Ruby Program to solve Circular Primes below number x - ruby
I'm working on project Euler #35. I am getting the wrong number returned and I can't find where I have done wrong!
def is_prime?(num)
(2..Math.sqrt(num)).each { |i| return false if num % i == 0}
true
end
def is_circular?(num)
len = num.to_s.length
return true if len == 1
(len - 1).times do
new_n = cycle(num)
break unless is_prime?(new_n)
end
end
def cycle(num)
ary = num.to_s.split("")
return ary.rotate!.join.to_i
end
def how_many
circulars = []
(2..1000000).each do |num|
if is_prime?(num) && is_circular?(num)
circulars << num
end
end
p circulars.count
end
how_many #=> 14426
The returned number is '14426'. I am only returning the circular primes, supposedly the correct answer is '55'
I have edited your code with few fixes in Ruby way. Your mistake was including corect set of [a, b, c] three times to count, instead of counting them as a one circular prime number. Your answer was correct, while 55 is the number of unique sets.
require 'prime'
class Euler35
def is_circular?(num)
circulars_for(num).all?{ |el| ::Prime.instance.prime?(el) }
end
def circulars_for(a)
a.to_s.split("").length.times.map{|el| a.to_s.split("").rotate(el).join.to_i }
end
def how_many
circulars = []
::Prime.each(1_000_000) do |num|
continue if circulars.include?(num)
if is_circular?(num)
circulars << circulars_for(num)
end
end
circulars.count
end
end
puts Euler35.new.how_many # => 55
Related
How do I fix a problem to call a function in Ruby?
I'm trying to use some ruby code that I've found in Github. I've downloaded the code and did the necessary imports the "requires" and tried to run it as it is described in the readme file on github repository. The code is the following:
In the file pcset_test.rb the code is the following:
require './pcset.rb'
require 'test/unit'
#
# When possible, test cases are adapted from
# Introduction to Post-Tonal Theory by Joseph N. Straus,
# unless obvious or otherwise noted.
#
class PCSetTest < Test::Unit::TestCase
def test_init
#assert_raise(ArgumentError) {PCSet.new []}
assert_raise(ArgumentError) {PCSet.new [1, 2, 3, 'string']}
assert_raise(ArgumentError) {PCSet.new "string"}
assert_raise(ArgumentError) {PCSet.new [1, 2, 3.6, 4]}
assert_equal([0, 1, 2, 9], PCSet.new([0, 1, 2, 33, 13]).pitches)
assert_equal([3, 2, 1, 11, 10, 0], PCSet.new_from_string('321bac').pitches)
assert_equal([0,2,4,5,7,11,9], PCSet.new([12,2,4,5,7,11,9]).pitches)
assert_nothing_raised() {PCSet.new []}
end
def test_inversion
end
def test_transposition
end
def test_multiplication
end
#
# set normal prime forte #
# 0,2,4,7,8,11 7,8,11,0,2,4 0,1,4,5,7,9 6-31
# 0,1,2,4,5,7,11 11,0,1,2,4,5,7 0,1,2,3,5,6,8 7-Z36
# 0,1,3,5,6,7,9,10,11 5,6,7,9,10,11,0,1,3 0,1,2,3,4,6,7,8,10 9-8
#
def test_normal_form
testPC = PCSet.new [0,4,8,9,11]
assert_kind_of(PCSet, testPC.normal_form)
assert_equal([8,9,11,0,4], testPC.normal_form.pitches)
assert_equal([10,1,4,6], PCSet.new([1,6,4,10]).normal_form.pitches)
assert_equal([2,4,8,10], PCSet.new([10,8,4,2]).normal_form.pitches)
assert_equal([7,8,11,0,2,4], PCSet.new([0,2,4,7,8,11]).normal_form.pitches)
assert_equal([11,0,1,2,4,5,7], PCSet.new([0,1,2,4,5,7,11]).normal_form.pitches)
assert_equal([5,6,7,9,10,11,0,1,3], PCSet.new([0,1,3,5,6,7,9,10,11]).normal_form.pitches)
end
def test_prime_form
assert_equal([0,1,2,6], PCSet.new([5,6,1,7]).prime.pitches)
assert_equal([0,1,4], PCSet.new([2,5,6]).prime.pitches)
assert_equal([0,1,4,5,7,9], PCSet.new([0,2,4,7,8,11]).prime.pitches)
assert_equal([0,1,2,3,5,6,8], PCSet.new([0,1,2,4,5,7,11]).prime.pitches)
assert_equal([0,1,2,3,4,6,7,8,10], PCSet.new([0,1,3,5,6,7,9,10,11]).prime.pitches)
end
def test_set_class
testPcs = PCSet.new([2,5,6])
testPrime = testPcs.prime
assert_equal([
[2,5,6], [3,6,7], [4,7,8], [5,8,9], [6,9,10], [7,10,11],
[8,11,0],[9,0,1], [10,1,2],[11,2,3],[0,3,4], [1,4,5],
[6,7,10],[7,8,11],[8,9,0], [9,10,1],[10,11,2],[11,0,3],
[0,1,4], [1,2,5], [2,3,6], [3,4,7], [4,5,8], [5,6,9]
].sort, PCSet.new([2,5,6]).set_class.map{|x| x.pitches})
assert_equal(testPcs.set_class.map{|x| x.pitches}, testPrime.set_class.map{|x| x.pitches})
end
def test_interval_vector
assert_equal([2,1,2,1,0,0], PCSet.new([0,1,3,4]).interval_vector)
assert_equal([2,5,4,3,6,1], PCSet.new([0,1,3,5,6,8,10]).interval_vector)
assert_equal([0,6,0,6,0,3], PCSet.new([0,2,4,6,8,10]).interval_vector)
end
def test_complement
assert_equal([6,7,8,9,10,11], PCSet.new([0,1,2,3,4,5]).complement.pitches)
assert_equal([3,4,5], PCSet.new([0,1,2], 6).complement.pitches)
end
#
# Test values from (Morris 1991), pages 105-111
# Citation:
# Morris. Class Notes for Atonal Music Theory
# Lebanon, NH. Frog Peak Music, 1991.
#
def test_invariance_vector
assert_equal([1,0,0,0,5,6,5,5],PCSet.new([0,2,5]).invariance_vector)
assert_equal([2,2,2,2,6,6,6,6],PCSet.new([0,1,6,7]).invariance_vector)
assert_equal([6,6,6,6,6,6,6,6],PCSet.new([0,2,4,6,8,10]).invariance_vector)
assert_equal([1,0,0,0,0,0,0,0],PCSet.new([0,1,2,3,4,5,8]).invariance_vector)
assert_equal([1,0,0,1,0,0,0,0],PCSet.new([0,1,2,3,5,6,8]).invariance_vector)
assert_equal([12,12,12,12,0,0,0,0],PCSet.new([0,1,2,3,4,5,6,7,8,9,10,11]).invariance_vector)
end
#
# Test values from (Huron 1994). Huron rounds, thus the 0.01 margin of error.
# Citation:
# Huron. Interval-Class Content in Equally Tempered Pitch-Class Sets:
# Common Scales Exhibit Optimum Tonal Consonance.
# Music Perception (1994) vol. 11 (3) pp. 289-305
#
def test_huron
h1 = PCSet.new([0,1,2,3,4,5,6,7,8,9,10,11]).huron
assert_in_delta(-0.2, h1[0], 0.01)
assert_in_delta(0.21, h1[1], 0.01)
h2 = PCSet.new([0,2,4,5,7,9,11]).huron
assert_in_delta(4.76, h2[0], 0.01)
assert_in_delta(0.62, h2[1], 0.01)
end
def test_coherence
end
end
And in the file pcset.rb the folloing code:
#
# => PCSet Class for Ruby
# => Beau Sievers
# => Hanover, Fall 2008.
#
#
# TODO: Make this a module to avoid namespace collisions.
# Lilypond and MusicXML output
#
include Math
def choose(n, k)
return [[]] if n.nil? || n.empty? && k == 0
return [] if n.nil? || n.empty? && k > 0
return [[]] if n.size > 0 && k == 0
c2 = n.clone
c2.pop
new_element = n.clone.pop
choose(c2, k) + append_all(choose(c2, k-1), new_element)
end
def append_all(lists, element)
lists.map { |l| l << element }
end
def array_to_binary(array)
array.inject(0) {|sum, n| sum + 2**n}
end
# the following method is horrifically inelegant
# but avoids monkey-patching.
# TODO: do this right, incl. error checking
def pearsons(x, y)
if !x.is_a?(Array) || !y.is_a?(Array) then raise StandardError, "x and y must be arrays", caller end
if x.size != y.size then raise StandardError, "x and y must be same size", caller end
sum_x = x.inject(0) {|sum, n| sum + n}
sum_y = y.inject(0) {|sum, n| sum + n}
sum_square_x = x.inject(0) {|sum, n| sum + n * n}
sum_square_y = y.inject(0) {|sum, n| sum + n * n}
xy = []
x.zip(y) {|a, b| xy.push(a * b)}
sum_xy = xy.inject(0) {|sum, n| sum + n}
num = sum_xy - ((sum_x * sum_y)/x.size)
den = Math.sqrt((sum_square_x - ((sum_x*sum_x)/x.size)) * (sum_square_y - ((sum_y*sum_y)/x.size)))
(num/den)
end
class PCSet
include Comparable
attr_reader :pitches, :base, :input
def initialize(pcarray, base = 12)
if pcarray.instance_of?(Array) && pcarray.all?{|pc| pc.instance_of?(Fixnum)}
#base, #input = base, pcarray
#pitches = pcarray.map{ |x| x % #base }.uniq
else
raise ArgumentError, "Improperly formatted PC array", caller
end
end
def PCSet.new_from_string(pcstring, base = 12)
if base > 36 then raise StandardError, "Use PCSet.new to create pcsets with a base larger than 36", caller end
pcarray = []
pcstring.downcase.split(//).each do |c|
if c <= 'z' and c >= '0' then pcarray.push(c.to_i(36)) end
end
PCSet.new pcarray, base
end
def <=>(pcs)
#pitches <=> pcs.pitches
end
def [](index)
#pitches[index]
end
# Intersection
def &(other)
PCSet.new #pitches & other.pitches
end
# Union
def |(other)
PCSet.new #pitches | other.pitches
end
def inspect
#pitches.inspect
end
def length
#pitches.length
end
def invert(axis = 0)
PCSet.new #pitches.map {|x| (axis-x) % #base}
end
def invert!(axis = 0)
#pitches.map! {|x| (axis-x) % #base}
end
def transpose(interval)
PCSet.new #pitches.map {|x| (x + interval) % #base}
end
def transpose!(interval)
#pitches.map! {|x| (x + interval) % #base}
end
def multiply(m = 5)
PCSet.new #pitches.map {|x| (x * m) % #base}
end
def multiply!(m = 5)
#pitches.map! {|x| (x * m) % #base}
end
def zero
transpose(-1 * #pitches[0])
end
def zero!
transpose!(-1 * #pitches[0])
end
def transpositions
(0..(#base-1)).to_a.map{|x| #pitches.map {|y| (y + x) % #base}}.sort.map {|x| PCSet.new x}
end
def transpositions_and_inversions(axis = 0)
transpositions + invert(axis).transpositions
end
#
# Normal form after Straus. Morris and AthenaCL do this differently.
#
def normal_form
tempar = #pitches.sort
arar = [] # [[1,4,7,8,10],[4,7,8,10,1], etc.] get each cyclic variation
tempar.each {arar.push PCSet.new(tempar.unshift(tempar.pop))}
most_left_compact(arar)
end
def normal_form!
#pitches = normal_form.pitches
end
def is_normal_form?
self.pitches == self.normal_form.pitches
end
def set_class
transpositions_and_inversions.map{|pcs| pcs.normal_form}.sort
end
def prime
most_left_compact([normal_form.zero, invert.normal_form.zero])
end
def prime!
self.pitches = self.prime.pitches
end
def is_prime?
self.pitches == self.prime.pitches
end
def complement
new_pitches = []
#base.times do |p|
if !#pitches.include? p then
new_pitches.push p
end
end
PCSet.new new_pitches
end
def full_interval_vector
pairs = choose(#pitches, 2) # choose every pc pair
intervals = pairs.map {|x| (x[1] - x[0]) % #base} # calculate every interval
i_vector = Array.new(#base-1).fill(0)
intervals.each {|x| i_vector[x-1] += 1} # count the intervals
i_vector
end
def interval_vector
i_vector = full_interval_vector
(0..((#base-1)/2)-1).each {|x| i_vector[x] += i_vector.pop}
i_vector
end
#
# Morris's invariance vector
#
def invariance_vector(m = 5)
t = transpositions.map!{|pcs| self & pcs}
ti = invert.transpositions.map!{|pcs| self & pcs}
tm = multiply(m).transpositions.map!{|pcs| self & pcs}
tmi = invert.multiply(m).transpositions.map!{|pcs| self & pcs}
tc = complement.transpositions.map!{|pcs| self & pcs}
tic = complement.invert.transpositions.map!{|pcs| self & pcs}
tmc = complement.multiply(m).transpositions.map!{|pcs| self & pcs}
tmic = complement.invert.multiply(m).transpositions.map!{|pcs| self & pcs}
[t, ti, tm, tmi, tc, tic, tmc, tmic].map{|x| x.reject{|pcs| pcs.pitches != #pitches}.length}
end
# Huron's aggregate dyadic consonance measure.
# Huron. Interval-Class Content in Equally Tempered Pitch-Class Sets:
# Common Scales Exhibit Optimum Tonal Consonance.
# Music Perception (1994) vol. 11 (3) pp. 289-305
def huron
if #base != 12 then raise StandardError, "PCSet.huron only makes sense for mod 12 pcsets", caller end
# m2/M7 M2/m7 m3/M6 M3/m6 P4/P5 A4/d5
huron_table = [-1.428, -0.582, 0.594, 0.386, 1.240, -0.453]
interval_consonance = []
interval_vector.zip(huron_table) {|x, y| interval_consonance.push(x * y) }
aggregate_dyadic_consonance = interval_consonance.inject {|sum, n| sum + n}
[aggregate_dyadic_consonance, pearsons(interval_vector, huron_table)]
end
#
# Balzano's vector of relations. Citation for all Balzano methods:
#
# Balzano. "The Pitch Set as a Level of Description for Studying Musical
# Pitch Perception" in Music, Mind, and Brain ed. Clynes. Plenum Press. 1982.
#
def vector_of_relations
(0..length-1).to_a.map do |i|
(0..length-1).to_a.map do |j|
(#pitches[(i + j) % length] - #pitches[i]) % #base
end
end
end
#
# Checks if the set satisfies Balzano's uniqueness.
#
def is_unique?
vector_of_relations.uniq.size == vector_of_relations.size
end
#
# Checks if the set satisfies Balzano's scalestep-semitone coherence.
# For all s[i] and s[i1]:
# j < k => v[i][j] < v[i1][k]
# Where j and k are scalestep-counting indices.
# And unless v[i][j] == 6 (a tritone), in which case the strict inequality is relaxed.
#
def is_coherent?
v = vector_of_relations
truth_array = []
all_pair_indices = choose((0..length-1).to_a, 2)
all_pair_indices.each do |i, i1|
all_pair_indices.each do |j, k|
if v[i][j] == 6
truth_array.push(v[i][j] <= v[i1][k])
else
truth_array.push(v[i][j] < v[i1][k])
end
if v[i1][j] == 6
truth_array.push(v[i1][j] <= v[i][k])
else
truth_array.push(v[i1][j] < v[i][k])
end
end
end
!truth_array.include?(false)
end
#
# Strict Balzano coherence, no inequality relaxation for tritones.
#
def is_strictly_coherent?
v = vector_of_relations
truth_array = []
all_pair_indices = choose((0..length-1).to_a, 2)
all_pair_indices.each do |i, i1|
all_pair_indices.each do |j, k|
truth_array.push(v[i][j] < v[i1][k])
truth_array.push(v[i1][j] < v[i][k])
end
end
!truth_array.include?(false)
end
def notes(middle_c = 0)
noteArray = ['C','C#','D','D#','E','F','F#','G','G#','A','A#','B']
if #base != 12 then raise StandardError, "PCSet.notes only makes sense for mod 12 pcsets", caller end
out_string = String.new
transpose(-middle_c).pitches.each do |p|
out_string += noteArray[p] + ", "
end
out_string.chop.chop
end
def info
print "modulo: #{#base}\n"
print "raw input: #{#input.inspect}\n"
print "pitch set: #{#pitches.inspect}\n"
print "notes: #{notes}\n"
print "normal: #{normal_form.inspect}\n"
print "prime: #{prime.inspect}\n"
print "interval vector: #{interval_vector.inspect}\n"
print "invariance vector: #{invariance_vector.inspect}\n"
print "huron ADC: #{huron[0]} pearsons: #{huron[1]}\n"
print "balzano coherence: "
if is_strictly_coherent?
print "strictly coherent\n"
elsif is_coherent?
print "coherent\n"
else
print "false\n"
end
end
# def lilypond
#
# end
#
# def musicXML
#
# end
###############################################################################
private
#
# Convert every pitch array to a binary representation, e.g.:
# [0,2,4,8,10] -> 010100010101
# 2^n: BA9876543210
# The smallest binary number is the most left-compact.
#
def most_left_compact(pcset_array)
if !pcset_array.all? {|pcs| pcs.length == pcset_array[0].length}
raise ArgumentError, "PCSet.most_left_compact: All PCSets must be of same cardinality", caller
end
zeroed_pitch_arrays = pcset_array.map {|pcs| pcs.zero.pitches}
binaries = zeroed_pitch_arrays.map {|array| array_to_binary(array)}
winners = []
binaries.each_with_index do |num, i|
if num == binaries.min then winners.push(pcset_array[i]) end
end
winners.sort[0]
end
end
I'm calling them as follows:
> my_pcset = PCSet.new([0,2,4,6,8,10])
> my_pcset2 = PCSet.new([1,5,9])
It shoud return:
> my_pcset = PCSet.new([0,2,4,6,8,10])
=> [0, 2, 4, 6, 8, 10]
> my_pcset2 = PCSet.new([1,5,9])
=> [1, 5, 9]
But is returning nothing.
The code is available on github
Thanks
Try this in terminal: irb -r ./path_to_directory/pcset.rb and then initialize the objects.
I think the documentation for the repo is bad as it does not explain how you should be running this. The result of my_pcset = PCSet.new([0,2,4,6,8,10]) should set my_pcset to an instance of a PCSet not an array, so these lines from the README file are confusing at best. 3. How to use it Make new PCSets: my_pcset = PCSet.new([0,2,4,6,8,10]) => [0, 2, 4, 6, 8, 10] my_pcset2 = PCSet.new([1,5,9]) => [1, 5, 9] Looking at the code, I see inspect has been delegated to #pitches def inspect #pitches.inspect end I think if you inspect my_pcset you will get the expected result. my_pcset = PCSet.new([0,2,4,6,8,10]) p my_pcset # will print [0, 2, 4, 6, 8, 10] or `my_pcset.inspect` will return what you are expecting.
How to optimize code - it works, but I know I'm missing much learning
The exercise I'm working on asks "Write a method, coprime?(num_1, num_2), that accepts two numbers as args. The method should return true if the only common divisor between the two numbers is 1."
I've written a method to complete the task, first by finding all the factors then sorting them and looking for duplicates. But I'm looking for suggestions on areas I should consider to optimize it.
The code works, but it is just not clean.
def factors(num)
return (1..num).select { |n| num % n == 0}
end
def coprime?(num_1, num_2)
num_1_factors = factors(num_1)
num_2_factors = factors(num_2)
all_factors = num_1_factors + num_2_factors
new = all_factors.sort
dups = 0
new.each_index do |i|
dups += 1 if new[i] == new[i+1]
end
if dups > 1
false
else
true
end
end
p coprime?(25, 12) # => true
p coprime?(7, 11) # => true
p coprime?(30, 9) # => false
p coprime?(6, 24) # => false
You could use Euclid's algorithm to find the GCD, then check whether it's 1. def gcd a, b while a % b != 0 a, b = b, a % b end return b end def coprime? a, b gcd(a, b) == 1 end p coprime?(25, 12) # => true p coprime?(7, 11) # => true p coprime?(30, 9) # => false p coprime?(6, 24) # => false```
You can just use Integer#gcd: def coprime?(num_1, num_2) num_1.gcd(num_2) == 1 end
You don't need to compare all the factors, just the prime ones. Ruby does come with a Prime class
require 'prime'
def prime_numbers(num_1, num_2)
Prime.each([num_1, num_2].max / 2).map(&:itself)
end
def factors(num, prime_numbers)
prime_numbers.select {|n| num % n == 0}
end
def coprime?(num_1, num_2)
prime_numbers = prime_numbers(num_1, num_2)
# & returns the intersection of 2 arrays (https://stackoverflow.com/a/5678143)
(factors(num_1, prime_numbers) & factors(num_2, prime_numbers)).length == 0
end
Euler 23 in Ruby
All right. I think I have the right idea to find the solution to Euler #23 (The one about finding the sum of all numbers that can't be expressed as the sum of two abundant numbers).
However, it is clear that one of my methods is too damn brutal.
How do you un-brute force this and make it work?
sum_of_two_abunds?(num, array) is the problematic method. I've tried pre-excluding certain numbers and it's still taking forever and I'm not even sure that it's giving the right answer.
def divsum(number)
divsum = 1
(2..Math.sqrt(number)).each {|i| divsum += i + number/i if number % i == 0}
divsum -= Math.sqrt(number) if Math.sqrt(number).integer?
divsum
end
def is_abundant?(num)
return true if divsum(num) > num
return false
end
def get_abundants(uptonum)
abundants = (12..uptonum).select {|int| is_abundant?(int)}
end
def sum_of_two_abunds?(num, array)
#abundant, and can be made from adding two abundant numbers.
array.each do |abun1|
array.each do |abun2|
current = abun1+abun2
break if current > num
return true if current == num
end
end
return false
end
def non_abundant_sum
ceiling = 28123
sum = (1..23).inject(:+) + (24..ceiling).select{|i| i < 945 && i % 2 != 0}.inject(:+)
numeri = (24..ceiling).to_a
numeri.delete_if {|i| i < 945 && i % 2 != 0}
numeri.delete_if {|i| i % 100 == 0}
abundants = get_abundants(ceiling)
numeri.each {|numerus| sum += numerus if sum_of_two_abunds?(numerus, abundants) == false}
return sum
end
start_time = Time.now
puts non_abundant_sum
#Not enough numbers getting excluded from the total.
duration = Time.now - start_time
puts "Took #{duration} s "
Solution 1
A simple way to make it a lot faster is to speed up your sum_of_two_abunds? method:
def sum_of_two_abunds?(num, array)
array.each do |abun1|
array.each do |abun2|
current = abun1+abun2
break if current > num
return true if current == num
end
end
return false
end
Instead of that inner loop, just ask the array whether it contains num - abun1:
def sum_of_two_abunds?(num, array)
array.each do |abun1|
return true if array.include?(num - abun1)
end
false
end
That's already faster than your Ruby code, since it's simpler and running faster C code. Also, now that that idea is clear, you can take advantage of the fact that the array is sorted and search num - abun1 with binary search:
def sum_of_two_abunds?(num, array)
array.each do |abun1|
return true if array.bsearch { |x| num - abun1 <=> x }
end
false
end
And making that Rubyish:
def sum_of_two_abunds?(num, array)
array.any? do |abun1|
array.bsearch { |x| num - abun1 <=> x }
end
end
Now you can get rid of your own special case optimizations and fix your incorrect divsum (which for example claims that divsum(4) is 5 ... you should really compare against a naive implementation that doesn't try any square root optimizations).
And then it should finish in well under a minute (about 11 seconds on my PC).
Solution 2
Or you could instead ditch sum_of_two_abunds? entirely and just create all sums of two abundants and nullify their contribution to the sum:
def non_abundant_sum
ceiling = 28123
abundants = get_abundants(ceiling)
numeri = (0..ceiling).to_a
abundants.each { |a| abundants.each { |b| numeri[a + b] = 0 } }
numeri.compact.sum
end
That runs on my PC in about 3 seconds.
Turning a method into an enumerable method
I rewrote the map method: def my_map(input, &block) mod_input = [] x = -1 while x < input.length - 1 x = x + 1 if block == nil return input break end mod_input.push(block.call(input[x])) end return mod_input end I need to call this code as I would call map or reverse. Does anyone know the syntax for that?
Are you asking how you put a method into a module? That's trivial:
module Enumerable
def my_map(&block)
mod_input = []
x = -1
while x < length - 1
x = x + 1
if block == nil
return self
break
end
mod_input.push(block.call(self[x]))
end
return mod_input
end
end
[1, 2, 3, 4, 5].my_map(&2.method(:*))
# => [2, 4, 6, 8, 10]
Or are you asking how to make your method an Enumerable method? That's more involved: your method currently uses many methods that are not part of the Enumerable API. So, even if you make it a member of the Enumerable module, it won't be an Enumerable method. Enumerable methods can only use each or other Enumerable methods. You use length and [] both of which are not part of the Enumerable interface, for example, Set doesn't respond to [].
This would be a possible implementation, using the Enumerable#inject method:
module Enumerable
def my_map
return enum_for(__method__) unless block_given?
inject([]) {|res, el| res << yield(el) }
end
end
[1, 2, 3, 4, 5].my_map(&2.method(:*))
# => [2, 4, 6, 8, 10]
A less elegant implementation using each
module Enumerable
def my_map
return enum_for(__method__) unless block_given?
[].tap {|res| each {|el| res << yield(el) }}
end
end
[1, 2, 3, 4, 5].my_map(&2.method(:*))
# => [2, 4, 6, 8, 10]
Note that apart from being simply wrong, your code is very un-idiomatic. There is also dead code in there.
the break is dead code: the method returns in the line just before it, therefore the break will never be executed. You can just get rid of it.
def my_map(&block)
mod_input = []
x = -1
while x < length - 1
x = x + 1
if block == nil
return self
end
mod_input.push(block.call(self[x]))
end
return mod_input
end
Now that we have gotten rid of the break, we can convert the conditional into a guard-style statement modifier conditional.
def my_map(&block)
mod_input = []
x = -1
while x < length - 1
x = x + 1
return self if block == nil
mod_input.push(block.call(self[x]))
end
return mod_input
end
It also doesn't make sense that it is in the middle of the loop. It should be at the beginning of the method.
def my_map(&block)
return self if block == nil
mod_input = []
x = -1
while x < length - 1
x = x + 1
mod_input.push(block.call(self[x]))
end
return mod_input
end
Instead of comparing an object against nil, you should just ask it whether it is nil?: block.nil?
def my_map(&block)
return self if block.nil?
mod_input = []
x = -1
while x < length - 1
x = x + 1
mod_input.push(block.call(self[x]))
end
return mod_input
end
Ruby is an expression-oriented language, the value of the last expression that is evaluated in a method body is the return value of that method body, there is no need for an explicit return.
def my_map(&block)
return self if block.nil?
mod_input = []
x = -1
while x < length - 1
x = x + 1
mod_input.push(block.call(self[x]))
end
mod_input
end
x = x + 1 is more idiomatically written x += 1.
def my_map(&block)
return self if block.nil?
mod_input = []
x = -1
while x < length - 1
x += 1
mod_input.push(block.call(self[x]))
end
mod_input
end
Instead of Array#push with a single argument it is more idiomatic to use Array#<<.
def my_map(&block)
return self if block.nil?
mod_input = []
x = -1
while x < length - 1
x += 1
mod_input << block.call(self[x])
end
mod_input
end
Instead of Proc#call, you can use the .() syntactic sugar.
def my_map(&block)
return self if block.nil?
mod_input = []
x = -1
while x < length - 1
x += 1
mod_input << block.(self[x])
end
mod_input
end
If you don't want to store, pass on or otherwise manipulate the block as an object, there is no need to capture it as a Proc. Just use block_given? and yield instead.
def my_map
return self unless block_given?
mod_input = []
x = -1
while x < length - 1
x += 1
mod_input << yield(self[x])
end
mod_input
end
This one is opinionated. You could move incrementing the counter into the condition.
def my_map
return self unless block_given?
mod_input = []
x = -1
while (x += 1) < length
mod_input << yield(self[x])
end
mod_input
end
And then use the statement modifier form.
def my_map
return self unless block_given?
mod_input = []
x = -1
mod_input << yield(self[x]) while (x += 1) < length
mod_input
end
Also, your variable names could use some improvements. For example, what does mod_input even mean? All I can see is that it is what you output, so why does it even have "input" in its name? And what is x?
def my_map
return self unless block_given?
result = []
index = -1
result << yield(self[index]) while (index += 1) < length
result
end
This whole sequence of initializing a variable, then mutating the object assigned to that variable and lastly returning the object can be simplified by using the K Combinator, which is available in Ruby as Object#tap.
def my_map
return self unless block_given?
[].tap {|result|
index = -1
result << yield(self[index]) while (index += 1) < length
}
end
The entire while loop is useless. It's just re-implementing Array#each, which is a) unnecessary because Array#each already exists, and b) means that your my_map method will only work with Arrays but not other Enumerables (for example Set or Enumerator). So, let's just use each instead.
def my_map
return self unless block_given?
[].tap {|result|
each {|element|
result << yield(element)
}
}
end
Now it starts to look like Ruby code! What you had before was more like BASIC written in Ruby syntax.
This pattern of first creating a result object, then modifying that result object based on each element of a collection and in the end returning the result is very common, and it even has a fancy mathematical name: Catamorphism, although in the programming world, we usually call it fold or reduce. In Ruby, it is called Enumerable#inject.
def my_map
return self unless block_given?
inject([]) {|result, element|
result << yield(element)
}
end
That return self is strange. map is supposed to return a new object! You don't return a new object, you return the same object. Let's fix that.
def my_map
return dup unless block_given?
inject([]) {|result, element|
result << yield(element)
}
end
And actually, map is also supposed to return an Array, but you return whatever it is that you called map on.
def my_map
return to_a unless block_given?
inject([]) {|result, element|
result << yield(element)
}
end
But really, if you look at the documentation of Enumerable#map, you will find that it returns an Enumerator and not an Array when called without a block.
def my_map
return enum_for(:my_map) unless block_given?
inject([]) {|result, element|
result << yield(element)
}
end
And lastly, we can get rid of the hardcoded method name using the Kernel#__method__ method.
def my_map
return enum_for(__method__) unless block_given?
inject([]) {|result, element|
result << yield(element)
}
end
Now, that looks a lot better!
class Array def my_map(&block) # your code, replacing `input` with `self` end end The code itself is not really idiomatic Ruby - while is very rarely used for iteration over collections, and if you don't need to pass a block somewhere else, it is generally cleaner to use block_given? instead of block.nil? (let alone block == nil), and yield input[x] instead of block.call(input[x]).
more ruby way of doing project euler #2
I'm trying to learn Ruby, and am going through some of the Project Euler problems. I solved problem number two as such:
def fib(n)
return n if n < 2
vals = [0, 1]
n.times do
vals.push(vals[-1]+vals[-2])
end
return vals.last
end
i = 1
s = 0
while((v = fib(i)) < 4_000_000)
s+=v if v%2==0
i+=1
end
puts s
While that works, it seems not very ruby-ish—I couldn't come up with any good purely Ruby answer like I could with the first one ( puts (0..999).inject{ |sum, n| n%3==0||n%5==0 ? sum : sum+n }).
For a nice solution, why don't you create a Fibonacci number generator, like Prime or the Triangular example I gave here.
From this, you can use the nice Enumerable methods to handle the problem. You might want to wonder if there is any pattern to the even Fibonacci numbers too.
Edit your question to post your solution...
Note: there are more efficient ways than enumerating them, but they require more math, won't be as clear as this and would only shine if the 4 million was much higher.
As demas' has posted a solution, here's a cleaned up version:
class Fibo
class << self
include Enumerable
def each
return to_enum unless block_given?
a = 0; b = 1
loop do
a, b = b, a + b
yield a
end
end
end
end
puts Fibo.take_while { |i| i < 4000000 }.
select(&:even?).
inject(:+)
My version based on Marc-André Lafortune's answer:
class Some
#a = 1
#b = 2
class << self
include Enumerable
def each
1.upto(Float::INFINITY) do |i|
#a, #b = #b, #a + #b
yield #b
end
end
end
end
puts Some.take_while { |i| i < 4000000 }.select { |n| n%2 ==0 }
.inject(0) { |sum, item| sum + item } + 2
def fib first, second, sum = 1,2,0 while second < 4000000 sum += second if second.even? first, second = second, first + second end puts sum end
You don't need return vals.last. You can just do vals.last, because Ruby will return the last expression (I think that's the correct term) by default.
fibs = [0,1]
begin
fibs.push(fibs[-1]+fibs[-2])
end while not fibs[-1]+fibs[-2]>4000000
puts fibs.inject{ |sum, n| n%2==0 ? sum+n : sum }
Here's what I got. I really don't see a need to wrap this in a class. You could in a larger program surely, but in a single small script I find that to just create additional instructions for the interpreter. You could select even, instead of rejecting odd but its pretty much the same thing.
fib = Enumerator.new do |y|
a = b = 1
loop do
y << a
a, b = b, a + b
end
end
puts fib.take_while{|i| i < 4000000}
.reject{|x| x.odd?}
.inject(:+)
That's my approach. I know it can be less lines of code, but maybe you can take something from it. class Fib def first #p0 = 0 #p1 = 1 1 end def next r = if #p1 == 1 2 else #p0 + #p1 end #p0 = #p1 #p1 = r r end end c = Fib.new f = c.first r = 0 while (f=c.next) < 4_000_000 r += f if f%2==0 end puts r
I am new to Ruby, but here is the answer I came up with.
x=1
y=2
array = [1,2]
dar = []
begin
z = x + y
if z % 2 == 0
a = z
dar << a
end
x = y
y = z
array << z
end while z < 4000000
dar.inject {:+}
puts "#{dar.sum}"
def fib_nums(num)
array = [1, 2]
sum = 0
until array[-2] > num
array.push(array[-1] + array[-2])
end
array.each{|x| sum += x if x.even?}
sum
end