@Sherlock9 I wouldn't worry too much about understanding dislike of golfing languages. It's a common response which most people move away from after seeing that they aren't a short cut to victory and still require creativity. Those who don't move away from it and don't express a good reason why probably don't have one.
@JesterTran The result is just showing the order in which the nodes will be visited. First A, then all the nodes directly connected to A, then all the nodes 2 steps away, and so on
"Breadth First" means that you look at all the close nodes before you move on to any of the more distant nodes. This contrasts with "Depth First" where you follow a single path all the way to the end before you try the next path.
@JesterTran The result isn't the answer, it is just showing the order in which you will search. Perhaps "result" isn't the best name for it...
If you check all the nodes which are 1 step away, then all the nodes which are 2 steps away, then while checking all the nodes which are 3 steps away you find Z, then you know Z is 3 steps away. There can't be a shorter route because otherwise you would have found it while checking the 1 steps and 2 steps
Matthias Goergens has a 25,604-character (down from the original 63,993-character) regex to match numbers divisible by 7, but that includes a lot of fluff: redundant parentheses, distribution (xx|xy|yx|yy rather than [xy]{2}) and other issues, though I'm sure a fresh start would be helpful in sav...
Which of the below would count as a duplicate of this question? A. regex for divisibility by 3 B. regex for divisibility by 7 in binary C. regex for divisibility by 3 in binary
Result is written as one long list which isn't very informative. If you write it down separately: all the nodes 1 step away, then 2 steps away,... then it will be clearer what it means
@KennyLau If answers to the original cannot be trivially converted to answers to the new one then it isn't a duplicate, but may or may not be popular depending on how fresh the task seems
most regex flavours aren't really regular expressions in the theoretical sense, because many regex features can match languages which aren't regular. in fact, something as simple as a backreference means that the regex is no longer regular, so there's no corresponding DFA.
.NET's balancing groups are a lot more powerful than that in fact.
@Sherlock9 Still not following. You seem to be saying that both directions are allowed, but then also that search would be limited to one sequence of forward transformation followed by a sequence of backward transformations? I'm not sure which moves are allowed when yet.
@KennyLau I can't tell. I don't know how divisibility by 3 or 7 works in binary, but divisibility by 3 in decimal would probably be a lot easier than divisibility by 7.
@Sherlock9 it's not clear that every optimal path has to go via 1 though (and seems unlikely). I'd assume that the shortest paths will wiggle up and down on larger values to take shortcuts
@KennyLau it's so short that I don't think they'll be able to save anything
I finished coding Brachylog's transpiler in Prolog, this was way less challenging than I thought it would, and now Brachylog's grammar is much more powerful
Lesson learned: don't use Java to hack an esolang :p
it's also a bit like asking "how would you solve this with character classes". when character classes are useless in shortening it, why would you use them?
seems a bit like standard BFS path-finding challenge, of which we've had quite a lot on various types of graphs
it's an interesting problem, in that I'd be interested in the shortest paths between various numbers, but I'm not sure it adds a lot to PPCG as a challenge
maybe fastest-code? I wonder if there's any interesting shortcuts one can take?
Given three mutually tangent circles, we can always find two more circles which are tangent to all three of those. These two are called Apollonian circles. Note that one of the Apollonian circles might actually be around the three initial circles.
Starting from three tangent circles, we can crea...
Some years ago I created a regexp for divisibility by 7 for binary numbers:
/^0*((1((0(01|1(001*0)*11)*(00|1(001*0)*10))|1(01*0(1(10)*11)*0)*01*0(1(10)*11)*1(10)*0)*(1(01*0(1(10)*11)*0)*1|0(01|1(001*0)*11)*1(001*0)*01))0*)+$/
I never found a good way to minimize regex' automatically, so this o...
honestly, I prefer Riker's upvoted solution. we always score answers by the bytes in string representation of the source code. it would only be consistent to score Box-256 by the length of the shortest string which you can feed into its "load" feature.
@Quill Could you write a blog post on when vars take on the exact same value in JS, and which value types do this? E.g. a = b = [] means any changes done to one is done to the other, but a = b = 0 has a and b independent.
Produce only the name of a programming language using only the facilities of the language itself. The name of the language can only be invoked indirectly, i.e., from non custom environment variables, built-in functions or methods or the interpreter/compiler of the language. Output should be the n...
.NET regex, 119 118 105 bytes
^(?>(?=[1468](?<4>)|)(?=[2569](?<4>){2}|)([3-6]()|\d)((?<-2>)(){3}|){7}((?<-4>){7}|(?<2-4>)|){9})+$(?!\2)
111 characters disallowing initial 0s:
^(?!0.)(?>(?=[1468](?<4>)|)(?=[2569](?<4>){2}|)([3-6]()|\d)((?<-2>)(){3}|){7}((?<-4>){7}|(?<2-4>)|){9})+$(?!\2)
113 ...
Some years ago I created a regexp for divisibility by 7 for binary numbers:
/^0*((1((0(01|1(001*0)*11)*(00|1(001*0)*10))|1(01*0(1(10)*11)*0)*01*0(1(10)*11)*1(10)*0)*(1(01*0(1(10)*11)*0)*1|0(01|1(001*0)*11)*1(001*0)*01))0*)+$/
I never found a good way to minimize regex' automatically, so this o...
