My code in lisp is as follows:
(defun solve-hanoi(from) (hanoi (length from) from '() '()))
(defun hanoi(height from to aux) (when (>= height 1)
(hanoi (- height 1) from aux to)
(format t "~%Move ~a from ~a to ~a" (nth 0 from) from to)
(push (pop from) to)
(hanoi (- height 1) aux to from)))
I am new to lisp and have NO clue as to what I am doing wrong.
Help with this would be GREATLY appreciated since I've been at this for hours.
Thanks.
The recursive algorithm is:
To move n discs from peg A to peg C:
1. move n−1 discs from A to B. This leaves disc n alone on peg A
2. move disc n from A to C
3. move n−1 discs from B to C so they sit on disc n
(From http://en.wikipedia.org/wiki/Tower_of_Hanoi#Recursive_solution)
Since the meaning of A, B and C (your from, aux and to) are relative to the current iteration and keep changing, it's easier not to pass the state of the game around and trying to understand what it means, but to simply generate solving instructions.
To implement the algorithm above in this way, you need the following inside your (when (>= height 1):
1. Recursive call with n-1, swapping B and C. You got this one right already.
2. Print info about the move, for instance (format t "~%Move ~a to ~a" from to).
3. Recursive call with n-1, swapping A and B. You got this one right too.
Then change your (solve-hanoi) so that it takes the number of disks on the first rod as argument, and calls (hanoi) with this number and whatever names you want for the rods, for instance (hanoi 4 'A 'B 'C) or (hanoi 4 1 2 3) for 4 disks.
Related
A node is a
(define-struct node (left right))
A leafy binary tree (LBT) is one of
; - 'leaf
; - (make-node LBT LBT)
I have to design a function that takes in a natural number n and creates a list of all leafy binary trees that have height n.
So for example, a height of 2 should return:
`(list
(make-node 'leaf (make-node 'leaf 'leaf))
(make-node (make-node 'leaf 'leaf) 'leaf)
(make-node (make-node 'leaf 'leaf) (make-node 'leaf 'leaf)))`
I've been stuck on this assignment for days now. The best I could come up with was:
`(define (lbt-list n)
(cond [(= 0 n) 'leaf]
[(= 1 n) (cons (make-node (lbt-list (- n 1)) (lbt-list (- n 1))) empty)]
[else (list (make-node (first (lbt-list (- n 1))) (lbt-list (- n n)))
(make-node (lbt-list (- n n)) (first (lbt-list (- n 1))))
(make-node (first (lbt-list (- n 1))) (first (lbt-list (- n 1)))))]))`
The problem involves recursion and I'm just not sure how to fully code out this function. Any help would be appreciated!
You are trying to make nodes out of lists; the only valid arguments to make-node are leafs and nodes. Instead, you should take the results of lbt-list and use each of its elements to make each of the nodes for the list to be returned.
Also, you never use the rest of any returned lbt-list.
To start, the reason why you're only getting three trees is that your else clause calls list with exactly three instances of make-node. This might be a little easier to see if you make (= n 1) your base case, and explicitly use 'leaf wherever you need it, instead of (lbt-list (- n n)).
I think you'll benefit from stepping back from the code for a second and asking yourself what mathematical products you're trying to generate. In particular, how do you get the n-depth binary trees from the (n-1)-depth binary trees?
It's simplest to break it into three parts. Here are two big hints:
is make-node commutative?
how is the level-on-level generation related to a Cartesian product?
Finally, helper functions are your friend here - they'll go a long way towards dividing the mental burden of generating the list into smaller pieces. I wrote three helper functions when I threw together a solution, but you might be able to make do with two.
I got a little problem here. I want to make the dragon curve using Racket.
First, I want to make a list with the turns of the given order of the dragon curve.
For example: The order 3 would give me the list: (list 'R 'R 'L 'R 'R 'L 'L).
L means a 90 degree turn to the left and R means a 90 degree turn to the right.
The algorithm for generating the list by a given order is:
First order is always a right turn (list 'R)
the next order is the previous order plus the element (list 'R) plus the previous order, where the middle symbol is been replaced by an 'L.
So, the second order would be (list 'R 'R 'L)
But I don't really know, how to write this 'algorithm' as (recursive) code.
;;number -> list
;; number 'n' is the order of the dragon curve.
;; (dragon-code 3) should make: (list 'R 'R 'L 'R 'R 'L 'L)
(define (dragon-code n)
(cond
[(zero? n) empty]
[else
I would be thankful for every hint! :)
A literal translation of the text gives:
(define (dragon order)
(if (= order 1)
(list 'R)
(append (dragon (- order 1))
(list 'R)
(replace-middle-with-L (dragon (- order 1))))))
Implement replace-middle-with-L and test with (dragon 3).
I'm starting out with Clojure and, despite having an understanding of recursion, am having trouble thinking of the "right" way to build a lazy-seq for the following function:
I want to build a list of all the frequencies starting from middle C. My first element would be 120 (the frequency of middle C). To get the second element, I'd multiply the first element, 120, by 1.059463 to get 127.13556. To the get the third I'd multiply the second element, 127.13556, by 1.059463, etc etc...
What's the best way to do this in Clojure?
You can use the iterate function for that.
(iterate #(* % 1.059463) 120)
If you are planning to expand this into something more complicated, then you would create a function that recursive calls itself inside a call to lazy-seq. (This is what iterate does internally.)
(defn increasing-frequencies
([] (increasing-frequencies 120))
([freq]
(cons freq (lazy-seq (increasing-frequencies (* freq 1.059463))))))
(nth (increasing-frequencies) 2) ;; => 134.69542180428002
If you start to use this in a tight loop, you may also want to generate a chunked lazy seq. This will pre-calculate the next few elements, instead of one by one.
(defn chunked-increasing-frequencies
([] (chunked-increasing-frequencies 120))
([freq]
(lazy-seq
(let [b (chunk-buffer 32)]
(loop [i freq c 0]
(if (< c 32)
(do
(chunk-append b i)
(recur (* i 1.059463) (inc c)))
(chunk-cons (chunk b) (chunked-increasing-frequencies i))))))))
Note: I would advise against doing this until you have measured a performance problem related to calculating individual elements.
(defn get-frequencies []
(iterate #(* 1.059463 %) 120))
See iterate
Or, if you want to use lazy-seq explicitly, you can do this:
(defn get-frequencies-hard []
(cons
120
(lazy-seq
(map #(* 1.059463 %) (get-frequencies-hard)))))
Which will cons 120 to a lazy seq of every value applied to the map function.
I am fairly new to lisp and this is one of the practice problems.
First of all, this problem is from simply scheme. I am not sure how to answer this.
The purpose of this question is to write the function, count-odd that takes a sentence as its input and count how many odd digits are contained in it as shown below:
(count-odd'(234 556 4 10 97))
6
or
(count-odd '(24680 42 88))
0
If possible, how would you be able to do it, using higher order functions, or recursion or both - whatever gets the job done.
I'll give you a few pointers, not a full solution:
First of all, I see 2 distinct ways of doing this, recursion or higher order functions + recursion. For this case, I think straight recursion is easier to grok.
So we'll want a function which takes in a list and does stuff, so
(define count-odd
(lambda (ls) SOMETHING))
So this is recursive, so we'd want to split the list
(define count-odd
(lambda (ls)
(let ((head (car ls)) (rest (cdr ls)))
SOMETHING)))
Now this has a problem, it's an error for an empty list (eg (count-odd '())), but I'll let you figure out how to fix that. Hint, check out scheme's case expression, it makes it easy to check and deal with an empty list
Now something is our recursion so for something something like:
(+ (if (is-odd head) 1 0) (Figure out how many odds are in rest))
That should give you something to start on. If you have any specific questions later, feel free to post more questions.
Please take first into consideration the other answer guide so that you try to do it by yourself. The following is a different way of solving it. Here is a tested full solution:
(define (count-odd num_list)
(if (null? num_list)
0
(+ (num_odds (car num_list)) (count-odd (cdr num_list)))))
(define (num_odds number)
(if (zero? number)
0
(+ (if (odd? number) 1 0) (num_odds (quotient number 10)))))
Both procedures are recursive.
count-odd keeps getting the first element of a list and passing it to num_odds until there is no element left in the list (that is the base case, a null list).
num_odds gets the amount of odd digits of a number. To do so, always asks if the number is odd in which case it will add 1, otherwise 0. Then the number is divided by 10 to remove the least significant digit (which determines if the number is odd or even) and is passed as argument to a new call. The process repeats until the number is zero (base case).
Try to solve the problem by hand using only recursion before jumping to a higher-order solution; for that, I'd suggest to take a look at the other answers. After you have done that, aim for a practical solution using the tools at your disposal - I would divide the problem in two parts.
First, how to split a positive integer in a list of its digits; this is a recursive procedure over the input number. There are several ways to do this - by first converting the number to a string, or by using arithmetic operations to extract the digits, to name a few. I'll use the later, with a tail-recursive implementation:
(define (split-digits n)
(let loop ((n n)
(acc '()))
(if (< n 10)
(cons n acc)
(loop (quotient n 10)
(cons (remainder n 10) acc)))))
With this, we can solve the problem in terms of higher-order functions, the structure of the solution mirrors the mental process used to solve the problem by hand:
First, we iterate over all the numbers in the input list (using map)
Split each number in the digits that compose it (using split-digits)
Count how many of those digits are odd, this gives a partial solution for just one number (using count)
Add all the partial solutions in the list returned by map (using apply)
This is how it looks:
(define (count-odd lst)
(apply +
(map (lambda (x)
(count odd? (split-digits x)))
lst)))
Don't be confused if some of the other solutions look strange. Simply Scheme uses non-standard definitions for first and butfirst. Here is a solution, that I hope follows Simply Scheme friendly.
Here is one strategy to solve the problem:
turn the number into a list of digits
transform into a list of zero and ones (zero=even, one=odd)
add the numbers in the list
Example: 123 -> '(1 2 3) -> '(1 0 1) -> 2
(define (digit? x)
(<= 0 x 9))
(define (number->digits x)
(if (digit? x)
(list x)
(cons (remainder x 10)
(number->digits (quotient x 10)))))
(define (digit->zero/one d)
(if (even? d) 0 1))
(define (digits->zero/ones ds)
(map digit->zero/one ds))
(define (add-numbers xs)
(if (null? xs)
0
(+ (first xs)
(add-numbers (butfirst xs)))))
(define (count-odds x)
(add-numbers
(digits->zero/ones
(number->digits x))))
The above is untested, so you might need to fix a few typos.
I think this is a good way, too.
(define (count-odd sequence)
(length (filter odd? sequence)))
(define (odd? num)
(= (remainder num 2) 1))
(count-odd '(234 556 4 10 97))
Hope this will help~
The (length sequence) will return the sequence's length,
(filter proc sequence) will return a sequence that contains all the elements satisfy the proc.
And you can define a function called (odd? num)
I read a lot of documentation about Clojure (and shall need to read it again) and read several Clojure questions here on SO to get a "feel" of the language. Besides a few tiny functions in elisp I've never written in any Lisp language before. I wrote my first project Euler solution in Clojure and before going further I'd like to better understand something about map and reduce.
Using a lambda, I ended up with the following (to sum all multiple of either 3 or 5 or both between 1 and 1000 inclusive):
(reduce + (map #(if (or (= 0 (mod %1 3)) (= 0 (mod %1 5))) %1 0) (range 1 1000)))
I put it on one line because I wrote it on the REPL (and it gives the correct solution).
Without the lambda, I wrote this:
(defn val [x] (if (or (= 0 (mod x 3)) (= 0 (mod x 5))) x 0))
And then I compute the solution doing this:
(reduce + (map val (range 1 1000)))
In both cases, my question concerns what the map should return, before doing the reduce. After doing the map I noticed I ended up with a list looking like this: (0 0 3 0 5 6 ...).
I tried removing the '0' at the end of the val definition but then I received a list made of (nil nil 3 nil 5 6 etc.). I don't know if the nil are an issue or not. I figured out that I was going to sum while doing a fold-left anyway so that the zero weren't really an issue.
But still: what's a sensible map to return? (0 0 3 0 5 6 ...) or (nil nil 3 nil 5 6...) or (3 5 6 ...) (how would I go about this last one?) or something else?
Should I "filter out" the zeroes / nils and if so how?
I know I'm asking a basic question but map/reduce is obviously something I'll be using a lot so any help is welcome.
It sounds like you already have an intuative undestanding of the need to seperate mapping concerns form the reducing It's perfectly natural to have data produced by map that is not used by the reduce. infact using the fact that zero is the identity value for addition make this even more elegant.
mappings job is to produce the new data (in this case 3 5 or "ignore")
reduces job is to decide what to include and to produce the final result.
what you started with is idiomatic clojure and there is no need to complicate it any more,
so this next example is just to illustrate the point of having map decide what to include:
(reduce #(if-not (zero? %1) (+ %1 %2) %2) (map val (range 10)))
in this contrived example the reduce function ignores the zeros. In typical real world code if the idea was as simple as filtering out some value then people tend to just use the filter function
(reduce + (filter #(not (zero? %)) (map val (range 10))))
you can also just start with filter and skip the map:
(reduce + (filter #(or (zero? (rem % 3)) (zero? (rem % 5))) (range 10)))
The watchword is clarity.
Use filter, not map. Then you don't have to choose a null
value that you later have to decide not to act on.
Naming the filtering/mapping function can help. Do so with let
or letfn, not defn, unless you have use for the function elsewhere.
Acting on this advice brings us to ...
(let [divides-by-3-or-5? (fn [n] (or (zero? (mod n 3)) (zero? (mod n 5))))]
(reduce + (filter divides-by-3-or-5? (range 1 1000))))
You may want to stop here for now.
This reads well, but the divides-by-3-or-5? function sticks in the throat. Change the factors and we need a completely new function. And that repeated phrase (zero? (mod n ...)) jars. So ...
We want a function, that - given a list (or other collection) of possible factors - tells us whether any of them apply to a given number. In other words, we want
a function of a collection of numbers - the possible factors - ...
that returns a function of one number - the candidate - ...
that tells us whether the candidate is divisible by any of the possible factors.
One such function is
(fn [ns] (fn [n] (some (fn [x] (zero? (mod n x))) ns)))
... which we can employ thus
(let [divides-by-any? (fn [ns] (fn [n] (some (fn [x] (zero? (mod n x))) ns)))]
(reduce + (filter (divides-by-any? [3 5]) (range 1 1000))))
Notes
This "improvement" has made the program a little slower.
divides-by-any? might prove useful enough to be promoted to a
defn.
If the operation were critical, you could consider stripping out
redundant factors. For example [2 3 6] could be reduced to [6].
If the operation were really critical, and the factors were supplied
as constants, you could consider creating the filter function with a
macro that went back to using or.
This is a bit of a shaggy-dog story, but it recounts the thoughts prompted by the problem you refer to.
In your case I would use keep instead of map. It is similar to map except that it keeps only the non-nil values.