I decided to incorporate recursions into my code and for that I want to train using it. I have a recursive function which finds the max element in the given array.
def maxx(lst):
if len(lst) == 1:
return lst[0]
else:
m = maxx(lst[1:])
return m if m > lst[0] else lst[0]
maxx([1,3,2])
> 3
I tried to draw a sequence of events and I can't get why the code above works?
Because according to what I see it should be 2 not 3. Where is my mistake? Is there a method that correctly expands the recursion and thus helps you to understand its flow?
Depth 0:
lst = [1,3,2]
m = maxx([3,2])
Depth 1:
lst = [3,2]
m = maxx([2])
Depth 2:
lst = [2]
returning 2
Back to Depth 1:
lst = [3,2]
m = 2
return m (2) if m > lst[0] (3) else return lst[0] (3)
returning 3
Back to Depth 0:
lst = [1,3,2]
m = 3
return m (3) if m > lst[0] (1) else return lst[0] (1)
returning 3
The reply by Alfabravo is correct, I think that's where you lost track of lst[0] as you went back up the tree.
Related
imagine if i had a list ([1,2,2,3,2]) and i want to find how many times does the number 2 appear in the list, how would i do it recursively, do i set my counter to 0 in my base case? How do i make the list count my desired item.
def rc_count(L, x):
if len(L)==0:
n=0
return n
else:
rc_count(L[1:], x)
if L[0]==x:
n+=1
return n
it says local variable n referenced before assignment so where should i put my counter
I do not really se the reason for this recursive function. However this is how you would do it:
def rc_count(l,x):
n = 0
if not l:
return n
else:
n = re(l[1:], x)
if l[0] == x:
n+=1
return n
Here rc_count([1,2,2,2,2,3,4,5],2) will return 4
First, I have a function called permeability.
# permeabiliy function
# L is short for the Lable
mu_0 = 4 * pi * 10^(-7);
mu_r_core = 50;
mu_r_air = 1;
L = Int16;
function permeability(L)
if L in 1:4
if L !== 3
return mu = mu_r_air * mu_0
else
return mu = mu_r_core * mu_0
end
else
println("null") #print output in a new line
end
end
Then, I have a matrix called domain, which is shown below,
domain
2392-element Array{Int16,1}:
1
1
3
1
...
When I called permeability(domain[3]), the output is,
L = domain[3]
permeability(L)
the output is
1.2566370614359177e-6
However, when I simply called permeability(3),
permeability(3)
the output is
6.283185307179588e-5
So, it seems that the value passed from matrix domain is just "1", but in this case, domain[3] should be 3 and the result should be the same in these 2 cases.
Can someone please tell me where I was wrong?
The problem is that your array stores Int16 while 3 is an Int64. L!==3 requires L to be of the same type as 3 ie Int64. You wanted instead L!=3. Your confusion probably comes from the fact that != is the inverse of == while !== is the inverse of ===
I'm using NLopt for a constrained maximization problem. Regardless of the algorithm or start values, the optimization program is force stopped even before the first iteration (or so I assume because it gives me the initial value). I've attached my code here. I'm trying to find probabilities attached to a grid such that a function is maximized under some constraints. Any help is appreciated.
uk = x -> x^0.5
function objective(u,p,grd)
-p'*u.(grd)
end
function c3(grd,p)
c =[]
d =[]
for i=1:length(grd)
push!(c,quadgk(x -> (i-x)*(x <= i ? 1 : 0),0,1)[1])
push!(d,sum(p[1:i]'*(grd[1:i] .- grd[i])))
end
return append!(d-c,-p)
end
function c4(grd,p)
return (grd .* p)-quadgk(x,0,1)
end
grd = n -> collect(0:1/n:1)
opt = Opt(:LD_SLSQP,11)
inequality_constraint!(opt, p -> c3(grd(10),p))
inequality_constraint!(opt, p -> -p)
equality_constraint!(opt, p -> sum(p)-1)
equality_constraint!(opt, p -> c4(grd(10),p))
opt.min_objective = p -> objective(-uk, p, grd(10))
k = push!(ones(11)*(1/11))
(minf,minx,ret) = optimize(opt, k)
I'm not a julia developer, but I only know this, if you need exit before complete the loop for is not your best choice, you need do a while with a sentinel variable.
here you have an article that explain you how while with sentinels works
and here you have a julia example changing your for to a while with a sentinel that exit after the third loop
i = 1
third = 0
while i < length(grd) && third != 1
# of course you need change this, it is only an example that will exit in the 3 loop
if i == 3
third = 1
end
push!(c,quadgk(x -> (i-x)*(x <= i ? 1 : 0),0,1)[1])
push!(d,sum(p[1:i]'*(grd[1:i] .- grd[i])))
i += 1
end
I am experimenting with the functional programming paradigm in R. I have defined a function that sums a sequence of integers from n to m. When I use sum() the function returns the expected result:
sumRange <- function(n, m) {
if (n <= m) {
return(sum(n, sumRange((n + 1), m)))
}
}
sumRange(1, 10)
# [1] 55
However, when I use the + operator the function returns numeric(0):
sumRange <- function(n, m) {
if (n <= m) {
return(n + sumRange((n + 1), m))
}
}
sumRange(1, 10)
# numeric(0)
Why does the operator + not work in this recursive function? Is there a way to rewrite the function so that it does?
The issue is that you never specify an else condition, hence at the end of the recursion it appears that R is returning NULL when the if condition fails. Returning 0 as the else condition fixes your problem:
sumRange <- function(n, m) return(ifelse (n <= m, (n + sumRange((n+1), m)), 0))
sumRange(1, 10)
[1] 55
Note that this is essentially defining a base case for your recursion. A base case, when hit, ends the recursion and causes the calls on the stack to be unwound.
To see the issue with the way you phrased your code, try writing out your function explicitly:
sumRange <- function(n, m) {
if (n <= m) {
return(n + sumRange((n+1), m))
}
// but what gets returned if n > m ?
// this is undefined behavior
}
I'm not an R guru, but my understanding is that R was written in C, and C might allow a recursion like this with no else condition. But the behavior is not well defined and you should not be relying on it.
Demo
If there is no return (using a explicit or implicit return statement) is executed, then R functions seems to return a NULL object.
If you add numerical value to a this object, it will simply return numeric(0).
So, what happens in the second case is that when n reaches 11, it returns a NULL object, and goes back adding values to it. But NULL + 10 + 9 .. = numeric(0).
Check this with
no_ret <- function ()
{
# just return nothing
}
obj <- no_ret()
obj
# NULL
class(obj)
# "NULL
new_obj <- obj + 10
new_obj
# numeric(0)
When the first function is executed, the what the sum statement get is
a vector with a NULL in it. For example,
vec <- c(NULL, 10, 9,...) which is actually vec <- c(10, 9, ...), so you get the expected outcome.
> c(NULL, 10:1)
[1] 10 9 8 7 6 5 4 3 2 1
> sum(NULL, 10:1)
[1] 55
> NULL + 10:1
integer(0)
I'm new to Julia and learning use of Map, reduce, filter.
It is becoming very hard for me to comprehend how it can replace for and while loops.
For ex for below code, I would like to replace for loop
function addMultiplesOf3And5(N::Int)
sumOfMultiples = 0
if(N == 3)
return sumOfMultiples + N
end
for i = 3:N-1
if(i % 3 == 0 && i % 5 == 0)
continue
elseif(i % 3 == 0)
sumOfMultiples += i
elseif(i % 5 == 0)
sumOfMultiples += i
end
end
return sumOfMultiples
end
I would really appreciate the help.
Update :
This is what I did after going through tutorials
function addMultiplesOf3And5(N::Int)
array = range(1,N-1)
return reduce(+, map(x -> multiples_of_3_Or_5(x), array))
end
function multiples_of_3_Or_5(n)
if(n % 3 == 0 && n % 5 == 0)
return 0
elseif(n % 3 == 0)
return n
elseif(n % 5 == 0)
return n
else
return 0
end
end
Final:
function addMultiplesOf3And5(N::Int)
array = range(1,N-1)
return reduce(+, filter(x -> ((x%3==0)$(x%5==0)), array))
end
To understand how you can replace your 'for loop + if block' code with 'map / reduce / filter' you need to know exactly how they work and why they might be chosen instead.
1. The map function
map is a function that takes a function variable and a list as arguments, and returns a new list, where each element is the result of applying the function to each element of the old list. So for example if your variable f refers to a function f(x) = x + 5 you defined earlier, and you have a list L=[1,2,3,4,5], then map(f, L) will return [f(L[1]), f(L[2]), f(L[3]), f(L[4]), f(L[5])]
So if you have code like:
f(x) = x + 5;
L = [1,2,3,4,5];
A = zeros(5);
for i in L
A[i] = f(i);
end
You could rewrite this as a mapping operation like so:
A = map(x -> x + 5, [1,2,3,4,5]);
2. The reduce function
reduce takes a binary function variable (i.e. a function that takes two arguments) and a list as arguments. What it does is probably best explained by an example. Calling reduce with the + operator, and list [1,2,3,4,5] will do the following:
Step 1: [1, 2, 3, 4, 5] % : 5 elements
Step 2: [1+2, 3, 4, 5] % [3,3,4,5] : 4 elements
Step 3: [3+3, 4, 5] % [6, 4, 5] : 3 elements
Step 4: [6+4, 5] % [10, 5] : 2 elements
Step 5: [10+5] % [15] : 1 elements
result: 15
i.e. we have reduced the list to a single result by successively applying the binary function to the first pair of elements, consuming the list little by little.
So if you have code like:
f(x,y) = x + y
L = [1,2,3,4,5];
A = L[1];
for i in 2:length(L)
A = f(A, L[i])
end
you could rewrite this as a reduce operation like so:
A = reduce(x,y -> x+y, [1,2,3,4,5])
3. The filter function
filter takes a predicate function (e.g. iseven, isnull, ==, or anything that takes an argument and performs a test on it, resulting in true or false) and a list, tests each element of the list with the function and returns a new list that only contains the elements that pass that test. e.g.
filter(iseven, [1,2,3,4,5]) # returns [2,4]
The answer to your problem
If I understand correctly, addMultiplesOf3And5 takes a number N (e.g. 20), and does the following:
filter out all the elements that can be divided by either 3 or 5 from the list [1,2,3,...,20]
successively add all elements of the resulting list together using a reduce function.
You should be able to use the above to figure out the exact code :)
Not sure what the function in the question is supposed to calculate, but:
addMult3or5(N) = N==3 ? 3 : sum(filter(x->((x%3==0)$(x%5==0)),3:N-1))
calculates the same thing.
sum is a a reduce-like function for the + operation.
Hope this helps clarify.
Also, $ is the exclusive-or operation in Julia.