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Make Project Euler 27 solution idiomatic Ruby
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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
I learned to program in Java and C# and in my free time I am using Ruby because it is fun. Unfortunately I have the feeling that I write Ruby code like I am making Java or C# code. I have learned to use regex instead of strings for comparison, using each instead of for-loops, keeping method names lowercase, and how to use code blocks (which are quite similar to lambda expressions in C#).
I would love to improve the Rubiness of my code and I hope you would be willing to give me one or more pointers on a piece of code I made. It answers Project Euler problem 27.
class Integer
def prime?
return false if self < 1
2.upto(Math.sqrt(self)) do |i|
return false if self % i == 0
end
true
end
end
def get_amount_of_primes_from_quadratic_formula(a,b)
primes = []
still_all_primes = true
n = 0
while still_all_primes
result = n**2 + a*n + b
if result.prime? then
primes << result
else
still_all_primes = false
end
n += 1
end
primes.size
end
def get_product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values()
max_product = 0
max_primes = 0
-999.upto(1000) do |a|
-999.upto(1000) do |b|
primes = get_amount_of_primes_from_quadratic_formula(a,b)
if primes > max_primes then
max_primes = primes
max_product = a*b
end
end
end
max_product
end
start = Time.now
answer = get_product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values()
puts "The answer is #{answer} and it took #{Time.now-start} seconds."
I think I can improve on the if-then statement and write it more concise, and also the two "upto-loops" where I first declare the variables max_primes
and max_product
can be written in a more Ruby-way I am sure.
I would be very grateful if you could let me know how to write more like Ruby!
Links that ask similar questions which I am reading in the meantime:
- Advice on making ruby code more ruby-like
- Ruby method return values - how to be more Ruby-like?
- How to make my script nicer / more Ruby?
ruby primes mathematics programming-challenge
$endgroup$
add a comment |
$begingroup$
I learned to program in Java and C# and in my free time I am using Ruby because it is fun. Unfortunately I have the feeling that I write Ruby code like I am making Java or C# code. I have learned to use regex instead of strings for comparison, using each instead of for-loops, keeping method names lowercase, and how to use code blocks (which are quite similar to lambda expressions in C#).
I would love to improve the Rubiness of my code and I hope you would be willing to give me one or more pointers on a piece of code I made. It answers Project Euler problem 27.
class Integer
def prime?
return false if self < 1
2.upto(Math.sqrt(self)) do |i|
return false if self % i == 0
end
true
end
end
def get_amount_of_primes_from_quadratic_formula(a,b)
primes = []
still_all_primes = true
n = 0
while still_all_primes
result = n**2 + a*n + b
if result.prime? then
primes << result
else
still_all_primes = false
end
n += 1
end
primes.size
end
def get_product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values()
max_product = 0
max_primes = 0
-999.upto(1000) do |a|
-999.upto(1000) do |b|
primes = get_amount_of_primes_from_quadratic_formula(a,b)
if primes > max_primes then
max_primes = primes
max_product = a*b
end
end
end
max_product
end
start = Time.now
answer = get_product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values()
puts "The answer is #{answer} and it took #{Time.now-start} seconds."
I think I can improve on the if-then statement and write it more concise, and also the two "upto-loops" where I first declare the variables max_primes
and max_product
can be written in a more Ruby-way I am sure.
I would be very grateful if you could let me know how to write more like Ruby!
Links that ask similar questions which I am reading in the meantime:
- Advice on making ruby code more ruby-like
- Ruby method return values - how to be more Ruby-like?
- How to make my script nicer / more Ruby?
ruby primes mathematics programming-challenge
$endgroup$
add a comment |
$begingroup$
I learned to program in Java and C# and in my free time I am using Ruby because it is fun. Unfortunately I have the feeling that I write Ruby code like I am making Java or C# code. I have learned to use regex instead of strings for comparison, using each instead of for-loops, keeping method names lowercase, and how to use code blocks (which are quite similar to lambda expressions in C#).
I would love to improve the Rubiness of my code and I hope you would be willing to give me one or more pointers on a piece of code I made. It answers Project Euler problem 27.
class Integer
def prime?
return false if self < 1
2.upto(Math.sqrt(self)) do |i|
return false if self % i == 0
end
true
end
end
def get_amount_of_primes_from_quadratic_formula(a,b)
primes = []
still_all_primes = true
n = 0
while still_all_primes
result = n**2 + a*n + b
if result.prime? then
primes << result
else
still_all_primes = false
end
n += 1
end
primes.size
end
def get_product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values()
max_product = 0
max_primes = 0
-999.upto(1000) do |a|
-999.upto(1000) do |b|
primes = get_amount_of_primes_from_quadratic_formula(a,b)
if primes > max_primes then
max_primes = primes
max_product = a*b
end
end
end
max_product
end
start = Time.now
answer = get_product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values()
puts "The answer is #{answer} and it took #{Time.now-start} seconds."
I think I can improve on the if-then statement and write it more concise, and also the two "upto-loops" where I first declare the variables max_primes
and max_product
can be written in a more Ruby-way I am sure.
I would be very grateful if you could let me know how to write more like Ruby!
Links that ask similar questions which I am reading in the meantime:
- Advice on making ruby code more ruby-like
- Ruby method return values - how to be more Ruby-like?
- How to make my script nicer / more Ruby?
ruby primes mathematics programming-challenge
$endgroup$
I learned to program in Java and C# and in my free time I am using Ruby because it is fun. Unfortunately I have the feeling that I write Ruby code like I am making Java or C# code. I have learned to use regex instead of strings for comparison, using each instead of for-loops, keeping method names lowercase, and how to use code blocks (which are quite similar to lambda expressions in C#).
I would love to improve the Rubiness of my code and I hope you would be willing to give me one or more pointers on a piece of code I made. It answers Project Euler problem 27.
class Integer
def prime?
return false if self < 1
2.upto(Math.sqrt(self)) do |i|
return false if self % i == 0
end
true
end
end
def get_amount_of_primes_from_quadratic_formula(a,b)
primes = []
still_all_primes = true
n = 0
while still_all_primes
result = n**2 + a*n + b
if result.prime? then
primes << result
else
still_all_primes = false
end
n += 1
end
primes.size
end
def get_product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values()
max_product = 0
max_primes = 0
-999.upto(1000) do |a|
-999.upto(1000) do |b|
primes = get_amount_of_primes_from_quadratic_formula(a,b)
if primes > max_primes then
max_primes = primes
max_product = a*b
end
end
end
max_product
end
start = Time.now
answer = get_product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values()
puts "The answer is #{answer} and it took #{Time.now-start} seconds."
I think I can improve on the if-then statement and write it more concise, and also the two "upto-loops" where I first declare the variables max_primes
and max_product
can be written in a more Ruby-way I am sure.
I would be very grateful if you could let me know how to write more like Ruby!
Links that ask similar questions which I am reading in the meantime:
- Advice on making ruby code more ruby-like
- Ruby method return values - how to be more Ruby-like?
- How to make my script nicer / more Ruby?
ruby primes mathematics programming-challenge
ruby primes mathematics programming-challenge
edited Apr 13 '17 at 12:40
Community♦
1
1
asked Apr 19 '14 at 16:10
Erwin RooijakkersErwin Rooijakkers
26927
26927
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Succinctness
As rubyists, we love being succinct, and we love playing with enumerations.
You will see very few literal false
and true
in ruby code, as well as very few explicit return
calls.
For example:
Instead of writing return false if self < 1
we will prefer to compound the condition to self >= 1 && ...
which will do the same thing, but we "save" return false
.
The power of Enumeration
Ruby has a very powerful Enumerable
, and is used widely, often more than once in a line (using method chaining).
For example:
2.upto(Math.sqrt(self)) do |i|
return false if self % i == 0
end
Here you check if any of the numbers in the range are a divisor for self
, and break if there is any. A more ruby way of doing it will be:
return false if 2.upto(Math.sqrt(self)).any? { |i| self % i == 0 }
We'll also prefer to more succinct range syntax (2..Math.sqrt(self))
, which is simply shorter...
So now, our def prime?
method could be reduced to a one-liner:
class Integer
def prime?
self > 1 && !(2..Math.sqrt(self)).any? { |i| self % i == 0 }
end
end
Mapping
Anywhere in the code I see the following pattern:
result = []
some_loop do
result << something
end
A red flag is raised, and I look for a way to use map
to do the same thing:
result = some_loop.map { something }
Your code goes over all the non-negative integers, and takes counts how many of them result in a prime, until the first non-prime.
"All the non-negative integers" can be expressed in ruby as (0..Float::INFINITY)
, so we can write:
(0..Float::INFINITY).map { |n| n**2 + a*n + b }.take_while { |result| result.prime? }.count
This code takes each integer, maps it into the result of n**2 + a*n + b
, takes all the results until they are no longer prime, and counts how many are there.
Cool! Right? The only problem with the code above, is that it will take infinity to complete it, as it takes all the numbers and maps them, and then checks for how many to take.
To solve this problem ruby now has...
Lazy Enumerables
As of ruby 2.0, lazy enumerables allows you to calculate values in an infinite stream only as needed.
To solve the problem above, all we need to do now is to add the lazy
operator on the range:
(0..Float::INFINITY).lazy.map { |n| n**2 + a*n + b }.take_while(&:prime?).count
And we have another one-liner!
Everything is an enumerable
So you want to save on your "upto-loops"? Let's do it!
You want to enumerate over each pair of numbers from -999
to 1000
, so what you actually want is to have a long matrix of those pairs:
[[-999, -999], [-999, -998],...,[1000, 1000]].do_something_smart
To do that, you can use product
:
(-999..1000).to_a.product((-999..1000).to_a)
But since both a
and b
have the same range, we can even DRY this up, and use repeated_permutation
:
(-999..1000).to_a.repeated_permutation(2)
Both of these solutions will give you the needed matrix, so we can move on the see what we should do with it...
We want to get the coeffiecients that produce the number of primes, so let's do just that:
a, b = (-999..1000).to_a.repeated_permutation(2).max_by { |a, b| get_amount_of_primes_from_quadratic_formula(a,b) }
Now all we need to do is multiply them with each other!
Method naming
Your names are very verbose, which is a good thing, but ruby idiom frowns upon get_
prefixes. Also, prefer using verbs already in the language (count
) over those which are not in the language (amount_of
)
So now the code will look like:
class Integer
def prime?
self > 1 && !(2..Math.sqrt(self)).any? { |i| self % i == 0 }
end
end
def count_quadratic_formula_primes(a,b)
(0..Float::INFINITY).lazy.map { |n| n**2 + a*n + b }.take_while(&:prime?).count
end
def product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values()
a, b = (-999..1000).to_a.repeated_permutation(2).max_by { |a, b| count_quadratic_formula_primes(a,b) }
a * b
end
start = Time.now
answer = product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values
puts "The answer is #{answer} and it took #{Time.now-start} seconds."
15 lines of hard-core ruby-style code!
Enjoy!
Update
It seems that lazy
adds considerable overhead to the performance of the code. So it is not advisable to use it.
Fortunately this works:
(0..Float::INFINITY).take_while { |n| (n**2 + a*n + b).prime? }.count
My code still runs ~2 times slower than the original (ends in 18 seconds), but it is more reasonable than with lazy
...
$endgroup$
1
$begingroup$
also, you caninject
your a*b :)
$endgroup$
– gaussblurinc
Apr 20 '14 at 7:37
$begingroup$
I like your code better, but it takes 130 seconds, while the original code takes 10. The prime function is a beauty. I suspect that the count_quadratic_formula_primes method is slow.
$endgroup$
– Erwin Rooijakkers
Apr 20 '14 at 14:58
$begingroup$
@user2609980 - yes, apparentlylazy
add a lot of overhead... see my update
$endgroup$
– Uri Agassi
Apr 20 '14 at 18:08
3
$begingroup$
user2609980, I'm surprised at the difference in execution times, but I'd pay no heed to that at this stage of your Ruby education. Uri has covered a wide swath of ground in his answer, acquainting you with typical Ruby coding style, the addition of a method to an existing Ruby class (prime?
), the use of powerful enumeratorsmap
,product
,permutation
,any?
andtake_while
from theEnumerable
module, andup_to
from theInteger
class, and Ruby's newlazy
operator. He has also nicely explained his reasons for coding it the way he has. Great answer, Uri.
$endgroup$
– Cary Swoveland
Apr 20 '14 at 18:24
$begingroup$
@CarySwoveland I am indeed very happy with the answer :-).
$endgroup$
– Erwin Rooijakkers
Apr 21 '14 at 10:06
|
show 2 more comments
$begingroup$
I advise you to read a very good article.Red or Blue pill, Neo. Ruby vs. Python - which will you choose for your Backend? Good luck in your endeavors
New contributor
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Succinctness
As rubyists, we love being succinct, and we love playing with enumerations.
You will see very few literal false
and true
in ruby code, as well as very few explicit return
calls.
For example:
Instead of writing return false if self < 1
we will prefer to compound the condition to self >= 1 && ...
which will do the same thing, but we "save" return false
.
The power of Enumeration
Ruby has a very powerful Enumerable
, and is used widely, often more than once in a line (using method chaining).
For example:
2.upto(Math.sqrt(self)) do |i|
return false if self % i == 0
end
Here you check if any of the numbers in the range are a divisor for self
, and break if there is any. A more ruby way of doing it will be:
return false if 2.upto(Math.sqrt(self)).any? { |i| self % i == 0 }
We'll also prefer to more succinct range syntax (2..Math.sqrt(self))
, which is simply shorter...
So now, our def prime?
method could be reduced to a one-liner:
class Integer
def prime?
self > 1 && !(2..Math.sqrt(self)).any? { |i| self % i == 0 }
end
end
Mapping
Anywhere in the code I see the following pattern:
result = []
some_loop do
result << something
end
A red flag is raised, and I look for a way to use map
to do the same thing:
result = some_loop.map { something }
Your code goes over all the non-negative integers, and takes counts how many of them result in a prime, until the first non-prime.
"All the non-negative integers" can be expressed in ruby as (0..Float::INFINITY)
, so we can write:
(0..Float::INFINITY).map { |n| n**2 + a*n + b }.take_while { |result| result.prime? }.count
This code takes each integer, maps it into the result of n**2 + a*n + b
, takes all the results until they are no longer prime, and counts how many are there.
Cool! Right? The only problem with the code above, is that it will take infinity to complete it, as it takes all the numbers and maps them, and then checks for how many to take.
To solve this problem ruby now has...
Lazy Enumerables
As of ruby 2.0, lazy enumerables allows you to calculate values in an infinite stream only as needed.
To solve the problem above, all we need to do now is to add the lazy
operator on the range:
(0..Float::INFINITY).lazy.map { |n| n**2 + a*n + b }.take_while(&:prime?).count
And we have another one-liner!
Everything is an enumerable
So you want to save on your "upto-loops"? Let's do it!
You want to enumerate over each pair of numbers from -999
to 1000
, so what you actually want is to have a long matrix of those pairs:
[[-999, -999], [-999, -998],...,[1000, 1000]].do_something_smart
To do that, you can use product
:
(-999..1000).to_a.product((-999..1000).to_a)
But since both a
and b
have the same range, we can even DRY this up, and use repeated_permutation
:
(-999..1000).to_a.repeated_permutation(2)
Both of these solutions will give you the needed matrix, so we can move on the see what we should do with it...
We want to get the coeffiecients that produce the number of primes, so let's do just that:
a, b = (-999..1000).to_a.repeated_permutation(2).max_by { |a, b| get_amount_of_primes_from_quadratic_formula(a,b) }
Now all we need to do is multiply them with each other!
Method naming
Your names are very verbose, which is a good thing, but ruby idiom frowns upon get_
prefixes. Also, prefer using verbs already in the language (count
) over those which are not in the language (amount_of
)
So now the code will look like:
class Integer
def prime?
self > 1 && !(2..Math.sqrt(self)).any? { |i| self % i == 0 }
end
end
def count_quadratic_formula_primes(a,b)
(0..Float::INFINITY).lazy.map { |n| n**2 + a*n + b }.take_while(&:prime?).count
end
def product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values()
a, b = (-999..1000).to_a.repeated_permutation(2).max_by { |a, b| count_quadratic_formula_primes(a,b) }
a * b
end
start = Time.now
answer = product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values
puts "The answer is #{answer} and it took #{Time.now-start} seconds."
15 lines of hard-core ruby-style code!
Enjoy!
Update
It seems that lazy
adds considerable overhead to the performance of the code. So it is not advisable to use it.
Fortunately this works:
(0..Float::INFINITY).take_while { |n| (n**2 + a*n + b).prime? }.count
My code still runs ~2 times slower than the original (ends in 18 seconds), but it is more reasonable than with lazy
...
$endgroup$
1
$begingroup$
also, you caninject
your a*b :)
$endgroup$
– gaussblurinc
Apr 20 '14 at 7:37
$begingroup$
I like your code better, but it takes 130 seconds, while the original code takes 10. The prime function is a beauty. I suspect that the count_quadratic_formula_primes method is slow.
$endgroup$
– Erwin Rooijakkers
Apr 20 '14 at 14:58
$begingroup$
@user2609980 - yes, apparentlylazy
add a lot of overhead... see my update
$endgroup$
– Uri Agassi
Apr 20 '14 at 18:08
3
$begingroup$
user2609980, I'm surprised at the difference in execution times, but I'd pay no heed to that at this stage of your Ruby education. Uri has covered a wide swath of ground in his answer, acquainting you with typical Ruby coding style, the addition of a method to an existing Ruby class (prime?
), the use of powerful enumeratorsmap
,product
,permutation
,any?
andtake_while
from theEnumerable
module, andup_to
from theInteger
class, and Ruby's newlazy
operator. He has also nicely explained his reasons for coding it the way he has. Great answer, Uri.
$endgroup$
– Cary Swoveland
Apr 20 '14 at 18:24
$begingroup$
@CarySwoveland I am indeed very happy with the answer :-).
$endgroup$
– Erwin Rooijakkers
Apr 21 '14 at 10:06
|
show 2 more comments
$begingroup$
Succinctness
As rubyists, we love being succinct, and we love playing with enumerations.
You will see very few literal false
and true
in ruby code, as well as very few explicit return
calls.
For example:
Instead of writing return false if self < 1
we will prefer to compound the condition to self >= 1 && ...
which will do the same thing, but we "save" return false
.
The power of Enumeration
Ruby has a very powerful Enumerable
, and is used widely, often more than once in a line (using method chaining).
For example:
2.upto(Math.sqrt(self)) do |i|
return false if self % i == 0
end
Here you check if any of the numbers in the range are a divisor for self
, and break if there is any. A more ruby way of doing it will be:
return false if 2.upto(Math.sqrt(self)).any? { |i| self % i == 0 }
We'll also prefer to more succinct range syntax (2..Math.sqrt(self))
, which is simply shorter...
So now, our def prime?
method could be reduced to a one-liner:
class Integer
def prime?
self > 1 && !(2..Math.sqrt(self)).any? { |i| self % i == 0 }
end
end
Mapping
Anywhere in the code I see the following pattern:
result = []
some_loop do
result << something
end
A red flag is raised, and I look for a way to use map
to do the same thing:
result = some_loop.map { something }
Your code goes over all the non-negative integers, and takes counts how many of them result in a prime, until the first non-prime.
"All the non-negative integers" can be expressed in ruby as (0..Float::INFINITY)
, so we can write:
(0..Float::INFINITY).map { |n| n**2 + a*n + b }.take_while { |result| result.prime? }.count
This code takes each integer, maps it into the result of n**2 + a*n + b
, takes all the results until they are no longer prime, and counts how many are there.
Cool! Right? The only problem with the code above, is that it will take infinity to complete it, as it takes all the numbers and maps them, and then checks for how many to take.
To solve this problem ruby now has...
Lazy Enumerables
As of ruby 2.0, lazy enumerables allows you to calculate values in an infinite stream only as needed.
To solve the problem above, all we need to do now is to add the lazy
operator on the range:
(0..Float::INFINITY).lazy.map { |n| n**2 + a*n + b }.take_while(&:prime?).count
And we have another one-liner!
Everything is an enumerable
So you want to save on your "upto-loops"? Let's do it!
You want to enumerate over each pair of numbers from -999
to 1000
, so what you actually want is to have a long matrix of those pairs:
[[-999, -999], [-999, -998],...,[1000, 1000]].do_something_smart
To do that, you can use product
:
(-999..1000).to_a.product((-999..1000).to_a)
But since both a
and b
have the same range, we can even DRY this up, and use repeated_permutation
:
(-999..1000).to_a.repeated_permutation(2)
Both of these solutions will give you the needed matrix, so we can move on the see what we should do with it...
We want to get the coeffiecients that produce the number of primes, so let's do just that:
a, b = (-999..1000).to_a.repeated_permutation(2).max_by { |a, b| get_amount_of_primes_from_quadratic_formula(a,b) }
Now all we need to do is multiply them with each other!
Method naming
Your names are very verbose, which is a good thing, but ruby idiom frowns upon get_
prefixes. Also, prefer using verbs already in the language (count
) over those which are not in the language (amount_of
)
So now the code will look like:
class Integer
def prime?
self > 1 && !(2..Math.sqrt(self)).any? { |i| self % i == 0 }
end
end
def count_quadratic_formula_primes(a,b)
(0..Float::INFINITY).lazy.map { |n| n**2 + a*n + b }.take_while(&:prime?).count
end
def product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values()
a, b = (-999..1000).to_a.repeated_permutation(2).max_by { |a, b| count_quadratic_formula_primes(a,b) }
a * b
end
start = Time.now
answer = product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values
puts "The answer is #{answer} and it took #{Time.now-start} seconds."
15 lines of hard-core ruby-style code!
Enjoy!
Update
It seems that lazy
adds considerable overhead to the performance of the code. So it is not advisable to use it.
Fortunately this works:
(0..Float::INFINITY).take_while { |n| (n**2 + a*n + b).prime? }.count
My code still runs ~2 times slower than the original (ends in 18 seconds), but it is more reasonable than with lazy
...
$endgroup$
1
$begingroup$
also, you caninject
your a*b :)
$endgroup$
– gaussblurinc
Apr 20 '14 at 7:37
$begingroup$
I like your code better, but it takes 130 seconds, while the original code takes 10. The prime function is a beauty. I suspect that the count_quadratic_formula_primes method is slow.
$endgroup$
– Erwin Rooijakkers
Apr 20 '14 at 14:58
$begingroup$
@user2609980 - yes, apparentlylazy
add a lot of overhead... see my update
$endgroup$
– Uri Agassi
Apr 20 '14 at 18:08
3
$begingroup$
user2609980, I'm surprised at the difference in execution times, but I'd pay no heed to that at this stage of your Ruby education. Uri has covered a wide swath of ground in his answer, acquainting you with typical Ruby coding style, the addition of a method to an existing Ruby class (prime?
), the use of powerful enumeratorsmap
,product
,permutation
,any?
andtake_while
from theEnumerable
module, andup_to
from theInteger
class, and Ruby's newlazy
operator. He has also nicely explained his reasons for coding it the way he has. Great answer, Uri.
$endgroup$
– Cary Swoveland
Apr 20 '14 at 18:24
$begingroup$
@CarySwoveland I am indeed very happy with the answer :-).
$endgroup$
– Erwin Rooijakkers
Apr 21 '14 at 10:06
|
show 2 more comments
$begingroup$
Succinctness
As rubyists, we love being succinct, and we love playing with enumerations.
You will see very few literal false
and true
in ruby code, as well as very few explicit return
calls.
For example:
Instead of writing return false if self < 1
we will prefer to compound the condition to self >= 1 && ...
which will do the same thing, but we "save" return false
.
The power of Enumeration
Ruby has a very powerful Enumerable
, and is used widely, often more than once in a line (using method chaining).
For example:
2.upto(Math.sqrt(self)) do |i|
return false if self % i == 0
end
Here you check if any of the numbers in the range are a divisor for self
, and break if there is any. A more ruby way of doing it will be:
return false if 2.upto(Math.sqrt(self)).any? { |i| self % i == 0 }
We'll also prefer to more succinct range syntax (2..Math.sqrt(self))
, which is simply shorter...
So now, our def prime?
method could be reduced to a one-liner:
class Integer
def prime?
self > 1 && !(2..Math.sqrt(self)).any? { |i| self % i == 0 }
end
end
Mapping
Anywhere in the code I see the following pattern:
result = []
some_loop do
result << something
end
A red flag is raised, and I look for a way to use map
to do the same thing:
result = some_loop.map { something }
Your code goes over all the non-negative integers, and takes counts how many of them result in a prime, until the first non-prime.
"All the non-negative integers" can be expressed in ruby as (0..Float::INFINITY)
, so we can write:
(0..Float::INFINITY).map { |n| n**2 + a*n + b }.take_while { |result| result.prime? }.count
This code takes each integer, maps it into the result of n**2 + a*n + b
, takes all the results until they are no longer prime, and counts how many are there.
Cool! Right? The only problem with the code above, is that it will take infinity to complete it, as it takes all the numbers and maps them, and then checks for how many to take.
To solve this problem ruby now has...
Lazy Enumerables
As of ruby 2.0, lazy enumerables allows you to calculate values in an infinite stream only as needed.
To solve the problem above, all we need to do now is to add the lazy
operator on the range:
(0..Float::INFINITY).lazy.map { |n| n**2 + a*n + b }.take_while(&:prime?).count
And we have another one-liner!
Everything is an enumerable
So you want to save on your "upto-loops"? Let's do it!
You want to enumerate over each pair of numbers from -999
to 1000
, so what you actually want is to have a long matrix of those pairs:
[[-999, -999], [-999, -998],...,[1000, 1000]].do_something_smart
To do that, you can use product
:
(-999..1000).to_a.product((-999..1000).to_a)
But since both a
and b
have the same range, we can even DRY this up, and use repeated_permutation
:
(-999..1000).to_a.repeated_permutation(2)
Both of these solutions will give you the needed matrix, so we can move on the see what we should do with it...
We want to get the coeffiecients that produce the number of primes, so let's do just that:
a, b = (-999..1000).to_a.repeated_permutation(2).max_by { |a, b| get_amount_of_primes_from_quadratic_formula(a,b) }
Now all we need to do is multiply them with each other!
Method naming
Your names are very verbose, which is a good thing, but ruby idiom frowns upon get_
prefixes. Also, prefer using verbs already in the language (count
) over those which are not in the language (amount_of
)
So now the code will look like:
class Integer
def prime?
self > 1 && !(2..Math.sqrt(self)).any? { |i| self % i == 0 }
end
end
def count_quadratic_formula_primes(a,b)
(0..Float::INFINITY).lazy.map { |n| n**2 + a*n + b }.take_while(&:prime?).count
end
def product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values()
a, b = (-999..1000).to_a.repeated_permutation(2).max_by { |a, b| count_quadratic_formula_primes(a,b) }
a * b
end
start = Time.now
answer = product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values
puts "The answer is #{answer} and it took #{Time.now-start} seconds."
15 lines of hard-core ruby-style code!
Enjoy!
Update
It seems that lazy
adds considerable overhead to the performance of the code. So it is not advisable to use it.
Fortunately this works:
(0..Float::INFINITY).take_while { |n| (n**2 + a*n + b).prime? }.count
My code still runs ~2 times slower than the original (ends in 18 seconds), but it is more reasonable than with lazy
...
$endgroup$
Succinctness
As rubyists, we love being succinct, and we love playing with enumerations.
You will see very few literal false
and true
in ruby code, as well as very few explicit return
calls.
For example:
Instead of writing return false if self < 1
we will prefer to compound the condition to self >= 1 && ...
which will do the same thing, but we "save" return false
.
The power of Enumeration
Ruby has a very powerful Enumerable
, and is used widely, often more than once in a line (using method chaining).
For example:
2.upto(Math.sqrt(self)) do |i|
return false if self % i == 0
end
Here you check if any of the numbers in the range are a divisor for self
, and break if there is any. A more ruby way of doing it will be:
return false if 2.upto(Math.sqrt(self)).any? { |i| self % i == 0 }
We'll also prefer to more succinct range syntax (2..Math.sqrt(self))
, which is simply shorter...
So now, our def prime?
method could be reduced to a one-liner:
class Integer
def prime?
self > 1 && !(2..Math.sqrt(self)).any? { |i| self % i == 0 }
end
end
Mapping
Anywhere in the code I see the following pattern:
result = []
some_loop do
result << something
end
A red flag is raised, and I look for a way to use map
to do the same thing:
result = some_loop.map { something }
Your code goes over all the non-negative integers, and takes counts how many of them result in a prime, until the first non-prime.
"All the non-negative integers" can be expressed in ruby as (0..Float::INFINITY)
, so we can write:
(0..Float::INFINITY).map { |n| n**2 + a*n + b }.take_while { |result| result.prime? }.count
This code takes each integer, maps it into the result of n**2 + a*n + b
, takes all the results until they are no longer prime, and counts how many are there.
Cool! Right? The only problem with the code above, is that it will take infinity to complete it, as it takes all the numbers and maps them, and then checks for how many to take.
To solve this problem ruby now has...
Lazy Enumerables
As of ruby 2.0, lazy enumerables allows you to calculate values in an infinite stream only as needed.
To solve the problem above, all we need to do now is to add the lazy
operator on the range:
(0..Float::INFINITY).lazy.map { |n| n**2 + a*n + b }.take_while(&:prime?).count
And we have another one-liner!
Everything is an enumerable
So you want to save on your "upto-loops"? Let's do it!
You want to enumerate over each pair of numbers from -999
to 1000
, so what you actually want is to have a long matrix of those pairs:
[[-999, -999], [-999, -998],...,[1000, 1000]].do_something_smart
To do that, you can use product
:
(-999..1000).to_a.product((-999..1000).to_a)
But since both a
and b
have the same range, we can even DRY this up, and use repeated_permutation
:
(-999..1000).to_a.repeated_permutation(2)
Both of these solutions will give you the needed matrix, so we can move on the see what we should do with it...
We want to get the coeffiecients that produce the number of primes, so let's do just that:
a, b = (-999..1000).to_a.repeated_permutation(2).max_by { |a, b| get_amount_of_primes_from_quadratic_formula(a,b) }
Now all we need to do is multiply them with each other!
Method naming
Your names are very verbose, which is a good thing, but ruby idiom frowns upon get_
prefixes. Also, prefer using verbs already in the language (count
) over those which are not in the language (amount_of
)
So now the code will look like:
class Integer
def prime?
self > 1 && !(2..Math.sqrt(self)).any? { |i| self % i == 0 }
end
end
def count_quadratic_formula_primes(a,b)
(0..Float::INFINITY).lazy.map { |n| n**2 + a*n + b }.take_while(&:prime?).count
end
def product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values()
a, b = (-999..1000).to_a.repeated_permutation(2).max_by { |a, b| count_quadratic_formula_primes(a,b) }
a * b
end
start = Time.now
answer = product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values
puts "The answer is #{answer} and it took #{Time.now-start} seconds."
15 lines of hard-core ruby-style code!
Enjoy!
Update
It seems that lazy
adds considerable overhead to the performance of the code. So it is not advisable to use it.
Fortunately this works:
(0..Float::INFINITY).take_while { |n| (n**2 + a*n + b).prime? }.count
My code still runs ~2 times slower than the original (ends in 18 seconds), but it is more reasonable than with lazy
...
edited Apr 20 '14 at 18:07
answered Apr 20 '14 at 7:02
Uri AgassiUri Agassi
6,37211246
6,37211246
1
$begingroup$
also, you caninject
your a*b :)
$endgroup$
– gaussblurinc
Apr 20 '14 at 7:37
$begingroup$
I like your code better, but it takes 130 seconds, while the original code takes 10. The prime function is a beauty. I suspect that the count_quadratic_formula_primes method is slow.
$endgroup$
– Erwin Rooijakkers
Apr 20 '14 at 14:58
$begingroup$
@user2609980 - yes, apparentlylazy
add a lot of overhead... see my update
$endgroup$
– Uri Agassi
Apr 20 '14 at 18:08
3
$begingroup$
user2609980, I'm surprised at the difference in execution times, but I'd pay no heed to that at this stage of your Ruby education. Uri has covered a wide swath of ground in his answer, acquainting you with typical Ruby coding style, the addition of a method to an existing Ruby class (prime?
), the use of powerful enumeratorsmap
,product
,permutation
,any?
andtake_while
from theEnumerable
module, andup_to
from theInteger
class, and Ruby's newlazy
operator. He has also nicely explained his reasons for coding it the way he has. Great answer, Uri.
$endgroup$
– Cary Swoveland
Apr 20 '14 at 18:24
$begingroup$
@CarySwoveland I am indeed very happy with the answer :-).
$endgroup$
– Erwin Rooijakkers
Apr 21 '14 at 10:06
|
show 2 more comments
1
$begingroup$
also, you caninject
your a*b :)
$endgroup$
– gaussblurinc
Apr 20 '14 at 7:37
$begingroup$
I like your code better, but it takes 130 seconds, while the original code takes 10. The prime function is a beauty. I suspect that the count_quadratic_formula_primes method is slow.
$endgroup$
– Erwin Rooijakkers
Apr 20 '14 at 14:58
$begingroup$
@user2609980 - yes, apparentlylazy
add a lot of overhead... see my update
$endgroup$
– Uri Agassi
Apr 20 '14 at 18:08
3
$begingroup$
user2609980, I'm surprised at the difference in execution times, but I'd pay no heed to that at this stage of your Ruby education. Uri has covered a wide swath of ground in his answer, acquainting you with typical Ruby coding style, the addition of a method to an existing Ruby class (prime?
), the use of powerful enumeratorsmap
,product
,permutation
,any?
andtake_while
from theEnumerable
module, andup_to
from theInteger
class, and Ruby's newlazy
operator. He has also nicely explained his reasons for coding it the way he has. Great answer, Uri.
$endgroup$
– Cary Swoveland
Apr 20 '14 at 18:24
$begingroup$
@CarySwoveland I am indeed very happy with the answer :-).
$endgroup$
– Erwin Rooijakkers
Apr 21 '14 at 10:06
1
1
$begingroup$
also, you can
inject
your a*b :)$endgroup$
– gaussblurinc
Apr 20 '14 at 7:37
$begingroup$
also, you can
inject
your a*b :)$endgroup$
– gaussblurinc
Apr 20 '14 at 7:37
$begingroup$
I like your code better, but it takes 130 seconds, while the original code takes 10. The prime function is a beauty. I suspect that the count_quadratic_formula_primes method is slow.
$endgroup$
– Erwin Rooijakkers
Apr 20 '14 at 14:58
$begingroup$
I like your code better, but it takes 130 seconds, while the original code takes 10. The prime function is a beauty. I suspect that the count_quadratic_formula_primes method is slow.
$endgroup$
– Erwin Rooijakkers
Apr 20 '14 at 14:58
$begingroup$
@user2609980 - yes, apparently
lazy
add a lot of overhead... see my update$endgroup$
– Uri Agassi
Apr 20 '14 at 18:08
$begingroup$
@user2609980 - yes, apparently
lazy
add a lot of overhead... see my update$endgroup$
– Uri Agassi
Apr 20 '14 at 18:08
3
3
$begingroup$
user2609980, I'm surprised at the difference in execution times, but I'd pay no heed to that at this stage of your Ruby education. Uri has covered a wide swath of ground in his answer, acquainting you with typical Ruby coding style, the addition of a method to an existing Ruby class (
prime?
), the use of powerful enumerators map
, product
, permutation
, any?
and take_while
from the Enumerable
module, and up_to
from the Integer
class, and Ruby's new lazy
operator. He has also nicely explained his reasons for coding it the way he has. Great answer, Uri.$endgroup$
– Cary Swoveland
Apr 20 '14 at 18:24
$begingroup$
user2609980, I'm surprised at the difference in execution times, but I'd pay no heed to that at this stage of your Ruby education. Uri has covered a wide swath of ground in his answer, acquainting you with typical Ruby coding style, the addition of a method to an existing Ruby class (
prime?
), the use of powerful enumerators map
, product
, permutation
, any?
and take_while
from the Enumerable
module, and up_to
from the Integer
class, and Ruby's new lazy
operator. He has also nicely explained his reasons for coding it the way he has. Great answer, Uri.$endgroup$
– Cary Swoveland
Apr 20 '14 at 18:24
$begingroup$
@CarySwoveland I am indeed very happy with the answer :-).
$endgroup$
– Erwin Rooijakkers
Apr 21 '14 at 10:06
$begingroup$
@CarySwoveland I am indeed very happy with the answer :-).
$endgroup$
– Erwin Rooijakkers
Apr 21 '14 at 10:06
|
show 2 more comments
$begingroup$
I advise you to read a very good article.Red or Blue pill, Neo. Ruby vs. Python - which will you choose for your Backend? Good luck in your endeavors
New contributor
$endgroup$
add a comment |
$begingroup$
I advise you to read a very good article.Red or Blue pill, Neo. Ruby vs. Python - which will you choose for your Backend? Good luck in your endeavors
New contributor
$endgroup$
add a comment |
$begingroup$
I advise you to read a very good article.Red or Blue pill, Neo. Ruby vs. Python - which will you choose for your Backend? Good luck in your endeavors
New contributor
$endgroup$
I advise you to read a very good article.Red or Blue pill, Neo. Ruby vs. Python - which will you choose for your Backend? Good luck in your endeavors
New contributor
New contributor
answered 8 mins ago
Vladyslav AfrinVladyslav Afrin
11
11
New contributor
New contributor
add a comment |
add a comment |
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