I can think of some very convoluted methods with loops and nested loops to solve this problem but I'm trying to be more professional than that.
My scenario is that I need to enter a section of code every ten percent but it isn't quite working as expected. It is entering the code about every percent which is due to my code but I lack the knowledge to know how to change it.
int currentPercent = Math.Truncate((current * 100M) / total);
//avoid divide by zero error
if (currentPercent > 0)
{
if (IsDivisible(100, currentPercent))
{
....my code that works fine other than coming in too many times
}
}
Helper referenced above where the trouble is:
private bool IsDivisible(int x, int y)
{
return (x % y) == 0;
}
So obviously it works as it should. Mod eliminates currentPercent of 3 but 1 & 2 pass when really I don't want a true value until currentPercent = 10 and then not again till 20...etc.
Thank you and my apologies for the elementary question
Mod will only catch exact occurrences of your interval. Try keeping track of your next milestone, you'll be less likely to miss them.
const int cycles = 100;
const int interval = 10;
int nextPercent = interval;
for (int index = 0; index <= cycles; index++)
{
int currentPercent = (index * 100) / cycles;
if (currentPercent >= nextPercent)
{
nextPercent = currentPercent - (currentPercent % interval) + interval;
}
}
I might misunderstand you, but it seems like you're trying to do something extremely simple more complex than it needs to be. What about this?
for (int i = 1; i <= 100; i++)
{
if (i % 10 == 0)
{
// Here, you can do what you want - this will happen
// every ten iterations ("percent")
}
}
Or, if your entire code enters from somewhere else (so no loop in this scope), the important part is the i % 10 == 0.
if (IsDivisible(100, currentPercent))
{
....my code that works fine other than coming in too many times
}
try changing that 100 to a 10. And I think your x and y are also backwards.
You can try a few sample operations using google calculator.
(20 mod 10) = 0
Not sure if I fully understand, but I think this is what you want? You also reversed the order of modulo in your code (100 mod percent, rather than the other way around):
int currentPercent = current * 100 / total;
if (currentPercent % 10 == 0)
{
// your code here, every 10%, starting at 0%
}
Note that code this way only works properly if you are guaranteed to hit every percentage-mark. If you could, say, skip from 19% to 21% then you'll need to keep track of which percentage the previous time was to see if you went over a 10% mark.
try this:
for (int percent = 1; percent <= 100; percent++)
{
if (percent % 10 == 0)
{
//code goes here
}
}
Depending on how you increment your % value, this may or may not work % 10 == 0. For example jumping from 89 to 91 % would effectively skip the code execution. You should store last executed value, 80 in this case. Then check if interval is >= 10, so 90 would work, as well as 91.
Related
I'm working on this:
Write a function, persistence, that takes in a positive parameter num
and returns its multiplicative persistence, which is the number of
times you must multiply the digits in num until you reach a single
digit.
For example:
persistence(39) == 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
// and 4 has only one digit
persistence(999) == 4 // because 9*9*9 = 729, 7*2*9 = 126,
// 1*2*6 = 12, and finally 1*2 = 2
persistence(4) == 0 // because 4 is already a one-digit number
This is what I tried:
public static int Persistence(long n)
{
List<long> listofints = new List<long>();
while (n > 0)
{
listofints.Add(n % 10);
n /= 10;
}
listofints.Reverse();
// list of a splited number
int[] arr = new int[listofints.Count];
for (int i = 0; i < listofints.Count; i++)
{
arr[i] = (int)listofints[i];
}
//list to array
int pro = 1;
for (int i = 0; i < arr.Length; i++)
{
pro *= arr[i];
}
// multiply each number
return pro;
}
I have a problem with understanding recursion - probably there is a place to use it. Can some1 give me advice not a solution, how to deal with that?
It looks like you've got the complete function to process one iteration. Now all you need to do is add the recursion. At the end of the function call Persistence again with the result of the first iteration as the parameter.
Persistence(pro);
This will recursively call your function passing the result of each iteration as the parameter to the next iteration.
Finally, you need to add some code to determine when you should stop the recursion, so you only want to call Persistence(pro) if your condition is true. This way, when your condition becomes false you'll stop the recursion.
if (some stop condition is true)
{
Persistence(pro);
}
Let me take a stab at explaining when you should consider using a recursive method.
Example of Factorial: Factorial of n is found by multiplying 1*2*3*4*..*n.
Suppose you want to find out what the factorial of a number is. For finding the answer, you can write a foreach loop that keeys multiplying a number with the next number and the next number until it reaches 0. Once you reach 0, you are done, you'll return your result.
Instead of using loops, you can use Recursion because the process at "each" step is the same. Multiply the first number with the result of the next, result of the next is found by multiplying that next number with the result of the next and so on.
5 * (result of rest)
4 * (result of rest )
3 * (result of rest)
...
1 (factorial of 0 is 1).---> Last Statement.
In this case, if we are doing recursion, we have a terminator of the sequence, the last statement where we know for a fact that factorial of 0 = 1. So, we can write this like,
FactorialOf(5) = return 5 * FactorialOf(4) = 120 (5 * 24)
FactorialOf(4) = return 4 * FactorialOf(3) = 24 (4 * 6)
FactorialOf(3) = return 3 * FactorialOf(2) = 6 (3 * 2)
FactorialOf(2) = return 2 * FactorialOf(1) = 2 (2 * 1)
FactorialOf(1) = return 1 * FactorialOf(0) = 1 (1 * 1)
FactorialOf(0) = Known -> 1.
So, it would make sense to use the same method over and over and once we get to our terminator, we stop and start going back up the tree. Each statement that called the FactorialOf would start returning numbers until it reaches all the way to the top. At the top, we will have our answer.
Your case of Persistence
It calls for recursive method as well as you are taking the result and doing the same process on it each time.
Persistence(39) (not single) = return 1 + Persistence(3 * 9 = 27) = 3
Persistence(27) (not single) = return 1 + Persistence(2 * 7 = 14) = 2
Persistence(14) (not single) = return 1 + Persistence(1 * 4 = 4) = 1
Persistence(4) (single digit) = Known -> 0 // Terminator.
At the end of the day, if you have same process performed after each calculation / processing with a termination, you can most likely find a way to use recursion for that process.
You definitely can invoke your multiplication call recursively.
You will need initial sate (0 multiplications) and keep calling your method until you reach your stop condition. Then you return the last iteration you've got up to as your result and pass it through all the way up:
int persistence(int input, int count = 0) {} // this is how I would define the method
// and this is how I see the control flowing
var result = persistence(input: 39, count: 0) {
//write a code that derives 27 out of 39
//then keep calling persistence() again, incrementing the iteration count with each invocation
return persistence(input: 27, count: 1) {
return persistence(input: 14, count: 2) {
return persistence(input: 4, count: 3) {
return 3
}
}
}
}
the above is obviously not a real code, but I'm hoping that illustrates the point well enough for you to explore it further
Designing a simple recursive solution usually involves two steps:
- Identify the trivial base case to which you can calculate the answer easily.
- Figure out how to turn a complex case to a simpler one, in a way that quickly approaches the base case.
In your problem:
- Any single-digit number has a simple solution, which is persistence = 1.
- Multiplying all digits of a number produces a smaller number, and we know that the persistence of the bigger number is greater than the persistence of the smaller number by exactly one.
That should bring you to your solution. All you need to do is understand the above and write that in C#. There are only a few modifications that you need to make in your existing code. I won't give you a ready solution as that kinda defeats the purpose of the exercise, doesn't it. If you encounter technical problems with codifying your solution into C#, you're welcome to ask another question.
public int PerRec(int n)
{
string numS = n.ToString();
if(numS.Length == 1)
return 0;
var number = numS.ToArray().Select(x => int.Parse(x.ToString())).Aggregate((a,b) => a*b);
return PerRec(number) + 1;
}
For every recursion, you should have a stop condition(a single digit in this case).
The idea here is taking your input and convert it to string to calculate that length. If it is 1 then you return 0
Then you need to do your transformation. Take all the digits from the string representation(in this case from the char array, parse all of them, after getting the IEnumerable<int>, multiply each digit to calculate the next parameter for your recursion call.
The final result is the new recursion call + 1 (which represents the previous transformation)
You can do this step in different ways:
var number = numS.ToArray().Select(x => int.Parse(x.ToString())).Aggregate((a,b) => a*b);
convert numS into an array of char calling ToArray()
iterate over the collection and convert each char into its integer representation and save it into an array or a list
iterate over the int list multiplying all the digits to have the next number for your recursion
Hope this helps
public static int Persistence(long n)
{
if (n < 10) // handle the trivial cases - stop condition
{
return 0;
}
long pro = 1; // int may not be big enough, use long instead
while (n > 0) // simplify the problem by one level
{
pro *= n % 10;
n /= 10;
}
return 1 + Persistence(pro); // 1 = one level solved, call the same function for the rest
}
It is the classic recursion usage. You handle the basic cases, simplify the problem by one level and then use the same function again - that is the recursion.
You can rewrite the recursion into loops if you wish, you always can.
I have a small problem. My code is this one :
int c = 0;
int i = 0;
int a = 28;
while (i < a) {
i++;
if (i % a == 0) {
c += i;
Console.WriteLine(i.ToString());
}
}
Why does the string i is displayed only once, after the end of the while ? It should be displayed a times.
Your help will be appreciated !
Your if condition is opposite it should be:
if (a % i == 0)
Currently you are trying to do remainder division with i % a and it will only meet the condition when i reaches 28, so you get the output once.
% is for modulus division, which basically divides by the number and gives you back the remainder. When you're loop reaches 28 it divides it by 28 and the resulting remainder is 0. This only happens once "when your loop reaches 28".
It would help if you told us what was printed out. I guess it is 28 because
i % a
returns the reminder of
i / a
(i divided by a) and it is only 0 when i is equal to a, i.e., 28.
I was adapting a simple prime-number generation one-liner from Scala to C# (mentioned in a comment on this blog by its author). I came up with the following:
int NextPrime(int from)
{
while(true)
{
n++;
if (!Enumerable.Range(2, (int)Math.Sqrt(n) - 1).Any((i) => n % i == 0))
return n;
}
}
It works, returning the same results I'd get from running the code referenced in the blog. In fact, it works fairly quickly. In LinqPad, it generated the 100,000th prime in about 1 second. Out of curiosity, I rewrote it without Enumerable.Range() and Any():
int NextPrimeB(int from)
{
while(true)
{
n++;
bool hasFactor = false;
for (int i = 2; i <= (int)Math.Sqrt(n); i++)
{
if (n % i == 0) hasFactor = true;
}
if (!hasFactor) return n;
}
}
Intuitively, I'd expect them to either run at the same speed, or even for the latter to run a little faster. In actuality, computing the same value (100,000th prime) with the second method, takes 12 seconds - It's a staggering difference.
So what's going on here? There must be fundamentally something extra happening in the second approach that's eating up CPU cycles, or some optimization going on the background of the Linq examples. Anybody know why?
For every iteration of the for loop, you are finding the square root of n. Cache it instead.
int root = (int)Math.Sqrt(n);
for (int i = 2; i <= root; i++)
And as other have mentioned, break the for loop as soon as you find a factor.
The LINQ version short circuits, your loop does not. By this I mean that when you have determined that a particular integer is in fact a factor the LINQ code stops, returns it, and then moves on. Your code keeps looping until it's done.
If you change the for to include that short circuit, you should see similar performance:
int NextPrimeB(int from)
{
while(true)
{
n++;
for (int i = 2; i <= (int)Math.Sqrt(n); i++)
{
if (n % i == 0) return n;;
}
}
}
It looks like this is the culprit:
for (int i = 2; i <= (int)Math.Sqrt(n); i++)
{
if (n % i == 0) hasFactor = true;
}
You should exit the loop once you find a factor:
if (n % i == 0){
hasFactor = true;
break;
}
And as other have pointed out, move the Math.Sqrt call outside the loop to avoid calling it each cycle.
Enumerable.Any takes an early out if the condition is successful while your loop does not.
The enumeration of source is stopped as soon as the result can be determined.
This is an example of a bad benchmark. Try modifying your loop and see the difference:
if (n % i == 0) { hasFactor = true; break; }
}
throw new InvalidOperationException("Cannot satisfy criteria.");
In the name of optimization, you can be a little more clever about this by avoiding even numbers after 2:
if (n % 2 != 0)
{
int quux = (int)Math.Sqrt(n);
for (int i = 3; i <= quux; i += 2)
{
if (n % i == 0) return n;
}
}
There are some other ways to optimize prime searches, but this is one of the easier to do and has a large payoff.
Edit: you may want to consider using (int)Math.Sqrt(n) + 1. FP functions + round-down could potentially cause you to miss a square of a large prime number.
At least part of the problem is the number of times Math.Sqrt is executed. In the LINQ query this is executed once but in the loop example it's executed N times. Try pulling that out into a local and reprofiling the application. That will give you a more representative break down
int limit = (int)Math.Sqrt(n);
for (int i = 2; i <= limit; i++)
I manually adjust the thread count:
if (items.Count == 0) { threads = 0; }
else if (items.Count < 1 * hundred) { threads = 1; }
else if (items.Count < 3 * hundred) { threads = 2; }
else if (items.Count < 5 * hundred) { threads = 4; }
else if (items.Count < 10 * hundred) { threads = 8; }
else if (items.Count < 20 * hundred) { threads = 11; }
else if (items.Count < 30 * hundred) { threads = 15; }
else if (items.Count < 50 * hundred) { threads = 30; }
else threads = 40;
I need a function that returns the necessary/optimized thread count.
Ok, now forget above. I need a graph curve to plot. I give the coords, function plots the curve. Imagine the point(0,0) and point(5,5) -in (x,y) form. It should be straight line. So then I can measure x for y=3.
What happens if I give the points (0,0), (2,3), (8,10), (15,30) and (30,50). It will be a curve like thing. Now can I calculate x for given y or vice versa?
I think you get the idea. Should I use MathLab or could it be done in C#?
You're looking for curve fitting, or the derivation of a function describing a curve from a set of data points. If you're looking to do this once, from a constant set of data, Matlab would do the job just fine. If you want to do this dynamically, there are libraries and algorithms out there.
Review the Wikipedia article on linear regression. The least squares approach mentioned in that article is pretty common. Look around, and you'll find libraries and code samples using that approach.
You can probably make that run faster by reordering the tests (and using nested if). But that's not a smooth function, there's not likely to be any simpler description.
Or are you trying to find a smooth function that passes near those points?
You could use a linear regression; you would get something like this:
So I would probably encode it in C# like this:
int threads = (int) Math.Ceiling(0.0056*items.Count + 0.5);
I used Math.Ceiling to ensure that you don’t get 0 when the input isn’t 0. Of course, this function gives you 1 even if the input is 0; if that matters, you can always catch that as a special case, or use Math.Round instead.
However, this means the number of threads will go up continuously. It will not level out at 40. If that’s what you want, you might need to research different kinds of regression.
I'm trying to refactor this algorithm to make it faster. What would be the first refactoring here for speed?
public int GetHowManyFactors(int numberToCheck)
{
// we know 1 is a factor and the numberToCheck
int factorCount = 2;
// start from 2 as we know 1 is a factor, and less than as numberToCheck is a factor
for (int i = 2; i < numberToCheck; i++)
{
if (numberToCheck % i == 0)
factorCount++;
}
return factorCount;
}
The first optimization you could make is that you only need to check up to the square root of the number. This is because factors come in pairs where one is less than the square root and the other is greater.
One exception to this is if n is an exact square then its square root is a factor of n but not part of a pair.
For example if your number is 30 the factors are in these pairs:
1 x 30
2 x 15
3 x 10
5 x 6
So you don't need to check any numbers higher than 5 because all the other factors can already be deduced to exist once you find the corresponding small factor in the pair.
Here is one way to do it in C#:
public int GetFactorCount(int numberToCheck)
{
int factorCount = 0;
int sqrt = (int)Math.Ceiling(Math.Sqrt(numberToCheck));
// Start from 1 as we want our method to also work when numberToCheck is 0 or 1.
for (int i = 1; i < sqrt; i++)
{
if (numberToCheck % i == 0)
{
factorCount += 2; // We found a pair of factors.
}
}
// Check if our number is an exact square.
if (sqrt * sqrt == numberToCheck)
{
factorCount++;
}
return factorCount;
}
There are other approaches you could use that are faster but you might find that this is already fast enough for your needs, especially if you only need it to work with 32-bit integers.
Reducing the bound of how high you have to go as you could knowingly stop at the square root of the number, though this does carry the caution of picking out squares that would have the odd number of factors, but it does help reduce how often the loop has to be executed.
Looks like there is a lengthy discussion about this exact topic here: Algorithm to calculate the number of divisors of a given number
Hope this helps
The first thing to notice is that it suffices to find all of the prime factors. Once you have these it's easy to find the number of total divisors: for each prime, add 1 to the number of times it appears and multiply these together. So for 12 = 2 * 2 * 3 you have (2 + 1) * (1 + 1) = 3 * 2 = 6 factors.
The next thing follows from the first: when you find a factor, divide it out so that the resulting number is smaller. When you combine this with the fact that you need only check to the square root of the current number this is a huge improvement. For example, consider N = 10714293844487412. Naively it would take N steps. Checking up to its square root takes sqrt(N) or about 100 million steps. But since the factors 2, 2, 3, and 953 are discovered early on you actually only need to check to one million -- a 100x improvement!
Another improvement: you don't need to check every number to see if it divides your number, just the primes. If it's more convenient you can use 2 and the odd numbers, or 2, 3, and the numbers 6n-1 and 6n+1 (a basic wheel sieve).
Here's another nice improvement. If you can quickly determine whether a number is prime, you can reduce the need for division even further. Suppose, after removing small factors, you have 120528291333090808192969. Even checking up to its square root will take a long time -- 300 billion steps. But a Miller-Rabin test (very fast -- maybe 10 to 20 nanoseconds) will show that this number is composite. How does this help? It means that if you check up to its cube root and find no factors, then there are exactly two primes left. If the number is a square, its factors are prime; if the number is not a square, the numbers are distinct primes. This means you can multiply your 'running total' by 3 or 4, respectively, to get the final answer -- even without knowing the factors! This can make more of a difference than you'd guess: the number of steps needed drops from 300 billion to just 50 million, a 6000-fold improvement!
The only trouble with the above is that Miller-Rabin can only prove that numbers are composite; if it's given a prime it can't prove that the number is prime. In that case you may wish to write a primality-proving function to spare yourself the effort of factoring to the square root of the number. (Alternately, you could just do a few more Miller-Rabin tests, if you would be satisfied with high confidence that your answer is correct rather than a proof that it is. If a number passes 15 tests then it's composite with probability less than 1 in a billion.)
You can limit the upper limit of your FOR loop to numberToCheck / 2
Start your loop counter at 2 (if your number is even) or 3 (for odd values). This should allow you to check every other number dropping your loop count by another 50%.
public int GetHowManyFactors(int numberToCheck)
{
// we know 1 is a factor and the numberToCheck
int factorCount = 2;
int i = 2 + ( numberToCheck % 2 ); //start at 2 (or 3 if numberToCheck is odd)
for( ; i < numberToCheck / 2; i+=2)
{
if (numberToCheck % i == 0)
factorCount++;
}
return factorCount;
}
Well if you are going to use this function a lot you can use modified algorithm of Eratosthenes http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes and store answars for a interval 1 to Max in array. It will run IntializeArray() once and after it will return answers in 0(1).
const int Max =1000000;
int arr [] = new int [Max+1];
public void InitializeArray()
{
for(int i=1;i<=Max;++i)
arr[i]=1;//1 is factor for everyone
for(int i=2;i<=Max;++i)
for(int j=i;i<=Max;i+=j)
++arr[j];
}
public int GetHowManyFactors(int numberToCheck)
{
return arr[numberToCheck];
}
But if you are not going to use this function a lot I think best solution is to check unitll square root.
Note: I have corrected my code!
An easy to implement algorithm that will bring you much farther than trial division is Pollard Rho
Here is a Java implementation, that should be easy to adapt to C#: http://www.cs.princeton.edu/introcs/78crypto/PollardRho.java.html
https://codility.com/demo/results/demoAAW2WH-MGF/
public int solution(int n) {
var counter = 0;
if (n == 1) return 1;
counter = 2; //1 and itself
int sqrtPoint = (Int32)(Math.Truncate(Math.Sqrt(n)));
for (int i = 2; i <= sqrtPoint; i++)
{
if (n % i == 0)
{
counter += 2; // We found a pair of factors.
}
}
// Check if our number is an exact square.
if (sqrtPoint * sqrtPoint == n)
{
counter -=1;
}
return counter;
}
Codility Python 100 %
Here is solution in python with little explanation-
def solution(N):
"""
Problem Statement can be found here-
https://app.codility.com/demo/results/trainingJNNRF6-VG4/
Codility 100%
Idea is count decedent factor in single travers. ie. if 24 is divisible by 4 then it is also divisible by 8
Traverse only up to square root of number ie. in case of 24, 4*4 < 24 but 5*5!<24 so loop through only i*i<N
"""
print(N)
count = 0
i = 1
while i * i <= N:
if N % i == 0:
print()
print("Divisible by " + str(i))
if i * i == N:
count += 1
print("Count increase by one " + str(count))
else:
count += 2
print("Also divisible by " + str(int(N / i)))
print("Count increase by two count " + str(count))
i += 1
return count
Example by run-
if __name__ == '__main__':
# result = solution(24)
# result = solution(35)
result = solution(1)
print("")
print("Solution " + str(result))
"""
Example1-
24
Divisible by 1
Also divisible by 24
Count increase by two count 2
Divisible by 2
Also divisible by 12
Count increase by two count 4
Divisible by 3
Also divisible by 8
Count increase by two count 6
Divisible by 4
Also divisible by 6
Count increase by two count 8
Solution 8
Example2-
35
Divisible by 1
Also divisible by 35
Count increase by two count 2
Divisible by 5
Also divisible by 7
Count increase by two count 4
Solution 4
Example3-
1
Divisible by 1
Count increase by one 1
Solution 1
"""
Github link
I got pretty good results with complexity of O(sqrt(N)).
if (N == 1) return 1;
int divisors = 0;
int max = N;
for (int div = 1; div < max; div++) {
if (N % div == 0) {
divisors++;
if (div != N/div) {
divisors++;
}
}
if (N/div < max) {
max = N/div;
}
}
return divisors;
Python Implementation
Score 100% https://app.codility.com/demo/results/trainingJ78AK2-DZ5/
import math;
def solution(N):
# write your code in Python 3.6
NumberFactor=2; #one and the number itself
if(N==1):
return 1;
if(N==2):
return 2;
squareN=int(math.sqrt(N)) +1;
#print(squareN)
for elem in range (2,squareN):
if(N%elem==0):
NumberFactor+=2;
if( (squareN-1) * (squareN-1) ==N):
NumberFactor-=1;
return NumberFactor