Draw a Signal strength indicator

Question

Given the signal level L in the range 0-4. Your program should draw the following ASCII art.

When L=0:

               ....
               ....
          .... ....
          .... ....
     .... .... ....
     .... .... ....
.... .... .... ....
.... .... .... ....
.... .... .... ....

When L=1:

               ....
               ....
          .... ....
          .... ....
     .... .... ....
     .... .... ....
oooo .... .... ....
oooo .... .... ....
oooo .... .... ....

When L=2:

               ....
               ....
          .... ....
          .... ....
     oooo .... ....
     oooo .... ....
oooo oooo .... ....
oooo oooo .... ....
oooo oooo .... ....

And etc. You can change the characters o and . to others, it's up to you. Trailing spaces are allowed.

This is a code-golf challenge so the shortest solution wins.

That would include allowing trailing spaces on each line, correct? — SuperJedi224, Feb 24, 2022 at 1:50

Jonah · Accepted Answer · 2022-02-24 03:39:22Z

4

J, ⁴⁷ 46 bytes

' .o'{~[:|.@|:[:,/(4,&0@#"+3+2*i.4)#"+1+4{.$&1

Try it online!

edited Feb 24, 2022 at 3:39

answered Feb 24, 2022 at 1:40

Jonah

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Graham · Accepted Answer · 2022-02-24 08:06:37Z

2

APL+WIN, 48 bytes

Prompts for n

⊖⍉⊃((n↑m)⍴¨⊂⊂4⍴'0'),((-4-n←⎕)↑m←1+2×⍳4)⍴¨⊂⊂4⍴','

Try it online! Thanks to Dyalog Classic

answered Feb 24, 2022 at 8:06

Graham

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pxeger · Accepted Answer · 2022-02-24 16:01:39Z

2

Python, 75 bytes

def f(n,j=4):print(*(" .o"[(i>=j)*~(n<i)]*4for i in b""));1/j;f(n,j-.5)

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Finishes with an exception. The b"" string contains the literal bytes 0x01 0x02 0x03 0x04, but they're invisible here.

Could probably be improved with some more tweaking.

76 bytes

def f(n,j=4):print(*(" .o"[(i>=j)*~(n<i)]*4for i in b""));j>0!=f(n,j-.5)

Attempt This Online!

No error.

edited Feb 24, 2022 at 16:01

answered Feb 24, 2022 at 9:17

pxeger

23.3k4 gold badges52 silver badges135 bronze badges

\$\begingroup\$ Ok how do you know when to use lambda or when to use regular function? Also what is that b"..." string doing \$\endgroup\$
– Aiden Chow
Feb 24, 2022 at 10:07
\$\begingroup\$ In this case the function doesn't return anything, it prints to stdout, so it's shorter to use def. Here the b"" string acts as equivalent to [1,2,3,4], but it's shorter: see this tip. \$\endgroup\$
– pxeger
Feb 24, 2022 at 10:10
\$\begingroup\$ @Neil I'm not sure whether you're joking, but in either case "substandard" doesn't come across very nice. Also, ATO and TIO both generate markdown which renders as those 4 literal bytes, but they're invisible in both cases (I was wrong; it's not SE markdown's fault), so I changed it to the clearer \1\2\3\4. You're welcome to submit constructive feedback on the GitHub repository: github.com/attempt-this-online/attempt-this-online \$\endgroup\$
– pxeger
Feb 24, 2022 at 15:47
\$\begingroup\$ Thanks for correcting your post. \$\endgroup\$
– Neil
Feb 24, 2022 at 19:36
\$\begingroup\$ I checked my poor shoddy inferior imperfect inadequate second-rate unacceptable thesaurus but it didn't come up with a better synonym. Any suggestions? \$\endgroup\$
– Neil
Feb 25, 2022 at 19:42

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emanresu A · Accepted Answer · 2022-02-24 18:44:38Z

2

Vyxal, 17 bytes

4ɾ≥vS4ɾd›*¶+4*⁋↵§

Try it Online!

-3 thanks to Seggan.

4ɾ≥               # Create a map of ones and zeroes corresponding to the bar characters.
   vS             # Stringify those
     4ɾd›         # Create the range [3, 5, 7, 9]
         *        # And repeat each char of those
          ¶+      # Append a newline to each
            4*    # Repeat each four times
              ⁋↵  # Turn into a list of lines
                § # Rotate 90°

edited Feb 24, 2022 at 18:44

answered Feb 24, 2022 at 2:07

emanresu A

35.6k5 gold badges83 silver badges200 bronze badges

\$\begingroup\$ Vyxal doesn't have something to repeat an integer as string? The challenge allows other characters, so no need for the ‛.o$İ if you can interpret the 0/1 as strings so the * repeats it instead of multiply. I see there is also ẋ to repeat, but that doesn't seem to work either. \$\endgroup\$
– Kevin Cruijssen
Feb 24, 2022 at 10:59
1

\$\begingroup\$ 17 bytes @KevinCruijssen yeah, vS works \$\endgroup\$
– Seggan
Feb 24, 2022 at 15:39

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Neil · Accepted Answer · 2022-02-24 19:36:06Z

Charcoal, 17 bytes

↑Ｅ⁴Ｅ⁴×§.o›Ｉθι⁺³⊗ι

Try it online! Link is to verbose version of code. Explanation:

  ⁴                 Literal integer `4`
 Ｅ                  Map over implicit range
    ⁴               Literal integer `4`
   Ｅ                Map over implicit range
       .o           Literal string `.o`
      §             Indexed by
           θ        Input `L`
          Ｉ         Cast to integer
         ›          Is greater than
            ι       Outer value
     ×              Repeated by
              ³     Literal integer `3`
             ⁺      Plus
                ι   Outer value
               ⊗    Doubled
↑                   Print rotated 90° anticlockwise

Charcoal prints each array element in its own row (or column, if rotated as it is here) and automatically double-spaces nested arrays when printing them.

14 bytes by using the custom characters 1 and 0 instead of o and .:

↑Ｅ⁴Ｅ⁴⭆⁺³⊗ι›Ｉθι

Try it online! Link is to verbose version of code. Explanation:

  ⁴             Literal integer `4`
 Ｅ              Map over implicit range
    ⁴           Literal integer `4`
   Ｅ            Map over implicit range
        ³       Literal integer `3`
       ⁺        Plus
          ι     Outer value
         ⊗      Doubled
     ⭆          Map over implicit range and join
            θ   Input `L`
           Ｉ    Cast to integer
          ›     Is greater than
             ι  Outer value
↑               Print rotated 90° anticlockwise

The challenge allows other characters than .o, so you can simply use the 01 for 15 bytes. — Kevin Cruijssen, Feb 24, 2022 at 11:03
@KevinCruijssen I guess I could add a 14 byte alternative to show how it should be done using 0 and 1... — Neil, Feb 24, 2022 at 13:13

tsh · Accepted Answer · 2022-02-25 02:21:43Z

2

Python 3, 76 bytes

f=lambda L,i=0:i<36and". o"[i%4+i/8<3or-(i%4<L)]*4+" \n"[i%4>2]+f(L,i+1)or""

Try it online!

-2 bytes by dingledooper

Python 3, 91 bytes

f=lambda L:L<45and" ".join(4*" o."[j+j+L//5>5and~(j<L%5)]for j in[0,1,2,3])+"\n"+f(L+5)or""

Try it online!

edited Feb 25, 2022 at 2:21

answered Feb 24, 2022 at 2:40

tsh

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\$\begingroup\$ 77 bytes \$\endgroup\$
– dingledooper
Feb 24, 2022 at 20:30
\$\begingroup\$ @dingledooper Good idea. But it could be shorter by changing to /8 instead of *8. \$\endgroup\$
– tsh
Feb 25, 2022 at 2:22

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Kevin Cruijssen · Accepted Answer · 2022-02-24 10:35:54Z

05AB1E (legacy), 18 bytes

4L‹ā·>×ε4иø}ζR€.B»

Uses 01 instead of o. respectively.

Try it online or verify all test cases.

Explanation:

Uses the legacy version of 05AB1E because it can zip/transpose on a list of strings, where the new version would require a character-matrix.

4L      # Push list [1,2,3,4]
  ‹     # Check for each whether it's larger than the (implicit) input
   ā    # Push a list in the range [1,length] (without popping): [1,2,3,4]
    ·   # Double each: [2,4,6,8]
     >  # Increase each by 1: [3,5,7,9]
      × # Repeat the earlier checks that many times as strings
        #  (e.g. input=2 → [0,0,1,1] → ["000","00000","1111111","111111111"])
ε       # Map over each string:
 4и     #  Repeat it 4 times as list
   ø    #  Zip/transpose; swapping rows/columns
}ζ      # After the map: zip/transpose again, with implicit " " as filler
  R     # Reverse this list of lists
   €    # Map over each inner list:
    .B  #  Box: expand the single filler space to length 4 by adding trailing
        #  spaces based on the string with the longest length in the list
      » # Join each inner list by spaces; and then each string by newlines
        # (after which the result is output implicitly)

G B · Accepted Answer · 2022-02-24 12:45:33Z

1

Ruby, 65 bytes

->n{(0..8).map{|x|(0..3).map{|y|(y<3-x/2?" ":y<n ??o:?.)*3}*' '}}

Try it online!

answered Feb 24, 2022 at 12:45

G B

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ovs · Accepted Answer · 2022-02-24 15:13:01Z

1

APL (Dyalog Unicode), 26 bytes

Uses 1/0 instead of o/..

3 5 7 9,.{¯9 5↑⍺4⍴⍕⍵}4↑⍴∘1

Try it online!

⍴∘1: Reshape of 1. Creates a vector of L 1's.
4↑: Take 4. Extend to length 4 by appending 0's.

,.{...}: Inner product. Pair elements from the computed boolean vector and 3 5 7 9 and call the inner dfn on it. Reduce the resulting list by concatenating character matrices horizontally.
⍕⍵: Format. Convert the 1 or 0 to a string.
⍺4⍴: Reshape into a ⍺×4 character matrix by replicating the digit.
¯9 5↑: Extend to a 9×5 matrix by padding with spaces to the top and right.

edited Feb 24, 2022 at 15:13

answered Feb 24, 2022 at 14:42

ovs

58.2k3 gold badges42 silver badges154 bronze badges

\$\begingroup\$ Can you add a quick explanation? I understand most of it but not all and want to see if I can steal any ideas to improve my J answer. \$\endgroup\$
– Jonah
Feb 24, 2022 at 15:03
1

\$\begingroup\$ @Jonah Added a bit of text. Reshape and Overtake do most of the work here, the inner product combines a reduction (,/) and ¨ to save a byte. \$\endgroup\$
– ovs
Feb 24, 2022 at 15:16

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AZTECCO · Accepted Answer · 2022-02-24 21:43:01Z

1

C (gcc), 78 bytes

j,i;f(x){for(i=180;--i;putchar(j?i%5&&i/35+j/5<5?j/5+x>3?79:46:32:10))j=i%20;}

Try it online!

Pretty straightforward approach:
often times in C is convenient to iterate backwards the entire volume so we start with i=lines x width.
Each time we print the proper character based on div modulus values.

Luckily I found that I could save some Bytes by using i%35 instead of i%40 which would require some extra operations to fit the right shape of each bar.

edited Feb 24, 2022 at 21:43

answered Feb 24, 2022 at 19:16

AZTECCO

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Jonathan Allan · Accepted Answer · 2022-02-25 00:22:32Z

Jelly, 15 bytes

4RðḤ‘ṁ"@>z⁶Ṛx4Y

Prints with o as 0 and . as 1.

Try it online!

How?

4RðḤ‘x"@>z⁶Ṛx4Y - Main Link: integer, L
4               - four
 R              - range -> [1,2,3,4]
  ð             - new dyadic chain - f(X=[1,2,3,4], L)
   Ḥ            - double X          -> [2,4,6,8]
    ‘           - increment         -> [3,5,7,9]
        >       - X greater than L? -> [1>L,2>L,3>L,4>L]
       @        - with swapped arguments - f([1>L,2>L,3>L,4>L], [3,5,7,9]):
      "         -   zip with:
     x          -     times -> [[1>L,1>L,1>L],[2>L,2>L,2>L,2>L,2>L],[3>L,...],[4>L...]]
         z⁶     - transpose with filler of space characters
           Ṛ    - reverse
            x4  - times four (vectorises)
              Y - join with newline characters
                - implicit, smashing print

SuperJedi224 · Accepted Answer · 2022-02-25 00:41:26Z

1

Javascript, 99 bytes

x=>"0004003402341234"[r="replace"](/\d{4}/g,"$&\n$&\n")[r](/\d/g,y=>+y?x<y?".... ":"oooo ":"     ")

(Didn't it use to be $0 for this and not $&?)

edited Feb 25, 2022 at 0:41

answered Feb 24, 2022 at 2:53

SuperJedi224

11.9k1 gold badge44 silver badges67 bronze badges

\$\begingroup\$ The + is unnecessary for -1 byte \$\endgroup\$
– emanresu A
Feb 24, 2022 at 6:13
\$\begingroup\$ @tsh fixed, I think \$\endgroup\$
– SuperJedi224
Feb 24, 2022 at 12:21

Add a comment |

tsh · Accepted Answer · 2022-02-25 02:20:36Z

1

JavaScript (Node.js), 73 bytes

L=>(g=i=>i--?` . 
 o`[i%20?i%5&&i/20+2*(x=i/5&3)<9|x+L&4:3]+g(i):'')(180)

Try it online!

-1 byte by Arnauld

JavaScript (Node.js), 90 bytes

L=>(g=i=>i>8?'':[1,2,3,4].map(j=>' o.'[j+j+i<8?0:j>L?2:1].repeat(4)).join` `+`
`+g(++i))``

Try it online!

edited Feb 25, 2022 at 2:20

answered Feb 24, 2022 at 1:57

tsh

34k2 gold badges33 silver badges123 bronze badges

\$\begingroup\$ I was just about to post my own js answer and you beat it by half \$\endgroup\$
– SuperJedi224
Feb 24, 2022 at 2:01
1

\$\begingroup\$ @SuperJedi224 Don't worry about posting not so well-golfed code. Idea behind could still be interesting somehow. \$\endgroup\$
– tsh
Feb 24, 2022 at 2:35
\$\begingroup\$ 73 bytes \$\endgroup\$
– Arnauld
Feb 24, 2022 at 12:47

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Aiden Chow · Accepted Answer · 2022-02-26 05:42:30Z

1

Python 3.8 (pre-release), 104 100 93 bytes

lambda n:[f'{"oooo "*(b:=n-4+(k:=min(i,7)//2+1))+".... "*(k-max(0,b)):>20}'for i in range(9)]

Try it online!

Returns a list of strings, allowed here.

Thanks everyone for all the tips on my previous python answer. I used some of those tips here!

edited Feb 26, 2022 at 5:42

answered Feb 24, 2022 at 2:06

Aiden Chow

12.7k1 gold badge14 silver badges53 bronze badges

\$\begingroup\$ @DialFrost yeah, i don't even know how tsh's answer even works. recursion is way out of my league :P \$\endgroup\$
– Aiden Chow
Feb 24, 2022 at 7:01
\$\begingroup\$ tsh's answer builds the 36 parts individually. \$\endgroup\$
– ophact
Feb 24, 2022 at 9:28

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Lynn · Accepted Answer · 2022-02-27 00:00:09Z

1

Python 3, 76 bytes

lambda n:"".join(". o"[i<~i%4*8or-(i%4<n)]*4+"   \n"[i%4]for i in range(36))

Try it online!

answered Feb 27, 2022 at 0:00

Lynn

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nununoisy · Accepted Answer · 2022-02-24 19:14:31Z

brainfuck, 544 bytes

+[-->---[-<]>]>[<<+>>>+<-]<-[>--<-------]>+[<+>-]>[<+>-]+++++++[<++>-]>+++++[<++>-]>-[>+<-----]>---<,>[<->-]<[>>+<<-[>>>>+<<<<-[>>>>>>+<<<<<<-[>>>>>>>>+<<<<<<<<-]]]]++[<+++++[<<<<...>>>>-]>->>>>>>>>[<]<<<<<<<<<<<....[>]<.>>]++[<+++++[<<<<..>>>>-]>->>>>>>[<]<<<<<<<<<....[<]>.[>]>>>>>>>>>[<]<<<<<<<<<<<....[>]<.>>]++[<<<<<.....>>>>>->>>>[<]<<<<<<<....[<]>.[>]>>>>>>>[<]<<<<<<<<<....[<]>.[>]>>>>>>>>>[<]<<<<<<<<<<<....[>]<.>>]+++[->>[<]<<<<<....[<]>.[>]>>>>>[<]<<<<<<<....[<]>.[>]>>>>>>>[<]<<<<<<<<<....[<]>.[>]>>>>>>>>>[<]<<<<<<<<<<<....[>]<.>>]

Try it online!

Builds each line in segments. Ungolfed:

memory layout
0    1    2    3    4    5    6    7    8    9    10   11   12   13
' '  dot  'o'  \n   cnt  inp  0    Nge1 0    Nge2 0    Nge3 0    Nge4

constants
space = 32
+[-->---[-<]>]>
[<<+>>>+<-]<
o = 111
-[>--<-------]>+
[<+>-]
dot = 46
>[<+>-]
+++++++[<++>-]
\n = 10
>+++++[<++>-]>
get input
-[>+<-----]>---<,>[<->-]<#

check 0
[
  >>+<<
  check 1
  -[
    >>>>+<<<<
    check 2
    -[
     >>>>>>+<<<<<<
     check 3
      -[
        >>>>>>>>+<<<<<<<<
        check 4
        -
      ]
    ]
  ]
]#
first 2 lines
++[
  15 spaces
  <+++++[<<<<...>>>>-]>-
  4 bar segment
  >>>>>>>>[<]<<<<<<<<<<<....
  [>]<.>>
]
next 2 lines
++[
  10 spaces
  <+++++[<<<<..>>>>-]>-
  3 bar segment
  >>>>>>[<]<<<<<<<<<....
  space
  [<]>.[>]>
  4 bar segment
  >>>>>>>>[<]<<<<<<<<<<<....
  [>]<.>>
]
next 2 lines
++[
  5 spaces
  <<<<<.....>>>>>-
  2 bar segment
  >>>>[<]<<<<<<<....
  space
  [<]>.[>]>
  3 bar segment
  >>>>>>[<]<<<<<<<<<....
  space
  [<]>.[>]>
  4 bar segment
  >>>>>>>>[<]<<<<<<<<<<<....
  [>]<.>>
]
last 3 lines
+++[
  -
  1 bar segment
  >>[<]<<<<<....
  space
  [<]>.[>]>
  2 bar segment
  >>>>[<]<<<<<<<....
  space
  [<]>.[>]>
  3 bar segment
  >>>>>>[<]<<<<<<<<<....
  space
  [<]>.[>]>
  4 bar segment
  >>>>>>>>[<]<<<<<<<<<<<....
  [>]<.>>
]

It could probably be shorter with a complex loop to handle generating each line but I couldn't get something like that to work.

Kevin Cruijssen · Accepted Answer · 2022-02-24 20:15:33Z

Java 11, 105 102 bytes

n->{for(int i=36;i-->0;)System.out.print((i/4>8-i%4*2?" ":n<i%4?".":"o").repeat(4)+(i%4<1?"\n":" "));}

Could probably be golfed slightly more by removing the .repeat(4) and looping 144 times instead, but I don't have the time to figure it out for now.

Try it online.

Explanation:

n->{                   // Method with integer parameter and no return-type
  for(int i=36;i-->0;) //  Loop `i` in the range (36,0]:
    System.out.print(  //   Print:
       (i/4            //    If `i` integer-divided by 4
        >              //    is larger than
        8-i%4*2?       //    8 minus `i` modulo-4 times 2
         " "           //     Use a space
        :n<i%4?        //    Else-if the input is smaller than `i` modulo-4:
         "."           //     Use a dot
        :              //    Else:
         "o"           //     Use an O
       ).repeat(4)     //    Repeat this character 4 times
       +(i%4<1?        //    Once every four iterations:
         "\n"          //     Append a newline
        :              //    Every other iteration:
         " "));        //     Append a space instead

Stack Exchange Network

Draw a Signal strength indicator

17 Answers 17

J, ⁴⁷ 46 bytes

APL+WIN, 48 bytes

Python, 75 bytes

76 bytes

Vyxal, 17 bytes

Charcoal, 17 bytes

Python 3, 76 bytes

Python 3, 91 bytes

05AB1E (legacy), 18 bytes

Ruby, 65 bytes

APL (Dyalog Unicode), 26 bytes

C (gcc), 78 bytes

Jelly, 15 bytes

How?

Javascript, 99 bytes

JavaScript (Node.js), 73 bytes

JavaScript (Node.js), 90 bytes

Python 3.8 (pre-release), 104 100 93 bytes

Python 3, 76 bytes

brainfuck, 544 bytes

Java 11, 105 102 bytes

Your Answer

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code-golf
ascii-art
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Draw a Signal strength indicator

17 Answers 17

J, 47 46 bytes

APL+WIN, 48 bytes

Python, 75 bytes

76 bytes

Vyxal, 17 bytes

Charcoal, 17 bytes

Python 3, 76 bytes

Python 3, 91 bytes

05AB1E (legacy), 18 bytes

Ruby, 65 bytes

APL (Dyalog Unicode), 26 bytes

C (gcc), 78 bytes

Jelly, 15 bytes

How?

Javascript, 99 bytes

JavaScript (Node.js), 73 bytes

JavaScript (Node.js), 90 bytes

Python 3.8 (pre-release), 104 100 93 bytes

Python 3, 76 bytes

brainfuck, 544 bytes

Java 11, 105 102 bytes

Your Answer

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J, ⁴⁷ 46 bytes

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