Welcome! This is a hint/walkthrough guide for my Escape Simulator maps, handily combined in one location. Click on the headers on the right to jump to the room you would like hints on. Click the images below to be taken to their workshop pages. Each section will have hints first and a walkthrough at the end. Thanks for playing!

Currently this guide covers:
The Barred Room https://steamhost.cn/steamcommunity_com/sharedfiles/filedetails/?id=2636195694
Pharaoh's Maze https://steamhost.cn/steamcommunity_com/sharedfiles/filedetails/?id=2641579050
Sudoku Puzzle Cabinet https://steamhost.cn/steamcommunity_com/sharedfiles/filedetails/?id=2642575023
Sudoku Puzzle Rooms https://steamhost.cn/steamcommunity_com/sharedfiles/filedetails/?id=2653042148
Sudoku in Spaaaace! https://steamhost.cn/steamcommunity_com/sharedfiles/filedetails/?id=2661724780
Spooky Sudoku https://steamhost.cn/steamcommunity_com/sharedfiles/filedetails/?id=2679953625
The Maze of Eyes (2P) https://steamhost.cn/steamcommunity_com/sharedfiles/filedetails/?id=2659930237
Voxas Puzzle Room https://steamhost.cn/steamcommunity_com/sharedfiles/filedetails/?id=2680820195
Testing Chamber https://steamhost.cn/steamcommunity_com/sharedfiles/filedetails/?id=2686253355
Watch Your Step https://steamhost.cn/steamcommunity_com/sharedfiles/filedetails/?id=2690331360
Tomb of the Sudoku https://steamhost.cn/steamcommunity_com/sharedfiles/filedetails/?id=2696750817
Sudoku Dungeon https://steamhost.cn/steamcommunity_com/sharedfiles/filedetails/?id=2728110891
Sudoku with (Former) Friends! https://steamhost.cn/steamcommunity_com/sharedfiles/filedetails/?id=2777340803

Welcome to The Barred Room. The goal is to find all 5 keys to exit. Hints below.




Hint 1: There are 2 papers in the room that hold the key to the code

Hint 2 : Check the order of the planets on the desk

Hint 3 : The planet order is Earth, Jupiter, Mars, Saturn


Answer: Earth is the 3rd planet, Jupiter = 5, Mars = 4, Saturn = 6, the code is 3546




Hint 1 : The pattern on the box is very similar to something else in the room...

Hint 2 : Not all of the poker chips are required, but certain colors are

Hint 3 : Check under the rug

Hint 4 : Use the black poker chip to align the code correctly on the box

Answer:


Hint 1 : Sink the battleship!

Hint 2 : Check the floor, bookcase, center cabinet, and the desktop

Answer : Place all 4 red pegs into the battleship on the board


Hint 1 : The message about misses is the key to this lock

Hint 2 : Check the battleship board

Hint 3 : Associate the misses on the board with letters and numbers


Answer : The code is DEAD. Using the board, the miss on D column is in row 5, E is row 1, A is row 6, thus the code is 5165



Hint 1 : The dominoes are just used to convey numbers, they are not used as dominoes

Hint 2 : White = 5, Black = 10

Answer : The black dominoes sum to 10 and the white dominoes sum to 5. The first domino is has a 2 and a blank, so the blank is 8 (to sum to 10). The second is a white 3, so the code number is 2, etc. The final code is 82741

Welcome to the Pharaoh's Maze, a sliding wall puzzle room. To open the chest you will need to find a key and the 3 gemstones.

Most puzzles in this room are confined to their immediate area, so you will not need to search far for the answers to the puzzles. Also, every room shown on the map is reachable.







Hint 1 : The cups that start will balls are the key

Hint 2 : The item count is important, but not the quantity


Answer : Blue is odd and red is even


Hint 1 : Check this square and the adjacent for the 4 statues you need

Hint 2 : The mirror is a hint on where the statues go


Answer : If you number the possible statue slots from left to right (6 total), guy with crossed arms goes in slot 2, statue with bird head and ball on a stick in slot 3, statue with bird head and scroll in slot 4, statue with jackal head in slot 6



Hint 1 : The blocks against the wall are the key

Hint 2 : What would the blocks look like if the fallen ones were stacked correctly?

Answer : The number of blocks on each row is the combination, so 1234


Hint 1 : Check all the draggable objects

Answer : Drag the Jackal over to the pressure plate under the vase



Hint 1 : Solve the cup puzzle to find it


Hint 1 : Search the room with the gold statue with a red bowl

Answer : There is a button on top of the statue that opens the wall


Hint 1 : The gem is hidden behind a sliding wall near the window

Hint 2 : The sliding wall it is hidden behind is one that has already been triggered

Answer : Remove the gold ball from the cup at the start to close the wall leading to the cup puzzle


Hint 1 : Find the red gem to open the door

Hint 2 : There is a button in the room beyond the circle door

Answer : One of the red squares on the chest opens the wall, the key is in the urn

Welcome to the Sudoku Puzzle Cabinet, where the puzzles are themed off of sudoku and some of its variants. This guide does not cover the tutorial puzzle, just the cabinet itself. Solve all 4 puzzles on the cabinet to win. Note that the block storage cabinet is there to help you keep track of numbers, it is not part of a puzzle itself.

This guide also assumes that you either know the thermo and XV sudoku variants or did the tutorial puzzle that teaches them.

Note on labeling sudoku boxes. The boxes are labeled starting at the top left, with box 1, top center is box 2, top right is box 3, etc.


Hint 1 : Take a close look at the dial

Hint 2 : Note that there are 6 spaces on the thermo

Answer : The dial is in the center of the thermo. If you arrange the numbers in the thermo it reads 8 7 6 4 3 1, thus the code is 7643



Hint 1 : It is easier to check what things can be instead of what they cannot be

Hint 2 : Each circled number only has 1 option

Answer : Digit 1 sees 2,8,9 in the column, 1,4,5 in the row, and 6,7 in the box, which only leaves 3 as a possible option.
Digit 2 sees 2,3,4,6,7 in the column, 5,8 in the row, and 9 in the box, leaving 1.
Digit 3 sees 1,4,6,7 in the column, 2,3,9 in the row, 8 in the box, leaving 5
Digit 4 sees 3,5 in the column, 1,2,4,7,9 in the row, 8 in the box, leaving 6
The final code is 3 1 5 6


Hint 1 : The blue border represents the outline of the sudoku, telling you which boxes of the sudoku each cube face represents

Hint 2 : The left cube is box 1 and 4 (top left and center left) The right cube is box 5 and 6 (center and center right)

Hint 3 : Start with the 3rd digit

Hint 4 : Remember that a V in sudoku only has 2 options, 1/4 and 2/3

Answer : The key to digit 3 is the V in box 6. These cells only have one option due to the 2 in box 5, which is 1/4. This means that digit 3 cannot be 1,2, or 4, leaving only 3, since digit 3 is also on a V.
Digit 2 is also on a V, and since digit 3 already uses 2/3, it must be a 1/4. The 4 in box 5 makes digit 2 a 4
Digit 1 is on an X with a 1, due to digit 2 being a 4, thus digit 1 is a 9
Digit 4 only has 1 option left. It sees 1,2,3,4,9 in the box, 6 and 5 from the row, and an 8 from the X in box 5 (attached to the 2), leaving only 7
The final code is 9 4 3 7


Hint 1 : Digits 1 and 2 can be deduced with only the information in their respective boxes

Hint 2 : Digit 3 can be deduced with only the information in its row

Hint 3 : For digit 4, consider the X dominoes and which digits on them are high and which are low

Hint 4 : For digit 4, in its box, digits 1 and 9 only have 1 option

Answer : The thermo takes up the whole box and thus contains all the digits from 1 to 9, which means digit 1 is 4
All cells in box 2 are on an X or V except for digit 2. 5 is the only digit that cannot be on an X or V, so digit 2 is 5
The only place a 9 can go on a thermo is the end, and since every row must contain a 9 this makes digit 3 a 9
Digit 4 = 8 is the short answer, the logic path to get there follows:

1 and 9 are very limited in the box. If 1 is on an X, it must be paired with a 9 and vice versa. If either 1 or 9 is on an X in this box, it would break the thermo, by putting the 9 or 1 in the middle of it. This means that 9 is on the end of the thermo and 1 is on the bulb (due to the 1 in box 4). The 1 in the bulb is on a V, which means it is paired with a 4. That leaves us with 2,3,5,6,7,8. If 6 is placed on an X, it must be paired with a 4, which is already placed in the box elsewhere, which means the 6 must be in the top center or top right box on the thermo. As there are only 2 numbers between 6 and 9, and the numbers along the thermo must increase, the 6 can only be in the top right, leaving an 8 for digit 4

The final code for puzzle 4 is 4 5 9 8

This is the second puzzle room based on Sudoku Variants. Solve the cells with a yellow corner to proceed and complete the puzzle. Note on cell and box labels: Each room has the box number on the wall for reference. The start with 1 in the top left, and proceed left to right, top to bottom. Rows and columns start at the top left cell in box 1. The cell labelled r1c1 is in the top left corner of the sudoku grid, while r9c9 is in the bottom right.

Don't forget that you do not need to solve all of the cells immediately to proceed (and in fact, you cannot solve them all), just the ones with yellow corners

Variant Rules:
A blue palindrome line reads the same from the front as it does from the back.
i.e. 3 6 4 5 4 6 3 or 9 1 9

A white kropki dot between 2 cells indicates those cells are consecutive
A black kropki dot between 2 cells indicates those cells have a 1 to 2 ratio.


Hint 1 : I suggest starting with the digits 1,2, and 9 in box 5

Hint 2 : The black kropki dot in box 4 limits what can be on the palindrome line


Answer : The only spots available for the digit 9 in box 5 are in row 6 (the 9 in box 4 prevents it from being on the palindrome) This places the 9 on a set of 3 consecutive digits, which must therefore be 7,8,9. Either way you arrange them, 8 is in the middle.
The digits 1 and 2 cannot be in row 5, box 5, because that would place them on the palindrome that enters box 6, leaving duplicates in the box. This places the 1 and 2 in row 5 instead in box 4, on the palindrome. Since a 1 on a black kropki dot always pairs with a 2, it cannot go in the middle of box 4. This makes the palindrome read 1 2 5 2 1, and leaves a 4 paired with the 2 on the black dot



Hint 1 : Start with the 3 black dots in the corner

Hint 2 : When deciding what can go on the palindrome, don't forget to check the other end for conflicts

Answer : A chain of 3 black kropki dots (in the same box) can only be 1-2-4 or 2-4-8. Since the 8 is already in the box, the top left corner must be 1-2-4, and the 1 in box 4 sets the placement.
The 5 can only go in the middle, since putting it on the palindrome either conflicts with the 5 in box 4 or box 5. The 6 can only go in r3c1 (bottom left), as placing it in r2c3 (middle right) would mirror it into box 4, and clash with the 6 in box 5


Hint 1 : You should be able to complete box 4 now. Now each column in box 7 only has 3 options left. Start with the middle

Hint 2 : The right column in box 7 is limited by what can be on the other end of the palindrome

Hint 3 : Check the 3 in box 8

Answer : Finishing box 4 leave each column in box 7 with only 3 options. Column 2 (box 7,middle column) is 3,6,7. A 7 can never be on a black dot and the 6 in box 8 places the 3 and 6 in box 7.
Column 3 (box 7, right) is 1,2,4. The 2 in box 8 prevents it from being on the palindrome in box 7 and the 1 in box 5 places the 1 and 4 on the palindrome in boxes 7 and 8.
The 3 in box 8 can only go in r9c6 (bottom right). This forces the 5 onto the white kropki dot, pairing up with the 6. The 6 in box 5 sets the placement.


Hint 1 : Start with the kropki dots in box 9

Hint 2 : You should be able to complete box 5 -> box 9 -> box 8

Hint 3 : The answer below is an image of the final layout.


Answer :

This is the third puzzle room based on Sudoku Variants. Solve the cells/boxes with a yellow corner to proceed to the next step and complete the puzzle. Note on cell and box labels: Each box is numbered from the top and increases from left to right, with box 1 in the top left, and box 9 in the bottom right. Boxes are labelled in their bottom left corners.

This puzzle starts you on box 8 (bottom center). Do not forget that you do not need to solve all cells immediately to proceed.

Variant rules:

For Quadruples: All numbers in the white circle must appear in the 4 adjacent cells. These can repeat if allowed by the rules of basic sudoku

For Maximums: A grey cell with black arrows facing outward must contain a digit larger than all orthogonally adjacent cells (that are not also Maximums)
For Minimums: A grey cell with white arrows facing inward must contain a digit smaller than all orthogonally adjacent cells (that are not also Minimums)


Hint 1: The circles have both a shared digit and mutually excluded digit

Answer : Since 4 must be adjacent to both circles, it goes in the center box. Since each circle has 4 digits that must surround it and 9 is not one of those, it must go in the bottom left cell


Hint 1: Where does 9 go in box 5?

Hint 2: Where can the 1 in box 5 go?

Hint 3: The 1 in box 5 affects box 8, giving you the location of the 2 in box 5

Hint 4: The 3 4 and 5 in box 5 are now limited

Answer: 9 must go in the center column of box 5 (due to the 9 in boxes 3 and 8) and since it can never be next to a maximum, it goes in the center box. The minimum next to the (1,2) quad forces the 1 into the minimum as it can never be adjacent to a minimum. The 1 in box 5 gives you the location of the 1 in box 8, which in turn gives a 2/3 pair in box 8. This 2 and 3 pair forces the 2 in box 5. The 4 in box 5 can only go in one location (middle right) due to the quadruples. The 8’s in boxes 8 and 3 place the 8 in box 5, leaving only 1 spot for the 3(bottom right) and 5(top right).


Hint 1: Start with the two (4,5,9) quadruples

Hint 2: Consider the maximum cell in box 7 and what can go in the top right cell of box 7

Hint 3 : What digits that are left over can go in the maximum?

Answer: Since the shared cell between the two (2,5,9) quadruples is a minimum cell, it must be a 4. This gives two 5/9 pairs to the left and right of the 4. The 5/9 in the top right cell of box 7 is adjacent to a maximum, therefore it cannot be a 9 and is a 5. The only remaining option in box 7 for the maximum is 8, (as it must be greater than 7)


Hint 1: How does the (1) quadruple affect the (1,2,3) quadruple options?

Hint 2: The minimum cell should be placeable now

Hint 3: The top left and top center cells should be placeable

Hint 4: Don’t forget to check back on box 4 and 7 for cells you can place now

Answer: The (1) quadruple in box 1 forces the 1 for the (1,2,3) quadruple into the top middle of box 4. This leaves the 2 and 3 for the (1,2,3) quadruple and the minimum cell forces them to 2 in the bottom left and 3 in the bottom center. In box 1, the 9 (top center) and then 8 (top left) can be placed by basic sudoku rules. 5 (middle center) and then 6 (middle left) can be placed by sudoku.


Hint 1: You should be able to complete boxes 5, 7, and 8 at this point

Hint 2: Start with the Minimums in box 3 and the (8,9) quadruple in box 9

Hint 3: Check box 6: What are the options for the middle row?

Hint 4: Check box 3: It should be completable.

Hint 5: See hint below for image of final grid
Answer:

Welcome to the Maze of Eyes. Find the ladder at the end to escape! This room was created with 2 players in mind and you need at least 2 to play. It is possible to play with more. You will probably need to take some notes to keep track of certain information.

For the purposes of this guide, the player who starts with the red eye will be the red player and the person who start with the blue eye will be the blue player.
The colors I use to represent each eye are:
Each eye has a different pupil shape if the colors are giving you trouble

Hint to get started: Make sure you take the eyes with you when you can, you may need them later.

Answer: Place the eye from the cup/tankard into the skull

Hint 1: The red player should search around their area carefully
Hint 2: The blue player should know which eyes to use
Answer: On the red side, there is a lever hidden next to the barrels. This opens the wall for the blue player to look out the window. The eyes out the window should be placed in the skeleton’s skull, green on the left, purple on the right

Hint 1: The eyes next to the coins on the table/bench are a hint as to their value and how that value is represented
Hint 2: The eyes on the chest are the key to the coin code
Hint 3: Red Eye = 1, Blue Eye = 2, green must be deduced
Answer: G=gold coin, S=silver coins, the code of red blue green red, or 1 2 X 1, gives: G S G G S G G G S G, so Green = 3. This opens the chest on the blue side. The revealed skulls indicate which eyes to use to proceed

Hint 1: Check out the window
Hint 2: The 2 codes must be combined to get the entire sequence of coins
Hint 3: All 3 walls of coins make up one sequence. Don’t forget that there must always be a silver coin in between the gold coins
Answer: G=gold coin, S=silver coins. The code is blue blue green purple yellow blue. Blue is known to be 2, so the coins on the blue player side start as G G S G G S. Since the first 3 coins on the middle wall are G G S, this means the last coin on the blue side must be G, otherwise the code would be blue blue blue, making the full code on blue side G G S G G S G.
On the red players side, the last digit of the code is blue, which represents 2. The fourth eye of the code is purple(4), which is represented by the coins in the middle room. This leaves yellow in between the purple 4 and the blue 2. There is already 1 coin after the purple set, leaving only 5 as a valid option for yellow. The red side code is therefore: G G G G S G G.

Hint 1: Each side has 4 eyes to find
Hint 2: Each eye shows what the opposite player should look for
Hint 3: Each eye also shows where eyes should be placed
Answer : Blue Side: Black eye goes on the chessboard, pink eye goes in the box under the bookshelf with 6 books, blue eye goes in the bucket between 2 anvils, purple eye goes in the bucket next to the bookshelf with 4 bags on the bottom.
Red Side: Grey eye goes in the bucket next to 7 bottles, green eye goes in the tankard next to 3 stacked crates, yellow eye goes in the tankard under the 5 hammers, red eye goes in the bucket on the throne

Hint 1: Each skull’s directions are for the opposite player
Memory/Time:
Hint 2: You should have what number each eye represents from previous puzzles
Hint 3: The plaque under the skull gives the order for each side (which is different)
Answer: Time is the eyes from 1 to 8, Memory is the eyes from 8 to 1
Eye Numbers:
Red = 1
Blue = 2
Green = 3
Purple = 4
Yellow = 5
Pink = 6
Grey = 7
Black = 8
Twins:
Hint 1: The board is linked, so pressing a button on one board also presses on the other players board
Hint 2: Each player has half the clues, and should input the information they know is complete first
Answer:
G S S G G
S G G S S
S G G G G
G S G S G
S G S G G

Chaos:
Hint 1: If every digit must be at least 4 away from the adjacent digits, then there are 2 numbers that are very restricted in the set of numbers from 1 to 8
Hint 2: Consider the parity of the sequence, i.e. which digits must be from 1-4(and are low) and which digits must be 5-8(and are high)
Hint 3: Consider the high numbers that are left over and what low numbers are allowed next to them
Answer: The digits 4 and 5 can only be next to 8 and 1 respectively, as 4 digits away in the opposite direction are 0 and 9, therefore they must go on the ends, with their valid digits next to them. This gives 4 8 x x x x 1 5. Of the high digits left, 7 can be next to the digits (1,2,3) but 6 can only be next to the digits (1,2). If 6 is surrounded by 1 and 2, it must go next to the 1, with the 2 on the opposite side. This gives 4 8 x x 2 6 1 5. 3 and 7 then slot in for a final solution of 4 8 3 7 2 6 1 5

Law:
Hint 1: Start with where the digit 7 can go
Hint 2: With digit 7 and 6 in mind, where can 8 go? (Don’t forget clue 1 and 2)
Answer: Short answer is 4 8 1 3 6 2 5 7. Explanation follows:
The digit 7 is at minimum spot 5, due to the given 4 and the rule about odd digits increasing (4 1 3 5 7). Since the 2 must be adjacent to the 5, this makes digit 7s minimum spot 6 (of 8). Since 6 is exactly 3 away from 7 (and it cannot go after it, as there are only 2 spots left), this moves 7 to a minimum of spot 7. The only even digit left that could be in spot 8 is digit 8, however, rule 3 does not allow consecutive digits, so this forces digit 7 into spot 8, and digit 6 into spot 5(3 away). This gives us (4 x x x 6 x x 7)
The digit 6 cannot be surrounded by odd digits, and 8 must be at least 2 away, so the digit 2 must be adjacent to 6. Digit 5 must be adjacent to digit 2, due to rule 5. This gives us (4 x x x 6 2 5 7).
The digit 8 must at least 2 away from 6 and cannot be surrounded by odd digits, thus it goes next to 4. The remaining 1 and 3 slot in as they must increase from left to right, giving a final solution of 4 8 1 3 6 2 5 7

This is the fourth puzzle room based on Sudoku Variants. Solve the cells/boxes with a yellow corner to proceed to the next step and complete the puzzle. Click on the BOX button to show the marking buttons. Boxes are labeled on the top left and will show a green circle if the box is correct

This puzzle starts you on box 9 (bottom right). Do not forget that you do not need to solve all cells immediately to proceed.

Variant rules:

For XV: Cells separated by an X sum to 10, Cells separated by a V sum to 5

For Between Lines: Digits on the line must be between the digits in the circles on the end of the line


Hint 1: Consider where the 1 can go

Hint 2: Don’t forget that for whatever digit you place on the line, there must be one larger and one smaller digit on the ends

Answer : A V pair has 2 options: 1/4 and 2/3. A 1/4 pair does not work because it would either put a 1 on the line or a 9 on the line (due to the X). This means the V is a 2/3 pair. A 2 or 3 on an X pairs with an 8 or 7. You cannot put the 8 on the line because there are no digits left in the box that are larger than 8. So the 3 cells needed are 7, 3, 2


Hint 1: Start with the V and its possibilities

Hint 2: Consider the X in relation to the V possibilities

Hint 3: Don’t forget the between lines rule!

Answer: The V must be a 1/4 pair due to box 9. Since a 1 on the X pairs with a 9 and that is blocked by the 9 in box 9, the X is a 4/6 pair. All of the other cells in the box are part of the between line, which means the largest and smallest of the digits that are left go in the circles (2 and 9) The 9 in box 9 sets their locations



Hint 1: Some digits cannot go on a line

Hint 2: Where can 1 go in box 7?

Answer: Since neither 1 or 9 can go on a between line and they cannot go in the bottom row due to box 9, the 2 yellow marked cells must be 1 and 9


Hint 1: Where can the 5 go in box 4?

Hint 2: Where can the 2 go in box 7?

Hint 3: The 2 in box 7 limits what can go on the V in box 4

Hint 4: The X in box 4 at top left only has 1 option left

Answer: The 5 must go in the middle since it can never go on an X or V and the between line is too long for a 5 to be on the end. Since the 2 in box 7 is on the between line, the 1 must be in one of the circles in box 4. This gives you a 1 4 6 on the right side of box 4. The X on the left side of box 4 can only be a 3/7 pair now, leaving a 2/8 pair for the top left X.


Hint 1: You should know all 3 of the possibilities of the left column of box 1

Hint 2: Consider where 1 can go in box 1

Hint 3: What are the possibilities for the middle cell when considering hints 1 and 2

Answer: The left column in box 1 contains 4/5/6. Since the 1 in box 1 must be in the center column (but not the middle) this means that 9 is in one of the circles. This greatly restricts what can be on the line to be between 6 and 9 and there is one other circle to fill. This places a 7 in the middle, with the right 2 circles being 8 or 9


Hint 1: Don’t forget to check back to previous areas to see if you can place more digits!

Hint 2: Start with box 5 when you are done with hint 1
Hint 3: The X in Box 2 should be placable

Hint 4: Where does 1 go in box 2?

Hint 5: You should be able to complete most of box 5 with basic sudoku rules

Answer: The 9 in box 5 only has 1 place it can go (middle) and this gives a 1 on the X. The 1 in box 2 only has one place it can go (top left) and this gives the 2/8 and 1/9 pairs in box 1. The left X in box 2 cannot be 3/7 (from box1) or 4/6 (from box 8) or 1/9, which leaves 2/8. In box 5, the 2 is placable, followed by the 6, 4, 5, and 8. This gives a 3/7 pair in column 4 (bottom left of box 2 and top left of box 5). The final X in box 2 therefore cannot be 2/8, 1/9, or 3/7 and is 4/6





Hint 1: Start with the V in box 3 and consider how it affects box 1

Hint 2: You should be able to complete box 1, 2 and 8 now

Hint 3: It’s all sudoku from here!

Answer: See image below for final answer

This room is based on a pencil puzzle ruleset by Eric Fox called Voxas.

This puzzle starts you on box 15 (bottom center). Box numbers are row/column, so box 15 is row 1 column 5. The whole room is 9x9.

Voxas rules:

Divide the 9x9 room into 1x2 or 1x3 cells. No other cell size is allowed/

Cells separated by a white wall have the same size AND the same orientation.

Cells separated by a black wall have NEITHER the same size nor orientation.

Cells separated by an orange wall have EITHER the same size OR orientation but not both.

Click on the blue walls to raise or lower them. The door at the end will open when all cells are correct.



Some general solving hints:

Hint 1: Since all cells are at least 1x2, the starting walls can force a direction

Hint 2: It can be helpful to raise walls you know must be there due to orientation even if you do not know if it is 1x2 or 1x3

Hint 3: Keep an eye out for situations that would leave a 1x1 as that is not allowed

Hint 4: You may need to move to other given walls if you get stuck in one section

Unfortunately giving specific hints for this was going to be a massive mess, but feel free to hit me up on the Escape Simulator discord if you have questions, I am ZekTheMad on there as well.



Answer: See hint below for image of final grid (grey dots represent orange walls)

Welcome to the testing chamber, where you will solve a series of logic puzzles to exit!

Note: Orthogonal means the 4 cells around a cell vertically or horizontally. The green cells surrounding the 4 in the example below are all orthogonally adjacent. The green cells surrounding the 8 in the example below are all adjacent (but not necessarily orthogonal).

The green path around the 7 and 5 in the example below both show incorrect paths. The 7 path meets itself diagonally and the 5 path doubles back on itself, creating a 2x2 of green.




Rules: Make an orthogonal blue path from one blue square to the other. A clue in the grid indicated how many cells of the path are in the adjacent cells (including diagonally). The path cannot meet itself, even diagonally, and you cannot create a 2x2 block of blue cells.

Note that for this puzzle, the non-blue cells can be either orange or grey.

Hint 1: Start with the left 4

Answer : This one is pretty short, so here is an image of the answer:


Rules: Same as puzzle 1 but orange this time

Hint 1: Start with the 6 on the right

Hint 2: It can be helpful to fill in the spaces which must be orange (such as with the 6 on the left) and work from there

Answer: Image of the answer:


Rules: Same as puzzle 1

Hint 1: A 1 with this ruleset is very restricted

Hint 2: I suggest marking spaces that CANNOT be blue, as this will limit the path

Hint 3 : The path next to the 2 forces the path at the 3 on the right

Answer: Image of the answer:


Rules:
Clues represent the total number of orange cells that are connected in a straight line both vertically and horizontally, including itself. All clues count as orange cells.
All orange cells must form one orthogonally connected area.

All other cells are blue. Blue cells must connect orthogonally to the edge of the grid.



This is a type of pencil puzzle called a cave. In this case the blue cells are the cave walls and must touch the border and the orange cells are the cave itself and must be orthogonally connected. Here is an example puzzle of this type (with green cells being the cave)













Tip: It may help to set all cells to grey at the start.
Tip2: When a clue is completed you can set the next horizontal and vertical cells that it sees as blue

Puzzle link if you want to try it separate from the game: https://puzz.link/p?cave/8/6/3i7h7h5g2h25q5j3m2i543

Hint 1: Don’t forget that clue cells count as cave and are orange. With that in mind, start with the 5/4/3 in the bottom right

Hint 2: 2’s are very restricted in this type of puzzle

Hint 3: An interesting quirk of these types of puzzles is that you can never have a checkerboard pattern, as that would isolate either a blue or orange section

Hint 4: Don’t forget to make sure that any blue cells can get back to the border

Answer: Image of the answer:


Same rules as puzzle 4

Puzzle link if you want to try it separate from the game: https://puzz.link/p?cave/8/8/i4g3g4h4l9g6g8h5k8v3k2i3g8g6h

Hint 1: This is a repeat hint from 4 if you skipped it: An interesting quirk of these types of puzzles is that you can never have a checkerboard pattern, as that would isolate either a blue or orange section

Hint 2: I suggest starting with the 2/6 pair and the 8 nearby

Hint 3: The 9 6 8 in row 3 are limited

Hint 4: The 8 in column 7 should be limited at this point

Answer: Image of the solution below:

You find yourself stuck in a tomb! Carefully navigate the chambers to escape



Activate all 4 statues in the correct order to open the door

Hint 1: The exact number each statue represents is not required, only their relative values

Hint 2: The order is from largest to smallest

Hint 3: Blue+White=Orange+Black is the key to the order of white/orange

Answer : Blue > Black > Orange > White Since blue is greater than orange+white and black is equal to orange+white, blue is greater than black. Since blue > black, if you add some number to each to make them equal, you will have to add a smaller number to blue.


All scarab tiles must be colored in to open the door.

Answer: If we consider the sun on the scarab to be north, starting from the closest tile go: N, N, W, S, S, W, N, N,


All scarab tiles must be colored in, in addition to the statues being pressed in the correct order
Note: There are multiple ways to activate all the scarab tiles but you will need to be at the exit door when done

Hint 1: There are 3 clues to find for the order of the statues

Hint 2: The exact number each statue represents is not required, only their relative values

Statue Order Answer: Orange > Blue > Black > White. Since Orange > Blue (from the clue) and Blue > Black and White, then 3xBlack will be greater than 3xWhite

Hint 3 : Since your path across the tiles must cross itself, it can be helpful to double back on tiles you know you must cross again

Answer: This is one possible solution:



All scarab tiles must be colored in, in addition to the statues being pressed in the correct order
Note: There are multiple ways to activate all the scarab tiles but you will need to be at the exit door when done
Note 2: One of the statues is purple. Due to lighting in the game it sometimes looks pink. I refer to it as purple below, but if you see a statue that you think is pink, it is supposed to be purple.

Hint 1: There are 6 statue clues to find

Hint 2: The black and purple clue is an important restriction

Hint 3: The previous hint 2 + the clue that says Black must be between White and Blue are what defines the order

Statue Order Answer: White | Black | Blue | Orange | Purple | Green
There are 6 statues and green must be last, so there are 5 spaces left. Blue and orange must be adjacent. Black must have at least 2 spaces between itself and Purple. Since there are only 5 spaces left and blue/orange are together, Black and Purple must be on either side of the blue/orange pair (in order for there to be at least 2 spaces in between). Black is between white and blue/orange. Blue and Orange's order is set because Blue must have at least 2 spaces in between itself and Green. This means the order is either White | Black | Blue | Orange | Purple or Purple | Orange | Blue | Black | White and purple cannot be first.

Hint 4: You will need to enter the room with the white/purple statues twice, so leave yourself a path to get back

Hint 5: Tiles only trigger when you enter the space, so if you have a solo tile you can step on it and then step back to the previous tile

There are likely multiple ways to walk the path but here is one:

1st half of the path answer: Image of the answer:

2nd half of the path answer: Image of the answer:

This is the fourth puzzle room based on Sudoku Variants. Solve the cells/boxes with an orange corner to proceed to the next step and solve all of the boxes to complete the puzzle. Note on cell and box labels: Each box is numbered from the top and increases from left to right, with box 1 in the top left, and box 9 in the bottom right. Boxes are labelled in their bottom left corners.

This puzzle starts you on box 8 (bottom center). Do not forget that you do not need to solve all cells immediately to proceed.

Variant rules:

For X-Sums: A clue outside the grid indicates the sum of the first “X” digits in that row/column where “X” is the first digit.
For example: If the clue outside the grid is a 7, then we must find a set of digits that sum to 7.

There are several ways to sum to 7: 1+6, 2+5, 3+4, and 7 by itself

The digit 7 by itself does not work: This would make “X” 7 and we would then need 6 more digits to sum to 7.

The digits 1 and 6 do not work: While 1+6 does sum to 7, “X” would either be 1(and then the sum of the first “X” digits would be 1) or “X” would be 6 and we cannot make 6 different digits sum to 7.

The digits 3 and 4 do not work for the same reason 1 and 6 do not work

So the only solutions for a 7 clue would be 2 and 5, with 2 being in the first cell making it “X”. Then the sum of the first “X” cells would then be 2 and 5 which equals 7.

There are some examples of this in the side room.


For Killer Sudoku: A clue in the top left corner of a cage indicates the sum of the digits in the cage. Digits may NOT repeat within a cage. (← This is important)

The secret!: The sum of all the digits from 1 to 9 is 45


Hint 1: There is only one way to make 10 with 4 digits

Hint 2: There is only one way to make 17 with 2 digits

Hint 3: The X-Sums clues outside the grid give you the yellow cell digits

Answer : Since the 10 cage has 4 cells it must contain 1/2/3/4. The 17 cage must be 8/9. The middle X-Sums 10 clue therefore must have a 1/2/3/4 as the first digit and an 8/9 as the second. Since 1 does not work as the first digit of the X-Sum, we must use 2, which gives us 2 8 9 in the middle column of the box (from bottom to top). This leaves 1/3/4 in the right column to use for the other X-Sums 10 clue. 1 and 3 do not work with the digits we have, so it must be 4 in the first cell and a 1/3 pair in the two cells above it.


Hint 1: The bottom right cell of box 5 is part of the clue from the previous box

Hint 2: A two cell 16 cage only has 1 option

Answer: 2 must go in the bottom right cell to make the x-sum clue complete. (There is already a 1/3/4 in the column) A two cell 16 cage must be 7/9 (since 8/8 is not an option) and the 9 in box 8 sets the order.


Hint 1: Don’t forget the secret!

Hint 2: Where can 9 go in box 3?

Hint 3 : The middle right cell (in the middle 8 cage) is very limited now

Hint 4: The top middle 10 x-sum sets the bottom left 8 cage

Answer: The 35 x-sum clue in the right column lines up with the 10 clue at the bottom of the grid. Add them together and you get 45, which means the 35 clue takes up the rest of the column that the 10 clue does not. This means the “X” for the 35 clue is 5 (since the 10 clue had an “X” of 4). The 9 is now limited to the right column and you cannot place a 9 in an 8 cage, so it goes in the bottom right. Now the middle right cell of box 3 can only be a 7 or 6. Placing a 6 would cause a 2 to be placed in the middle where it would clash with box 8. So the middle 8 cage is a 1/7. The 10 x-sum clue makes the bottom middle cell a 6, which will be paired with a 2 in the 8 cage.


Hint 1: Don’t forget the secret!

Hint 2: There are only 2 options for the “X” cell of an 11 x-sum clue

Hint 3: Don’t forget to check the possibilities for the rest of the 9 cage

Answer: The “X” for the 44 clue must be 8, since all of the digits from 1 to 9 sum to 45, we will need at least 8 digits to sum to 44. The 11 x-sum clue has 2 options for “X”: a 2 or 3. While you can sum to 11 in four cells you cannot do it with a 4. Using a 2 would place a 9 in the 8 cage, which does not work, so the “X” is a 3. The part of the 9 cage that is in box 8 only has 3 options: 5/6/7. A 7 would mean we need two 1’s, which is not allowed, a 6 would need a 1/2 pair, which is blocked by the 2 in box 5. This leaves 5, which means we need a 1/3 pair in the rest of the cage, which we can place due to the 3 in box 4.


Hint 1: Start with the 6 x-sum clue

Hint 2: What are the options for “X” for the 27 x-sum clue?

Hint 3: For the 27 clue, it can be helpful to subtract out the numbers you know must be included and work with the remainder

Answer: A 6 x-sum can only start with 2 or 3 and the 2 in box 8 conflicts. The 27 clue has a few options for “X”. A 4 would work (9+8+6+4) but the 9 is already in the row and is too far away to be included in the first 4 cells. A 5 is already in the row, so it cannot be used. A 7 is not possible since the lowest value 7 digits can be is 28, so that leaves 6. With the 6 in place, we have a 6/5/9 already in the row, which sums to 20. This leaves 7 left to make in 3 cells, which can only be 1/2/4. The top middle cell of the 20 cage can only be a 1 or 2 and the top cell of the 10 cage can only be a 1 or 3. This leaves 4 as the top right cell of box 7, a 1 in the 10 cage (because we do not need a 3) and a 2 in the top of the 20 cage.


Hint 1: Don’t forget to check back on other boxes to see if you can fill in more numbers (Boxes 2, 4, 7 and 8 should be completable)

Hint 2: Don’t forget that digits cannot repeat in a cage

Answer: Since the 29 cage has an 8 and 4 in it already, we need to make 17 with the remaining 3 cells. The middle column of box 1 is a 5/8/9 triple, but the 8 cannot go in the cage (since there is already an 8 there), so the 5/8/9 in the middle column are placable. This gives a 3 as the last digit of the 29 cage and the other 2 digits of the right column in box 1 are placeable (6/7)





Here are a couple hints for each of the last 3 boxes. The final solution will be below

Box 9:

Hint 1: Due to the rest of the digits in the bottom row, there are only 3 options for the “X” cell for both 21 clues

Hint 2: The opposite side of a 40 x-sum clue is limited


Box 6:

Hint 1: All of the digits outside the cages should be placable (due to the limited options inside the cages)

Box 3:

Hint 1: The “X” of a 40 x-sum clue only has 2 options

Hint 2: Where can the 9 go?


Answer: See hint below for image of final grid

This is the fifth puzzle room based on Sudoku Variants. Solve the cells/boxes with a yellow corner to proceed to the next step and complete the puzzle. Click on the gold coin to show the marking buttons. Note that this map does not have the green circles in the corners of the boxes, they are on the final door

This puzzle starts you on box 5 (middle). Do not forget that you do not need to solve all cells immediately to proceed.

Variant rules:

For Thermo: Cells along an orange thermometer line must increase from the bulb end. Digits may not repeat on a thermo

For German Whispers: Digits along a blue line must have a difference of at least 5. Digits may repeat if allowed by other rules.

Note: In many hints I refer to low and high digits. For the purposes of this guide, low digits are 1/2/3/4 and high digits are 6/7/8/9. 5 is neither.


Hint 1: Consider where the low and high digits can go on the blue line in box 5 in relation to the thermo

Hint 2: Once you find the low and high digits on the blue line, the last 4 digits of the thermo are forced

Hint 3: On a german whispers line, the digits 1 and 9 are the least restricted and the digits 4 and 6 are the most restricted

Answer : In box 5, the blue line is of length 6, so it either starts with a low digit and ends with a high digit or vice versa. The end of the blue line that starts at the bulb of the thermo cannot be a high digit because it is too long to start from 6 and increase to the end of the thermo. This forces the other end of the blue line to be 6/7/8/9 and the thermo makes this a 6, with a 7/8/9 along the rest of the thermo. A 6 on a german whispers line must be adjacent to a 1. This leaves 2/3/4 as the possible low digits. If you try to use a 4 in either spot on the blue line then you would need a 9, which is blocked by the 9 in box 2. This leaves 2/3 as the only options and you cannot place a 7 between a 2 and 3, so this must be an 8.


Hint 1: Once again, start with trying to figure out the low/high digits on the blue line

Hint 2: One low and one high digit must be on the blue line where it enters box 2 and this will define the rest of the line

Hint 3: The thermo should give you the required digits now

Answer: Where the blue line extends into box 2, we need a high and a low digit. The only high digit left in the box is 6. Since the 6 must be next to a 1 and the 1 cannot go on the end due to the 1 in box 5, the 6 goes on the end with a 1 next to it. This means the intersection of the blue and orange lines in box 1 must be a low digit and 1/2 are the only low digits available. 1 cannot go on the middle of a thermo, so the 2 yellow boxes are 2 and 1.



Hint 1: Check for high/low on the X of blue lines

Hint 2: Consider where the 2 and 3 can go in box 4

Answer: The X of blue german whispers lines is either 1 low digit and 4 high, or 1 high and 4 low. Since there is a 4 already in the box, it cannot be 4 low digits on the X. The only low digit that can go in the middle is 1. Since the 4 ends of the X are high digits, the 2/3 must go in the middle row of the box. This leaves only one place for 5.


Hint 1: What digits can go on the middle thermo and how does that affect the blue line?

Hint 2: Consider what happens to box 4 if both 2 and 3 are on the blue line

Hint 3: The remaining digits for this box force the beginning and end of the left thermo

Answer: The center thermo must have digits from 6/7/8/9. This means the blue line cannot have 2 high digits as this would not allow enough digits for the center thermo. This makes the center of the blue line a high digit with 2 low digits around it. Those low digits are from 2/3/4. If both low digits are from 2/3 then box 4 is broken as there would be no room for both 2 and 3. This means 4 is forced onto the blue line and the only digit that 4 can be next to is a 9. This sets the center thermo as 6/7/8. The left thermo is from the digits 1 to 5 so 1 is the bulb end and 5 is the tail end. Click on the lit torch to open the wall


Hint 1: Where can 1 go in box 8 and 9?

Hint 2: Where can 9 go in box 5 and 8?

Hint 3: Check for 8s

Hint 4: Where can 6 go in box 8?

Answer: 1 must go in the right column of box 8 and 1 can only go on a bulb of a thermo. The 9 in box 5 must go in the left column, forcing the 9 in box 8 into the right column. A 9 can only go on the end of a thermo. 8 only has 1 position left in box 8 and this forces the 6 in box 8 into the bottom left corner. The 3/4/5/7 can be placed to finish the box. The 1 and 8 in box 9 are now forced and the 6 is the only digit left in that row. Click on the green wheel to open the secret wall


Hint 1: Don’t forget to check back to previous areas to see if you can place more digits! You should be able to complete boxes 1,7, and 8

Hint 2: The end of the blue line is box 6 that pokes into box 9 is limited. This sets the high/low sequence of the line

Hint 3: The 3 cell thermo in box 3 should be completable now

Hint 4: This should set most of the blue line in box 6

Hint 5: See answer below for picture of final grid

Answer:

This is a 2 player (or more) puzzle room based on Sudoku Variants. Solve the cells/boxes with a yellow corner to proceed to the next step and complete the puzzle.

This puzzle starts each group on opposite sides of the sudoku grid and you must work together and communicate to solve the puzzle

Variant rules:

For Odds/Evens: A cell with a circle must be odd. A cell with a square must be even

For Diagonals: Digits cannot repeat along the marked diagonals. Or, said another way, each diagonal must contain the digits 1 to 9.

The hints on this one are pretty light as most of the difficulty will be communication.


Hint 1: Count the odds and evens in each column


Answer : The 7 in blue and the middle force the 7 in orange to be on the left side. Since there are already 4 marked or known odd digits in this column, the 7 must go on the circle in orange. The same logic applies to the 6 in blue.The 1/5 pair in blue and the 2/8 pair in orange gives you the 2 in blue and the 1 and 4 in orange


Hint 1: Counting odd/even quantities will be a recurring theme in this puzzle

Hint 2: Don’t forget that digits cannot repeat on the diagonal.

Hint 3: Where can the 6 go on the purple diagonal?

Answer: The 6 on the purple diagonal can only go in box 7, due to the locations of the 6 in boxes 3 and 5. The 6 in box 8 forces the 6 in box 7 to the top right corner. The even cell in box 7 now only has the option of being an 8. Since all of the odds on the purple diagonal are marked, the rest are even. This puts a 2 on the diagonal in box 3, which then gives you the 4 on the diagonal. The only options left on the diagonal are 3 and 5, and the 5 in box 3 fixes the order.


Hint 1: Start with the box 4 even cell possibilities and the box 6 odd possibilities

Answer: In box 4, the even cell in the top right can only be a 4, as it can see a 2, 6, and 8. Likewise, in box 6, the odd cell at the top middle can only be a 9. This row can then be finished off with the 5 and 6. The 2 remaining even digits in box 6 can now be placed. This leaves 1/3/7 as possibilities for the bottom row of boxes 4 and 6 and 1 cannot be in box 4, forcing the 1 into box 6 (bottom right). The middle row of box 4 is now 2/6/9, which is placable.


Hint 1: This one is pretty straightforward, just focus on the diagonal possibilities

Answer: The middle cell of box 1 can only be a 2 as it sees the 4/6/8 in the column. Similarly, the bottom right cell in box 9 can only be a 4. This gives you the 8 in box 9 and the 6 win box 1 (on the green diagonal). The last 2 cells on the diagonal are 7 and 9, and the middle cell in box 9 can only be a 7.


Hint 1: Its all regular sudoku from here! See answer spoiler below for final grid


Answer: