# Archive for the ‘Average’ Category

This article discusses (a problem that I recently solved on codility ).

The core of the problem is the following:
Given two non negative integers N and M, , the task is to check whether they have the same set of prime divisors.
A prime divisor of an integer P is a prime d s.t.  for some positive . You are given up to  of such queries, and should return the total number of them that evaluates to true.

For instance given if  and  then our function should return *true* because the set of prime divisor of  is equal the
the set of primal divisor of  i.e.  while for  and  the function should return *false*.

## Smart (not enough) Brute Force approach

An easy approach in solving this problem could be computing by bruteforce the set of prime divisors of both numbers and compare them, and return true if they contains the same elements.
In order to achieve that we should:

1. compute the set of primes up to . There are approximately  (Read here for a more info on this topic https://primes.utm.edu/howmany.html)
2. Using the Sieve of Έρατοσθένης approach we are able to compute it in  time and space (see the following code).
3. 
void getPrimesSieve(vector<long long>&primes, const ll n) {
using ll = long;
vector<bool> nums(n + 1, true);
primes.push_back(2);
for (ll i = 3; i * i < n + 1; i += 2) {
if (nums[i]) {
for (ll j = i * i; j < n + 1; j += 2 * i) {
nums[j] = false;
}
}
}

for (ll i = 3 ; i <= n ; i += 2)
if (nums[i])
primes.push_back(i);
}

4. With the set of prime numbers in place then we can start finding the primal divisors. The following function computes the set of unique prime divisors of  and stores them in .
5. bool primalDivisors( long p , vector<long>& F, vector<long>& primes) {
int i = 0;
int c = 0;
if (i >= primes[primes.size()-1])
return false;
while (p > 1) {
if (p % primes[i] == 0)
{
c++;
p /= primes[i];
}
else
{
if (c > 0) {
F.push_back(primes[i]);
c = 0;
}
i++;
}
}
if (c > 0)
F.push_back(primes[i]);

return true;
}

6. Finally we can iterate among all the queries and count the number of pairs returning *true*.
7. int solution(vector<int>& A,vector<int>& B){
int LIM = -1;
int ans = 0;
LIM = max( *max_element(A.begin(), A.end()) ,*max_element(B.begin(), B.end()) ) ;
vector<int> primes;
getPrimesSieve(primes, LIM);

for (unsigned i = 0; i < A.size(); ++i)
{
vector<int> divA;
vector<int> divB;
primalDivisors(A[i] , divA, primes);
primalDivisors(B[i] , divB, primes);

//same set of divisors
if( divA == divB)
ans++;

}
return ans;
}


Altough this approach is correct, it is too slow. There are  primes up to . Since the primalDivisors function gets called up to  times and it has linear complexity then the overall time complexity is:  which translates to !

## Fast Approach using GCD

For two number to share the same set  of prime divisors it must be the case that  and . Checking if a number has a set of prime divisors that is subset of another is easy. In order for two number to have the same set of prime divisors the following must hold:
 and  i.e. their factorization only differ in the exponents.
Since clearly the  we can divide  by it obtaining .  cannot have a larget set of factors than  and hence its set of prime divisor is still contained into 's one. We can keep repeating this process until  becomes one.
Dividing by the  eliminates the common part of among the two number. And if eventually the dividend becomes , that means that the common part was the only part present.

As an example consider  and ,  and . Repeting the process leads to  and  and  and .
This shows that the set of prime divisor of M is contained in the set of prime divisor of N.
If we can show that the set of N is included in M's one, than the two sets must be equal. It is sufficient to do the same process with roles of N and M swapped.
If apply the same process to swapping the values of M and N from the previous example we obtain that  and ,  and .

If there is a divisor that is not common to both number then the process will eventually isolate it.
Imagine that  and . All common divisors will be cancelled out by repeting the division by the GCD leaving  and .

Finally, the following is a possible implementation of the overall previous idea, which is also much simpler and cleaner than the one presented in the previous section.

bool containedSet(int M, int N){
int g = __gcd(min(N,M), max(N,M));
while(g != 1){
M /= g;
g = __gcd(min(N,M), max(N,M));
}
return M==1;
}

int solution(vector<int>& A,vector<int>& B){
int ans = 0;
for (unsigned i = 0; i < A.size(); ++i)
{
//divisor A contained in divisor of B
//divisor A contained in divisor of A
//implies divisor A =  divisor of B
if(containedSet(A[i],B[i]) && containedSet(B[i],A[i]))
ans++;

}
return ans;
}


# Weighted Tree Vertex Cover Problem

Vertex cover of graph  is defined as  s.t.  . In other word a subset of the vertices such that all vertices are incident to a vertex in the vertex cover.
We will derive an algorithm for finding the weight of a minimal (yes is not unique) vertex cover for a subclass of graphs i.e. tree (which are acyclic graphs with the property that only one path between each vertex exists).

Remember that vertex cover for graphs is an NP-Complete (NP-hard and NP, hard at least as all NP problems and it is an NP problem itself) problem i.e. no deterministic polynomial tyme algorithm was discovered (if you discover one, contact me, we will be millionaire).

# Tree Vertex Cover - Problem Definition

Given a weighted tree  with  write an algorithm for computing a vertex cover with minimum weight i.e. V' is a vertex cover and the sum of the weight of its element is minimal.

The following is the tree structure that we will use throughout the article.

template<typename T,int DEGREE>
struct node{
array<node*,DEGREE> children;

T data;
int weight;

node(const T& v , int _weight) : weight(_weight){
data=v;
mincover=0;
}
};



What if the weight is equal to the degree of the node?

The first observation we can make is that the root node can weather be or not be in the vertex cover. If we include it in the solution then we are sure that all the edges from it to its children have been covered and to solve the problem we need only to compute the cover of its children (which is a simpler problem). Read On…

# List Cycle Detection

Linked list cycle detection problem is a very instructive and fun problem to reason about. This article will state the problem and explains how we can solve it efficiently while giving some insight on the underlying maths.

Well, list can gets corrupted and some node can be linked by one or more nodes (as in the following figure). This could lead to never ending traversal of the list. So it make sense to solve the following problem:

Cicular List

## List Cycle Detection - Problem Statement

1. Given  a linked list detect (if any) check if the list is circular i.e. contains a cycle
2. Find the starting node of the cycle (the node with two inward arrows in the figure)

The problem is easily solvable in  time and space considering that we can visit the list from the head and store in a separate list the visited nodes. As the visit continues we check if the node we are examining was previously visited (is that node contained in the support list?). If yes the list is circular and that node is the start point of the cycle. If we reach the tail of the list then the list is not circular. We can lower the complexity of this approach down to time using a more efficient support (set) data structure (like a tree set). But we can do much better, and the rest of the article will show how to obtain  time and .

## List Cycle Detection - Floyd’s algorithm

We use the fact that like clock's hands things iterating on a cicle will meet at some point in the future. Consider two runner  with velocities starting running from the same point in a circular stadium. They will meet again when the slower runner reach the starting point for the second time (Why? By the time the slower one has completed half circle the faster has completed a complete cycle and by the time the slower finishes his run, arriving at the starting point again, the faster has completed a second entire cycle).

Things are a bit more complicated in the List Cycle Detection problems because the iterators do not necessarily start from the same point. Consider two iterators  with velocities   respectively. Suppose the cycle has length  and that it starts at node number  When the slower iterator reaches  the faster is at location . How many iteration  it will take before they meet? And at which node?

The situation is described by the following congruence

• 
• 
• 
• 
• 

which has solution . This means that they will meet after  iteration of the slower iterator and since it starts from  this means that they will meet at  nodes before the beginning of the cycle. We can use that fact to count  nodes from the beginning of the list. Once the iterators matche in the cycle, we can move the fast iterator back to the beginning of the list and iterate forward one node per stepo with both iterators until they match again (They are far away  nodes from the beginning of the cycle, so they will meet exaclty in that point.

Let's consider now the case when . This means that by the time the slower iterator reaches the beginning of the cycle the faster one has completed more that a cycle. What will be the starting point for the faster one? We argue that once p reaches A q is at node 2A but since A > n this means that some it will be at position A + (A mod(n)). We can now use similar argument to the previous case and write:

• 
• 
• 
•  since 

which has solution . This means that the meetpoint is  nodes before the beginning of the cycle. If we do the sape operation as the previous case we obtain the sme result. Itrators wil meet at the beginning of the cycle. Why? Well advancing  makes  cycles possibly several times ( remember that   ) and it will clearly stops at .

In other words the slower pointer is at first  at node number . We can write where . After  advancing steps it will be at location   Since  the result follows.

As an example consider in which we have a list with a cycle of length  and starts at node number . The first part of the algorithm tells us that the nodes will meet at node . Moving the fast pointer back to te head of the list and iterating one node per time both iterators will lead the slower point to node:

• 12 again after advancing of 4 nodes
• 12 again after advancing of 4 nodes
• 10 advancing of the remaining 2 nodes.