Looking for opinions of puzzle break in alternatives

[Merdock]

I recently created and posted a puzzle online. I was looking back at it and kind of regret how I set it up.

I was wondering if someone can take a look at it, as well an alternative, and give me their opinion. These are basically the same puzzles and the differences are in the break-in. The break-in in my opinion, is easiest in my original puzzle, and less obvious in the new one. The break-in revolves around determining the very center digit. In the first puzzle, with some trial and error, you can fairly easily guess the center digit. This is not how I wanted that digit found. I think the second forces you to determine this digit in the method that I wanted. The first puzzle uses a lot of variables for the killer cage sums, the updated one uses no variables.

I did make a note in the first version that clarifies that guessing the center digit is not required, which kind of indicates that that's not the method I want the solver using.

I would appreciate any opinions, and if you want to know how to break it in, I can provide that information. I'm not super familiar with posting here, so I don't know how to include text that reveals itself once you click on it. My apologies for that. 

Original: https://sudokupad.app/9sxe7q81rb

Updated: https://sudokupad.app/zkh3hp1vqs

[Schachus]

in the original version, I'd say there is no trial and error required at all. Looking at the bottom box, x^2+x should be at most 24, so x<5 and (x^2+x) + (x+5) should be at least 21, so x>3. Now you found x.
In the second version, first of all, I wouldnt guess I have to try and find the centre digit, and secondly it isnt obvious to me, how I could find it, even now that you pointed out I should try and find it (and knowing what it should be from the other version). So I like the original better^^

Edit:I guess I see what you want in the second maybe? If I do law of leftowers on the two diagonals with the blue lines, I can find the sum of the two diagonals withoutthe center and subtracting the square diagonally above the center once. That gives me the sum of the centre square and the one diagonally above. Is that the intended way to start? Still dont know how to isolate the centre digit though. Edit 2: The first number I find with this method was the bottom right number in the centre box . I can see now that I should have just looked at one diagonal and then when I find the center with that, start looking at the other diagonal. I was looking at both diagonals together and then looked at the result (which was sum of two numbers, to compare with the that diagonal, but that gave me the 1 first, which in turn gives an 8 above it. the center 4 is still not clear to me). I kept solving now, found the center as my fifth digit (using sukoku on the 6s to resolve the sum of 10 I mentioned first).

[Merdock]

in the original version, I'd say there is no trial and error required at all. Looking at the bottom box, x^2+x should be at most 24, so x<5 and (x^2+x) + (x+5) should be at least 21, so x>3. Now you found x.
In the second version, first of all, I wouldnt guess I have to try and find the centre digit, and secondly it isnt obvious to me, how I could find it, even now that you pointed out I should try and find it (and knowing what it should be from the other version). So I like the original better^^

Edit:I guess I see what you want in the second maybe? If I do law of leftowers on the two diagonals with the blue lines, I can find the sum of the two diagonals withoutthe center and subtracting the square diagonally above the center once. That gives me the sum of the centre square and the one diagonally above. Is that the intended way to start? Still dont know how to isolate the centre digit though. Edit 2: The first number I find with this method was the bottom right number in the centre box . I can see now that I should have just looked at one diagonal and then when I find the center with that, start looking at the other diagonal. I was looking at both diagonals together and then looked at the result (which was sum of two numbers, to compare with the that diagonal, but that gave me the 1 first, which in turn gives an 8 above it. the center 4 is still not clear to me). I kept solving now, found the center as my fifth digit (using sukoku on the 6s to resolve the sum of 10 I mentioned first).

Thanks for the response and the attempts to find the break-in. Double thanks for finding a different break-in. My intended method was using set equivalence theory. 

The first set is the positive diagonal, negative diagonal, row 5, and column 5. (R5c5 is counted 4 times)

The second set is column 4, column 6, row 4, and row 6.
(R4c4, r4c6, r6c4, and r6c6 are each counted twice)

R4c5, r5c4, r5c6, and r6c4 can be removed from both sets. 

R4c4, r4c6, r6c4, and r6c6 can be removed from the first set, and once from the second set

R3c5 and r4c4 can be removed from their respective sets because they're the same digit. The same is true for r5c3 & r6c4 and r6c6 and r5c7.

Then r1c1, r2c2, and r3c3 have the same sum as r1c4, r2c4, and r3c4. We can remove them from their sets, and both sets have the same total (keeping in mind r5c5 is counted 4 times. This can be done with the other similar sets of cells going around the grid. 

Ultimately, your left with only the cells in the cages and the very center cell. 

The sum of the first set is 47+4(r5c5) and the sum of the second set is 63. Solving for r5c5 gets you it's digit.