(defn fib [n]
(if ((= n 0) 0)
((= n 1) 1)
(:else (+ (fib (- n 1))
(fib (- n 2))))))
(fib 10)
ClassCastException java.lang.Boolean cannot be cast to clojure.lang.IFn
The same exception with the following.
(defn A [x y]
(cond ((= y 0) 0)
((= x 0) (* 2 y))
((= y 1) 2)
(:else (A (- x 1) (A x (- y 1))))))
(A 1 10)
Whats wrong with this unable to understand, please explain ?
You were so close!
(defn A [x y]
(cond (= y 0) 0
(= x 0) (* 2 y)
(= y 1) 2
:else (A (- x 1)
(A x (- y 1)))))
You simply had too many parentheses wrapping the cond forms.
Works fine now:
user=> (A 1 10)
1024
There are some similar issues in your recursive fib function. Pay careful attention to indentation - this will always help you see where your issue lies.
In this particular case the exception ClassCastException java.lang.Boolean cannot be cast to clojure.lang.IFnis being thrown by this line:
((= n 1) 1)
... because (= n 1) is being evaluated to either Boolean true or false, and because this resulting boolean is in the first position of the ((= n 1) 1) form, it means that Clojure will attempt to call the boolean as a function (clojure.lang.IFn).
This is what Clojure is really seeing:
(true 1)
Which is why Clojure is trying to cast a Boolean as an IFn. IFn is a Java interface which represents a callable function.
I hope that makes sense.
Related
I'm new to CHICKEN and Scheme. In my quest to understanding tail recursion, I wrote:
(define (recsum x) (recsum-tail x 0))
(define (recsum-tail x accum)
(if (= x 0)
accum
(recsum-tail (- x 1) (+ x accum))))
This does what I expect it to. However, this seems a little repetitive; having an optional argument should make this neater. So I tried:
(define (recsum x . y)
(let ((accum (car y)))
(if (= x 0)
accum
(recsum (- x 1) (+ x accum)))))
However, in CHICKEN (and maybe in other scheme implementations), car cannot be used against ():
Error: (car) bad argument type: ()
Is there another way to implement optional function arguments, specifically in CHICKEN 5?
I think you're looking for a named let, not for optional procedure arguments. It's a simple way to define a helper procedure with (possibly) extra parameters that you can initialize as required:
(define (recsum x)
(let recsum-tail ((x x) (accum 0))
(if (= x 0)
accum
(recsum-tail (- x 1) (+ x accum)))))
Of course, we can also implement it with varargs - but I don't think this looks as elegant:
(define (recsum x . y)
(let ((accum (if (null? y) 0 (car y))))
(if (= x 0)
accum
(recsum (- x 1) (+ x accum)))))
Either way, it works as expected:
(recsum 10)
=> 55
Chicken has optional arguments. You can do it like this:
(define (sum n #!optional (acc 0))
(if (= n 0)
acc
(sum (- n 1) (+ acc n))))
However I will vote against using this as it is non standard Scheme. Chicken say they support SRFI-89: Optional positional and named parameters, but it seems it's an earlier version and the egg needs to be redone. Anyway when it is re-applied this should work:
;;chicken-install srfi-89 # install the egg
(use srfi-89) ; imports the egg
(define (sum n (acc 0))
(if (= n 0)
acc
(sum (- n 1) (+ acc n))))
Also your idea of using rest arguments work. However keep in mind that the procedure then will build a pair on the heap for each iteration:
(define (sum n . acc-lst)
(define acc
(if (null? acc-lst)
0
(car acc-lst)))
(if (= n 0)
acc
(sum (- n 1) (+ acc n))))
All of these leak internal information. Sometimes it's part of the public contract to have an optional parameter, but in this case it is to avoid writing a few more lines. Usually you don't want someone to pass a second argument and you should keep the internals private. The better way would be to use named let and keep the public contract as is.
(define (sum n)
(let loop ((n n) (acc 0))
(if (= n 0)
acc
(loop (- n 1) (+ acc n))))
I'm starting to get to grips with Lisp and I'm trying to write a procedure to approximate pi using the Leibniz formula at the moment; I think I'm close but I'm not sure how to proceed. The current behavior is that it makes the first calculation correctly but then the program terminates and displays the number '1'. I'm unsure if I can call a defined function recursively like this,
;;; R5RS
(define (pi-get n)
(pi 0 1 n 0))
(define (pi sum a n count)
;;; if n == 0, 0
(if (= n 0) 0)
;;; if count % 2 == 1, + ... else -, if count == n, sum
(cond ((< count n)
(cond ((= (modulo count 2) 1)
(pi (+ sum (pi-calc (+ 2 a))) (+ a 2) n (+ count 1)))
(pi
(- sum (pi-calc (+ 2 a))) (+ a 2) n (+ count 1))))))
(define (pi-calc a)
(/ 1.0 a))
Apologies if this is a little unreadable, I'm just learning Lisp a few weeks now and I'm not sure what normal formatting would be for the language. I've added a few comments to hopefully help.
As Sylwester mentioned it turned out to be a mistake on my part with syntax.
;;; R5RS
(define (pi-get n)
(pi 1 1 n 0))
(define (pi sum a n count)
(if (= n 0) 0)
(cond ((< count n)
(cond ((= (modulo count 2) 1)
(pi (+ sum (pi-calc (+ 2 a))) (+ a 2) n (+ count 1)))
((= (modulo count 2) 0)
(pi (- sum (pi-calc (+ 2 a))) (+ a 2) n (+ count 1))))
(display (* 4 sum)) (newline))))
(define (pi-calc a)
(/ 1.0 a))
Here is what I have done so far:
(define sumOdd
(lambda(n)
(cond((> n 0)1)
((odd? n) (* (sumOdd n (-(* 2 n) 1)
output would look something like this:
(sumOdd 1) ==> 1
(sumOdd 4) ==> 1 + 3 + 5 + 7 ==> 16
(sumOdd 5) ==> 1 + 3 + 5 + 7 + 9 ==> 25
This is what I am trying to get it to do: find the sum of the first N odd positive integers
I can not think of a way to only add the odd numbers.
To elaborate further on the sum-odds problem, you might solve it in terms of more abstract procedures that in combination accumulates the desired answer. This isn't necessarily the easiest solution, but it is interesting and captures some more general patterns that are common when processing list structures:
; the list of integers from n to m
(define (make-numbers n m)
(if (= n m) (list n) ; the sequence m..m is (m)
(cons n ; accumulate n to
(make-numbers (+ n 1) m)))) ; the sequence n+1..m
; the list of items satisfying predicate
(define (filter pred lst)
(if (null? lst) '() ; nothing filtered is nothing
(if (pred (car lst)) ; (car lst) is satisfactory
(cons (car lst) ; accumulate item (car lst)
(filter pred (cdr lst))) ; to the filtering of rest
(filter pred (cdr lst))))) ; skip item (car lst)
; the result of combining list items with procedure
(define (build-value proc base lst)
(if (null? lst) base ; building nothing is the base
(proc (car lst) ; apply procedure to (car lst)
(build-value proc base (cdr lst))))) ; and to the building of rest
; the sum of n first odds
(define (sum-odds n)
(if (negative? n) #f ; negatives aren't defined
(build-value + ; build values with +
0 ; build with 0 in base case
(filter odd? ; filter out even numbers
(make-numbers 1 n))))) ; make numbers 1..n
Hope this answer was interesting and not too confusing.
Let's think about a couple of cases:
1) What should (sumOdd 5) return? Well, it should return 5 + 3 + 1 = 9.
2) What should (sumOdd 6) return? Well, that also returns 5 + 3 + 1 = 9.
Now, we can write this algorithm a lot of ways, but here's one way I've decided to think about it:
We're going to write a recursive function, starting at n, and counting down. If n is odd, we want to add n to our running total, and then count down by 2. Why am I counting down by 2? Because if n is odd, n - 2 is also odd. Otherwise, if n is even, I do not want to add anything. I want to make sure that I keep recursing, however, so that I get to an odd number. How do I get to the next odd number, counting down from an even number? I subtract 1. And I do this, counting down until n is <= 0. I do not want to add anything to my running total then, so I return 0. Here is what that algorithm looks like:
(define sumOdd
(lambda (n)
(cond ((<= n 0) 0)
((odd? n) (+ n (sumOdd (- n 2))))
(else (sumOdd (- n 1))))))
If it helps you, here is a more explicit example of a slightly different algorithm:
(define sumOdd
(lambda (n)
(cond ((<= n 0) 0)
((odd? n) (+ n (sumOdd (- n 1))))
((even? n) (+ 0 (sumOdd (- n 1))))))) ; note that (even? n) can be replaced by `else' (if its not odd, it is even), and that (+ 0 ..) can also be left out
EDIT:
I see that the problem has changed just a bit. To sum the first N positive odd integers, there are a couple of options.
First option: Math!
(define sumOdd (lambda (n) (* n n)))
Second option: Recursion. There are lots of ways to accomplish this. You could generate a list of 2*n and use the procedures above, for example.
You need to have 2 variables, one which keep counter of how many odd numbers are still to be added and another to hold the current odd number which gets increment by 2 after being used in addition:
(define (sum-odd n)
(define (proc current start)
(if (= current 0)
0
(+ start (proc (- current 1) (+ start 2)) )))
(proc n 1))
Here is a nice tail recursive implementation:
(define (sumOdd n)
(let summing ((total 0) (count 0) (next 1))
(cond ((= count n) total)
((odd? next) (summing (+ total next)
(+ count 1)
(+ next 1)))
(else (summing total count (+ next 1))))))
Even shorter tail-recursive version:
(define (sumOdd n)
(let loop ((sum 0) (n n) (val 1))
(if (= n 0)
sum
(loop (+ sum val) (- n 1) (+ val 2)))))
I have the following 2 functions that I wish to combine into one:
(defun fib (n)
(if (= n 0) 0 (fib-r n 0 1)))
(defun fib-r (n a b)
(if (= n 1) b (fib-r (- n 1) b (+ a b))))
I would like to have just one function, so I tried something like this:
(defun fib (n)
(let ((f0 (lambda (n) (if (= n 0) 0 (funcall f1 n 0 1))))
(f1 (lambda (a b n) (if (= n 1) b (funcall f1 (- n 1) b (+ a b))))))
(funcall f0 n)))
however this is not working. The exact error is *** - IF: variable F1 has no value
I'm a beginner as far as LISP goes, so I'd appreciate a clear answer to the following question: how do you write a recursive lambda function in lisp?
Thanks.
LET conceptually binds the variables at the same time, using the same enclosing environment to evaluate the expressions. Use LABELS instead, that also binds the symbols f0 and f1 in the function namespace:
(defun fib (n)
(labels ((f0 (n) (if (= n 0) 0 (f1 n 0 1)))
(f1 (a b n) (if (= n 1) b (f1 (- n 1) b (+ a b)))))
(f0 n)))
You can use Graham's alambda as an alternative to labels:
(defun fib (n)
(funcall (alambda (n a b)
(cond ((= n 0) 0)
((= n 1) b)
(t (self (- n 1) b (+ a b)))))
n 0 1))
Or... you could look at the problem a bit differently: Use Norvig's defun-memo macro (automatic memoization), and a non-tail-recursive version of fib, to define a fib function that doesn't even need a helper function, more directly expresses the mathematical description of the fib sequence, and (I think) is at least as efficient as the tail recursive version, and after multiple calls, becomes even more efficient than the tail-recursive version.
(defun-memo fib (n)
(cond ((= n 0) 0)
((= n 1) 1)
(t (+ (fib (- n 1))
(fib (- n 2))))))
You can try something like this as well
(defun fib-r (n &optional (a 0) (b 1) )
(cond
((= n 0) 0)
((= n 1) b)
(T (fib-r (- n 1) b (+ a b)))))
Pros: You don't have to build a wrapper function. Cond constructt takes care of if-then-elseif scenarios. You call this on REPL as (fib-r 10) => 55
Cons: If user supplies values to a and b, and if these values are not 0 and 1, you wont get correct answer
I want to program a function to find C(n,k) using tail recursion, and I would greatly appreciate your help.
I have reached this:
(defun tail-recursive-binomial (n k)
(cond ((or (< n k) (< k 0)) NIL)
((or (= k 0) (= n k)) 1)
(T (* (tail-recursive-binomial (- n 1) (- k 1)) (/ n k)))))
Using the following property of the binomial coefficients.
But I don't know how to make the recursive call to be the last instruction executed by each instance, since there the last one is the product. I have been trying it by using an auxiliary function, which I think is the only way, but I haven't found a solution.
As starblue suggests, use a recursive auxiliary function:
(defun binom (n k)
(if (or (< n k) (< k 0))
NIL ; there are better ways to handle errors in Lisp
(binom-r n k 1)))
;; acc is an accumulator variable
(defun binom-r (n k acc)
(if (or (= k 0) (= n k))
acc
(binom-r (- n 1) (- k 1) (* acc (/ n k)))))
Or, give the main function an optional accumulator argument with a default value of 1 (the recursive base case):
(defun binom (n k &optional (acc 1))
(cond ((or (< n k) (< k 0)) NIL)
((or (= k 0) (= n k)) acc)
(T (binom (- n 1) (- k 1) (* acc (/ n k))))))
The latter option is slightly less efficient, since the error condition is checked in every recursive call.
You need an auxiliary function with an extra argument, which you use for computing and passing the result.