I have some things to say about the word game "Wordle" that became popular in 2022, along with related games that are based on (roughly) the same dictionary (Quordle, Octordle, etc.) I do like to============================================================================== Projects for the future including looking to see how the claims are modified for the other games (with different word lists), or when allowing any valid starting word from the Wordle input list Does our "metric" really satisfy the triangle inequality? Need more examples of triangles! Check whether a 4 DAILY + 1 ADMX combo can give 25 letters. Need to talk in THREE about why the informal ranking is or isn't supported by the experience in TWO , since I only have time to examine the "top" of the list.playthem, but as a mathematician I wanted to do a thoroughanalysisof some questions that arose as I played. What I present here is (mostly) a discussion of the "good" sets of words drawn from the original 2315-word Wordle vocabulary list. A set of words is "good" if, when used as a starting set of entries in the games, it enables a human guess all the words in a small number of turns. The starting set must help win all the "subgames" in a compound game; the player then switches to "hard mode" in each of the sub-games one at a time, guessing only words that are consistent with the clues (the colored tiles that result). I wish to find and to compare those initial lists of starting words. I will use both exhaustive searches and clever optimization techniques to find "discriminating" sets of words: sets for which each resulting array of green/yellow/gray tiles matches relatively few vocabulary words. If I accomplish nothing else with these sets of words, at least I will have generated some great passwords! I am hardly the first person to apply a thorough mathematical analysis to Wordle. Some open forums with information include Stack Exchange and Reddit. Laurent Poirrier has collected information about "optimal" algorithms for playing the games. When applied to Wordle itself, those alternatives are more "efficient" than anything I propose here, at least in the sense that the average number of guesses will be higher playing the way(s) that I propose here. But beginning with a fixed, good starting list is bothsimpler for a human player(many fewer branching rules required), andbetter suited for the compound games, and those are the criteria of interest to me. If I ever want to waste more time on this project I really should re-write from scratch the software routines I used, rerun everything efficiently and to completion, and summarize the findings in a clear, complete, and concise way. Yeah, sure. Please let me know of corrections or additions to this document. -- dave (rusin@math.utexas.edu)Index of sections:

============================================================================== Some initial caveats first: 1. All comments here are about playing Wordle in "easy mode". (I don't even know what "hard mode" would mean for a compound game.) 2. All my analyses are built upon the word lists in the version of Wordle that was a simple web page in February 2022 (before purchase by the New York Times). In particular, (almost) all uses of the word "word" here mean "one of the original 2315 possible answers to a Wordle puzzle". (I do make a few comments below that refer to the larger, 12972-word, list of acceptable inputs to Wordle, but I have made little effort to update them in response to NYT's enlargment of that set in Summer 2022.) I have gathered together a long list of comments about the word list(s) that I recommend to a person who actually wants to play the games well. (It is important to know what words are, or are not, potential Wordle answers!) Some of the compound games use slightly different word lists; these are discussed only briefly. 3. In original Wordle, the daily hidden words were presented in a particular (random-looking) order; since November 2022 they are chosen by a "curator" at the Times. Our model of the games assumes instead that the words in the word list are chosen at random, with uniform probability, to be hidden each time we play. (This appears to be the mode of play in the compound games, at least in "practice mode", although as far as I can determine the multiple hidden words are chosen to be distinct.) 4. In some cases I am stating claims of optimality or completeness. The proofs I give are mostly just sketches that can be fleshed out by the reader if interested. The only parts that do not amount to a simple case-by-case computer check have to do with the computation of covering sets (which I did with linear-programming/ optimization software Gurobi). I have written up a brief introduction to that technique available here. The key ideas are (a) to cement ideas of "nearness" or "similarity" in the word list, (b) to identify sets of "most-similar" words that will be problematic late in the game, (c) to compute for each such set the collection of words to play that will help avoid these problematic sets, then (d) to find a "cover" for these collections --- a set of words that intersects all or most of these collections. ============================================================================== INTRODUCTION: HOW DO PEOPLE PLAY WORDLE? Ask a frequent Wordler their strategy and you'll get a variety of answers. "I just pick a random word to start with and run with it"; "I start with ADIEU to get a lot of vowels"; "I read somewhere that it's best to start with CRANE + SPILT". To me, these sound like only Phase 1 of a strategy: all these answers specify a certain number "a" of fixed starting words (a=0, a=1, and a=2 respectively). (I should note parenthetically that the second answer came from someone who, unlike me, is willing to guess a word that's not a Wordle answer-word, and the third answer came from someone who, like most of us, is playing Wordle's "easy mode".) This Phase 1 is about gathering information about what the hidden word might be. And it's important for people who (like me) play the compound games, because the hope is that these first "a" moves will simultaneously reveal a lot of information about the multiple Wordle subgames. But then comes Phase 2: how do we use the information gained? At some point, people start to enter guesses of what the hidden word might be; unlike using SPLIT after CRANE, most people at some point in the game start to enter only words that might actually be the answer word (i.e., they unconsciously switch to Wordle's "hard mode"). They'll spend some number "b" of guesses in this mode trying to guess the right word. They will try to have a+b no larger than 6, to win the game. A person playing Wordle does not need to think of the distinction between Phase 1 and Phase 2. But it does become more important when playing the compound games like Quordle (with N=4 subgames) in which we are in essence playing Phase 1 just once for all N of the subgames and then carrying out Phase 2 separately for each of them. Thus the total number of moves would be not a+b but rather a + N b . As N increases, it becomes more and more important to keep b small, even if it means a has to be a bit larger. In other words, we want to find sets of words for Phase 1 that are really good at determining what the hidden word(s) might be, so that we spend very little time after that actually pondering the clues left after Phase 1. Considering the optimal solutions we find in this document, the expected number of steps to solve an N-fold compound game need not be higher than the lowest of these: 6 + N 5 + 1.0058 N 4 + 1.0298 N 3 + 1.1682 N 2 + 1.6590 N These are merely upper bounds, but the pattern is clear: for sufficiently large values of N it can be more efficient to use a larger starting set. ---------------- Let's see how we can analyze just how good a starting set is. As we shall see, in order to be able to compare different starting sets, it will be important to know not just what words a player starts with but also what exactly they will do after those words are entered. Let's follow one person whose Phase 1 has a=3: they use the three starting words LOATH+MURKY+SPINE. It's a good start! But now what? Consider what this person might do when the colored tiles show up as in each of these examples. lower case = yellow tile = right letter, wrong place; UPPER CASE = green tile = right letter, right place. Play along here: what would YOU do in each case? LOATH + MURKY + SPINE 1) .O..h .u... s...E 2) ..... ..... ....E 3) ..A.. ..rK. ....E 4) ....H .ur.. s.... 5) l.A.. ...k. ...N. 6) ...t. ..r.. ..i.e 7) .o... ..r.. .p..E 8) ..... ..r.. .pI.E 9) .o... ..r.. ...n. 10) LOA.. m...Y ..... 11) ..a.. ..r.. ....e 12) ..at. .u... ...N. I hope for Example #1 you decided the word was "HOUSE". You're right! That's the only word consistent with those clues, and you might as well enter it on your next turn and win. Example #2 is much harder but it turns out there is only one Wordle word that matches this pattern: "WEDGE". Most people, I think, would need a hint to figure this out, and that's fine, but of course it costs a turn. In what follows, we will assume the player is perspicacious enough to spot the right word without hints, when there is only one (and more generally we will assume the player can list all the possible words consistent with the hints). This is NOT realistic; from time to time we will discuss ways to make things easier for the player. But it illustrates why a person needs to really know the word list! It turns out these first two examples are pretty representative of what this player will face: Of all the words in the Wordle dictionary, 1364 (59%) are uniquely identified by the colored tiles that result from playing LOATH+MURKY+SPINE. But for the rest, the colored tiles have only indicated that the hidden word is one of a "cluster" of similar-looking words. In Example #3, there's a good chance you see the "_rake" and so you'd enter "brake" right away. Again, not a bad plan but it turns out "drake" is also a Wordle word. This is common: we think we know the word, so why not enter it? But then we discover the word we entered is only one of several possibilities; in this example we have a 50-50 chance of getting the word right, even if we *do* know the two possibilities. The same would happen in Example #4: this time there's a better chance you recognize both "brush" and "crush" as possibilities, and they are the only ones, so what else is there to do but enter one or the other and hope for the best; half the time you'll win on the first guess, half the time on the second. With a bit of effort you might figure out that when Example #5 shows up, the word is either BLANK, CLANK, or FLANK. So what do we do now? Most people would probably enter these words one at a time, especially if at first only one or two of the possibilities comes to mind. (Really? "clank"?!) If that's what you choose to do, you'll get the right word in either 1, 2, or 3 more moves, each with probability 1/3. Playing this way --- simply entering the first matching word that comes to mind --- we might call "guess-at-will" mode, or (since we have now slipped into playing "hard mode"!) we might call it "free-form hard mode". The situation in Example #6 is a little different. The possible words now are REFIT, RIVET, and TIGER. But this is a better situation for the player than example 5! No matter which of the three we choose to enter as our fourth word, if it's wrong we will get enough new information to tell which of the other two is the hidden word. So in reality we're playing the same way, but have better odds: still a 1/3 chance of winning on move 4 but then a 2/3 chance of winning on move 5. This now leads us to look at Example #7: there are again three possible answers: GROPE, PROBE, and PROVE. But this time the situation is a mix of the previous two. If we guess GROPE on move 4, then we have no additional information to distinguish whether PROBE or PROVE is the right word. If instead we guess one of the other two, then we *do* get that information and can surely win on move 5. So in this case, the player has two choices: it's- Some caveats
- Introduction: what are we doing in this document?
- The best starting sets of six (and more!), and why these are interesting
- Best starting quintuples and waltzing nymphs
- Best starting quadruples: everyone can win at Wordle
- Best starting triples (by various measures)
- Best (and possibly best) starting pairs
- The best single word to start with
- Concluding remarks
simplerjust to continue to guess whatever seems to fit, continuing in freeform hard mode. But it's moreefficientto use a "guided hard-mode", in which (in addition to memorizing the three starting words) the player memorizes that playing PROBE when it's possible to do so is the better thing to do. In that last example, taking the effort to remember an additional rule has only a small payoff, but the same principle applies in more important cases. Example #8 shows an array that could signal any of GRIPE, PRICE, PRIDE, or PRIZE. It's quite possible that a player using freeform guessing would guess GRIPE first, which unfortunately would give no information about which other word is the hidden one (if it's not GRIPE itself) and then no matter which other of the four words we try next, we never get more information about the remaining candidates when we guess wrong. In this case the player could definitely run out of turns and lose. By contrast, if the player takes pains to remember to keep an eye out for PRIZE and play it when it's possible to do so, then he or she will definitely win by move 6 whichever of the other three is the hidden word. So in our analysis of starting wordsets, we will draw a distinction between their value to a player guessing freely each time, and the value to a player who remembers preferential members of key clusters. There's one more option that a clever player can use, and it's again illustrated by example 5. For a player not committed to hard mode at all, there's no reason the player could not enter, say, BRACE as the fourth word. Depending on whether the B, the C, or neither gets a colored tile, the player knows right away which is the right word and can enter it as the fifth word, and win. So this manner of play --- this "out of the box thinking" --- can reduce themaximumnumber of turns that it will take to resolve a situation like example 5. That can mean the difference between winning and losing! Out-of-the-box mode can also reduce theaveragenumber of turns needed (i.e. the expected value of this random variable). Example #9 demonstrates this: there are five words that could possibly be that day's hidden word: BROWN, CROWN, DROWN, FROWN, and GROWN. It's pretty clear that both freeform and guided hardmode can take up to five moves to get the right answer, with the player losing the game. But if the player enters BADGE for the fourth word, for example, then there will be a clear signal whether or not BROWN, DROWN, or GROWN is the hidden word and can be entered to win on move 5; otherwise the word is either CROWN or FROWN, and we can enter one of them on move 5 and if necessary enter the other on move 6 to win. So the maximum number of guesses needed drops from 5 to 3, and the expected number drops from 3.00 to 2.40. That's a significant improvement, but it does come at the cost of the player having to memorize more steps to their algorithm (i.e. to remember that if the word could be BROWN, then it's best to play BADGE). With more options to choose from, it's not surprising that we can often find out-of-cluster words that trim the set of candidates in a cluster more often than using in-cluster ("preferred") words. In an attempt to use fewer moves it's tempting to look outside the cluster more often. I prefer not to do so when an in-cluster word is available for two reasons. First, the special rules to resolve a problematic cluster take half as much memory this way! Secondly, when used in the compound games, the preferred-word rules continue to be just as useful even when previously-solved subgames remove some candidates from a cluster (e.g. the rule "play PRIZE whenever it's a candidate" continues to be at least as good a move as making a random selection among the candidates, irrespective of how many candidates are left in the same cluster as PRIZE). On the other hand, using up a move to enter an out-of-cluster word could be *less* efficient than choosing a random candidate, after some of the candidates are removed (e.g. if DROWN and GROWN have already been eliminated, it is a waste of a move to enter BADGE). So we will generally assume the player will NOT reach for an out-of-cluster word if at least one of the words in the cluster will lead to a guaranteed win. In practice -- particularly in compound games -- humans will find it can be handy to employ those tactics if they can spot them on the fly (potentially even using words that are not on the shorter Wordle answer-list) if the player is close to running out of moves. But we will avoid discussion of such ad-hoc strategies. ---------------- For a player who has just entered LOATH, MURKY, and SPINE there are 1,715 possible ways the colored tiles can then appear. (We've just worked through 9 of them.) Fully 80% of them indicate precisely one word and the player has an easy win on move 4. But those other 20% can be tricky, as we have seen. It's easy enough to write a computer program to alert us to all of them and outline potential responses, but if we wish to answer the question of how good this starting set is, we have to know just *how* the player intends to proceed in those other 20% of the cases! In my analyses, I will assume players play out "Phase 2" in one of two ways: (1) Simply use a guess-at-will strategy. With a six-move limit, that may mean accepting the possibility of a loss; we might want to compute that probability. Or we can imagine the freedom to continue playing more moves until victory, and then we can ask for the probability distribution: what is the probability that the player will win after 1, 2, 3, ... moves. From that we could compute theexpectednumber of moves until a win. (2) Use the same guess-at-will strategy in general, but by pre-computing startegies for the possible tricky situations, the player will use a "preferred" right answer (like PRIZE for Example 8) when necessary to do so to ensure a win. And (only) if no such preferred word exists, the player will use an "out-of-the-box" solution (like playing BADGE when BROWN is indicated by the clues, in Example 9). In either case, I would expect the player to revert to free-form guessing in the very next move. There can also be situations in which neither a "preferred" word nor an "out of the box" solution exists. That turns out not to happen with LOATH + MURKY + SPINE but it occurs for example with LEARN + STICK + DOUGH: when the hidden word could be "batch", it could also be any of batch, catch, hatch, match, patch, watch It turns out that no matter what word you enter next, from the entire Wordle answer list, you could STILL have a set of at least 3 words that produce the same colored-tile patterns. In fact, just this once, I even checked all 14,853 currently-allowed Wordle input words, and every one of them still leaves a set of three of them unseparated. (I have to say I was surprised by this!) That doesn't mean the player cannot still win by the 6th move, but now he or she must use BOTH moves 4 and 5 to gain enough information to be able confidently to enter the correct word, finally, on move 6. For example the player would have to know in advance to enter (say) BAWDY on move 4 and CHAMP on move 5 . So that's another rule this player has to remember: BATCH -> BAWDY + CHAMP ; using it, the player will get enough information to know what to enter for move 6. Even a reasonably good starting set might need a "two-word strategy" like this for a couple of its most problematic clusters of similar words that yield the same tile patterns, but the best starting sets resolve every cluster with just "preferred" and (single) "out-of-the-box" moves. There is one more playing option we might consider, as we try to model what the player might do when armed with a fixed starting word-set. Consider the player using LOATH + MURKY + SPINE who gets the colored-tile pattern in Example #10. It is already clear after the first two words have been entered that the word must be LOAMY, so there is no point to entering SPINE. This example is especially obvious but more generally there is no point to entering SPINE if the first two words have already revealed five different letters in the hidden word; since SPINE does not repeat any of the letters in LOATH and MURKY, we know in advance that the response to SPINE would just be five gray tiles, giving us no new information. Even if only four colored tiles had shown from LOATH and MURKY, we could probably skip entering SPINE; the same might be true if we had gotten three green tiles from LOATH and MURKY. The point is that in such cases it is likely that there are only very few words --- maybe just one --- that fit these unusually helpful sets of clues. A dedicated player might even make a list in advance of any problematic situations that could arise from skipping SPINE when there are four colored tiles from LOATH and MURKY, and then devise separate strategies for those cases. We will not pursue this line of inquiry very far because it is less useful in a compound game. Suppose, for example, a person is playing Dordle (N=2) and after just LOATH and MURKY are entered the player sees LOA..+m...Y in one subgame but .....+..... in the other. Surely they can enter LOAMY next to win the first subgame, but this will give no new information in the other, so inevitably the player will use SPINE again anyway. More generally, it only makes sense to abandon the intended list of starting words if *all* of the subgames have already given abundant clues in response to the first couple of words. This certainly can happen, especially in Wordle itself (N=1) but it becomes increasingly rare as N increases. ---------------- To complete this introduction, we can now summarize the prospects for the player who begins with LOATH + MURKY + SPINE . (1) The guess-at-will strategy cannot guarantee success. Example #11 is among the worst: from those tiles all we know is that the hidden word is one of thesetwelve: bread, cedar, debar, dread, eager, gazer, racer, rebar, wafer, wager, waver, zebra An unlucky guesser could take as many as six more guesses to find the hidden word, even if properly following all the additional clues that come from earlier incorrect guesses, and even if only guessing legitimate Wordle answer-words. A player who enters these three starter words and then follows this process can expect to guess the hidden word on the very next turn 74.1% of the time. On subsequent moves the player will win (21.1%, 3.90%, 0.78%, 0.12%, 0.002%) of the time. In particular that means losing standard Wordle 0.90% of the time. If it is possible to keep guessing through move 9, the expected number of moves needed is 4.317. (2) On the other hand, the player who does not want to face defeat (but still wants to begin with LOATH + MURKY + SPINE ) has the option of using special rules for tricky situations. As it turns out, of the 1715 possible ways the colored tiles can result from this starting word-set, 25 of them can lead to defeat if we just guess at will any word consistent with the clues. Of these, 13 cases can still give a victory if we play a "preferred" word in the cluster of consistent words (like PRIZE) but the other 12 require an out-of-cluster word (like BADGE) to ensure success. (See e.g. Example #12, which could indicate any of daunt, gaunt, jaunt, taunt, or vaunt as an answer. But the player could enter JUDGE and then be sure of a victory after just 1 or 2 more moves.) For a player who follows this strategy, I calculate that the probability of completing the puzzle on moves 4, 5, and 6 to be 73.56%, 22.88%, and 3.56% respectively, for an average of 4.300 moves to win (and a 100% chance of winning by the sixth move). This represents an improvement over the guess-at-will strategy, but comes at the expense of having a more complicated set of rules of play. Those probabilities apply to Wordle itself but they point to some statistics for the compound games too. If the player is playing a compound game built from N Wordle subgames, then (if sufficiently many moves are permitted), the player will first enter the three starting words, and then in any of the N subgames can expect to require 1.317 additional moves to win by using a guess-at-will strategy (or 1.300 additional moves to win by memorizing what to do with the 25 anomalous clusters). That would mean the expected number of total moves is 3 + 1.317 N. Well... it *would* mean this expected number of moves if (after the three starter words) the subsequent guesses are applied only to one subgame at a time. In practice, this is not true -- the additional words entered for the first subgame may give additional clues in the second and subsequent subgames, so the expected number of moves is surely smaller than 3 + 1.317 N . Nonetheless, this gives an upper bound on the expected number of moves for e.g. Quordle, and more broadly shows the relative importance of the two phases of Wordle-solving. For Wordle itself (N=1), it is only the combined number of moves taken for both the fixed initial guesses and then the guess-at-will phase. But for increasingly large values of N, the size "a" of the starter set (here, a=3) becomes less important than the set's effectiveness in approaching a solution. For example, we will see later that there is a four-word starting set that completes the puzzle on the very next move 97.02% of the time and on the following move 2.98% of the time, so the expected number of moves is 4 + 1.0298 for Wordle and no more than 4 + 1.0298 N for a compound game. Already for Quordle this is a smaller number than 3 + 1.317 N. In short: for Wordle itself (and the smaller compound games) it may be inefficient to fix more than one or two starting words to be used every day, but for the larger compound games, it may indeed be more efficient to begin with a larger fixed starting set. Also note that, while we have a way to turn LOATH + MURKY + SPINE into a 100%-winning strategy for Wordle, it doesn't guarantee success for the compound games. Our refined strategy (2) will surely winWordlewith no more than 3 moves after the starting set, but for an N-fold compound game that means the maximum number of moves needed could be 3 + 3N, in the (rare) case that all the subgames force the player to use three additional guesses to discover the word. (Not only would this happen at most .0356^N of the time, according an earlier paragraph, but it would require that none of the words entered to complete any subgame offer any succor in any of the other subgames -- a very rare situation!) How else could one offer more information about how the strategies will fare in the compound cases? Surely failure is possible for N>1 even if it is impossible for N=1. Presumably one could (at least for very small values of N) itemize a catalogue of the *combinations* of tile patterns that could lead to a loss and perhaps find ways to circumvent them, as we did above with "preferred" and "out-of-the-box" moves, but I have not tried to do so. I have tried to run computer simulations of thousands of randomly-selected Quordle games to see how the different starting sets compare, but it is not clear how representative these are, since there are over one trillion different Quordle games, and many more for Octordle, etc. So in the end, is LOATH+MURKY+SPINE a good Phase-1 strategy? That's a matter of opinion. Memorizing the 13 "preferred" words, plus recognizing the other 12 difficult clusters and remembering the out-of-cluster word that resolves them, might be a bit much. Figuring them out on the fly is not impossible, but taxing. Sticking with mode (1) is simple, and if you're a gambler who is a "lucky guesser", that may be sufficient. And, well, maybe it's more fun too, even if it means the occasional loss. It's a personal decision, of course. But we can provide comparable data for other word sets, and let each player make a decision separately. ---------------- So to summarize all this notation: we will analyze some sets of "a" words to be entered at the start of an "N"-subgame compound Wordle game. Based on the colored tiles that result, the 2,315 Wordle answer words will be split into "clusters", and seeing the cluster in which the hidden word is contained, the player will try to discover which word it is by either making a random guess, or by using a memorized "preferred" member of the cluster, or by using a pre-computed "out-of-the-box" word that separates the cluster into smaller clusters, or (in extreme cases) by entering pre-determined words over the next *two* moves to resolve ambiguity. In order to compare different starting sets, we can tabulate, for each such starting set, the number and sizes of the clusters; the probabilities of completing the puzzle after 1, 2, 3, ... additional moves; and the amount of information the player must memorize to carry out their algorithm. We can look for measures of how likely it is that the collections of colored tiles will be very sparse (does the player need a hint?) or abundant (can we guess the hidden word before using all the starter words?) After all these measurements, we can decide for ourselves which is most important, and then determine which starting set is "best" Very well, then, let's look at some very good starting sets of Wordle words. We start with sets of a=6 words, then progress down to smaller values of a. ============================================================================== SIX (AND UP) The six-word set [catty, frond, rumba, spill, verge, whack] is nearly perfect for playing these games. It completely distinguishes all 2315 Wordle words (despite not including j,q,x, nor z !). That is, any two different hidden Wordle words will generate different patterns of green/yellow/gray tiles when these six words are entered. So if we enter these words as Phase 1, then Phase 2 is simply: enter the hidden word and win; there is no need to wonder about the player's behaviour. There is no guessing nor branching in this routine. After the a=6 starting words are entered, the probability of finishing on the next move is 100%. So this six-word set is nearly ideal for the N-fold compound Wordle games: all of them will be completed successfully in N+6 turns, 100% of the time. Sadly, most of these games (Wordle itself is the case N=1, then Dordle, Quordle, etc.) allow only N+5 turns to finish the game, so this starting set has a 0% chance of actually winning the game. One exception is Octordle's variant that requires the player to solve 8 simultaneous Wordle subgames *in order*; because of the extra level of difficulty, that game allows the player 15 guesses to try to win. This starting set of 6 words permits the player to win every such game in only 14 guesses! (Here I use the fact that all the answers-words for Octordle are among the 2315 Wordle answers.) *Almost* serving as another application is Sexaginta-quattuordle, which gives the player 70 moves to guess 64 words --- just enough to use this starting sextet. In fact, starting with this sextet gives anexcellentway to play this compound game, since then the player need only scan the first six rows of the crowded display for each subgame. Unfortunately, the word list for 64ordle is significantly larger than that of Wordle. so in 64ordle there are pairs that are not distinguished this sextent: edged, egged sided, sized dozed, oozed boded, boxed dazed, jaded waded, waxed including some pairs involving Wordle words* : unzip*, unpin bonus*, bosun mummy*, yummy organ*, argon so the player would not be assured a win with this sextet. Indeed no sextet will suffice for the full set of 64ordle words; I suspect there are many septets that will suffice to distinguish every one of the game's answer words, but with only 70 moves allowed, a starting septet will not allow the player a victory. The sextet also definitely fails (slightly) for Dordle. whose solution-list includes the words UNPIN, YUMMY, and ARGON; the sextet does not distinguish these from UNZIP, MUMMY, and ORGAN, respectively. Since the Dordle word list also *deletes* some of Wordle's words, there may be a different discriminating sextet for Dordle. I have not looked for one. Of course, this discriminating sextet also works for those compound games whose solution set is a subset of Wordle's: Quordle, Octordle, and Duotrigordle. But it's of no practical value since those games do not allow N+6 words to be entered. It is comparatively easy to get more 6-word sets that *almost* split the whole wordset, and thus it is easy to get many sets of 7 that do. But it turns out that this six-word set is the UNIQUE sextet of Wordle answer-words that splits the Wordle dictionary into singletons in the way I have described! (I have to say I was very surprised by this!) Of course, even when playing this sextet, the player still has to do some thinking to *recognize* the hidden word each day; knowing that it is unique, and knowing a few letters in it, are not quite the end of the story. Of the 30 letter tiles shown after the six words are entered, the player may see no greens at all and as few as four yellow tiles (both {b,i,n,o} and {i,n,p,u} can occur) and it can take some effort to realize the hidden words are "inbox" and "unzip". A novice player who wants to practice recognizing Wordle words might want to play with this set, since it allows only one Wordle-correct word to be built from any set of clues. You could, for example, enter these words into the sequential version of sedecordle (because it allows you to enter so many words) and then practice recognizing Wordle words. If you want to have a set of words that has the same property as the Magic Six, but includes all 26 letters, you'll need at least eight starting words. One such solution is [cross, equip, expel, flank, jumbo, razor, vodka, wight] How much help do you need? We can even flag for each letter that's ever doubled, though to do so you'll need at least 17 starting words, e.g. [affix, booby, ditto, jazzy, kappa, kayak, mimic, occur, penny, piggy, queue, radar, shush, slyly, undid, vivid, widow] Oh, there are some tripled letters too; if you want those flagged as well you'll need at least 20 words, e.g. [bobby, cocoa, daddy, error, fluff, heath, jazzy, knack, leggy, mamma, melee, ninny, pixie, puppy, queue, sassy, slyly, tatty, vivid, widow] At that point, you know not only which letters appear but how many of each there are! With 20 words entered, Sedecordle is leaving you space for just one guess, but you have literally nothing to do but to permute the letters in yellow! (On average, 3.74 of the tiles are already green; only abort, acorn, adorn, avian, axial, offal have no green tiles and force you to consider all 120 permutations of the five yellow tiles.) I can even provide a starting word-set that relieves the player ofallthought! We do so by finding an (optimal) solution to the game Kilordle, which requires solving N=1000 Wordle games at once. (In Kilordle, it is not necessary to *enter* each correct word, merely to get a green tile in each of its five columns. As an additional assist, the 1000 subgames are sorted to present first the ones that are closest to completion by some metric, with the completed subgames removed from view.) But in fact we can ignore the given subgames completely! Just treat this as requiring a list of words that contains each letter in each position -- 130 tasks. Actually 5 of the tasks are never presented in a Wordle game (e.g. there is no word with an x as the first letter). When solving kilordle manually, I typically need to enter about 100 words. Certainly 26 words would be a minimum because we would, among all the subgames, eventually need to enter every letter in column 2. So the minimum number of words needed, to be sure to solve every round of kilordle, is between 26 and 125. I found manually a set of 36 words that did the trick. But by using optimization software I discovered that the minimum is actually 35. A sample solution is [above, affix, askew, banjo, bayou, civic, debug, eject, epoxy, ethos, evoke, extra, fritz, globe, howdy, igloo, imply, jazzy, known, leggy, maxim, nymph, ozone, pique, quasi, rajah, scrub, skimp, squad, tweak, udder, vinyl, whiff, yacht, zesty] Other 35-word kilordle solutions exist but all of them must contain "pique", which is the unique word having q in position 3. They must likewise all contain "bayou" (u5), and either "banjo" or "ninja" (j4), "azure" or "ozone" (z2), "eject" or "fjord" (j2), etc. I already fiddled with the list to remove words I didn't care for ("squib", "waxen", "twixt",...); I'm not sure what I would consider the most normal-sounding list of 35 words would be. So not only does this set of words solve every game of Kilordle, it gives a "simple" way to solve Wordle: just enter all 36 of these words, and then locate the green tiles in each column to form a 5-letter word! Of course, we're now waaay past the 6-word Wordle limit... (If you don't mind using words like "embog" and "jambu" then the minimum drops to 30, using the 12000-word list of possible Wordle inputs. A solution was posted to reddit by user "k3and". As the Times increases the pool of acceptable input words, the size of a minimal winning Kilordle set can decrease.) Mathematicians might want to click here for a short description of how I first found the 6-word set; You'll probably want to read about measures of word similarity first. The point is that we can talk in a meaningful way about what it means for two words to be "close" to each other. (Mathematically, we can impose a metric on the set of these words, and all our searches for optimal word sets focus on finding words that are within a small distance of each other, then ensuring that Phase 1 leaves us with ways to distinguish those words.) The claims of minimality for the 8-, 17-, 20-, and 35-word sets are proved by covering-set arguments and computations using Gurobi. ============================================================================== FIVE With five 5-letter words we can hope to include all but one letter of the English alphabet, and sure enough this is possible. One example that comes immediately to mind contains all the letters but j : [waqfs, vozhd, blunk, cimex, grypt] Hah hah, just kidding, that's a bit ridiculous. Not one of those five words is in the Wordle wordlist, although all five of them are accepted asinputin a Wordle game. We can get as many as three wordlist words into the set and still have 25 letters: [waltz, fjord, chunk; vibex, gymps] (This one misses only q .) But having two words outside the basic Wordle wordlist is the provable(*) minimum, if you hope to include 25 of the 26 letters, and the only two such sets are this one and [waltz, fjord, nymph; vibex, gucks] and neither of these is particularly great as an opening play. (The first one leaves 41 pairs of words and 4 triples undistinguished; the second is similar. So neither will guarantee you a win of the game in 6 moves.) [ (*) UPDATE: I re-verified this after the NYT increased the set of Wordle inputs. Simply compare the list of 20-letter quadruples of Wordle words to the new list of acceptable Wordle inputs; in every case there is non-empty intersection. Although, let's be honest, who would know better if I announced that [ fjord, nymph, squib, waltz ; gveck ] was a 25-letter quintuple that used only one word from the list of acceptable inputs?... ] A five-word set with 25 distinct letters is impossible if it includes only zero or one non-wordlist words; the best we can do is 24 distinct letters. Before I turn to the 24-letter sets formed only from answer-list words, let me mention one example that does include just ONE non-Wordle word, which I will do because it's actually a reasonable word. It's (almost) a sentence, or at least a headline: [quick, waltz, vexes, fjord, nymph] (It's got two e's while missing b and g . "Vexes" is not in the Wordle wordlist, being a third-person singular form of a verb.) Entertaining though it may be, it's not as perfect for Wordle as the sextet in the previous section: the colored tiles returned from these five words are not sufficient to distinguish "error" from "gorge", "blast" from "stall", etc. ---------------- That's the last word set I analyzed that uses non-Wordle words; in everything that follows, I only consider sets of words from the Wordle solution-word list. Of all the 5-word sets made of Wordle answer words, none have 25 distinct letters. There are 58 5-word sets with 24 different letters. Four of them include one word with a repeated letter, e.g. [blitz, chump, fjord, gawky, seven] and the rest have a pair of words sharing a letter, e.g. [coven, fjord, gawky, plumb, sixth] As starting sets for playing Wordle and the other hgames, I would argue that these two are each the best in their class. But despite revealing nearly all the letters in the hidden word (they miss q,x and q,z respectively) they still don't quite pin down the word unambiguously: using the first one to get clues we would not be able to distinguish odder, order, and rodeo from each other as candidates for the missing word, and there are 26 additional pairs of words that cannot be distinguished. The second quintuple has no unresolved triples but does have 31 indistinguishable pairs. With so many letters revealed, these quints do give a human player lots of help figuring out the missing word. And they make a great start for the N-fold games with large N, because the average number "b" of words that are used in Phase 2 is so small: the second of the quints terminates by the next (sixth) move 98.66% of the time, and in the remaining 1.34% of the cases it finishes after just b=2 moves. (There's really no question about mode of play here: each set of 25 colored tiles corresponds to only one or two possible words.) The first quint is a little different, though. That one triple {odder, order, rodeo} does benefit from preferentially choosing odder or order instead of rodeo. So we can run two analyses: using a guess-at-will strategy, the distribution is 98.79% of games end with (0 or) 1 additional move 1.20% end with 2 additional moves 0.0014% end with 3 Or we can use instead a "guided hard mode" strategy: if we choose to play "order" when it is consistent with the clues, the distribution changes only little, but in an important way: 98.79% of games end with (0 or) 1 additional move 1.21% end with 2 additional moves none need 3 (or more) additional moves. ---------------- Next we set aside the desire to include 24 different letters, and just look for ANY good set of five words. Is there any five-word set that's as good as the six-word set of the previous section --- one that always narrows down the set of possibilities to just one word (and thus guarantees a win by move 6)? The answer is provably "no". I have an elementary argument that explains why no such perfect quint exists. But it's faster to simply use the Linear Programming techniques already described in this document, especially since (a) That technique also proves we cannot even find a perfect quintuple among the much larger set of Wordle's list of recognized inputs, (even including the additional words added Aug 2022); and (b) LP techniques helped me find the "tough pairs" that I used to create the elementary argument, anyway. (The LP techniques are also used to prove the uniqueness of the six-word set in the previous section, and that uniqueness in turn trivially proves that no five-word set can detect every hidden word unambiguously.) Nonetheless, some really good sets of five words do exist. ---------------- A provably most-efficient-possible 5-word set is: [blank, chump, goody, river, swift] (No jqxz; has double r, double o, and two i) Laurent Poirrier found this one and we have proved that no other quintuple is better in the sense that this quint can distinguish all the Wordle answer words except for these 11 pairs: [ample, maple] [booby, boozy] [bugle, bulge] [chili, chill] [eagle, legal] [gauge, gauze] [jaunt, taunt] [lemon, melon] [pasty, patsy] [skate, stake] [testy, zesty] So there is no need for anything but free guessing --- if the hidden word can only be one of these 22 words, we might as well flip a coin to pick one in the right pair. Thus the distribution of games is that 99.52% of games complete with 1 more move 0.58% complete on the second move (after the first 5). The only other quintuple that is equally efficient is [bawdy, clove, furor, might, spank] (No jqxz; has double r, two o, and two a); it has the very same distribution of numbers of moves needed. If we sequentially ran N independent random processes, 99.52% of which finished after 1 step and the remainder after 2 steps, then the number of steps needed to complete all of them could be anywhere between N and 2N ; the probability that exactly k of these N processes took that second step to complete would be binomial(N,k) (.9952)^(N-k) (.0058)^k, and the expected number of steps taken would be 1.0058 N . Thatalmostmodels what happens with an N-fold compound Wordle game like Quordle (N=4) : the expected total number of moves needed, if we begin with either of the starting quintuples above, would be 5 + 1.0058 NIFthe N subgames were independent. But they're not! Suppose for example we are using the first of these two quintuples. If after entering those words we have concluded that in one of the N subgames the hidden word could either be "ample" or "maple", and we also know that the hidden word in another subgame is either "bugle" or "bulge", then we should indeed flip a coin to enter either "ample" or "maple", but then (by looking to see whether the L is yellow or green) we would know whether the other hidden word is "bugle" or "bulge". As it turns out, for EVERY one of the 11 pairs at least one of the words in the pair can resolve at least one of the other 10 ambiguous pairs. In fact, the compound games which contain exactly two subgames in which the hidden words lie in (different) ambiguous pairs can involve 55 different combinations of two of these ambiguous pairs, and of these 55 combinations, 24 allow us to guess a word from one pair that will resolve the question of which word from the other pair is the hidden word in that other subgame. In fact, it is impossible for a game to require more than 10 + N steps to complete (far fewer than 5 + 2N except for the smallest N) because so many of the ambiguous pairs include words to guess preferentially so as to discover the hidden words in other pairs! (A maximal example includes subgames with the hidden words chili gauge jaunt lemon skate Such a game would require five coin tosses, which if they are all unlucky would cost us 10 moves to win, after the initial quintuple is entered.) Since the subgames are very likely NOT independent, then, we can only conclude that the expected number of moves to complete an N-fold compound game, when starting with one of these two quintuples, isat most5 + 1.0058 N . If we are playing an N-fold compound game that allows only N+5 moves to complete the game, then after the starting quintuple is entered, we must finish every one of the subgames with just one move (each). That would happen with probability (0.9952)^NIFthe games were independent but as above we notice that the earlier subgames can provide additional information to help resolve the later ones. So in fact the probability of success is for all N at least 0.9952^5 = 0.97623 (i.e. a 97.623% chance of winning), assuming the player resolves any of the 11 ambiguous cases in an advantageous order. ---------------- When I claim that the previous two starting quintiuples are optimal, what I mean is that they minimize the number of pairs of words that are not distinguished from each other, and consequently they minimize the expected number of moves needed until a win. The proof of their optimality comes from searching for sets of five words that maximize the pairs split among a select list of pairs of similar words. Searching for maximizing quints in that way allows us to discover other quints that are nearly as good. The next few close contenders for "best" starting quintuple, all of which happen to contain all letters except j,q,x, and z, are these: [bawdy, furor, month, speck, vigil] will win on move 6 99.48% of the time, and on move 7 every other time. Nothing is any better than free-form guessing. This quint leaves 10 pairs and a triple unresolved. [flock, haven, rugby, swept, timid] has exactly the same distribution of moves, but this one leaves 12 pairs unresolved (and no triples). [bawdy, chump, front, skill, verge] finishes by move six 99.35% of the time, and on move 7 every other time. No need to think about different strategies or modes; all this quint leaves undecided is 15 pairs. [batty, champ, furor, slink, wedge] which leaves undecided 15 pairs plus a triple (the same triple [bobby, booby, boozy] as a previous case) for a 6-move success rate of 99.27%. (This last one might be slightly easier for a human to use because it most frequently returns four or more yellow or green tiles, although how to compare the quints on this score is not clear because the repeated letters in the quint mean there may be less information from yellow tiles than meets the eye.) For the curious: these are the six quints that cover the largest numbers among the 919 hardest splitting sets, that is, I checked the 919 hardest pairs of words to differentiate, and these quints covered the most --- at least 907 of them --- and moreover these quints *did* distinguish any pair of words that wasn't on this list of 919 tough pairs. (FOr completeness' sake, I ran a similar test with the much more accommodating set of 14,853 currently-allowed input words for Wordle. There exist (multiple) sets of five words which successfully distinguish all the Wordle answer-words except for 8 pairs, and 8 is the minimal number of failures. For example, for spill verge dumbo fawny chott all the clusters are singletons except for these eight pairs: algae/glaze crock/crook dried/drier husky/hussy liken/linen odder/order piper/riper rebar/zebra It is easy to find many starting quintuples which give success rates over 99%, but never 100%, so perhaps we are just splitting hairs here. But one quint of note is [carve, sight, downy, plumb, fetal] which extends the primary quadruple of the next section to reduce that quadruple's ambiguity --- all that's left are 20 ambiguous pairs and two triples. There's no need for any strategy; even the triples can be guessed at will. 98.96% of the games finish by move 6, 1.94% finish on move 7 Also note that this quint uses 20 different letters in the first four words, potentially helping the player discover the hidden word more quickly. So to summarize, we have found several quintuples of starting words that *almost* always enable us to know the hidden word and enter it as move 6 and win --- but not one of them allows us to win 100% of the time. This is about to change... ============================================================================== FOUR Using the right starting set, we can guarantee a win of Wordle. With a four-word starting set, it is conceivable one could win *before* using up all the moves allowed by Wordle's rules. Indeed we will see in the next section that a perfect player can always win Wordle in at most 5 moves. But in order to do so with a four-word starting set, that quadruple would have to unambiguouslyidentifythe hidden word every time. As discussed in the previous section, that's not even possible with FIVE starting words, let alone with four. So instead we look for sets of four starting words that can guarantee a win by move 6, i.e. with TWO rounds of guessing after the four starting words are entered. That's flexibility we did not have in the previous section, and it turns out to be just what the doctor ordered. We start with: Most useful 4-word set, ideal for casual players who want a 100% win rate: [carve, downy, plumb, sight] This set has 20 different letters (all but k, f, and the rare j,q,x,z) which gives the human player a lot of information about the hidden word. Then, in 94.25% of all cases, there is only one possible word which can be entered on move 5. The uncertain clusters consist of 5 triples of words and 59 pairs. Guessing a word from each set will add wins on move 5 in an additional 2.76% of the cases; and in the remaining 2.98% of cases there is a guaranteed win on move 6. So we always win, and over time take an average of 5.0298 moves to do so. With this starting set, the player can finish using "freeform hard mode" --- there is no need to use the other modes --- and will finish on move five 97.02% of the time and on move six the rest of the time. As in the previous section we can at leastestimatethe performance for compound games: an independence assumption would make the expected number of moves be 4 + 1.0298 N , and the fact that the subgames need not be independent only serves to lower the expected number of moves. If I have run the tabulation correctly, there are 45147 sets of four Wordle words that use 20 different letters. None of them appears to be better than the one above, but some are reasonably good. Many permit a win just by guessing (after the four words are entered), i.e. finishing with "freeform hard mode", but they have slightly worse distributions of numbers of moves needed compared to the quad above. For example after entering [carve, downy, fight, slump] we have only 4 ambiguous triples, but 63 unsplit pairs, leaving a probability distribution of ending on move five 96.93% of the time, and on move six 3.07% of the time, so an average of 5.037 moves to win. Other quads that are good (by various measures) include [burst, champ, dingy, vowel] [brawl, coven, dumpy, sight] [covet, gland, shrub, wimpy] etc. They all involve a handful of undifferentiated triples and dozens of undifferentiated pairs; none of the triples require preferential guessing or out-of-the-box thinking. All four of these sets end in a win by the 6th word, and in fact the fraction of the time a 6th move is needed is small: 3.02%, 3.02%, 3.07% respectively. The set [bawdy, flung, porch, smite] was suggested to me by a friend when I was first introduced to Wordle. It's nearly as good by the above measures (not quite!) though it does have the advantage that there are fewer words for which we get only a couple of yellow/green tiles. This doesn't technically make it better for distinguishing the words, but it does make it easier for a human player to deduce what the possible matching words are! The particular Achilles heel for this quadruple is the set of 3 words jaunt,taunt,vaunt which will yield the same tile colors. In this case in order to be sure to win by turn 6, the player cannot continue in hard mode but must turn to "out-of-the-box" mode of play: s/he must enter either "jetty" or "trove" on move 5 to get enough additional information to know which of the three candidates is the actual word of the day. (There are three other triples and 69 pairs which are also undifferentiated after the first four words are entered, but in those cases we can simply guess at will and be sure to win by move 6.) Using just free-form guessing we will win in five moves 96.67% of the time; six moves 3.28%; 7 moves 0.04% But using out-of-the-box mode for that one cluster, we would change those numbers to 96.63%, 3.37%, and 0% -- it's still the same 5.0337 expected moves, but (crucially!) we would always be able to finish by move 6. The 4-word set [bugle, champ, downy, first] (again with 20 letters, this time missing k, v, and jqxz) has a slightly higher ability to identify words unambiguously; it leaves undifferentiated only 50 pairs and only 4 triples, but also one quad. ([piper, riper, viper], [jaunt, taunt, vaunt], [eater, extra, taker], [bobby, booby, boozy], and [skate, stake, state, stave]). These larger sets prevent us from winning by move 6 if we use a "freeform hard-mode" style of playing: the distribution of play times is 97.37% of games end by move 5, 2.52% end on move 6, 0.12% end on move 7 Using "guided hard-mode" allows us to finish one of those sets by move 6 (play SKATE if you can) but two more require "out-of-the-box mode" to finish that fast: if JAUNT fits, play JETTY; and if PIPER fits, play PARER. Using those three rules, the probability distribution changes to 97.28% win by move 5, 2.72% win on move 6. So this quadruple is slightly more efficient than the others, but we comes at a price of extra complexity: we must remember the three additional rules Prefer: SKATE Replace: JAUNT->JETTY, PIPER->PARER An interesting also-ran in this category is: [blast, midge, porch, funky] which extends an excellent starting triple that we will meet in the next section; so if you start to play Wordle thinking you will just use the first three, and then you get stuck or otherwise want to bail out, you could just use funky as your fourth word. This quadrule would still leave 66 pairs to be split, and 9 triples. We can guarantee a win by move 6 if we steer clear of free guessing in four of those triples. A suitable set of rules is Prefer: EAGER, FEVER Replace: JAUNT->JETTY, CATCH->CLOWN which gives game probabilites of 96.29% by step 5, 3.71% on step 6. There may be more efficient 4-word sets that involve fewer than 20 letters; I haven't found any yet. (And I observe that these may be harder to use for people who don't know the wordlist well.) Here is one example that at least comes close: [champ, flown, rugby, steed] This one leaves 90 ambiguous clusters: 87 pairs and 3 triples. The distributions for free-guessing mode would be 95.98% by move 5, 3.92% on move 6, 0.10% on move 7 We can guarantee a win by move 6 with rules for the three triples: only one has a preferred word (play STARK if it fits); the other two require an out-of-cluster word (if JAUNT fits, play JETTY; if STAKE fits, play EVOKE). Then we can replace those probabilities with 95.90% by move 5, 4.10% on move 6 (never on move 7 or later) for an average of 5.04 moves. A search is underway for other, "better" four-word starting sets, but already we can prove that any quadruple of Wordle solution words will leave dozens of pairs undifferentiated, possibly in sets of three or more (as illustrated in the examples above). ============================================================================== THREE This is a long section because there are many starting triples that are "good" for various reasons, so no single one can be called "best", With a set of three starting words, we can surely win by move 6, but in practice this can be tricky. After just three initial words there are at least 11 letters that will not have been tested, so the player must do more sleuthing; e.g. it is quite possible that after three initial guesses the player has seen nothing but grey tiles! Still, by choosing anappropriatestarting set of three words, one can hope to have a 100% win rate at Wordle. After all, we have already seen in the last section that we can win 100% of the time starting with CARVE + DOWNY + PLUMB; surely with the freedom to choose something other than SIGHT next, we should be able to ask for something more than just a 100% success rate in 6 moves. At the very least we should be able to arrange a lower average number of moves until a win. What else might we ask for? What are we willing to give up? How do we decide that one or another three-word starting set is "better" than another, or even "the best"? The question of what is a good three-word starting set arises periodically on the Reddit forum. I fashioned a detailed response analyzing many of the starting triples that had been proposed. In this document, we can put that analysis into context. What we will see is that trying to reduce the expected number of moves needed to win will introduce more complexity in our algorithms. To be precise, what we had in the previous section was a starting set [carve,downy,plumb,sight] that had two features: (A) It wins 100% of the time within two more moves. (B) It requires no decision-making besides guessing freely among candidates. In this section we could hope for a THREE-word starting set with both those properties. After all, it *is* known that there are algorithms to win Wordle in just 5 moves (although the best algorithms that do that to my knowledge all require significant amounts of branching). Sadly, I can prove that no set of three words can have BOTH properties (A) and (B). In fact, I am pretty sure that (B) alone is impossible (more on this below). But we can find starting triples that have property (A). ---------------- I have found that there are exactly 261 starting triples with which every game can be won by move 5. For each of these triples, there will be clusters of words (signalled by the pattern of the 15 colored tiles) that could lead to a loss if we simply play with a guess-at-will mode, that is, we will have to map out some preferred words or out-of-cluster words to use in those cases. (See the examples of PROBE and BRACE in the Introduction.) Unfortunately, for each of these starting triples, in order to achieve goal (A), we need at least 54 rules of these two types, which is perhaps too many for a human to execute while playing a game. Which is "best" among these 261 triples is a matter of taste but [blast, midge, porch] is certainly a good choice. It minimizes the expected number of moves (4.18) by maximizing the probability of a win by move 4 (1888/2315 = 81.6%), and among all these starting triples it reveals the highest number (1597 --- about two-thirds) of words with certainty. There are also 414 more words that come in pairs, such that if we guess one of the pair on move 4 and it's not the hidden word, then the other is surely right and we can enter this one on move 5 and win. The remaining 304 words are grouped in 84 other clusters, each cluster containing between 3 and 10 words, that are still ambiguous. For 29 of the clusters, free-form hard mode suffices: we may simply guess any word in the cluster on move 4; if that's not the hidden word, we will get enough information from those colored tiles to know unambiguously which other word in the cluster is the right one. But the other 55 clusters require extra care: there *are* words to enter on move 4 which will give enough information to allow a win on move 5, but we must choose them carefully. In 39 of the cases, a word from within the cluster will do (e.g. {arena, freak, raven, wafer, waver, wreak} is such a cluster; "wafer" is the one choice that will work). In the other 16 clusters we need a word not in the cluster (e.g. one such cluster is {jaunt, taunt, vaunt}; in order to guarantee a win by move 5 the only words in the Wordle answer-list that can be entered on move 4 are "jetty" and "trove"). So in toto we have 55 such rules that must be memorized. A sample algorithm using this information might be this: After blast+midge+porch, enter any word consistent with the clues, except * If the word COULD be any of the following 39 words, then play it: allow, antic, awake, award, bevel, crown, dizzy, dowdy, dried, drone, eater, enter, equal, fauna, fatty, fewer, filly, finer, folly, funky, jelly, kitty, liner, mafia, otter, relax, safer, seize, sever, shown, skate, skulk, swash, taste, testy, udder, unfed, value, wafer * If the word COULD be any of the following 16 first-halves, play the second half: [anger, gawky], [catch, crown], [cinch, crown], [crane, ozone], [fatal, fella], [field, gawky], [fight, frown], [fizzy, ozone], [focal, fella], [forth, crown], [fudge, funky], [jaunt, jetty], [major, jetty], [rower, gawky], [snoop, frown], [stoke, funky] Then (if the hidden word has not already been played) there is only one Wordle word consistent with the clues; play it on move 5 and win. Other strategies exist; for example, for every one of the 55 problematic clusters --- indeed forall but fourof the 291 non-singleton clusters! --- one or more of the following ten words will split the cluster completely. Make a table of which of these words you wish to use to resolve each of the 55 clusters to obtain your win-by-move-5 algorithm: [crown, fewer, filly, funky, gawky, jetty, navel, skate, spunk, tawny] (The cardinalities of the sets of ten words here and the seven used in the previous algorithm are minimal, as determined by Gurobi.) With this given starting triple ("Phase 1") these different algorithms to complete the daily puzzle ("Phase 2") can have different expected numbers of moves, and different probabilities of success on moves 3, 4, and 5, but the numbers do not range very far. Note that by using this 3-word starter set on an N-fold compound game, we can solve all N of the subgames in at worst 3+2N moves. In that worst case, this is more than the number of moves typically allowed in compound games, which is N+5. But when N=2, the two are equal, meaning we have a guaranteed winning strategy for 2-fold Wordle. Unfortunately, Dordle uses a different wordset than Wordle, that is, Dordle is not exactly a 2-fold Wordle, so one does not have an a priori guarantee that this algorithm will work for Dordle. But as it turns out, itdoesstill work, with minor modifications. Change the preferred word (Rule 1) to this set of 35: allow, assay, awake, awash, awful, crown, dowdy, drone, eater, enjoy, enter, fatty, fever, finer, folly, funky, goner, jawed, kneed, lefty, newly, otter, relax, sally, seize, sever, skate, skier, skulk, snipe, testy, tower, value, viper, wafer and change the set of out-of-cluster moves to this set of 20 pairs: [anger, wagon], [catch, clown], [cinch, awful], [crane, anvil], [dizzy, dozen], [fatal, awful], [field, awful], [fifty, flank], [fight, flown], [focal, fever], [forth, awful], [foyer, gawky], [fudge, fauna], [jaunt, jetty], [liner, anvil], [lower, anvil], [major, agony], [snoop, flown], [staff, bonus], [stoke, ankle] Then each half of the Dordle game will definitely end within 3+2 moves, i.e. the whole game will end within 3+2+2=7 moves. This starting triple can also be used, more easily, to win in Wordle by move 6. That is, a player who initially intends to follow this algorithm so as to win by move 5, may decide during the play that it would be sufficient to win by move 6, and then need not remember all the 55 special cases listed above; the only ones needed are the preferred words CROWN FAUNA and FEWER, and the out-of-cluster pairs [fight, frown], [fudge, funky], [rower, gawky], [snoop, frown] Alternatively only in the first pair is it still necessary to go outside the cluster; if we are willing to wait until move 6 to win, we can instead use the first elements of the other three pairs as a preferred word. This algorithm will still complete by move 5 98.74% of the time, and has an only slightly higher expected number of moves -- 4.197 -- than the complicated, win-by-move-5 algorithm. And we have already mentioned a third alternative in the previous section: we can consistently play FUNKY on move 4 and then follow rules for four special cases; but this gives a significantly higher expected number of moves: 5.037. Finally, for basic Wordle we may return to a point made in the Introduction. If we enter only BLAST + PORCH, on about one-fourth of the days we will see 4 or 5 colored tiles, or 3 greens, or 2 greens and a yellow. In most of those cases we can still win by move 5 by simple guessingwithoutentering MIDGE! We need only watch for the following words, to be used as preferred members of their cluster: [blade, blond, brain, ditch, graft, gulch, mouth, plain, swash] and if the word could be STORE or CATCH, play HYMEN. Doing so will lower the expected number of moves to 3.9 . (We can similarly avoid MIDGE in thecompoundgames, but this analysis applies only ifallthe subgames show such a favorable return from just BLAST + PORCH, which becomes increasingly rare as the number of subgames increases.) This long analysis of BLAST + MIDGE + PORCH can be repeated for each of the other 260 starting triples that have property (A). I have not done so, but have collected some data about those triples and invite a discussion of which others are, by some measure, better than this one. ---------------- Now, what about property (B)? Surely it would be convenient to have a startingtriplethat worked as well as the startingquadof CARVE + DOWNY + PLUMB + SIGHT : just enter the starting set and keep guessing words that are consistent with the clues. I believe I can prove that no such triple exists when using only words from the Wordle answer-list. Just for this search, though, I also looked at the longer 14853-word list of valid input words. In order to speed things up I made the reasonable, but not ironclad, assumption that such a triple would involve 15 distinct letters. (This permitted me to doing a preliminary compression to the 5,649 sets of five distinct letters that are involved in those words, and then to non-intersecting triples of such letter-sets.) If I have done the search properly, I can report that no such perfect triple exists: for every (15-letter) triple of allowed input words, there is at least one cluster for which the guess-at-will strategy can lead to a loss in standard 6-move Wordle. I did also look for near-misses, though, and found a couple of triples for which there is only one bad cluster. The best is BONDS + GLAMP + FECHT Each of the words {skate,stake,stare,state,stave} will turn yellow the S, A, E, and T tiles, and obviously a guess-at-will strategy would for example allow the player to guess them in reverse alphabetical order, which would be a loss if the hidden word were "stake". So in this case the player must remember (only) one additional rule: "prefer SKATE". Also having just one bad cluster is TECHS + GLAMP + ROWND. For this triple, the cluster {berry, eerie, ferry, fever, jerky, verve} can again lead to a loss from random guessing (e.g. the sequence verve, jerky, ferry, berry) but again the loss can be prevented by playing the preferred word BERRY on move 4 if that is the cluster indicated after the initial triple. (For technical reasons I consider BONDS+GLAMP+FECHT to be the better of the two. Finding these good triples amounts to making sure they never (or rarely) permit quadruples like {berry, ferry, jerky, eerie} to be together in a cluster after the initial triple of words is entered. I assembled a list of tens of thousands of these problematic quadruples and then developed mechanisms to detect starting triples that broke most of these quads apart. The first triple I listed only missed its one quadruple "STA_E". The other one actually missed two: both {berry, ferry, jerky, eerie} and {berry, ferry, jerky, verve} are problematic. This is a minor distinction of course. The first triple is also better than the second in the sense that we only have to invoke our special rule ("guess SKATE") on five days out of a 6.5-year cycle, as opposed to needing the special rule ("guess BERRY") on six days per cycle!) I did not find any other starting triples that involved only a single non-singleton cluster. Both GLAND+ROMPS+FECHT and GLAND+ROMPS+WECHT involve just two (and each of them actually leaves three problematic quadruples unseparated). I make no claim about whether other equally-good triples exist. (My method of sorting was only designed to make sure I didn't miss any starting triples that guaranteed success *just by guessing* (withzerospecial rules like "use SKATE"), and so I am fairly confident that such a triple does not exist; but along the way I had to branch though decision trees to trim the candidate pool, and a starting triple that was a "near miss" might not have been good enough to survive an early-stage pruning.) ---------------- Among triples of words drawn from the more limited (and more reasonable!) Wordle answer-list, it appears that the minimum number of problematic clusters is three, that is, every starting triple requires the player to remember at least three additional rules if he or she wishes to ensure a win by move 6. I believe there are just two that accomplish this just with preferred words: [blond, girth, swamp]: prefer CATER, ROCKY, and TUTOR [blond, right, swamp]: prefer CATER, SKATE, and FORCE If the player chooses randomly from possible words at each stage, except when using the three preferred words for the fourth move, then he or she will win the game by the fourth move 73.22% of the time; on the fifth move 24.33% of the time; and on the sixth move 2.45% of the time, for an expected number of moves of 4.292 . (The numbers for the second triple are just a bit different: 72.87%, 24.85%, 2.28%, and 4.294 moves on average.) The first of these triples actually leaves seven of the forbidden quadruples (mentioned earlier) intact; the second leaves eight unbroken. These two triples were found waiting in the list of 261 triples that allow a win by move 5, that is, if instead of just the three preferred words we were to memorize dozens of rules, then we could force a win one move earlier even in those 2%-3% of the cases when the game would go to the sixth round when we ignore those dozens of moves. Compared to the triple BLAST + MIDGE + PORCH that we discussed earlier, these two are less good for trying to finish in 5 moves, but simpler for trying to finish in 6 moves. Other starting triples come very close to meeting goal (B): [choir, swept, gland] also has just three tricky clusters, but only two of them can be resolved with preferred words (merry and mayor); the last requires an out-of-cluster solution (FOYER -> FAVOR). Similar remarks apply to [copse, drawn, light] [blimp, cedar, ghost] [force, glint, swamp] [glint, peach, sword] the last of which requires TWO out-of-cluster words (BATCH->CLIMB, BOXER->ROCKY) as well as one preferred in-cluster word (MAYOR). I believe there are no other starting triples that lead to just three problematic clusters. There's also [blimp, cedar, thong] [blend, match, sprig] which can be played using only preferred words, yet need four of them (kitty, leave, rower, skate for the first; fudge, leaky, outer, rower for the second) because there are four problematic clusters. Only one of these triples can guarantee a win by the fifth move, but they all can come close: for example just by following the four rules, the last will finish on move four 80.09% of the time, taking an average of 4.212 moves. (Actually in 16% of the cases, we can confidently guess the hidden word after just blend+match, which lowers our average to around 4.0 --- assuming we are sure that we can recognize those cases when they occur!) The other triples have numbers that are similar. ---------------- So we have found starting triples that come as close as possible to the goal of having a simple mode of play; but such triples tend to take more moves to win. Before that, we found starting triples that never need more than 5 moves to win; but they all require complex sets of rules to achieve this. It may be better to find triples that are somehow intermediate between these extremes. And, since the player now hasthreemore moves after the starting set is entered, we have more latitude to devise different sets of rules for the triples; so we will need to describe both the "good" starting triples, AND the algorithm we will use after entering them. We will see there aremanyreasonable starting triples and many ways to play, so no matter how we will choose to judge when a triple is "best", there will likely be many triples that are very close in quality. In short: in this section it will be hard to name a "winner" starting triple! Using only the words in the Wordle answer list, there are over 2 billion sets of three words that we might potentially consider as starting sets. As we discussed in the sections on five- and four-word starting sets, it is certainly true that some starting sets with repeated letters are good. In fact, 46 of the 261 starting triples that can lead to a guaranteed win by move 5 have repeated letters! (For example [crump, doubt, salve] repeats "u" but lacks both i and o (and y)! ) Some of the sets people reported on Reddit that they enjoy using are reasonably good and duplicate letters, e.g. [blind, stare, wimpy] and [colon, right, speed]. But generally speaking it seems prudent to focus on sets of three Wordle words that include 15 different letters. That reduces the number of candidate triples to 1,243,026. (I have listed them all in a 26Mb zipped file.) This is small enough a set that it is possible to run some quick preliminary computations on all of them and then run longer analyses on the most promising among them. I have done so (and continue to do so as I write this) but now the question arises: what exactly do we want to measure about each candidate? How do we judge which is better or worse? For each triple, I could measure the following: 1. The number of clusters it creates. (This happens to be the same as (the number of Wordle words, 2315) * (probability of guessing the right word on the very next guess), so higher is better.) 2. In more detail we can count the number of clusters of different sizes. Generally we prefer more smaller clusters, e.g. more singletons, and fewer larger ones, e.g. a low maximum cluster size. 3. The total numbers of green and yellow tiles that are produced by the triple across a 2315-day (6.5-year) cycle. Whether the hidden word is uniquely identified by a set of tiles or not, it is difficult for a human to realize what the hidden word might be if he or she sees mostly gray tiles! So we want these numbers to be high. 4. The number of clusters for which guessing randomly could lead to a loss; specifically we could count the number that can be resolved by using a preferred word in the cluster; the number that can be resolved by a single word (but not one in the cluster); and the others, that require a two-word or other compound strategy. Of course to actually USE a starting triple, we would also have to know what these words ARE. 5. The probability distribution of numbers of moves until victory, if the player simply turns to freeform guessing after the initial triple is entered. From this in particular we can compute the probability of a loss when using this strategy ("loss" = "more than 6 moves used") and the average number of moves used. 6. The probability distribution of numbers of moves until victory, assuming the player follows a pre-determined strategy. I would assume the player follows "guess-at-will", "best preferred word", and "best out-of-the-box word" strategies in that order --- whichever guarantees a win by move 6. From this we can then determine the expected number of moves until victory. 7. As a measure of the difficulty of proceeding after the initial triple, it is helpful to measure the number of times per cycle that the player must begin with fewer than three colored tiles when deciding which is the cluster containing the day's hidden word. 8. As a measure of how often one can abandon a word in the chosen triple, we can count how often a player has an abundant set of clues after just two of the words have been entered. (It would also be relevant to know just how complicated the algorithm would have to be if we try to branch off at that point, but I will not carry out that analysis for many starting triples!) 9. As an assist to the stumped player, it would be helpful to list the words that best complement the starting triple: to find the words that introduce the greatest number of additional letters in case the player is already stymied at move 4. 10. It would be fairly straightforward to run a simulation to see how a starting set would fare for Quordle and other compound games. Whether this simulation accurately represents rare behaviour is unclear. (Note that statistics 3 and 9 depend only on the 15 letters involved, not the actual words, so these can be precomputed and recycled for multiple triples.) Ultimately it is up to the player to choose how to weight these numbers but having done so, the list of tested starting triples can be sorted. I would like to calculate these numbers at least for "best" of the 1.2million triples that involve 15 different letters. (I should probably do likewise at least for the 46 triples that allow a guaranteed win by move 5, even if I am only testing them for their suitability for a win by move 6.) Until I have processed more data to include here, I will direct the reader to the partial analysis I have already done: I computed #1 and #3 for all 1.2million 15-letter triples in this zipped file, and I computed assorted other statistics for some "promising" triples in this Reddit post. But already it is clear that many, many triples score well by some measures, and almost inevitably they are noticeably lacking in others. Thus there is not likely to be a clear "best" starting triple, but rather various contenders for the title depending on what the player values most. Here are some notable triples to consider. We have already mentioned [blast, midge, porch] It has the lowest expected number of moves (4.197) and the highest rate of completion-by-move-5 (98.74%) We also mentioned [blond, girth, swamp] It's an excellent choice for someone who simply wants to use freeform guessing after the initial three words are played. It's actually kind of hard to make such bad guesses that you lose after 6 entries: the expected number of losses in an entire 2315-word cycle is 0.446, i.e. you could reasonably expect to go *14 years* between losses! (But yes, it can happen: if the initial 3 words leave you with just yellow A,R, and T tiles, you might guess "treat"; then if you get a yellow E too, you might go for "extra"; seeing the T go green you might then guess "cater"; but then you'd lose if the hidden word is "after"). Close behind are [blond, right, swamp] and [blend, right, scamp]. "Best" by several standards is [bland, copse, right] 1. It is the triple with the lowest expected length of the game; it will finish with an average of 4.1682 moves if the player follows the decision tree I have indicated. (Just slightly longer averages can be expected from [midge, porch, slant], [blimp, dance, short], and [bland, copse, mirth].) 2. This is also the triple with the largest number of clusters. It spreads the 2315 Wordle words into 1954 different clusters, each identified by its unique set of colored tiles when these three words are played. That includes 1689 singletons (words that are already uniquely identified after the starting triple). Second place by this measure, but perhaps overall better, is [blimp, dance, short] (Obviously the singletons and the clusters of sizes 2 and 3 can be resolved by move 6. The extra rules cover the four largest clusters, of sizes 10,7,6,5, and 4. That leaves only one cluster of size 5 and 13 of size 4 which fortunately can all be solved by freeform guessing.) 3. This triple has the highest probability of finishing on the very next move (84.36%) (Actually this statement is logically equivalent to statement (2)! ) 4. Experimental simulations suggest this is the best set with which to start Quordle, in order to minimize the expected length of the game (about 7.420 moves per game, and winning Quordle over 99.5% of the time.) (To win in six moves with BLAND+COPSE+RIGHT, you can preferentially use MAJOR MOWER FEWER and WATER, and if FIGHT fits, play AWFUL. To win with BLIMP+DANCE+SHORT you can preferentially play ROVER, SKATE, FOLLY, FREER, and WAGER; if GAUNT fits, play GRAVY.) Dual to the most-on-move-4 record is the least-on-move-6. The lowest I've found is with [brown, midst, place] which only gets to a sixth move 0.82% of the time. (Second place goes to [blown, caper, midst]. The triple [blond, midge, porch] also scores very well.) You can use preferred words {gayer,skate,other,gavel,earth,rover,relay} If the word could be FOLLY, play FIGHT; if it could be TAUNT, play THONG Every one of the 15-letter triples leaves at least one cluster with six or more elements (and I would imagine that's true of triples with repeated letters, too!). There are 283 triples whose largest clusters have only 6 members. Probably the best of these is [blimp, dance, worst] it leaves just one cluster of six (if you get just a yellow E and a yellow R then the hidden word is one of {every, fever, freer, queer, query, refer} Comparably good is [bland, comet, sprig] which also has only one such cluster. These two triples are also the ones in this pool that immediately identify the most words (i.e. they leave the largest numbers of singleton "clusters"). And they are the two with the smallest average sizes of clusters. (To win with BLIMP+DANCE+WORST you can preferentially play ROVER and SKATE; if JOLLY fits, play FIGHT, and if JAUNT fits, play TIGHT. To win with SPRIG+COMET+BLAND, you need 6 preferences {filly, folly, fewer, hater, witch, value} and 2 out-of-box rules HAUNT->VOUCH, FOYER-> GAWKY). If you REALLY want to avoid days when you get the worst hints about the hidden word, I might suggest one of these triples: [handy, slice, tumor], [handy, lemur, stoic], [duchy, merit, salon] In each case there is only one day in the whole 7-year cycle when you will get just a single colored tile (it will be a green A because the hidden word will be "kappa") and only 15 days when you get just two yellow tiles. In the opposite direction, if you want the starting triple to give you the greatest likelihood of getting four or five colored tiles, from just two of the three words, I've got a couple of good ones: the best I've seen will put you in that position almost one game out of every three: [close, train; dumpy], [scone, trial; dumpy] (In all the good examples I found, the weak word is interchangeably dumpy, jumpy, or pudgy.) Similar examples can be made from almost any disjoint PAIR of words that provide a lot of yellow and green tiles. But apart from the one measure that's being optimized in this paragraph, these triples do not fare especially well. A personal favorite, scoring highly on several scales, is [blond, march, spite] It has one of the highest rates of completion on move 4 (84.06%) and a a high frequency of offering good hints right away (on one-third of all days it reveals 4 or more colored tiles, or three colored tiles with more greens than yellows.) To play, use the out-of-cluster solution GAUNT -> GUAVA and the preferred words {eager, rower, fewer, jolly, otter}. I also recommend [blond, parse, wight] It uses all 9 of the most common letters; it allows the player to win by the fourth move in 80.3% of the cases just by guessing the first word that comes to mind, and in fact by just guessing that way the player will win in 99.95% of all games. In order to handle the remaining cases one need only remember to use local mayor otter mover fella when possible. At that point the player will need the 6th move only 1.8% of the time. (That sixth move can also be avoided, by remembering 38 "preferred" words and 24 substitution pairs.) A few other triples that I have found to have a good mix of virtues include these: [blend, right, scamp] [crawl, fight, spend] [glide, spawn, throb] All of them can be resolved with just a few preferred words, respectively {local, water, judge, fewer, voter} {gamer, baker, mayor, outer, berry} {match, mower, mayor, outer, mover, dryer} ============================================================================== TWO It is very popular to ask about "best" pairs of words to start a Wordle game. What goes unsaid is what we discovered in the previous section: very much depends on how exactly the player intends to proceed, and on what the player values when comparing one starting pair over another. We have already found startingtriplesthat allow us to complete every Wordle game in 5 moves, while it is known that there isnoalgorithm that can successfully complete every Wordle game in 4 moves. So it is not clear what else we want to accomplish with a startingpairthat we have not already accomplished with larger starting sets, except to reduce the number of moves. (We could measure this with the expected number of moves, or the frequency of winning on moves 3,4,5, or 6, etc.) At the same time we must accept the fact that the algorithms to improve performace will of necessity be more complicated than the ones we have already encountered. Moreover, there is now an even greater likelihood that the starting set (having now at most 10 letters in it) will give us paltry information about what the hidden word might be. When we looked at five-word starting sets, we discovered that having repeated letters was theonlychoice; among four-word starting sets it was acompetitivechoice, and among three-word starting sets, having repeated letters was anuncommonchoice. Now, among two-word starting sets, we expect that having repeated letters will be apoorchoice. So in what follows we will restrict our attention to pairs with no repeated letters. It turns out that there are exactly 196,175 such pairs of Wordle words. I have made a list of all of them, along with some basic data about each. (It is sorted informally to reflect a notion of expected "quality". At the top is [salon, trice]; at the bottom is [inbox, jumpy].) But what is there about any of them that would cause us to call it better or worse than another? Here are some possible considerations. First of all, we could ask that the pair give us the best chance to get a solution on the very next move (move 3). It is an elementary probability exercise to see that the probability of success on move 3 is K/S, where S is the number of possible words (S=2315) and K is the number of clusters into which they are separated by the starting words. So we look for the pair that partitions the dictionary into the most clusters. Of all the 10-letter starting pairs, the ones that produce the most clusters (K=1071 of them) are PRICE + SLANT and CRANE + SPILT. Play these and then guess a word consistent with the clues; there is a 46% chance your guess will be correct. (The average size of a cluster is S/K, so these are also the starting pairs that have the smallest average cluster size, 2.1615.) More generally, we want our starting pair to sort the dictionary into a "large" number of "small" clusters. The smallest (singleton) clusters are best of course: these are the words we know unambiguously. So we can ask for the pair that has the most clusters of cardinality 1. That pair is again PRICE + SLANT (with 634; CRANE + SPILT is second, with 631.) Looking forward to the remaining moves, we might instead want to insist that there be no "large" clusters. Sadly, every pair leaves some clusters of size 16 or more --- no matter what starting pair is played, there will be days on which there are 16 or more words that are consistent with the clues they provide. The only pairs whose largest clusters have "only" 16 elements are SCALD + TENOR CLONE + STAIR NOSEY + TRAIL They have, respectively, two, three, and four clusters of that size. (For example, if after the first pair is played, the E and R tiles go yellow while the other eight are gray, then the hidden word that day could be any of bribe, brief, every, fibre, fiery, grief, grime, gripe, prime, prize, puree, purge, query, rhyme, rupee, where An all-gray color display indicates the hidden word is one of the sixteen Wordle words that lack s,c,a,l,d,t,e,n,o, and r.) Not all clusters are equally tricky of course. Since (after the starting pair) we have four moves in which to guess the hidden word, a cluster of four or fewer words will surely allow us a win. So we may ask for the pair which puts the most words into those small clusters (i.e. for which the fewest words lie in clusters of size 5 or more). The prize now goes to SALON + TRICE, which has 1,543 words in these "guaranteed" clusters. (Or we can count the clusters rather than the words; PRICE + SLANT has 91.5% of its clusters "small", just beating second-place CRANE + SPILT which has 91.4%) There is another way to look at the situation just after we have played our starting triple: how hard is it (for a human!) to come up with a word that matches the clues provided by the starting pair? It's obviously easier when there are many green and yellow tiles, green being better. Of course this will vary by day; we can look at the average numbers of those tiles or equivalently, the total number that appear over the entire 2315-day Wordle cycle. The pair that will show the most greens is CRONY + SLATE, with 2692. I considered weighting the yellows as half as valuable as a green; with that weighting CRONY + SLATE is still the best (it will show 4131 yellows over the cycle, too) but if we value the yellows more heavily the ranking changes. At the extreme of counting yellows and greens equally, there is a 13-way tie. Indeed it is not hard to show that the total number of colored tiles that will appear over a Wordle cycle depends only on the letters used. Thus all of these pairs [route, slain] [route, snail] [louse, train] [sonar, utile] [solar, unite] [outer, slain] [solar, untie] [outer, snail] [arose, unlit] [arose, until] [alien, torus] [noise, ultra] [arson, utile] will score equally (they will produce 7062 colored tiles) because they are made of the same letters: a,e,i,o,u and l,n,r,s,t . These happen to be the 10 most-used letters in Wordle if we count by *words containing the letter*. If instead we multiply-count any repeat letters within a word, then "c" would replace "u" in this list; and as it happens the words containing aeioclnrst are (tied for) second in this ranking, with 7053 colored tiles. My list of all the 10-letter pairs is sorted by an ad-hoc measure that combines several of the ideas discussed so far. By that criterion the top pairs include some that are by now familiar: [salon, trice] [cairn, stole] [price, slant] [close, train] [crane, spilt] ... ---------------- In order to analyze and rank the starting pairs any further, we would have to know how the player intends to proceed after playing the starting pair. I will investigate two procedures the player might follow. First let us suppose the player follows a guess-at-will strategy. As it turns out, for every one of the starting pairs there is then a nonzero possibility of losing (depending on what the hidden word is, and the ways that the coin flips as the player picks words that are consistent with the clues). So one of the ways of ranking the starting pairs is to compare the probability of a loss when following this strategy. For this search I resorted to some heuristics to trim the search space a bit, so it is *possible* that there can be a better pair, but these appear to be the best. Shown here is the computed (not experimental) rate of failure when following a guess-at-will strategy after starting with each pair: [spend, trawl] 0.0024206777 (1 fail per 413 days --- about 5.6 per six-year cycle) [blond, tramp] 0.0024243336 [blend, tramp] 0.0024699297 [bland, swept] 0.0025154240 [scold, tramp] 0.0025959420 [blend, stamp] 0.0026618621 [blend, swamp] 0.0026902371 [bland, trump] 0.0026914026 Interestingly the pattern continues even further down the list: the "best" pairs by this standard only use words with a single vowel, in the middle. While these pairs have the lowest failure rate, they tend in general to take longer to achieve a win. For example the games that start with these pairs will typically be able to guess the hidden word on move 3 less than one-third of the time -- many fewer than the 46% success rate we have seen for other pairs. As noted already, this strategy can result in a loss, no matter which starting pair is used. But we can imagine allowing the game to continue for more than 6 turns, until a win is inevitably found. Counting the 7-move and 8-move (etc.) games too, we can compute the average number of moves needed for a win for each pair. The starting pairs with the lowest average numnbers of moves, following the guess-at-will strategy, are 3.700452018, [crane, spilt] 3.702213793, [price, slant] 3.712982832, [crane, split] 3.713728776, [cried, slant] 3.719044188, [print, scale] ---------------- A player who does not want to allow a loss would have to follow a different strategy. In this document we have consistently proposed one alternative: to keep things as simple as possible, we would assume the player would examine the tiles after the starting pair is played, to determine the cluster in which the hidden word lies. Then: (1) If that cluster will surely find the hidden word by move 6 using a guess-at-will strategy, the player will pursue that. (2) If not, but if one "preferred" word in the cluster will give enough extra information to finish by move 6, then play it on move 3. (3) if not, but a non-cluster word will separate the cluster into smaller clusters that can each be resolved by move 6 using guess-at-will, then play that word on move 3. If in (2) or (3) there are multiple candidate words to play, then play the one that gives the best probability distribution. Unfortunately it seems that for most pairs there are other clusters that do not fall under any of these three categories. We have seen that there can be two-word solutions in those cases, or a recursive use of "preferred" words. I have done some by-hand analyses for these cases but only for limited examples. Now, how shall we pick a 'best" starting pair to use with this strategy? Unlike the previous subsection, the chance for a loss is alwayszero; but now there is something new to optimize: the difficulty of the algorithm. For none of the starting pairs is it possible to have ashortlist of additional rules (as e.g. the three we provided for CHOIR + SWEPT + GLAND), But we can do our best! The starting pair with the simplest algorithm that I have found is [blond, spite] It will lead to a guaranteed win by move 6, with an average number of moves that is below 4.0, if we simply follow rules (2) and (3) for some exceptional clusters. But there are 23 of them: twenty-three combinations of yellow and green tiles after these two words have been entered, that signal a cluster of possible words that cannot be resolved just by freeform guessing. The set of rules to cover these tricky clusters is then more difficult to memorize than anything we did with starting triples! In 21 clusters we will use the "preferred" words aider, cater, chain, charm, chart, crave, crest, fifth, folly, girly, grill, legal, mayor, money, scary, scree, shrew, stark, trait, twang, wager The other two clusters require out-of-cluster moves on move 3: if the word could be "found", then play "wharf" if the word could be "mover", then play "rocky" Using these 23 rules will guarantee success by move 6, taking on average 3.8237 moves. Among the many starting pairs I have so far studied, none has fewer than 23 rules for exceptional clusters like this, so BLOND + SPITE is "best" by this criterion for "simplicity". (It has the added virtue that using the word MARCH on move 3 resolves many of the 23 clusters by move 6, that is, if forget some of the 23 rules while playing, we could simply revert to the six rules already discussed for this startingtriplein the previous section.) I have found one other starting pair that creates only 23 tricky clusters: BLOND + TRACE. To play this pair, use the 19 preferred words agape, amiss, amuse, cinch, favor, fetal, foist, folly, gaunt, gipsy, grape, hairy, impel, palsy, serif, serve, setup, shift, swill and the four out-of-cluster moves: if the word could be "catch", then play "champ" if the word could be "found", then play "swamp" if the word could be "gamer", then play "gawky" if the word could be "mover", then play "whisk" Using these 23 rules will guarantee success by move 6, taking on average 3.7461 moves. An interesting candidate for "simplest" is GLAND + SWEPT . This starting pair leaves 35 clusters that require special treatment; 34 of them can be resolved using a preferred member of the cluster, and the last (the one including BERRY) can be resolved using the out-of cluster word ROCKY. But alternatively we can use CHOIR for 33 of the first 34 --- all except the one containing [arbor, armor, favor, major, mayor, razor] since we noted in the previous section that CHOIR + GLAND + SWEPT is a good starting triple, having itself just three difficult clusters, this setbeing one of them, (It can be resolved using MAYOR as a preferred word.) In other words we have a "simple" algorithm for winning Wordle: Start with GLAND + SWEPT and see which cluster contains the day's word. play ROCKY if the cluster contains BERRY, play MAYOR if the cluster contains MAYOR, play CHOIR if the cluster is any of the other problematic ones, Otherwise, guess at will. But this is a cheat! To use this algorithm we mustrecognizethose 33 clusters as they arise, which is no easier than remembering the preferred words that signal them. Instead, this algorithm is essentially a variant of that used with the starting triple: Start with GLAND + SWEPT. If the word could be BERRY, play ROCKY; then guess-at-will Otherwise play CHOIR. Then If the word could be MAYOR, play MAYOR; then guess-at-will Otherwise guess-at-will. The other natural criterion to use is the average number of moves until a win. Alex Selby's web page advocates CRANE + SPILT, which we have seen is also good by other criteria. This pair allows a win by move 6 with most of the 1071 clusters dispatched by free-form guessing. There are 25 clusters which can be handled by using a preferred member: above, allow, awake, batch, bevel, blade, corer, dingy, ditch, ditty, dogma, dumpy, earth, foist, goody, gouge, grade, haven, marry, merge, otter, sewer, vomit, wager, women An additional four clusters require out-of-the-cluster solutions: billy->BAWDY, bound->BAWDY, bully->FJORD, daunt->JUDGE But then in addition, there is the largest cluster (containing 33 words), and in this one we cannot pin down the hidden word among them with certainty with any single word played on move 3 (neither from within nor outside the cluster). A double-word substitution will suffice: play ROWDY and JUMBO on moves 3 and 4, and this will determine the hidden word uniquely, to be played on move 5, except we have to toss a coin to pick between {roger,rover} and between {every, ferry}, which could force us to defer victory until move 6. In the previous subsection we noted this starting pair used fewer moves on average (3.7005) than any other starting pair. More precisely, if we use this starting pair and simply resort to a guess-at-will strategy, we would discover the hidden word by move 6 about 99.505% of the time: 46.3% of the time by move 3, 40.4% on move 4, 10.9% on move 5, and 1.9% on move 6. But now, following the recipe above will guarantee a win even in the remaining cases, with a lower average of 3.6590 moves per day. (We now finish on move 3 only 46.1% of the time, but a higher 43.1% of the time on move 4 as the longer tail of the old probability vector is moved toward shorter games by the complicated rules of play.) Clearly an average game length of 3.6590 is better than anything we obtained using fixed startingtriples. But when applied to an N-fold compound game, this would imply (an upper bound for) the length of the game being 2 + 1.6590 N. For N=1 this is clearly better than say the bound 3 + 1.1682 N which we obtained in the previous section. But already for N=2 the advantage is nearly lost; so even for Dordle it is not clear that we are better off with the best starting pair than we would be with one of our good starting triples. It is unlikely to be better for Quordle and beyond. We have also already met another promisinhg starting pair: PRICE + SLANT. This one can win Wordle 100% of the time if we use these 24 preferred words: awake, badge, berry, bowel, corer, dingy, ditty, dogma, dough, eater, field, foist, gouge, grade, haven, legal, marry, merge, modal, otter, sewer, vomit, wager, women and employ these 5 out-of-cluster pairs: [batch, climb], [billy, howdy], [bound, thumb], [bulky, fjord], [daunt, dough] However, the cluster containing "berry" is large (33 elements) and even if we preferentially play berry on move 3, there are still two large sub-clusters that could lead to a loss unless we play preferred members of each of THEM, too: *after* playing "berry" on move 3, if the word could (still) be "roger" or "mover", then play those words on move 4. (So that's a "recursive guided freeform hard mode". WHew.) With this algorithm, Wordle is definitely won by move 6, taking an average of 3.6618 moves. AS far as I have computed to date (several thousand of the most promising starting pairs), the pairs which use the fewest moves on average, while guaranteeing a win by move 6 using this style of procedure, start with 3.6591 [crane, spilt] 3.6618 [price, slant] 3.6739 [crane, split] 3.6757 [cried, slant] 3.6771 [print, scale] ... I have not finished an exhaustive search of word-pairs but I have looked at all pairs drawn from what I consider the "better half" of all Wordle words. I am running a background process at home that sifts through promising pairs; for each one it is necessary to identify the problematic clusters and to find in- or out-of-cluster words that can resolve them, if possible; I can also search for procedures to resolve the clusters which cannot be won with these tools, and then compute the probability distribution showing the frequencies with which this algorithm will end on moves 3, 4, 5, or 6. Over time I may use these results to update this section. But it seems clear that these procedures to guarantee a win with a particular starting pair are inevitably too complicated to actually be used by a human, and unlikely to be useful for the compound games. ============================================================================== ONE As far as the best single starting word, by various criteria that might be trace, or brute or chant, or raise or arise, or filet or parse, or dealt, or... Many people have weighed in, using different rubrics to assess the choices. YMMV: by now the reader understands that we cannot answer the question of "What is the best starting word?" without deciding how to measure quality, and without fixing an algorithm that the player will follow after entering that first word. (Most answers to this question assume the player "will play optimally" but that seems unrealistic unless a simple algorithm is stated clearly and then followed by the player!) ============================================================================== CONCLUSION(?) So ... what does all this tell us about how to play Wordle and the compound games? One conclusion, surely, is that what we choose to do will depend on what we want consider important. We have seen in the examples that the most important goals may be at odds. We want to win as often as possible; we want to keep our numbers of moves used as low as possible; we want to follow a procedure that is as simple as possible; and along the way we appreciate not having to come up with good moves in the absence of concrete hints. In order to say something more definitive, I ran some simulations of Wordle and the compound games, playing with the starting sets discussed in the text, and following the intended procedure (using free-form guessing whenever possible; playing preferred words when necessary; and using out-of-cluster words only when that is the only option). I tried a couple dozen of the starting sets from the text, and used them to play Wordle, Quordle, Octordle, Sedecordle, and Sexaginta-quattuordle, as well as idealized Dordle and Duotrigordle (versions that would use the same word list as Wordle). More precisely, I simulated the "sequential" versions of these games, in which the player must complete the subgames in order. The primary statistics to collect are (a) the average numbers of moves used to win (assuming no limit to the number of moves allowed) and (b) the frequency of failure (taking more than N+5 moves to complete an N-subgame game). In more detail, I kept track of the distribution of the numbers of moves used. Looking over the numbers, we can draw some general conclusions. 1. The primary determinant of these numbers is the number of words in the starting set. The two-word starting sets take lower numbers of moves, but have the highest rates of failure; the four-word starting sets are just the opposite. 2. For two-word starting sets, the average number of moves increases rapidly with the number N of subgames; for larger starting sets the effect is less pronounced. The overall result is what we predicted in the introduction: when N is small (up to N=3), a one- or two-word starting set is likely to be best; in an intermediate range (N=4 through perhaps N=100?) a three-word starting set will probably be best, and for very large N we might expect the four-word starting sets to win. 3. The average number of moves taken in an N-fold compound game is obviously at least N . But what is interesting is just how much larger than N it is: the excess over N increases as N increases --- but only up to a point, and then starts to decrease! (The maximum in each table is usually around N=16 or N=32). I suppose the explanation is that as the number of subgames grows very large, the set of hints available in the later subgames grows quite abundant from the words played in the earlier subgames, and so it becomes more and more likely in later games that we can immediately deduce the hidden word in one try --- or that the new hidden word was already one of the guesses in a previous subgame! 4. Players wishing to keep their "streak" alive should definitely use the larger starter sets. For example it is possible to win at Quordle over 99.8% of the time using a 4-word starting set; the failure rate for a 2-word starting set is over ten times as large. 5. For starting sets of a given size, it can be difficult to observe a meaningful distinction between their "success rates". I tested quite a few 3-word starting sets and while there is some noticeable difference between the starting triples that were optimal by one criterion and the triples that were optimal by a different criterion, many of the "generally good" triples had very similar tables. 6. "User error" is likely to be a large enough problem to obliterate the differences in many cases. For example, I have personally played a few hundred games of Quordle recently using the spite+march+blond opening; from the tables I see that about two-thirds of them should have completed with me immediately entering the answer in every one of the four subgames. I know I have done so more than half the time, but not 2/3! I fail to notice a yellow tile, I type "blind" instead of "blond", I forget to use the preferred words, or I just draw a blank! ---------------- To repeat the highlights, the "best" starting sets that I have found, of various sizes, are catty, frond, rumba, spill, verge, whack blank, chump, goody, river, swift carve, downy, plumb, sight bland, copse, right crane, spilt But who am I to say? :-) I will continue to process the datasets that I have constructed, and intend to update this document when something new pops up. In the mean time, I welcome corrections and suggestions for further investigation. Now, how about moving on to a nice game of Nerdle, hmm? :-) --dave rusin@math.utexas.edu