You’re being hit by filter‘s laziness. Change (filter …) to (doall (filter …)) in your recur form and the problem should go away. A more in-depth explanation: The call to filter returns a lazy seq, which materialises actual elements of the filtered seq as required. As written, your code stacks filter upon filter upon filter…, …
This question and its answers appear to date back to Ruby 1.8.x, which used the C stack. Ruby 1.9.x and later use a VM that has its own stack. In Ruby 2.0.0 and later, the size of the VM stack can be controlled via the RUBY_THREAD_VM_STACK_SIZE environment variable.
There is an interesting way to do it that does rely neither on rebinding nor the behavior of def. The main trick is to go around the limitations of recursion by passing a function as an argument to itself: (defn make-fibo [y] (let [fib (fn [mem-fib x] (let [fib (fn [a] (mem-fib mem-fib a))] (if …
You can’t access Function2 inside of it if it is in the line where you declare it. The reason is that you’re not referring to a function but to a variable (whose type is a function) and it will be accessible only after the declaration. Quoting from Spec: Declarations and scope: The scope of a …
The problem was not with the recursion. Your function did call itself recursively, it just didn’t yield values outside. When you call helper(), you get an iterator as a return value, but you wanted the iterated values of that iterator to be yielded. If you want to yield recursively, you need to yield *. Try …
First, to your original question, just be aware that if you’re working with very long strings, you don’t want to be making exact copies minus a single letter each time you make a function call. So you should favor using indexes or verify that your language of choice isn’t making copies behind the scenes. Second, …
You are looking for the set intersection of all the sublists, and the data type you should use for set operations is a set: result = set(p[0]) for s in p[1:]: result.intersection_update(s) print result
Think about the difference between the auto version and the fully specified type version. The auto keyword infers its type from whatever it’s initialized with, but what you’re initializing it with needs to know what its type is (in this case, the lambda closure needs to know the types it’s capturing). Something of a chicken-and-egg …
You can use recursion to explore every “branch” that emerges when every possible letter containing the proper initial character is added to a running list: words = [‘giraffe’, ‘elephant’, ‘ant’, ‘tiger’, ‘racoon’, ‘cat’, ‘hedgehog’, ‘mouse’] def get_results(_start, _current, _seen): if all(c in _seen for c in words if c[0] == _start[-1]): yield _current else: for …
There’s nothing wrong with direct, “ad-hoc” solution. It can be clean enough too. Just notice that spiral is built from segments. And you can get next segment from current one rotating it by 90 degrees. And each two rotations, length of segment grows by 1. edit Illustration, those segments numbered … 11 10 7 7 …