Shell script to print "*" pattern

Shell Programming and Scripting
1

Plz tel me how to print the following pattern using shell script??

   *
  ***
 *****
*******
 *****
  ***
   *

and

    *
*********
    *
2

Use following :

echo \*
 
i.e :
echo \*
echo \***
echo \****
echo \******
echo \****
echo \***
echo \*

:b:

3

one (bash) way:

#  for x in 1 3 5 7 5 3 1; do printf "%.$((3-$x/2))d" 0|tr "0" " ";printf "%.${x}d\n" 0|tr "0" "*"; done
   *
  ***
 *****
*******
 *****
  ***
   *
4

Diamond long version :rolleyes:

perl -e 'printf("%s%s\n%s%s\n%s%s\n%s\n%s%s\n%s%s\n%s%s\n"," "x3,"*"x1," "x2,"*"x3," ","*"x5,"*"x7," ","*"x5," "x2,"*"x3," "x3,"*"x1)'

Diamond short version :cool:

perl -e 'for (3,2,1,0,1,2,3) {print " "x$_."*"x(7-2*$_)."\n";}'

Division sign:

perl -e 'for (4,0,4) {print " "x$_."*"x(9-2*$_)."\n";}'
5 
cat<<EOF
   *
  ***
 *****
*******
 *****
  ***
   *
EOF