Question is a user gives 3 sides and identifies triangles, like equilateral, isosceles and scalene. Here is my coding, I don't know why gives any sides that always show up "invalid". I think it's logic wrong, but I can't figure out.
puts "please input the length of 3 sides:"
a = gets.chomp.to_i
b = gets.chomp.to_i
c = gets.chomp.to_i
if a + b <= c
puts "invalid"
elsif a <= 0 || b <= 0 || c <= 0
puts "invalid"
else
if a == b && b == c
puts"equilateral triangle"
elsif a == b
puts"isosceles triangle"
else
puts"scalene triangle"
end
end
The fact that your code always prints "invalid" makes me think that input is passed in on one line instead of being on separate lines. For example, when the input is:
50 50 50
instead of getting 50 in all three variables you would get 50 in a and 0 in b, c. This is because gets takes in an entire line instead of taking one value.
In such an event, this is what you need:
a, b, c = gets.split.map{ |value| value.to_i }
A better more effective way to do this is to store the values of the triangle sides into a hash first, the value of of each triangle side will be the keys, and the value of each key can be the repeats. This will work with strings too.
Here is an Example.
# First you get an array, you can use gets.chomp as string and split to array, whichever way you choose, but in the end we end up with an array, and we pass the array to the method.
def triangle_type(arr)
# Create new empty hash
repeated_sides = Hash.new(0)
# First make sure the array is only a length of three. (this is optional)
if arr.length == 3
# Iterate through each value in the array and store it to to a hash to find duplicates
arr.each do |x|
repeated_sides[x] += 1
end
# sort the hash by it's values in descending order, for logic to work later.
repeated_sides = repeated_sides.sort_by {|k,v| v}.reverse.to_h
# uncomment this below to see the duplicate sides hash
#puts "#{repeated_sides}"
# Iterate through the sorted hash, apply logic starting from highest and first value the iterator will find.
repeated_sides.each do |k,v|
return v == 3 ? 'Equilateral Triangle' : v == 2 ? 'Isosceles Triangle' : 'Scalene Triangle'
end
end
# Return Not a triangle if the condition fails
return 'Not a triangle'
end
# Test with integers
puts triangle_type([4,1,2,5]) # output: Not a triangle
puts triangle_type([3,3,3]) # output: Equilateral Triangle
puts triangle_type([4,3,3]) # output: Isosceles Triangle
puts triangle_type([4,2,3]) # output: Scalene Triangle
# Test with strings
puts triangle_type(['4','1','2','5']) # output: Not a triangle
puts triangle_type(['3','3','3']) # output: Equilateral Triangle
puts triangle_type(['4','3','3']) # output: Isosceles Triangle
puts triangle_type(['4','2','3']) # output: Scalene Triangle
puts triangle_type(['a','a','a']) # output: Equilateral Triangle
puts triangle_type(['a','c','c']) # output: Isosceles Triangle
puts triangle_type(['a','b','c']) # output: Scalene Triangle
Skipping user inputs, since I can not reproduce the error (even if Unihedron found a fix) there is still a problem with the logic.
When the input is a = 1000, b = 1, c = 1, the result is "scalene triangle", but it should return "invalid". Below a fix I suggest.
Let's store the input in an array (already converted into integer or float):
sides = [a, b, c]
First you need to check that all sides are positive:
sides.all? { |x| x > 0 }
Then, check that the sum of two sides is greater than the other:
sides.combination(2).map{ |x| x.sum }.zip(sides.reverse).all? { |xy, z| xy > z }
Finally (I'm missing something?), to pick the triangle denomination you can use an hash accessing it by sides.uniq result:
triangle_kinds = {1 => 'equilateral', 2 => 'isosceles', 3 => 'scalene'}
triangle_kinds[sides.uniq.size]
Used the following methods over array (enumerable):
https://ruby-doc.org/core-2.5.1/Enumerable.html#method-i-all-3F
https://ruby-doc.org/core-2.5.1/Array.html#method-i-combination
https://ruby-doc.org/core-2.5.1/Array.html#method-i-map
https://ruby-doc.org/core-2.5.1/Array.html#method-i-zip
https://ruby-doc.org/core-2.5.1/Array.html#method-i-reverse
https://ruby-doc.org/core-2.5.1/Array.html#method-i-sum
https://ruby-doc.org/core-2.5.1/Array.html#method-i-uniq
Related
I am doing some data migration, comparing tables between a legacy and a new database. I have a loop that raises exceptions when two arrays do not have the same size.
array1.zip(array2).each do |ar1, ar2|
# some code here
end
I want to know how to exit the loop when we reach the same size of the two arrays.
The loop breaks when reaches the last element of the first array zipped.
array1 = ['a', 'b', 'c', 'd']
array2 = ['x', 'y', 'z']
array1.zip(array2).each do |ar1, ar2|
puts "#{ar1} -- #{ar2}"
end
puts "-"*10
array2.zip(array1).each do |ar2, ar1|
puts "#{ar1} -- #{ar2}"
end
You could swap the variables if the first array is bigger:
array1, array2 = array2, array1 if array1.size > array2.size
array1.zip(array2).each do |ar1, ar2|
puts "#{ar1} -- #{ar2}"
end
If you just want to check if data are the same and don't want keep track of data origin.
[array1.size, array1.size].min.each do |i|
# code here referencing array1[i] and array2[i]
end
Given arrays:
a = %w[ 1 2 3 10 ]
b = %w[ 1 4 5 1 ]
c = %w[ 5 4 3 ]
If you want to compare two arrays for length:
a.length == b.length
# => true
a.length == c.length
# => false
If you want to compare that the elements in the arrays are of the same length and that the arrays are the same length:
def equal_size_elements(a, b)
return false unless a.length == b.length
a.zip(b).all? do |_a, _b|
_a.length == _b.length
end
end
Where that checks if all of the elements have different lengths because if they all match then it's good, otherwise not good. That method will halt iterating as soon as it finds a mismatch because at that point they can't all pass.
I found a solution, maybe there is better one than mine, I am just a beginner in ruby :
j = 0
array1.zip(array2).each do |ar1, ar2|
j += 1
break if [array1.size,array2.size].min == j
....
end
Just a bit better than your solution:
array1.zip(array2).each_with_index do |zipped, index|
break if index == [array1.size, array2.size].min
puts zipped.first
puts zipped.last
puts
end
I have this code:
require 'set'
N, K = gets.split().map{ |v| v.to_i }
set = Set.new
numbers = gets.split().map{ |v| v.to_i }
pairs = 0
N.times do |i|
set.add(numbers[i])
end
set.each{ |value| pairs += set.include?(value+K) ? 1 : 0 }
puts pairs
But when I put N and K, return this error:
`+': nil can't be coerced into Fixnum (TypeError)
I should convert or other thing? Thanks!
Do you want to count the number of pairs in your set that are separated by K?
Your code works when you input 3 1 followed by 1 2 3. It answers 2.
First, you really should describe a bit more what your goal is.
Then, no need to input N. It just should be the size of your set.
Write an example of the desired input before calling gets
Here's a possible implementation :
require 'set'
puts 'Please type the integers of your set, separated by a space. Example : 1 2 3'
numbers = Set.new(gets.split.map{ |v| v.to_i})
# N=numbers.size # But you don't need it
puts 'Which pair difference are you looking for? Example : 1'
k = gets.to_i
pairs = numbers.select{|value| numbers.include?(value+k)}
count = pairs.size
puts "#{count} pair(s) found :"
pairs.each{|first_value|
puts format("(%d, %d)",first_value,first_value+k)
}
# Please type the integers of your set, separated by a space. Example : 1 2 3
# 1 2 3
# Which pair difference are you looking for? Example : 1
# 1
# 2 pair(s) found :
# (1, 2)
# (2, 3)
Have the function PermutationStep (num) take the num parameter being passed and return the next number greater than num using the same digits. For example: if num is 123 return 132, if it's 12453 return 12534. If a number has no greater permutations, return -1 (ie. 999)
Here's my code. I'd like to sort an array of large integers in numerical order. Using the regular sort method doesn't give the right order for some numbers. Is there a sort_by structure that I can replace 'sort' with in my code below?
def PermutationStep(num)
num = num.to_s.split('').map {|i| i.to_i}
permutations = num.permutation.to_a.sort #<= I want to sort by numerical value here
permutations.each_with_index do |n, idx|
if n == num
if n == permutations[-1]
return -1
else
return permutations[idx+1].join.to_i
end
end
end
end
For example, 11121. When I run the code it gives me 11121.I want the next highest permutation, which should be 12111.
Also, when I try { |a,b| b <=> a }, I also get errors.
You can pass a block to sort.
num.permutation.to_a.sort { |x, y| x.to_i <=> y.to_i }
This SO thread may be of some assistance: How does Array#sort work when a block is passed?
num.permutation.to_a is an array of arrays, not an array of integers, which causes the result not what you expected.
Actually you don't need to sort since you only need the minimum integer that is bigger than the input.
def PermutationStep(num)
nums = num.to_s.split('')
permutations = nums.permutation.map{|a| a.join.to_i}
permutations.keep_if{|n| n > num}.min || -1
end
puts PermutationStep(11121) # 11211
puts PermutationStep(999) # -1
Call to_i before your sort the permutations. Once that is done, sort the array an pick the first element greater than your number:
def PermutationStep(num)
numbers = num.to_s.split('')
permutations = numbers.permutation.map { |p| p.join.to_i }.sort
permutations.detect { |p| p > num } || -1
end
You don't need to consider permutations of digits to obtain the next higher number.
Consider the number 126531.
Going from right to left, we look for the first decrease in the digits. That would be 2 < 6. Clearly we cannot obtain a higher number by permuting only the digits after the 2, but we can obtain a higher number merely by swapping 2 and 6. This will not be the next higher number, however.
We therefore look for the smallest digit to the right of 2 that is greater than 2, which would be 3. Clearly, the next higher number will begin 13 and will have the remaining digits ordered smallest to largest. Therefore, the next higher number will be 131256.
You can easily see that the next higher number for 123 is 132, and for 12453 is 12534.
The proof that procedure is correct is easily established by induction, first showing that it is correct for numbers with two digits, then assuming it is correct for numbers with n>=2 digits, showing it is correct for numbers with n+1 digits.
It can be easily implemented in code:
def next_highest(n)
a = n.to_s.reverse.split('').map(&:to_i)
last = -Float::INFINITY
x,ndx = a.each_with_index.find { |d,i| res = d<last; last=d; res }
return nil unless x
swap_val = a[ndx]
swap_ndx = (0...ndx).select { |i| a[i] > swap_val }.min_by{ |i| a[i] }
a[ndx], a[swap_ndx] = a[swap_ndx], swap_val
a[0...ndx] = a[0...ndx].sort.reverse
a.join.reverse
end
next_highest(126531) #=> "131256"
next_highest(109876543210) #=> "110023456789"
can you have a ruby for loop that has two indexes?
ie:
for i,j in 0..100
do something
end
Can't find anything in google
EDIT: Adding in more details
I need to compare two different arrays like such
Index: Array1: Array2:
0 a a
1 a b
2 a b
3 a b
4 b b
5 c b
6 d b
7 d b
8 e c
9 e d
10 e d
11 e
12 e
But knowing that they both have the same items (abcde)
This is my logic in pseudo, lets assume this whole thing is inside a loop
#tese two if states are for handling end-of-array cases
If Array1[index_a1] == nil
Errors += Array1[index_a1-1]
break
If Array2[index_a1] == nil
Errors += Array2[index_a2-1]
break
#this is for handling mismach
If Array1[index_a1] != Array2[index_a2]
Errors += Array1[index_a1-1] #of course, first entry of array will always be same
if Array1[index_a1] != Array1[index_a1 - 1]
index_a2++ until Array1[index_a1] == Array2[index_a2]
index_a2 -=1 (these two lines are for the loop's sake in next iteration)
index_a1 -=1
if Array2[index_a2] != Array2[index_a2 - 1]
index_a1++ until Array1[index_a1] == Array2[index_a2]
index_a2 -=1 (these two lines are for the loop's sake in next iteration)
index_a1 -=1
In a nutshell, in the example above,
Errors looks like this
a,b,e
As c and d are good.
You could iterate over two arrays using Enumerators instead of numerical indices. This example iterates over a1 and a2 simultaneously, echoing the first word in a2 that starts with the corresponding letter in a1, skipping duplicates in a2:
a1 = ["a", "b", "c", "d"]
a2 = ["apple", "angst", "banana", "clipper", "crazy", "dizzy"]
e2 = a2.each
a1.each do |letter|
puts e2.next
e2.next while e2.peek.start_with?(letter) rescue nil
end
(It assumes all letters in a1 have at least one word in a2 and that both are sorted -- but you get the idea.)
The for loop is not the best way to approach iterating over an array in Ruby. With the clarification of your question, I think you have a few possibly strategies.
You have two arrays, a and b.
If both arrays are the same length:
a.each_index do |index|
if a[index] == b[index]
do something
else
do something else
end
end
This also works if A is shorter than B.
If you don't know which one is shorter, you could write something like:
controlArray = a.length < b.length ? a : b to assign the controlArray, the use controlArray.each_index. Or you could use (0..[a.length, b.length].min).each{|index| ...} to accomplish the same thing.
Looking over your edit to your question, I think I can rephrase it like this: given an array with duplicates, how can I obtain a count of each item in each array and compare the counts? In your case, I think the easiest way to do that would be like this:
a = [:a,:a,:a,:b,:b,:c,:c,:d,:e,:e,:e]
b = [:a,:a,:b,:b,:b,:c,:c,:c,:d,:e,:e,:e]
not_alike = []
a.uniq.each{|value| not_alike << value if a.count(value) != b.count(value)}
not_alike
Running that code gives me [:a,:b,:c].
If it is possible that a does not contain every symbol, then you will need to have an array which just contains the symbols and use that instead of a.uniq, and another and statement in the conditional could deal with nil or 0 counts.
the two arrays are praticatly the same except for a few elements that i have to skip in either/or every once in a while
Instead of skipping during iterating, could you pre-select the non-skippable ones?
a.select{ ... }.zip( b.select{ ... } ).each do |a1,b1|
# a1 is an entry from a's subset
# b1 is the paired entry bfrom b's subset
end
I am trying to create a ruby program where three numbers are entered and their sum is taken but if any numbers are the same they don't count toward the sum.
example (4,5,4) = 5
My problem is with my expressions. If i enter the same number I get multiple output for various combination. example enter 5,5,5 = 15,5,0
if a != b or c then
puts a+b+c
elsif b != a or c then
puts a+b+c
elsif c != a or b then
puts a+b+c
end
if a == b then
puts c
elsif a == c then
puts b
elsif b == c then
puts a
end
if a == b and c then
puts 0
elsif b == a and c then
puts 0
elsif c == a and b then
puts 0
end
Solving it with two beautiful self explanatory one-liners
array = [a,b,c]
array = array.keep_if {|item| array.count(item) == 1 }
array.inject(0){|sum,item| sum + item}
-The first line creates an array with your parameters.
-The second line only keep the items whose count equals to 1 (remove the ones that appear more than one time), and store that on the array.
-The third line sums all the remaining elements.
VoilĂ , the ruby way :)