There are \(2^{N}\) possible sets of cards on the table. It is sufficient to determine for each of them whether the first or second player will win if these cards is on the table right before the first player makes a move.

This can be solved with the technique so-called bit DP.
Let \(dp[S]\) = true if the first player will win if the cards in \(S\) are on the table right before the first player makes a move, and false otherwise.
The transitions are as follows.

For a set \(S\) of cards, if \(dp[S \setminus \{i,j\}]\) is false for some pair \(i,j \in S\) such that \(A_i=A_j\) or \(B_i=B_j\), then \(dp[S]\) is true; otherwise, it is false.

The final answer is Takahashi if \(dp[S]\) for \(S=\{1,2,\ldots,N\}\) is true and Aoki otherwise.

The number of states is \(O(2^{N})\) and each transition costs \(O(N^{2})\) time, so the total complexity is \(O(2^{N} {N}^{2})\), which is fast enough.

Sample code (C++):

#include <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cin >> n;
    vector<int> a(n), b(n);
    for (int i = 0; i < n; ++i)
        cin >> a[i] >> b[i];
    vector<int> dp(1 << n, -1);
    dp[0] = 0;
    for (int i = 1; i < (1 << n); ++i) {
        bool f = false;
        for (int j = 0; j < n; ++j)
            for (int k = j + 1; k < n; ++k) {
                if (((i >> j) & 1) && ((i >> k) & 1)) {
                    if ((a[j] == a[k] || b[j] == b[k]) && dp[i ^ (1 << j) ^ (1 << k)] == 0) {
                        f = true;
                    }
                }
            }
        dp[i] = f;
    }
    cout << (dp.back() ? "Takahashi" : "Aoki") << endl;
}