(Analysis by Nick Wu)

For this problem, the figurine is small enough and there are few enough pieces that we can check every possible pair of figurines along with every possible pair of ways to shift the two pieces.

Because the board has size N, we can shift either vertically or horizontally in either direction by at most N-1.

Here is my Java solution.

import java.io.*;
import java.util.*;
public class bcsB {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new FileReader("bcs.in"));
PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("bcs.out")));
StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int k = Integer.parseInt(st.nextToken());
boolean[][] goal = read(br, n);
boolean[][][] grids = new boolean[k][n][n];
for(int i = 0; i < k; i++) {
grids[i] = read(br, n);
}
for(int i = 0; i < k; i++) {
for(int j = i+1; j < k; j++) {
for(int idx = -n+1; idx <= n-1; idx++) {
for(int idy = -n+1; idy <= n-1; idy++) {
for(int jdx = -n+1; jdx <= n-1; jdx++) {
for(int jdy = -n+1; jdy <= n-1; jdy++) {
boolean good = true;
for(int x = 0; good && x < n; x++) {
for(int y = 0; good && y < n; y++) {
boolean iLoc = get(grids[i], idx + x, idy + y);
boolean jLoc = get(grids[j], jdx + x, jdy + y);
if(iLoc && jLoc) {
good = false;
}
if(goal[x][y] != (iLoc || jLoc)) {
good = false;
}
}
}
if(good) {
pw.println((i+1) + " " + (j+1));
}
}
}
}
}
}
}
pw.close();
}

public static boolean get(boolean[][] grid, int x, int y) {
return x >= 0 && x < grid.length && y >= 0 && y < grid[x].length && grid[x][y];
}

public static boolean[][] read(BufferedReader br, int n) throws IOException {
boolean[][] grid = new boolean[n][n];
for(int i = 0; i < n; i++) {
String s = br.readLine();
for(int j = 0; j < n; j++) {
grid[i][j] = s.charAt(j) == '#';
}
}
return grid;
}

}