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Let's get this out of the way first. You may be aware of how drasticly different sequels could be from the original source material. Zelda 2, Castlevania 2, Super Mario Brothers 2 for given values of what Super Mario Brothers 2 is.... And Final Fantasy 2 is no different.

While every Final Fantasy sequel does its own thing, this one is so different that it immediately split off into its own thing. In other words to quote Chokes McGee, "SaGa, y'all!"

SaGa games are nortorious for their esoteric bullshit mechanics. Apocryphycally any time lead designer Akitoshi Kawazu could figure out what was happening, he'd demand it be muddled further. But that was over 30 years ago, we've got tools and experience to dismantle the game and figure out what's going on under the hood. And once you know what levers to pull the game goes from daunting to enjoyable.

So without further ado, let's discuss the mechanics.


First, you'll want to change your menu settings from Default to Memory.


I mean Sorted. Stupid localization quirks. Anyway you'll thank me later for using Sorted.


So for Greater Than Killer SuDoKu SaGa you'll notice that there are no squares prepopulated. No safety net. You alone must figure out what goes where. Draw the rest of the owl? How quaint, draw the whole genome.

We focus now on any squares with greater than or less than signs on them. So for the upper left corner, you can narrow the 5 squares down to these possible values. The lower right square can be 5 or greater. The one above it can only be 4 through 8, as the squares around it eliminate the chances of it being 1,2,3 or 9.


We repeat the process in every other square with restrictions on them.


Now there's a lot of squares that we narrowed their values down a little bit. Everything else can be 1-9, which I keep blank to keep from getting too overwhelmed. The next step now is to start looking for rows or columns that have one square that has one square with one unique number, usually 1 or 9. Here we find one single square in this column with a 1.


This means that any other square in this lower left 3x3 section can't be 1. And since the center square in this grid is surrounded by 3 other squares that can't be 1, that means they can be no less than 2,3 and 4, meaning it can be no less than 5, and up to 9.


So the third bottom row can't have any other 1s. And since we already ruled out the middle row in the lower right grid from having a 1, it means that they must be on the bottom...


Which means the middle grid must have its 1 in the middle row, and there's only one spot it can be.


Here.


Which eliminates all others in this column from being a 1. The middle 3x3 square already excluded 1s from its middle column, so 1 has to be in the middle column of the top somewhere.


Here, top dead center.


We can now fill in the top row, anything that was blank now can be 2-9.


And now things quickly snap into place. SaGa goes in bursts where a lot happens, and then nothing happens. The one at top dead center excludes 1s from being in the 2nd right column in the top grid, and math already excluded it from that column in the grid below it. So 1 has to be in that column in the bottom grid. And since all the other squares where 1 could be now can be no less than 2, and all of them are feeding towards the upper right square, it must be 9.


So we can populate the right column with possibilities, and exclude the 9s from any other spot in the 3rd row from the bottom.


We begin poking again, looking for places where there is only one 9 or 1 in a column, row or grid. Like this 9 for example.


The 9 means that this square can now only be 7 or 8 (the square next to it can only be 6 or 7, and then the squares above and below it 3,4 or 5 and so on. for now I'll start abriding that process).


9 must be in either the upper or lower right of the middle grid, which excludes it from that column in the bottom center grid. The 9 must be in the bottom row of that grid. which narrows the location for the 9 in the lower left grid.


And back to ruling out the other 9s in that column.


Now things get a bit more difficult, we have to start eliminating options that aren't just a simple single option. For example, 1 can be in this or the square to the left. Because no other square in this grid can be, which means all the other 1s in that row are eliminated.


This now eliminates all 1s from the rightmost column but the top grid.


Which is a dead end for us for now.


But we have another place to attack. The 2 empty squares in the middle can only be 1 or 9. Which in hindsight now doesn't actually help besides just filling in squares.


But here we see that this middle column only has one possible spot for a 2.


Which eliminates any other 2s in that row.


Which leaves only one 2 in the left column...


And only 1 is less than 2...



Which leads to another more complicated technique. Note that there is a square with the possibility of 6, 7, or 8. Then there's squares in that same row with 6 or 7, and 7 or 8. These mean that any other squares in that column can't be those numbers.


At first it doesn't look like it has solved anything...


But now there is no other place in the center block that can be 8.


And that narrows this down to 7.


And now 6.


I will eventually figure out that 7 shouldn't be greater than 6. But in my defense I live near colleges that decide that apparently 14 is equal to 10. I guess it is a big 10 but still.


Not pictured: several minutes just glaring at random rows and columns.


Finally here it is.


Which narrows down the 3 and the 4.


The 5 and 6 in the second column from the right gives us the 7.


And the 8.


And that's enough process of elimination to solve the whole right from center grid


Just cleaning up some stuff...


Abd some more cleanup. This was just ruling out the 4 in this stack...


1 can't be in that square.


And 2 can't be in any other square but here.


2 squares in this row can be 3 or 6, so the other 2 can't.


Other than looking to clean up 9s in this column I'm unsure why I wanted to point this one out?


And with this 9 we only have to figure out 2 more... That being said I don't figure this out yet.


I'm over here now apparently and about to clear up the superfluous 7s and 8s.


This rules out the 6s and 7s...


And then the three squares in the column with 3s, 4s and 5s rules out them out of the other squares in the column.





The 2,7 and 8 in the column eliminate the 2,7s and 8s in the other squares in the top center grid.


4 can't be greater than 4...


6 can't be here.


The top right grid has to have a 6 in the leftmost column, which means that the rightmost column has to be 1,2 or 5.


Some more elimination...


Some more, more elimination.


More...


Running out of things to eliminate here!


Finally I realize this has to be 9.


Even if you consider a very big 10 to be comparable to 14, 7 can't possibly be greater than 5.


At most it can be equal to it I guess.


A quick detour from eliminating 2s from this section... only one thing can be greater than 7 or 8.


I'm looking at solving that 9 in the upper right...


But that's apparently something for a later time, off I go to the lower left corner?


This is a quick fire solution that the bottom dead center grid only has one place to put the 9. There is only one place in that column to put the 7, and with the 9 gone only 8 is greater than 7.


The bottom row has 4 squares that contain 2,3,4 and 5.


Which means the remaining 2 squares can't have them.


Now off we go over here I guess.


This means something...


It means all the 9s have been accounted for!


This solves the 3 in the lower right.


And then the 3 at bottom dead center is solved too. But what's more important is cleaning up this column, and eliminating the 2, 5 and 6s from the other messier blocks there.


Then clean up the 3s and 4s in this grid. Ignore the 8 that we can solve below that for now I guess.


Instead we can clean up the 3s, 4s and 5s in this row.


This solves a 1. I could wrap up all the 1s right now but if you've been paying attention I can't keep focused to save my life.


Instead we solve the 6, since 5 has to be below it somewhere.


Cleaning up the 6 gives us 2 squares that can have 7 or 8, and 2 other squares with 2 and 5.


Which gives me the last 1... in a much more complicated way than it needed to be.


However with clearing out all that stuff in the upper left means that the lower left 6 has to be here.


Cleaning up some extra 8s in the left middle grid.


3s in the left are narrowed down to 2 squares each in the top 2 grids, and one in the lower left.


Which forces this square to be 4.


2 squares of 7 and 8s cleans up some more in the column.


But enough about that over here to start picking away at the middle grids.


Things are going much quicker now that the coin flips between 2 numbers are solved.






All that remains is to zig zag between squares as they solve themselves.




It is only now that I realize that this is the first release of FF2's mechanics

Oh well, its important to see this anyway so you know the baseline chicanery that we're dealing with.


So let's discuss the modern mechanics. Even if doing so will push this update from March 31st to March 32nd. Now in addition to some random comparitive functions we have blocks that (may) have a total sum that must be contained inside them.


Every 3x3 square still has to have 1-9 in them, which means that each grid must contain a grand sum of 45. 8+13+14+7 is 42. This means something.This means everything Or at least it means that the odd square out must be 3. Which means that the second half of the block containing 8 must be 5.

Some things to consider for quick eliminations: A 2 block of 8 can only be 1 and 7, 2 and 6 or 3 and 5. No chance of 4 and 4 since there can't be doubles. 2 blocks of 13 can only be 4 and 9, 6 and 7, or 5 and 8. 2 blocks of 14 can only be 5 and 9 or 6 and 8. Since either option of 14 contains a 5 or an 8, that means that the 13 in this section can't be 5 and 8. A 2 square block of 7 can be 1 and 6, 2 and 5 or 3 and 4. If that 3 wasn't over there killing that option.


2 blocks of 5 can only be 1 and 4 or 2 and 3. This means that the bottom center set of squares has 1,2,3 and 4 locked down. Also, a value of 10 over 4 blocks is the same thing: 1,2,3 and 4 must be in them. A 29 value block of 4 must be 5, 7, 8 and 9. A 30 would be 6,7 8 and 9.


Even if there wasn't the 5 hovering up at the top preventing this from being 5, the block of 29 to its right would mean that 5 couldn't exist here, since either the little tail of that block would be a 5 and would rule out all 5s in that row... or the 5 exists in the center of the block and would prevent it from being here either. And the 1,2,3 and 4 aren't possible because of the 2 5 blocks.


Making it a 6, and the 1 to complete the set for 7. The 6 in the row eliminates some possibilities in the 9 and 14 block highlighted here. As does the presence of the 1. Eventually you may notice the 4 squares that must contain 5, 7 8 and 9 in the bottom row, which will require the remaining squares to be only 2, 3 and 4.


But that's for later I guess. Off to the upper left again I guess. Since the 3 removed some options for both the 7 block and the 8 block they can be further excluded. 7 can either be 2 and 5, or 1 and 6, either option contains the 2 or 6 which means that the 8 must only be 1 or 7. Which means that the 7 block can only be 2 or 5.


The 5 restricts the 14 block to 6 and 8. Which restricts the 13 block to the 4 and 9 set.


Which then eliminates the possibility of the 15 block down below from being 6 and 9.


Oh yeah at some point I did some math to figure out that the unlabeled block above the 1 had to be 11 (45-10-9-14-1). Since the 11 block can only be 2 and 9 or 4 and 7, that means that the 9 block can't be 2 and 7. And the presence of the 6 eliminates the 3 and 6 set.


And 1 and 9 are eliminated.


There was a different way to start how I did the process of elimination here. I could have started with ruling out the 2s and 5s in the second column from the left... instead I went with the remaining blank tiles in the 3rd column being limited to 4,5 and 7. Which eliminated 5 from the 2nd column. Which quickly cascaded all those 5 squares in place.


And instead of following through with the rest of the second column from the left I go off elsewhere again. This 7 block immediately can't be 2 or 5 because of the 5 in the grid. But also because of the 5 block at the bottom being 1 and 4 or 2 and 3 that prevents the 7 block being 3 and 4.


Which means that the 7 block had to be 1 and 6 which eliminated the 5 block from being 1 and 4. The 3 in the lower left grid prevented 3 from being in the top block. Also the presence of 8 in the bottom row prevents 8 from being in the bottom 3 squares of the 29 block.


Which means that the 5 in the 9 block in the left column can't be on the bottom.


What, did you think I was going to solve the 7 in the upper left square? Why would I do that when I have 2 ways to solve it? No its time to do something else. Let's instead do some comparative function stuff I guess. The block above the 13 can be 14, 15, 16 or 17. Which means it can only contain 5, 6, 7, 8 and 9... but the 5 over to the right excludes it from any squae. And the 6 above and the 8 to the left excludes them from the right square.


Now let's start ruling things out in this unlabeled block at the top. 1, 5, 6, 7 and 9 are prevented from being in the 3 left squares in the block, which means they can only be 2,3, 4 and 8.


I could do some further eliminations here but lets go over here now I guess. The 4 block of 18 already has 2 and 3 called for. So we need to solve for the remaining 13. The presence of a 5 and 8 in one column prevents if from being anything but 4 and 9 or 6 and 7.


That sure needed to be done now huh, let's go back to what I was doing apparently. 8 can't be in the middle column since its at the bottom. And the bottom square can't be 2 since its over to the left.


I now take the time to clean out the left square in that block, it can't be 3 or 8. Does that mean I can exclude the square to the right of the 8 from being 3? Of course I can don't be silly. Will I? Eventually probably.


Instead lets solve for that long 25 block in the middle. There's a lot going on here to exclude but off hand these are the important ones. It must contain a 1 and a 6. It can't contain 2 or 8. It can't contain 5, 7 and 9, it can only contain 2 or fewer of those options. This eliminates 3 from the top option and restricts the remaining blocks to be 1,5 6 and 9 which solves the bottom row's 7.


Another rapid string of solutions pop up. There are 3 blocks that can only contain 5, 7 and 9 in the 4th row from the right. This eliminates all other options for the 13 block to be 4 on the left and 9 on the right. Which solves the 5 in the bottom row and the 9 next to it and solves the remaning 5 and 7s in that column and the column next to it.


Back to the block that has to be greater than 13... the 7 on the left prevents it from being 6 (since that would be equal to 13) or 7 (which is prevented from duplicates).


Time to start filling in some potential values in the right squares. I also solve the remaining blocks in the lower left square. 8 is put in the only place it can.


I'm not clear why I decided to go over here now, The squares in the upper right square that I filled in can't have 6, because 14 over in the upper left and the 7 at top dead center both contain 6s.

Also because 3 can't exist in the 3rd row from the right in the middle grid, it means that 3 can only exist in the 3rd row from the right in the upper grid.


Which since there are 5 blocks that contain 1,6,7,8 and 9 that means that eliminates those possibilities from this square.


Now for some more complicated bullshit. Akitoshi Kawazu is laughing his ass off right now. This highlighted square can't be 7, because that would mean that that block with the 5 would be 12, which isn't less than 3,5 or 7 (or 10 but then that'd be 2 7s in the column). It can't be 6 either for the same reasons. So that block with the 5 can only be a total sum of 6 or 7, which eliminates 1 and 2 from the block with the 3.


This column has no other possible location for the 2 any more. Which solves the 13 block.


We now solve the block that is greater than 7. 7 can't be greater than 7 so it has to be 10.


Which solves the 8 in the upper left, the 7 in top dead center. The 14 in the left. and otherwise clears out a bunch of other squares.


These 2 squares can only be 1,5 or 6.


Which solves the 8 in the left column. The lower right grid is solved by eliminating 1s and 4s.


This solves the remaining bit of the 18 block, which solves the rest of the bottom squares, aside from that stubborn 2 and 3.


We now move to solve this block which is less than or equal to 13. We already have a 7 to work with, so the remaining blocks can't be more than 6. This immediately excludes 6 on its own, and since 3 and 4 are set in the other square 5 can't be used either. Which solves the 6 and 5 squares.


And draw the rest of the owl.


See? FF2 isn't that complicated, now is it?