[SOLVED] generating series of numbers of sequence

vjramana · 03-19-2011, 01:58 AM

I am trying to generate sequence of number in the order of :

1
2
3

4
5
6

7
8
9

10
11
12

13
14
15

Where by after three lines of numbers there must be a blank line.

I tried to write this using bash script as below:

Code:

for i in {1..5}; do #lipid
#  echo $i
	
		for j in {1..3}; do #atoms in lipid
		
			echo $[$j*$i ]
			
		done
		echo ""  
	

done

but I get a sequence as below
1
2
3

2
4
6

3
6
9

4
8
12

5
10
15

Can anyone help me?
Thanks in advance

acid_kewpie · 03-19-2011, 02:27 AM

well that's just a basic math problem, no scripting issue in the slightest, just logic. This sounds a lot like homework so I'm not going to give you the answer, but to get your approach to work you just need to think more about the maths you're doing. Using $i, $j and the maximum size of j each time you can do this easily. I would also say that the range for i is not ideal, and things might be clearer if you start it at a different number.

Disillusionist · 03-19-2011, 02:35 AM

Go back to basics, and look at what you are doing:

Code:

for i in {1..5}; do #lipid
#  echo $i
	
		for j in {1..3}; do #atoms in lipid
		
			echo $[$j*$i ]
                      # Do you really want to multiply i by j?
			
		done
		echo ""  
	

done

What you want to do is:

create a sequence of numbers 1..15
check if iteration is an exact multiple of 3 and display a new line if it is

HTH

colucix · 03-19-2011, 02:38 AM

Just an aside note: the usage of the $[...] operator is deprecated and it has been replaced by the arithmetic operator ((...)).

kurumi · 03-19-2011, 03:24 AM

Here's a Ruby(1.91+) one-liner for you

Code:

$ ruby -e '(1..15).to_a.each_slice(3){|x| puts x;puts}'

vjramana · 03-19-2011, 10:03 AM

Dear, acid_kewpie

Thanks for the hint. Anyway this in not my homework.
Well after some try I got it as I want. For sharing purpose I would like post mine here.

Quote:

#!/bin/bash

for k in {0..12..3}; do
for n in {1..3}; do
echo $(($k+$n))
done
echo " "
done

You can give suggestion if there is any. I can improve with other way of doing things.

For Mr Kurumi, Thanks for your suggestion as well. It works. But I am more on bash.

theNbomr · 03-19-2011, 10:03 AM

In these cases, I tend to make one sequence, and on each iteration, test whether that special case is true. In this case, the 'special case' is 'evenly divisible by three', which suggests the use of a modulo operator. Does bash have one of those? If not, it shouldn't be hard to craft one as a function.
--- rod.

grail · 03-19-2011, 10:34 AM

Quote:

Does bash have one of those?

Yep

Code:

#!/bin/bash

for (( i=1;i<=15;i++))
do
    if ((  i % 3 ))
    then
        echo $i
    else
        echo -e "$i\n"
    fi
done