[NEW] Square-1 Method: JQ Method v2

[JackTriton]

Hi, this is Jack Triton
Recently, I bought mf8 Square-1 and got around 20 seconds with this method (which I would never been able to reach with Vandenbergh)
I call this method JQ method last week, got no reactions so I dig up and find several problems
Now I can say this method is fully functional as speed method

JQ-Method
This method goes along something like this

1. Cube Shape, CSP(to avoid parity)
2. First Block (Forming Corner - Edge - Corner set on DL, color can be anything)
3. CPSB (Corner Permute Second Block) [6]
4. Pairing Edges (Pair edges with opposite one) [3]
5. L6PE (Last 6 Paired Edges) [1 Parity case]

Pros

• Like Roux, Screw or Lin, having low number of algorithms (at least 15 algorithms) results in faster recognition of cases
• CSP is not necessary (although to fasten the method, still needs to be implemented)
• Easy to look-a-head cases, especially for CPSB and Pairing Edges
• Relying on 90 degree movements mostly after First Block

Cons

• Block building can be difficult for beginners to get used to it
• While CPSB, forming corner-edge set would need practices to get used to it

First Block

After shaping the cube, you need to create corner - edge - corner set on DL: which is basic thing for Lin and Screw​

Color on above showing that the color of first block can be anything if you use D layer pieces​

CPSB [6]

First you have to form RBD corner-edge set on UFL and RFD corner on DBL​

Easy to say, it's the same state as doing (1, 0) / (3, 0) / from solved state​

​

After forming, do latter algorithms corresponding how the color on face relates on the other 3 corners​

​

​

Same color on both sides : ​

(4, -3) / (0, -3) / (3, 0) / (-3, 0) / (3, 6) / (-1, 0)​

(1, 0) U D' / D' / U / U' / U D2 / (-1, 0)​

​

Same color on R :​

(1, 0) / (-3, 0) / (-1, 0)​

(1, 0) / U' / (-1, 0)​

​

Same color on B :​

(1, 0) / (0, -3) / (0, -3) / (3, 0) / (-3, 0) / (-3, 6) / (-1, 0)​

(1, 0) / D' / D' / U / U' / U' D2 / (-1, 0)​

​

​

Opposite color on both sides :​

(1, -3) / (3, 0) / (-3, 0) / (-3, 3) / (-1, 0)​

(1, 0) D' / U / U' / U' D / (-1, 0)​

​

Opposite color on R :​

(4, 6) / (3, 0) / (0, 3) / (6, 3) / (6, -3) / (-1, 0)
(1, 0) U D2 / U / D / U2 D / U2 D' / (-1, 0)​

​

​

Opposite color on B :​

(4, 0) / (0, 3) / (3, 6) / (0, 3) / (0, 3) / (-1, 0)​

(1, 0) U / D / U D2 / D / D / (-1, 0)​

Doing this solves corner permutation and Second Block​

Pairing Edges [3]

Pairing Edge consists only 3 algorithms and the goal is to pair each edges with opposite edge​

For example, White-Orange Edge can be paired with White-Red Edge​

​

For images, edges with the same color are pair​

​

UF and UL are pair and UB and DF are pair (1 adjacent) :​

(1, 0) / (-3, 0) / (-3, 0) / (-4, -1) / (-2, 1) / (-3, 0) /​

(1, 0) / 2( U' / ) U' (-1, -1) / U' (1, 1) / U' /​

​

UF and UB are pair and UR and DF are pair (1 opposite) :​

(4, 0) / (3, 0) / (-3, 0) / (-1, -1) / (4, 1) / (-3, 0) /​

(1, 0) 2( U / ) U' / (-1, -1) / U (1, 1) / U' /​

​

UF and UR are pair and UB and UL are pair (2 adjacent) :​

(1, 0) / (2, -1) / (4, 1) / (3, 0) / (-3, 0) / (5, -1) / (-2, 1) /​

(1, 0) / U (-1, -1) / U (1, 1) / U / U' / U2 (-1, -1) / U' (1, 1) /​

Now you have paired all edges and I suggest to solve middle layer orient on this phase​

L6PE [1 Parity]

The last of all is L6PE​

This consists only 1 parity algorithm which you can avoid by doing CSP​

Other than that, you can slot every single paired edges using (1, 0) / (-1, -1) / (0, 1)​

​

Parity (UL and UR)​

​

/ (3, 3) / (-1, 0) /(2, -4) / (4, -2) / (0, -2) / (-4, 2) / (1, -5) / (3, 0) / (3, 3) / (3, 0)
Thanks for reading! I hope this will help you to solve Square-1 much easier
If you’re interested, please leave a comment and share your ideas!

 

[Caleb Valenzuela]

why did you call i the JQ method?

 

[JackTriton]

why did you call i the JQ method?

Initially I wanted to call it Jack method based off by my name but I thought it’s too normal to call it so I just came up the name with shorting Jack’s Square-1 Method

 

[Caleb Valenzuela]

Initially I wanted to call it Jack method based off by my name but I thought it’s too normal to call it so I just came up the name with shorting Jack’s Square-1 Method

what about naming it the JT method? (stands for Jack Triton’s Square-1 method)
edit: JTQ makes more sense but it’s longer

 

[JackTriton]

Example solve:

SCR:
(-5,0)/ (5,2)/ (-5,1)/ (5,-4)/ (6,-3)/ (-5,-3)/ (3,0)/ (-4,-5)/ (-3,0)/ (-4,-1)/ (-4,0)/ (-4,0)/

Solution:
(2, 0) / (0, 3) / (1, 2) / (-3, 6) / -// Cube Shape
(0, -1) / (1, -5) / (-1, -3) -// First Block
(0, -1) / (-3, 0) / (1, -1) / -// Recognize case
/ (0, -3) / (0, -3) / (3, 0) / (-3, 0) / (3, 6) / (6, 0) / (-1, 0) -// CPSB
(-2, 0) / (3, 0) / (-3, 0) / (-1, -1) / (4, 1) / (-3, 0) / -// Pairing Edges
(3, 0) / (-1, -1) / (3, 0) / (1, 1) / (2, 3) / (3, 3) / (-1, 0) / (2, -4) / (4, -2) / (0, -2) / (-4, 2) / (1, -5) / (3, 0) / (3, 3) / (3, 0) -// L6PE

 

[JackTriton]

what about naming it the JT method? (stands for Jack Triton’s Square-1 method)
edit: JTQ makes more sense but it’s longer

JT in Japan is Japan Train thing so since I live in Japan, I wanted to avoid that
Also, I have several methods for other cubes so I wanted to make contrast with them as well

 

[abunickabhi]

Hi, this is Jack Triton
Recently, I bought mf8 Square-1 and got around 20 seconds with this method (which I would never been able to reach with Vandenbergh)
I call this method JQ method last week, got no reactions so I dig up and find several problems
Now I can say this method is fully functional as speed method

JQ-Method
This method goes along something like this

1. Cube Shape, CSP(to avoid parity)
2. First Block (Forming Corner - Edge - Corner set on DL, color can be anything)
3. CPSB (Corner Permute Second Block) [6]
4. Pairing Edges (Pair edges with opposite one) [3]
5. L6PE (Last 6 Paired Edges) [1 Parity case]

Pros

• Like Roux, Screw or Lin, having low number of algorithms (at least 15 algorithms) results in faster recognition of cases
• CSP is not necessary (although to fasten the method, still needs to be implemented)
• Easy to look-a-head cases, especially for CPSB and Pairing Edges
• Relying on 90 degree movements mostly after First Block

Cons

• Block building can be difficult for beginners to get used to it
• While CPSB, forming corner-edge set would need practices to get used to it

First Block
View attachment 36266

After shaping the cube, you need to create corner - edge - corner set on DL: which is basic thing for Lin and Screw​

Color on above showing that the color of first block can be anything if you use D layer pieces​

CPSB [6]
View attachment 36267

First you have to form RBD corner-edge set on UFL and RFD corner on DBL​

Easy to say, it's the same state as doing (1, 0) / (3, 0) / from solved state​

​

After forming, do latter algorithms corresponding how the color on face relates on the other 3 corners​

​

View attachment 36268​

Same color on both sides : ​

(4, -3) / (0, -3) / (3, 0) / (-3, 0) / (3, 6) / (-1, 0)​

(1, 0) U D' / D' / U / U' / U D2 / (-1, 0)​

View attachment 36269​

Same color on R :​

(1, 0) / (-3, 0) / (-1, 0)​

(1, 0) / U' / (-1, 0)​

View attachment 36270​

Same color on B :​

(1, 0) / (0, -3) / (0, -3) / (3, 0) / (-3, 0) / (-3, 6) / (-1, 0)​

(1, 0) / D' / D' / U / U' / U' D2 / (-1, 0)​

​

View attachment 36271​

Opposite color on both sides :​

(1, -3) / (3, 0) / (-3, 0) / (-3, 3) / (-1, 0)​

(1, 0) D' / U / U' / U' D / (-1, 0)​

View attachment 36272​

Opposite color on R :​

(1, -3) / (3, 0) / (-3, 0) / (-3, 3) / (-1, 0)​

(1, 0) D' / U / U' / U' D / (-1, 0)​

​

View attachment 36273​

Opposite color on B :​

(4, 0) / (0, 3) / (3, 6) / (0, 3) / (0, 3) / (-1, 0)​

(1, 0) U / D / U D2 / D / D / (-1, 0)​

Doing this solves corner permutation and Second Block​

Pairing Edges [3]
View attachment 36291

Pairing Edge consists only 3 algorithms and the goal is to pair each edges with opposite edge​

For example, White-Orange Edge can be paired with White-Red Edge​

​

For images, edges with the same color are pair​

View attachment 36292​

UF and UL are pair and UB and DF are pair (1 adjacent) :​

(1, 0) / (-3, 0) / (-3, 0) / (-4, -1) / (-2, 1) / (-3, 0) /​

(1, 0) / 2( U' / ) U' (-1, -1) / U' (1, 1) / U' /​

View attachment 36293​

UF and UB are pair and UR and DF are pair (1 opposite) :​

(4, 0) / (3, 0) / (-3, 0) / (-1, -1) / (4, 1) / (-3, 0) /​

(1, 0) 2( U / ) U' / (-1, -1) / U (1, 1) / U' /​

View attachment 36294​

UF and UR are pair and UB and UL are pair (2 adjacent) :​

(1, 0) / (2, -1) / (4, 1) / (3, 0) / (-3, 0) / (5, -1) / (-2, 1) /​

(1, 0) / U (-1, -1) / U (1, 1) / U / U' / U2 (-1, -1) / U' (1, 1) /​

Now you have paired all edges and I suggest to solve middle layer orient on this phase​

L6PE [1 Parity]

The last of all is L6PE​

This consists only 1 parity algorithm which you can avoid by doing CSP​

Other than that, you can slot every single paired edges using (1, 0) / (-1, -1) / (0, 1)​

​

Parity (UL and UR)​

View attachment 36296​

/ (3, 3) / (-1, 0) /(2, -4) / (4, -2) / (0, -2) / (-4, 2) / (1, -5) / (3, 0) / (3, 3) / (3, 0)
Thanks for reading! I hope this will help you to solve Square-1 much easier
If you’re interested, please leave a comment and share your ideas!

Interesting method, thanks for sharing yo!

 

[xyzzy]

Relying on 90 degree movements mostly after First Block

This isn't really a pro of this specific method because almost every square-1 method has this feature too… It's kind of the whole point of doing cubeshape first.

------

Rather than doing CP together with SB, there's a pretty substantial slice count reduction if you split and rejoin the steps like this:
1. Cubeshape (or CSP)
2. First block
3. Second block (absolutely no influencing whatsoever needed here!)
4. CP + pairing edges (16 algs)
5. L6PE

Original JQ v2 method:
CPSB takes 26/6 = 4.333 slices on average, but always takes at least 2 slices, so effectively this solves CP in 2.333 slices.
Edge pairing takes 86/15 = 5.733 slices on average.
Total: 8.067 slices.

Modified method:
CP + edge pairing takes 550/90 = 6.111 slices on average.

Realistically, this gives more freedom in building the second block as well, so the true move count reduction should be even greater.

Unfortunately, in the process of labbing this, I was reminded that I hate learning squan algs so I'll never use either the original method or my modification of it…

Spoiler: algs and blah

I used Batch Solver to gen the algs, restricted to a move set where the puzzle never leaves cubeshape.

Puzzle definition:

U: (UF UL UB UR) (UFR UFL UBL UBR)
S: (UR DR) (UB DF) (UFR DBR) (UBR DFR)
D: (DF DR DB DL) (DFR DBR DBL DFL)
Z: (UR DR) (UF DB) (UFR DBR) (UBR DFR)

t: (UL UR) (UFR UBR)
n: (UL UR) (UFR UBL)

setupa: (UL UB UR)
setupb: (UF DB) (UL UR)
setupc: (UF UL) (UR DB)

S is to slice while U is offset; Z is to slice while D is offset.

---

solved CP:
bottom paired, top paired:
SKIP

bottom paired, top not paired:
UF-UR, UB-UL
1,0 / D' / -1,0 / U / 1,0 / D / -1,0 / -3,0 /
S U S Z U' S U' S U S Z U' S U
(U perm)

bottom not paired, top paired:
UL-UR, UF-DF
S U S U' S Z U S U' S

bottom not paired, top not paired:
UL-UB, UR-DF
S U S U S Z U' S U' S
(1-1 EO)

---

Every case here can be done in 6 slices.

adj CP (4/6 = 2/3):
bottom paired, top paired:
/ U / U' D' / D / 4,0 / U / U' D' / D / -1,0
(T perm)

bottom paired, top not paired:
UF-UR, UB-UL
1,0 / U / D' / U / U' / U' D / -1,0
(J perm)

UF-UL, UB-UR
/ U / D' / U / U' / U' D /
(J perm)

bottom not paired, top paired:
M2 into J perm (+2 slices)

UF-UB, UR-DF
S U Z U' Z U' D Z U' Z U Z D'

UF-UB, UR-DB
D2 then above
Z U S D' S U S U' S U' D S

UL-UR, UF-DB
U S U D' Z U Z U' Z D Z U' Z

UL-UR, UF-DF
D2 then above

bottom not paired, top not paired:
UL-UB, UR-DF
U' Z U' D Z U S D' Z D S D' S
D Z U' Z U Z D' S U' Z U' Z U Z

UL-UF, UR-DF
Z U S U' Z U' D S U' S U S D'
D Z U Z U' Z U' D S U' Z U Z

UR-UF, UL-DF
S U S D' Z U S U' Z U' D Z

UR-UB, UL-DB
S U S U' Z U' D S U' S U S

---

diag CP (1/6):
bottom paired, top paired:
/ U D / U' / U D / U' / U D /
(N perm)

bottom paired, top not paired:
UF-UL, UB-UR
/ U / U' D' / D / -5,0 / U / U' D' / D / -1,0
(Y perm)

bottom not paired, top paired:
UL-UR, UF-DF
S U S U' Z D S D S U' S U S

bottom not paired, top not paired:
UL-UF, UR-DB
D S U S U' S D Z D S U' S U S

 

[JackTriton]

This isn't really a pro of this specific method because almost every square-1 method has this feature too… It's kind of the whole point of doing cubeshape first.

------

Rather than doing CP together with SB, there's a pretty substantial slice count reduction if you split and rejoin the steps like this:
1. Cubeshape (or CSP)
2. First block
3. Second block (absolutely no influencing whatsoever needed here!)
4. CP + pairing edges (16 algs)
5. L6PE

Original JQ v2 method:
CPSB takes 26/6 = 4.333 slices on average, but always takes at least 2 slices, so effectively this solves CP in 2.333 slices.
Edge pairing takes 86/15 = 5.733 slices on average.
Total: 8.067 slices.

Modified method:
CP + edge pairing takes 550/90 = 6.111 slices on average.

Realistically, this gives more freedom in building the second block as well, so the true move count reduction should be even greater.

Unfortunately, in the process of labbing this, I was reminded that I hate learning squan algs so I'll never use either the original method or my modification of it…

Spoiler: algs and blah

I used Batch Solver to gen the algs, restricted to a move set where the puzzle never leaves cubeshape.

Puzzle definition:

U: (UF UL UB UR) (UFR UFL UBL UBR)
S: (UR DR) (UB DF) (UFR DBR) (UBR DFR)
D: (DF DR DB DL) (DFR DBR DBL DFL)
Z: (UR DR) (UF DB) (UFR DBR) (UBR DFR)

t: (UL UR) (UFR UBR)
n: (UL UR) (UFR UBL)

setupa: (UL UB UR)
setupb: (UF DB) (UL UR)
setupc: (UF UL) (UR DB)

S is to slice while U is offset; Z is to slice while D is offset.

---

solved CP:
bottom paired, top paired:
SKIP

bottom paired, top not paired:
UF-UR, UB-UL
1,0 / D' / -1,0 / U / 1,0 / D / -1,0 / -3,0 /
S U S Z U' S U' S U S Z U' S U
(U perm)

bottom not paired, top paired:
UL-UR, UF-DF
S U S U' S Z U S U' S

bottom not paired, top not paired:
UL-UB, UR-DF
S U S U S Z U' S U' S
(1-1 EO)

---

Every case here can be done in 6 slices.

adj CP (4/6 = 2/3):
bottom paired, top paired:
/ U / U' D' / D / 4,0 / U / U' D' / D / -1,0
(T perm)

bottom paired, top not paired:
UF-UR, UB-UL
1,0 / U / D' / U / U' / U' D / -1,0
(J perm)

UF-UL, UB-UR
/ U / D' / U / U' / U' D /
(J perm)

bottom not paired, top paired:
M2 into J perm (+2 slices)

UF-UB, UR-DF
S U Z U' Z U' D Z U' Z U Z D'

UF-UB, UR-DB
D2 then above
Z U S D' S U S U' S U' D S

UL-UR, UF-DB
U S U D' Z U Z U' Z D Z U' Z

UL-UR, UF-DF
D2 then above

bottom not paired, top not paired:
UL-UB, UR-DF
U' Z U' D Z U S D' Z D S D' S
D Z U' Z U Z D' S U' Z U' Z U Z

UL-UF, UR-DF
Z U S U' Z U' D S U' S U S D'
D Z U Z U' Z U' D S U' Z U Z

UR-UF, UL-DF
S U S D' Z U S U' Z U' D Z

UR-UB, UL-DB
S U S U' Z U' D S U' S U S

---

diag CP (1/6):
bottom paired, top paired:
/ U D / U' / U D / U' / U D /
(N perm)

bottom paired, top not paired:
UF-UL, UB-UR
/ U / U' D' / D / -5,0 / U / U' D' / D / -1,0
(Y perm)

bottom not paired, top paired:
UL-UR, UF-DF
S U S U' Z D S D S U' S U S

bottom not paired, top not paired:
UL-UF, UR-DB
D S U S U' S D Z D S U' S U S

Thanks for the alternative!

I originally calculate those as well in my alg note but there is 1 problem with the CPPE comparing to CPSB

On CPPE, you have to remember longer and larger sets of algs and recognize by both edges and corners

So what I did was to separate these two, combining CP and Second Block so that solver can separate corner solving and edge solving

 

[JackTriton]

Pairing Edge case recognition can be done by counting how many white edges on top and how non-white edge on top relates

Assume each case labeled as 1,2,3 and skip

If 2 white edges are on top and adjoining, if non-white edge relates, then it is case 3
If not, it is case 1

If 2 white edges are on top and paired, if non-white edge relates, then it is skip
If not, then it is case 2

If 1 white edge is on top, if non-white edge adjoining, then it is case 1
If non-white edge is paired, then it is case 2

And finally, if no white edges are on top, if non-white edge adjoining, then it is case 3
If non-white edge is paired, then it is skip

 

[JackTriton]

Advanced: Full CPSB
Forms CPSB in any SB situation

All algorithms written HERE!