Treasure Map Drawing Bot

Question

You're organizing a treasure hunt for your friends. To conduct things more easily, you want to draw a map of all locations where you hid the precious objects.

Input

Any form of input denoting a list of points consisting of (nonnegative) x- and y-coordinate, 0 0 being the upper left corner is permitted (Note: You may also use 1-based indexing in your answer, please comment on that if you do). Example:

1 2
3 0
0 1

Challenge

Your function or program should be able to construct a map denoting every given location with an x where the mark is found in row y + 1 and column x + 1 in the output. Unmarked locations are represented with a . The map also consists of a frame where the corners are +s, the vertical lines are |s and the horizontal lines are -s. Your solution should output the smallest possible frame. Map for the input example given above:

+----+
|   x|
|x   |
| x  |
+----+

Possible Test Cases

---

"0 0"
=>
+-+
|x|
+-+

---

"0 10
 5 5
 10 0"
=>
+-----------+
|          x|
|           |
|           |
|           |
|           |
|     x     |
|           |
|           |
|           |
|           |
|x          |
+-----------+

---

""
=>
++
++

---

"0 0
 0 2
 2 0"
=>
+---+
|x x|
|   |
|x  |
+---+

---

Of course, this is code-golf, meaning that the solution with the lowest byte count wins! Explanations of your solution are encouraged.

Answer 1

J, 37 34 bytes

0<@|:' x'{~((i.@]e.#.~)1+>./) ::#:

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                       1+>./          maximum for each coordinate + 1
             i.@]                     make an array with these dimensions filled with 0..x*y
                                      /* if the input is empty, 
                                         1+>./ is negative infinity
                                         and i.@] throws an error  */
                   #.~                mixed base conversion of input
                 e.                   replace the elements of i.@]
                                        with 1 if it's present in the
                                        converted input, 0 otherwise
           (                ) ::      if there's an error do the other thing instead
                                #:    "to binary", for empty input this returns a 0x0 matrix
0<@|:' x'{~                           index into character string, transpose and put in a box

Answer 2

JavaScript (ES6), 150 bytes

Takes input as a list of 1-indexed coordinates in [x,y] format. Returns a string.

a=>(g=w=>y<h?' |-+x'[4*a.some(a=>a+''==[x,y])|2*(-~y%h<2)|++x%w<2]+[`
`[x=x<w?x:+!++y]]+g(w):'')((M=i=>Math.max(2,...a.map(a=>a[i]+2)))(x=y=0),h=M(1))

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Answer 3

Haskell, 127 123 bytes

This defines the operator (!) which takes a list of x-coordinates and a list of the corresponding y-coordinates:

x!y|l<-'+':('-'<$m x)++"+"=unlines$l:['|':[last$' ':['x'|(i,j)`elem`zip x y]|i<-m x]++"|"|j<-m y]++[l];m x=[1..maximum$0:x]

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Ungolfed/Explanation

The helper function m expects a list and returns indices (1-based) up to the maximum, if the list is empty it returns []:

m x | null x    = []
    | otherwise = [1 .. maximum x]

The actual operator (!) is just a list-comprehension, traversing all the coordinates and choosing a or x character, which gets joined with newlines:

x ! y
  -- construct the top and bottom line
  | l <- "+" ++ replicate (maximum (0:x)) '-' ++ "+"
  -- join the list-comprehension with new-lines
  = unlines $ 
  -- prepend the top line
      [l]
  -- the actual map:
    -- begin the line with | and add the correct chars for each coordinate
      ++ [ "|" ++ [ if (i,j) `elem` zip x y then 'x' else ' '
    -- "loop" over all x-coordinates
                 | i <- m x
                 ]
    -- end the line with a |
           ++ "|"
    -- "loop" over all y-coordinates
         | j <- m y
         ]
  -- append the bottom line
      ++ [l]

Answer 4

Canvas, 22 bytes

ø╶｛X；┤╋｝ｌ|＊ｅＬ┤-×+ｅ：└∔∔

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Takes 1-indexed inputs.

Finally decided to fix a bug that's been annoying me for ages and golfed this down to 21 bytes.

Explanation (half-ASCII-fied for monospace):

ø╶{X;┤╋}l|*eL┤-×+e:└++  full program, implicitly outputting ToS at the end
ø                       push an empty Canvas - the map
 ╶{    }                for each array in the input array
   X                      push "X"
    ;┤                    and push the two coordinates separately on the stack
      ╋                   and overlap the "X" there in the map
        l               get the vertical length of the map
         |*             repeat "|" vertically that many times
           e            encase the map in two of those vertical bars
            L           get the horizontal length of the map
             ┤          subtract 2 (leave place for the "+"es)
              -×        repeat "-" that many times
                +e      encase that line in "+"es
                  :└    push a copy of that below the map
                    ++  and join the 3 items vertically

Answer 5

Python 2, 151 140 138 bytes

-2 bytes thanks to Jo King.

Input is 1-indexed.

m=input()
w,h=map(max,zip((0,0),*m))
b=['+'+'-'*w+'+']
M=b+['|'+' '*w+'|']*h+b
for x,y in m:M[y]=M[y][:x]+'x'+M[y][x+1:]
print'\n'.join(M)

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Answer 6

Charcoal, 37 bytes

≔Ｅ²⁺²⌈Ｅθ§λιηＢ⊟⮌η⊟ηＦθ«Ｊ⊟⮌ι⊟ιx

Try it online! Link is to verbose version of code. 1-indexed. Explanation:

¿¬ＬθＵＲ²+«

Special-case empty input by drawing a 2x2 rectangle of +s.

≔Ｅ²⁺²⌈Ｅθ§λιη

Transpose the input, take the maximum of each column (now row) and add 2 to get the box size in Charcoal co-ordinates.

Ｂ⊟⮌η⊟η

Draw the box.

Ｆθ«

Loop over each co-ordinate.

Ｊ⊟⮌ι⊟ι

Jump to its position.

x

Mark with a cross.

Answer 7

Stax, 32 31 24 bytes

╩╠ee%╙æM■↓^⌐╧ΩΓ¡c¥èf¢○ [

Run and debug it

Takes 0-based indices as array of [y, x] pairs.

Explanation:

zs'X&|<cM%'-*'+|S]s{'||Smn++m Unpacked program, implicit input
zs                            Tuck empty array under input
  'X                          Push "X"
    &                         Assign element at all indices (create map)
                                As the indexing arrays are an array of arrays, treat them as a path to navigate a multidimensional array.
                                Extend array if needed.
     |<                       Left-align all to the length of the longest.
       cM%                    Copy, transpose, length (width)
          '-*                 Repeat "-"
             '+|S             Surround with "+"
                 ]s           Make a singleton and tuck it below the map
                   {    m     Map:
                    '||S        Surround with "|"
                         n++  Surround with the above/below border (built above)
                            m Map:
                                Implicit output

Answer 8

R, 133 125 122 bytes

function(m)cat(z<-c("+",rep("-",u<-max(m[,1])),"+","
"),rbind("|",`[<-`(matrix(" ",u,max(m[,2])),m,"x"),"|","
"),z,sep="")

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1-indexed. Takes a matrix as argument. Saved 8 bytes thanks to digEmAll, 3 thanks to Giuseppe! Explanation (earlier version of code):

function(m){                           #x and y are the 1st and 2nd col of m
s=matrix(32,u<-max(m[,1]),max(m[,2]))  #s (treasure map) has dim max(x), max(y) 
s[m]=120                               #place the X's on the map
cat(                                   #print:
    z<-c("+",rep("-",u),"+","\n"),     #the top line
    intToUtf8(rbind(124,s,124,13)),    #the map
    z,                                 #the bottom line.
    sep="")
}

Answer 9

coords taken of the format [y,x]

JavaScript (Node.js), 191 184 bytes

c=f=a=>{a.map(([y,x])=>(c[M<++y?M=y:y]=c[y]||[])[m<++x?m=x:x]="x",M=m=0)
m++
M++
s=""
for(i=0;i<=M;s+=`
`,i++)for(j=0;j<=m;j++)s+=(c[i]||0)[j]||(j%m?i%M?" ":"-":i%M?"|":"+") 
return s}

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Answer 10

JavaScript, 180 bytes

F = 

s=>s.map(([x,y])=>(t[y]=t[Y<y?Y=y:y]||[])[X<x?X=x:x]='x',t=[X=Y=0])&&[...t,0].map((_,y)=>[...Array(X+2)].map((_,x)=>[(t[y]||0)[x]||' ',...'-|+'][!(y%~Y)+2*!(x%~X)]).join``).join`
`


console.log(F([[1,11],[6,6],[11,1]]))

Answer 11

C (gcc), 246 234 bytes

Thanks to ceilingcat for the suggestion.

Zero-indexed. The function takes a list of co-ordinates and buffer, finds the maximum x and y values, fills the buffer with spaces, generates the frame, and then plots the 'x's.

f(int*a,char*c){int*b=a,x,y=x=-1,i=0;for(;~*b;*++b>y?y=*b:0,++b)*b>x?x=*b:0;for(x+=4,y+=3,memset(c,32,x*y);++i<x;c[i]=c[y*x-i]=45);for(i=0;i<y;c[x*++i-1]=10*(i<=y))c[x*i]=c[x*i+x-2]=i&&y/i?124:43;for(b=a;~*b;b+=2)c[*b+1-~b[1]*x]='x';}

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Answer 12

05AB1E, 44 42 bytes

ζεZ}>`UX'-×'+.ø©,F'|NVXF¹YN‚.å„ xè}'|J,}®,

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---

 ζεZ}>`                                     # Push the max of X and Y to the stack +1.
       UX                                   # Store the max X.
         '-×'+.ø©,                          # Print the top border.
                  F                     }   # From 0 to Y...
                   '|                       # Push left border.
                     NV                     # Store current Y in Y.
                       XF          }        # From 0 to X...
                         ¹                  # Push input.
                          YN‚               # Group current X and Y.
                             .å             # Exists in original input ? 1 : 0
                               „ xè         # Exists ? 'X' : ' '
                                    '|J,    # Right border, join, print.
                                         ®, # Print bottom border.

X and Y may be reversed, didn't know if that mattered at all.

---

I think I have this in less bytes, but we'll see... Nope.

ζεZ}>`D'-×'+.øUð×'|.øs.D)X.ø©svy>`s®sUXès'xsǝXǝ}

Answer 13

Java 10, 238 223 220 219 bytes

c->{var r="";int w=0,h=0,x,y;for(var l:c){w=(x=l.get(0))>w?x:w;h=(y=l.get(1))>h?y:h;}for(w++,h-=x=-1;x++<w;r+="\n")for(y=-1;y++<h;)r+=x%w+y%h<1?"+":x%w<1?"-":y%h<1?"|":(c+r).contains("["+x+", "+y+"]")?"x":" ";return r;}

-3 bytes thanks to @ceilingcat.

1-indexed coordinates.

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Explanation:

c->{                      // Method with 2D Lists as parameter and String return-type
  var r="";               //  Result-String, starting empty
  int w=0,h=0,            //  Width and height, starting at 0
      x,y;                //  Temp x,y coordinates
  for(var l:c){           //  Loop over the Inner Lists containing the coordinates
    w=(x=l.get(0))>w?x:w; //   Determine width based on max x-coordinate
    h=(y=l.get(1))>h?y:h;}//   Determine height based on max y-coordinate
  for(w++,h-=             //  Increase both the width and height by 1
      x=-1;x++<w;         //  Loop `x` in the range [0, width]:
      r+="\n")            //    After every iteration: append a new-line to the result
    for(y=-1;y++<h;)      //   Inner loop `y` in the range [0, height]:
      r+=                 //    Append the following character to the result-String:
         x%w+y%h<1?       //    If it's one of the corners:
          "+"             //     Append "+"
         :x%w<1?          //    Else-if it's the top or bottom row:
          "-"             //     Append "-"
         :y%h<1?          //    Else-if it's the right or left column:
          "|"             //     Append "|"
         :(c+r).contains("["+x+", "+y+"]")? 
                          //    Else-if the current `x,y` is part of the input-coordinates:
          "x"             //     Append "x"
         :                //    Else:
          " ";            //     Append " "
  return r;}              //  Return the result-String

Answer 14

C (gcc), 229 220 216 bytes

-9 bytes thanks to ceilingcat.

Zero-indexed. Takes coordinates as list of numbers, where even numbers are X and odd numbers are Y.

X,Y,i,j,k,x,z;f(l,n)int*l;{for(X=Y=0,i=n*=2;i--;X=fmax(l[i],X))Y=fmax(l[i--],Y);n&&X++-Y++;for(--i;i++<Y;puts(""))for(j=-1;j<=X;z=i<0|i==Y,putchar(j++<0|j>X?z?43:'|':x?z?45:32:'x'))for(x=k=n;k--;)x*=l[k--]-i|l[k]-j;}

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