It is sufficient to check the conditions described in the problem statement for all $i$. Once such $i$ is found, you can output Yes on the spot and terminate the program. Be careful of the range of the loop.

Sample Code (C++)

#include <iostream>
using namespace std;
int main(){
	int n;
	cin >> n;
	int a[100][2];
	for (int i=0; i<n; i++){
		cin >> a[i][0] >> a[i][1];
	}
	// Be careful when the loop ends
	for (int i=0; i<n-2; i++){
		if(a[i][0]==a[i][1] && a[i+1][0]==a[i+1][1] && a[i+2][0]==a[i+2][1]){
			cout << "Yes";
			return 0;
		}
	}
	cout << "No";
}

#include <iostream>
using namespace std;

int main(){
	int n;
	cin >> n;
	int a[100][2];
	for (int i=0; i<n; i++){
		cin >> a[i][0] >> a[i][1];
	}
	// Be careful when the loop ends
	for (int i=0; i<n-2; i++){
		if(a[i][0]==a[i][1] && a[i+1][0]==a[i+1][1] && a[i+2][0]==a[i+2][1]){
			cout << "Yes";
			return 0;
		}
	}
	cout << "No";
}