You can use a JSON reviver to build an ID to object map and to identify object references. Then, after reviving, you can rewrite the properties that are references to point to the referent object. // A heuristic to identify references that appear as string property values. function looksLikeUid(s) {...

graph,ocaml,self-reference,nfa,cyclic

You can create data structures with cycles without explicit references, using lazy types or functions. Indeed, both of them hides some form of mutability. Here is an example of a simplest lazy structure, that is more complex than a list type 'a tree = 'a tr Lazy.t and 'a tr...

java,compiler-errors,observer-pattern,cyclic

Use the publish-subscribe pattern where each of your classes is both a publisher and a subscriber. interface Publisher { void addSubscriber(Subscriber sub); } interface Subscriber { void handle (Object event); } class A implements Publisher, Subscriber {...} class B implements Publisher, Subscriber {...} If you don't mind using some JavaFX...

math,sequence,discrete-mathematics,cyclic

Another way to approach this problem. Note that F(n) = F(n - 1) - F(n - 2) is the same as F(n) - F(n - 1) + F(n - 2) = 0 which makes it a linear difference equation. Such equations have fundamental solutions a^n where a is a root...

You can use the Stringbuilder class to append the String from index 0. http://docs.oracle.com/javase/6/docs/api/java/lang/StringBuilder.html...

java,string,boolean,string-length,cyclic

You don't need to do that loop. You can simply test cyclic strings like this: public Boolean IsCyclicWord(String s1, String s2) { return s1.length() == s2.length() && (s1 + s1).contains(s2); } How this works: Consider your example: s1 = "picture"; s2 = "icturep"; For cyclic string, first character of one...