Question
Eduard Lebedyuk · Apr 5, 2018

Drawing boxes

Hello, Community!

Here's an interesting task I found on the Internet.

Problem description

Write a method that would draw a box of a specified size.

The goal is to write the shortest method.

Here's a method signature (it can't be modified):

ClassMethod main(s As %Integer = 10)

And call sample:

>do ##class(ITPlanet.Task4).main(5)
#####
## ##
# # #
## ##
#####

>do ##class(ITPlanet.Task4).main(10)
##########
##      ##
# #    # #
#  #  #  #
#   ##   #
#   ##   #
#  #  #  #
# #    # #
##      ##
##########
 
>do ##class(ITPlanet.Task4).main(20)
####################
##                ##
# #              # #
#  #            #  #
#   #          #   #
#    #        #    #
#     #      #     #
#      #    #      #
#       #  #       #
#        ##        #
#        ##        #
#       #  #       #
#      #    #      #
#     #      #     #
#    #        #    #
#   #          #   #
#  #            #  #
# #              # #
##                ##
####################

Also, here's the code to  check your result length:

ClassMethod length(class = {$classname()}, method = "main") As %Integer
{
    #dim methodObj As %Dictionary.MethodDefinition
    set methodObj = ##class(%Dictionary.MethodDefinition).IDKEYOpen(class, method)
    quit methodObj.Implementation.Size
}

My best result is 91 76 so far.

What's yours? ;)

2
0 495
Discussion (15)4
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Sorry to be late. I was busy today:

Thanks to  @Eduard Lebedyuk

I did it in a traditional one-liner: 79  77 char.   + 4 char extra to read the size.

I'll wrap it into a method later to see how much waste of space this generates. devil

5 min. later:
OK. Method consumes 3 char. extra {} to enclose it + blank at the start   =>>> 82   80  ~3.9% overhead

s=1...4 looks odd but it improves.

f j=1:1:15 zw j d ##class(DC.size).main(j) 

j=1
#
j=2
##
##
j=3
###
###
###
j=4
####
####
####
####
j=5
#####
## ##
# # #
## ##
#####
j=6
######
##  ##
# ## #
# ## #
##  ##
######
j=7
#######
##   ##
# # # #
#  #  #
# # # #
##   ##
#######
j=8
########
##    ##
# #  # #
#  ##  #
#  ##  #
# #  # #
##    ##
########

That's mine:

ClassMethod main(As %Integer = 10)
r=1:1:! c=1:1:$s(c=1!(r=1)!(r=s)!(c=s)!(c=r)!(s-r+1=c):"#",1:" ")}
i=1:1:! j=1:1:$c(i=1!(i=s)!(j=1)!(j=s)!(j=i)!(s-i+1=j)*3+32)

My cheeky solution was 

r=1:1:! c=1:1:w:r#s<2!(c#s<2)!(c=r)!(s-r+1=c) "#" " "

I'm not sure if it will copy/paste properly, but there's a backspace character in the string before the #  laugh I considered it cheeky because despite looking correct in the terminal it's definitely not if you compared it byte for byte wink

Nice! And you can shave off 3 characters by using $EXTRACT instead of $SELECT

f i=1:1:s w ! f j=1:1:s w $e(" #",j#s<2!(i#s<2)!(j=i)!(s-i+1=j)+1)

I like that    

$lf($lb(1,s,y,s-y+1),x)

My best result is 77 72 69 67 so far.

Class ITPlanet.Task4 Abstract ]
{

ClassMethod main(As %Integer 10)
{
 x=1:1:y=1:1:w $c(y#s<2!$lf($lb(1,s,y,s-y+1),x)*3+32)
}

ClassMethod length(
  class = {$classname()},
  method "main"As %Integer CodeMode = expression ]
{
##class(%Dictionary.MethodDefinition).IDKEYOpen(classmethod).Implementation.Size
}

}

USER>##class(ITPlanet.Task4).length()
67
USER>##class(ITPlanet.Task4).main(11)

###########
##       ##
# #     # #
#  #   #  #
#   # #   #
#    #    #
#   # #   #
#  #   #  #
# #     # #
##       ##
###########

FOO>d ##class(FOO.Boxes).main(20)
 
####################
##                ##
# #              # #
#  #            #  #
#   #          #   #
#    #        #    #
#     #      #     #
#      #    #      #
#       #  #       #
#        ##        #
#        ##        #
#       #  #       #
#      #    #      #
#     #      #     #
#    #        #    #
#   #          #   #
#  #            #  #
# #              # #
##                ##
####################


FOO>w ##class(FOO.Boxes).length()
75

Still thinking, seems like it might go a bit smaller...

70 or 73, depending on how cheeky you'll let me be wink

edit: 69 for my cheeky solution, the other one was the same as Zenkov's below but I wasted a character on a for bracket because I foolishly put my newline on the other side of the $s()

Let's finish this competition this Wednesday (so it would be a week) and publish our best efforts then.

ClassMethod main(As %Integer = 10)
 i=1:1:! j=1:1:$S(j#s<2!(i#s<2)!(j=i)!(s-i+1=j):"#",1:" "}

My solution and how I got there.

I started with this:

 r=1:1:s{c=1:1:s{r=1!(r=s)!(c=1)!(c=s)!(r=c)!(s=(c+r-1)){"#"}else{" "c=s{!}}}

First improvement was thanks to @Robert.Cemper who suggested moving i c=s{w !}}} into a first for:

 r=1:1:! c=1:1:r=1!(r=s)!(c=1)!(c=s)!(r=c)!(s=(c+r-1)){"#"}else{" "}

Finally got the idea of using $lb/$lf to get my best result of 76:

f r=1:1:s w ! f c=1:1:s w $s($lf($lb(c,r,r=s,c=s,c=r,r+c-s),1):"#",1:" ")

Some other ideas that didn't pan out.

First of all I thought about replacing $lf($lb)) with $f() but -1 and 1x numbers became a problem:

f r=1:1:s w ! f c=1:1:s w $s($f(c_r_(r=s)_(c=s)_(c=r)_$replace(r+c-s,-1,""),1):"#",1:" ")

Other idea was using $translate:

f r=1:1:s w ! f c=1:1:s w $tr(''$lf($lb(c,r,r=s,c=s,c=r,r+c-s),1),10,"# ")

Interestingly if we allow the box to be made of any symbols, some other solutions became possible. For example binary box (63 symbols):

f r=1:1:s w ! f c=1:1:s w '$lf($lb(c,r,r=s,c=s,c=r,r+c-s),1)