Would this string work as string?

Question

Write a program that takes a single line string that you can assume will only contain the characters /_‾. (That's forward and backward slash, underline and overline. You can use ~ in place of overline if you need since overline is not convenient ASCII.)

For example, one possible input is:

__/‾‾/_/‾

Your program needs to output a truthy or falsy value depending on whether the left edge of the string is "connected", so to speak, to the right edge of the string via the lines of the characters. So, if the kerning was a bit less, there would be a solid black (albeit kinky) line all the way from the left edge to the right, like an unbroken piece of string or twine.

The output for the above example would be true because the edges are connected:

example path

To be clear on the connections:

/ connects on its bottom left and top right

connects on its top left and bottom right

_ connects on its bottom left and bottom right

‾ (or ~) connects on its top left and top right

Also:

It doesn't matter whether the edges of the string started on the top or the bottom, it only matters that they connect horizontally through the whole length of the string.

You can assume the input string is non-empty, and of course just one line.

Here are some more examples followed by 1 (truthy) if they are connected or 0 (falsy) if not:

__/‾‾/_/‾

1



_

1





1



/

1



‾

1



___

1



/

1



//

1



//

1



‾‾‾

1



\

0



‾‾

1



_‾

0



‾_

0



_____/

1



/\/\___

0



/__/‾‾

1



______/_____

1



‾‾‾‾‾‾\_____

0



‾‾‾‾‾‾______

1



_____/‾‾‾‾‾

1



___/‾‾‾___/‾‾‾

1



_/__

0



_/_

1



////////////

1



____________________

1



‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

1



‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾/

0



‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

1



/‾/‾___/|_|/__/‾‾

0

The shortest code is the winner.

Are the characters specified in your challenge the only ones that will appear in the string? — 1 hour ago

histocrathistocrat 19.1k43172 · Accepted Answer · 2019-03-18 15:57:45Z

6

Ruby -n, 30 bytes

p !/[_\][\‾]|[/‾][_/]/

Try it online!

Reduces all of the string-breaking sequences to two cases using Regex character classes.

answered 1 hour ago

histocrat

19.1k43172

2

$begingroup$

You can save 4 bytes by using ~ instead of ‾. I'm not sure if it matters for this challenge, since the character count is the same.

$endgroup$

– iamnotmaynard

36 mins ago

add a comment |

score 5 · Accepted Answer · 2019-03-18 15:49:51Z

5

JavaScript (ES6), 45 bytes

The naive way.

s=>!///|\\|_~|~_|~/|_\|/_|\~/.test(s)

Try it online!

edited 1 hour ago

answered 1 hour ago

Arnauld

78.9k795327

$begingroup$

So this is checking all the invalid pairings, making sure they don't exist in the string? Smart.

$endgroup$

– Discrete Games

1 hour ago

$begingroup$

@DiscreteGames Yes, exactly. (Except that I forgot 2 of them. Now fixed.)

$endgroup$

– Arnauld

1 hour ago

add a comment |

Henry THenry T 1716 · Accepted Answer · 2019-03-18 15:41:01Z

3

Python 3, 126 bytes

lambda s,d={'‾':'‾','_':'/_','/':'‾','\':'/_'}:len(s)<2or all([s[i+1] in d[s[i]]for i in range(len(s)-1)if s[i]in d])

answered 1 hour ago

Henry T

1716

add a comment |

Jonathan FrechJonathan Frech 6,41311040 · Accepted Answer · 2019-03-18 16:04:09Z

3

C (gcc), 93 bytes

w,o,r;k(char*_){for(r=0;w=*_,o=*++_;)r|=w-126&&w>47?w-95&&w-92?0:o>47&&o-95:o-92&&o<126;_=r;}

Try it online!

answered 1 hour ago

Jonathan Frech

6,41311040

add a comment |

score 3 · Accepted Answer · 2019-03-18 16:34:11Z

05AB1E, 29 bytes

„_~SD2×s:Çü-т+•6_üê{â†•ƵΔвåO_

It sounded shorter in my head.. Will try to golf it down from here.

Try it online or verify all test cases.

Explanation:"

„_~S                          # Push the characters ["_","~"]

    D2×                       # Duplicate it, and increase each to size 2: ["__","~~"]

       s:                     # Swap and replace all "__" with "_" and all "~~" with "~"

                              #  in the (implicit) input-string

         Ç                    # Convert the remaining characters to unicode values

          ü-                  # Calculate the difference between each pair

            т+                # Add 100 to each

              •6_üê{â†•       # Push compressed integer 1781179816800959

                       ƵΔ     # Push compressed integer 180

                         в    # Convert the larger integer to Base-180 as list: 

                              #  [52,66,69,100,103,131,179]

                          å   # Check for each if it's in the difference-list

                              # (1 if present; 0 if not)

                           O  # Sum the truthy values

                            _ # Check if this sum is exactly 0 (1 if 0; 0 otherwise)

                              # (and output this result implicitly)

See this 05AB1E tip of mine (sections How to comrpess large integers? and How to compress integer lists?) to understand why •6_üê{â†• is 1781179816800959, ƵΔ is 180 and •6_üê{â†•ƵΔв is [52,66,69,100,103,131,179].

Additional explanation:

There are 16 ($2^4$) possible pairs of characters we have to verify. If we convert each character to its unicode value, and calculate the differences, we would get these differences. Because compressed integer lists in 05AB1E has to have positive integers only, I add 100 to each. The invalid pairs and their corresponding values are then: ["/_", 52]; ["~", 66], ["_~", 69], ["//", 100], ["\", 100], ["_", 103], ["~_", 131], ["~/", 179], which is why I have the compressed integer list in my code containing these values.

Since __ and ~~ will just like // and \ result in 0 (or 100 after I add 100), I first remove any adjacent duplicates of ~ and _ in the input-string, before calculating and verifying the pair-differences.

score 2 · Accepted Answer · 2019-03-18 16:48:15Z

Python 3, 80 78 bytes

I don't realy do many python code golfs but I thought I could give it a try

-2 bytes: realised not(any()) is the same as all(not()) and could move the not into the r-string

def f(x):*l,=map(r'_/~'.find,x);return 1-any((i^j//2)%2for i,j in zip(l,l[1:]))

Try it online!

Python 3.8 (pre-release), 71 bytes

I wanted to try out the new := expression assignment

lambda x:all((i^j//2)%2for i,j in zip(l:=[*map(r'~_/'.find,x)],l[1:]))

Try it online!

score 1 · Accepted Answer · 2019-03-18 17:00:33Z

Jelly, 14 bytes

Supports (and expects) the ‾ character in the input string. Returns $0$ or $1$.

O*Ɲ×12%29%15ḂẠ

Try it online!

O*Ɲ×12%29%15ḂẠ  - main link, taking a string          e.g. "_/"

O               - get ASCII codes                     -->  [92, 95, 47]

 *Ɲ             - exponentiation on all pairs         -->  [92**95, 95**97]

   ×12          - multiply by 12                      -->  [(92**95)*12, (95**97)*12]

      %29       - modulo 29                           -->  [11, 24]

         %15    - modulo 15                           -->  [11, 9]

            Ḃ   - least significant bit (i.e. parity) -->  [1, 1]

             Ạ  - all values equal to 1?              -->  1

ElcanElcan 32115 · Accepted Answer · 2019-03-18 16:48:14Z

0

Dart, 94 bytes

f(s)=>!(r'//,\,~/,_,~_,_~,/_,~'.split(',').map((t)=>s.contains(t)).fold(false,(p,e)=>p|e));

Try it online!

answered 21 mins ago

Elcan

32115

add a comment |

histocrathistocrat 19.1k43172 · Accepted Answer · 2019-03-18 15:57:45Z

6

Ruby -n, 30 bytes

p !/[_\][\‾]|[/‾][_/]/

Try it online!

Reduces all of the string-breaking sequences to two cases using Regex character classes.

answered 1 hour ago

histocrat

19.1k43172

2

$begingroup$

You can save 4 bytes by using ~ instead of ‾. I'm not sure if it matters for this challenge, since the character count is the same.

$endgroup$

– iamnotmaynard

36 mins ago

add a comment |

score 5 · Accepted Answer · 2019-03-18 15:49:51Z

5

JavaScript (ES6), 45 bytes

The naive way.

s=>!///|\\|_~|~_|~/|_\|/_|\~/.test(s)

Try it online!

edited 1 hour ago

answered 1 hour ago

Arnauld

78.9k795327

$begingroup$

So this is checking all the invalid pairings, making sure they don't exist in the string? Smart.

$endgroup$

– Discrete Games

1 hour ago

$begingroup$

@DiscreteGames Yes, exactly. (Except that I forgot 2 of them. Now fixed.)

$endgroup$

– Arnauld

1 hour ago

add a comment |

Henry THenry T 1716 · Accepted Answer · 2019-03-18 15:41:01Z

3

Python 3, 126 bytes

lambda s,d={'‾':'‾','_':'/_','/':'‾','\':'/_'}:len(s)<2or all([s[i+1] in d[s[i]]for i in range(len(s)-1)if s[i]in d])

answered 1 hour ago

Henry T

1716

add a comment |

Jonathan FrechJonathan Frech 6,41311040 · Accepted Answer · 2019-03-18 16:04:09Z

3

C (gcc), 93 bytes

w,o,r;k(char*_){for(r=0;w=*_,o=*++_;)r|=w-126&&w>47?w-95&&w-92?0:o>47&&o-95:o-92&&o<126;_=r;}

Try it online!

answered 1 hour ago

Jonathan Frech

6,41311040

add a comment |

score 3 · Accepted Answer · 2019-03-18 16:34:11Z

05AB1E, 29 bytes

„_~SD2×s:Çü-т+•6_üê{â†•ƵΔвåO_

It sounded shorter in my head.. Will try to golf it down from here.

Try it online or verify all test cases.

Explanation:"

„_~S                          # Push the characters ["_","~"]

    D2×                       # Duplicate it, and increase each to size 2: ["__","~~"]

       s:                     # Swap and replace all "__" with "_" and all "~~" with "~"

                              #  in the (implicit) input-string

         Ç                    # Convert the remaining characters to unicode values

          ü-                  # Calculate the difference between each pair

            т+                # Add 100 to each

              •6_üê{â†•       # Push compressed integer 1781179816800959

                       ƵΔ     # Push compressed integer 180

                         в    # Convert the larger integer to Base-180 as list: 

                              #  [52,66,69,100,103,131,179]

                          å   # Check for each if it's in the difference-list

                              # (1 if present; 0 if not)

                           O  # Sum the truthy values

                            _ # Check if this sum is exactly 0 (1 if 0; 0 otherwise)

                              # (and output this result implicitly)

See this 05AB1E tip of mine (sections How to comrpess large integers? and How to compress integer lists?) to understand why •6_üê{â†• is 1781179816800959, ƵΔ is 180 and •6_üê{â†•ƵΔв is [52,66,69,100,103,131,179].

Additional explanation:

There are 16 ($2^4$) possible pairs of characters we have to verify. If we convert each character to its unicode value, and calculate the differences, we would get these differences. Because compressed integer lists in 05AB1E has to have positive integers only, I add 100 to each. The invalid pairs and their corresponding values are then: ["/_", 52]; ["~", 66], ["_~", 69], ["//", 100], ["\", 100], ["_", 103], ["~_", 131], ["~/", 179], which is why I have the compressed integer list in my code containing these values.

Since __ and ~~ will just like // and \ result in 0 (or 100 after I add 100), I first remove any adjacent duplicates of ~ and _ in the input-string, before calculating and verifying the pair-differences.

score 2 · Accepted Answer · 2019-03-18 16:48:15Z

Python 3, 80 78 bytes

I don't realy do many python code golfs but I thought I could give it a try

-2 bytes: realised not(any()) is the same as all(not()) and could move the not into the r-string

def f(x):*l,=map(r'_/~'.find,x);return 1-any((i^j//2)%2for i,j in zip(l,l[1:]))

Try it online!

Python 3.8 (pre-release), 71 bytes

I wanted to try out the new := expression assignment

lambda x:all((i^j//2)%2for i,j in zip(l:=[*map(r'~_/'.find,x)],l[1:]))

Try it online!

score 1 · Accepted Answer · 2019-03-18 17:00:33Z

Jelly, 14 bytes

Supports (and expects) the ‾ character in the input string. Returns $0$ or $1$.

O*Ɲ×12%29%15ḂẠ

Try it online!

O*Ɲ×12%29%15ḂẠ  - main link, taking a string          e.g. "_/"

O               - get ASCII codes                     -->  [92, 95, 47]

 *Ɲ             - exponentiation on all pairs         -->  [92**95, 95**97]

   ×12          - multiply by 12                      -->  [(92**95)*12, (95**97)*12]

      %29       - modulo 29                           -->  [11, 24]

         %15    - modulo 15                           -->  [11, 9]

            Ḃ   - least significant bit (i.e. parity) -->  [1, 1]

             Ạ  - all values equal to 1?              -->  1

ElcanElcan 32115 · Accepted Answer · 2019-03-18 16:48:14Z

0

Dart, 94 bytes

f(s)=>!(r'//,\,~/,_,~_,_~,/_,~'.split(',').map((t)=>s.contains(t)).fold(false,(p,e)=>p|e));

Try it online!

answered 21 mins ago

Elcan

32115

add a comment |

Label

Cari Blog Ini

Arsip Blog

Laporkan Penyalahgunaan

Mengenai Saya

Senin, 18 Maret 2019

Would this string work as string?

8 Answers 8

Ruby -n, 30 bytes

JavaScript (ES6), 45 bytes

Python 3, 126 bytes

C (gcc), 93 bytes

05AB1E, 29 bytes

Python 3, 80 78 bytes

Python 3.8 (pre-release), 71 bytes

Jelly, 14 bytes

Dart, 94 bytes

Your Answer

Sign up or log in

Post as a guest

Post as a guest

8 Answers 8

8 Answers 8

Ruby -n, 30 bytes

Ruby -n, 30 bytes

Ruby -n, 30 bytes

Ruby -n, 30 bytes

JavaScript (ES6), 45 bytes

JavaScript (ES6), 45 bytes

JavaScript (ES6), 45 bytes

JavaScript (ES6), 45 bytes

Python 3, 126 bytes

Python 3, 126 bytes

Python 3, 126 bytes

Python 3, 126 bytes

C (gcc), 93 bytes

C (gcc), 93 bytes

C (gcc), 93 bytes

C (gcc), 93 bytes

05AB1E, 29 bytes

05AB1E, 29 bytes

05AB1E, 29 bytes

05AB1E, 29 bytes

Python 3, 80 78 bytes

Python 3.8 (pre-release), 71 bytes

Python 3, 80 78 bytes

Python 3.8 (pre-release), 71 bytes

Python 3, 80 78 bytes

Python 3.8 (pre-release), 71 bytes

Python 3, 80 78 bytes

Python 3.8 (pre-release), 71 bytes

Jelly, 14 bytes

Jelly, 14 bytes

Jelly, 14 bytes

Jelly, 14 bytes

Dart, 94 bytes

Dart, 94 bytes

Dart, 94 bytes

Dart, 94 bytes

Sign up or log in

Post as a guest

Post as a guest

Sign up or log in

Post as a guest

Sign up or log in

Post as a guest

Sign up or log in

Post as a guest

0 komentar:

Posting Komentar

Categories

Popular Posts

8 Answers

8

8 Answers

8

8 Answers

8