Efficient Letter Boxed Solver Algorithm Guide

Model the Letter Boxed / letter box puzzle as a state (bitmask_of_used_letters, last_letter). BFS / bitmask DP on that state space gives provably minimal-word solutions; with aggressive preprocessing (filtering, collapsing words by (first,last,mask), grouping by first-letter) the search for typical L≈12 puzzles is small and fast. If the dictionary or branching factor is large, use A*/IDA* with admissible heuristics (for example ceil(remaining_letters / max_new_letters_per_word) and a fractional set-cover relaxation) and apply reachability/dominance pruning; use ILP/CP only to certify minimality when needed.

Contents

• State-space model for Letter Boxed
• Admissible heuristics for Letter Boxed solvers
• Pruning & reductions for letter box solvers
• Exact vs approximate methods and trade-offs
• Implementation tips & complexity estimates
• Worked example & minimal-solution trace
• Sources
• Conclusion

State-space model for Letter Boxed

Why a bitmask? Because the goal is to cover a fixed, small set of letters (typically L = 12). Encode which letters have already been used as an L-bit integer; encode the chain constraint with a second small component: the index of the last letter produced by the most-recent word. Together (mask, last) is a compact, lossless state that fully determines what legal words may follow and whether the goal (mask == all_mask) is reached.

Concrete model

• Map the L letters used in the puzzle to indices 0…L-1.
• mask: integer in [0, 2^L) where bit i = 1 iff letter i has been used somewhere so far.
• last: integer in [0, L) equal to the index of the last letter of the last-placed word.
• Initial states: push one state for every legal single word w: (mask_w, last_w) with cost 1.
• Transition rule: from (mask, last) you may play any word w whose first letter equals last; next state is (mask | mask_w, last_w) and cost increases by 1.

Provable optimality

• Every transition corresponds to adding exactly one word. If you run an unweighted BFS on the implicit directed graph of states described above, the first time you reach any state with mask == all_mask you have found a sequence with the minimal possible number of words (BFS expands states by increasing number of words). That provably yields an optimal minimal-word solution.

Complexity (basic)

• Number of distinct states ≤ L * 2^L.
• For L = 12 that’s 12 * 4096 = 49,152 states — small.
• Time depends on transitions: per state you iterate words that start with that state.last. If wordsByFirst[last] lists are short (after pruning) this is fast; worst-case it can be large but is usually manageable.
• Memory: dist array sized L * 2^L (bytes or small ints) plus predecessor info to reconstruct a path.

Reference (SEO / interest): search traffic shows interest in “letter box” puzzles, e.g. Yandex Wordstat - ‘letter box’. That’s useful for SEO but the solver design is independent of those counts.

Minimal BFS sketch (Python-like pseudocode)

from collections import deque

def bfs_min_words(words_by_first, all_mask, L):
 Nmask = 1 << L
 # flatten state id: state_id = last * Nmask + mask
 INF = 255
 dist = [INF] * (Nmask * L)
 prev = [-1] * (Nmask * L) # previous state id
 prev_word = [-1] * (Nmask * L) # word id used to get here
 q = deque()

 # initialize with every single-word state
 for w in all_words:
 idx = w.last * Nmask + w.mask
 if dist[idx] > 1:
 dist[idx] = 1
 prev[idx] = -2 # start marker
 prev_word[idx] = w.id
 q.append(idx)

 while q:
 s = q.popleft()
 mask = s & (Nmask - 1)
 last = s >> L_bits # or s // Nmask
 if mask == all_mask:
 return reconstruct_path(s, prev, prev_word)
 for w in words_by_first[last]:
 nm = mask | w.mask
 nid = w.last * Nmask + nm
 if dist[nid] == INF:
 dist[nid] = dist[s] + 1
 prev[nid] = s
 prev_word[nid] = w.id
 q.append(nid)
 return None

Admissible heuristics for Letter Boxed solvers

If BFS is too big because the dictionary is huge, guided search (A* / IDA*) helps — but only if your heuristic h(state) is admissible (never overestimates the remaining number of words). Here are practical admissible bounds you can compute cheaply or precompute.

1. Simple per-word capacity bound (very cheap)

• remaining = popcount(all_mask & ~mask)
• global_max = max_{w} popcount(mask_w) (maximum distinct letters any single word can cover)
• per_last_max[last] = max_{w with first==last} popcount(mask_w)
• admissible lower bound: h1 = ceil(remaining / global_max) and h1_last = ceil(remaining / per_last_max[last])
• Use max(h1, h1_last). It’s admissible because no word can add more than these maxima.

2. Fractional set-cover relaxation (stronger, more expensive)

• Formulate remaining letters as elements and candidate words (ignoring chaining constraints) as sets; solve the linear program:
  minimize sum x_w
  subject to for each remaining letter ℓ: sum_{w contains ℓ} x_w ≥ 1, 0 ≤ x_w ≤ 1.
• The LP optimum is a lower bound on integer set cover (so admissible). For L ≤ 12 you can precompute this fractional bound for every subset of letters offline (2^L LPs) and store results; lookup is O(1) at search time. If you can’t precompute, solve a small LP on-demand for a node (still feasible if candidate word count is small).

3. Co-occurrence / component bound (cheap and often effective)

• Build an undirected co-occurrence graph on remaining letters: connect letters if they appear together in some single allowed word. Let c = number of connected components. Because a single word can only cover letters in a single component of this graph, you need at least c words. So h2 = c is admissible.

4. Reachability bound (chain-aware, cheap)

• Build directed letter graph with edges u→v if there exists a word that starts at letter u and ends at v and that contains at least one remaining letter. From current state last, some remaining letters may be unreachable by any sequence of transitions — if so, prune the state. Otherwise, the number of strongly connected components or reachable components can give a small lower bound.

Combine heuristics

• Use h(state) = max(h1_last, h2, fractional_bound(mask)) to tighten A*/IDA* pruning. Always take the maximum of admissible bounds — the result is still admissible.

(SEO note: interest in “letterboxed” appears in search data; see Yandex Wordstat - ‘letterboxed’.)

Pruning & reductions for letter box solvers

Real wins come from reducing branching before search begins.

1. Collapse equivalent words (massive reduction)

• For DP/BFS that only uses (first, last, mask), any two words with identical (first, last, mask) are indistinguishable. Keep only one representative.
• For fixed (first, last) keep only masks that are maximal under set-inclusion: sort masks by popcount desc and discard any mask that is a subset of an earlier one. Rationale: a strictly smaller mask never helps if a strictly larger mask starts and ends the same way.

2. Group by first letter

• Build wordsByFirst[first] lists of (mask, last, popcount, id). Iteration over neighbors becomes a single tight inner loop.

3. Remove inert cycles

• If a word neither adds new letters (mask_w ⊆ mask) nor changes last (last == first), it has no effect and can be discarded.
• If it neither adds letters and doesn’t help reach new last letters that enable future coverage, it’s useless. Use reachability pruning to detect and discard perpetual-no-op words.

4. Limit per-(first,last) branching (practical cap)

• After maximal-mask pruning, you may still have many candidates. In practice keep only the top-K masks per (first,last) sorted by new-letter count (K = 4…16 depending on speed/memory trade-off). This is a heuristic reduction that preserves most optimal solutions; if you need a guaranteed optimal result, use K = all (but first try smaller K to get good upper bounds).

5. Transposition table / best-g pruning

• Store the best (smallest) g found so far for each (mask, last); prune any new visit with g’ ≥ best_g[mask,last]. This is standard DP/BFS bookkeeping.

6. Use greedy upper bound early

• Before exact search run a fast greedy set-cover (pick words that cover the largest number of uncovered letters but respecting chaining by simple correction) to obtain an upper bound UB on number of words. Use branch-and-bound: prune any node with g + h >= UB. An early strong UB dramatically reduces search.

7. Chain reachability pruning

• Precompute the directed letter graph and reject any state from which not all remaining letters are reachable. This cheap test catches impossible branches early.

(Another relevant keyword hit: Yandex Wordstat - ‘letter boxed solver’.)

Exact vs approximate methods and trade-offs

• BFS / Bitmask DP (recommended default): exact, simple, fast for typical L ≤ 12 after the reductions above. If you implement the collapse-by-(first,last,mask) step, the practical branching factor plummets and BFS finishes quickly.

• A* (heuristic best-first): applies when dictionary is large and BFS explores too many states. Requires good admissible heuristics. A* reduces expansions but uses more memory. Use max(…) of the admissible bounds from the previous section.

• IDA* (iterative deepening A*): same pruning power as A* with far less memory, but it recomputes nodes. Use when memory is constrained and branching factor acceptable.

• CP / ILP: Use when you want a certificate of optimality and BFS/A* becomes awkward. Practical pattern: iterate K = 1…UB and for each K solve a feasibility CP model that assigns K ordered positions to words with chaining constraints and coverage constraints. Modern CP solvers (OR-Tools CP-SAT) can solve small K quickly; this approach is easy to encode and verifies optimality when K small (typical minimal K is small).

• Greedy / beam search / local search: very fast, often produces near-optimal solutions. Use this to get a tight UB before exact search. Beam search with beam width 50–100 is a good trade-off.

Summary recommendation

• For most cases: implement collapse + BFS/bitmask DP. Try a quick greedy to get UB. If BFS stalls, switch to A*/IDA* with the combined heuristics above. Use CP/ILP only to certify or when you cannot implement a specialized DP efficiently.

Implementation tips & complexity estimates

Practical engineering shortcuts that matter:

1. Letter -> bit mapping

• Build letter_to_idx map from puzzle letters to 0…L-1.
• Word mask = sum(1 << letter_to_idx[c]) for unique letters c in word.

2. Compact state indexing

• Use state_id = last * (1<<L) + mask (or mask * L + last). Pre-allocate arrays sized L * (1<<L) — numeric indexing is faster than nested dicts.

3. Dist / predecessor storage

• dist array: use uint8/uint16 depending on expected max word count — saves memory and speeds comparisons.
• predecessor arrays store previous state id and the word id used for reconstruction.

4. Popcount and bit ops

• Use builtin popcount: in Python 3.8+ use int.bit_count(); in C/C++ use __builtin_popcount. Use these heavily for heuristics.

5. Precompute and store:

• words_by_first: lists of Word objects (mask,last,popcount,id).
• per_last_max_new: max popcount over words_by_first[last].
• For each (first,last) keep only maximal masks (collapse step).

6. Choose language

• Implement core search in C++/Rust for speed if you expect many hard instances. Python is fine for quick prototyping and many practical puzzles when optimized (use local variables, built-in bit ops, and arrays rather than nested dicts).

7. Resource estimates (typical)

• L = 12 => states = 49k. If average out-degree after pruning is 100, transitions ≈ 4.9M — feasible in C++ in <1s; in well-written Python perhaps a few seconds.
• Memory for dist (uint8): ≈ 50 KB. Predecessors and indices add a few hundred KB. So space is not the bottleneck; CPU/branching is.

8. Use upper bounds and pruning early

• Get a greedy solution first. Plug UB into A*/branch-and-bound to prune aggressively.

9. Reconstructing solution

• Save prev_state and prev_word per visited state. When you hit mask==all_mask, reconstruct backward to get the sequence of words.

Worked example & minimal-solution trace

Small illustrative trace (letters A,B,C,D mapped 0…3)

• Words (first,last,mask): w1=(A,D,{A,B}), w2=(D,C,{C,D}), w3=(C,A,{A,C}), w4=(A,B,{B}) — all legal after adjacency filtering.
• all_mask = bits for {A,B,C,D} = 0b1111
• Init: push states
• s1 = (mask=0b0011,last=D) from w1, cost=1
• s2 = (0b1100,last=C) from w2, cost=1
• s3 = (0b0101,last=A) from w3, cost=1
• …
• BFS expansion:
• From s1 (0b0011,last=D) you can play any word starting with D (w2) → new mask 0b1111,last=C → goal found at depth 2.
• Minimal solution: [w1, w2] of length 2. BFS guarantees no 1-word solution exists, therefore 2 is minimal.

This toy example shows how BFS on (mask,last) finds the minimal number of words without searching word sequences explicitly.

Sources

• Yandex Wordstat — requestPhrase=“letterboxed” — keyword-count output used for SEO keywords.
• Yandex Wordstat — requestPhrase=“letter boxed” — keyword-count output used for SEO keywords.
• Yandex Wordstat — requestPhrase=“letterboxed solver” — no topRequests returned; recorded for coverage.
• Yandex Wordstat — requestPhrase=“letter boxed solver” — no topRequests returned; recorded for coverage.
• Yandex Wordstat — requestPhrase=“letterboxed solution” — no topRequests returned; recorded for coverage.

Conclusion

For practical, provably optimal Letter Boxed / letter box solvers start with a bitmask+last state model and run BFS / bitmask DP after collapsing equivalent words by (first,last,mask). That approach is exact and usually fastest for typical puzzles (L ≈ 12). If the dictionary or branching overwhelms BFS, add admissible heuristics (ceil(remaining / max_new_per_word), co-occurrence components, or a precomputed fractional set-cover bound) and switch to A*/IDA*; use greedy/beam search to get tight upper bounds and CP/ILP only when you need a certificate. Implement with compact integer masks, wordsByFirst lists, maximal-mask collapsing and reachability pruning to get provably minimal solutions quickly.