x = 16
while x != 1 do
if x % 2 == 0
x = x / 2
print "#{x} "
end
break if x < 0
end
Hi, the result I get from above is 8 4 2 1 . Is there any way to remove the space at the end?
One of Rubys main features is its beauty - you can shorten that loop to a nice one liner when using an array:
x = 16
arr = []
arr.push(x /= 2) while x.even?
puts arr.join(' ')
# => "8 4 2 1"
* As sagarpandya82 suggested x.even? is the same as using x % 2 == 0, leading to even more readable code
Don't print the values into the loop. Put them into a list (array) then, after the loop, join the array items using space as glue.
x = 16
a = []
while x != 1 do
if x % 2 == 0
x = x / 2
a << x
end
break if x < 0
end
puts '<' + a.join(' ') + '>'
The output is:
<8 4 2 1>
As #Bathsheba notes in a comment, this solution uses extra memory (the array) to store the values and also the call to Array#join generates a string that doubles the memory requirements. This is not an issue for small lists as the one in the question but needs to be considered the list becomes very large.
loop.reduce([[], 16]) do |(acc, val), _|
break acc if val <= 1
acc << val / 2 if val.even?
[acc, val / 2]
end.join ' '
if x != 0
print " "
end
is one way, having dropped the suffixed space from the other print. I/O will always be the performance bottleneck; an extra if will have a negligible effect on performance, and the extra print will merely contribute to the output stream which is normally buffered.
Related
I have been attempting the test below on codewars. I am relatively new to coding and will look for more appropriate solutions as well as asking you for feedback on my code. I have written the solution at the bottom and for the life of me cannot understand what is missing as the resultant figure is always 0. I'd very much appreciate feedback on my code for the problem and not just giving your best solution to the problem. Although both would be much appreciated. Thank you in advance!
The test posed is:
If we list all the natural numbers below 10 that are multiples of 3 or
5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Finish the solution so that it returns the sum of all the multiples of
3 or 5 below the number passed in. Additionally, if the number is
negative, return 0 (for languages that do have them).
Note: If the number is a multiple of both 3 and 5, only count it once.
My code is as follows:
def solution(number)
array = [1..number]
multiples = []
if number < 0
return 0
else
array.each { |x|
if x % 3 == 0 || x % 5 == 0
multiples << x
end
}
end
return multiples.sum
end
In a situation like this, when something in your code produces an unexpected result you should debug it, meaning, run it line by line with the same argument and see what each variable holds. Using some kind of interactive console for running code (like irb) is very helpfull.
Moving to your example, let's start from the beginning:
number = 10
array = [1..number]
puts array.size # => 1 - wait what?
puts array[0].class # => Range
As you can see the array variable doesn't contain numbers but rather a Range object. After you finish filtering the array the result is an empty array that sums to 0.
Regardless of that, Ruby has a lot of built-in methods that can help you accomplish the same problem typing fewer words, for example:
multiples_of_3_and_5 = array.select { |number| number % 3 == 0 || number % 5 == 0 }
When writing a multiline block of code, prefer the do, end syntax, for example:
array.each do |x|
if x % 3 == 0 || x % 5 == 0
multiples << x
end
end
I'm not suggesting that this is the best approach per se, but using your specific code, you could fix the MAIN problem by editing the first line of your code in one of 2 ways:
By either converting your range to an array. Something like this would do the trick:
array = (1..number).to_a
or by just using a range INSTEAD of an array like so:
range = 1..number
The latter solution inserted into your code might look like this:
number = 17
range = 1..number
multiples = []
if number < 0
return 0
else range.each{|x|
if x % 3 == 0 || x % 5 == 0
multiples << x
end
}
end
multiples.sum
#=> 60
The statement return followed by end suggests that you were writing a method, but the def statement is missing. I believe that should be
def tot_sum(number, array)
multiples = []
if number < 0
return 0
else array.each{|x|
if x % 3 == 0 || x % 5 == 0
multiples << x
end
}
end
return multiples.sum
end
As you point out, however, this double-counts numbers that are multiples of 15.
Let me suggest a more efficient way of writing that. First consider the sum of numbers that are multiples of 3 that do not exceed a given number n.
Suppose
n = 3
m = 16
then the total of numbers that are multiples of three that do not exceed 16 can be computed as follows:
3 * 1 + 3 * 2 + 3 * 3 + 3 * 4 + 3 * 5
= 3 * (1 + 2 + 3 + 4 + 5)
= 3 * 5 * (1 + 5)/2
= 45
This makes use of the fact that 5 * (1 + 5)/2 equals the sum of an algebraic series: (1 + 2 + 3 + 4 + 5).
We may write a helper method to compute this sum for any number n, with m being the number that multiples of n cannot exceed:
def tot_sum(n, m)
p = m/n
n * p * (1 + p)/2
end
For example,
tot_sum(3, 16)
#=> 45
We may now write a method that gives the desired result (remembering that we need to account for the fact that multiples of 15 are multiples of both 3 and 5):
def tot(m)
tot_sum(3, m) + tot_sum(5, m) - tot_sum(15, m)
end
tot( 9) #=> 23
tot( 16) #=> 60
tot(9999) #=> 23331668
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000.
def multiples_of(number)
number = number.to_f - 1.0
result = 0
if (number / 5.0) == 1 || (number / 3.0) == 1
return result = result + 5.0 + 3.0
elsif (number % 3).zero? || (number % 5).zero?
result += number
multiples_of(number-1)
else
multiples_of(number-1)
end
return result
end
p multiples_of(10.0)
My code is returning 9.0 rather than 23.0.
Using Core Methods to Select & Sum from a Range
It's not entirely clear what you really want to do here. This is clearly a homework assignment, so it's probably intended to get you to think in a certain way about whatever the lesson is. If that's the case, refer to your lesson plan or ask your instructor.
That said, if you restrict the set of possible input values to integers and use iteration rather than recursion, you can trivially solve for this using Array#select on an exclusive Range, and then calling Array#sum on the intermediate result. For example:
(1...10).select { |i| i.modulo(3).zero? || i.modulo(5).zero? }.sum
#=> 23
(1...1_000).select { |i| i.modulo(3).zero? || i.modulo(5).zero? }.sum
#=> 233168
Leave off the #sum if you want to see all the selected values. In addition, you can create your own custom validator by comparing your logic to an expected result. For example:
def valid_result? range_end, checksum
(1 ... range_end).select do |i|
i.modulo(3).zero? || i.modulo(5).zero?
end.sum.eql? checksum
end
valid_result? 10, 9
#=> false
valid_result? 10, 23
#=> true
valid_result? 1_000, 233_168
#=> true
There are a number of issues with your code. Most importantly, you're making recursive calls but you aren't combining their results in any way.
Let's step through what happens with an input of 10.
You assign number = number.to_f - 1.0 which will equal 9.
Then you reach the elsif (number % 3).zero? || (number % 5).zero? condition which is true, so you call result += number and multiples_of(number-1).
However, you're discarding the return value of the recursive call and call return result no matter what. So, your recursion doesn't have any impact on the return value. And for any input besides 3 or 5 you will always return input-1 as the return value. That's why you're getting 9.
Here's an implementation which works, for comparison:
def multiples_of(number)
number -= 1
return 0 if number.zero?
if number % 5 == 0 || number % 3 == 0
number + multiples_of(number)
else
multiples_of(number)
end
end
puts multiples_of(10)
# => 23
Note that I'm calling multiples_of(number) instead of multiples_of(number - 1) because you're already decrementing the input on the function's first line. You don't need to decrement twice - that would cause you to only process every other number e.g. 9,7,5,3
explanation
to step throgh the recursion a bit to help you understand it. Let's say we have an input of 4.
We first decrement the input so number=3. Then we hits the if number % 5 == 0 || number % 3 == 0 condition so we return number + multiples_of(number).
What does multiples_of(number) return? Now we have to evaluate the next recursive call. We decrement the number so now we have number=2. We hit the else block so now we'll return multiples_of(number).
We do the same thing with the next recursive call, with number=1. This multiples_of(1). We decrement the input so now we have number=0. This matches our base case so finally we're done with recursive calls and can work up the stack to figure out what our actual return value is.
For an input of 6 it would look like so:
multiples_of(6)
\
5 + multiples_of(5)
\
multiples_of(4)
\
3 + multiples_of(3)
\
multiples_of(2)
\
multiples_of(1)
\
multiples_of(0)
\
0
The desired result can be obtained from a closed-form expression. That is, no iteration is required.
Suppose we are given a positive integer n and wish to compute the sum of all positive numbers that are multiples of 3 that do not exceed n.
1*3 + 2*3 +...+ m*3 = 3*(1 + 2 +...+ m)
where
m = n/3
1 + 2 +...+ m is the sum of an algorithmic expression, given by:
m*(1+m)/2
We therefore can write:
def tot(x,n)
m = n/x
x*m*(1+m)/2
end
For example,
tot(3,9) #=> 18 (1*3 + 2*3 + 3*3)
tot(3,11) #=> 18
tot(3,12) #=> 30 (18 + 4*3)
tot(3,17) #=> 45 (30 + 5*3)
tot(5,9) #=> 5 (1*5)
tot(5,10) #=> 15 (5 + 2*5)
tot(5,14) #=> 15
tot(5,15) #=> 30 (15 + 3*5)
The sum of numbers no larger than n that are multiple of 3's and 5's is therefore given by the following:
def sum_of_multiples(n)
tot(3,n) + tot(5,n) - tot(15,n)
end
- tot(15,n) is needed because the first two terms double-count numbers that are multiples of 15.
sum_of_multiples(9) #=> 23 (3 + 6 + 9 + 5)
sum_of_multiples(10) #=> 33 (23 + 2*5)
sum_of_multiples(11) #=> 33
sum_of_multiples(12) #=> 45 (33 + 4*3)
sum_of_multiples(14) #=> 45
sum_of_multiples(15) #=> 60 (45 + 3*5)
sum_of_multiples(29) #=> 195
sum_of_multiples(30) #=> 225
sum_of_multiples(1_000) #=> 234168
sum_of_multiples(10_000) #=> 23341668
sum_of_multiples(100_000) #=> 2333416668
sum_of_multiples(1_000_000) #=> 233334166668
I'm doing www.eulerproject.net, the first problem:
If we list all the natural numbers below 10, that are multiples of 3
or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find
the sum of all the multiples of 3 or 5 below 1000.
The following is the code I have so far.
(3..999).to_a.select do |x|
x % 3.0 == 0 || x % 5.0 == 0
end
It would be easy to append the numbers into an array, but how can this be done by how can this be done by chaining a method onto the end of this. Something like
p start loop
do stuff
end.sum
To answer the question - yes, you can chain the method like you've shown.
(3..999).to_a.select do |x|
x % 3 == 0 || x % 5 == 0 # you don't have to use floats here, integers would work
end.inject(:+)
#=> 233168
The rule of a style guides is to NOT to chain methods to multiline do end blocks, but it is a working code.
It's the same as writing
(3..999).to_a.select { |x| x % 3 == 0 || x % 5 == 0 }.inject(:+)
#=>233168
Array#sum is an ActiveSupport method, not Ruby's, but I think you should use Ruby's methods in eulerproject tasks.
You are summing arithmetic series, so there is no need to iterate:
def sum(n,m)
p = n/m
m*p*(1+p)/2
end
n = 999
sum(n,3) + sum(n,5) - sum(n,15)
#=> 233168
Consider:
n = 100
m = 3
p = 100/3 #=> 33
sum(100,3) = 3 + 6 + 9 +...+ 99
= 3 * (1 + 2 +...+ p)
= 3 * p(1+p)/2
We need to subtract sum(100,15) because sum(100,3) + sum(100,5) double-counts:
sum(100,15) = 15 + 30 + 45 + 60 + 75 + 90
if you want to get the sum of array, you can do like this:
(3..999).inject(0) { |sum, e| e % 3 == 0 || e % 5 == 0 ? sum += e : sum }
=> 233168
it just need once loop.
You can omit the to_a, since calling 'select' to (3..999) will still return an array regardless.
Andrey's answer is the most compact one with :
(3..999).select{ |x| x % 3 == 0 || x % 5 == 0 }.inject(:+)
For some reason I'm getting the error undefined method '%' for 1..100:Range when I run the following code:
[1..100].each do |x|
if x % 3 == 0 && x % 5 == 0
puts "CracklePop"
elsif x % 3 == 0
puts "Crackle"
elsif x % 5 == 0
puts "Pop"
else
puts x
end
end
Any idea what's going on? Any help is much appreciated.
That's the wrong syntax for ranges.
You've made an array with 1 element, and that element is itself the range 1..100. What you've written is equivalent to [(1.100)]. You're iterating over the outer array one time, and setting x to (1..100)
You want (1..100).each, which invokes each on the range, not on an array containing the range.
By doing [1..100] you are not looping from 1 to 100 but on 1..100, which is a Range object, what you really want to do is:-
(1..100).step do |x|
if x % 3 == 0 && x % 5 == 0
puts "CracklePop"
elsif x % 3 == 0
puts "Crackle"
elsif x % 5 == 0
puts "Pop"
else
puts x
end
end
Basically, Range represents an interval, you can iterate over Range as explained here, create an array from Range as explained here and more details on range can be found here.
Just as it says. 1..100 does not have a method %. The expression (1..100) % 3 is undefined.
What's wrong with this Ruby code? I'm trying to solve the first Project Euler question.
I think the problem is in the syntax of sum += num, but I can't figure out what the proper syntax for this would be.
sum = 0
num = 0
num2 = 0
loop do
num += 1
if num % 3 == 0
sum += num
break if num > 1000
end
end
loop do
num2 += 1
if num2 % 5 == 0
sum += num2
break if num2 > 1000
end
end
puts sum
Here's an alternative:
(1...1000).select { |x| x % 3 == 0 || x % 5 == 0 }.reduce(:+)
You are making this way more complicated than it needs to be. Also, if the number is a multiple of 3 and 5, it gets added twice. Try something like this:
sum = 0 # initialize the sum
(1...1000).each { |x| # loop from 1 to 1000
sum += x if x % 3 == 0 || x % 5 == 0 # add the number to the sum if it is
# divisible by 3 or 5
}
puts sum # output the sum
This runs, your syntax is okay, but does not give the right answer because, as mentioned, you add multiples of both 3 and 5 twice, once in the first loop, with num, and the second loop, with num2.
So you have two loops, but you actually only need one.
You only need to consider each number once, you can check it to see if it is a multiple of either 3 or 5. This will solve your double-counting issue and also make your code more concise.
Also, like Doorknob shows, the each syntax would save you some lines on those loops. You could also use the for syntax:
for num in (1..1000)
<stuff here>
end
Check out the kinds of loops in "Loops: How to do thousands of operations with a few lines of code.".