
    # The arcane "(((lower/2).floor)*2+1)" performs this task
    (((lower/2).floor)*2+1).step(upper,2) {|i|
        if prime?(i) == true
            print i.to_s + " "
        end
    }
end
end

Then when I enter
a=primes.new
a.show_primes(1000000,1000100) I get
1000003 1000033 1000037 1000039 1000081 1000099 1000001
Where is that trailing 1000001 coming from? It is not a prime number and
in fact is the lower limit.
If I enter the same methods outside of class Primes and enter
show_primes(1000000,1000100) I don’t get the lower limit at the end of
the printed values.

I am running version 1.18.4 using scite in Ubuntu.


#2

Hi Charles,

I entered the code in Ruby 1.8.5 in Windows, using Scintilla IDE :-))
(the joke is that Scintilla-Scite is not an IDE at all!) it gave the
error:

ruby prime.rb
prime.rb:34: undefined local variable or method `primes’ for main:Object
(NameError)

Then changed

a=primes.new
a.show_primes(1000000,1000100)

to

a=Primes::new
a.show_primes(1000000,1000100)

The good result appeared :slight_smile:

ruby prime.rb
1000003 1000033 1000037 1000039 1000081 1000099 >Exit code: 0

Don’t know how you had not an error instead of the wrong result
before…
For that you need a Ruby guru :slight_smile:

Kind Regards

J. Augusto


#3

Hi –

On Wed, 27 Dec 2006, Charles A Gray wrote:

Then when I enter
a=primes.new

(As another respondent pointed out, you need Primes.new.)

a.show_primes(1000000,1000100) I get
1000003 1000033 1000037 1000039 1000081 1000099 1000001
Where is that trailing 1000001 coming from? It is not a prime number and
in fact is the lower limit.
If I enter the same methods outside of class Primes and enter
show_primes(1000000,1000100) I don’t get the lower limit at the end of
the printed values.

I am running version 1.18.4 using scite in Ubuntu.

I can’t duplicate the problem, I’m afraid.

David


#4

On Dec 26, 2006, at 8:20 PM, Charles A Gray wrote:

    if number % 2 == 0
end
end

the printed values.

I am running version 1.18.4 using scite in Ubuntu.

Charles G. – Phoenix, AZ; Where you can bake the chill out of your
bones

The value of the #step method is the initial value so that is being
returned from a.show_primes and presumably printed by whatever you’re
using to interpret your statements (like irb):

With your code in a file names “primes.rb”

$ irb -rprimes

a=Primes.new
=> #Primes:0x6e5d88

a.show_primes(1_000_000, 1_000_100)
1000003 1000033 1000037 1000039 1000081 1000099 => 1000001

a.show_primes(1_000_000, 1_000_100); nil
1000003 1000033 1000037 1000039 1000081 1000099 => nil

a.show_primes(1_000_000, 1_000_100); puts “”
1000003 1000033 1000037 1000039 1000081 1000099
=> nil

Note that the value of the last expression is displayed by irb
itself. (In the last example, the puts supplies a newline and the
value of puts as an expression is nil.)

To see the documentation for the step method (after first trying
Fixnum#step and Integer#step rather than looking it up in the
pickaxe :wink:

$ ri -T Numeric#step
----------------------------------------------------------- Numeric#step
num.step(limit, step ) {|i| block } => num

  Invokes _block_ with the sequence of numbers starting at _num_,
  incremented by _step_ on each call. The loop finishes when the
  value to be passed to the block is greater than _limit_ (if _step_
  is positive) or less than _limit_ (if _step_ is negative). If all
  the arguments are integers, the loop operates using an integer
  counter. If any of the arguments are floating point numbers, all
  are converted to floats, and the loop is executed _floor(n +
  n*epsilon)+ 1_ times, where _n = (limit - num)/step_. Otherwise,
  the loop starts at _num_, uses either the +<+ or +>+ operator to
  compare the counter against _limit_, and increments itself using
  the +++ operator.

     1.step(10, 2) { |i| print i, " " }
     Math::E.step(Math::PI, 0.2) { |f| print f, " " }

  _produces:_

     1 3 5 7 9
     2.71828182845905 2.91828182845905 3.11828182845905

-Rob

Rob B. http://agileconsultingllc.com
removed_email_address@domain.invalid


#5

On Wed, 2006-12-27 at 12:07 +0900, removed_email_address@domain.invalid wrote:

Then when I enter
a=primes.new

(As another respondent pointed out, you need Primes.new.)

That was a typo. I actually had a=Primes.new. Which Ruby are you
running?


#6

On Dec 26, 2006, at 10:07 PM, removed_email_address@domain.invalid wrote:

1000003 1000033 1000037 1000039 1000081 1000099 1000001
I can’t duplicate the problem, I’m afraid.

David


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Charles,

I was prepared to be disheartened when I saw that David had responded
ahead of me, but I have a solution!

$ irb -f --noprompt -rprimes
a=Primes.new
#Primes:0x1d5084
a.show_primes(1_000_000, 1_000_100)
1000003 1000033 1000037 1000039 1000081 1000099 1000001
exit

Since your commands didn’t show the typical prompts from irb, I went
poking around and came up with this. The ‘-f’ suppresses the
~/.irbrc (which I have), --noprompt ought to be obvious, and -rprimes
requires the primes.rb file with your original Primes class definition.

-Rob

Rob B. http://agileconsultingllc.com
removed_email_address@domain.invalid


#7

On Wed, 2006-12-27 at 12:23 +0900, Rob B. wrote:

    limit = Math.sqrt(number).ceil
    end
        end

If I enter the same methods outside of class Primes and enter
The value of the #step method is the initial value so that is being
1000003 1000033 1000037 1000039 1000081 1000099 => 1000001
To see the documentation for the step method (after first trying
incremented by step on each call. The loop finishes when the
the +++ operator.
-Rob
Rob, I went back and added the puts for a blank line and got the same
results as before, namely:

a=Primes.new
p a.prime?(1000001)
p a.show_primes(1000000,1000100);puts “”

produced:

ruby primes.rb
false
1000003 1000033 1000037 1000039 1000081 1000099 1000001

Exit code: 0

I am still running it in scite. As well as being a Ruby newby, I am a
Linux newby and haven’t figured out how to run irb in Linux.

I am going to reboot into windows and see what happens there.


#8

On Wed, 2006-12-27 at 11:51 +0900, Jose Augusto wrote:

For that you need a Ruby guru :slight_smile:

Kind Regards

J. Augusto

Tahnks for the reply a=Primes.new ans a=Primes::new both give me the
same incorret result


#9

Rob B. wrote:

The value of the #step method is the initial value so that is being
returned from a.show_primes and presumably printed by whatever you’re
using to interpret your statements (like irb):

As a simple example:

irb(main):001:0> puts 1.step( 2 ){ print "a " }
a a 1
=> nil

In the above, the block is called twice, and prints out 'a ’ each time.
Then the return value of the step method (the initial value) it passed
to the puts function, which prints it out.

You’re seeing your first non-even value as the return, since the step
method is the last call in your show_primes method.

One other comment - the pairing of methods you have inside your class
seems odd, given Ruby’s built-in classes for numbers. Instead of your
class wrapper, I might personally place those methods as:

class Integer
def is_prime?
# …
end
end

def show_primes( lower, upper )
end

Although, since I’m personally not a big fan of ‘global’ functions, I
might even do:

class Integer
def self.show_primes( lower, upper )
# …
end
end


#10

Charles A Gray wrote:

On Wed, 2006-12-27 at 11:51 +0900, Jose Augusto wrote:

For that you need a Ruby guru :slight_smile:

Kind Regards

J. Augusto

Tahnks for the reply a=Primes.new ans a=Primes::new both give me the
same incorret result

Try this:

class Primes
def prime?( number )
( ( ( number / 2 ).floor ) * 2 + 1 )
end
end

a = Primes.new
a.prime?( 1000000 )

The last variable assigned is being returned from a.prime? (in this case
an anonymous variable). This is how Ruby returns a value from a method.

Hope this helps.

Joe


#11

On Dec 26, 2006, at 11:09 PM, Charles A Gray wrote:

I am still running it in scite. As well as being a Ruby newby, I am a
Linux newby and haven’t figured out how to run irb in Linux.

Opening a terminal and typing irb should just work assuming ruby is
installed in a standard fashion. Where to find the terminal will
vary a bit by distro. Worst case, you should be able to hit Ctrl+Alt
+F1 and jump to a text terminal on any distro (should be Ctrl+Alt+F7
to get you back to X windows, possibly F11). Have fun!
-Mat