I'm Haruo. My pleasure is solving SPOJ in Common Lisp(CLISP). Today I solved Classical/Balk! but in SBCL not CLISP. My CLISP submit failed due to runtime error (NZEC).
I hope my code becomes more sophisticated. Today's problem is just a chance. Please the following my code and tell me your refactoring strategy. I trust you.
https://github.com/haruo-wakakusa/SPOJ-ClispAnswers/blob/0978813be14b536bc3402f8238f9336a54a04346/20040508_adrian_b.lisp
Haruo
Take for example get-x-depth-for-yz-grid.
(defun get-x-depth-for-yz-grid (planes//yz-plane grid)
(let ((planes (get-planes-including-yz-grid-in planes//yz-plane grid)))
(unless (evenp (length planes))
(error "error in get-x-depth-for-yz-grid"))
(sort planes (lambda (p1 p2) (< (caar p1) (caar p2))))
(do* ((rest planes (cddr rest)) (res 0))
((null rest) res)
(incf res (- (caar (second rest)) (caar (first rest)))))))
style -> ERROR can be replaced by ASSERT.
possible bug -> SORT is possibly destructive -> make sure you have a fresh list consed!. If it is already fresh allocated by get-planes-including-yz-grid-in, then we don't need that.
bug -> SORT returns a sorted list. The sorted list is possibly not a side-effect. -> use the returned value
style -> DO replaced with LOOP.
style -> meaning of CAAR unclear. Find better naming or use other data structures.
(defun get-x-depth-for-yz-grid (planes//yz-plane grid)
(let ((planes (get-planes-including-yz-grid-in planes//yz-plane grid)))
(assert (evenp (length planes)) (planes)
"error in get-x-depth-for-yz-grid")
(setf planes (sort (copy-list planes) #'< :key #'caar))
(loop for (p1 p2) on planes by #'cddr
sum (- (caar p2) (caar p1)))))
Some documentation makes a bigger improvement than refactoring.
Your -> macro will confuse sbcl’s type inference. You should have (-> x) expand into x, and (-> x y...) into (let (($ x)) (-> y...))
You should learn to use loop and use it in more places. dolist with extra mutation is not great
In a lot of places you should use destructuring-bind instead of eg (rest (rest )). You’re also inconsistent as sometimes you’d write (cddr...) for that instead.
Your block* suffers from many problems:
It uses (let (foo) (setf foo...)) which trips up sbcl type inference.
The name block* implies that the various bindings are scoped in a way that they may refer to those previously defined things but actually all initial value may refer to any variable or function name and if that variable has not been initialised then it evaluates to nil.
The style of defining lots of functions inside another function when they can be outside is more typical of scheme (which has syntax for it) than Common Lisp.
get-x-y-and-z-ranges really needs to use loop. I think it’s wrong too: the lists are different lengths.
You need to define some accessor functions instead of using first, etc. Maybe even a struct(!)
(sort foo) might destroy foo. You need to do (setf foo (sort foo)).
There’s basically no reason to use do. Use loop.
You should probably use :key in a few places.
You write defvar but I think you mean defparameter
*t* is a stupid name
Most names are bad and don’t seem to tell me what is going on.
I may be an idiot but I can’t tell at all what your program is doing. It could probably do with a lot of work
I am able to set a default argument and do a regular recursion with it, but for some reason I cannot do with recur for tail optimization... I keep getting an java.lang.UnsupportedOperationException: nth not supported on this type: Long error.
For example, for a Tail Call Factorial, here is what works, but isn't optimized for tail call recursion and will fail for large recursion stacks.
(defn foo [n & [optional]]
(if (= n 0) (or optional 1)
(foo (dec n) (*' (or optional 1) n))))
And I call this by (foo 3)
And when I try this to get TCO, I get the unsupported operation error...
(defn foo [n & [optional]]
(if (= n 0) (or optional 1)
(recur (dec n) (*' (or optional 1) n))))
And I call this one the same way (foo 3)
Why is this difference causing an error? How exactly would I be able to do TCO with optional default arguments?
Thank you!
EDIT:
and when I try to take out the (or optional 1) in the recursion call and make it just optional , i get a null exception error... Which makes sense.
This also does not get fixed when I try to remove the ' from *' in the recursion call
EDIT: I would also prefer to do this without loop as well
It is a known issue:
Recur doesn't re-enter the function, it just goes back to the top (the vararging doesn't happen again) ... recur with a collection and you will be fine.
I personally feel it should either be mentioned in the recur docstring, or at least appear in the doc. Takes a bit of digging to understand what's happening (I had to check Clojure compiler source along with the compiled classes.)
Why is this difference causing an error?
In short, it's trying to destructure a Long, which it can't
Straight foo call
Takes n arguments
Automatically puts everything after the first argument (n) into a seq behind the scenes, which can be destructured
recur call to foo
Takes exactly 2 arguments
First argument: n
Second argument: Something seqable with the rest of the arguments
How exactly would I be able to do TCO with optional default arguments?
Simply wrap the second argument to recur like so:
(defn foo [n & [optional]]
(if (= n 0) (or optional 1)
(recur (dec n) [(*' (or optional 1) n)])))
(foo 3)
;;=> 6
Recommendations
Although he didn't answer your questions, #DanielCompton's recommendation is the way to go to completely avoid the problem in the first place in a clearer and more efficient way
You can give a function multiple different arities. This might be what you're after?
(defn foo
([n]
(foo n 1))
([n optional]
(if (= n 0)
(or optional 1)
(recur (dec n) (*' (or optional 1) n)))))
I don't quite understand why there is an error, but recur wouldn't normally be used in a function with optional arguments.
Edit: after reading the other answer links, I understand the problem now. recur doesn't destructure the rest args like it does when you call the function. If you recur with a collection as the second arg, it will work, but it is probably still better to be explicit with two different arities:
(defn foo [n & [optional]]
(if (= n 0)
(or optional 1)
(recur (dec n) [(*' (or optional 1) n)])))
While working in the repl is there a way to specify the maximum times to recur before the repl will automatically end the evaluation of an expression. As an example, suppose the following function:
(defn looping []
(loop [n 1]
(recur (inc n))))
(looping)
Is there a way to instruct the repl to give up after 100 levels of recursion? Something similar to print-level.
I respectfully hope that I'm not ignoring the spirit of your question, but why not simply use a when expression? It's nice and succinct and wouldn't change the body of your function much at all (1 extra line and a closing paren).
Whilst I don't believe what you want exists, it would be trivial to implement your own:
(def ^:dynamic *recur-limit* 100)
(defn looping []
(loop [n 1]
(when (< n *recur-limit*)
(recur (inc n)))))
Presumably this hasn't been added to the language because it's easy to construct what you need with the existing language primitives; apart from that, if the facility did exist but was 'invisible', it could cause an awful lot of confusion and bugs because code wouldn't always behave in a predictable and referentially transparent manner.
I am brand spanking new to scheme. To the extent that tonight is the first time I've ever played around with it aside from what a powerpoint slide explained to me in class. I have to write a scheme program that takes a user input operator and then performs that operation on the numbers that follow it. (in other words implement my own version of schemes built-in '+', '*', etc operators). The catch is that I have to use recursion. Trying to maneuver my way around this scheme syntax is making it very difficult for me to even figure out where to start.
So, I decided to start with some code that at least recursively sums up values entered by the user (without worrying about recognizing the user inputting an operator and parentheses). I just plain can't figure out how to make it work. Here's what I'm trying:
(define run (lambda (x)
(cond
((eqv? x "a") (display x) )
(else (+ x (run(read))))
)))
(run (read))
The condition to check if x equals "a" was intended to be a way for me to break the recursion right now. In the final version, the input will be in between a set of parenthesis, but I thought I'd cross that bridge when I come to it. Anyways, the code just keeps accepting input, and never stops. So, the condition to display x is never getting called I guess. I know I'm probably doing everything wrong, and I would really appreciate some pointers.
EDIT: OK, I'm at least starting to realize that something isn't making sense. Such as my displaying x won't actually give the sum since I'm not storing the sum in x in any way. However, I still don't understand why the code isn't at least stopping when I enter the letter a.
EDIT 2: I've got it to work but only when the condition is equal to 1. How can I get it to recognize an input of a char or better yet a right parenthesis to get it to end the recursion just like it is with the 1? :
(define run (lambda (x)
(cond
((eq? x 1) 0)
(else (+ x (run(read))))
)))
(run (read))
End-Of-File (EOF) marks the end of input. Scheme has a procedure eof-object? for this. Your code is basically correct:
(define (run x)
(if (eof-object? x)
0
(+ x (run (read)))))
You end the run using Control-D. If for some reason you don't have access to eof-object? just use a special character like X. In that case your test would be (eq? x #\X)
(define (addition-loop)
(let ((answer (read)))
(if (number? answer)
(+ answer (addition-loop))
0)))
This will break as soon as you enter something that isn't a number.
In addition as you can see, you can have a function of zero arguments.
Anyways the problem with your code is when you enter a into read, it's value it actually the same a 'a or (quote a), rather than "a".
(eqv? 'a "a")
;Value: #f
I've solved 45 problems from 4clojure.com and I noticed a recurring problem in the way I try to solve some problems using recursion and accumulators.
I'll try to explain the best I can what I'm doing to end up with fugly solutions hoping that some Clojurers would "get" what I'm not getting.
For example, problem 34 asks to write a function (without using range) taking two integers as arguments and creates a range (without using range). Simply put you do (... 1 7) and you get (1 2 3 4 5 6).
Now this question is not about solving this particular problem.
What if I want to solve this using recursion and an accumulator?
My thought process goes like this:
I need to write a function taking two arguments, I start with (fn [x y] )
I'll need to recurse and I'll need to keep track of a list, I'll use an accumulator, so I write a 2nd function inside the first one taking an additional argument:
(fn
[x y]
((fn g [x y acc] ...)
x
y
'())
(apparently I can't properly format that Clojure code on SO!?)
Here I'm already not sure I'm doing it correctly: the first function must take exactly two integer arguments (not my call) and I'm not sure: if I want to use an accumulator, can I use an accumulator without creating a nested function?
Then I want to conj, but I cannot do:
(conj 0 1)
so I do weird things to make sure I've got a sequence first and I end up with this:
(fn
[x y]
((fn g [x y acc] (if (= x y) y (conj (conj acc (g (inc x) y acc)) x)))
x
y
'()))
But then this produce this:
(1 (2 (3 4)))
Instead of this:
(1 2 3 4)
So I end up doing an additional flatten and it works but it is totally ugly.
I'm beginning to understand a few things and I'm even starting, in some cases, to "think" in a more clojuresque way but I've got a problem writing the solution.
For example here I decided:
to use an accumulator
to recurse by incrementing x until it reaches y
But I end up with the monstrosity above.
There are a lot of way to solve this problem and, once again, it's not what I'm after.
What I'm after is how, after I decided to cons/conj, use an accumulator, and recurse, I can end up with this (not written by me):
#(loop [i %1
acc nil]
(if (<= %2 i)
(reverse acc)
(recur (inc i) (cons i acc))))
Instead of this:
((fn
f
[x y]
(flatten
((fn
g
[x y acc]
(if (= x y) acc (conj (conj acc (g (inc x) y acc)) x)))
x
y
'())))
1
4)
I take it's a start to be able to solve a few problems but I'm a bit disappointed by the ugly solutions I tend to produce...
i think there are a couple of things to learn here.
first, a kind of general rule - recursive functions typically have a natural order, and adding an accumulator reverses that. you can see that because when a "normal" (without accumulator) recursive function runs, it does some work to calculate a value, then recurses to generate the tail of the list, finally ending with an empty list. in contrast, with an accumulator, you start with the empty list and add things to the front - it's growing in the other direction.
so typically, when you add an accumulator, you get a reversed order.
now often this doesn't matter. for example, if you're generating not a sequence but a value that is the repeated application of a commutative operator (like addition or multiplication). then you get the same answer either way.
but in your case, it is going to matter. you're going to get the list backwards:
(defn my-range-0 [lo hi] ; normal recursive solution
(if (= lo hi)
nil
(cons lo (my-range-0 (inc lo) hi))))
(deftest test-my-range-1
(is (= '(0 1 2) (my-range-0 0 3))))
(defn my-range-1 ; with an accumulator
([lo hi] (my-range-1 lo hi nil))
([lo hi acc]
(if (= lo hi)
acc
(recur (inc lo) hi (cons lo acc)))))
(deftest test-my-range-1
(is (= '(2 1 0) (my-range-1 0 3)))) ; oops! backwards!
and often the best you can do to fix this is just reverse that list at the end.
but here there's an alternative - we can actually do the work backwards. instead of incrementing the low limit you can decrement the high limit:
(defn my-range-2
([lo hi] (my-range-2 lo hi nil))
([lo hi acc]
(if (= lo hi)
acc
(let [hi (dec hi)]
(recur lo hi (cons hi acc))))))
(deftest test-my-range-2
(is (= '(0 1 2) (my-range-2 0 3)))) ; back to the original order
[note - there's another way of reversing things below; i didn't structure my argument very well]
second, as you can see in my-range-1 and my-range-2, a nice way of writing a function with an accumulator is as a function with two different sets of arguments. that gives you a very clean (imho) implementation without the need for nested functions.
also you have some more general questions about sequences, conj and the like. here clojure is kind-of messy, but also useful. above i've been giving a very traditional view with cons based lists. but clojure encourages you to use other sequences. and unlike cons lists, vectors grow to the right, not the left. so another way to reverse that result is to use a vector:
(defn my-range-3 ; this looks like my-range-1
([lo hi] (my-range-3 lo hi []))
([lo hi acc]
(if (= lo hi)
acc
(recur (inc lo) hi (conj acc lo)))))
(deftest test-my-range-3 ; except that it works right!
(is (= [0 1 2] (my-range-3 0 3))))
here conj is adding to the right. i didn't use conj in my-range-1, so here it is re-written to be clearer:
(defn my-range-4 ; my-range-1 written using conj instead of cons
([lo hi] (my-range-4 lo hi nil))
([lo hi acc]
(if (= lo hi)
acc
(recur (inc lo) hi (conj acc lo)))))
(deftest test-my-range-4
(is (= '(2 1 0) (my-range-4 0 3))))
note that this code looks very similar to my-range-3 but the result is backwards because we're starting with an empty list, not an empty vector. in both cases, conj adds the new element in the "natural" position. for a vector that's to the right, but for a list it's to the left.
and it just occurred to me that you may not really understand what a list is. basically a cons creates a box containing two things (its arguments). the first is the contents and the second is the rest of the list. so the list (1 2 3) is basically (cons 1 (cons 2 (cons 3 nil))). in contrast, the vector [1 2 3] works more like an array (although i think it's implemented using a tree).
so conj is a bit confusing because the way it works depends on the first argument. for a list, it calls cons and so adds things to the left. but for a vector it extends the array(-like thing) to the right. also, note that conj takes an existing sequence as first arg, and thing to add as second, while cons is the reverse (thing to add comes first).
all the above code available at https://github.com/andrewcooke/clojure-lab
update: i rewrote the tests so that the expected result is a quoted list in the cases where the code generates a list. = will compare lists and vectors and return true if the content is the same, but making it explicit shows more clearly what you're actually getting in each case. note that '(0 1 2) with a ' in front is just like (list 0 1 2) - the ' stops the list from being evaluated (without it, 0 would be treated as a command).
After reading all that, I'm still not sure why you'd need an accumulator.
((fn r [a b]
(if (<= a b)
(cons a (r (inc a) b))))
2 4)
=> (2 3 4)
seems like a pretty intuitive recursive solution. the only thing I'd change in "real" code is to use lazy-seq so that you won't run out of stack for large ranges.
how I got to that solution:
When you're thinking of using recursion, I find it helps to try and state the problem with the fewest possible terms you can think up, and try to hand off as much "work" to the recursion itself.
In particular, if you suspect you can drop one or more arguments/variables, that is usually the way to go - at least if you want the code to be easy to understand and debug; sometimes you end up compromising simplicity in favor of execution speed or reducing memory usage.
In this case, what I thought when I started writing was: "the first argument to the function is also the start element of the range, and the last argument is the last element". Recursive thinking is something you kind of have to train yourself to do, but a fairly obvious solution then is to say: a range [a, b] is a sequence starting with element a followed by a range of [a + 1, b]. So ranges can indeed be described recursively. The code I wrote is pretty much a direct implementation of that idea.
addendum:
I've found that when writing functional code, accumulators (and indexes) are best avoided. Some problems require them, but if you can find a way to get rid of them, you're usually better off if you do.
addendum 2:
Regarding recursive functions and lists/sequences, the most useful way to think when writing that kind of code is to state your problem in terms of "the first item (head) of a list" and "the rest of the list (tail)".
I cannot add to the already good answers you have received, but I will answer in general. As you go through the Clojure learning process, you may find that many but not all solutions can be solved using Clojure built-ins, like map and also thinking of problems in terms of sequences. This doesn't mean you should not solve things recursively, but you will hear -- and I believe it to be wise advice -- that Clojure recursion is for solving very low level problems you cannot solve another way.
I happen to do a lot of .csv file processing, and recently received a comment that nth creates dependencies. It does, and use of maps can allow me to get at elements for comparison by name and not position.
I'm not going to throw out the code that uses nth with clojure-csv parsed data in two small applications already in production. But I'm going to think about things in a more sequency way the next time.
It is difficult to learn from books that talk about vectors and nth, loop .. recur and so on, and then realize learning Clojure grows you forward from there.
One of the things I have found that is good about learning Clojure, is the community is respectful and helpful. After all, they're helping someone whose first learning language was Fortran IV on a CDC Cyber with punch cards, and whose first commercial programming language was PL/I.
If I solved this using an accumulator I would do something like:
user=> (defn my-range [lb up c]
(if (= lb up)
c
(recur (inc lb) up (conj c lb))))
#'user/my-range
then call it with
#(my-range % %2 [])
Of course, I'd use letfn or something to get around not having defn available.
So yes, you do need an inner function to use the accumulator approach.
My thought process is that once I'm done the answer I want to return will be in the accumulator. (That contrasts with your solution, where you do a lot of work on finding the ending-condition.) So I look for my ending-condition and if I've reached it, I return the accumulator. Otherwise I tack on the next item to the accumulator and recur for a smaller case. So there are only 2 things to figure out, what the end-condition is, and what I want to put in the accumulator.
Using a vector helps a lot because conj will append to it and there's no need to use reverse.
I'm on 4clojure too, btw. I've been busy so I've fallen behind lately.
It looks like your question is more about "how to learn" then a technical/code problem. You end up writing that kind of code because from whatever way or source you learned programming in general or Clojure in specific has created a "neural highway" in your brain that makes you thinking about the solutions in this particular way and you end up writing code like this. Basically whenever you face any problem (in this particular case recursion and/or accumulation) you end up using that "neural highway" and always come up with that kind of code .
The solution for getting rid of this "neural highway" is to stop writing code for the moment, keep that keyboard away and start reading a lot of existing clojure code (from existing solutions of 4clojure problem to open source projects on github) and think about it deeply (even read a function 2-3 times to really let it settle down in your brain). This way you would end up destroying your existing "neural highway" (which produce the code that you write now) and will create a new "neural highway" that would produce the beautiful and idiomatic Clojure code. Also, try not to jump to typing code as soon as you saw a problem, rather give yourself some time to think clearly and deeply about the problem and solutions.