Recursively counting divisors of a number - function

I have a question, more on the theoretical side. I want to make a recursive function that counts all (not only prime) different divisors of a given natural number.
For example with f(0)=0 (per Def.), f(3) = 2, f(6) = 4, f(16) = 5 etc.
Theoretically, how could I do that?
Thanks.


If I understand correctly, you only want to COUNT them, not to collect them, right?
A second assumption is that you don't want to count only independent divisors (i.e. you want to count "2", "3" but not "6").
If this is the case, the algorithm shown in Sean's answer can be simplified significantly:
You don't need the array divisorList but only a counter,
You as soon as you find a divisor, you can reduce the max limit of the loop by the result of dividing the root number by the divisor (e.g. if your root number is 900 and 2 is the first divisor, you can set the limit of the loop to 450; then, when checking 3 you will reduce the limit to 150 and so on).
EDIT:
After thinking a little bit more, here is the correct algorithm:
Assume that the number is "N". Then, you already start with a count of 2 (i.e. 1 and N),
You then check if N divides by 2; if it does, you need to add 2 to the count (i.e. 2 and N/2),
You then change the limit of the loop to N/2,
Test if dividing by 3 yields an integer; if it does, you add 2 to the count (i.e. 3 and N/3) and reduce the limit to N/3,
Test 4...
Test 5...
...
In Pseudo-code:
var Limit = N ;
Count = 2 ;
for (I = 2 ; I < Limit ; I++) {
if (N/I is integer) {
Count = Count + 3 ;
Limit = N / I ;
} ;
} ;
Note: I don't know which language you are programming, so you need to verify if your language allows you to change the limit of the loop. If it does not, you can include an EXIT-LOOP condition (e.g. if I >= Limit then exit loop).
Hope this resolves your problem.


public static ArrayList<int> recursiveDivisors(int num)
{
ArrayList<int> divisorList = new ArrayList<int>();
for (int i = 1; i <= num; i++)
{
if (num % i == 0)
divisorList.add(i)
}
return divisorList;
}
Something like this?
Returns all divisors in a divisor array list
EDIT: Not recursive

Related

Finding prime numbers up till a number

I am trying to list down all the prime numbers up till a specific number e.g. 1000. The code gets slower as the number increase. I am pretty sure it is because of the for loop where (number -1) is checked by all the prime_factors. Need some advise how I can decrease the processing time of the code for larger numbers. Thanks
import time
t0 = time.time()
prime_list = [2]
number = 0
is_not_prime = 0
count = 0
while number < 1000:
print(number)
for i in range (2,number):
count = 0
if (number%i) == 0:
is_not_prime = 1
if is_not_prime == 1:
for j in range (0,len(prime_list)):
if(number-1)%prime_list[j] != 0:
count += 1
if count == len(prime_list):
prime_list.append(number-1)
is_not_prime = 0
count = 0
break
number += 1
print(prime_list)
t1 = time.time()
total = t1-t0
print(total)

Your solution, on top of being confusing, is very inefficient - O(n^3). Please, use the Sieve of Eratosthenes. Also, learn how to use booleans.
Something like this (not optimal, just a mock-up). Essentially, you start with a list of all numbers, 1-1000. Then, you remove ones that are the multiple of something.
amount = 1000
numbers = range(1, amount)
i = 1
while i < len(numbers):
n = i + 1
while n < len(numbers):
if numbers[n] % numbers[i] == 0:
numbers.pop(n)
else:
n += 1
i += 1
print(numbers)
Finally, I was able to answer because your question isn't language-specific, but please tag the question with the language you're using in the example.

Writing Fibonacci Sequence Elegantly Python

I am trying to improve my programming skills by writing functions in multiple ways, this teaches me new ways of writing code but also understanding other people's style of writing code. Below is a function that calculates the sum of all even numbers in a fibonacci sequence up to the max value. Do you have any recommendations on writing this algorithm differently, maybe more compactly or more pythonic?
def calcFibonacciSumOfEvenOnly():
MAX_VALUE = 4000000
sumOfEven = 0
prev = 1
curr = 2
while curr <= MAX_VALUE:
if curr % 2 == 0:
sumOfEven += curr
temp = curr
curr += prev
prev = temp
return sumOfEven
I do not want to write this function recursively since I know it takes up a lot of memory even though it is quite simple to write.

You can use a generator to produce even numbers of a fibonacci sequence up to the given max value, and then obtain the sum of the generated numbers:
def even_fibs_up_to(m):
a, b = 0, 1
while a <= m:
if a % 2 == 0:
yield a
a, b = b, a + b
So that:
print(sum(even_fibs_up_to(50)))
would output: 44 (0 + 2 + 8 + 34 = 44)

Arduino - waiting on function causes a number pile-up

I have a function filledFunction() that returns a float filled:
float filledFunction(){
if (FreqMeasure.available()) {
sum = sum + FreqMeasure.read();
count = count + 1;
if (count > 30) {
frequency = FreqMeasure.countToFrequency(sum / count);
a = frequency * x;
b = exp (a);
c = w * b;
d = frequency * z;
e = exp (d);
f = y * e;
float filled = c + f;
sum = 0;
count = 0;
return filled;
}
}
}
When I call this function with
while (1){
fillLevel = filledFunction();
int tofill = 500 - fillLevel;
Serial.print("fillLevel: ");
Serial.println(fillLevel);
Serial.print("tofill: ");
Serial.println(tofill);
The serial monitor should output two numbers that add up to 500 named fillLevel and tofill. Instead I get a repeating sequence of similar values:
http://i.imgur.com/Y9Wu8P2.png
The First two values are correct (410.93 + 89 = 500), but the following 60ish values are unknown to me, and do not belong there.
I am using an arduino nano

The filledFunction() function only returns a value if FreqMeasure.available() returns true AND count > 30. As stated in the answers to this question the C89, C99 and C11 standards all say that the default return value of a function is undefined (that is if the function completes without executing a return statement). Which really means that anything could happen, such as outputting arbitrary numbers.
In addition, the output that you're seeing starts off 'correct' with one of the numbers subtracted from 500, even when they have weird values such as 11699.00 and -11199 (which equals 500 - 11699.00). However, lower down in the output this seems to break down and the reason is that on Arduino Nano an int can only hold numbers up to 32767 and therefore the result of the subtraction is too big and 'overflows' to be a large negative number.
Fixing the filledFunction() function to explicitly return a value even if FreqMeasure.available() is false or count <= 30 and ensuring that it can't return a number greater than 500 will likely solve these issues.

What way to "rank" dataset Mysql

Situation is as follows:
I have a database with 40.000 cities. Those cities have certain types of properties with an value.
For example "mountains" or "beaches". If a city has lots of mountains the value for mountain will be high if there are less mountains the number is lower.
Table with city name and properties and values:
With that, I have a table with the avarage values of all those properties.
What I need to happen: I want the user search for a city with has one or multiple properties, find the best match and attach a score from 0 - 100 to it.
The way I do this is as follow:
1. I first get the 25%, 50% and 70% values for the properties:
_var_[property]_25 = [integer]
_var_[property]_50 = [integer]
_var_[property]_70 = [integer]
2. Then I need to use this algorithm:
_var_user_search_for_properties = [mountain,beach]
_var_max_property_percentage = 100 / [properties user search for]
_var_match_percentage = 0
for each _var_user_search_for_properties
if [property] < _var_[property]_25 then
_var_match_percentage += _var_max_property_percentage
elseif [property] < _var_[property]_50 then
_var_match_percentage += _var_max_property_percentage / 4 * 3
elseif [property] < _var_[property]_75 then
_var_match_percentage += _var_max_property_percentage / 4 * 2
elseif [property] < 0 then
_var_match_percentage += _var_max_property_percentage / 4 * 1
end if
next
order all rows by _var_match_percentage desc
The question is: is it posible to do this with MySQL?
How do I calculate this "match percentage" with it?
Or wil it be faster to get all the rows and indexes out of the database and loop them all trough .NET?

If the percentages can be stored in the database, you could try MySQL's LIMIT clause. See http://www.mysqltutorial.org/mysql-limit.aspx.

Finding closest match in collection of numbers [closed]

Closed. This question needs to be more focused. It is not currently accepting answers.
Closed 8 years ago.
Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
So I got asked today what was the best way to find the closes match within a collection.
For example, you've got an array like this:
1, 3, 8, 10, 13, ...
What number is closest to 4?
Collection is numerical, unordered and can be anything. Same with the number to match.
Lets see what we can come up with, from the various languages of choice.

11 bytes in J:
C=:0{]/:|#-
Examples:
>> a =: 1 3 8 10 13
>> 4 C a
3
>> 11 C a
10
>> 12 C a
13
my breakdown for the layman:
0{ First element of
] the right argument
/: sorted by
| absolute value
# of
- subtraction

Shorter Python: 41 chars
f=lambda a,l:min(l,key=lambda x:abs(x-a))

My attempt in python:
def closest(target, collection) :
return min((abs(target - i), i) for i in collection)[1]

Groovy 28B
f={a,n->a.min{(it-n).abs()}}

Some C# Linq ones... too many ways to do this!
decimal[] nums = { 1, 3, 8, 12 };
decimal target = 4;
var close1 = (from n in nums orderby Math.Abs(n-target) select n).First();
var close2 = nums.OrderBy(n => Math.Abs(n - target)).First();
Console.WriteLine("{0} and {1}", close1, close2);
Even more ways if you use a list instead, since plain ol arrays have no .Sort()

Assuming that the values start in a table called T with a column called N, and we are looking for the value 4 then in Oracle SQL it takes 59 characters:
select*from(select*from t order by abs(n-4))where rownum=1
I've used select * to reduce the whitespace requirements.

Because I actually needed to do this, here is my PHP
$match = 33;
$set = array(1,2,3,5,8,13,21,34,55,89,144,233,377,610);
foreach ($set as $fib)
{
$diff[$fib] = (int) abs($match - $fib);
}
$fibs = array_flip($diff);
$closest = $fibs[min($diff)];
echo $closest;

PostgreSQL:
select n from tbl order by abs(4 - n) limit 1
In the case where two records share the same value for "abs(4 - id)" the output would be in-determinant and perhaps not a constant. To fix that I suggest something like the untested guess:
select n from tbl order by abs(4 - n) + 0.5 * 4 > n limit 1;
This solution provides performance on the order of O(N log N), where O(log N) is possible for example: https://stackoverflow.com/a/8900318/1153319

Ruby like Python has a min method for Enumerable so you don't need to do a sort.
def c(value, t_array)
t_array.min{|a,b| (value-a).abs <=> (value-b).abs }
end
ar = [1, 3, 8, 10, 13]
t = 4
c(t, ar) = 3

Language: C, Char count: 79
c(int v,int*a,int A){int n=*a;for(;--A;++a)n=abs(v-*a)<abs(v-n)?*a:n;return n;}
Signature:
int closest(int value, int *array, int array_size);
Usage:
main()
{
int a[5] = {1, 3, 8, 10, 13};
printf("%d\n", c(4, a, 5));
}

Scala (62 chars), based on the idea of the J and Ruby solutions:
def c(l:List[Int],n:Int)=l.sort((a,b)=>(a-n).abs<(b-n).abs)(0)
Usage:
println(c(List(1,3,8,10,13),4))

PostgreSQL:
This was pointed out by RhodiumToad on FreeNode and has performance on the order of O(log N)., much better then the other PostgreSQL answer here.
select * from ((select * from tbl where id <= 4
order by id desc limit 1) union
(select * from tbl where id >= 4
order by id limit 1)) s order by abs(4 - id) limit 1;
Both of the conditionals should be "or equal to" for much better handling of the id exists case. This also has handling in the case where two records share the same value for "abs(4 - id)" then that other PostgreSQL answer here.

The above code doesn't works for floating numbers.
So here's my revised php code for that.
function find_closest($match, $set=array()) {
foreach ($set as $fib) {
$diff[$fib] = abs($match - $fib);
}
return array_search(min($diff), $diff);
}
$set = array('2.3', '3.4', '3.56', '4.05', '5.5', '5.67');
echo find_closest(3.85, $set); //return 4.05

Python by me and https://stackoverflow.com/users/29253/igorgue based on some of the other answers here. Only 34 characters:
min([(abs(t-x), x) for x in a])[1]

Haskell entry (tested):
import Data.List
near4 = head . sortBy (\n1 n2 -> abs (n1-4) `compare` abs (n2-4))
Sorts the list by putting numbers closer to 4 near the the front. head takes the first element (closest to 4).

Ruby
def c(r,t)
r.sort{|a,b|(a-t).abs<=>(b-t).abs}[0]
end
Not the most efficient method, but pretty short.

returns only one number:
var arr = new int[] { 1, 3, 8, 10, 13 };
int numToMatch = 4;
Console.WriteLine("{0}",
arr.OrderBy(n => Math.Abs(numToMatch - n)).ElementAt(0));

returns only one number:
var arr = new int[] { 1, 3, 8, 10, 13 };
int numToMatch = 4;
Console.WriteLine("{0}",
arr.Select(n => new{n, diff = Math.Abs(numToMatch - n) }).OrderBy(x => x.diff).ElementAt(0).n);

Perl -- 66 chars:
perl -e 'for(qw/1 3 8 10 13/){$d=($_-4)**2; $c=$_ if not $x or $d<$x;$x=$d;}print $c;'

EDITED = in the for loop
int Closest(int val, int[] arr)
{
int index = 0;
for (int i = 0; i < arr.Length; i++)
if (Math.Abs(arr[i] - val) < Math.Abs(arr[index] - val))
index = i;
return arr[index];
}

Here's another Haskell answer:
import Control.Arrow
near4 = snd . minimum . map (abs . subtract 4 &&& id)

Haskell, 60 characters -
f a=head.Data.List.sortBy(compare`Data.Function.on`abs.(a-))

Kdb+, 23B:
C:{x first iasc abs x-}
Usage:
q)a:10?20
q)a
12 8 10 1 9 11 5 6 1 5
q)C[a]4
5

Python, not sure how to format code, and not sure if code will run as is, but it's logic should work, and there maybe builtins that do it anyways...
list = [1,4,10,20]
num = 7
for lower in list:
if lower <= num:
lowest = lower #closest lowest number
for higher in list:
if higher >= num:
highest = higher #closest highest number
if highest - num > num - lowest: # compares the differences
closer_num = highest
else:
closer_num = lowest

In Java Use a Navigable Map
NavigableMap <Integer, Integer>navMap = new ConcurrentSkipListMap<Integer, Integer>();
navMap.put(15000, 3);
navMap.put(8000, 1);
navMap.put(12000, 2);
System.out.println("Entry <= 12500:"+navMap.floorEntry(12500).getKey());
System.out.println("Entry <= 12000:"+navMap.floorEntry(12000).getKey());
System.out.println("Entry > 12000:"+navMap.higherEntry(12000).getKey());

int numberToMatch = 4;
var closestMatches = new List<int>();
closestMatches.Add(arr[0]); // closest tentatively
int closestDifference = Math.Abs(numberToMatch - arr[0]);
for(int i = 1; i < arr.Length; i++)
{
int difference = Math.Abs(numberToMatch - arr[i]);
if (difference < closestDifference)
{
closestMatches.Clear();
closestMatches.Add(arr[i]);
closestDifference = difference;
}
else if (difference == closestDifference)
{
closestMatches.Add(arr[i]);
}
}
Console.WriteLine("Closest Matches");
foreach(int x in closestMatches) Console.WriteLine("{0}", x);

Some of you don't seem to be reading that the list is unordered (although with the example as it is I can understand your confusion). In Java:
public int closest(int needle, int haystack[]) { // yes i've been doing PHP lately
assert haystack != null;
assert haystack.length; > 0;
int ret = haystack[0];
int diff = Math.abs(ret - needle);
for (int i=1; i<haystack.length; i++) {
if (ret != haystack[i]) {
int newdiff = Math.abs(haystack[i] - needle);
if (newdiff < diff) {
ret = haystack[i];
diff = newdiff;
}
}
}
return ret;
}
Not exactly terse but hey its Java.

Common Lisp using iterate library.
(defun closest-match (list n)
(iter (for i in list)
(finding i minimizing (abs (- i n)))

41 characters in F#:
let C x = Seq.min_by (fun n -> abs(n-x))
as in
#light
let l = [1;3;8;10;13]
let C x = Seq.min_by (fun n -> abs(n-x))
printfn "%d" (C 4 l) // 3
printfn "%d" (C 11 l) // 10
printfn "%d" (C 12 l) // 13

Ruby. One pass-through. Handles negative numbers nicely. Perhaps not very short, but certainly pretty.
class Array
def closest int
diff = int-self[0]; best = self[0]
each {|i|
if (int-i).abs < diff.abs
best = i; diff = int-i
end
}
best
end
end
puts [1,3,8,10,13].closest 4