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[Extended] Dyalog APL, 4 bytes* *10 bytes, 4 unique '÷'÷⍥≢'÷÷' This computes the length of the string '÷' divided by the length of the string '÷÷' Alternate solution (extended only) (6 unique...
Answer
#3: Post edited
by
RubenVerg
·
2023-08-04T09:09:31Z (over 1 year ago)
-1 unique
# [Extended] Dyalog APL, 5 bytes**12 bytes, 5 unique- ```apl
(≢'÷')÷≢'÷÷'- ```
- This computes the length of the string `'÷'` divided by the length of the string `'÷÷'`
- ## Alternate solution (extended only) (6 unique, 6 bytes)
- ```apl
- ⊢÷+⍨×∞
- ```
- This computes
- \[
- \left(\lambda x.\frac{x}{x+x}\right) \mathrm{sign}\infty
- \]
- # [Extended] Dyalog APL, 4 bytes*
- *10 bytes, 4 unique
- ```apl
- '÷'÷⍥≢'÷÷'
- ```
- This computes the length of the string `'÷'` divided by the length of the string `'÷÷'`
- ## Alternate solution (extended only) (6 unique, 6 bytes)
- ```apl
- ⊢÷+⍨×∞
- ```
- This computes
- \[
- \left(\lambda x.\frac{x}{x+x}\right) \mathrm{sign}\infty
- \]
#2: Post edited
by
RubenVerg
·
2023-07-07T13:18:07Z (over 1 year ago)
does the leaderboard like this more?
# [Extended] Dyalog APL, 5 unique (12 bytes)- ```apl
- (≢'÷')÷≢'÷÷'
- ```
- This computes the length of the string `'÷'` divided by the length of the string `'÷÷'`
- ## Alternate solution (extended only) (6 unique, 6 bytes)
- ```apl
- ⊢÷+⍨×∞
- ```
- This computes
- \[
- \left(\lambda x.\frac{x}{x+x}\right) \mathrm{sign}\infty
- \]
- # [Extended] Dyalog APL, 5 bytes*
- *12 bytes, 5 unique
- ```apl
- (≢'÷')÷≢'÷÷'
- ```
- This computes the length of the string `'÷'` divided by the length of the string `'÷÷'`
- ## Alternate solution (extended only) (6 unique, 6 bytes)
- ```apl
- ⊢÷+⍨×∞
- ```
- This computes
- \[
- \left(\lambda x.\frac{x}{x+x}\right) \mathrm{sign}\infty
- \]
#1: Initial revision
by
RubenVerg
·
2023-07-07T13:16:38Z (over 1 year ago)
# [Extended] Dyalog APL, 5 unique (12 bytes)
```apl
(≢'÷')÷≢'÷÷'
```
This computes the length of the string `'÷'` divided by the length of the string `'÷÷'`
## Alternate solution (extended only) (6 unique, 6 bytes)
```apl
⊢÷+⍨×∞
```
This computes
\[
\left(\lambda x.\frac{x}{x+x}\right) \mathrm{sign}\infty
\]