I am using Ruby 1.9.3. I have done pattern program like follow:
n = 1
while n <= 5
n.downto 1 do |i|
print "* "
end
puts
n += 1
end
Output of above program is like follow:
*
* *
* * *
* * * *
* * * * *
Now I am trying to do pattern program like follow:
*
* *
* * *
* * * *
* * * * *
I am not getting idea how can I do it?
Could anyone help me on this?
Thank you.
You can use rjust:
n = 1
while n <= 5
puts "* " * n
n += 1
end
*
* *
* * *
* * * *
* * * * *
n = 1
while n <= 5
puts ("* " * n).rjust(10)
n += 1
end
*
* *
* * *
* * * *
* * * * *
A shortened version of this will be:
5.times { |i| puts ('* ' * (i+1)) }
and
5.times { |i| puts ('* ' * (i+1)).rjust(10) }
You can do:
1.upto 5 do |n|
print ' ' * (5-n)
print '* ' * n
puts
end
Here's another way:
def print_two_ways(n, spaces=0)
arr = Array.new(n) { |i| Array.new(n) { |j| (i >= j) ? '*' : ' ' } }
print_matrix(arr, spaces)
puts
print_matrix(arr.map(&:reverse), spaces)
end
def print_matrix(arr, spaces = 0)
sep = ' '*(spaces)
arr.each { |r| puts "#{r.join(sep)}" }
end
print_two_ways(5)
*
**
***
****
*****
*
**
***
****
*****
print_two_ways(5,1)
*
* *
* * *
* * * *
* * * * *
*
* *
* * *
* * * *
* * * * *
a = 5
b = 1
while a>0
while b<=5
puts "*"*b
b = b+1
a = a-1
end
end
*
**
***
****
*****
Here, a is 5 and b is 1.
In the first iteration of the outer while loop, a is 1 and the inner while loop is inside the body of the outer while loop. So, the inner while loop will be executed and "*"1 ( b is 1 ) i.e "" will be printed and b will become 2 and a will become 4.
Here is simple set variable and use all programs.
# Set vars
n = 4 # Set number of rows
br = "\n" * 2
# Simple loop
puts "Right triangle:#{br}"
for i in 1..n do
puts "* " * i
end
puts br
=begin
simple loop result:
*
* *
* * *
* * * *
=end
# Countdown loop
puts "Inverted right triangle:#{br}"
n.downto(0) do
puts "* " * n
n -= 1
end
puts br
=begin
countdown loop result:
* * * *
* * *
* *
*
=end
# Function loop
puts "Inverted pyramid:#{br}"
n = 4 # Reset number of rows
for i in 1..n do
# Use a func to reduce repetition
def printer(var, str)
print "#{str}" * (2 * var - 1)
end
printer(i, " ")
printer(n, "* ")
print "\n"
n -= 1
end
puts br
=begin
function loop result:
* * * * * * *
* * * * *
* * *
*
=end
# Count up loop
puts "Close pyramid:#{br}"
n = 4 # Set number of rows
i = 1
1.upto(n) do
#n.times do
# print ' '
#end
print ' ' * n
#(2 * i - 1).times do
# print '*'
#end
print '*' * (2 * i -1)
print "\n"
n -= 1
i += 1
end
print br
=begin
count up loop result:
*
***
*****
*******
=end
I'm still new to Ruby, but this was my solution. Anything wrong with doing it this way?
def staircase(n)
sum_tot = n
n.times do
sum_tot-= 1
space = n - sum_tot
puts ('#' * space).rjust(n)
end
end
Related
How to print rectangle like below using Ruby:
* = = = *
* = = = *
* * * * *
* = = = *
* = = = *
in this case, length of row and column are same and must odd.
Example:
r = rectangle(5)
should print :
* = = = *
* = = = *
* * * * *
* = = = *
* = = = *
and if :
r = rectangle(7)
should print:
* = = = = = *
* = = = = = *
* * * * * * *
* = = = = = *
* = = = = = *
* * * * * * *
* = = = = = *
Thanks in advance.
I'd start by creating the two lines: (there are plenty of ways to achieve this)
size = 7
a = Array.new(size, '*').fill('=', 1..-2).join(' ') #=> "* = = = = = *"
b = Array.new(size, '*').join(' ') #=> "* * * * * * *"
Then I'd define a repeating pattern:
pattern = [a, a, b].cycle
Finally, I'd print the pattern size times:
puts pattern.take(size)
* = = = = = *
* = = = = = *
* * * * * * *
* = = = = = *
* = = = = = *
* * * * * * *
* = = = = = *
[SOLVE BY ME] :D
def square(n)
begin
if n.odd?
1.upto(n) do | row |
if row % 3 != 0
puts "#{'*'} #{'= ' * (n - 2)}#{'*'}"
else
puts "#{'* ' * n}"
end
end
else
puts 'Must odd number!'
end
rescue
puts 'Must integer number!'
end
end
square(5)
Output:
* = = = *
* = = = *
* * * * *
* = = = *
* = = = *
square(7)
Output:
* = = = = = *
* = = = = = *
* * * * * * *
* = = = = = *
* = = = = = *
* * * * * * *
* = = = = = *
square(8)
Output:
Must odd number!
square(8.5)
Output:
Must integer number!
square('blabla')
Output:
Must integer number!
def rectangle(n)
puts("-----------------For #{n}--------------------")
if n % 2 == 1
for i in (1..n)
for j in (1..n)
if j == 1 || j == n || 0 == i % 3
print "* "
else
print "= "
end
end
print("\n")
end
end
end
rectangle(3)
rectangle(5)
rectangle(7)
rectangle(9)
rectangle(11)
-----------------For 3--------------------
* = *
* = *
* * *
-----------------For 5--------------------
* = = = *
* = = = *
* * * * *
* = = = *
* = = = *
-----------------For 7--------------------
* = = = = = *
* = = = = = *
* * * * * * *
* = = = = = *
* = = = = = *
* * * * * * *
* = = = = = *
-----------------For 9--------------------
* = = = = = = = *
* = = = = = = = *
* * * * * * * * *
* = = = = = = = *
* = = = = = = = *
* * * * * * * * *
* = = = = = = = *
* = = = = = = = *
* * * * * * * * *
-----------------For 11--------------------
* = = = = = = = = = *
* = = = = = = = = = *
* * * * * * * * * * *
* = = = = = = = = = *
* = = = = = = = = = *
* * * * * * * * * * *
* = = = = = = = = = *
* = = = = = = = = = *
* * * * * * * * * * *
* = = = = = = = = = *
* = = = = = = = = = *
Just for fun modify #spn answer
def rectangle(count)
return 'Must odd number more than 1' unless count.is_a?(Integer) && count.odd? && count > 1
Array.new(count) { |index| (index + 1) % 3 == 0 ?
"#{'* ' * count}".chomp(" ") :
"#{'*'} #{'= ' * (count - 2)}#{'*'}" }.join("\n")
end
And now
puts rectangle(2) # will print Must odd number more than 1
puts rectangle("asdf") # will print Must odd number more than 1
puts rectangle(9) # will print:
* = = = = = = = *
* = = = = = = = *
* * * * * * * * *
* = = = = = = = *
* = = = = = = = *
* * * * * * * * *
* = = = = = = = *
* = = = = = = = *
* * * * * * * * *
It's better to avoid puts in the methods. So you can use them again in web, telegrambots, etc. Also it's not good idea to duplicate Ruby exceptions by your own (but the same) messages.
Let's first define a method to construct the two types of lines.
def make_line(n, mid_char)
['*', *[mid_char]*(n-2), '*'].join(' ')
end
make_line(5, '*') #=> "* * * * *"
make_line(5, '#') #=> "* # # # *"
Now create a method to draw the lines in the desired pattern. The ith line (base zero) is comprised of stars and spaces only if (i+1) % 3 equals zero; else it is the line containing pound signs as well.
def draw(n)
all_stars = make_line(n, '*')
two_stars = make_line(n, '#')
n.times { |i| puts ((i+1) % 3).zero? ? all_stars : two_stars }
end
draw 5
* # # # *
* # # # *
* * * * *
* # # # *
* # # # *
draw 6
* # # # # *
* # # # # *
* * * * * *
* # # # # *
* # # # # *
* * * * * *
draw 7
* # # # # # *
* # # # # # *
* * * * * * *
* # # # # # *
* # # # # # *
* * * * * * *
* # # # # # *
I want to print triangle & pyramid "*" using for loop.
can somebody help me on this?
Output like :
*****
****
***
**
*
and
*
**
***
****
*****
1: Print triangle using While loop
n = 5
while n >= 1
puts "* " * n
n = n - 1
end
* * * * *
* * * *
* * *
* *
*
n = 1
while n <= 5
puts ("* " * n).rjust(10)
n += 1
end
*
* *
* * *
* * * *
* * * * *
2: Print pyramid using loop
n = 4 # Set number of rows
i = 1
1.upto(n) do
print ' ' * n
print '*' * (2 * i - 1)
print "\n"
n -= 1
i += 1
end
*
***
*****
*******
5.downto(1).each{|n| puts ("*" * n).ljust(5)}
1.upto(5).each{|n| puts ("*" * n).rjust(5)}
I have no coding experience other than this book
Programming Swift! Swift 2 Kindle Edition
by Nick Smith (Author)
I am currently at Chapter
5.3 Nested FOR LOOPS
// NESTED FOR LOOP #2
This code -
for var a = 0; a < 11; a++ {
print("")
for var b = 0; b < a; b++ {
print("*", terminator: " ")
}
}
GENERATES THIS PATTERN...
Now [after several/ 4 hours 'odd'] I SIMPLY CAN'T WORK OUT HOW TO CHANGE THE ABOVE 'simple' [if you know how] CODE TO GENERATE THIS PATTERN??
I (think I) can see Outer and Inner loops I just can't work out the rest!?? I have tried every variation I can think of!?? (and am aware that just doing 'permutations' doesn't mean I have true understanding of what I am trying to do!...)
Tried using --operators and changing [most/ all] values [but 'permutations' is a limited method]
I feel like a total fool but figure if it's the very first time I've seen this stuff maybe it's not so bad, these things take learning!??
Help (the answer LOL) would be GREATLY appreciated 😬 😬 😬
for var a = 10; a > 0; a-- {
for var b = 0; b < a; b++ {
print("*", terminator: " ")
}
print()
}
prints
* * * * * * * * * * *
* * * * * * * * * *
* * * * * * * * *
* * * * * * * *
* * * * * * *
* * * * * *
* * * * *
* * * *
* * *
* *
*
UPDATE for nowadays Swift syntax, with the same functionality
for a in stride(from: 10, through: 0, by: -1) {
for _ in stride(from: 0, to: a, by: 1) {
print("*", terminator: " ")
}
print()
}
How to do this systematically: If you want to got for example user3441734's output: There are 11 lines. We number the lines from 0 to 10. So we have a loop that sets line to the values 0 to 10.
for var line = 0; line < 11; ++line
Next, what do we want to print in each line? In line 0 we want to print 11 * characters. In line 10 we want to print 1 star character. The number of stars is 11 - line. How do I get the expression 11 - line? The number of stars goes down as line goes up, so it must be something - line. When line = 0 there must be 11 stars, so something - 0 = 11, and something = 11. So the first line in the loop:
let starcount = 11 - line
Then we want to print (star count) times a star and a space character, follow by starting a new line.
for var star = 0; star < starcount; ++star {
print ("*", terminator: " ")
}
print ("")
All together:
for var line = 0; line < 11; ++line {
let starcount = 11 - line
for var star = 0; start < star count; ++star {
print ("*", terminator: " ")
}
print ("")
}
And we simplify the loops a bit:
for var line in 0 ..< 11 {
let starcount = 11 - line
for var star in 0 ..< starcount {
print ("*", terminator: " ")
}
print ("")
}
If you wanted a different pattern, all you have to do is change the number 11 if the number of lines is different, and change the calculation of starcount. Actually it would be better to have a variable for linecount as well, so changing for a different pattern is even easier:
let linecount = 11
for var line in 0 ..< line count {
let starcount = linecount - line
for var star in 0 ..< starcount {
print ("*", terminator: " ")
}
print ("")
}
I have written an application which will write square with diagonal (from left side) - output:
+ * * * *
* + * * *
* * + * *
* * * + *
* * * * +
Code for first application:
PROGRAM cycle4;
USES CRT;
VAR a,r,s:INTEGER;
BEGIN
CLRSCR;
WRITE (‘Enter the number of lines :‘) ;
READLN(a);
FOR r:= 1 TO a DO
BEGIN
FOR s:=1 TO a DO
IF r = s THEN WRITE(‘+‘)
ELSE WRITE(‘*‘) ;
WRITELN;
END;
READLN;
END.
And now I have to create an application which will write square with diagonal (from right side) - output:
* * * * +
* * * + *
* * + * *
* + * * *
+ * * * *
But I don't know how can I write it. Can you help me?
Thanks :)
The line of code which defines the position of + sign is that:
IF r = s THEN WRITE(‘+‘)
and this is the only line you need to change:
IF r + s = a + 1 THEN WRITE(‘+‘)
I think this should work, check with Pascal compiler, haven't used it for about 10 years :)
Make programm, that asks from user his name and number of lines for the incoming shape.
For example.
Peeter and 6, the example wil be :
Peeter shape:
* * * * * *
* *
* *
* *
* *
* * * * * *
So my code is:
def square():
row = int(input('Insert number of rows'))
line = len(input('Insert your name'))
for r in range(1, (row+1), 1):
if r == 1 or r == rida:
for v in range(1, (line+1), 1):
print('*', end=' ')
else:
print('*', end=' ')
for v in range(2, row, 1):
print(' ', end=' ')
print('*', end=' ')
print()
square()
But the problem is, it gives me syntaxerror and I am out of ideas how to fix it.
Thank you for your help and time for advance.
Nika