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In Liar Wordle, standard rules apply except each row contains exactly one clue of the wrong colour. For instance if we consider only the first row the correct colours could be Y-B-B-B-Y and the correct word would be BENCH. But that would break several other rows.

Green = correct letter in correct position

Yellow = correct letter in wrong position (assuming it’s not counterfeited by a duplicate letter)

Black = Incorrect letter.

Your task is to work out the five-letter word and deduce which of the blobs are impostors.

wordle

Transcription:

There are 6 imposters among us

C U R V E 🟨
D I S C O 🟩
F I S H Y 🟩 🟨
K N E E L 🟨 🟨 🟩
Q U I E T 🟨 🟨 🟨
V A P O R 🟩 🟨
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  • $\begingroup$ If you liked this puzzle then you may also like this programming competition that just started: topcoder.com/challenges/… $\endgroup$ Feb 9 at 23:56
  • $\begingroup$ I created a version of this game a while back, before seeing this post. I hope that some people find it enjoyable. word-lie.glitch.me $\endgroup$
    – Dave R
    Apr 10 at 15:15

2 Answers 2

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The (EDIT: formerly) accepted answer works its way to the solution by going mostly "backwards" by excluding possibilities until only a few are left. But it's also possible to start with a set theoretic observation and run most of the logic in the "forward" direction. Like so:

Starting with the rows with more than 1 colourful clue, and noting that the rows can have at most one impostor each, we know that the answer word must contain

  • at least one of F/S
  • at least two of N/E/L
  • at least two of U/E/T
  • at least one of V/O

Since the only overlap is at the letter E, the bottom 4 rows contain a colourful clue for 9 distinct letters. But this is the absolute maximum! We can have at most 5 distinct letters in the answer word, and the four rows can contain at most 4 impostors. Therefore, five distinct answer letters and 4 impostors must be exactly what we are seeing. So we can deduce all of this:

  1. the answer word doesn't contain any repeated letters
  2. in the bottom 4 rows, both the clues for the letter E are legit. (Otherwise we'd need a fifth impostor in only four rows)
  3. we can replace the "at least" with "exactly" in all of the above sentences. (Otherwise we'd run out of distinct letters in the answer word)
  4. all the impostors on the bottom 4 rows are colourful, and would be black instead if they were not lying. (The impostor letters cannot exist in the solution word)
  5. only the mentioned letters can occur in the solution word, so all black clues for any other letters are automatically legit. (Turns out, we won't actually need to use this deduction at any point. Leaving it in for completeness' sake though.)

enter image description here

From here, we can concentrate on the "S" clue in "FISHY". It is either legit (S exists in the solution, but isn't the 3rd letter) or an impostor that would be black if it wasn't lying (see deduction 4 above). In either case we know for sure that

the green S clue in DISCO is an impostor.

With that deduction out of the way, it's all straightforward:

  • DISCO's "no O" clue is legit
  • Therefore VAPOR's O clue is an impostor, making the V clue legit.
  • This means the "no V" clue in CURVE is an impostor, so the "no U" clue is legit
  • Therefore the U clue in QUIET is the impostor, and the T clue is legit, and finally
  • Since the green V clue in VAPOR is legit, the green F clue at the same position in FISHY must be that row's impostor

enter image description here

At this point, we have identified all the impostors except the N or L in KNEEL. This ambiguity comes as no big surprise, since those are the only occurrences of either letter in the puzzle.

To get forward, then, we must solve the Wordle. We know the solution has either the letters "ENSTV" or "ELSTV", and it must begin with "VE". (There are enough known-legit E clues to deduce the E's actual position.)

Checking the dictionary, there is only one word that matches these criteria, and it's

VENTS,

which is definitely a topically suitable way for catching the final impostor clue,

the green L in "KNEEL".

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  • 3
    $\begingroup$ +1! I figured there was a way to solve it without finding contradictions to assumptions, but I couldn't find it - this certainly is a much cleaner solution! XD $\endgroup$
    – samm82
    Feb 5 at 16:01
  • 2
    $\begingroup$ Yes, I am convinced this answer is better than the original, and therefore deserving of the Green Tick - much closer to my intended solve path and more votes at the time of writing. $\endgroup$
    – happystar
    Feb 6 at 8:21
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    $\begingroup$ @happystar Sounds good by me! $\endgroup$
    – samm82
    Feb 6 at 16:29
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Notation: a number 'XY' denotes the blob in the Xth row and the Yth column, and the grid uses x's for crewmates, o's for impostors, dashes for unidentified, and @ for pairs where at least one is a crewmate. Word info is given as "_____ [known letters being filled in] (<yellow letters>) no: <excluded letters>". Hopefully this makes sense!

Step 1: Identifying Inconsistencies

One of 14 and 61 must be incorrect. Therefore, there is no r in the word.
One of 23 and 33 must be incorrect. Therefore, there is no i in the word.
One of 12 and 52 must be incorrect. Therefore, there is an e in the word.
One of 25 and 64 must be incorrect. Therefore, at least one of these conditions is true: there is a v in the 1st place and/or an s in the 3rd.

This gives us this grid with the following letters/word knowledge:

--x--
-x@--
-x---
-----
--x--
@---x

_____ (e) no: ir

Step 2: First Assumption

Assuming 12 and 14 are both correct, 52 and 61 must both be incorrect. Therefore, the letters {eto} are in the word and there is an s in the 3rd position. This means that 33 is incorrect, implying there is an f in first position. This means we have the word info "f_st_ (oe)". The final word then must be "foste", or else the first row doesn't have an incorrect clue. This isn't a word, so either 12 or 14 is incorrect, as is either 52 or 61. We do learn, however, that there is no c, and e is not in the 5th position.

x-x-x
-x@x-
-x---
-----
--x--
@---x

_____ (e) no: cir

Step 3: A More Useful Assumption

Assuming 23 is correct implies that s is in the 3rd place, f is in the 1st, and o is in word, but not in the 4th place. This means that 25 is incorrect, which implies there is no d in the word. Since we now "know" 4/5 of the letters [f_s__ (oe)], one coloured clue in row 4 is wrong. The full set of letters becomes {fsonel}, which means both 52 and 55 must be incorrect. This is impossible, so 23 is incorrect, which implies there is no d or o in the word.

x-x-x
xxoxx
-x---
-----
--x--
----x

_____ (e) no: cdior

Step 4: Cleaning Up

Since there's no o, 54 is incorrect, meaning that v is in the first position and there is no a or p. Since v is in first position, 31 is incorrect, implying that there is no h or y, but there is an s in the word (but not in the 3rd place). Since v is in first position, 14 is incorrect, and e is not in the 5th position.

xxxox
xxoxx
oxxxx
-----
--x--
xxxox

v____ (es) no: ahikpqry [only one e]

Step 5: Too Many Letters

Any black clue being incorrect would lead to too many letters, therefore they're all correct. Since we have three letters already {ves}, the remaining two rows provide {nlut} (the e included in the letters we already have) - this means the two "e" clues are correct and the e is in the 2nd place.

xxxox
xxoxx
oxxxx
x-xx-
x-xx-
xxxox

ve___ (s) no: ahikpqry

Step 6: Finishing Off with Assumptions

Assuming the l clue is correct, we have the word info 've--l (nsu) no: ahiprty'. Therefore, the possible "words" including s (but not in the 3rd position) are 'vensl' and 'veusl', which aren't real. Therefore, the l clue is incorrect.
Assuming the u clue is correct, possible words meeting our criteria are 'vensu', 'venus', 'veuns', and 'veusn'. 'Venus' is close but is a proper noun; therefore, the u clue is incorrect.

xxxox
xxoxx
oxxxx
xxxxo
xoxxx
xxxox

ve___ (nstu) no: ahiklpqry

Final Answer:

Omitting the non-words 'venst', 'vetns', and 'vetsn', we are left with 'VENTS', a perfect word for six sussy impostors! 😳😳😳

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