if i run the code, it will stop and not do anything and i am unable to type. seems to be an infinite loop.
the problem seems to be the end until loop, however if i take that out, my condition will not be met.
can anyone find a solution? i have tried all the loops that i can think of.
/. 2d array board ./
board = Array.new(10) { Array.new(10, 0) }
/. printing board ./
if board.count(5) != 5 && board.count(4) != 4 && board.count(3) != 3
for i in 0..9
for j in 0..9
board[i][j] = 0
end
end
aircraftcoord1 = (rand*10).floor
aircraftcoord2 = (rand 6).floor
aircraftalign = rand
if aircraftalign < 0.5
for i in 0..4
board[aircraftcoord2+i][aircraftcoord1] = 5
end
else
for i in 0..4
board[aircraftcoord1][aircraftcoord2+i] = 5
end
end
cruisercoord1 = (rand*10).floor
cruisercoord2 = (rand 7).floor
cruiseralign = rand
if cruiseralign < 0.5
for i in 0..3
board[cruisercoord2+i][cruisercoord1] = 4
end
else
for i in 0..3
board[cruisercoord1][cruisercoord2+i] = 4
end
end
destroyercoord1 = (rand*10).floor
destroyercoord2 = (rand 8).floor
destroyeralign = rand
if destroyeralign < 0.5
for i in 0..2
board[destroyercoord2+i][destroyercoord1] = 3
end
else
for i in 0..2
board[destroyercoord1][destroyercoord2+i] = 3
end
end
end until board.count(5) == 5 && board.count(4) == 4 && board.count(3) == 3
print " "
for i in 0..9
print i
end
puts
for i in 0..9
print i
for j in 0..9
print board[i][j]
end
puts
end
The line board.count(5) == 5 ... will never be true because board is a two-dimensional array. I can't tell what the condition should be, but it could look something like:
board[5].count(5) == 5
Related
Invoice numbers are numeric only with any number of digits. To format one correctly, group the digits in group of three plus a group of any remainder, but never leave one digit by itself, unless it's a one digit number. Eg these are all correct formatting
123
12-34
6
783-907-23-45
And these are not
123-4
98-456
There's one more catch user input is passed directly to the function and you never know what characters users might type. Ignore any part of the input that is not digit
Invoice.format_number should always return a string
module Invoice
def self.format_number(str)
return ""
end
end
puts Invoice.format_number("ab1234")
What I have tried
1st approach
arr = []
str.chars.each do |elem|
val = elem =~ /\A[-+]?[0-9]*\.?[0-9]+\Z/
arr << elem if val == 0
end
num_of_digits = arr.length
pairs_of_two = 0
pairs_of_three = 0
if num_of_digits > 5
while num_of_digits > 0 do
break if num_of_digits <= 3
if num_of_digits >= 3 && (num_of_digits % 3 == 0 || num_of_digits % 3 == 2)
pairs_of_three += 1
num_of_digits -= 3
elsif num_of_digits % 2 == 0 || num_of_digits % 2 == 1
pairs_of_two += 1
num_of_digits -= 2
end
end
end
2nd approach
arr = []
str.chars.each do |elem|
val = elem =~ /\A[-+]?[0-9]*\.?[0-9]+\Z/
arr << elem if val == 0
end
len = arr.length - 1
if arr.length > 4
str = ""
i = 0
while i < len do
if arr[i..i+3].length == 4
str << arr[i..i+2].join + "-"
i += 3
elsif arr[i..i+2].length == 3
str << arr[i..i+1].join + "-"
i += 2
elsif arr[i..i+1].length == 2
str << arr[i..i+1].join
i += 2
elsif !arr[i].nil?
str << arr[i]
i += 1
end
end
puts str
else
if arr.length <= 3
puts arr.join
else
puts arr[0..1].join + "-" + arr[2..3].join
end
end
But none of them is correct
Here is the function invoice_number in python
def invoice_number(invoice):
s = ''.join(x for x in invoice if x <= '9' and x >= '0')
n = len(s)
if n <= 3:
return s
w = ''
i = 0
while i + 3 <= n:
for j in range(0, 3):
w += s[i + j]
i += 3
w += ('-')
m = n - i
if m == 0: return w[:-1]
if m == 1: return w[:m-3] + '-' + s[-2:]
return w + s[i:]
Testing
print(invoice_number('1234567'))
print(invoice_number('12345678'))
print(invoice_number('abc123456789'))
print(invoice_number('1234abc5678xyz9foobar'))
123-45-67
123-456-78
123-456-789
123-456-789
Eliminating non-digits is easy with re. For your format, the key is to figure our the "right" splitting indices.
Here is a try:
import re
def splits(n, k):
idx = [(i, min(n, i+k)) for i in range(0, n, k)]
if len(idx) > 1:
(a, b), (c, d) = idx[-2:]
if d - c < 2:
idx[-2:] = [(a, b - 1), (c - 1, d)]
return idx
def myformat(s):
s = re.sub(r'[^0-9]+', '', s)
parts = [s[a:b] for a, b in splits(len(s), 3)]
return '-'.join(parts)
Tests:
>>> myformat('123')
123
>>> myformat('1234')
12-34
>>> myformat('6')
6
>>> myformat('7839072345')
783-907-23-45
As the question was asked for ruby, adding solution for ruby. (The inspiration of the code is mostly from #yuri answer)
def format_invoice(invoice)
# only numbers are allowed
invoice = invoice.tr("^0-9","")
#puts invoice
return invoice if(invoice.length <= 3)
formatted_invoice = ''
i = 0
# Loop to divide the invoice in group of 3
while i + 3 <= invoice.length do
for j in 0..2 do
formatted_invoice += invoice[i + j]
end
i += 3
formatted_invoice += ('-')
end
m = invoice.length - i
return formatted_invoice[0..-2] if m == 0
return formatted_invoice[0..m-4] + '-' + invoice[-2..-1] if m == 1
return formatted_invoice + invoice[i..-1]
end
Testing
puts format_invoice('abc1') # 1
puts format_invoice('abc123') # 123
puts format_invoice('abc123A4') # 12-34
puts format_invoice('1234567') # 123-45-67
puts format_invoice('12345678') # 123-456-78
puts format_invoice('abc123456789') # 123-456-789
puts format_invoice('1234a#c5678xyz9foobar') # 123-456-789
I need output of Floyd triangle like:
1
0 1
1 0 1
0 1 0 1
I tried. I didn't get it exactly. Can anyone explain the logic?
This is the code I tried:
k = 0
for i in 1..5
for j in 1..5
if (i%2)==0;
k = (j%2==0) ? 1:0;
else;
k = (j%2==0) ? 0:1;
puts k,'';
end
end
puts
end
The main issue here is that in order to get the "triangle" shape of your output, you need your inner loop to increment from 1 to i instead of 1 to 5.
k = 0
for i in 1..5
for j in 1..i
if (i%2)==0
k = j + 1
else
k = j
end
print "#{k%2} "
end
puts
end
Here's a one line approach:
5.times {|line| puts (line + 1).times.with_object(""){|num, str| (num + line).even? ? (str << " 1 ") : (str << " 0 ") } }
to make it more clear:
lines = 5
lines.times do |line|
str = ""
line = line + 1 # 5.times runs from 0 to 4 and we need 1 to 5
line.times do |num|
# the condition is a bit different because I changes the code a bit
if (line + num).even?
str << " 0 "
else
str << " 1 "
end
end
puts str
end
Alright the following should work, but i hope it's readable. If you need more explanation or have specific questions let me know
i = 1
while i <= 4 do
if i%2 > 0
output = 1
else
output = 0
end
j = 1
while j <= i do
print( "#{output} " )
if output == 1
output = 0
else
output = 1
end
j+=1
end
print( "\n" )
i+=1
end
You can try following code for output you are expecting:
k = 0
for i in 1..4
for j in 1..i // inner loop code runs i times for each outer loop iteration
if (i%2)==0;
k = (j%2==0) ? 1:0;
else;
k = (j%2==0) ? 0:1;
end
print k,' ';
end
puts
end
Click Here to see output.
You can also get idea about for loops through this link.
The prefered ruby way:
layers = 4 # Change to as many layers as you want
layers.times do |i| # i starts from 0
(i + 1).times do |j| # j also starts from 0
print (i + j + 1) & 1, ' '
end
puts
end
The for way:
layers = 4
for i in 0...layers
for j in 0...(i + 1)
print (i + j + 1) & 1, ' '
end
puts
end
I have a very basic question about Ruby loops.
This program as written returns the ith prime number +1 (ie the example should return 17). I know I could simply return cand-1, but I was wondering what the "Ruby way" of checking if the answer has been found at the bottom of the while loop and only incrementing if it hasn't.
def ith_prime(i)
pI = 0 # primes index
divs = []
cand = 2
until pI == i do
if divs.find { |div| cand%div == 0 } == nil
divs << cand
pI += 1
end
cand += 1
end
cand
end
puts ith_prime(7)
> 18
I use loop instead of while or until most of the time. This way I can put the exit condition anywhere in the loop.
I would write it like that (if I understood the problem correctly):
def ith_prime(i)
pI = 0 # primes index
divs = []
cand = 2
loop do
unless divs.find { |div| cand%div == 0 }
divs << cand
pI += 1
end
break if pI == i
cand += 1
end
cand
end
I'm trying out to have a block in a while and begin statements in Ruby, but I get a syntax error. Any other way to implement it?
Here's what I want to accomplish
(1..limit).each { |i|
while (true) do |n|
x = n * (i%n)
puts n if n%i != 0
break if x.even? && !x.zero?
n += 1
end
}
while is a keyword, so you do not need the block. Your code should be:
(1..limit).each { |i|
while (true)
x = n * (i%n)
puts n if n%i != 0
break if x.even? && !x.zero?
n += 1
end
}
But you are requesting a block variable form the while statement. Variable names inside the pipes are for variables passed to your block containing information from whatever calls your block. I will assume that n is supposed to increment. Here is a working version:
(1..limit).each { |i|
n = 0
while (true)
x = n * (i%n)
puts n if n%i != 0
break if x.even? && !x.zero?
n += 1
end
}
If you really need the code in a block, you could create one and then call it, like this (ruby 1.9 only):
(1..limit).each { |i|
n = 0
while (true)
-> do
x = n * (i%n)
puts n if n%i != 0
break if x.even? && !x.zero?
n += 1
end.()
end
}
By the way, here is a cleaner version:
(1..limit).each do |i|
n = 0
loop do
x = n * (i % n)
puts n if n % i != 0
break if x.even? and !x.zero?
n += 1
end
end
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