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## Question

1) What is the minimum number of squares that would be sufficient to create each of the following patterns?

2) Twenty-seven identical cubes are placed together to form the object below. If one of the cubes is displaced, four different shapes are possible: one in which the missing cube is at the corner of the stack, one where it is in the middle of an edge of the stack, one in which it is in the middle of a side of the stack, and one in which it is at the core of the stack. If two of the small cubes are removed rather than one, how many different shapes are possible?

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For question number 1), I'm not sure that I understand the problem correctly.

Are we overlapping squares over each other to make this shape or what?

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For question number 1), I'm not sure that I understand the problem correctly.

Sorry, allow me to clarify.

Example:

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My count for the green in number 1

13 Method:

For ease of explanation, use 4 units for size of each side fo final figure

place 2 unit squares in each corner = 4

place 1 unit squares along left side = 4

place 1 unit square in upper right corner of lower left 2 unit square = 1

place 1 unit square in lower left and upper right corners of upper right 2 unit square =2

place center square at 45 degree angle (side length = square root of 2) = 1

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Just realized that it can be done in less:

I overlooked some duplication of sides change my solution to 10

method:

For ease of explanation, use 4 units for size of each side for final figure

place 2 unit square in center = 1

place 2 unit squares along left side=2

place 1 unit squares along left side = 4

place 1 unit square upper right corner =1

place center square at 45 degree angle (side length = square root of 2) = 1

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I could do the green pattern in number 1 with 10 squares.

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For the cube in number 2

that mirror image patterns do not count, I can identify 13 different patterns.

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I have a question to the original poster. Can you add a square, but then erase a line. For the green puzzle the bottom right corner is potentially three unit squares, but their inside lines are erased. Is this a potential solution.

Second question. Are the squares just an empty brace of 4 lines, or are they filled in objects that cover up any lines below them. Again for the green puzzle, there is a square rotated 45 degrees in the middle. If there was a line below that rotated square, and then I put the rotated square on top of that line, is that line now covered up.

I assume that part of the puzzle is figuring out how to get it to look identical.

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i could be wrong but i got up to

90 forms for the cube one

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sry the spoiler didn't work cause i forgot the title.... but actually i just thought of smth

i just realized that there is more .. i don't have time right now to think about it decently but i think it is 258 forms.. but i got the second conclusion of the top of my head so i could have miss-thought about smth...plz tell me your thoughts

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Solution for puzzle 1: 11 squares. I think that the solution from thoughtful fellow that uses 10 squares is missing a horizontal line between square 8 and square 12

From left to right, top to bottom there are 16 positions.

1. a 4x4 square covering all

2. a 2x2 square covering 6,7,10,11

3. a 2x2 square covering 1,2,5,6

4. a 2x2 square covering 9,10,13,14

5. a square covering 1

6. a square covering 5

7. a square covering 9

8. a square covering 13

9. a 2x2 square rotated 45 degrees covering 6,7,10,11

10. a square covering 4

11. a square covering 8

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I'm not sure why you are talking about 258 forms, but the original post says to use the least number of squares.

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solution for puzzle 2: 12 steps ( and not because this puzzle will drive you to drink)

1. 5x5 square

2. 3x3 in 13,14,15,18,19,20,23,24,25

3. 3x3 in 1,2,3,6,7,8,11,12,13

4. 1x1 in 3

5. 1x1 in 10

6. 1x1 in 11

7. 1x1 in 13

8. 1x1 in 15

9. 1x1 in 17

10. 1x1 in 23

11. 2x2 rotated 45 degrees in 1,2,6,7

12. 2x2 rotated 45 degrees in 19,20,24,25

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For the cube puzzle, I was a bit confused by the word "displaced", which would mean to me, "moved over", so a cube would stick out somewhere. But then you used the word "removed", which means to me that it was replaced by a hole. Assuming that we're talking about removal...

Denote the kinds of locations as follows:

C = cube center (the core of the 3x3)

F = face center

E = edge center

X = corner

Omitting reflections and rotations, I believe the possibilities are:

* If one is C, then F, E, and X (3 cases)

* else, if one is F, then 7 cases

--F adjacent (ie sharing an edge)

--F opposite (not sharing an edge)

--E same face

--X same face

--E adjacent (ie connecting this face to opposite)

--X opposite face

--E opposite face (ie edge on opposite face)

* Else, if one is E (5 cases)

--E adjacent (ie sharing a vertex)

--E opposite (ie not sharing a vertex)

--C same edge

--C adjacent edge (ie edge sharing a vertex)

--C non-adjacent edge (ie edge not sharing a vertex)

* else if one is X (2 cases)

--X adjacent (ie sharing an edge)

--X non-adjacent (ie not sharing an edge)

So I count 17 cases

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Sorry revised solution to puzzle 1 to eliminate 1 step down to 10

From left to right, top to bottom there are 16 positions.

1. a 4x4 square covering all

2. a 2x2 square covering 6,7,10,11

3. a 2x2 square covering 3,4,7,8

4. a 2x2 square covering 9,10,13,14

5. a square covering 1

6. a square covering 5

7. a square covering 9

8. a square covering 13

9. a 2x2 square rotated 45 degrees covering 6,7,10,11

10. a square covering 4

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For the cube puzzle, I was a bit confused by the word "displaced", which would mean to me, "moved over", so a cube would stick out somewhere. But then you used the word "removed", which means to me that it was replaced by a hole. Assuming that we're talking about removal...

Denote the kinds of locations as follows:

C = cube center (the core of the 3x3)

F = face center

E = edge center

X = corner

Omitting reflections and rotations, I believe the possibilities are:

* If one is C, then F, E, and X (3 cases)

* else, if one is F, then 7 cases

--F adjacent (ie sharing an edge)

--F opposite (not sharing an edge)

--E same face

--X same face

--E adjacent (ie connecting this face to opposite)

--X opposite face

--E opposite face (ie edge on opposite face)

* Else, if one is E (5 cases)

--E adjacent (ie sharing a vertex)

--E opposite (ie not sharing a vertex)

--C same edge

--C adjacent edge (ie edge sharing a vertex)

--C non-adjacent edge (ie edge not sharing a vertex)

* else if one is X (2 cases)

--X adjacent (ie sharing an edge)

--X non-adjacent (ie not sharing an edge)

So I count 17 cases

...just based on X. If you remove any X, there are 3 possible other X's you could remove to make a different shape:

1. X adjacent on a single face

2. X diagonal on a single face (not-adjacent, but on the same face)

3. X opposite on the cube (sharing no faces). <--I think you missed this one

Edited by Molly Mae
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Way to go, Molly! I agree. 18 cases for the cube.

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For the cube in number 2

that mirror image patterns do not count, I can identify 13 different patterns.

Solution for puzzle 1: 11 squares. I think that the solution from thoughtful fellow that uses 10 squares is missing a horizontal line between square 8 and square 12

From left to right, top to bottom there are 16 positions.

1. a 4x4 square covering all

2. a 2x2 square covering 6,7,10,11

3. a 2x2 square covering 1,2,5,6

4. a 2x2 square covering 9,10,13,14

5. a square covering 1

6. a square covering 5

7. a square covering 9

8. a square covering 13

9. a 2x2 square rotated 45 degrees covering 6,7,10,11

10. a square covering 4

11. a square covering 8

aamph, you are correct, I made two errors. First the omitted line that you mentioned but also note that the left side only required three 1 unit squares so I had another redundancy. Changing the solution accordingly to 3 along left side (top, bottom and either middle ones. and adding 1, 1 unit square to cover 8 and provide missing line leaves 10 as the number required.

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Way to go, Molly! I agree. 18 cases for the cube.

There are 4 edge-edge cases:

1) parallel sharing face

2) parallel not sharing face

3) sharing edge

4) not parallel and not sharing edge

I think you missed it.

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aamph, you are correct, I made two errors. First the omitted line that you mentioned but also note that the left side only required three 1 unit squares so I had another redundancy. Changing the solution accordingly to 3 along left side (top, bottom and either middle ones. and adding 1, 1 unit square to cover 8 and provide missing line leaves 10 as the number required.

Oh good point. Not sure if you read my revised solution, but I eliminated another step there so it is now down to 9 steps. I'll post the revised solution.

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I made an assumption that instead of needing to put the 2x2 unit squares along the left edge, I could put them in the top right and bottom left corners, but I think this fails.

There is a line that must not exist between 2 and 6.

I need to rethink this puzzle

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Oh good point. Not sure if you read my revised solution, but I eliminated another step there so it is now down to 9 steps. I'll post the revised solution.

s

Thank you for keeping me thinking. I was wrong about reducing to 3 1 unit squares on left side since left 2 unit squares covers the lines needed to complete the square. Examining your revised solution for 10, I, too realized that a line remains between 2 & 6 that should not be there. You have inspired me to find another solution for 10 squares.

Solution for 10 squares:

1. 4 unit square = 1

2. 1 unit squares covering 1, 5, 9, 13 = 4

3. 2 unit square covering 11, 12, 15, 16 = 1

4. 1 unit square covering 3 = 1

5 1 unit square covering 8 = 1

6. 1 unit square covering 11 = 1

6. 45 degree square in center = 1

1+4+1+1+1+1+1=10

Posted, then read post #21 from aamph, edited to account for aamph's statement.

Edited by thoughtfulfellow
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There are 4 edge-edge cases:

1) parallel sharing face

2) parallel not sharing face

3) sharing edge

4) not parallel and not sharing edge

I think you missed it.

The name "witzar" reminds me of some Germanic root meaning "knowledge", although perhaps it also includes "vision". I see your point. If I were to number the blocks, bottom face 1-3 on top, 4-6 in middle, 7-9 on bottom, then the next layer 10-18, and the top layer 19-27, then I see that you are observing that I missed two edge-edge cases:

given that 16 is an edge center, the other edge center could be:

(case 1) 10 or 18

(case 2) 22, 4, 26, 8

(case 3) 20, 24, 2, 6

(case 4) 12

Edited by CaptainEd
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I just wrote a program to count unique shapes.

I get 22.

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I just wrote a program to count unique shapes.

Can you improve your program to print solutions as well?

I've found 20 and was pretty sure there are no more.

So now I'm curious.

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Can you improve your program to print solutions as well?

I've found 20 and was pretty sure there are no more.

So now I'm curious.

Sure

OK, here's how I number the cubes in the faces which are listed from front to back:

```
Face 1 (front):
1  2  3
4  5  6
7  8  9
Face 2 (middle):
10 11 12
13 14 15
16 17 18
Face 3 (back):
19 20 21
22 23 24
25 26 27
[code]
Here are the equivalence classes in terms of which pair of cubes are missing:
Class # 1 (24 in class -- cumulative=24):
(1,2)(1,4)(1,10)(2,3)(3,6)(3,12)(4,7)(6,9)(7,8)(7,16)(8,9)(9,18)(10,19)(12,21)(16,25)(18,27)(19,20)(19,22)(20,21)(21,24)(22,25)(24,27)(25,26)(26,27)
Class # 2 (12 in class -- cumulative=36):
(1,3)(1,7)(1,19)(3,9)(3,21)(7,9)(7,25)(9,27)(19,21)(19,25)(21,27)(25,27)
Class # 3 (24 in class -- cumulative=60):
(1,5)(1,11)(1,13)(3,5)(3,11)(3,15)(5,7)(5,9)(7,13)(7,17)(9,15)(9,17)(11,19)(11,21)(13,19)(13,25)(15,21)(15,27)(17,25)(17,27)(19,23)(21,23)(23,25)(23,27)
Class # 4 (24 in class -- cumulative=84):
(1,6)(1,16)(1,20)(2,7)(2,21)(3,8)(3,10)(3,24)(4,9)(4,19)(6,27)(7,18)(7,22)(8,25)(9,12)(9,26)(10,25)(12,19)(16,27)(18,21)(19,26)(20,27)(21,22)(24,25)
Class # 5 (24 in class -- cumulative=108):
(1,8)(1,12)(1,22)(2,9)(2,19)(3,4)(3,18)(3,20)(4,25)(6,7)(6,21)(7,10)(7,26)(8,27)(9,16)(9,24)(10,21)(12,27)(16,19)(18,25)(19,24)(20,25)(21,26)(22,27)
Class # 6 (12 in class -- cumulative=120):
(1,9)(1,21)(1,25)(3,7)(3,19)(3,27)(7,19)(7,27)(9,21)(9,25)(19,27)(21,25)
Class # 7 (8 in class -- cumulative=128):
(1,14)(3,14)(7,14)(9,14)(14,19)(14,21)(14,25)(14,27)
Class # 8 (24 in class -- cumulative=152):
(1,15)(1,17)(1,23)(3,13)(3,17)(3,23)(5,19)(5,21)(5,25)(5,27)(7,11)(7,15)(7,23)(9,11)(9,13)(9,23)(11,25)(11,27)(13,21)(13,27)(15,19)(15,25)(17,19)(17,21)
Class # 9 (24 in class -- cumulative=176):
(1,18)(1,24)(1,26)(2,25)(2,27)(3,16)(3,22)(3,26)(4,21)(4,27)(6,19)(6,25)(7,12)(7,20)(7,24)(8,19)(8,21)(9,10)(9,20)(9,22)(10,27)(12,25)(16,21)(18,19)
Class #10 (4 in class -- cumulative=180):
(1,27)(3,25)(7,21)(9,19)
Class #11 (24 in class -- cumulative=204):
(2,4)(2,6)(2,10)(2,12)(4,8)(4,10)(4,16)(6,8)(6,12)(6,18)(8,16)(8,18)(10,20)(10,22)(12,20)(12,24)(16,22)(16,26)(18,24)(18,26)(20,22)(20,24)(22,26)(24,26)
Class #12 (24 in class -- cumulative=228):
(2,5)(2,11)(4,5)(4,13)(5,6)(5,8)(6,15)(8,17)(10,11)(10,13)(11,12)(11,20)(12,15)(13,16)(13,22)(15,18)(15,24)(16,17)(17,18)(17,26)(20,23)(22,23)(23,24)(23,26)
Class #13 (12 in class -- cumulative=240):
(2,8)(2,20)(4,6)(4,22)(6,24)(8,26)(10,12)(10,16)(12,18)(16,18)(20,26)(22,24)
Class #14 (24 in class -- cumulative=264):
(2,13)(2,15)(4,11)(4,17)(5,10)(5,12)(5,16)(5,18)(6,11)(6,17)(8,13)(8,15)(10,23)(11,22)(11,24)(12,23)(13,20)(13,26)(15,20)(15,26)(16,23)(17,22)(17,24)(18,23)
Class #15 (12 in class -- cumulative=276):
(2,14)(4,14)(6,14)(8,14)(10,14)(12,14)(14,16)(14,18)(14,20)(14,22)(14,24)(14,26)
Class #16 (12 in class -- cumulative=288):
(2,16)(2,24)(4,18)(4,20)(6,10)(6,26)(8,12)(8,22)(10,26)(12,22)(16,24)(18,20)
Class #17 (24 in class -- cumulative=312):
(2,17)(2,23)(4,15)(4,23)(5,20)(5,22)(5,24)(5,26)(6,13)(6,23)(8,11)(8,23)(10,15)(10,17)(11,16)(11,18)(11,26)(12,13)(12,17)(13,18)(13,24)(15,16)(15,22)(17,20)
Class #18 (12 in class -- cumulative=324):
(2,18)(2,22)(4,12)(4,26)(6,16)(6,20)(8,10)(8,24)(10,24)(12,26)(16,20)(18,22)
Class #19 (6 in class -- cumulative=330):
(2,26)(4,24)(6,22)(8,20)(10,18)(12,16)
Class #20 (12 in class -- cumulative=342):
(5,11)(5,13)(5,15)(5,17)(11,13)(11,15)(11,23)(13,17)(13,23)(15,17)(15,23)(17,23)
Class #21 (6 in class -- cumulative=348):
(5,14)(11,14)(13,14)(14,15)(14,17)(14,23)
Class #22 (3 in class -- cumulative=351):
(5,23)(11,17)(13,15)
[/spoiler]```

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