replaced http://codegolf.stackexchange.com/ with https://codegolf.stackexchange.com/
JavaScript (ES6), (削除) 237 (削除ここまで) (削除) 210 (削除ここまで) (削除) 204 (削除ここまで) (削除) 188 (削除ここまで) (削除) 182 (削除ここまで) 178 bytes
Credit to @Downgoat @Downgoat for saving 16 bytes in the 188-byte revision
Update: I had a brainwave and reduced the first operation on s to a single map call instead of two separate ones
s=>(r=" .':",a=[],s=[...s].map(c=>(t=('00'+r.indexOf(c).toString(2)).slice(-2),a.push(t[0]),t[1])),a.splice(0,0,s.shift()),s.push(a.pop()),a.map((v,i)=>r[+('0b'+v+s[i])]).join``)
###Pretty Print & Explanation
s => (
r = " .':", // Map of characters to their (numerical) binary representations (e.g. r[0b10] = "'")
a = [], // extra array needed
// Spread `s` into an array
s = [...s].map(c => (
// Map each character to a `0`-padded string representation of a binary number, storing in `t`
t = ('00' + r.indexOf(c).toString(2)).slice(-2)),
// Put the first character of `t` into `a`
a.push(t[0]),
// Keep the second character for `s`
t[1]
)),
// Put the first character of `s` in the first index of `a`
a.splice(0,0,s.shift()),
// Append the last character of `a` to `s`
s.push(a.pop(),
// Rejoin the characters, alternating from `a` to `s`, representing the rotated matrix, and map them back to their string representation
// Use implicit conversion of a binary number string using +'0b<num>'
a.map((v,i) => r[+('0b' + v + s[i])]).join``
)
JavaScript (ES6), (削除) 237 (削除ここまで) (削除) 210 (削除ここまで) (削除) 204 (削除ここまで) (削除) 188 (削除ここまで) (削除) 182 (削除ここまで) 178 bytes
Credit to @Downgoat for saving 16 bytes in the 188-byte revision
Update: I had a brainwave and reduced the first operation on s to a single map call instead of two separate ones
s=>(r=" .':",a=[],s=[...s].map(c=>(t=('00'+r.indexOf(c).toString(2)).slice(-2),a.push(t[0]),t[1])),a.splice(0,0,s.shift()),s.push(a.pop()),a.map((v,i)=>r[+('0b'+v+s[i])]).join``)
###Pretty Print & Explanation
s => (
r = " .':", // Map of characters to their (numerical) binary representations (e.g. r[0b10] = "'")
a = [], // extra array needed
// Spread `s` into an array
s = [...s].map(c => (
// Map each character to a `0`-padded string representation of a binary number, storing in `t`
t = ('00' + r.indexOf(c).toString(2)).slice(-2)),
// Put the first character of `t` into `a`
a.push(t[0]),
// Keep the second character for `s`
t[1]
)),
// Put the first character of `s` in the first index of `a`
a.splice(0,0,s.shift()),
// Append the last character of `a` to `s`
s.push(a.pop(),
// Rejoin the characters, alternating from `a` to `s`, representing the rotated matrix, and map them back to their string representation
// Use implicit conversion of a binary number string using +'0b<num>'
a.map((v,i) => r[+('0b' + v + s[i])]).join``
)
JavaScript (ES6), (削除) 237 (削除ここまで) (削除) 210 (削除ここまで) (削除) 204 (削除ここまで) (削除) 188 (削除ここまで) (削除) 182 (削除ここまで) 178 bytes
Credit to @Downgoat for saving 16 bytes in the 188-byte revision
Update: I had a brainwave and reduced the first operation on s to a single map call instead of two separate ones
s=>(r=" .':",a=[],s=[...s].map(c=>(t=('00'+r.indexOf(c).toString(2)).slice(-2),a.push(t[0]),t[1])),a.splice(0,0,s.shift()),s.push(a.pop()),a.map((v,i)=>r[+('0b'+v+s[i])]).join``)
###Pretty Print & Explanation
s => (
r = " .':", // Map of characters to their (numerical) binary representations (e.g. r[0b10] = "'")
a = [], // extra array needed
// Spread `s` into an array
s = [...s].map(c => (
// Map each character to a `0`-padded string representation of a binary number, storing in `t`
t = ('00' + r.indexOf(c).toString(2)).slice(-2)),
// Put the first character of `t` into `a`
a.push(t[0]),
// Keep the second character for `s`
t[1]
)),
// Put the first character of `s` in the first index of `a`
a.splice(0,0,s.shift()),
// Append the last character of `a` to `s`
s.push(a.pop(),
// Rejoin the characters, alternating from `a` to `s`, representing the rotated matrix, and map them back to their string representation
// Use implicit conversion of a binary number string using +'0b<num>'
a.map((v,i) => r[+('0b' + v + s[i])]).join``
)
JavaScript (ES6), (削除) 237 (削除ここまで) (削除) 210 (削除ここまで) (削除) 204 (削除ここまで) (削除) 188 (削除ここまで) 182(削除) 182 (削除ここまで) 178 bytes
Credit to @Downgoat for saving 16 bytes in the 188-byte revision
Update: I had a brainwave and reduced the first operation on s to a single map call instead of two separate ones
s=>(r=" .':",a=[],s=[...s].map(c=>(t=('00'+r.indexOf(c).toString(2)).slice(-2),a.push(t[0]),t[1])),a.splice(0,0,s.shift()),s.push(a.pop()),a.map((v,i)=>r[parseInt=>r[+(v+s[i],2'0b'+v+s[i])]).join``)
###Pretty Print & Explanation
s => (
r = " .':", // Map of characters to their (numerical) binary representations (e.g. r[0b10] = "'")
a = [], // extra array needed
// Spread `s` into an array
s = [...s].map(c => (
// Map each character to a `0`-padded string representation of a binary number, storing in `t`
t = ('00' + r.indexOf(c).toString(2)).slice(-2)),
// Put the first character of `t` into `a`
a.push(t[0]),
// Keep the second character for `s`
t[1]
)),
// Put the first character of `s` in the first index of `a`
a.splice(0,0,s.shift()),
// Append the last character of `a` to `s`
s.push(a.pop(),
// Rejoin the characters, alternating from `a` to `s`, representing the rotated matrix, and map them back to their string representation
// Use implicit conversion of a binary number string using +'0b<num>'
a.map((v,i) => r[parseIntr[+('0b' + v + s[i],2)]).join``
)
JavaScript (ES6), (削除) 237 (削除ここまで) (削除) 210 (削除ここまで) (削除) 204 (削除ここまで) (削除) 188 (削除ここまで) 182 bytes
Credit to @Downgoat for saving 16 bytes in the 188-byte revision
Update: I had a brainwave and reduced the first operation on s to a single map call instead of two separate ones
s=>(r=" .':",a=[],s=[...s].map(c=>(t=('00'+r.indexOf(c).toString(2)).slice(-2),a.push(t[0]),t[1])),a.splice(0,0,s.shift()),s.push(a.pop()),a.map((v,i)=>r[parseInt(v+s[i],2)]).join``)
###Pretty Print & Explanation
s => (
r = " .':", // Map of characters to their (numerical) binary representations (e.g. r[0b10] = "'")
a = [], // extra array needed
// Spread `s` into an array
s = [...s].map(c => (
// Map each character to a `0`-padded string representation of a binary number, storing in `t`
t = ('00' + r.indexOf(c).toString(2)).slice(-2)),
// Put the first character of `t` into `a`
a.push(t[0]),
// Keep the second character for `s`
t[1]
)),
// Put the first character of `s` in the first index of `a`
a.splice(0,0,s.shift()),
// Append the last character of `a` to `s`
s.push(a.pop(),
// Rejoin the characters, alternating from `a` to `s`, representing the rotated matrix, and map them back to their string representation
a.map((v,i) => r[parseInt(v + s[i],2)]).join``
)
JavaScript (ES6), (削除) 237 (削除ここまで) (削除) 210 (削除ここまで) (削除) 204 (削除ここまで) (削除) 188 (削除ここまで) (削除) 182 (削除ここまで) 178 bytes
Credit to @Downgoat for saving 16 bytes in the 188-byte revision
Update: I had a brainwave and reduced the first operation on s to a single map call instead of two separate ones
s=>(r=" .':",a=[],s=[...s].map(c=>(t=('00'+r.indexOf(c).toString(2)).slice(-2),a.push(t[0]),t[1])),a.splice(0,0,s.shift()),s.push(a.pop()),a.map((v,i)=>r[+('0b'+v+s[i])]).join``)
###Pretty Print & Explanation
s => (
r = " .':", // Map of characters to their (numerical) binary representations (e.g. r[0b10] = "'")
a = [], // extra array needed
// Spread `s` into an array
s = [...s].map(c => (
// Map each character to a `0`-padded string representation of a binary number, storing in `t`
t = ('00' + r.indexOf(c).toString(2)).slice(-2)),
// Put the first character of `t` into `a`
a.push(t[0]),
// Keep the second character for `s`
t[1]
)),
// Put the first character of `s` in the first index of `a`
a.splice(0,0,s.shift()),
// Append the last character of `a` to `s`
s.push(a.pop(),
// Rejoin the characters, alternating from `a` to `s`, representing the rotated matrix, and map them back to their string representation
// Use implicit conversion of a binary number string using +'0b<num>'
a.map((v,i) => r[+('0b' + v + s[i])]).join``
)
JavaScript (ES6), (削除) 237 (削除ここまで) (削除) 210 (削除ここまで) (削除) 204 (削除ここまで) 188(削除) 188 (削除ここまで) 182 bytes
Credit to @Downgoat for saving 16 bytes in the latest188-byte revision
Update: I had a brainwave and reduced the first operation on s to a single map call instead of two separate ones
s=>(r=" .':",a=[],s=[...s].map(c=>(t=('00'+r.indexOf(c).toString(2)).slice(-2)).map(n=>(,a.push(n[0]t[0]),n[1]t[1])),a.splice(0,0,s.shift()),s.push(a.pop()),a.map((v,i)=>r[parseInt(v+s[i],2)]).join``)
###Pretty Print & Explanation
s => (
r = " .':", // Map of characters to their (numerical) binary representations (e.g. r[0b10] = "'")
a = [], // extra array needed
// Spread `s` into an array, mapping s = [...s].map(c => (
// Map each character to a `0`-padded string representation of a binary number, storing in `t`
s = [...s].map(ct =>= ('00' + r.indexOf(c).toString(2)).slice(-2)),
// Put the first character of `t` into `a` and keep the second charactera.push(t[0]),
for `s`
// Keep the second character for `s`
.map(nt[1]
=> (a.push(n[0]),n[1])),
// Put the first character of `s` in the first index of `a`
a.splice(0,0,s.shift()),
// Append the last character of `a` to `s`
s.push(a.pop(),
// Rejoin the characters, alternating from `a` to `s`, representing the rotated matrix, and map them back to their string representation
a.map((v,i) => r[parseInt(v + s[i],2)]).join``
)
JavaScript (ES6), (削除) 237 (削除ここまで) (削除) 210 (削除ここまで) (削除) 204 (削除ここまで) 188 bytes
Credit to @Downgoat for saving 16 bytes in the latest revision
s=>(r=" .':",a=[],s=[...s].map(c=>('00'+r.indexOf(c).toString(2)).slice(-2)).map(n=>(a.push(n[0]),n[1])),a.splice(0,0,s.shift()),s.push(a.pop()),a.map((v,i)=>r[parseInt(v+s[i],2)]).join``)
###Pretty Print & Explanation
s => (
r = " .':", // Map of characters to their (numerical) binary representations (e.g. r[0b10] = "'")
a = [], // extra array needed
// Spread `s` into an array, mapping each character to a `0`-padded string representation of a binary number
s = [...s].map(c => ('00' + r.indexOf(c).toString(2)).slice(-2))
// Put the first character into `a` and keep the second character for `s`
.map(n => (a.push(n[0]),n[1])),
// Put the first character of `s` in the first index of `a`
a.splice(0,0,s.shift()),
// Append the last character of `a` to `s`
s.push(a.pop(),
// Rejoin the characters, alternating from `a` to `s`, representing the rotated matrix, and map them back to their string representation
a.map((v,i) => r[parseInt(v + s[i],2)]).join``
)
JavaScript (ES6), (削除) 237 (削除ここまで) (削除) 210 (削除ここまで) (削除) 204 (削除ここまで) (削除) 188 (削除ここまで) 182 bytes
Credit to @Downgoat for saving 16 bytes in the 188-byte revision
Update: I had a brainwave and reduced the first operation on s to a single map call instead of two separate ones
s=>(r=" .':",a=[],s=[...s].map(c=>(t=('00'+r.indexOf(c).toString(2)).slice(-2),a.push(t[0]),t[1])),a.splice(0,0,s.shift()),s.push(a.pop()),a.map((v,i)=>r[parseInt(v+s[i],2)]).join``)
###Pretty Print & Explanation
s => (
r = " .':", // Map of characters to their (numerical) binary representations (e.g. r[0b10] = "'")
a = [], // extra array needed
// Spread `s` into an array s = [...s].map(c => (
// Map each character to a `0`-padded string representation of a binary number, storing in `t`
t = ('00' + r.indexOf(c).toString(2)).slice(-2)),
// Put the first character of `t` into `a` a.push(t[0]),
// Keep the second character for `s`
t[1]
)),
// Put the first character of `s` in the first index of `a`
a.splice(0,0,s.shift()),
// Append the last character of `a` to `s`
s.push(a.pop(),
// Rejoin the characters, alternating from `a` to `s`, representing the rotated matrix, and map them back to their string representation
a.map((v,i) => r[parseInt(v + s[i],2)]).join``
)
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still attempting syntax highlighting, not sure I'm going to get it to work.
RevanProdigalKnight
- 111
- 4
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attempted (and failed) to add language syntax highlighting - doesn't work here?
RevanProdigalKnight
- 111
- 4
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