Sunday, December 25, 2016

Steps to return to {1, 2, ..n} with specified movements - GeeksforGeeks


Steps to return to {1, 2, ..n} with specified movements - GeeksforGeeks
Given an array moves[] that contains a permutation of first n natural numbers, each element of this array represents a movement that is each element shows an index where the element goes after each step. Now following these steps we need to tell after how many steps array [1..n] returns back to [1..N]. A step is defined as follows, after one step each element will be moved to position defined by moves array indices.

Input  : moves[] = [4, 5, 1, 3, 2]
Output : 6
Explanation:
We need to consider an array of first 5 
natural numbers, i.e., arr[] = {1, 2, 3, 4, 5} as
size of moves[] is 5.
Now we one by one move elements of arr[] using 
given moves.
      moves[] = [4, 5, 1, 3, 2]
 arr[] = [1, 2, 3, 4, 5] 
In step 1, we move 1 to position 4, 2 to position
5, 3 to position 1, 4 to position 3 and 5 to 
position 2.
After step 1: arr[] = [3, 5, 4, 1, 2]
In step 2, we move 3 to position 4, 5 to position
5, 4 to position 1, 1 to position 3 and 2 to 
position 2
After step 2: arr[] = [4, 2, 1, 3, 5]
After step 3: arr[] = [1, 5, 3, 4, 2]
After step 4: arr[] = [3, 2, 4, 1, 5]
After step 5: arr[] = [4, 5, 1, 3, 2]
After step 6: arr[] = [1, 2, 3, 4, 5]
So we can reach to initial array in 6 steps, 
this is the minimum steps for reverting to 
the initial configuration of array.
We can solve this problem by observing a pattern among the sequences which are formed. A particular set of element of moves array, forms a cycle. As in above moves array example [4, 1, 3] and [5, 2] are two such sets. These two cycles are independent.
[4, 1, 3] causes [1, 3, 4] -> [3, 4, 1] -> [4, 1, 3] 
-> [1, 3, 4] -> [3, 4, 1] -> [4, 1, 3] -> [1, 3, 4]

[5, 2] causes [2, 5] -> [5, 2] -> [2, 5] -> [5, 2]
 -> [2, 5] -> [5, 2] -> [2, 5]
We can see from above changes that a cycle of length 3, takes 3 steps to reach to same state and cycle of length 2, takes 2 steps to reach to same state. In general if a cycle has length N then after N steps we can reach to same state.
Now if given moves array has just one cycle, then we can reach to starting state in number of moves equal to total elements in array but if it has more than 1 cycle then all of them move their elements independently and all of them will reach to starting state after x number of moves where x should be divisible by all cycle lengths, as smallest x which divides all cycle length is their LCM, we remain with finding LCM of all cycle lengths.
Cycle lengths can be found by visiting elements one by one, starting from any element we will move until we reach to starting element and we will count number of elements in this process which will be the corresponding cycle length.
int lcm(int a, int b)
{
    return (a * b) / __gcd(a, b);
}
 
// Method returns minimum number of steps to
// return to initial array
int getMinStepsToSort(int moves[], int N)
{
    //  initially all cells are unvisited
    bool visit[N];
    memset(visit, false, sizeof(visit));
 
    //  looping over all elements to get
    // various cycle
    int steps = 1;
    for (int i = 0; i < N; i++)
    {
        // if already visited, that means it
        // was a part of some cycle
        if(visit[i])
            continue;
 
        int cycleLen = 0;
 
        // Looping among cycle elements,  -1 is
        // for coverting value to 0-index based
        for (int j = i; !visit[j]; j = moves[j] - 1)
        {
            cycleLen++;
            visit[j] = true;
        }
 
        // Take the lcm of current result and
        // new cycle length
        steps = lcm(steps, cycleLen);
    }
    return steps;
}
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