# 2x2x2 Scramble Math Question

#### reThinking the Cube

##### Member
Here is the well known distribution table for min#of moves to solve 2x2x2.

n f
0
0 1
1 9
2 54
3 321
4 1847
5 9992
6 50136
7 227536
8 870072
9 1887748
10 623800
11 2644
12 0
13 0
14 0
Total 3674160

How many of these 3,674,160 positions would yield a scrambled cube with no more than one correctly placed piece? (simultaneously - nothing else solved except for 1 arbitrary reference corner).

+10 Bonus: a table (min#of moves to solve) like the one above for just those positions.
.

#### qqwref

##### Member
I expect Lucas to come up with a Mathematica one-liner in an hour or two, so I'm not going to try to calculate this myself. It's an interesting problem, though.

#### ianini

##### Member
I expect Lucas to come up with a Mathematica one-liner in an hour or two, so I'm not going to try to calculate this myself. It's an interesting problem, though.
I actually thought you were going to be the one to answer this question.

#### TomZ

##### Member
The distribution of positions on the 2x2x2 that have at least 2/8 pieces solved is:

Code:
#0: 1 position
#1: 9 positions
#2: 54 positions
#3: 165 positions
#4: 827 positions
#5: 3740 positions
#6: 17478 positions
#7: 69586 positions
#8: 250422 positions
#9: 520180 positions
#10: 177808 positions
#11: 1240 positions
#12: 0 positions
That means that 1041510/3674160 or around 28% of the positions of the 2x2x2 have at least two pieces solved.

#### cmhardw

That means that 1041510/3674160 or around 28% of the positions of the 2x2x2 have at least two pieces solved.
I just wanted to post that I don't agree with this calculation for how many positions there are with at least two solved pieces. If you count the number of positions where exactly 1 corner is solved, essentially a derangement of the other pieces after the 1st corner is rotated to be solved, you get:

7!*3^6 - (7 C 1)*6!*3^5 + (7 C 2)*5!*3^4 - (7 C 3)*4!*3^3 + (7 C 4)*3!*3^2 - (7 C 5)*2!*3^1 + 1

= 2632645

(Number of positions with exactly 1 solved piece) + (Number of positions with at least two solved pieces) = Total number of solved positions

Keep in mind that for a 2x2x2 the total number of positions is calculated by "solving" one corner as a reference point. Therefore:

2632645 + 1041510 = 3674155 /= 3674160

Where are the missing 5 positions? Was this a simple miscalculation? I will admit my ignorance in saying that I can't tell if the error is mine, or in that table you posted.

--edit--
I think I found the error, and it was mine.

Derangements of the other 7 corners should be:
7!*3^6 - (7 C 1)*6!*3^5 + (7 C 2)*5!*3^4 - (7 C 3)*4!*3^3 + (7 C 4)*3!*3^2 - (7 C 5)*2!*3^1 + (7 C 6)*1!*3^0 - 1

= 2632650

And therefore:
2632650 + 1041510 = 3674160

And I agree with the calculation.

-------------------

Chris

Last edited:

#### qqwref

##### Member
I think calculations of this type would be disregarding positions in which two pieces (neither of which are the fixed piece) are solved relative to each other, though. I think the best way to do this would be to just exhaustively generate and provide an optimal length for each position satisfying this property.

#### reThinking the Cube

##### Member
I think calculations of this type would be disregarding positions in which two pieces (neither of which are the fixed piece) are solved relative to each other, though.
MG is correct. There are 8 possibilities (not just 1) that can be used for the solved reference piece. For ANY one of them, if another piece is found that is solved relative to that reference, then the position as a whole contains more than 1 solved piece by cube rotations alone.
.

#### cuBerBruce

##### Member
I think calculations of this type would be disregarding positions in which two pieces (neither of which are the fixed piece) are solved relative to each other, though.
MG is correct. There are 8 possibilities (not just 1) that can be used for the solved reference piece. For ANY one of them, if another piece is found that is solved relative to that reference, then the position as a whole contains more than 1 solved piece by cube rotations alone.
.
I get the following distribution:

Code:
moves   n=1      n=2      n=3      n=4      n=5      n=6      n=8   totals   n >= 2

0        0        0        0        0        0        0        1        1        1
1        0        0        0        9        0        0        0        9        9
2        0       54        0        0        0        0        0       54       54
3       48      186        0       87        0        0        0      321      273
4      264      956      528       99        0        0        0     1847     1583
5     1092     6302     1944      654        0        0        0     9992     8900
6     7092    35976     6408      660        0        0        0    50136    43044
7    44904   151900    26592     3948      192        0        0   227536   182632
8   186864   577424    99688     5328      768        0        0   870072   683208
9   451728  1235424   186288    13524      784        0        0  1887748  1436020
10   151512   401360    62496     7932      384      116        0   623800   472288
11      240     1776      328      276        0       24        0     2644     2404
------  -------   ------    -----      ---      ---        -  -------  -------
843744  2411358   384272    32517     2128      140        1  3674160  2830416

Edit: average number of moves added, per Stefan's request
8.8199   8.7447   8.6894   8.6794   8.6391  10.1714   0.0000   8.7556   8.7364

Last edited:

#### reThinking the Cube

##### Member
I think calculations of this type would be disregarding positions in which two pieces (neither of which are the fixed piece) are solved relative to each other, though.
MG is correct. There are 8 possibilities (not just 1) that can be used for the solved reference piece. For ANY one of them, if another piece is found that is solved relative to that reference, then the position as a whole contains more than 1 solved piece by cube rotations alone.
.
I get the following distribution:

Code:
moves   n=1      n=2      n=3      n=4      n=5      n=6      n=8   totals   n >= 2

0        0        0        0        0        0        0        1        1        1
1        0        0        0        9        0        0        0        9        9
2        0       54        0        0        0        0        0       54       54
3       48      186        0       87        0        0        0      321      273
4      264      956      528       99        0        0        0     1847     1583
5     1092     6302     1944      654        0        0        0     9992     8900
6     7092    35976     6408      660        0        0        0    50136    43044
7    44904   151900    26592     3948      192        0        0   227536   182632
8   186864   577424    99688     5328      768        0        0   870072   683208
9   [B]451728[/B]  1235424   186288    13524      784        0        0  1887748  1436020
10   151512   401360    62496     7932      384      116        0   623800   472288
11      240     1776      328      276        0       24        0     2644     2404
------  -------   ------    -----      ---      ---        -  -------  -------
843744  2411358   384272    32517     2128      140        1  3674160  2830416

Edit: average number of moves added, per Stefan's request
8.8199   8.7447   8.6894   8.6794   8.6391  10.1714   0.0000   8.7556   8.7364
+10, and I owe you a case of your favorite Bruce! Very nice, I must have spent over an hour just looking/staring at this table. Pretending that all of the scrambles for 2x2x2 came from the bolded item above, I was wondering if it would be possible to reduce those, by rotations, reflections, inverses, and 1 face turn conjugations to get a bare bones minimum# of algs that would have to be learned to one-look solve just those 451,728 positions?

#### cuBerBruce

##### Member
+10, and I owe you a case of your favorite Bruce! Very nice, I must have spent over an hour just looking/staring at this table. Pretending that all of the scrambles for 2x2x2 came from the bolded item above, I was wondering if it would be possible to reduce those, by rotations, reflections, inverses, and 1 face turn conjugations to get a bare bones minimum# of algs that would have to be learned to one-look solve just those 451,728 positions?
There are a total of 40296 equivalence classes when reducing the 2x2x2 positions by symmetry and antisymmetry (that is, with cube rotations, reflections, and inverses). The case 451728 positions for n=1, distance=9 correspond to 4820 of these equivalence classes. I also calculated that the 1235424 positions for n=2 distance=9 correspond to 13353 of these equivalence classes. (I have not used conjugation by 1 turn ("face" turns???).)

I note that my "n" is the number of solved pieces including the piece used as a reference, and the reference is chosen that maximizes the number of solved pieces. So n=1 means that only the reference piece is considered solved regardless of which piece is chosen as the reference. n=2 means there is a choice of reference piece such that 1 other piece is solved with respect to it.

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#### reThinking the Cube

##### Member
There are a total of 40296 equivalence classes when reducing the 2x2x2 positions by symmetry and antisymmetry (that is, with cube rotations, reflections, and inverses). The case 451728 positions for n=1, distance=9 correspond to 4820 of these equivalence classes.
Bruce, can you include equivalence class numbers along with position counts? I am curious to see the equiv.#'s working backwards for moves=0,1,2,3,4...,

Also, is it possible for you to determine how many of these can be solved with a 2-generator alg?

#### miniGOINGS

##### Member
Also, is it possible for you to determine how many of these can be solved with a 2-generator alg?
I 2-Gen alg would only work if 2 corners are solved next to each other, and corner permutation is correct, which I believe is 1/3. I would also be interested in this.

#### cuBerBruce

##### Member
There are a total of 40296 equivalence classes when reducing the 2x2x2 positions by symmetry and antisymmetry (that is, with cube rotations, reflections, and inverses). The case 451728 positions for n=1, distance=9 correspond to 4820 of these equivalence classes.
Bruce, can you include equivalence class numbers along with position counts? I am curious to see the equiv.#'s working backwards for moves=0,1,2,3,4...,

Also, is it possible for you to determine how many of these can be solved with a 2-generator alg?
The table for symmetry/antisymmetry equivalence classes is:

Code:
moves  n=1     n=2     n=3     n=4     n=5     n=6     n=8
0       0       0       0       0       0       0       1
1       0       0       0       2       0       0       0
2       0       4       0       0       0       0       0
3       3       7       0       4       0       0       0
4       7      24       7       3       0       0       0
5      24      98      23      12       0       0       0
6      93     434      72      13       0       0       0
7     511    1710     293      59       3       0       0
8    2002    6255    1073      71       9       0       0
9    4820   13353    2038     187      12       0       0
10    1642    4501     726     129       9       6       0
11       3      36       8       7       0       2       0
Also, is it possible for you to determine how many of these can be solved with a 2-generator alg?
I 2-Gen alg would only work if 2 corners are solved next to each other, and corner permutation is correct, which I believe is 1/3. I would also be interested in this.
If 2 adjacent corners are correctly solved with respect to each other, there are 6! = 720 permutations for the last 6 pieces of which I believe 1/6 are solvable turning only the layers not containing the two solved pieces.

Some positions may have more than one pair of correctly solved corners. Which pair is chosen to be fixed may affect whether or not the position is solvable turning only the layers not containing the chosen pair.

(My phone & internet service has been out for about 49 hours so my responses may be sporadic until Verizon fixes my connection.)

EDIT:
2-gen information

First I simply calculated all the 2-gen positions. I give results for both <U,R> positions only (solved within <U,R>), and for all 2-gen positions when allowing any of the 12 choices for the two layers to be turned. I note that the total number of 2-gen solvable positions is a prime number (316699).

Code:
moves    <U,R>  all 2-gen

0         1         1
1         6         9
2        18        54
3        53       267
4       148      1308
5       400      4440
6       910     10536
7      1882     22008
8      3276     38088
9      4628     51124
10      6198     65828
11      6325     67104
12      4352     46392
13       941      9348
14        22       192
-----    ------
29160    316699
Next, I give the results (number of 2-gen solvable positions) with respect to the number of solved pieces and moves required to solve (3-gen or 2-gen).

Code:
Number of 2-gen solvable positions with respect to 3-gen distance and number of solved pieces.

moves   n=1     n=2     n=3     n=4     n=5     n=6     n=8       totals
(3-gen)
0       0       0       0       0       0       0       1            1
1       0       0       0       9       0       0       0            9
2       0      54       0       0       0       0       0           54
3       0     180       0      87       0       0       0          267
4       0     684     528      96       0       0       0         1308
5       0    2568    1296     624       0       0       0         4488
6       0    9240    2352     528       0       0       0        12120
7       0   24288    7104    2820       0       0       0        34212
8       0   62888   26472    1824      48       0       0        91232
9       0  100528   35784    3996      48       0       0       140356
10       0   22416    8208    1680      16      56       0        32376
11       0     144      96      24       0      12       0          276
-  ------   -----   -----     ---      --       -       ------
totals    0  222990   81840   11688     112      68       1       316699

Number of 2-gen solvable positions with respect to 2-gen distance and number of solved pieces.

moves   n=1     n=2     n=3     n=4     n=5     n=6     n=8       totals
(2-gen)
0       0       0       0       0       0       0       1            1
1       0       0       0       9       0       0       0            9
2       0      54       0       0       0       0       0           54
3       0     180       0      87       0       0       0          267
4       0     684     528      96       0       0       0         1308
5       0    2520    1296     624       0       0       0         4440
6       0    8232    1968     336       0       0       0        10536
7       0   15432    4944    1632       0       0       0        22008
8       0   25440   12024     576      48       0       0        38088
9       0   36784   12624    1668      48       0       0        51124
10       0   45548   18720    1512      16      32       0        65828
11       0   48096   16584    2388       0      36       0        67104
12       0   33456   10872    2064       0       0       0        46392
13       0    6468    2184     696       0       0       0         9348
14       0      96      96       0       0       0       0          192
-  ------   -----   -----     ---      --       -       ------
0  222990   81840   11688     112      68       1       316699

Last edited:

#### reThinking the Cube

##### Member
Nice work! These tables are very interesting.

Bruce, could you generate for the #of moves away, all 3674160 are from any n>=4 position? That would reveal the potential for human transitions into the easier known cases.

The table for symmetry/antisymmetry equivalence classes is:

Code:
moves  n=1     n=2     n=3     n=4     n=5     n=6     n=8
0       0       0       0       0       0       0       1
1       0       0       0       2       0       0       0
2       0       4       0       0       0       0       0
3       3       7       0       4       0       0       0
4       7      24       7       3       0       0       0
5      24      98      23      12       0       0       0
6      93     434      72      13       0       0       0
7     511    1710     293      59       3       0       0
8    2002    6255    1073      71       9       0       0
9    4820   13353    2038     187      12       0       0
10    1642    4501     726     129       9       6       0
11       3      36       8       7       0       2       0
Any chance of getting diagrams, or specific cycle descriptions for those n=4,5,6 above?
.

#### cuBerBruce

##### Member
Nice work! These tables are very interesting.

Bruce, could you generate for the #of moves away, all 3674160 are from any n>=4 position? That would reveal the potential for human transitions into the easier known cases.

The table for symmetry/antisymmetry equivalence classes is:

Code:
moves  n=1     n=2     n=3     n=4     n=5     n=6     n=8
0       0       0       0       0       0       0       1
1       0       0       0       2       0       0       0
2       0       4       0       0       0       0       0
3       3       7       0       4       0       0       0
4       7      24       7       3       0       0       0
5      24      98      23      12       0       0       0
6      93     434      72      13       0       0       0
7     511    1710     293      59       3       0       0
8    2002    6255    1073      71       9       0       0
9    4820   13353    2038     187      12       0       0
10    1642    4501     726     129       9       6       0
11       3      36       8       7       0       2       0
Any chance of getting diagrams, or specific cycle descriptions for those n=4,5,6 above?
.
OK, I have created a file with all the distinct cases (with respect to symmetry and antisymmetry) with 4 or more pieces solved (excluding the trivial case of all 8 solved).

I provide the cycle structure of one representative for each case. Unfortunately, the choice of representative is rather arbitrary. Cycles can be of one of two types: "oriented" cycles and "misoriented" cycles. For "oriented" cycles, the cycle length is equal to the number of cubies in the cycle. For "misoriented" cycles, the cycle length is 3 times the number of cubies in the cycle. For "misoriented" cycles, the full cycle is not included - only up to the point where the initial cubie is repeated (but with different "orientation"). An ellipsis ("...") is used to indicate the rest of the cycle is omitted.

#### Attachments

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#### reThinking the Cube

##### Member
Nice work! These tables are very interesting.

Bruce, could you generate for the #of moves away, all 3674160 are from any n>=4 position? That would reveal the potential for human transitions into the easier known cases.

The table for symmetry/antisymmetry equivalence classes is:

Code:
moves  n=1     n=2     n=3     n=4     n=5     n=6     n=8
0       0       0       0       0       0       0       1
1       0       0       0       2       0       0       0
2       0       4       0       0       0       0       0
3       3       7       0       4       0       0       0
4       7      24       7       3       0       0       0
5      24      98      23      12       0       0       0
6      93     434      72      13       0       0       0
7     511    1710     293      59       3       0       0
8    2002    6255    1073      71       9       0       0
9    4820   13353    2038     187      12       0       0
10    1642    4501     726     129       9       6       0
11       3      36       8       7       0       2       0
Any chance of getting diagrams, or specific cycle descriptions for those n=4,5,6 above?
.
OK, I have created a file with all the distinct cases (with respect to symmetry and antisymmetry) with 4 or more pieces solved (excluding the trivial case of all 8 solved).

I provide the cycle structure of one representative for each case. Unfortunately, the choice of representative is rather arbitrary. Cycles can be of one of two types: "oriented" cycles and "misoriented" cycles. For "oriented" cycles, the cycle length is equal to the number of cubies in the cycle. For "misoriented" cycles, the cycle length is 3 times the number of cubies in the cycle. For "misoriented" cycles, the full cycle is not included - only up to the point where the initial cubie is repeated (but with different "orientation"). An ellipsis ("...") is used to indicate the rest of the cycle is omitted.
Cycle structure file is remarkable. Great work once again, Bruce. Here is my 2-cents on how the case representations can be made less arbitrary, and maybe further reduced by 1-turn conjugation.

n = 6, 10 moves
(DFR FRD ...)(DRB BDR ...)
(DLF LFD ...)(DRB BDR ...)
(ULB LBU ...)(DFR RDF ...)
(DFR DRB)
(DLF RBD)
(ULB DFR)

n = 6, 11 moves
(DFR RBD)
(DLF DRB)

The bold ones above can be made equivalent to their respective n=6 cases, IF 1-turn conjugate moves are also allowed. It is interesting to note that the cross-cube swap (ULB DFR) can be oriented with just cube rotations, so that only 1 case is needed for all 5 of those corner 2-cycles. {(DFR DRB),(DLF RBD),(ULB DFR),(DFR RBD),(DLF DRB)} Likewise the other 3 cases (corner twisters) can be reduced to 1 - (ULB LBU ...)(DFR RDF ...)

The reductions should be even more dramatic for n=4,5. At first, I wasn't sure if a system could be adopted to (non-arbitrarily) conjugate these for the purpose of representation, but I have an idea now. With n=4 cubes there are 6 patterns for the locations of the remaining unsolved pieces.

LL = (UFR, ULF, UBL, URB)
TL = (UFR, ULF, DRF, URB)
FL = (UBL, ULF, DRF, URB)
XL = (UBL, ULF, DRF, DBR)
SL = (UFR, ULF, DBR, URB)
ZL = (UFR, DRF, UBL, URB)

These unsolved piece patterns can all be made equivalent to each other by some 1-2 turn conjugate, but 2-turns would make actual case recognition very confusing. So for example, with LL as the base pattern, then the TL "Tripod" cases would 1st require 2-move conjugate turns (i.e. R' U' z') to form some equivalent LL case. Too messy to make sense of. But note that the SL and ZL patterns can always be conjugated to ANY of the others with just 1-turn! And since SL is by equivalencies = ZL, ALL the n>=4 cases could be represented by forcing the unsolved pieces by at most a 1-turn conjugation, +cube rotations, to occupy ONLY the locations in ZL = (UFR, DRF, UBL, URB). Maybe this is viable.

..

Last edited:

#### TCUBER

##### Member
The distribution of positions on the 2x2x2 that have at least 2/8 pieces solved is:

Code:
#0: 1 position
#1: 9 positions
#2: 54 positions
#3: 165 positions
#4: 827 positions
#5: 3740 positions
#6: 17478 positions
#7: 69586 positions
#8: 250422 positions
#9: 520180 positions
#10: 177808 positions
#11: 1240 positions
#12: 0 positions
That means that 1041510/3674160 or around 28% of the positions of the 2x2x2 have at least two pieces solved.
It is 2-45!*8
So I have to agree with you

#### cuBerBruce

##### Member
The reductions should be even more dramatic for n=4,5. At first, I wasn't sure if a system could be adopted to (non-arbitrarily) conjugate these for the purpose of representation, but I have an idea now. With n=4 cubes there are 6 patterns for the locations of the remaining unsolved pieces.

LL = (UFR, ULF, UBL, URB)
TL = (UFR, ULF, DRF, URB)
FL = (UBL, ULF, DRF, URB)
XL = (UBL, ULF, DRF, DBR)
SL = (UFR, ULF, DBR, URB)
ZL = (UFR, DRF, UBL, URB)

These unsolved piece patterns can all be made equivalent to each other by some 1-2 turn conjugate, but 2-turns would make actual case recognition very confusing. So for example, with LL as the base pattern, then the TL "Tripod" cases would 1st require 2-move conjugate turns (i.e. R' U' z') to form some equivalent LL case. Too messy to make sense of. But note that the SL and ZL patterns can always be conjugated to ANY of the others with just 1-turn! And since SL is by equivalencies = ZL, ALL the n>=4 cases could be represented by forcing the unsolved pieces by at most a 1-turn conjugation, +cube rotations, to occupy ONLY the locations in ZL = (UFR, DRF, UBL, URB). Maybe this is viable.

..
I note that you're missing the (ULF, URB, DRF, DLB) case, and that case is only adjacent to XL. I also note that SL and ZL are mirrors of each other, so they are really the same case when reducing by symmetries of the cube.

So I can probably sort out the positions in the file, particular which ones are SL/ZL and the (ULF, URB, DRF, DLB) case, and similarly sort out for the n=5 and n=6 positions.

For n = 5, I get three cases, one of which is adjacent to the other two.

Bruce, could you generate for the #of moves away, all 3674160 are from any n>=4 position? That would reveal the potential for human transitions into the easier known cases.
It was rather easy to alter my code to treat all positions with 4 or more pieces solved with respect to each other as if they were "solved," and perform a breadth-first search to calculate how many moves are required to get to this (funny notion of) "solved" from each possible state.

Code:
Number of moves required to get at least 4 cubies of the 2x2x2 solved
with respect to each other, in terms of number of positions and
number of positions as reduced by symmetry and antisymmetry.

moves   positions    reduced
-----   ---------    -------
0         34786        529
1        247092       2863
2        961942      10606
3       1884540      20389
4        544192       5884
5          1608         25
-------      -----
total     3674160      40296
The 25 cases requiring 5 moves correspond to these positions (in cycle notation).

Code:
(UFL LUF ...)(URF LBU LFD FRD RFU ...)(UBR RBD)
(UFL RUB DRB BUL FUR LFD RDF)
(UFL BUL UBR URF)(DLF DRB LFD ...)(DFR RDF ...)
(UFL FLU ...)(URF DFR DRB LFD LBU UBR FUR ...)
(UFL FRD ULB FLU ...)(URF DLF UBR)(DRB BDR ...)
(UFL URF BDR FLU ...)(UBR LBU LFD BRU ...)(DFR FRD ...)
(UFL BUL FDL UBR URF BDR FLU ...)(DFR RDF ...)
(UFL FDL UBR FLU ...)(URF RFU ...)(ULB DRB DFR LBU ...)
(UFL FRD BRU FLU ...)(URF FDL ULB RFU ...)(DRB RBD ...)
(UFL BRU RFU FLU ...)(ULB RDF LBU ...)(DLF FDL ...)(DRB BDR ...)
(UFL RBD LBU FUR RDF FLU ...)(DLF FDL ...)
(UFL RUB RFU FLU ...)(ULB RDF BUL ...)(DLF FDL ...)(DRB RBD ...)
(UFL DRB LUF ...)(URF BRU)(ULB LBU ...)(DLF DFR)
(UFL UBR LUF ...)(URF RDF DRB FDL FUR ...)(ULB BUL ...)
(UFL BRU LUF ...)(URF DRB FDL FUR ...)(ULB FRD BUL ...)
(UFL FDL BDR UBR FLU ...)(URF RFU ...)(ULB BUL ...)(DFR RDF ...)
(UFL DLF BDR RUB)(ULB LBU ...)(DFR RDF ...)
(UFL RDF)(UBR DLF)(ULB BDR)
(UFL LUF ...)(URF BUL RDF FDL RFU ...)(UBR RBD)
(URF FUR ...)(ULB FRD LBU ...)(DLF FDL ...)(DRB RBD ...)
(UFL LBU LUF ...)(UBR BRU ...)(DLF FDL ...)(DFR BDR FRD ...)
(UFL DRB)(UBR BUL DLF FRD)
(UFL LBU RDF FDL BDR BRU LUF ...)(URF RFU ...)
(UFL LBU FRD LFD URF DRB RUB)
(UFL LBU RDF LFD URF DRB RUB)

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