A leaper is a category of fairy chess piece which moves by "jumping." A normal knight is a (1,2)-leaper, meaning each move involves moving a distance of 1 square in an orthogonal direction and 2 squares in the perpendicular direction.
.o.o.
o...o
..N..
o...o
.o.o.
There are many different leapers. The (1,3)-leaper is called the Long Knight, or Camel. Its move pattern looks like this:
..o.o..
.......
o.....o
...L...
o.....o
.......
..o.o..
There's also the (2,2) Alfil...
o...o
.....
..A..
.....
o...o
...and even the (0,1) Wazir.
.o.
oWo
.o.
Challenge
Given a pair of numbers as input, output the corresponding movement diagram. You may write a program or function, with input provided via STDIN/command-line or as an argument, and output provide by STDOUT or the return value. This is code-golf.
Input
Input will be a comma-separated list of two integers and an uppercase letter. The two integers will be in the range 0-7 (inclusive) and in non-decreasing order. Trailing newline optional for input and output.
Here are the inputs for the above four examples:
1,2,N
1,3,L
2,2,A
0,1,W
Output
Output will be a square-shaped multiline string. The capital letter will be placed in the center to represent the current location of the piece. The movement locations will be represented by either 4 or 8 lowercase os. All other spaces in the square will be filled with periods.
If the movement pattern is 0,0, output just the capital letter.
As a hint, if the second integer (the larger one) has value N, then the square will always have side length 2N+1. The os will always be on the perimeter of the square.
Additional Testcases
1,1,F
o.o
.F.
o.o
5,7,Q
..o.........o..
...............
o.............o
...............
...............
...............
...............
.......Q.......
...............
...............
...............
...............
o.............o
...............
..o.........o..
0,0,J
J
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\$\begingroup\$ Why is the camel testcase letter a 'L'? I know out doesn't matter but it might be helpful to change out to a 'C'. \$\endgroup\$Riker– Riker2016年01月09日 18:50:40 +00:00Commented Jan 9, 2016 at 18:50
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1\$\begingroup\$ @RikerW My choice wasn't completely arbitrary, that's its "standardized" letter. \$\endgroup\$PhiNotPi– PhiNotPi2016年01月09日 18:53:16 +00:00Commented Jan 9, 2016 at 18:53
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1\$\begingroup\$ Okay. That makes sense. \$\endgroup\$Riker– Riker2016年01月09日 18:54:07 +00:00Commented Jan 9, 2016 at 18:54
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1\$\begingroup\$ For a function, are 3 arguments ok or do you want a single string argument comma separated? \$\endgroup\$edc65– edc652016年03月06日 10:43:06 +00:00Commented Mar 6, 2016 at 10:43
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\$\begingroup\$ @PhiNotPi where do you get the "standard" code. L is more commonly Bishop in German - and two letter codes are used for the fairy pieces \$\endgroup\$Laska– Laska2022年12月21日 08:31:49 +00:00Commented Dec 21, 2022 at 8:31
4 Answers 4
Ruby,107
->a,b,n{(-b..b).map{|i|s='..'*b+?.
i%b==0&&(i==0?s[b]=n :s[b+a]=s[b-a]=?o)
i.abs==a&&s[0]=s[-1]=?o
puts s}}
Ungolfed in test program
f=->a,b,n{
(-b..b).map{|i| #iterate from -i to i (lines of grit)
s='..'*b+?. #make a string of 2n+1 .'s
i%b==0&&(i==0?s[b]=n :s[b+a]=s[b-a]=?o) #if i%b=0 plot the centre character (if i=0) or the o's on the top and bottom rows
i.abs==a&&s[0]=s[-1]=?o #if i.abs=a plot the o's in left and right columns
puts s #having substituted the .'s with o and centre as necessary, output the current line
}
}
a=gets.to_i
b=gets.to_i
n=gets.chop
f[a,b,n]
Pyth, 40 bytes
JASv<2zFZK+rH_1SHFY,LZKp?qJSY\o?sY\.ez)k
I'm learning Pyth! Try it out.
Explanation
J J =
A (G, H) =
S sorted(
v<2z eval(input[:-2]))
FZK+rH_1SH for Z in K = [H, H-1, ..., 0] + [1, 2, ..., H]:
FY,LZK for Y in [(Z, k) for k in K]:
p print the following value without newline:
?qJSY\o if J = sorted(Y): 'o'
?sY\. if sum(Y) != 0: '.'
ez else: input[-1]
) end for
k print newline
JavaScript (ES6), (削除) 163 (削除ここまで) (削除) 161 (削除ここまで) 145 bytes
(x,y,c,m=a=>`\n`+a.slice(1).reverse().join``+a.join``,a=Array(y+1).fill`.`,q=a.map(_=>[...a]))=>m(q.map(m,q[x][y]=q[y][x]='o',q[0][0]=c)).slice(2)
Where \n is the literal new line character. Works by generating the bottom right quadrant and mirroring it along both axes.
Edit: Saved 2 bytes thanks to @edc65.
(I got here via a duplicate question which allowed an array result which would have been 19 bytes shorter, but didn't guarantee nondecreasing order, which wasted 8 bytes.)
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\$\begingroup\$ You can save 3 bytes using a template string with no brackets for
filland a literal newline \$\endgroup\$edc65– edc652016年03月06日 10:53:55 +00:00Commented Mar 6, 2016 at 10:53 -
\$\begingroup\$ @edc65 I'd already tried to adjust for the literal newline (I always write "Where
\nis the literal new line character" when I do that) but thanks for the other tip. \$\endgroup\$Neil– Neil2016年03月06日 11:10:52 +00:00Commented Mar 6, 2016 at 11:10
JavaScript (ES6) 144 (削除) 150 (削除ここまで)
(a,b,c,g=Array(b-~b).fill`.`)=>(g=g.map(x=>[...g])).map(x=>x.join``,[b-a,b+a].map(t=>g[t][0]=g[0][t]=g[t][2*b]=g[2*b][t]='o'),g[b][b]=c).join`
`
Less golfed
(a,b,c)=> {
var g=Array(b*2+1).fill('.');
g=g.map(x=>[...g]);
var q=(t)=>
g[t][0] =
g[0][t] =
g[t][2*b] =
g[2*b][t] = 'o';
q(b+a);
q(b-a);
g[b][b] = c;
return g.map(x=> x.join('')).join('\n')
}
Test
f=(a,b,c,g=Array(b-~b).fill`.`)=>(g=g.map(x=>[...g])).map(x=>x.join``,[b-a,b+a].map(t=>g[t][0]=g[0][t]=g[t][2*b]=g[2*b][t]='o'),g[b][b]=c).join`
`
console.log=x=>O.textContent+=x+'\n'
t=`1,2,N
1,3,L
2,2,A
0,1,W
1,1,F
5,7,Q`.split`\n`
.forEach(t=>([x,y,z]=t.split`,`, console.log(t+'\n'+f(+x,+y,z)+'\n')))
<pre id=O></pre>