Draw a hexagon chain x long, each with side of y length
x - the length of the chain <= 50
y - the length of each side <= 50
x=1,y=1
_
/ \
\_/
x=4,y=1
_ _
/ \_/ \_
\_/ \_/ \
\_/ \_/
x=3,y=2
__ __
/ \ / \
/ \__/ \
\ / \ /
\__/ \__/
\ /
\__/
The shortest valid answer in bytes wins.
Leading and trailing newlines allowed.
Trailing white space allowed.
ＮθＦＮ«Ｍ∧﹪ι²⊗θ↓Ｐ×_θ←↖θ→↗θ×_θ↓↘θ←Ｐ↙θ↗
Try it online! Link is to verbose version of code. Takes the parameters in the order size, count. Explanation:
Ｎθ
Input the hexagon size.
ＦＮ«
Loop over the input number of hexagons.
Ｍ∧﹪ι²⊗θ↓
On alternate hexagons, move down a whole hexagon, so that the next hexagon is drawn to the lower right rather than the upper right.
Ｐ×_θ
Draw the bottom.
←↖θ
Draw the lower left side.
→↗θ
Draw the upper left side.
×_θ
Draw the top.
↓↘θ
Draw the upper right side.
←Ｐ↙θ
Draw the lower right side.
↗
Assume the next hexagon is to the upper right.
def f(n,w):
a=w*2
for j in range(1+w*3):print''.join([[[' ',[' /'[i%w==-j%w],' \\'[i%w==~-j%w]][i/a+~-j/w&1]][(j>0)*(i/w>=(j>a))*((i/w/n<2)or(n%-2<~-j/w<3-n%2))],' _'[(j+i/w%4/2*w)%a<(i<n*a)]][i/w%2]for i in range(-~n*a)])
def f(n,w):
c=2*w;r=[[' ']*(-~n*c)for _ in' '*(1+w*3)]
for i in range(w):
for j in range(n):a=i+j*2*w;b=j%2*w;r[b][w+a],r[c+b][w+i+j*c],r[b+w-i][a],r[b+c-i][a+c],r[b+w-~i][a],r[b-~i][a+c]=r'__//\\'
for l in r:print''.join(l)
ā.{e╚øΚe╔*ο⁴↔±┼┼╬±fe«*If2%e*I╬5
Explanation:
ā push an empty array - the canvas
.{ repeat input times
e╚ push a diagonal the length of the variable E (by default: next input)
øΚ prepend a line to it
e╔*ο push ["_"*E]
⁴ copy that diagonal
↔± and reverse it horizontally
┼┼ add the 3 parts together
╬± and palindromize vertically - one hexagon is finished
fe«*I push counter*E*2 + 1 (the counter is 0-based)
f2%e*I push counter%2 * E + 1
╬5 at [counter*E*2+1; counter%2*E+1] insert the hexagon in the canvas
+\#<*:0< vp93*p92+1*3:p91:&+1p
>1+:29g-!#@_>:1-19g+:19g/1-:2*49p2%!2*:1+59p19g*\19g%+69p01v
,>*19g3*79g-1-69g-!2*49g1+0g2%*79g69g-!49g0g2%*++4g,1+:39g-v
^^3*!-*g95g91+1g96!-g95%4p04p01:`\g90p05`0::/g91:p97%*4g91:_$55+
\/_
Takes input in currying syntax (y)(x).
y=>(F=(x,c=!(p=y-1,w=x*y*2+y,a=[...(' '.repeat(w++)+`
`).repeat(3*y+1)],g=(d,k)=>k!=y?g(d,-~k,a[p+=d]='_\\/'[c%3]):c++))=>x--?F(x,g(1),g(w+1),p++,g(w-1),g(-1),p+=w,g(~w),p--,g(1-w),p+=(c/6&1?w:-w)*y-w+y*2):a.join``)
let f =
y=>(F=(x,c=!(p=y-1,w=x*y*2+y,a=[...(' '.repeat(w++)+`
`).repeat(3*y+1)],g=(d,k)=>k!=y?g(d,-~k,a[p+=d]='_\\/'[c%3]):c++))=>x--?F(x,g(1),g(w+1),p++,g(w-1),g(-1),p+=w,g(~w),p--,g(1-w),p+=(c/6&1?w:-w)*y-w+y*2):a.join``)
O.innerText = f(3)(7)<pre id=O></pre>
ø╶｛╷⁷«×¹２％⁷×╵⁷⇵_× ⁷／∔×╬│╋
Explanation (some characters have been changed to look monospace):
ø╶{╷⁷«×¹2%⁷×╵⁷⇵_× ⁷/∔×╬│╋
ø push an empty canvas
╶{ for 1..input
╷ decrease (so this starts with 0)
⁷«× multiply by X*2; X coordinate of new hexagon done
¹2% push 0-indexed counter%2
⁷× multiply by X
╵ and increment; Y coordinate done
⁷⇵ push ceil(X/2), saving the remainder
_× repeat "_" that many times
⁷/ push " " and an ASCII diagonal with size X
∔ prepend verticall the space before the diagonal
done so there's space for the underscores
× and append the underscores horizontally to the diagonal
╬│ quad-palindromize with Y overlap of 1
and X overlap of the remainder taken before
╋ and at the before defined coords ((I-1)*X*2; (i%2)*X + 1)
insert the hexagon in the canvas