Class of rings between fields and euclidean domains (improvement of euclideanity)

I am new to algebraic number theory, arithmetic in domains and I want an answer to the following question (that might be naive/unusual):

According to Wikipedia, there is no class specified between fields and euclidean domains (see the chain of class inclusions). Going to the right of the chain, the ring has more arithmetic properties. I feel that there must be something there, to further distinguish between/classify euclidean domains (below are $3$ ideas). Are these ideas reasonable or is there a "better" class (or property) that "fills this gap", i.e. stronger than euclideanity, but weaker than a field?

(For example, Schreier domains are between integrally closed domains and GCD-domains.)

Research effort: I have only found $3$ examples/ideas (EDIT: the last two failed, and the first is known, due to Jodeit paper, see below), which are made in order to improve the arithmetic properties, but I don't know if these are "official", known classes, that might follow the standard pattern of class inclusions:

$1)$ $Q_1$ , shows when the quotient and remainder are uniquely determined. Let's call this U-Euclidean domains.

$2)$ $Q_2$ , this is a stronger case of U-Euclidean domains, because the euclidean function is a degree function, which gives uniqueness in the Euclidean division. In other words, $K$ and $K[X]$, where $K$ is a field, are U-Euclidean domains. (EDIT: Jodeit paper shows that these are the only U-Euclidean domains!)

$3)$ $Q_3$ , reveals that any euclidean function must take at least $3$ values, for a non-field euclidean domain. I am also aware of the Minimal Euclidean function . To be specific, during the construction of the minimal norm, in that partition via $R_n$, $R_0$ is the set of units, and $R_1$ is the set of all universal side divisors. If the domain is not a field, it follows from $Q_3$ that $R_1$ and $R_2$ are nonempty. An example where this partition is finite might be a solution to fill the gap. In any field $K$, $K=\{0\} \cup R_0$, indeed, finite. Finitude might come closer to fields, because in the case of the integers, even the Motzkin minimal function won't give a finite partition. Therefore, one can investigate the following situation (EDIT: I added the answer, there is no need to answer it):

Find an example of non-field euclidean domain such that $\exists k\geq 3$ with $R_i=\phi, \forall i\geq k$.

$A$: NO. There aren't non-field euclidean domains $R$ for the "finitude" described in $3)$. It is known that any euclidean function defined on $R$ is unbounded, including the minimal euclidean function $N$ described above ($N(x) \geq L(x)$, where $L$ is the factorization length function). Due to this, $R_i \neq \phi, \forall i\geq 0$. In fact, $Q_3$ can be improved by stating that any euclidean function must take infinitely many values (again, non-field case).

Solution 1:

I'm not sure I would agree that "filling the gap" is a very reasonable way to put it (I don't really think there is a gap to fill), but the class of local principal rings (so rings that are either fields or discrete valuation rings) is between fields and euclidean rings.