The final grid!
We've all seen the site logo before, before you may not have seen this variation!
It is one of the original site logos suggested at the graduation of the site!
- Starting cells
We can start with the easy cells. For the walkthrough cells filled in with value 1, I will use a light grey colour instead of white to differentiate between filled cells and as yet unused cells.
There are three easy cells: a corner cell with value 03, a border cell with value 05, and a central cell with value 08. These cells have 3, 5 and 8 neighbour cells respectfully, and hence must be completely surrounded by white cells with value 1. Filling these out gives a good start:
Now consider the following scenario:
The central cell, 13, has 7/8 of its neighbouring cells filled in, and hence we can deduce the value of the 8th cell that highlighted in red. In this case, it would be 13 - 7 = 6, which would mean the red cell is a dark grey cell. I will call this logical step the 8th Neighbour Deduction, or 8ND for future reference.
Using the 8ND on the current grid, we can fill out what we have to create several rectangles:
- The complicated bit - Simultaneous equations
There is now no obvious next step. To progress we will need to use a combination of (sort of) simultaneous equations and 8ND. Focus on the the empty space bottom middle.
From the 10 highlighted in blue, the value of the cells in green must be 10 - 3 = 7. As we know the green cells add up to 7, the purple 19 can then use 8ND to work out the value of the highlighted red cell. In this case it would be 19 - 6 - 4 - 7 = 2 which is a red cell.
This logic can be applied symmetrically on the opposite side to place a green cell too. Furthermore the same situation occurs slightly further up next to the two yellows and between the dark grey and yellow symmetrically on the opposite side:
The blue 26 tells us the two green cells are again equal to 7, so the purple 29 from 8ND tells us that the red 33 must be a 29 - 24 = 5 or a yellow cell. We now have this:
- Even more simultaneous equations
Another relatively simple deduction can be made next extending the new diagonal that's being created.
We know the green cells are 7, so the blue 15 from 8ND gives 15 - 9 = 6 or a dark grey cell for the red 20. Same on the other side.
Now consider the 10 next to the blue 15 above. It has 2 remaining neighbours that must add up to 2, so the neighbours must be a light grey 1. (Again, same on other side). Now we can make a bit of progress in the middle:
The green cells are 7, the blue 20 can then tell us the purple cells must add up to 20 - 12 = 8. And using this and 8ND on the yellow 19, the red 28 must be a 19 - 18 = 1. So now we have this:
- Final tricks
The 28 will be a key reference point now. Consider the 18 below it. It has 2 remaining neighbours, a 23 and 27 next to the 28, which must sum to 12, so they both must be dark grey 6s.
We can then use 8ND for the 19 and 23 neighbouring the 28 on it's bottom diagonals to solve the 34 and 36 as a red 2 and a dark grey 6. We now have the following at the bottom:
And pay attention to this bit:
The purple 30 tells us the 3 green cells add up to 13, and hence from 8ND on the blue 34, the red 26 must be 34 - 28 = a dark grey 6. The same symmetrically for the other side on the 31. 8ND can then be applied on the aforementioned central 28 cell to place a light grey 1 above it, and 8ND twice more with the 23 and 27 to fill out the middle a bit more:
- Solving a section
Some more simultaneous equations can finally help solve this section:
The blue 29 tell us the greens add up to 7, and the purple 34 then means by 8ND the red 25 is a 34-30 = yellow 5. We can then apply 8ND on the cyan 33 to place a yellow 5 as the purple 34. We can also do the same on the other side to place 3 cells, and then the central 27 can also be resolved by 8ND.
We are left with 4 upright lines of 2 cells at the bottom. Simply pick any of the unresolved cells, and it's pair can be solved by 8ND. We now have solved the bottom:
- Solving the sides
The sides can be solved simply with 8ND. For instance for the left side, starting bottom right of the unresolved cells, the yellow 28 places another yellow as the cell top left of it, and from there we have created a diagonal that can ripple across the space with repeated 8ND. Solving the sides gives:
- Finishing the job!
And from here we fill in the rest to get the final answer!