I am trying to create Pascal's Triangle using recursion. My code is:
(define (pascal n)
(cond
( (= n 1)
list '(1))
(else (append (list (pascal (- n 1))) (list(add '1 (coresublist (last (pascal (- n 1))))))
)))) ;appends the list from pascal n-1 to the new generated list
(define (add s lst) ;adds 1 to the beginning and end of the list
(append (list s) lst (list s))
)
(define (coresublist lst) ;adds the subsequent numbers, takes in n-1 list
(cond ((= (length lst) 1) empty)
(else
(cons (+ (first lst) (second lst)) (coresublist (cdr lst)))
)))
When I try to run it with:
(display(pascal 3))
I am getting an error that says:
length: contract violation
expected: list?
given: 1
I am looking for someone to help me fix this code (not write me entirely new code that does Pascal's Triangle). Thanks in advance! The output for pascal 3 should be:
(1) (1 1) (1 2 1)
We should start with the recursive definition for a value inside Pascals' triangle, which is usually expressed in terms of two parameters (row and column):
(define (pascal x y)
(if (or (zero? y) (= x y))
1
(+ (pascal (sub1 x) y)
(pascal (sub1 x) (sub1 y)))))
There are more efficient ways to implement it (see Wikipedia), but it will work fine for small values. After that, we just have to build the sublists. In Racket, this is straightforward using iterations, but feel free to implement it with explicit recursion if you wish:
(define (pascal-triangle n)
(for/list ([x (in-range 0 n)])
(for/list ([y (in-range 0 (add1 x))])
(pascal x y))))
It'll work as expected:
(pascal-triangle 3)
=> '((1) (1 1) (1 2 1))
Related
I'm creating a tail recursive function that evaluates a polynomial by passing a list of coefficients and an x value.
example: evaluate x^3 + 2x^2 + 5, so the user would pass the list '(5 0 2 1) and an x like 1 in a functional call (poly '(5 0 2 1) 1).
I can't figure out why i'm getting the following error:
if: bad syntax in: (if (null? (cdr lst)) (+ total (car lst))
eval-poly-tail-helper ((cdr lst) x (+ (* (expt x n) (car lst)) total)
(+ 1 n)))
(define (poly lst x)
(poly-assistant lst x 0 0))
(define (poly-assistant lst x total n)
(if (null? (cdr lst))
(+ total (car lst))
poly-assistant((cdr lst) x (+ (* (expt x n) (car lst)) total) (+ 1 n))))
You need a left paren before poly-assistant in the last line.
In Scheme, function applications start with a left paren. And if takes 2 or 3 operands.
Use a better editor (e.g. emacs) able to match parenthesis.
The two left parenthesis before cdr looks suspicious. You might need only one.
Learn to use your Scheme debugger, or at least add debugging prints.
Is there anyway to write the below code as to not have to make a list from which later on we have to car the element?
(define (square-it l)
(map (lambda (x) (* x x)) l))
(define (sum-it l)
(foldl + 0 l))
(define (sum-of-squares n)
(sum-it (square-it (numbers n))))
(define (square-of-sum n)
(square-it (*list* (sum-it (numbers n)))))
(- (*car* (square-of-sum 100)) (sum-of-squares 100))
As mentioned by #Sylwester, square-it is useful for squaring a list but not for squaring a single value, the inputs and outputs are different in each case, and sqr is the right procedure for squaring a single value. This should be enough to fix the problem:
(define (square-of-sum n)
(sqr (sum-it (numbers n))))
(- (square-of-sum 100) (sum-of-squares 100))
A simpler solution would be to use iterations and comprehensions and define each procedure independently. We can calculate the values directly over a range of values using only built-in procedures, there's no need to reinvent the wheel:
(define (sum-of-squares n)
(for/fold ([sum 0])
([i (in-range n)])
(+ sum (sqr i))))
(define (square-of-sum n)
(sqr (apply + (range n))))
I was asked to write a procedure that computes elements of Pascal's triangle by means of a recursive process. I may create a procedure that returns a single row in the triangle or a number within a particular row.
Here is my solution:
(define (f n)
(cond ((= n 1) '(1))
(else
(define (func i n l)
(if (> i n)
l
(func (+ i 1) n (cons (+ (convert (find (- i 1) (f (- n 1))))
(convert (find i (f (- n 1)))))
l))))
(func 1 n '()))))
(define (find n l)
(define (find i n a)
(if (or (null? a) (<= n 0))
'()
(if (>= i n)
(car a)
(find (+ i 1) n (cdr a)))))
(find 1 n l))
(define (convert l)
(if (null? l)
0
(+ l 0)))
This seems to work fine but it gets really inefficient to find elements of a larger row starting with (f 8). Is there a better procedure that solves this problem by means of a recursive process?
Also, how would I write it, if I want to use an iterative process (tail-recursion)?
There are several ways to optimize the algorithm, one of the best would be to use dynamic programming to efficiently calculate each value. Here is my own solution to a similar problem, which includes references to better understand this approach - it's a tail-recursive, iterative process. The key point is that it uses mutation operations for updating a vector of precomputed values, and it's a simple matter to adapt the implementation to print a list for a given row:
(define (f n)
(let ([table (make-vector n 1)])
(let outer ([i 1])
(when (< i n)
(let inner ([j 1] [previous 1])
(when (< j i)
(let ([current (vector-ref table j)])
(vector-set! table j (+ current previous))
(inner (add1 j) current))))
(outer (add1 i))))
(vector->list table)))
Alternatively, and borrowing from #Sylwester's solution we can write a purely functional tail-recursive iterative version that uses lists for storing the precomputed values; in my tests this is slower than the previous version:
(define (f n)
(define (aux tr tc prev acc)
(cond ((> tr n) '())
((and (= tc 1) (= tr n))
prev)
((= tc tr)
(aux (add1 tr) 1 (cons 1 acc) '(1)))
(else
(aux tr
(add1 tc)
(cdr prev)
(cons (+ (car prev) (cadr prev)) acc)))))
(if (= n 1)
'(1)
(aux 2 1 '(1 1) '(1))))
Either way it works as expected for larger inputs, it'll be fast for n values in the order of a couple of thousands:
(f 10)
=> '(1 9 36 84 126 126 84 36 9 1)
There are a number of soluitons presented already, and they do point out that usign dynamic programming is a good option here. I think that this can be written a bit more simply though. Here's what I'd do as a straightforward list-based solution. It's based on the observation that if row n is (a b c d e), then row n+1 is (a (+ a b) (+ b c) (+ c d) (+ d e) e). An easy easy to compute that is to iterate over the tails of (0 a b c d e) collecting ((+ 0 a) (+ a b) ... (+ d e) e).
(define (pascal n)
(let pascal ((n n) (row '(1)))
(if (= n 0) row
(pascal (- n 1)
(maplist (lambda (tail)
(if (null? (cdr tail)) 1
(+ (car tail)
(cadr tail))))
(cons 0 row))))))
(pascal 0) ;=> (1)
(pascal 1) ;=> (1 1)
(pascal 2) ;=> (1 2 1)
(pascal 3) ;=> (1 3 3 1)
(pascal 4) ;=> (1 4 6 4 1)
This made use of an auxiliary function maplist:
(define (maplist function list)
(if (null? list) list
(cons (function list)
(maplist function (cdr list)))))
(maplist reverse '(1 2 3))
;=> ((3 2 1) (3 2) (3))
I am writing a function to count the number of ways to make change in scheme given a list of denominations and an amount. My code is as follows, but it does not work as intended. Should I be using cons instead of the + operator? Should the third line down's base case be the empty list?
(define (change k l)
(cond ((= k 0) 1)
((or (< k 0) (null? l)) 0)
(else (+ (change k (cdr l))
(change (- k (car l))
(cdr l))))))
Test:
(change 11 (list 1 5 10 25))
If the returned value is just a number then forget about cons and '() for building the output, and only use car, cdr, null? for processing the input. Other than that, be aware that there's a small mistake in the last line of your code, here's the fixed version:
(define (change k l)
(cond ((= k 0) 1)
((or (< k 0) (null? l)) 0)
(else
(+ (change k (cdr l))
(change (- k (car l)) l))))) ; don't do (cdr l) here
Now it works as expected:
(change 11 (list 1 5 10 25))
=> 4
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)))))