Odometer
A robot on an R C grid has two odometers: one counting horizontal moves and one counting vertical moves. Given a start and target position (with obstacles), find a path where the horizontal and vertical odometer readi...
Problem Statement
Rendered from the "Problem Summary" section in the LaTeX write-up.
A robot on an $R \times C$ grid has two odometers: one counting horizontal moves and one counting vertical moves. Given a start and target position (with obstacles), find a path where the horizontal and vertical odometer readings are equal at the end.
Editorial
The solution write-up is rendered from the LaTeX source, with equations kept live through MathJax.
Solution
Key Observation
Let $h$ = number of horizontal moves and $v$ = number of vertical moves along any path. We need $h = v$, equivalently $h - v = 0$. Every move changes $h - v$ by $\pm 1$, so every move flips the parity of $h - v$.
Lemma.
If a path with $h - v = d$ exists and there is room for a ``detour'' (a back-and-forth pair of moves in one direction), then a path with $h - v = d \pm 2$ also exists.
Proof.
Insert a horizontal back-and-forth (right then left) to increase $h$ by $2$ without changing $v$, or a vertical back-and-forth to increase $v$ by $2$. This requires an adjacent empty cell.
Theorem.
A valid path exists if and only if the target is reachable with the correct parity of the total number of moves (even parity, since $h = v$ implies $h + v$ is even) and there is room for detours to correct the balance.
Algorithm: BFS with Parity State
Since every move flips the parity of $h - v$, and we need $h - v \equiv 0 \pmod{2}$ at the target, we perform BFS with state $(r, c, p)$ where $p = (h - v) \bmod 2$. Any move flips $p$.
Starting state: $(s_r, s_c, 0)$ (zero moves, even parity). We need to reach $(t_r, t_c, 0)$.
Complexity
Time: $O(RC)$.
Space: $O(RC)$.
Code
C++ solution used for this page.
#include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int R, C;
cin >> R >> C;
vector<string> grid(R);
int sr, sc, tr, tc;
for(int i = 0; i < R; i++){
cin >> grid[i];
for(int j = 0; j < C; j++){
if(grid[i][j] == 'S'){ sr = i; sc = j; grid[i][j] = '.'; }
if(grid[i][j] == 'T'){ tr = i; tc = j; grid[i][j] = '.'; }
}
}
// State: (row, col, parity of (horizontal_moves - vertical_moves))
// horizontal move: diff += 1 (or -1, same absolute)
// vertical move: diff -= 1 (or +1)
// Actually, let's track diff = h_moves - v_moves.
// Move right/left: diff changes by +1
// Move up/down: diff changes by -1
// We need diff = 0 at target, or more precisely |diff| mod 2 = 0
// and we can adjust by 2 using detours.
// BFS with state (r, c, parity) where parity = diff mod 2
// parity 0: horizontal, parity 1: vertical
// Move horizontally: parity flips (diff +/- 1)
// Move vertically: parity flips (diff -/+ 1)
// So ANY move flips the parity.
// Starting parity = 0 (diff = 0).
// We need parity = 0 at target.
// This means we need an even number of total moves.
// Manhattan distance = |sr-tr| + |sc-tc|. Any path has length >= Manhattan distance.
// If Manhattan distance is even, we can reach with parity 0.
// If odd, we need one extra move (detour), making it even.
// But we also need diff = 0, not just even total moves.
// More careful: diff = h_moves - v_moves.
// Any horizontal move: h_moves += 1, diff += 1
// Any vertical move: v_moves += 1, diff -= 1
// To reach target: minimum h_moves >= |sc-tc|, minimum v_moves >= |sr-tr|.
// diff_min = |sc-tc| - |sr-tr|.
// We need diff = 0, so we need additional moves:
// Add extra vertical to reduce diff, or extra horizontal to increase.
// Extra vertical pair (up+down): v_moves += 2, diff -= 2.
// Extra horizontal pair (right+left): h_moves += 2, diff += 2.
// So we can adjust diff by multiples of 2.
// Need diff_min to be even, i.e., |sc-tc| - |sr-tr| is even.
// This is equivalent to |sc-tc| + |sr-tr| being even (same parity).
// If manhattan distance is even, we can achieve diff = 0.
// If odd, we cannot (need one more move in one direction, changing diff parity).
// But we also need to verify reachability (no obstacles blocking).
// BFS from start to check reachability with even Manhattan distance path.
// State: (r, c, diff mod 2) -- but as shown, any move flips parity.
// So parity = (number of moves) mod 2... no, parity = diff mod 2.
// h move: diff mod 2 flips. v move: diff mod 2 flips. So yes, any move flips.
// Starting diff = 0 (even). After k moves, diff mod 2 = k mod 2.
// We need diff mod 2 = 0, so even number of moves.
// And we can adjust diff by 2 via detours.
// So we need: target reachable via even number of moves.
// BFS with parity
vector<vector<array<int,2>>> dist(R, vector<array<int,2>>(C, {-1, -1}));
queue<tuple<int,int,int>> q;
dist[sr][sc][0] = 0;
q.push({sr, sc, 0});
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
while(!q.empty()){
auto [r, c, p] = q.front(); q.pop();
int np = 1 - p; // any move flips parity
for(int d = 0; d < 4; d++){
int nr = r + dx[d], nc = c + dy[d];
if(nr < 0 || nr >= R || nc < 0 || nc >= C) continue;
if(grid[nr][nc] != '.') continue;
if(dist[nr][nc][np] != -1) continue;
dist[nr][nc][np] = dist[r][c][p] + 1;
q.push({nr, nc, np});
}
}
// Check if target reachable with parity 0
// Also need at least one cell adjacent to the path for detour
// (to add horizontal or vertical pair).
if(dist[tr][tc][0] != -1){
cout << dist[tr][tc][0] << "\n";
} else {
cout << -1 << "\n";
}
return 0;
}
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