Mimic Logic is a game about making you think, develop and apply strategies to solve logic puzzles. Following this guide, you won't develop any strategy, just apply them.

So if you want to think your own strategies, I suggest you to play the game without this guide.

There is no guide that can guarantee you will be able to open every non-mimic chest. That's because there are times when there is not enough information to tell which chest is a mimic and which isn't.

However the following tricks do guarantee that every chest you open is safe.

This guide is currently under development, so I'll probably add more thing in the future.

IMPORTANT: Not all tricks work for all dungeons, at least not without extra conditions (look at 'Tips and strategies for each dungeon' section for more info).

Validation: a sentence that says a chest or group of chests aren't mimics.
Accusation: a sentence that says a group of chests are mimics.
Direct accusation: a sentence that says a specific chest is a mimic.
Mimics remaining: refers to the mimics that are still unidentified.

Math:
The letter 'n' is used to refer to a variable number. This way I can specify rules that apply for different cases. If the rule says 'there are n mimics...', you have to replace n for the number of mimics in your puzzle.

Condition: A chest directly accuses another chest.
Conclusion: One must be a mimic and the other not.

Example:
Note: If the accusation is targeting a group, you can suspect there is at least one mimic among the involved chests (group + chest that accuses).

Condition: Two chests have opposite sentences.
Conclusion: One must be a mimic and the other not.

Example:

Condition: A chest accuses a group and it belongs to that same group.
Conclusion: That chest isn't a mimic.

Example:

Condition: Only one mimic remains and a chest validates another chest (or group of chests without containing known mimics).
Conclusion: Both the chest that validates and the validated chests are safe.

Example:
Note: This trick is really useful to speed run floors with only one mimic.

Condition: There are n mimics remaining and n + 1 chests have the same sentence.
Conclusion: The n + 1 chests are telling the truth.

Example:

Condition: There are n mimics remaining and a chest is known to be a mimic.
Conclusion: The rest of the puzzle can be played as if there were n - 1 mimics.

Example:
Explanation: There are two mimics and I know the chest of the center is a mimic, so I there is only one mimic among the rest. I can use that information to treat the puzzle as a one mimic puzzle.

Condition: There are n mimics remaining and there are two suspicious chests.
Conclusion: The rest of the puzzle can be played as if there were n - 1 mimics remaining (without trusting anything related to the suspicious chests).

Example:
Explanation:
The center chest is known to be a mimic, so the rest of the puzzle can be played as a 1 mimic puzzle. Since there is a contradiction, the remaining mimic is either the blue or the red chest. Thanks to that, I know the chests with green circles can't be mimics.

Note: If there are more than two suspected chests, you must contemplate both possible cases and keep track of how many chests lie on each possible case. This way you can subtract the liars count to n. At this point it can be a better idea to use the 'Grouping Strategy' or 'Proof By Contradiction'.

Sometimes there is not enough information to tell which chest is telling the truth and which one is lying. In these cases, it's better to skip both chests.

Example:
Note: A 'Blue Crystal' can be used to find out which isn't a mimic, but if the 50/50 is between a few chests, most of the time it's not worth it.

You must count how many validations gets every chest (self-validations doesn't count). I suggest you to use the right click to keep track of them.

"A chest is safe if its validation count is greater or equals to the amount of mimics remaining."
Note: This strategy is really powerful if there are low amount of mimics or high amount of validations. If you start a puzzle and see little or no validations, you probably won't get any value with this strategy.

The idea is to classify chests (as '1' or '2') by right-clicking their checkboxes.
Look for pairs of these types:

• A) If one is telling the truth, the other is lying. Mark one with '1' and the other with '2'.
  
• B) If one is telling the truth, the other is too. Mark both with the same number.

Tips for pair identification:
A pair belongs to A) if there is an accusation or contradiction.
A pair belongs to B) if there is a validation or coincidence.

Start with one pair, and extend it by pairing a classified chest with a non-classified one.

If there are n mimics, and n + 1 chests classified with the same number. All n + 1 chests are telling the truth, therefor all the chest with the other number are lying.

Example:
Explanation:

1. The yellow pair is type A, because the right mimic accuses the left one. => Mark chests with different numbers.
   
2. The orange pair is type A, because the blue mimic accuses the red one. => Mark the red one with a different number than the blue.
   
3. The blue pair is type B, because both sentences say the same. => Mark the left one with the same number than the right one.
   
4. The purple pair is type B, because the lower chest validates the upper one. => Mark the lower chest with the same number than the upper one.
   
5. There are 4 chests classified with '1', and only 2 mimics. => All the '1' chests are safe and the '2' chest is a mimic.

The next image serves as a guide on how to correctly group chests:
For each diagram, all possible values are evaluated based on a starting value of True and False for classified chests. If in all cases of a diagram the classified and unclassified chests end up with the same values, it's safe to say they are equivalent or opposite (in other words, it's safe to classify the unclassified chest).

To simplify, it's safe to classify if:

• The accusation/validation is direct.
  
• The accustaion/validation is to a group but the group is already classified.

Also, the chest that accuses/validates can be safely marked as True or False if:

• The chest accuses a group with different numbers => The chest is True.
  
• The chest validates a group with different numbers => The chest is False.

Note: This is probably the strongest strategy I have found. Works really well in high level dungeons.

This strategy consists in assuming a chest is either telling the truth or lying and continue the puzzle with that hypothesis.

If doing that leads you to a contradiction, for example, getting more mimics than it should be or two 'supposedly safe' chests contradicting each other, then you know the assumption you did at the start was wrong.

But what happens if after the assumption everything goes well? Then you lost your time, nothing guarantees your assumption was correct. You have to undo everything you did after the assumption. This is why it should be used as a last resource.

I recommend you to make assumptions on chests that validates/accuses a large group of chests.

• Chest validates a group: Assume it's telling the truth.
  
• Chest accuses a group: Assume it's lying.

There are only 3 mimics. Assuming the black chest is lying I get 4 => The assumption is wrong => The chest is telling the truth.

Note: If you tried solving the puzzle with 'O' and 'X' starting on the same chest and couldn't find a contradiction. You could check for chests that have the same value in both scenarios. If a chest's value doesn't change, that means that's the real value. This same info can be obtained when classifying in 'Grouping Strategy' with chests that accuse/validate a group with '1' and '2' (easier to notice).

This strategy uses multiple tricks. You can get more details of each trick in the 'Basic tricks' section.

• First look for any contradiction and mark them with numbers.
  
• Then look for accusations, and mark involved chests.

Now you know where mimics might be. It's time to apply the 'Discard' trick.
If you managed to guarantee all mimics are among the suspected chest, those free of suspicion are safe. In the other hand, if there are still mimics outside of the suspected groups, you can play the rest of the puzzle as if there were less mimics to find (ignoring the sentences related to the suspected chests)

Count how many direct accusations gets every chest (self-accusations means that chest isn't a mimic). You can use right click to keep track of them.

"A chest is mimic if its direct accusation count is greater than the amount of mimics remaining."
Note: This strategy uses Coincidence Trick, therefore all the chests that accused the mimic are safe.

All tricks and strategies above can be applied without any risk.

As with the Standard Dungeon, all tricks and strategies work here.

This dungeon requires you to always assume the max possible amount of mimics in order to apply tricks and strategies correctly. Also you can't assume the last n chests are mimics (n = mimic count), because there could be up to (n - 1) non-mimics among the last n chests.

The problem with this dungeon is you have to work with little information, because a lot of sentences talk about robbers and not mimics. Playing this dungeon you will face a lot of 50/50's (at least not of mimics, but robbers). Most of the time you can identify the mimics but not the robber.

The dungeon description says robbers steal all your gold but that's not true, they only steal 100G. It's not that bad. Also keep in mind you make a lot of gold selling unequipped gear and robbers can't steal gear.

So I recommend you to first identify mimics, then take a couple of seconds to see if you can identify the robber, and if you can't open all chest anyways.

Notes:

• Remember if a chest 'a' says a chest 'b' isn't a robber, that doesn't mean 'b' is safe to open, 'b' can be a mimic.
  
• Robbers tell the truth, so are harder to identify, unless a non-mimic accuses a chest of being a robber or a mimic says that chest isn't a robber.
  
• 'Blue Crystals' don't help to find non-robber chests. That item only shows you a chest isn't a mimic, but gives no information about robbers. However due to the lack of mimic information, is still useful.

In this dungeon, chests tell you the amount of mimics around them (8 positions), mimics will lie. By default the center chest have a normal sentence.

To solve these puzzles 'Proof by Contradiction' can be really useful. Assume one chest lies or tells the truth and work around that assumption. If you find a contradiction, the initial assumption was wrong, therefor you know the real value of that chest. If not, you have to try with a different value or chest.

The thing is, this can take ages if you randomly choose chests and try.

I suggest you to prioritize the next cases:
Chest with '3' in a corner: Assume it's safe. Mark as Mimics the 3 chests around it and look for contradictions in the remaining chests.
Chest with '0': Assume it's safe. Mark as safe all chest around it and look for contradictions.
Sentence chest validates a group: Assume it's safe. Mark the group as safe and look for contradictions.
Sentence chest accuses a group: Assume it's a Mimic. Mark the group as safe and look for contradictions.

Also if a sentence chest accuses another chest directly, one of them is a Mimic. You can use that information as an advantage.

Another useful tip is to see the '0' chests as 'There is no mimic among...' chests. You could try to apply 'True Validation' with them.

50/50's are common in this dungeon, I recommend to buy as many 'Blue Crystals' as possible, specially for the last floors.

Chests work differently in this dungeon:
Colors play an important role. Mimics tell the truth, but make all non-mimic of the same color lie. From now on, I'm going to refer to non-mimic that lie as possessed.

Example:
There are 3 blue and 3 red chests and only one mimic. If the mimic is red:

• All blue tell the truth.
  
• The mimic tells the truth.
  
• All red non-mimic are possessed.

Given this changes, old tricks don't work. Here you have some new tricks for this dungeon:

Self accusation: There are several ways a chest can self accuse...

• If a chest self accuses based on color, it must be a mimic.
  
• If a chest self accuses based on other criteria, you can only tell it's a mimic if its accusation is equivalent to a color based self accusation.

Equivalent accusation: An accusation 'a' is equivalent to accusation 'b' if 'a' accuses exactly the same group of chests as 'b'. Example: 'There is a mimic among the red boxes' and 'The top row contains a mimic' are equivalent only if all the red chests are in the top row and all the chest in the top row are red.

Self validation: If a chest self validates, it's safe.

Unique chests: If there is only one chest of a particular color, it must be telling the truth (this doesn't mean it's safe to open).

Mimics can't lie: If you know a chest is lying, it's safe to open.

Color validation: if a chest is telling the truth and it's not a mimic => All chests of that same color tell the truth and are safe to open.

Same color contradiction: If there is a contradiction between 2 chests of the same color, one must be a mimic. Therefor, all liars of that color are safe and all that tell the truth are mimics.

Coincidences: There are n mimics and...

• n chests of the same color tell the same. => Those chests tell the truth.
  
• (n + 1) chests of the same color tell the same. => Every chest of that color tells the truth and is safe to open.

(work in progress)

H) -> Hypothesis -> Statement that is assumed to be true in order to prove something else.
T) -> Thesis -> Is what is being attempted to be proved using the hypothesis

Value: Each chest has a value (True or False) that represents if is telling the truth or not.

---

H) A chest 'a' negates the value of a chest 'b'.
T) a's value is opposite to b's value.

Cases:
If a is True: a says b is False => b is False.
If a is False: a says b is False => b must be True.

In all cases 'a' and 'b' have opposite values. => Thesis verified ✔

---

H) A chest 'a' says 'x' and a chest 'b' says 'not x'.
T) a's value is opposite to b's value.

Cases:
If x is True:

• a says x => a says True
  
• b says not(x) => b says not(True) => b says False.

If x is False:

• a says x => a says False
  
• b says not(x) => b says not(False) => b says True.

In all cases 'a' and 'b' have opposite values. => Thesis verified ✔

---

H) There is a group of chests 'G'. A chest 'a' belongs to 'G'. 'a' says there is a chest with value = False in 'G'.
T) a's value must be True.

Proof by Contradiction (assume a's value is False):
If 'there is a chest with value = Flase in G' is False => All chests in G are telling the truth.
If all chest in G are telling the truth and 'a' belongs to G => a's value is True ❌Contradiction with assumption.

Conclusion: a's value must be True. => Thesis verified ✔

---

H) There is only 1 chest with value = False. Chest 'a' says the value of chest 'b' is True.
T) The value of 'a' and 'b' is True.

Proof by Contradiction (assume a's value is False):
If 'the value of chest 'b' is True' is False => The value of 'b' is False.
=> There are 2 chests with value = False, but Hypothesis says there is only one. ❌Contradiction with assumption.

Conclusion: a's value must be True. => b's value is True => Thesis verified ✔

---

H) There are only n chests with value = False. There are (n + 1) chests saying 'x'.
T) The value of all (n + 1) chests is True.

Proof by Contradiction (assume 'x' is False):
If 'x' is False => There are (n + 1) chests with value = False, but Hypothesis says there are only n chests with value = False. ❌Contradiction with assumption.

Conclusion: 'x' is True => The value of all (n + 1) chests is True => Thesis verified ✔

---

H) There is a group 'T' with 'm' chests. 'n' of those 'm' chests have value = False (n <= m). Chest 'a' belongs to 'T' and has value = False.
T) 'T' without 'a' has n - 1 chests with value = False.

'T' has n chests with value = False and 'a' has value = False. => If 'a' is removed from 'T', the amount of chests with value = False is reduced by 1. => 'T' without 'a' has n - 1 chests with value = False. => Thesis verified ✔

---

H) There is a group 'T' with 'm' chests. 'n' of those 'm' chests have value = False (n <= m). Chests 'a' and 'b' belongs to 'T' and one of them has value = False.
T) 'T' without 'a' and 'b' has n - 1 chests with value = False.

'T' has n chests with value = False and 'a' or 'b' has value = False. => If 'a' and 'b' are removed from 'T', the amount of chests with value = False is reduced by 1. => 'T' without 'a' and 'b' has n - 1 chests with value = False. => Thesis verified ✔

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H) There are only n chests with value = False. There are n chests saying 'a' has value = True. There is a chest 'a' and doesn't belong to those n chests.
T) The value of 'a' and all n chests is True.

Proof by Contradiction (assume a's value is False):
If a's value is False, all n chests that validates 'a' are lying => All n chests have value = False.
Therefor, there are (n + 1) chests with value = False ('a' + the n chests that validates 'a') ❌Contradiction with Hypothesis (There are only n chests with value = False).

Conclusion: a's value is True => The value of all n chests is True => Thesis verified ✔

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This strategy is similar to the 'Coincidences' trick, but more flexible. With 'Accusation' and 'Contradiction' it's possible to know 2 chests have opposite values. Also if a chest validates another one, both have the same value. Knowing this, it's possible to classify chests in two opposite groups and be sure if one group is True, the other must be False.

H) There are only n chests with value = False. There are 2 groups of chests 'A' and 'B'. All chests in each group have equivalent values and opposite to the chests on the other group. Group 'A' has (n + 1) chests.
T) All values in 'A' are True and all values in 'B' are False.

Proof by Contradiction (assume all values in 'A' are False):
If all values in 'A' are False, there are (n + 1) chests with value = False ❌Contradiction with Hypothesis (There are only n chests with value = False).

Conclusion: All values in 'A' are True and all values in 'B' are opposite => All values in 'B' a re False => Thesis verified ✔

If this guide was useful to you, consider leaving a like and fav. Any kind of support/feedback is much appreciated <3

Users that contributed with relevant tips/strategies in the comment section:

• TPAKTOP
  
• Herofighter
  
• Onion_Bubs