How do I add incrementally?
x = 1
while x < 365
x+=x
puts x
end
but this isn't right
Thanks!
Mathematically speaking, you're looking for the summation of n for n = 1 to y. According to WolframAlpha, this summation can be reduced to y(y+1)/2. Therefore, you can calculate this as:
x = 365*(365+1)/2 #=> 66795
No need for any loops. This code way more efficient; O(1) instead of O(n).
If I correctly understood what you need:
(1..365).reduce(0) { |memo, i| memo + i }
#⇒ 66795
or, in a short form (credits to #Jörg W. Mittag):
(1..365).reduce(:+)
Here we use standard reduce procedure on Range. First of all we construct range, containing integers from 1 up to 365. Then we iterate these integers, carrying the total (named memo according to ruby convention).
Related
I need to DRY this code but I don't know how.
I tried to dry the if condition but I don't know how to put the while in this.
def sum_with_while(min, max)
# CONSTRAINT: you should use a while..end structure
array = (min..max).to_a
sum = 0
count = 0
if min > max
return -1
else
while count < array.length
sum += array[count]
count += 1
end
end
return sum
end
Welcome to stack overflow!
Firstly, I should point out that "DRY" stands for "Don't Repeat Yourself". Since there's no repetition here, that's not really the problem with this code.
The biggest issue here is it's unrubyish. The ruby community has certain things it approves of, and certain things it avoids. That said, while loops are themselves considered bad ruby, so if you've been told to write it with a while loop, I'm guessing you're trying to get us to do your homework for you.
So I'm going to give you a couple of things to do a web search for that will help start you off:
ruby guard clauses - this will reduce your if-else-end into a simple if
ruby array pop - you can do while item = array.pop - since pop returns nil once the array is empty, you don't need a count. Again, bad ruby to do this... but maybe consider while array.any?
ruby implicit method return - generally we avoid commands we don't need
It's worth noting that using the techniques above, you can get the content of the method down to 7 reasonably readable lines. If you're allowed to use .inject or .sum instead of while, this whole method becomes 2 lines.
(as HP_hovercraft points out, the ternary operator reduces this down to 1 line. On production code, I'd be tempted to leave it as 2 lines for readability - but that's just personal preference)
You can put the whole thing in one line with a ternary:
def sum_with_while(min, max)
min > max ? -1 : [*(min..max)].inject(0){|sum,x| sum + x }
end
This is one option, cleaning up your code, see comments:
def sum_with_while(range) # pass a range
array = range.to_a
sum, count = 0, 0 # parallel assignment
while count < array.length
sum += array[count]
count += 1
end
sum # no need to return
end
sum_with_while(10..20)
#=> 165
More Rubyish:
(min..max).sum
Rule 1: Choose the right algorithm.
You wish to compute an arithmetic series.1
def sum_with_while(min, max)
max >= min ? (max-min+1)*(min+max)/2 : -1
end
sum_with_while(4, 4)
#=> 4
sum_with_while(4, 6)
#=> 15
sum_with_while(101, 9999999999999)
#=> 49999999999994999999994950
1. An arithmetic series is the sum of the elements of an arithmetic sequence. Each term of the latter is computed from the previous one by adding a fixed constant n (possibly negative). Heremax-min+1 is the number of terms in the sequence and (min+max)/2, if (min+max) is even, is the average of the values in the sequence. As (max-min+1)*(min+max) is even, this works when (min+max) is odd as well.
I am doing a ruby problem that wants a method to find all divisors of a number except itself with the output being a sorted array. If the number is prime, list that it is prime.
I am currently trying to teach myself recursion. Simple recursive problems like finding the factorial of a number is pretty basic to understand but I wanted to know if this particular problem could be done recursively. It seems it fits the criteria of one that could but I could not figure it out.
Example n = 15, divisors besides itself are [3,5].
My code that solved the problem.
require 'prime'
def divisors(n)
return "#{n} is prime" if Prime.prime?(n)
x = n/2
arr = []
until x == 1
arr << x if n % x == 0
x -= 1
end
arr.sort
end
Any help doing this recursively would be great or just letting me know it's not a problem that can be done this way would be helpful too.
def divisors(n, x=nil)
return "#{n} is prime" if Prime.prime?(n)
x ||= n/2
arr = []
return arr if x == 1
if n % x == 0
arr << x
end
(arr.concat divisors(n, x - 1)).sort
end
The function is refactored to handle three things:
the initial call (x ||= /2)
base cases (early returns)
iteration logic done through recursion.
An important thing is that the variable which changes during the iteration (x) is placed as a parameter for the method (with a default value, so it can essentially be used as a private parameter)
By the way, I personally found learning Elixir very helpful in understanding recursion. With pattern matching and multiple functional clauses, the initial call, base case, and iteration can be split into their own methods.
So I was running some benchmarks in Ruby 2.4.0 and realized that
(1...1000000000000000000000000000000).sum
calculates immediately whereas
(1...1000000000000000000000000000000).inject(:+)
takes so long that I just aborted the operation. I was under the impression that Range#sum was an alias for Range#inject(:+) but it seems like that is not true. So how does sum work, and why is it so much faster than inject(:+)?
N.B. The documentation for Enumerable#sum (which is implemented by Range) does not say anything about lazy evaluation or anything along those lines.
Short answer
For an integer range :
Enumerable#sum returns (range.max-range.min+1)*(range.max+range.min)/2
Enumerable#inject(:+) iterates over every element.
Theory
The sum of integers between 1 and n is called a triangular number, and is equal to n*(n+1)/2.
The sum of integers between n and m is the triangular number of m minus the triangular number of n-1, which is equal to m*(m+1)/2-n*(n-1)/2, and can be written (m-n+1)*(m+n)/2.
Enumerable#sum in Ruby 2.4
This property in used in Enumerable#sum for integer ranges :
if (RTEST(rb_range_values(obj, &beg, &end, &excl))) {
if (!memo.block_given && !memo.float_value &&
(FIXNUM_P(beg) || RB_TYPE_P(beg, T_BIGNUM)) &&
(FIXNUM_P(end) || RB_TYPE_P(end, T_BIGNUM))) {
return int_range_sum(beg, end, excl, memo.v);
}
}
int_range_sum looks like this :
VALUE a;
a = rb_int_plus(rb_int_minus(end, beg), LONG2FIX(1));
a = rb_int_mul(a, rb_int_plus(end, beg));
a = rb_int_idiv(a, LONG2FIX(2));
return rb_int_plus(init, a);
which is equivalent to:
(range.max-range.min+1)*(range.max+range.min)/2
the aforementioned equality!
Complexity
Thanks a lot to #k_g and #Hynek-Pichi-Vychodil for this part!
sum
(1...1000000000000000000000000000000).sum
requires three additions, a multiplication, a substraction and a division.
It's a constant number of operations, but multiplication is O((log n)²), so Enumerable#sum is O((log n)²) for an integer range.
inject
(1...1000000000000000000000000000000).inject(:+)
requires 999999999999999999999999999998 additions!
Addition is O(log n), so Enumerable#inject is O(n log n).
With 1E30 as input, inject with never return. The sun will explode long before!
Test
It's easy to check if Ruby Integers are being added :
module AdditionInspector
def +(b)
puts "Calculating #{self}+#{b}"
super
end
end
class Integer
prepend AdditionInspector
end
puts (1..5).sum
#=> 15
puts (1..5).inject(:+)
# Calculating 1+2
# Calculating 3+3
# Calculating 6+4
# Calculating 10+5
#=> 15
Indeed, from enum.c comments :
Enumerable#sum method may not respect method redefinition of "+"
methods such as Integer#+.
I'm trying to write a method that returns the nth prime number.
I've worked out a solution but the problem is in my method. I create a large array of numbers that seems to process super slow. (1..104729).to_a to be exact. I chose 104729 because the max n can be is 10000 and the 10000th integer is 104729. I'm looking for a way to optimize my method.
Is 104729 is too large a value? Is there a way to write this so that I'm not creating a large array?
Here's the method:
def PrimeMover(num)
def is_prime(x)
i = 0
nums = (2..x).to_a
while nums[i] < nums.max
if x % nums[i] != 0
i += 1
else
return false
end
end
return true
end
primes_arr = (3..104729).to_a.select {|y| is_prime(y)}
primes_arr[num]
end
require "prime"
def find_prime(nth)
Prime.take(nth).last
end
Combine Ruby's built-in prime library, and a lazy enumerator for performance:
require 'prime'
(1...100_000).lazy.select(&:prime?).take(100).to_a
Or simply, as highlighted by Arturo:
Prime.take(100)
You can use Ruby's built in #prime? method, which seems pretty efficient.
The code:
require 'prime'
primes_arr = (3..104729).to_a.select &:prime?
runs in 2-3 seconds on my machine, which I find somewhat acceptable.
If you need even better performance or if you really need to write your own method, try implementing the Sieve of Erathostenes. Here are some Ruby samples of that: http://rosettacode.org/wiki/Sieve_of_Eratosthenes#Ruby
Here's an optimal a trial division implementation of is_prime without relying on the Prime class:
A prime number is a whole number divisible only by 1 and itself, and 1 is not prime. So we want to know if x divides into anything less than x and greater than 1. So we start the count at 2, and we end at x - 1.
def prime?(x)
return false if x < 2
2.upto(x - 1) do |n|
return false if (x % n).zero?
end
true
end
As soon as x % n has a remainder, we can break the loop and say this number is not prime. This saves you from looping over the entire range. If all the possible numbers were exhausted, we know the number is prime.
This is still not optimal. For that you would need a sieve, or a different detection algorithm to trial division. But it's a big improvement on your code. Taking the nth up to you.
I'm trying to find all possible product of two 3-digit numbers. When I work with small ranges, I'm able to get an output in short amount of time but when the ranges are big, it seems to take really long time. Is there any way to to shorten the time to get the result?
The problem I'm working on is:
"A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
Find the largest palindrome made from the product of two 3-digit numbers."
a = []
for x in 100..999
for y in 100..999
num = (x * y)
unless a.include? num
a.push num
end
end
end
p a
This is going to compute 100 x 101 and 101 x 100 separately, even though they're not going to be pushed to the array since they're already in it.
I'm bad at math, but maybe every time x goes up, y's minimum range can go up since that one was just used? people who are better at math can tell me if this is going to start missing numbers.
z= 100
for x in 100..999
for y in z..999
num = (x * y)
unless a.include? num
a.push num
end
z = z+1
end
end
I think doing this might make the "unless a.include? num" line unnecessary, too.
Looking at your code a quick optimization you can make is to use a set rather than an array to store the already computed products.
Since a is an array, a.include?(num) will have to iterate through the entire list of elements before returning true / false.
If a were to be a set, a.include?(num) will return in sub linear time.
Example:
require 'set'
a = Set.new
for x in 100..999
for y in 100..999
num = (x * y)
unless a.include? num
a.add(num)
end
end
end
puts a.to_a.join(", ")
Moreover one of the nice properties of a set is that it only stores unique elements so the following would be equivalent:
require 'set'
a = Set.new
for x in 100..999
for y in 100..999
num = (x * y)
a.add(num)
end
end
puts a.to_a.join(", ")
What are you really trying to do, i.e. what is the original problem, and why do you need all of these products?
Are you printing every single one out? Is someone asking you for a concrete list of every single one?
If not, there is likely a better way to deal with this problem. For example, if all you wanted is to check if a number X will be an element in "that list of products", all you'd have to do is:
range = 100..999
range.any? { |i| range.include?(x / i) }