Rotating Bytes — Solution¶
Method and Contributions¶
Use the setter's reverse character BFS: turn edge labels into extra labeled blocks, maintain string order by stable binary grouping, and order reversed adjacency by predecessor vertex number. Codex (GPT-6) supplies proofs for all three objectives, storage sized to the actual input, and portable C++17 implementation. BFS and stable radix sorting are classical techniques.
Label Blocks and Character Distance¶
Replace each edge \(u\to v\) by a labeled block \(e\) and arcs \(u\to e\to v\) . Original vertices are also blocks, giving \(n+m\) blocks. Each stores its complete label, character position, visited flag and successor block from its first arrival.
Search backwards from destination one. Each step reads one character of the current label from right to left; after completing the block, enter its reversed neighbors. Each BFS round consumes exactly one character, so the round number is the suffix length read. No separate graph vertex is needed for every character: movement inside a block is fixed and needs only an index.
All labels are nonempty. A shortest route cannot contain an original-vertex cycle, since removing it reduces character length. This also ensures that a longer suffix to the destination can never help a shortest route.
Second Objective: Stable Sorting When Prepending Characters¶
Reading backwards does not mean comparing reversed strings. The characters read represent a suffix in its forward order. Each next character is prepended to that suffix.
Suppose one round's states are ordered by their represented suffixes. Strings in the next round have form \(cS\) , where \(c\) is the new character and \(S\) the preceding suffix. Stably split generated states by new character: process the zero group before the one group and retain the preceding state order within each group. This sorts by \((c,S)\) , exactly the new forward suffix. The initial state contains only the last character of the destination label, establishing the induction.
Use two queues for the current round and two for the next. A zero character is not a zero-cost edge: every character consumes one round.
Third Objective: Reverse Vertex-Number Order¶
For an original vertex \(v\) , order its reversed incoming edges by increasing original predecessor \(u\) . When entering an edge block, conceptually append that predecessor number to the reverse vertex sequence. Movement inside that edge block and its vertex block introduces no further branch.
Consider two optimal routes tied on total length and complete string. At any original vertex common to both, each suffix to the destination must be a shortest suffix for that vertex; replacing a longer suffix would shorten the complete route. Their suffix lengths are therefore equal. Since the complete strings agree, those equal-length terminal suffixes also have identical characters.
Follow the routes backwards to their first differing vertex number. Before that divergence, common vertices occur in the same character layers with the same suffix. Parallel edges preserving the same vertex sequence must also have equal lengths and labels in these tied optimal routes, or the common shortest suffixes would differ. Thus the first genuine vertex-number divergence occurs at the same state, where the smaller predecessor is enqueued first.
Afterwards, the two identical final strings have identical read suffixes at every character layer. Stable character grouping cannot reverse their relative order. The smaller reverse vertex sequence consequently finishes first. If one reverse sequence were a proper prefix of the other, continuing would add a nonempty label and could not preserve the tied optimal length.
This explains why varying label lengths do not invalidate the third objective: tied optimal routes have equal suffix distances at common original vertices. A later arrival at that same vertex cannot participate in the tie.
Why Mark a Block at First Arrival?¶
For the same block, arrival in an earlier round gives a shorter suffix. Within a round, queue order gives the smaller suffix string, and ties in characters are resolved by the smaller reverse vertex sequence. Remaining characters inside the block are identical. Any front half of a route reaching that block can use the better suffix, so a later arrival is dominated and may be discarded.
When comparing reverse vertex sequences for two suffixes at the same block, they end at the same original vertex. Simple shortest paths prevent one from being a proper prefix of the other. Appending the same front half therefore cannot reverse their order.
Record the first-arrival From pointer. From the first allowed start whose label is fully consumed, follow these pointers to one. Print each block's original forward label, then print only original-vertex numbers. This attains all three objectives. Do not stop upon merely entering a start block: its remaining characters still affect length and lexicographic order.
Adjacency Order and Complexity¶
Build the required order in linear time. Bucket original edges by their source, scan source numbers in decreasing order, and prepend the corresponding reversed edges to adjacency lists. Each original vertex then sees predecessors in increasing order. Parallel edges with identical endpoints may appear in either order: differing labels are distinguished by stable character grouping, while identical labels produce identical outputs.
Let \(L\) be the total character count of all vertex and edge labels. Every block is first entered once, every character is processed once, and every converted edge is examined once. Time and space are \(O(n+m+L)\) . Here labels have length at most five, so both are \(O(n+m)\) . The implementation stores five character slots per block instead of building a graph vertex per character; the four queues store block numbers and reuse capacity.
The reference program uses this reverse linear algorithm. An independent verification method first computes suffix distances with Dijkstra, filters the smallest forward string in the shortest-path graph, and uses matching-string reachability to greedily choose predecessors backwards. Separating the three objectives this way requires a heap and extra passes, raising worst-case time to \(O((n+m)\log n+L)\) , so it is not the preferred computation here.