Check if chords of a Circle are symmetric after some rotation
Given two Integers N and M, N indicating equidistant points on circumference of a circle and M indicating number of chords formed with those points. Also given is a vector of pairs C containing position of chords. The task is to rotate the circle by any degree, say X, where 0 < X < 360, and check if the chords of are still symmetric to the original circle.
Example:
Input: N = 12, M = 6, C = {{1, 3}, {3, 7}, {5, 7}, {7, 11}, {9, 11}, {11, 3}};
Output: YESOriginal
After Rotation
Input: N = 10, M = 3, C = {{1, 2}, {3, 2}, {7, 2}}
Output: NONo rotational symmetry possible
Naive Approach: Rotate for every distance K in the range[1, N] and check for each point [a, b] if the rotated point [a + K, b + K] exits.
If there exists any k then print YES else print NO
Time Complexity: O(N*M)
Efficient Approach: It is enough to check for the divisors of N.
Let us suppose if we rotate the image by K units then the whole image will be divided into N/K blocks. Then if K is not a divisor of N, there will be an asymmetric block of length less than K and the image will never be symmetric to the original figure.
So calculate all the divisors of N and check for each chord the rotated chord exists or not.
Below is the implementation of the above approach:
C++
// C++ Program to for the above approach#include <bits/stdc++.h>using namespace std;// Utility function to calculate// divisors of a number in O(sqrt(N))vector<int> calculateDivisors(int N){ vector<int> div; for (int i = 1; i * i <= N; i++) { if (N % i == 0) { div.push_back(i); if (N / i != i && i != 1) { div.push_back(N / i); } } } return div;}int checkRotationallySymmetric( vector<pair<int, int> > A, int N, int M){ // Maintain a set to check quickly // the presence of a chord set<pair<int, int> > st; for (int i = 0; i < M; i++) { --A[i].first, --A[i].second; if (A[i].first > A[i].second) { swap(A[i].first, A[i].second); } st.insert(A[i]); } // Calculate the divisors of N. vector<int> div = calculateDivisors(N); // Iterate through the divisors for (auto x : div) { bool exist = 1; for (int i = 0; i < M; i++) { int dx = (A[i].first + x) % N; int dy = (A[i].second + x) % N; if (dx > dy) { swap(dx, dy); } if (st.find({ dx, dy }) != st.end()) { // There exists a valid // chord after rotation } else { // There is no valid chord after rotation exist = false; break; } } // if there exist another chord after // rotation for every other chord print // YES and exit the function if (exist) { cout << "YES"; return 0; } } cout << "NO"; return 0;}// Driver Codeint main(){ int N = 12, M = 6; vector<pair<int, int> > C = { { 1, 3 }, { 3, 7 }, { 5, 7 }, { 7, 11 }, { 9, 11 }, { 11, 3 } }; checkRotationallySymmetric(C, N, M); return 0;} |
YES
Time Complexity: O(M*sqrt(N)*log M)
Space Complexity: O(M)
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