Background

MQTT (Message Queuing Telemetry Transport) is an ISO standard publish-subscribe-based messaging protocol (Wikipedia).

Each message has a topic, such as the following examples:

• myhome/groundfloor/livingroom/temperature


• USA/California/San Francisco/Silicon Valley


• 5ff4a2ce-e485-40f4-826c-b1a5d81be9b6/status


• Germany/Bavaria/car/2382340923453/latitude

MQTT clients may subscribe to message topics using wildcards:

• Single level: +
  


• All levels onward: #
  

For example, the subscription myhome/groundfloor/+/temperature would produce these results (non-conformances in bold):

✅ myhome/groundfloor/livingroom/temperature

✅ myhome/groundfloor/kitchen/temperature

❌ myhome/groundfloor/livingroom/brightness

❌ myhome/firstfloor/livingroom/temperature

❌ garage/groundfloor/fridge/temperature

Whereas the subscription +/groundfloor/# would produce these results:

✅ myhome/groundfloor/livingroom/temperature

✅ myhome/groundfloor/kitchen/brightness

✅ garage/groundfloor/fridge/temperature/more/specific/fields

❌ myhome/firstfloor/livingroom/temperature

❌ myhome/basement/corner/temperature

More info here.

The Task

Implement a function/program accepting two strings and returning a boolean. The first string is the subject topic, the second is the criteria topic. The criteria topic uses the subscription syntax detailed above. The function is truthy when the subject matches the criteria.

Rules for this task:

• Topics are ASCII


• There are no criteria fields beyond the # wildcard


• Wildcards do not appear in subject topics


• Number of subject fields >= number of criteria fields


• There are no 0-character fields nor leading or tailing forward slashes

Python 2, 85 84 80 92 89 bytes

lambda s,c:all(x in('+','#',y)for x,y in zip(c.split('/')+[0]*-c.find('#'),s.split('/')))

Try it online!

Thanks to Jonathan Allan and Value Ink for pointing out bugs.

Jelly, 20 bytes

ṣ€”/ZṖF”#eƊ¿œiÐḟ”+ZE

A monadic Link accepting a list of lists of characters, [topic, pattern], which returns 1 or 0 for match or no-match respectively.

Try it online! Or see a test-suite.

How?

ṣ€”/ZṖF”#eƊ¿œiÐḟ”+ZE - Link: list of lists of characters, [topic, pattern]
 € - for each:
ṣ - split at occurrences of:
 ”/ - '/' character
 Z - transpose (any excess of topic is kept)
 ¿ - while...
 Ɗ - ...condition: last three links as a monad:
 ”# - '#' character
 e - exists in:
 F - flatten
 Ṗ - ...do: pop the tail off
 Ðḟ - filter discard those for which:
 œi - first multi-dimensional index of: ([] if not found, which is falsey)
 ”+ - '+' character
 Z - transpose
 E - all equal?

Python 3, 72 bytes

lambda a,b:bool(re.match(b.translate(43:"[^/]+",35:".+"),a))
import re

Try it online!

This problem can be trivially simplified to a regex match, though another more interesting method may produce better results.

EDIT I came up with a 107-byte solution not using regex. I don't know if it can get shorter than 72 or maybe I'm just not seeing to correct approach to this. Just the split-zip structure seems to be too large though. Try It Online!

Ruby, 65 bytes

Regex solution. I added Regex.escape in case a criteria name just so happens to be something like com.java/string[]/n or something silly that would have regex pieces.

->s,cs=~/^#Regexp.escape(c).sub('#','.*').gsub'+','[^/]*'$/

Try it online!

Non-regex solution, 77 bytes

Uses a nice simple split, zip, and match technique. I developed this one first before realizing that even with Regex.escape the regex solution would've been shorter anyways.

->s,cs.split(?/).zip(c.split ?/).all?i,j

Try it online!