# Help with a program to determin leap years

So, I’m trying to go through the Teach Yourself Programming book by
Pragmatic Press and I am encountering a few hurdles. The chapter I am
working on now is asking me to create a program which will ask for a
start and end year and then calculate all leap years in that range.
The logic behind leap years (for those who need a refresher) is all
years divisible by for are leap years EXCEPT those that are divisible
by 100 UNLESS they are also divisible by 400. I am somewhat at a loss
for how to handle the logic for this…finding all numbers that are
divisible by 4 and removing those divisible by 100 is easy. Its adding
in that third condition which adds some of the removed numbers back
into the “true” group that I am having trouble with…or maybe I am
just not wrapping my mind around the problem well
enough…suggestions?

On Nov 7, 2006, at 12:15 PM, Shiloh M. wrote:

into the “true” group that I am having trouble with…or maybe I am
just not wrapping my mind around the problem well
enough…suggestions?

Here is simple way:

1. Write a is_leap? method.

def is_leap?(y)
return true if y % 400 == 0
return false if y % 100 == 0
return true if y % 4 == 0
false
end

1. .select { |y| is_leap?(y) }

For example,

(1895…1905).select { |y| is_leap?(y) } # => [1896, 1904]
(1995…2005).select { |y| is_leap?(y) } # => [1996, 2000, 2004]

Regards, Morton

-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1

• Morton G., 11/07/2006 09:09 PM:
1. Write a is_leap? method.

def is_leap?(y)
return true if y % 400 == 0
return false if y % 100 == 0
return true if y % 4 == 0
false
end

YAWOFOIAYIALY:

def is_leap?(y)
return true if y % 400 == 0
return false if y % 100 == 0
return y % 4 == 0
end

Jupp

YAWOFOIAYIALY: Yet another way of finding out if a year is a leap year.
-----BEGIN PGP SIGNATURE-----
Version: GnuPG v1.4.5 (GNU/Linux)

iD8DBQFFUPEXrhv7B2zGV08RAg/HAJ9GqtURKVlnjqFwH7UzRo2X+LMnXgCgoIG5
=N0Le
-----END PGP SIGNATURE-----

Hi –

On Wed, 8 Nov 2006, Shiloh M. wrote:

into the “true” group that I am having trouble with…or maybe I am
just not wrapping my mind around the problem well
enough…suggestions?

require ‘date’
Date.leap?(year) #

See Morton’s implementation. Here, just for fun, is another:

def leap?(year)
year % 4 == 0 unless (year % 100 == 0 unless year % 400 == 0)
end

It returns nil/true rather than false/true, so it’s a bit
non-slick. But I thought the semantics might be interesting.

David

On Nov 7, 2006, at 12:15 PM, Shiloh M. wrote:

into the “true” group that I am having trouble with…or maybe I am
just not wrapping my mind around the problem well
enough…suggestions?

The key point of all the methods proposed in this thread is: deal
with the years divisible by 400 first, the years divisible by 100
second, and the years divisible by 4 last of all.

Regards, Morton

AH! i was looking at it wrong…startin from validating divisible by 4
is the wrong direction…thank you all. I think I can make it work now
(many of you posted rather more advanced ways to do it. I am trying to
stick within the confines of the chapters ive done so far). Basically
I needed the way to approach the problem. Thanks all!

S.

On 11/7/06, [email protected] [email protected] wrote:

by 100 UNLESS they are also divisible by 400. I am somewhat at a loss
See Morton’s implementation. Here, just for fun, is another:

def leap?(year)
year % 4 == 0 unless (year % 100 == 0 unless year % 400 == 0)
end

It returns nil/true rather than false/true, so it’s a bit
non-slick. But I thought the semantics might be interesting.

You can fix that with the fun !! trick to convert nils back to false:
def leap?(year)
!!(year % 4 == 0 unless (year % 100 == 0 unless year % 400 == 0))
end

----- Original Message -----
From: [email protected]
To: “ruby-talk ML” [email protected]
Sent: Tuesday, November 07, 2006 3:33 PM
Subject: Re: Help with a program to determin leap years.

by 100 UNLESS they are also divisible by 400. I am somewhat at a loss
See Morton’s implementation. Here, just for fun, is another:

David A. Black | [email protected]
Author of “Ruby for Rails” [1] | Ruby/Rails training & consultancy [3]
DABlog (DAB’s Weblog) [2] | Co-director, Ruby Central, Inc. [4]
[1] Ruby for Rails | [3] http://www.rubypowerandlight.com
[2] http://dablog.rubypal.com | [4] http://www.rubycentral.org

what about small optimization (just for fun);
it works faster in most cases - non leap year happens more often:

def leap?(y)
(y%4).zero? && !(y%100).zero? || (y%400).zero?
end

sergey

# year % 4 == 0 unless (year % 100 == 0 unless year % 400 == 0)

heheh thanks, this one should go to my bag of tricks, a (first) example