3D Medusa takes Simple Colouring (“Single’s Chains”) to a new level. In rows, columns, and boxes, Simple Coloring looked for pairings of X.
The chains followed the same candidate number no matter where they went.
This is useful for keeping track of an elimination when you’ve written notes on a paper Sudoku for a certain number, but it restricts the strategy’s use. Up to the bi-value cells, which contain two different numbers, we extend the search.
Coloring has the devastating effect of demonstrating that utilizing only one color would solve the problem.
We don’t know which set it is yet, but if any of those cells turn out to be the solution, ALL the cells of the same color will be answered.
Rule 1 – Twice in a Cell
Rule 1 of 3D Medusa: Load Example or: From the Beginning, We eradicate in six distinct methods – six paradoxes. The first is seen in the illustration to the right. It makes no difference where you begin on the grid. I started with the 4s in row B in this example. As shown in the picture, we mentally construct a line between them by coloring one green and the other yellow. In B7, we may color the four yellow and the nine green in the third dimension because only two values remain in the cell.
Continue to hunt for prospects who are bi-valued and bi-located, and you’ll soon have a network of links. For example, the image of Medusa, whose head is a tangle of snakes, may have been associated with this method.
You could encounter a cell with the same color set twice after you’ve built up a web of connections alternating between two colors. H2 has been circled this. We can’t have a circumstance where two yellow numbers contend for the same cell since we know that yellow candidates can be true. Because this is a contradiction, we may conclude that no yellow numbers can be the solution!
The First rule is that if two candidates in a cell have the same color, that color can be deleted entirely, and the opposite color is all solutions.
Rule 1 merely said that the yellow candidates might be removed.
However, Steve Jacobs, a solver programmer, pointed out that all green candidates MUST be the solutions to their cells, which is a positive claim. Thus, the 1 in H1 is transformed into a Hidden Single. Because of the Medusa web’s binary either/or connections, this is true. Therefore, my solver will continue to remove just the yellow possibilities (hidden singles will be deleted in the following phase).
Still, pen and paper solvers should go ahead and fill in the ‘green’ solved cells. Rule 2 and Rule 6 are likewise covered by Jacobs’ corollary.
Note that in Simple Colouring, this rule does not apply because the same number does not occur twice in the same cell.
Try coloring any of the highlighted cells beginning from a different location as an exercise. You can wind up moving the colors around and making new connections as a result.
But, in this case, you’ll end up with two of the same color on H2. Nevertheless, this is an effective yet straightforward method.
Rule 2 – Twice in a Unit
Rule 2 of 3D Medusa: Load Example or: From the Beginning This guideline also applies to Simple Coloring. The premise is the same as the first rule, and only we’re looking for two-colored X instances in the same unit rather than two of the same color in the same cell.
Most of the relationships between bi-value and bi-location possibilities are colored between green and yellow in the example.
For example, two 7s in column 7 are ringed in red. Because neither can be true simultaneously, all yellow candidates may be eliminated, while all green candidates are solutions to their cells. (For example, three Medusa Rules 6 are required before Rules 2 may be used.)
Rule 3 of 3D Medusa: Load Example or: From the Beginning Because the pattern is a classic Nice Loop, if you had unticked 3D medusa in the solver, this example would have been identified by various subsequent algorithms especially Alternating Inference Chains. The numbers 3 and 7 alternate.
In a Nice Loop, it doesn’t matter where you start, but you can trace the on/off or green/blue around the loop. The numbers 3 and 7 appear twice in both units and cells.
However, 3D Medusa is about a network of links, not loops. This is the same structure as the previous one.
We know that either all of the blue candidates or all of the green candidates will be true. They cannot be solutions if there are any additional candidates in any cell with two colors. As a result, the number 8 may be eliminated from C2. This is an off-chain elimination in Nice Loop terminology.
Because it is limited to a single candidate number, a simple coloring cannot generate this elimination.
Rule 4 – Two colors ‘elsewhere’
Rule 4 of 3D Medusa: Load Example or: From the Beginning We can remove “off-chain” in a unit if we can eliminate “off-chain” in a cell. We are positive that ALL blues or ALL greens are the answer. As a result, those who can perceive both colors might be eliminated. By’see,’ we mean any candidates who have the same number as the blue/green link members. Because of the colored 6s down the row in B9 and along the column in H1, the 6 in B1 is gone. Similarly, the blue 6 in C2 and the green 6 in B9 indicate the number 6 in C8.
Rule 4 of 3D Medusa: Load Example or: From the Beginning
We find a cluster of 4s, 6s, and an eight a few steps later in the same problem using the same observation. I’ve been combining Rule 4 with the old Rule 5 since February 2015. I want to thank FallsOffRocks for pointing out that the previous rule covered all of the eliminations that Rules 4 did and was therefore redundant. So rule 5 (‘elsewhere’) has been incorporated into Rule 4 (‘along with a unit’), and Rules 6 and 7 have been decremented. This is a simplification! So there are just 6 Medusa Rules from now on. This has an impact on Simple Coloring as well.
Rule 5 – Two colors unit + cell
Rule 5 of 3D Medusa: Load Example or: From the Beginning, This sort of removal appears to be the most difficult, yet it is also the most prevalent. It’s worth keeping an eye out for. According to the regulation, It is possible to eliminate an uncolored candidate if it can see a colored candidate elsewhere (shares a unit) and an opposing colored candidate in its cell.
The colors green and blue may be discovered gazing along with a unit and within the same cell; therefore, it’s a mix of unit and cell. With four eliminations, the example on the right proves this.
The logic appeals to me a lot. 1 in E5 is an example. If 1 were the cell’s solution, it would remove a green one from E6 and a blue seven from its Cell in E5. Thus, we have a dilemma since we know that ALL blue or ALL green must be solutions.
Rule 6 – Cell Emptied by Color
Rule 6: From the Beginning or Load Example In the comments below, Anton Delprado has identified another technique to use 3D Medusa, which I’m happy to incorporate in the solver. It’s nearly like Rule 5 in reverse. Take any cell in the coloring that doesn’t have any colors and test whether all candidates can see the same color. If that color were the solution (and remember, it has to be all of one or all of the other), all of the candidates in that cell would be eliminated, leaving an empty cell!
The cell that captures this is highlighted in blue in Rule 6. The 2 and 5 in cell C1 are visible, as are the yellow 2 in cell C9 and the 5 in cell H1. (Eliminated cells are shown in yellow.) All candidates who are green in color are solutions for their cells.
Rule 6 when there are three candidates in a cell
Example of a Load or: From the Beginning
This problem contains a fantastic series of medusa calls that follow a variety of rules. It comes to a close with Rule 6. I included a second problem to demonstrate that the cell to which the Medusa net is being compared might contain any number of candidates. In C6, these are 4, 6, and 9. Although the green candidates have been colored yellow since they have been removed, the 4, 6, and 9 applicants can all see the same hue anywhere along the column or row. You can be sure that it will be one hue or the other, but never both at the same time. The solution looks for this method before Rule 5 since it is easier to notice and follows from Rule 4. But it’s too late now to renumber them. Anton, you’ve spotted something!
37 Rule 1 Eliminations: Load Example or: From the Beginning to round off this essay, I’d like to show you some unique puzzles created by Klaus Brenner, beginning with this 37-elimination Rule 1 Medusa, which entirely solves the puzzle from that point on. We move from 35 to 70 known numbers, and the rest is easy! 5 and 7 are two candidates in A1 who have the same color. As a result, all of them of that color may be eliminated.
However, the first problem is not easy, and there are a lot of stages to get to this giant medusa. This is unquestionably a difficult grade.
Load Example (Rule 5)
We get these ten eliminations using Rule 5 from a somewhat different problem with 22 clues.
Further discussion on 3D-Medusa
When two Cells in a region (Row, Column, or Square) are the only Cells in that region that contain a certain candidate, a Strong Link exists. If indeed the candidate isn’t the first cell’s solution, that must be the second cell’s solution, and conversely. If one of the candidates in a “Bi-Value” Cell is not the solution, the other candidate must be the solution. As a result, a Strong Link exists between the two possibilities for a “Bi-Value” Cell.
We may build a Chain of three Cells with the common cell in the middle of two Strong Links that share a common Cell. Because the concept of a Strong Link requires that the candidate in the cell at one end of the Chain is not the solution for that cell, it must be the solution for the next cell in the Chain. This, in turn, suggests that the candidate cannot be the answer for the third cell in the Chain for the same reason.
If we assume that the candidate in one of the Chain’s ends Cells is the solution for that cell, it cannot be the middle cell and must be the solution for the Chain’s other end Cell. If we imagine a longer Chain comprised entirely of Strong Links, the Nodes switch between one state and the next. This technique examines such chains and displays the states using colors. Because there are only two potential states, all candidates of the same hue must be the solution or cannot be the solution simultaneously.
Because two candidates cannot be the answer for the same cell simultaneously, the matching color cannot be the solution; thus, all candidates with this color must be removed. The answer is the other color.
Starting with candidate 3 in J3, candidates 1 and 3 in G3 are both colored yellow, as seen in the illustration. However, the yellow Chain cannot identify the answer since two separate candidates cannot solve the same cell. Therefore, all candidates colored in yellow may be removed, whereas the candidates colored in green are the solution.
If two Nodes in the Chain have the same color for the same candidate and belong to the same region (Row, Column, or Square), this color can’t be the solution since a candidate can’t be the solution for two cells in the same region; all candidates with this color may be discarded. So, as a result, the other color.
Candidate 3 is colored yellow in A9 and H9 in the illustration. Candidate 3 would answer two Cells in Column “9” if the yellow color identified the solution, which is impossible. As a result, the answer is identified by green, and all options colored in yellow may be dismissed.
However, this is a hypothetical situation. Iterations are used to construct the Chain. The iteration that “colors” the “second node” in the same unit comes before the one that “colors” the connected node with the other color. This signifies that the “second node” observed two candidates with different colors during the previous iteration (see below). Candidate 3 in H9, for example, sees a yellow candidate 3 in A9 and a green candidate 3 in H4 before being tinted in yellow.
As a result, before coloring candidate 3 in H9 in yellow, the rule “3d Medusa (Sees Two Different Colors)” would have been applied.
3d medusa (two colors in a cell):
The solution is one of the two colors. If two candidates in a Cell have different colors, one must be the answer for that cell. As a result, all other candidates in the cell may be discarded.
Candidate 1 or Candidate 5 must be the solution in case G9. None of the other choices in this cell can be the answer.
See Two Different Colors In 3d Medusa:
The solution is color, either one or the other. If an uncolored candidate can “see” (a Candidate may “see” another Candidate if both Candidates are from the same region) two candidates of different hues, one of them will be removed (we do not know yet by which one).
Candidate 7 in C9, for example, sees candidate 7 of various colors in A8 and C1. Therefore, candidate seven cannot be the solution in C9 since it must be the solution in one of these cells. Unit-cell elimination in 3d medusa: The solution is color, either one or the other. If an uncolored candidate can “see” a colored candidate with the same value and another candidate in its cell with the opposite color, it cannot be the solution and must be deleted.
Candidate 9 in C9 or Candidate 1 in F9 is the solution in the case. However, candidate 1 in C9 sees both candidates; thus, it can’t be the C9 answer.
3d Medusa (Emptying A Cell):
If an uncolored Cell can “see” Cells that each have one of the uncolored cell’s candidates, and if all of these candidates have the same color, this color cannot be the answer. It would exclude all candidates from the specific cell if it were the solution, which is impossible.
If candidate 2 was the answer in B4 and candidate 9 was the solution in E7, then E4 would exclude all potential possibilities.
3D Medusa Strategy
3D Medusa is a coloring technique that is comparable to traditional coloring. You can use it to remove candidates.
How to Use 3D Medusa
Simple coloring is how 3D Medusa begins. You select a candidate, assign it a color, and then assign the opposite color to additional candidates, alternating colors until eliminations are detected. 3D Medusa, on the other hand, is more difficult. Bi-value cells are used to extend the coloring over a larger network of possibilities. The 3D Medusa technique uses bi-value coloring similar to that used in XY-Chains.
You may begin 3D Medusa by assigning different colors to individuals who have a strong connection (see the page on X-Cycles for an explanation of strong links). Continue assigning colors and pursuing strong linkages until you’ve colored a candidate in a bi-value cell. After you’ve colored it, you may color the second contender in the bi-value cell the opposite color. It’s worth noting that a sudoku grid can contain numerous 3D medusas. If you don’t discover any eliminations when coloring a 3D medusa, you can start a new one with a candidate who hasn’t been colored in any prior 3D medusas.
After you’ve colored all of the cells in the network of strong linkages and bi-value cells that are connected, you may seek for candidates to delete using the following guidelines. In a cell, the same hue appears twice.
It’s a contradiction if the same color occurs twice in a cell in a 3D Medusa, and all candidates with that color should be discarded.
The ringed cell in the preceding example has two green possibilities, as you can see. Because the colors reflect two different approaches to solving the grid, either all green candidates or red candidates are accurate. However, there would have to be two numbers in the ringed cell for the green candidates to be accurate, which is impossible. As a result, none of the green candidates can be accurate; thus, they may all be eliminated. In a unit, the same color appears twice.
A contradiction exists when the same color shows twice in a row, column, or box (all numbers must be correct simultaneously); thus, all candidates of that color must be discarded.
The circled 3s in the example above are both red. Both of those 3s would have to be accurate if red was the proper color, but that is a contradiction. As a result, all red possibilities may be ruled out. In a cell with uncolored candidates, there are two colors.
If both colors of your 3D medusa appear in a cell with uncolored candidates, the uncolored candidates can be removed from that cell. Because you know that one of the colors must be accurate, all uncolored options in that cell must be incorrect.
The ringed cell in the example below has two possibilities with opposing colors. Therefore, the remaining candidates in that cell can be eliminated.
In a unit, there are two colors.
You can eliminate any uncolored copies of a candidate that the colored ones can view if they exist twice in a unit and are oppositely colored. An uncolored candidate can see two candidates with opposing colors.
If an uncolored candidate sees two of the same candidates, who are oppositely colored, that uncolored candidate can be removed. For example, you may delete two 2s in the example below since each one can see both a red and a green
In the same cell, an uncolored candidate can view a colored candidate and an oppositely colored candidate. It’s a little difficult to describe this one. You can delete a candidate if it can see a colored version of itself and shares a cell with a candidate (any number) with the opposite color. A green nine is in the same row as an uncolored 9 in the example below. However, the green nine is in a cell with a red number, which is the polar opposite of the uncolored 9. You can get rid of the uncolored 9 using this combination.
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