Chainword Solver — Solve Any Tile Chain

What Chainword Is

Chainword is a word game built around a rack of letter tiles. You are given the whole rack at once and you have to spend it on five-letter words. The words are not independent: each one has to begin on the final letter of the word before it. A ten-tile rack makes two words, a fifteen-tile rack makes three, and the daily puzzle is always the fifteen-tile version. Everyone who opens the daily gets the same rack, because it is generated from the date rather than from anything about your account.

The game is easy to confuse with a word ladder, where you change one letter at a time and the words stay the same length. There is no one-letter-at-a-time step here. You are not editing a word, you are spending tiles. It also is not an anagram game, even though the tiles look like an anagram rack, because the words are ordered and the joins are constrained. And it is not a crossword, because there are no clues and no grid. The only inputs are the letters you hold and the single rule that joins one word to the next.

That makes Chainword a small planning problem disguised as a vocabulary game. Knowing a lot of five-letter words helps, but the players who finish quickly are usually the ones who count letters before they type. A rack can contain a beautiful word that dooms the rest of the chain, and it can contain a dull word that makes the last two words fall into place. The solver on this page exists to show you both possibilities before you commit.

The Chain Rule, Worked Through

A Chainword answer is accepted when three things are true at once. Every word has exactly five letters. Every word after the first starts with the last letter of the word before it. Every tile in the rack appears in exactly one of the words, with no tile used twice and none left over. The game applies those checks separately, which matters when you are debugging a chain: a chain can have perfect joins and still fail because one tile is unused.

Here is a complete fifteen-tile chain, written the way the game reads it:

The rack for that answer is STONEEIGHTTHREE, fifteen letters in total. STONE ends on E, so EIGHT has to start on E. EIGHT ends on T, so THREE has to start on T. The chain stops there because the rack is empty, not because THREE ends on E. The final word is allowed to end on any letter at all. Players lose time looking for a fourth word that does not exist, or avoiding a strong final word because its last letter is awkward. The last letter only matters if something still has to follow it.

One detail that trips up almost everyone: the E at the end of STONE and the E at the start of EIGHT are two different tiles. The rack holds four Es, and this chain spends one in STONE, one in EIGHT and two in THREE. The rule is about the letter, not about reusing a physical tile. If you hold only one E, you cannot write a chain that appears to hand the same E from one word to the next unless the other word gets its E from somewhere else in the rack.

The game keeps a strike count, but it does not stop the run when the count reaches a number. A wrong word costs a strike and the puzzle continues. The validator also does not compare your words against the chain the puzzle generator had in mind. It checks the three conditions above, so any arrangement that satisfies them is a solved puzzle. That is why a solver can be useful without knowing the daily answer: there is no single hidden answer to recover.

How This Solver Reads Your Rack

The page loads a five-letter word list with 12,478 entries. That is the same acceptance list the Chainword game uses for validation, so a word returned here is a word the game recognises. The list includes plenty of obscure entries, which is why results can look strange even when they are legal. The solver does not filter for familiarity; it filters for the rules.

The first step is counting. The rack is reduced to twenty-six counts: three Es, one S, one T, and so on. A candidate word is usable only if every letter it needs is still available in the right quantity. This is stricter than checking that the letters appear somewhere in the rack. A word with two As cannot be played from a rack with one A, no matter how well it fits the chain.

The second step is indexing. Every word is filed under its first letter, because the join rule only ever asks one question: which words begin with this letter? The solver also stores a sorted-letter signature for each word, so the letters A, E, R and T become AERT. That signature is what makes the last word cheap to find. When one word remains, the solver does not loop over thousands of candidates. It sorts the leftover tiles into a signature and looks up only the words with exactly that signature, which is the same as asking which word spends the rest of the rack.

The third step is a depth-first search with backtracking. The solver places a word, removes its tiles, moves to the next required starting letter, and repeats. If a branch runs out of tiles or runs out of words that fit, it backs up and tries the next candidate. Failed states are remembered so the same dead end is not explored twice. A safety limit stops the search if it ever visits 250,000 states or runs for 2,500 milliseconds; when that happens the page says so instead of pretending the list is complete.

In practice the limit is not close. Fifteen-tile racks usually finish in a few milliseconds because the letter counts cut the branching down hard, and the signature lookup removes almost all of the work at the final word. The search is exhaustive within the limit, so if it reports 99 chains, it found all 99.

Solving From Either End

Most players start at the first word because that is the word the board asks for first. That is a habit, not a requirement. Nothing in the rules says your first guess has to be the first word of the chain. If you can see a strong word in the rack, you can decide where it belongs and build outward from there.

The optional starting word field is the direct version of this. Type the word you are confident about and the solver treats it as word one, removes its tiles, and searches only for chains that continue from its last letter. That turns a wide search into a narrow one and is the fastest way to test a hunch. If your hunch is wrong, the result is an honest empty state rather than a near miss.

The middle anchor is the more interesting technique, and it is the one to use when you have a fifteen-tile rack and no idea where to begin. Suppose you suspect EIGHT belongs in the puzzle. Put it in the middle and read the consequences. The word before it must end on E, and the word after it must start on T. Now the rack has ten tiles left and you need one word on each side. If the remaining tiles contain a clean E-ending word and a clean T-starting word that split the letters exactly, you have a chain. If they do not, EIGHT may still be in the puzzle, but it is probably the first or last word instead.

Working from the end is the same idea in reverse. Pick a word you like as the final word. Its first letter tells you what the previous word has to end on. That constraint is often more useful than a starting-letter constraint, because final words are free: they do not have to produce a usable next letter, so awkward endings are allowed there. A word ending in J, Q, V or Z is a terrible connector and a perfectly good finish. When a rack contains a rare letter, the final word is often the natural home for it.

There is a limit to this. The solver only accepts a fixed first word, so if you want to test a fixed final word you have to scan the returned chains for it or try likely openings by hand. The search itself always runs forward, because the join rule is directional. The hand method still works from the end; the tool just cannot be told to start there.

Reading a Rack: Repeated, Rare, and Ending Letters

The first thing to do with any rack is write it in alphabetical order with the duplicates visible. STONEEIGHTTHREE becomes EEEEGHINORSTTTH. That line is ugly, but it answers the questions that matter. How many Es are there? Four. Is there an S? One. Is there a V? No. Sorting takes ten seconds and prevents the most common self-inflicted error in the game, which is trying to spend a letter twice because it looked like one tile in your head.

Repeated letters are a resource and a constraint. A rack with four Es can support words like THREE and EERIE, but it cannot support four different words that each want two Es. A rack with no repeated letters is easier to divide, because every word takes five distinct tiles and the partition is cleaner. When you are stuck, count the vowels. Five-letter English words usually carry one or two vowels, occasionally three, and rarely four. A rack with eight vowels and seven consonants is telling you that some word has to be vowel-heavy, and those words are less common. A rack with two vowels and thirteen consonants is telling you the opposite: look for words with a Y doing vowel duty, or words built on a tight consonant cluster.

Rare letters deserve their own pass. J, Q, X, Z, V, K and W appear in far fewer five-letter words than S, C, T or P, and the words that contain them are memorable precisely because they are unusual. The important question is not "which word contains the Z" but "which word can contain the Z without wrecking the joins". A Z in the middle of a word is usually fine. A word that ends in Z needs a Z-starting successor, and there are very few of those. The same logic applies to J and V. If a rare letter has to be spent, spend it somewhere that leaves a common letter at the end of the word.

The last letter of each word is the connector, so endings matter more than beginnings when you are building forward. Common five-letter endings are -S, -E, -Y, -T, -R, -D and -L. Those are useful because the next word only needs to start on the same letter, and English has plenty of words starting with S, E, T, R, D and L. Endings like -CK, -SH, -CH, -NG and -GH are riskier: they hand the next word a two-letter sound that has to be spelled from its first letter, and the supply of K-starting, H-starting or G-starting words is thinner. That does not make those words unplayable. It makes them better choices for the final slot, where nothing has to follow.

Suffix families are worth recognising because they cluster. Words ending in -ER often leave you needing an R-starting word; words ending in -ED leave you needing an E-starting word; words ending in -LY leave you needing an L-starting word, and adverbs are rarer in five letters than they look. If your chain has two words ending in the same letter, the middle word has to start on that letter, which is fine, but it also has to end on a letter that another word can start on. A chain of three words has two joins, and both of them have to be satisfied. It is not enough for each word to be valid in isolation.

One practical habit: before you type anything, list the letters in the rack that have the largest first-letter families. S, C, P, T, B and M start a lot of five-letter words. If your rack can produce a word starting with one of those, you have options at that join. If every plausible word in your rack starts with a rare letter, the chain is fragile even if the words themselves are good. The solver will tell you whether it works, but the habit helps you choose a first word that leaves you somewhere to go.

Chain Traps That Eat Your Rack

Trap one: the dead-end ending. A word that ends on a letter with no usable successor kills the branch. This is obvious when the letter is Q or Z and much less obvious when the letter is a vowel that simply has no remaining word to spend. The fix is to check the outgoing letter before you fall in love with the word. In a three-word chain only the first two words need successors, so if you are choosing the last word you can ignore this entirely.

Trap two: burning a rare letter early. A single J or Z is a limited resource, and the word that spends it also spends four other tiles. If that word ends on a letter that forces an awkward second word, you have paid a high price for a flashy opening. Rare letters are often better held for the word that has no outgoing obligation. When you do spend one early, make sure the word ends on a common letter.

Trap three: double-counting a duplicate. If the rack shows one E and your candidate word has two, the word is not playable. The reverse error is just as common: a rack with two identical tiles looks like one letter on the board, and players avoid a word that is actually legal. Count both ways. The solver's rack field takes the full multiset, so typing the letters exactly as they appear is enough; it will not guess that you meant two Es.

Trap four: assuming a solver word is a game word. Here the lists match, so a returned word will be accepted by the validator. The trap is different: a legal but obscure word can make a chain that looks wrong to a human and feels like a bug. It is not a bug. The game accepts the word because it is in the list. If you want a chain that reads like English rather than like a dictionary dive, prefer the results built from common words and use the obscure ones only when you are stuck.

Trap five: forgetting the leftover tile. The most frustrating failure is a chain with perfect joins and one unused tile. The game checks the rack after you submit the last word, and it will reject the chain for leftover tiles even though every word was real. Count the letters you plan to spend before you type. Ten tiles must produce exactly two words, fifteen exactly three; there is no partial credit and no bonus for a longer chain.

Trap six: treating the last letter as a requirement. The final word does not need a successor. Players reject good final words because they end in an awkward letter, then spend ten minutes forcing a worse arrangement. If the first two joins work and the tiles are fully spent, the chain is done. Stop looking for a fourth word.

Strategy Without a Solver

Start by sorting the rack into counts. This is not busywork; it changes what you notice. Write the letters in alphabetical order and put a small number next to each duplicate. Then write the alphabet down one line and, for the letters you hold, note which of them start a word you can actually spell. That second list is your set of possible joins. If a letter appears in the rack but you cannot think of a word starting with it, it is a bad place to end a word.

Next, choose an anchor. If you see a word that uses a rare letter cleanly, start there. If you see a word that ends on a common letter, start there. If you see nothing, pick the two letters with the smallest first-letter families and try to build around them, because those are the constraints that will decide the puzzle. The anchor does not have to be the first word. It only has to be a word you trust.

Then build outward in both directions. From the anchor, ask what can precede it: a word ending on the anchor's first letter, using tiles that are still free. Ask what can follow it: a word starting on the anchor's last letter. For a three-word chain, one of those sides will have two words and the other will have one. Try the side with the tighter constraint first.

Keep a short candidate list rather than a long one. Three or four words per starting letter is enough. A long list encourages you to keep switching instead of finishing a branch, and the cost of switching is that you lose track of which tiles you have already assigned. If you are playing on paper, cross out the tiles as you spend them. If you are playing in the game, the rack display already dims used tiles; trust it and re-read it after every submission.

Before you submit, run the checklist in the next section. It takes about fifteen seconds and catches nearly every avoidable strike. The most common strike is not a bad word; it is a good word in the wrong position, or a chain that spends fourteen of fifteen tiles. The checklist is deliberately mechanical because the failure modes are mechanical.

Finally, practise on racks you have already solved. Replaying a finished chain and asking "what else could this rack have made?" is more instructive than starting a fresh puzzle, because you already know one valid partition and can compare alternatives against it. That is also exactly what this solver does, which makes it a decent training tool rather than only a shortcut.

Checking a Candidate Chain by Hand

Run these four checks in order. They are the same checks the game runs, and doing them in this order catches the cheapest mistakes first.

1. Length. Every word must be exactly five letters. Four or six letters is an instant rejection, regardless of how well the chain reads.
2. Joins. The first letter of word two must equal the last letter of word one. The first letter of word three must equal the last letter of word two. No other pair of letters is compared.
3. Spend. Write the rack as counts and subtract each word letter by letter. Every count must land on zero. A negative count means you used a tile you do not have; a positive count means you left one behind.
4. Freedom. The last word has no outgoing requirement. Do not reject a valid chain because the final letter is awkward.

If all four pass, the chain is valid under the game's rules. If the game rejects it anyway, the usual cause is a typo in a word you believed was in the list, or a rack that does not match the one on screen. Reload the puzzle and compare the tiles before assuming a rule has changed.

What the Solver Can and Cannot Do

The solver is deliberately narrow, and it is worth knowing exactly where the edges are.

It can find every complete chain for a 5, 10 or 15-tile rack within its safety limit; fix the first word and search only continuations; count all matches while showing only the number you asked for; and explain an empty result instead of inventing a near miss. It also tells you how long the search took and how many states it visited, so you can see whether the answer was exhaustive.

It cannot see your game board, your strikes or the words you have already tried. It does not know which tiles the game has dimmed. It cannot accept a wildcard or a blank; the rack must be letters A to Z. It does not search racks larger than fifteen tiles, because the official game does not use them and the search grows quickly. It does not rank results by how familiar the words are, and it does not model the game's scoring. And it cannot tell you "the" answer, because the game does not have one: any valid chain is accepted.

One more boundary worth stating plainly. The solver's 12,478-word list and the game's acceptance list are the same list, so a returned word will pass validation. But the daily puzzle is generated from a smaller curated answer pool, which means the solver may show you a chain the puzzle designer never intended. That is not a flaw in the result. It is a consequence of the game accepting any chain that follows the rules.

Related Chainword and Word Tools

If you want to play first and solve later, open the Chainword game. The Chainword rules guide walks through the modes and scoring, and today's Chainword hints give graded nudges before the full answer. For rack work outside this puzzle, the anagram solver handles unordered letters, the word finder filters by pattern, and the five-letter word lists browse the same vocabulary this page searches.

Play Chainword How to Play Hints Today