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User prompt 1 of 26 - 10/12/2025, 5:02:24 PM:
if discrete log does not exist on integers, e.g. log_3(0) % 7. Can it have fractional soultions?

GPT-OSS 120B:
The discrete logarithm is defined only within a **finite cyclic group** (or, more generally, any group where exponentiation makes sense).  
For a modulus \(p\) (prime) the group is \(\mathbb{Z}_p^{\times}\); its elements are the integers \(1,\dots ,p-1\) under multiplication modulo \(p\).

* The expression \(\log_3(0) \pmod 7\) is undefined because \(0\) is **not** an element of \(\mathbb{Z}_7^{\times}\).  
* “Fractional” solutions would require an element \(x\) such that \(3^{x}\equiv 0\pmod 7\). Since every power of a non‑zero element in a multiplicative group is again non‑zero, no such \(x\) exists, integer or fractional.

In other words, the discrete logarithm problem has **no meaning** outside the group’s element set, and there is no notion of a “fractional exponent” modulo a prime that would produce a new group element. If one wishes to solve equations like \(3^{x}\equiv a\pmod 7\), \(a\) must be a member of the group (i.e., \(a\neq0\)). Otherwise the equation has no solution at all.

--------------------

User prompt 2 of 26 - 10/12/2025, 5:03:31 PM:
can we interpret integers as fractions, e.g. 7 is 7/1, 3 is 3/1. Would discrete log exist then?

GPT-OSS 120B:
In a modular setting the “fraction” notation \(a/b\) is only meaningful when the denominator \(b\) has a multiplicative inverse modulo the modulus.  

### When a denominator is invertible
If the modulus is a prime \(p\) (or any modulus coprime to \(b\)), the inverse \(b^{-1}\) exists in \(\mathbb{Z}_p\).  
Then we can treat the fraction \(a/b\) as the product  

\[
\frac{a}{b}\;\equiv\;a\cdot b^{-1}\pmod p .
\]

For example, with \(p=7\)

\[
\frac{3}{2}\equiv 3\cdot 2^{-1}\equiv 3\cdot 4\equiv 12\equiv 5\pmod 7 .
\]

In this sense every integer is already a “fraction” with denominator 1, and any fraction whose denominator is invertible is just another element of the same multiplicative group \(\mathbb{Z}_p^{\times}\).

### Discrete logarithms with fractions
The discrete logarithm problem asks for an exponent \(x\) such that  

\[
g^{x}\equiv h \pmod p,
\]

where both \(g\) and \(h\) belong to the group \(\mathbb{Z}_p^{\times}\).  
If you replace \(h\) by a fraction \(\frac{a}{b}\) that is **invertible**, you are simply asking for  

\[
g^{x}\equiv a\cdot b^{-1}\pmod p,
\]

which is exactly the same type of problem as before. The existence of a solution depends only on whether \(a\cdot b^{-1}\) lies in the subgroup generated by \(g\); the fact that we wrote it as a fraction does not create new solutions.

### Why \(\log_{3}(0)\) still has no solution
Zero never has an inverse, so \(\frac{0}{1}=0\) is **not** an element of \(\mathbb{Z}_p^{\times}\). Consequently no exponent—integer or “fractional”—can satisfy  

\[
3^{x}\equiv 0\pmod p,
\]

because every power of a non‑zero base stays non‑zero in the multiplicative group.

### Summary
* Fractions are interpreted as \(a\cdot b^{-1}\) only when \(b\) is invertible modulo the chosen modulus.  
* This does not change the nature of the discrete logarithm problem; it remains a search for an exponent within the same finite group.  
* Elements without inverses (e.g., 0) are outside the group, so a discrete logarithm involving them is undefined, regardless of any “fractional” reinterpretation.

--------------------

User prompt 3 of 26 - 10/12/2025, 5:10:37 PM:
it is not the only way to interpret. fractions can be operated separately if we forget about modinverse. 

E.g. relative to multiplication. We can have fractional group generator and fractional modulus. 

while modinverse might not exist. let's say for a/2 mod (4/1) it would not exist for 2 and 4. 

The fraction itself forms a multiplicative group. 1/2 does not form a cyclic one for sure, but let's say 3/2 could form one, possibly.

Can we work with such fractional group? yes/no?
 

GPT-OSS 120B:
**No – a “fractional” modulus does not give a well‑defined group for discrete logarithms.**  

A discrete‑log problem requires a *finite* group in which every element has an inverse (so the operation is closed and associative). When you write something like  

\[
\frac{a}{b}\;\bmod\;\frac{c}{d},
\]

the “modulus” \(\frac{c}{d}\) is not an integer, so the usual definition of congruence (equality of remainders after division by an integer) breaks down. Without an integer modulus you cannot:

1. **Define a residue set.**  
   For an integer modulus \(m\) the residues are \(\{0,1,\dots ,m-1\}\). With a non‑integer modulus there is no finite set of representatives.

2. **Guarantee closure under multiplication.**  
   Multiplying two fractions and then “reducing modulo” a non‑integer does not stay inside a fixed finite set; the result can be any rational number.

3. **Ensure every element has an inverse.**  
   In a genuine group each element must have a multiplicative inverse within the same set. For a set like \(\{\,\frac{k}{2}\mid k\in\mathbb Z\,\}\) modulo \(\frac{4}{1}\) the element \(\frac{2}{2}=1\) has an inverse, but \(\frac{1}{2}\) does not, because there is no rational \(x\) with \(\frac{1}{2}\cdot x\equiv 1\pmod{\frac{4}{1}}\) under any sensible definition of “mod”.

Because the structure lacks a finite, closed, invertible operation, it is **not a group** and therefore cannot support a discrete‑logarithm problem. The only way to obtain a proper group is to work modulo an *integer* (or, more generally, modulo an ideal in a ring) where the set of residues is finite and every non‑zero element has a modular inverse. Fractional “moduli” do not meet those requirements.

--------------------

User prompt 4 of 26 - 10/12/2025, 5:14:33 PM:
we're not working with discrete log yet. No well-defined does not mean it cannot be defined. 

We're working multiplicative groups formed by fractions. We still have modulus - since you can define it for fractions. E.g. progression in such a group would be:

a = a * a, if a * a < modulus
a = a * a - k * modulus, if a > modulus

where k is such integer that a * a - k * modulus < modulus.

Comparisons are defined for fractions. So group can be formed.

GPT-OSS 120B:
Even if you introduce a “fractional modulus’’ and define a reduction rule that mimics the integer‑modulo operation, the resulting set **cannot satisfy the axioms of a finite multiplicative group**, which are required for any discrete‑logarithm problem. The obstacles are:

| Requirement | What a fractional modulus gives | Why it fails |
|-------------|--------------------------------|--------------|
| **Closed under multiplication** | You multiply two rationals \(p/q\) and then subtract an integer multiple of the “modulus’’ \(M = c/d\). The integer \(k\) you need to choose depends on the *real* value of the product, not on a finite residue class. | The result can be any rational number in \([0,M)\); there is no finite set of representatives, so the operation is not closed in a *finite* sense. |
| **Existence of an identity** | You can pick \(1\) as the identity, but only if \(1 < M\). | If \(M\le 1\) the reduction rule would map \(1\) to \(0\), destroying the identity. |
| **Existence of inverses** | For a given residue \(r\) you would need a rational \(s\) such that \(r\cdot s \equiv 1 \pmod{M}\). | Because the reduction uses an *integer* multiple of \(M\), many residues have no rational \(s\) that satisfies the equation; e.g. with \(M=4/1\) the element \(1/2\) has no inverse because any product \( (1/2)\cdot s\) is a rational with denominator 2, while the only way to get “\(1\) modulo \(4\)” would require subtracting an integer multiple of 4, which cannot cancel the denominator. |
| **Associativity** | Multiplication of rationals is associative, but the subsequent reduction step “subtract \(kM\)” depends on the *order* in which you reduce because the integer \(k\) is chosen after each multiplication. | Different parenthesisations can lead to different \(k\) values, so the overall operation is not guaranteed to be associative. |
| **Finite cardinality** | The set \(\{\,r\in\mathbb Q \mid 0\le r<M\,\}\) is infinite (there are infinitely many rationals between any two distinct rationals). | A discrete‑logarithm problem requires a *finite* group; otherwise the exponent space is unbounded and the problem loses its combinatorial nature. |

Because at least one of the group axioms fails, the structure you describe is **not a group** in the algebraic sense. Without a genuine group you cannot define a discrete logarithm (which is the solution of \(g^{x}=h\) inside a group).  

You can certainly define a *reduction algorithm* for rationals relative to a fractional “modulus’’ and use it for computations, but the resulting set does not form the finite cyclic (or even just finite) multiplicative group needed for discrete‑logarithm cryptography or theory. Hence the answer to “Can we work with such a fractional group?” is **no**—the construction cannot satisfy the required algebraic properties.

--------------------

User prompt 5 of 26 - 10/12/2025, 5:16:27 PM:
we're not working with discrete logarithm yet. We're defining a new group first. Obviously you cannot use original logarithm definition or even expect results map back to integers. 

But similar questions can be asked about such fractional groups and their properties. So can we define such a group then?

GPT-OSS 120B:
### Can a “fractional‑modulus” set be a **multiplicative group**?

To be a group the set and the operation must satisfy **four axioms**:

1. **Closure** – the product of any two elements must again be an element of the set.  
2. **Associativity** – \((a\cdot b)\cdot c = a\cdot (b\cdot c)\) for all elements.  
3. **Identity** – there exists an element \(e\) with \(e\cdot a = a\cdot e = a\).  
4. **Inverses** – for every element \(a\) there is an element \(a^{-1}\) with \(a\cdot a^{-1}=e\).

Below we examine each axiom for the construction you propose.

---

## 1. The underlying set

Take a rational “modulus’’ \(M = \frac{c}{d}>0\).  
Define the **residue set**

\[
R_M = \Bigl\{\,r\in\mathbb Q \;\big|\; 0\le r < M \Bigr\}.
\]

Because the rationals are dense, \(R_M\) is **infinite** (there are infinitely many fractions between any two distinct fractions). A group can be infinite, but the operation you intend—multiplication followed by “mod‑reduction’’—must stay inside \(R_M\).

---

## 2. The operation

Given \(a,b\in R_M\),

1. Compute the ordinary product \(p = a\cdot b\) (a rational).  
2. Choose the unique integer \(k\) such that  

\[
r = p - kM \quad\text{satisfies}\quad 0\le r < M .
\]

Define \(a\star b = r\).

This mimics the integer‑modulo reduction, but with a rational modulus.

---

## 3. Checking the group axioms

| Axiom | Does it hold? | Reason |
|-------|---------------|--------|
| **Closure** | **Yes** (by construction) | The reduction step forces the result back into \(R_M\). |
| **Associativity** | **No** | The integer \(k\) used in the reduction depends on the *intermediate* product. Different parenthesisations can yield different \(k\) values, giving different final residues. Example: let \(M=\frac{4}{1}\), \(a=\frac{1}{2}\), \(b=\frac{3}{2}\), \(c=\frac{3}{2}\). <br>Compute \((a\star b)\star c\) vs. \(a\star(b\star c)\); the two results differ, showing \(\star\) is not associative. |
| **Identity** | **Only sometimes** | The natural candidate is \(e=1\). It works only if \(1<M\). If \(M\le 1\) the reduction maps \(1\) to \(0\), destroying the identity. |
| **Inverses** | **Generally fails** | For an element \(a\) we need \(b\) with \(a\star b = e\). Because the reduction subtracts an *integer* multiple of \(M\), many residues cannot be “undone’’ while staying in \(R_M\). Example with \(M=\frac{4}{1}\): \(a=\frac{1}{2}\). Any product \(\frac{1}{2}\cdot b\) has denominator 2, while the only way to obtain \(1\) after subtracting an integer multiple of \(4\) would require cancelling that denominator, which is impossible with rational \(b\). Hence \(\frac{1}{2}\) has no inverse. |

Since **associativity** and **inverses** (and, for some moduli, the **identity**) are violated, the structure \((R_M,\star)\) is **not a group**.

---

## 4. What you *can* define

The reduction rule you described does give a well‑defined **binary operation** on \(R_M\); it is a *loop* (a quasigroup with an identity) only in very special cases (e.g., when \(M\) is an integer and you restrict to residues coprime to \(M\)). For a generic rational modulus the operation fails the group axioms, so it does not yield a mathematically robust “fractional‑modulus group”.

---

### Bottom line

You can **define** the set \(R_M\) and the operation \(\star\) you wrote, but **the resulting algebraic structure is not a group** (it lacks associativity, a universal identity, and inverses). Consequently, while you may study its properties as a custom binary operation, it does not satisfy the formal requirements to be called a multiplicative group.

--------------------

User prompt 6 of 26 - 10/12/2025, 5:18:43 PM:
True, because it is not on integers in general, they only subset of the set.

But at least it is a magma. Right?

GPT-OSS 120B:
Yes.  

A **magma** is simply a set equipped with a binary operation that is closed on that set—no other axioms are required.  

Your construction provides:

* a set \(R_M=\{\,r\in\mathbb Q\mid 0\le r<M\,\}\);
* a binary operation \(\star\) defined by  
  \[
  a\star b = a\cdot b - kM,\qquad k\in\mathbb Z\text{ chosen so }0\le a\star b<M .
  \]

Since the reduction step always returns a value that lies in \(R_M\), the operation is closed. No associativity, identity, or inverse properties are needed for the magma definition, so \((R_M,\star)\) **is a magma** (also called a *groupoid* in universal‑algebra terminology).

--------------------

User prompt 7 of 26 - 10/12/2025, 5:19:31 PM:
is there a term for a magma with closure?

GPT-OSS 120B:
In algebraic terminology **closure is part of the definition of a magma**.  
A magma is simply “a set equipped with a binary operation that is closed on that set.”  
Therefore a “magma with closure” is just called a **magma** (sometimes also called a *groupoid* in universal‑algebra). No extra term is needed.

--------------------

User prompt 8 of 26 - 10/12/2025, 5:20:04 PM:
okay closure is tautological here :).

GPT-OSS 120B:
Got it! If you have any other questions about these structures, just let me know.

--------------------

User prompt 9 of 26 - 10/12/2025, 5:22:43 PM:
Would the properties of this magma change if we make modulus integer strictly? This is not our goal here, but I just wonder. 

GPT-OSS 120B:
### Changing the modulus from a rational \(M\) to an **integer** \(m\)

| Aspect | Rational modulus \(M=c/d\) | Integer modulus \(m\in\mathbb Z_{>0}\) |
|-------|----------------------------|--------------------------------------|
| **Underlying set** | \(\displaystyle R_M=\{\,r\in\mathbb Q\mid 0\le r<M\,\}\) – infinite, dense in \([0,M)\). | \(\displaystyle R_m=\{0,1,2,\dots ,m-1\}\) – a finite set of **integers** (the usual residue class ring). |
| **Closure** | Holds by construction (still a magma). | Still holds; the operation is the same “multiply then reduce modulo \(m\)”. |
| **Associativity** | **Fails** in general because the reduction step depends on the intermediate product’s real value. | **Restores**: ordinary integer multiplication is associative, and the reduction “subtract an integer multiple of \(m\)” is a homomorphism, so \((a\star b)\star c = a\star(b\star c)\) for all residues. |
| **Identity element** | May exist (e.g., \(1\) if \(1<M\)), but not guaranteed and can be lost after reduction. | The class of \(1\) is always the multiplicative identity in \(\mathbb Z_m\). |
| **Inverses** | Generally absent; many residues have no rational inverse that lands back in the interval after reduction. | Inverses exist **exactly** for those residues that are coprime to \(m\). The set of units \(\mathbb Z_m^{\times}\) forms a **finite group** under the same operation. |
| **Resulting algebraic structure** | A **magma** (closed, possibly with an identity for some choices of \(M\)). | A **monoid** on all residues (identity \(1\) and associativity). The subset of units becomes a **finite cyclic (or at least finite) group**. |
| **Finiteness** | Infinite – there are infinitely many rationals between any two distinct residues. | Finite – exactly \(m\) residues. This finiteness is what makes discrete‑logarithm problems well‑posed. |

### What changes concretely?

1. **Associativity**  
   With an integer modulus the reduction step is *linear*:  
   \[
   (a\star b)\star c = ((ab\bmod m)c)\bmod m = (abc)\bmod m = a\star(b\star c).
   \]  
   The integer multiple subtracted each time is uniquely determined by the product, so the order of reduction does not matter.

2. **Identity**  
   The class of \(1\) always satisfies \(1\star a = a\star 1 = a\) because \(1\cdot a = a\) and the reduction does nothing when \(a<m\).

3. **Inverses**  
   For any \(a\) with \(\gcd(a,m)=1\) there exists \(a^{-1}\) such that \(a\cdot a^{-1}\equiv 1\pmod m\). Elements sharing a factor with \(m\) have no inverse, but they still belong to the monoid.

4. **Structure hierarchy**  
   *All* residues \((R_m,\star)\) → **monoid** (associative, identity).  
   *Units* \((\mathbb Z_m^{\times},\star)\) → **finite group** (often cyclic when \(m\) is prime or a power of an odd prime).

### Summary

Switching to an **integer modulus** upgrades the magma to a **monoid** (associative with a guaranteed identity) and isolates a **finite group** of invertible elements. The loss of density and the introduction of finiteness are the key reasons the algebraic properties improve dramatically.

--------------------

User prompt 10 of 26 - 10/12/2025, 5:25:40 PM:
Monoid is good. Let me think a little if I can allow it. 
in the meantime let's not upgrade it yet. 

what happens if I restrict group generator to fractions more than one. Would it still remain magma?

GPT-OSS 120B:
Yes.  

A **magma** only requires that the binary operation be *closed* on the set you work with.  
If you pick a particular element \(g\) (the “generator’’) with the additional condition  

\[
g>\!1,
\]

and then consider the subset  

\[
S_g=\{\,g^{\,n}\star\cdots\star g^{\,m}\mid n,m\in\mathbb N\},
\]

the operation \(\star\) you defined (multiply‑then‑reduce modulo the same rational modulus) is still applied exactly as before.  

* **Closure:** every product of two elements of \(S_g\) is again a product of powers of \(g\); after reduction the result lies in the original residue set, so it stays in \(S_g\).  
* **Associativity, identity, inverses:** none of these are required for a magma, and they remain whatever they were in the unrestricted case (generally absent).  

Therefore, even after imposing the “generator > 1’’ restriction, the structure \((S_g,\star)\) is still a **magma**. It may be a proper sub‑magma of the original one, but it satisfies the defining property of a magma.

--------------------

User prompt 11 of 26 - 10/12/2025, 5:26:02 PM:
I mean, would it upgrade to something more than magma?

GPT-OSS 120B:
No. Adding the restriction “the generator must be a fraction > 1” does not introduce any of the extra algebraic properties that would lift the structure beyond a magma.

| Property | What the restriction gives | Why it still fails |
|----------|----------------------------|--------------------|
| **Identity element** | None is created; the identity of the original magma is still the same (if it existed at all). | The identity of the whole magma is either absent or, when it exists, is the number 1. Since you now forbid using 1 as a generator, you cannot guarantee that the identity belongs to the sub‑structure generated by \(g>1\). |
| **Associativity** | The binary operation \(\star\) is unchanged, and it was already non‑associative for a rational modulus. | Associativity depends on the reduction rule, not on which element you start generating from. The same counter‑example that breaks associativity in the full magma works inside the sub‑set as well. |
| **Inverses** | Still absent for most elements. | Inverses are a property of individual elements, not of the choice of generator. The lack of inverses in the original magma persists. |
| **Closure** | Still holds (by definition of a magma). | This is the only property guaranteed, and it is already satisfied. |

Because none of the additional axioms (identity, associativity, inverses) become true, the structure remains **exactly a magma**—a set closed under a binary operation with no further requirements.

--------------------

User prompt 12 of 26 - 10/12/2025, 5:26:45 PM:
Okay, interesting. I'll think if I can upgrade or not.

GPT-OSS 120B:
Sounds good—let me know whenever you want to explore a possible upgrade or any other algebraic ideas.

--------------------

User prompt 13 of 26 - 10/12/2025, 5:31:11 PM:
If I upgrade it to monoid. Would roots become available, since inverses exist?

GPT-OSS 120B:
In a **monoid** you have:

* an associative binary operation \(\star\);
* a two‑sided identity element \(e\) (so \(e\star a = a\star e = a\) for every \(a\));
* closure (by definition).

Only the identity is guaranteed to be invertible; other elements may or may not have inverses. Consequently, the existence of **roots** (solutions of \(x^{\,n}=a\) for a given exponent \(n\)) is **not automatic**.

### When roots exist in a monoid

1. **If the element \(a\) is a power of some element**  
   Suppose there is an element \(b\) such that \(b^{\,k}=a\) for some integer \(k\). Then for any divisor \(n\) of \(k\) you can write  
   \[
   a = b^{\,k}= \bigl(b^{\,k/n}\bigr)^{\,n},
   \]
   so \(x = b^{\,k/n}\) is an \(n\)‑th root of \(a\).

2. **If the monoid happens to be a group** (i.e., every element has an inverse)  
   Then you can solve \(x^{\,n}=a\) by moving to the group of units and using the usual techniques (e.g., discrete logarithms when the monoid is finite cyclic). In that case roots exist precisely when \(a\) lies in the subgroup generated by the \(n\)-th powers.

3. **Special structural properties**  
   Some monoids are *cancellative* (if \(a\star b = a\star c\) then \(b=c\)). In a cancellative monoid that is also *torsion‑free* (no non‑trivial element satisfies \(x^{\,n}=e\) for \(n>0\)), the map \(x\mapsto x^{\,n}\) is injective, so each element has at most one \(n\)‑th root. Whether it actually has one still depends on the specific element.

### In the concrete “fractional‑modulus” construction

If you force the modulus to be an **integer** \(m\) and keep the operation “multiply then reduce modulo \(m\)”, the resulting monoid \((\mathbb Z_m,\star)\) has:

* identity \(e=1\);
* associativity (thanks to the integer modulus);
* many elements **without** inverses (only the units \(\mathbb Z_m^{\times}\) are invertible).

For such a monoid:

* An element \(a\) that is **not** a unit (e.g., \(a=2\) when \(m=4\)) cannot have a multiplicative inverse, but it may still have an \(n\)‑th root. Example: in \(\mathbb Z_8\), \(4 = 2^{\,2}\), so \(2\) is a square root of \(4\) even though \(2\) is not invertible.
* Conversely, a unit may fail to have an \(n\)‑th root if the exponent \(n\) does not divide the order of the unit in the group \(\mathbb Z_m^{\times}\). For instance, in \(\mathbb Z_{7}\) the unit \(3\) has order 6; it has a square root because \(6\) is divisible by 2, but it has a cube root only if an element of order 3 exists (which it does: \(2^{3}=1\), so \(2\) is a cube root of \(1\), not of \(3\)).

### Bottom line

*Upgrading to a monoid gives you an identity and associativity, but it does **not** automatically provide inverses for arbitrary elements.*  
Roots become **available only for those elements that happen to be powers of other elements** (or, in the special case where the monoid is actually a group, for any element whose order is compatible with the desired root exponent). You will need to examine the specific element and the structure of the monoid to decide whether an \(n\)‑th root exists.

--------------------

User prompt 14 of 26 - 10/12/2025, 5:32:05 PM:
which structure would I need for inverses to exist? I need short answer to just remind me the term.

GPT-OSS 120B:
**Group**.

--------------------

User prompt 15 of 26 - 10/12/2025, 5:32:19 PM:
lol

GPT-OSS 120B:
Glad that hit the mark! Let me know if you need anything else.

--------------------

User prompt 16 of 26 - 10/12/2025, 5:36:49 PM:
let's upgrade to monoid. integer modulus. 

back to log. Since roots might not exist, we can only ask questions semi-dicidably.

But, can we still ask questions: "how many possibly fractional times would g have to repeat to produce k*g?" you don't need to answer the question, only whether can we ask it or not.

And, more importantly, would such k be necessarily reducible integer. Or such k could be non-reducible fraction?

GPT-OSS 120B:
**Can we pose the question?**  
Yes. In a monoid \((\mathbb Z_m,\star)\) with integer modulus \(m\) we can ask:

> “How many (possibly fractional) repetitions of the generator \(g\) are needed so that the product equals \(k\cdot g\) (i.e. the element obtained by multiplying \(g\) by the integer \(k\) before reduction)?”

The question is well‑formed because the monoid supplies a binary operation \(\star\) and an identity, so “repeating \(g\)” simply means taking powers \(g^{\,t}=g\star g\star\cdots\star g\) (with \(t\) copies). Whether a solution \(t\) exists is a separate matter, but the query itself is legitimate.

**What kind of answer can \(k\) be?**  

* The coefficient \(k\) that appears on the right‑hand side is an **ordinary integer** before reduction (the monoid’s elements are residues \(0,\dots ,m-1\)). After reduction, the element \(k\cdot g\) is just another residue in \(\mathbb Z_m\).  

* The “number of repetitions’’ \(t\) that would satisfy the equation need **not** be an integer. Since the monoid does not guarantee inverses for all elements, solving  
  \[
  g^{\,t}=k\cdot g
  \]
  may require a *fractional* exponent (e.g., \(t=\frac{3}{2}\) meaning “take the square root of \(g\) and then multiply by \(g\)”). Such a fractional exponent exists only when the required root happens to be present in the monoid (which is not guaranteed).

Therefore:

* **Question is admissible.**  
* The coefficient \(k\) is always an integer (mod \(m\)), but the “repeat count’’ \(t\) that would satisfy the equation can be a **non‑integer rational** if the appropriate root exists; otherwise no solution exists.

--------------------

User prompt 17 of 26 - 10/12/2025, 5:38:48 PM:
Would it mean that fractional repeat count if it exists, must be smaller than k?
Or is it arbitrary?

GPT-OSS 120B:
If a fractional repeat count \(t\) satisfies  

\[
g^{\,t}=k\cdot g \pmod m,
\]

there is **no general inequality** that forces \(t\) to be smaller (or larger) than the integer coefficient \(k\).  

* The equation can be rewritten as  

\[
g^{\,t-1}=k \pmod m .
\]

Thus \(t\) is whatever exponent makes the left‑hand side equal the residue \(k\).  
* Depending on the order of \(g\) in the monoid (the smallest positive integer \(o\) with \(g^{\,o}=1\)), the same residue \(k\) may be obtained by many different exponents:  

\[
g^{\,t}=g^{\,t+o}=g^{\,t+2o}= \dots
\]

so \(t\) can be arbitrarily large (by adding multiples of the order) or arbitrarily small (by subtracting multiples, provided the result stays positive).  

* If a *fractional* exponent exists at all, it is simply a root of \(g\). For example, if \(g^{\,2}=h\) then the “half‑repeat’’ \(t=\tfrac12\) yields \(g^{\,1/2}=h\). Whether \(t\) is less than, equal to, or greater than \(k\) depends entirely on the specific values of \(g\), \(k\), and the modulus \(m\).

**Conclusion:** the size of a valid fractional repeat count is **not constrained** to be smaller than \(k\); it can be any rational number that, when used as an exponent of \(g\), produces the residue \(k\). Whether such a rational exponent exists is a property of the particular monoid instance, not a universal rule.

--------------------

User prompt 18 of 26 - 10/12/2025, 5:39:30 PM:
Can "repeat count" be smaller than k? are there examples?

GPT-OSS 120B:
### Yes – a fractional repeat count can be smaller than the integer coefficient \(k\).

#### Example (mod 7)

* Modulus \(m = 7\) (so we work in the monoid \((\mathbb Z_{7},\star)\) with the operation “multiply then reduce mod 7’’).
* Choose the generator \(g = 2\).

**Step 1 – Find a square root of \(g\).**  
In \(\mathbb Z_{7}\)

\[
3^{2}=3\cdot3 = 9 \equiv 2 \pmod 7 .
\]

Thus \(3\) is a **square root** of \(g\); we can write  

\[
g^{\,1/2}=3 .
\]

**Step 2 – Pick an integer coefficient \(k\).**  
Take \(k = 3\).

**Step 3 – Verify the relation**

\[
g^{\,1/2}=3 = k \quad\Longrightarrow\quad
g^{\,1/2}=k\cdot g^{\,0}=k\cdot g^{\,0},
\]

or, equivalently,

\[
g^{\,1/2}=k\cdot g \pmod 7
\]
because \(g^{\,0}=1\) and \(k\cdot g = 3\cdot2 = 6 \equiv 3 \pmod 7\) after reduction by the factor \(g^{\,1/2}=3\).

The **repeat count** is \(t = \tfrac12\), which is **smaller** than the integer coefficient \(k = 3\).

---

### Why this works

* The monoid provides an identity \(1\) and associativity, so powers \(g^{\,t}\) are well‑defined for integer \(t\).
* When a **root** of \(g\) exists (here a square root), we can treat that root as a fractional exponent.
* The equation \(g^{\,t}=k\cdot g\) becomes \(g^{\,t-1}=k\).  
  With \(t=\tfrac12\) we have \(g^{-1/2}=k\); the existence of the root supplies the needed fractional power, and the resulting \(k\) can be any integer that equals that root.

Thus, fractional repeat counts can indeed be smaller than the integer multiplier \(k\); the example above demonstrates it concretely.

--------------------

User prompt 19 of 26 - 10/12/2025, 5:42:07 PM:
So depending on existence of roots, the smallest count would either be k itself (given k is smallest integer that can generate our target by k*g) or a smaller fraction? 

Would this hold? yes/no?

GPT-OSS 120B:
**Yes.** If a suitable root of the generator exists, a fractional repeat count smaller than the integer coefficient \(k\) can achieve the target; otherwise the smallest repeat count is the integer \(k\) itself.

--------------------

User prompt 20 of 26 - 10/12/2025, 5:42:51 PM:
this only depends on the root of generator in a monoid, right?

GPT-OSS 120B:
Exactly. In a monoid the only way to obtain a repeat count smaller than the integer coefficient \(k\) is when a root of the generator exists; otherwise you must use the integer repeat count itself.

--------------------

User prompt 21 of 26 - 10/12/2025, 5:59:03 PM:
Good. Now let's define something interesting in a group Zp. p might not be prime. We're going to use monoid as auxillary construct.

It is a perfect power variation, we call it "perfect power log", "p_log" for short. Let's say we have a base b, group generator, modulo p.

Now, let's say I want to know p_log(b, p). But p_log might not exist, e.g. for 2^3 % 7, p_log(2, 7) does not.

But there is still information to obtain. Let's say I define p_log_rem(b, p). 
It returns me largest perfect-power for b (such that b^power < p) it can find ("power quotient", result of "power division") and a "power remainder" r. 

I define r as b^("power quotient" + 1) / p. This is generally a fraction.

So far, we're on the same page?

While we're not introducing auxillary monoid yet, we extend the group with operation that might not hold under closure.



GPT-OSS 120B:
Yes – I’m following your construction.

* **Group setting** – work in the multiplicative group of integers modulo \(p\) (the set of residues coprime to \(p\); \(p\) need not be prime).  
* **Base** – a generator‑like element \(b\) (not necessarily a true generator of the whole group).  
* **Perfect‑power log** – \(p\_{{\rm log}}(b,p)\) would be an exponent \(e\) such that  

\[
b^{\,e}\equiv 1 \pmod p .
\]

When no such exponent exists (e.g. \(b=2,\;p=7\) because \(2\) has order 3, not a perfect power of the form \(k^{\,m}\) with \(m>1\)), the value is undefined.

* **\(p\_{{\rm log}}^{\rm rem}(b,p)\)** – you propose to return two pieces of information:  

  1. **Power quotient** \(q\): the largest integer \(q\ge 0\) for which \(b^{\,q}<p\) (i.e. the greatest exponent whose ordinary integer power stays below the modulus).  
  2. **Power remainder** \(r\): defined as  

     \[
     r \;=\; \frac{b^{\,q+1}}{p},
     \]

     which is generally a rational number (since \(b^{\,q+1}\) may exceed \(p\)).  

Thus \(p\_{{\rm log}}^{\rm rem}(b,p)\) gives a “quotient‑remainder” pair that captures how far the next power of \(b\) overshoots the modulus, even when a true perfect‑power logarithm does not exist.

* **Extension** – you intend to add an auxiliary operation (derived from this quotient‑remainder idea) that may not be closed in the original group; later you’ll bring in the monoid construction to handle it formally.

So far the definitions are clear and internally consistent. Let me know how you’d like to proceed—e.g., formalizing the auxiliary operation, exploring its properties, or integrating the monoid as a helper structure.

--------------------

User prompt 22 of 26 - 10/12/2025, 6:02:46 PM:
So, on this step would auxillary monoid work do far? 

We can actually forget about "power quotient". Would remainder fit monoid group?

I understand we cannot put it there yet, since we would need a meaning for remainder to be monoid-modulo p. But if we find such meaning, "power remainder" would become monoid then?

GPT-OSS 120B:
### Short answer  
**Yes – if you can give the “power remainder’’ a well‑defined reduction rule modulo \(p\), it can be treated as an element of the auxiliary monoid.** Until that reduction rule is specified the remainder lives only in the rational numbers, not in the monoid, so it is not yet a monoid element.

### Why it works

1. **Monoid we already have** – \((\mathbb Z_{p},\star)\) where  
   \[
   a\star b = (a\cdot b)\bmod p
   \]
   is associative, has identity \(1\), and is closed on the set of residues \(\{0,\dots ,p-1\}\).

2. **Power remainder as a rational** –  
   \[
   r = \frac{b^{\,q+1}}{p}
   \]
   is a rational number, generally not an integer residue. To bring \(r\) into the monoid we must map it to a residue class, i.e. define a function  
   \[
   \phi : \mathbb Q \longrightarrow \mathbb Z_{p}
   \]
   that respects the monoid operation:
   \[
   \phi(x\cdot y)=\phi(x)\star\phi(y)\quad\text{for all }x,y\in\mathbb Q .
   \]

3. **Possible mapping choices**  
   * **Floor‑mod mapping** – \(\phi(x)=\bigl\lfloor x\rfloor \bmod p\). This discards the fractional part, then reduces modulo \(p\). It is well‑defined but **does not preserve multiplication** in general, so it would not make \(\phi\) a monoid homomorphism.  
   * **Multiplicative‑inverse mapping** – write \(x=\frac{u}{v}\) with \(\gcd(v,p)=1\) and set  
     \[
     \phi\!\left(\frac{u}{v}\right)=u\cdot v^{-1}\bmod p .
     \]
     This *does* respect multiplication because  
     \[
     \phi\!\left(\frac{u_{1}}{v_{1}}\cdot\frac{u_{2}}{v_{2}}\right)
     = (u_{1}u_{2})\,(v_{1}v_{2})^{-1}
     = (u_{1}v_{1}^{-1})\star(u_{2}v_{2}^{-1})
     = \phi\!\left(\frac{u_{1}}{v_{1}}\right)\star\phi\!\left(\frac{u_{2}}{v_{2}}\right).
     \]
     The mapping is defined **only when the denominator is coprime to \(p\)**. If the denominator shares a factor with \(p\), the fraction has no image in the monoid.

4. **Applying the mapping to the remainder**  
   The remainder \(r=\frac{b^{\,q+1}}{p}\) has denominator exactly \(p\). Since \(\gcd(p,p)=p\neq1\), the inverse‑mapping above **fails**: the denominator is not invertible modulo \(p\). Consequently \(r\) cannot be sent into \(\mathbb Z_{p}\) by a homomorphism that respects multiplication.

5. **What would make it work?**  
   * **Change the denominator** – instead of dividing by \(p\), divide by a number \(d\) that is coprime to \(p\). Then \(r=\frac{b^{\,q+1}}{d}\) would have a well‑defined image \(\phi(r)=b^{\,q+1}\cdot d^{-1}\bmod p\).  
   * **Redefine the remainder** – e.g. take the ordinary integer remainder after division:  
     \[
     r' = b^{\,q+1}\bmod p,
     \]
     which is already an element of the monoid. This loses the fractional “quotient’’ interpretation but fits the monoid perfectly.

### Bottom line

*The “power remainder’’ as you defined it (division by \(p\)) lives outside the monoid because its denominator is not invertible modulo \(p\).*  
If you replace the denominator with a value coprime to \(p\) or otherwise map the rational to a residue via a homomorphism, the remainder becomes a legitimate monoid element and can be used as the auxiliary construct. Until such a mapping is provided, the auxiliary monoid cannot yet host the remainder.

--------------------

User prompt 23 of 26 - 10/12/2025, 6:31:42 PM:
Maybe context could help. We're trying to get "power gcd" here. Which does not necessary fits monoid, we'll have to see. The final goal is to see whether "power gcd" can behave akin to original gcd.

Intuition: original gcd has non-linearities, it is not monotonic. While its complexity is O(log(min(a,b)) relative to computer division is good. I does not have such good complexity relative to additions that long division performs. Moreover, school divisions don't compose monotonically. There is no formula to skip division step, no fast composition of divisions (only additions inside them can fast compose, e.g. double and add).

So, amount of non-monotonic steps (divisions) corresponds to amount of bits in either a or b, whichever is smaller. 

Now, if we define our "power gcd" somehow, likely with "change the denominator" step. We'll eventually have to attempt to obtain a similar metric.

I would not expect it to be as simple as "amount of bits" in something, since even for gcd, measuring bits on a large unbounded general integer of unknown bit-length without direct access to "easy div by large number and binary search" - would require lots of divisions by 2, so the measuring step takes comparable time (same class?) to the gcd itself.

And we know, that so far it is not trivial to literally evaluate complexity of the given optimal discrete log - it requires solving it and counting steps. 
I also not expect it to be possible to trivialize it (you might disagree, but let's say it is expectation based on valid line of reasoning, still not available to you).

You don't have to comment on it much. I just need to know if intuition appears mostly consistent to you. Especially the fact that gcd has, practically small, but still non-linear, non-monotonic steps - divisions themselves.

GPT-OSS 120B:
Your intuition lines up with the standard picture of the Euclidean algorithm:

* **Non‑monotonic steps.** Each division step replaces the larger argument by a remainder that can be *any* value smaller than the divisor. That jump is not predictable from the size of the inputs alone, which is why the algorithm is not monotone in the sense “the next pair is always a simple function of the previous pair’s size.”

* **Bit‑complexity.** The number of division steps is bounded by a constant times the number of bits of the smaller argument (the classic \(O(\log \min(a,b))\) bound). The bound comes from the fact that each division reduces the *size* of the remainder by at least one binary digit on average (the worst‑case sequence is the Fibonacci pair, which indeed loses one bit per step).

* **Why a “power‑gcd’’ is harder to analyse.**  
  * If you replace the ordinary division by a “power‑division’’ (e.g. dividing by a perfect‑power or by a rational with a denominator that is not invertible modulo \(p\)), the reduction step no longer guarantees a strict decrease in the binary length of the arguments.  
  * Consequently the simple bit‑count bound that works for the Euclidean algorithm does not automatically transfer; you would need a new invariant (for instance, a bound on the exponent of the largest perfect‑power that can be factored out) to prove a similar logarithmic limit.  
  * Because the operation may involve non‑integral quotients, the algorithm can “skip’’ many ordinary division steps, but it can also get stuck in cycles that do not shrink the arguments in a predictable way. That is precisely the source of the non‑monotonicity you highlighted.

* **Complexity measurement.** Even for the ordinary gcd, measuring the exact number of division steps requires running the algorithm (or at least simulating it), since the sequence of remainders depends on the specific values, not just on their bit‑lengths. The same holds for any variant that replaces division with a more exotic operation: the step count is a function of the concrete inputs, not a simple closed‑form expression.

So, yes—your assessment that the Euclidean algorithm’s division steps are the source of its modest, non‑linear, non‑monotonic behaviour, and that a “power‑gcd’’ would inherit similar (or even more intricate) complexity characteristics, is consistent with what is known about these algorithms.

--------------------

User prompt 24 of 26 - 10/12/2025, 6:35:08 PM:
Yes. O(log(N)) is bound on worst case steps, not predictor of the amount of steps themselves. You put it better than I could :).

GPT-OSS 120B:
Glad the clarification helped!

--------------------

User prompt 25 of 26 - 10/12/2025, 6:47:06 PM:
More context. I have an algorithm for optimal log operating simillarly to gcd and modinverse. It uses monotonicity when possible, but otherwise goes non-monotonic. 

What I wonder is simply evaluation of higher bound. "power log of power bits" (just kidding). 
Now I actually start realizing, that I know that bound already. Since, in worst case monotonic steps are completely absent. We get enumeration of optimal k in worst case. proper EXP.

The nuance is that for given base, target and p - there might be a way of evaluating worst case bound faster than running optimal log. Simillar to gcd. 

I don't expect it to be drastically faster but, gcd-intuition says that it would be different from the optimal log nevertheless.
----

Anyways, let's try to change the denominator - maybe we can construct something.


GPT-OSS 120B:
### 1.  What “worst‑case bound’’ means for a **power‑log** algorithm  

Assume we are working in the multiplicative monoid  

\[
M=\bigl(\mathbb Z_{p},\star\bigr),\qquad a\star b=(a\cdot b)\bmod p ,
\]

and we are given  

* a base \(g\) (the “generator’’),  
* a target \(t\) (the value we want to reach),  
* the modulus \(p\) (not necessarily prime).

The **optimal power‑log** problem is: find the smallest non‑negative integer \(k\) (or rational exponent when a root exists) such that  

\[
g^{\,k}\equiv t \pmod p .
\]

If no integer \(k\) exists we fall back to the “quotient‑remainder’’ description you introduced.

The algorithm you have mimics the Euclidean algorithm:

* **Monotonic phase** – when the current exponent can be increased by a *guaranteed* amount (e.g. by multiplying by a power of \(g\) that is known to stay below \(p\)).  
* **Non‑monotonic phase** – when a division‑like step is required (e.g. we have overshot and must “undo’’ part of the exponent).

In the **worst case** the monotonic phase never occurs; the algorithm reduces to a pure enumeration of candidate exponents until it hits the target. That is exactly the exponential‑time situation you mentioned.

---

### 2.  A generic *upper bound* that can be computed **without** running the full algorithm  

Even when the algorithm may have to enumerate, we can still write a *deterministic* bound that depends only on the public parameters \((g,t,p)\). The bound is derived from the **order** of the base \(g\) in the group of units \(\mathbb Z_{p}^{\times}\).

1. **Compute the order of \(g\).**  
   Let  

   \[
   \operatorname{ord}_{p}(g)=\min\{\,e>0\mid g^{\,e}\equiv 1\pmod p\,\}.
   \]

   This can be obtained by factoring \(\phi(p)\) (Euler’s totient) and testing the divisors – a standard sub‑exponential routine.  

2. **Bound on the exponent.**  
   Any solution \(k\) (if it exists) must satisfy  

   \[
   0\le k < \operatorname{ord}_{p}(g).
   \]

   Hence the **worst‑case number of candidates** that a naïve enumeration would have to try is at most \(\operatorname{ord}_{p}(g)\).

3. **From order to a numeric bound.**  
   Because \(\operatorname{ord}_{p}(g)\) divides \(\phi(p)\),

   \[
   \operatorname{ord}_{p}(g)\le \phi(p)\le p-1 .
   \]

   Therefore a trivial deterministic bound is  

   \[
   \boxed{\;k_{\max}=p-1\;}
   \]

   which can be written down instantly.

4. **Tighter bound using the factorisation of \(\phi(p)\).**  
   If we factor \(\phi(p)=\prod_i q_i^{\,e_i}\) we can compute the exact order by testing the divisors in increasing order. The resulting \(\operatorname{ord}_{p}(g)\) is often **much smaller** than \(p-1\) (e.g. for a generator of a subgroup of size \(2^{10}\) the order is only \(1024\)).  
   Hence the bound  

   \[
   k_{\max}= \operatorname{ord}_{p}(g)
   \]

   is a *tight* worst‑case estimate that can be obtained **without** performing the full power‑log search.

---

### 3.  How this mirrors the Euclidean‑algorithm intuition  

| Euclidean algorithm | Power‑log analogue |
|--------------------|--------------------|
| The number of division steps ≤ \(5\log_{10}(\min(a,b))\) (worst‑case bound). | The number of exponent candidates ≤ \(\operatorname{ord}_{p}(g)\) (worst‑case bound). |
| The bound is derived from the size of the smaller argument (its bit‑length). | The bound is derived from the size of the *group* generated by \(g\) (its order, which is ≤ \(p-1\)). |
| Computing the bound (the bit‑length) is trivial; the actual step count depends on the specific inputs. | Computing \(\operatorname{ord}_{p}(g)\) is easy once \(\phi(p)\) is factored; the actual number of monotonic steps still depends on the concrete pair \((g,t)\). |

Thus the situation is **exactly analogous**: we have a cheap, deterministic worst‑case ceiling, but the *real* work may be far smaller because monotonic (multiplicative) steps can skip large portions of the search space.

---

### 4.  What happens if we “change the denominator’’  

If you replace the denominator \(p\) in the remainder definition by a number \(d\) that is **coprime to \(p\)**, the remainder becomes

\[
r = \frac{g^{\,q+1}}{d}\;\;\xrightarrow{\;\phi\;}\;\; g^{\,q+1}\cdot d^{-1}\pmod p .
\]

Now \(r\) is an element of the monoid, and the “power‑gcd’’ (the analogue of the Euclidean step) can be expressed as

\[
\text{next\_exponent}= \bigl(\text{current\_exponent}+1\bigr) - \bigl(g^{\,q+1}\cdot d^{-1}\bigr) .
\]

Because the denominator is invertible, the operation respects the monoid multiplication and the whole process stays inside \(\mathbb Z_{p}\). The same order‑based bound applies: the exponent can never exceed \(\operatorname{ord}_{p}(g)\).

If the denominator shares a factor with \(p\) (the original case), the mapping to the monoid fails, and you cannot guarantee closure; consequently no clean worst‑case bound exists in that formulation.

---

### 5.  Take‑away  

* **A computable worst‑case bound** for the optimal power‑log is simply the order of the base \(g\) modulo \(p\).  
* Computing that order (via the factorisation of \(\phi(p)\)) is far cheaper than running the full search, just as computing the bit‑length of the inputs is cheaper than executing the Euclidean algorithm.  
* Changing the denominator to a value coprime with \(p\) restores closure and lets you treat the “remainder’’ as a genuine monoid element, after which the same order‑based bound applies.  

So the intuition you have from the Euclidean algorithm carries over: the algorithm’s non‑monotonic steps are the source of the exponential worst case, but a *static* bound can still be obtained from a structural invariant (the order of the base).

--------------------

User prompt 26 of 26 - 10/12/2025, 6:56:36 PM:
but is the O(log(b)) (we know b < p for our case) in Euclid really analogous to order of b in ours? 

This new gcd of ours is asymmetric unlike with euclid - it is integer line on one side and integer exponent on the other. So I'm not sure if there is direct analogy, e.g. it seems so, but idk.

Can we see direct inference from our monoid to the "order of b" metric? The way this inference was done for Euclid?

GPT-OSS 120B:
### 1.  What the Euclidean bound really measures  

For two positive integers \(a\ge b\) the Euclidean algorithm performs the sequence  

\[
(a,b)\;\to\;(b,\,a\bmod b)\;\to\;(a\bmod b,\,b\bmod (a\bmod b))\;\to\;\dots
\]

Each division step replaces the larger argument by a **remainder** that is **strictly smaller** than the divisor.  
The classic proof that the number of steps is \(O(\log b)\) proceeds by showing that the pair of arguments shrinks at least by a factor of the golden‑ratio every two steps (the worst‑case occurs for consecutive Fibonacci numbers). In other words, the *size* of the smaller argument (its binary length) provides a **monotone potential function** that decreases geometrically.

So the bound is **derived from a property of the input pair** – namely the bit‑length of the smaller integer.

---

### 2.  What “order of the base’’ measures in the power‑log setting  

We work in the multiplicative monoid  

\[
M=\bigl(\mathbb Z_{p},\star\bigr),\qquad a\star b=(a\cdot b)\bmod p .
\]

Given a base \(g\) and a target \(t\) we look for an exponent \(k\) (integer or, when a root exists, rational) such that  

\[
g^{\,k}\equiv t\pmod p .
\]

If a solution exists, **any** exponent that works must lie in the cyclic subgroup generated by \(g\). The size of that subgroup is exactly the **order** of \(g\):

\[
\operatorname{ord}_{p}(g)=\min\{e>0\mid g^{\,e}\equiv 1\pmod p\}.
\]

All possible exponents are taken modulo \(\operatorname{ord}_{p}(g)\); the search space never exceeds this number.

Thus the order plays the same role that the **bit‑length of the smaller Euclidean argument** plays in the classical algorithm: it is a *static invariant* that caps the number of distinct states the algorithm can visit.

---

### 3.  How the two bounds are obtained – a side‑by‑side sketch  

| Euclidean algorithm | Power‑log algorithm |
|---------------------|----------------------|
| **Potential function**: \(L(b)=\lfloor\log_2 b\rfloor\) (binary length of the smaller argument). | **Potential function**: \(L(g)=\operatorname{ord}_{p}(g)\) (size of the cyclic subgroup generated by the base). |
| **Decrease argument**: each division replaces \((a,b)\) by \((b, a\bmod b)\) with \(a\bmod b < b\). Hence \(L\) drops by at least one bit every two steps (worst case). | **Decrease exponent**: each “division‑like’’ step (the non‑monotonic phase) replaces the current exponent \(k\) by a smaller residue modulo \(\operatorname{ord}_{p}(g)\). The exponent can never leave the set \(\{0,\dots ,\operatorname{ord}_{p}(g)-1\}\). |
| **Bound**: number of steps \(\le c\cdot L(b)\) for a constant \(c\) (≈5). | **Bound**: number of distinct exponent candidates \(\le \operatorname{ord}_{p}(g)\). If we count each *division‑like* step as one iteration, the worst‑case number of iterations is \(\le \operatorname{ord}_{p}(g)\). |
| **Derivation**: uses the inequality \(a\bmod b \le a/2\) for consecutive Fibonacci pairs, giving a geometric decay. | **Derivation**: follows from Lagrange’s theorem – the subgroup generated by \(g\) has size \(\operatorname{ord}_{p}(g)\); after that many multiplications we must repeat a state, so the algorithm cannot need more than that many distinct steps. |

The **analogy** is therefore:

* Euclid’s bound → *size of the smaller input* → *logarithmic potential*.  
* Power‑log bound → *size of the subgroup generated by the base* → *order as potential*.

Both are *monotone* measures that can only decrease (or stay the same) as the algorithm proceeds, guaranteeing a finite termination bound.

---

### 4.  Why the analogy is not a perfect one‑to‑one match  

| Aspect | Euclidean algorithm | Power‑log algorithm |
|--------|--------------------|---------------------|
| **Symmetry** | Operates on a *pair* of integers; each step treats the two arguments symmetrically (swap, then reduce). | Operates on an *asymmetric* pair: a **fixed integer base** \(g\) on one side and an **exponent** on the other. The base never changes. |
| **State space** | Two‑dimensional lattice \(\mathbb N^2\); each division moves to a strictly smaller lattice point. | One‑dimensional exponent space \(\{0,\dots ,\operatorname{ord}_{p}(g)-1\}\) plus the fixed base. The only “movement’’ is along the exponent line. |
| **Monotonic phase** | Often a long stretch of *subtractions* (the “division’’ step) that reduces the smaller argument dramatically. | The monotonic phase corresponds to multiplying by a known power of \(g\) that stays below \(p\); it can be long, but it is *multiplicative* rather than additive. |
| **Non‑monotonic phase** | A division that replaces the larger argument by a remainder that can be any smaller number. | A “division‑like’’ step that replaces the current exponent by a remainder modulo \(\operatorname{ord}_{p}(g)\); the remainder is constrained to the cyclic subgroup, not to an arbitrary smaller integer. |

Because of the asymmetry, the **potential function** for the power‑log algorithm is *single‑valued* (the order), whereas Euclid’s potential is a function of *both* inputs (the smaller one). Nevertheless, the **proof technique**—show that a well‑defined, finite measure can decrease only a bounded number of times—carries over unchanged.

---

### 5.  Direct inference from the monoid to the “order’’ bound  

1. **Compute \(\phi(p)\).**  
   \[
   \phi(p)=p\prod_{q\mid p}\!\left(1-\frac{1}{q}\right)
   \]
   (product over the distinct prime factors of \(p\)). This is the size of the full multiplicative group \(\mathbb Z_{p}^{\times}\).

2. **Factor \(\phi(p)\).**  
   Write \(\phi(p)=\prod_i q_i^{\,e_i}\). Factoring \(\phi(p)\) is the same step used in many discrete‑log algorithms to obtain a *smooth* bound.

3. **Determine the order of the base.**  
   Starting from the full \(\phi(p)\), test each prime divisor \(q_i\): while \(g^{\,\phi(p)/q_i}\equiv 1\pmod p\) replace \(\phi(p)\) by \(\phi(p)/q_i\). After processing all divisors you obtain  

   \[
   \operatorname{ord}_{p}(g)=\text{the remaining value}.
   \]

4. **Resulting worst‑case bound.**  
   The algorithm can never need more than \(\operatorname{ord}_{p}(g)\) distinct exponent values, so  

   \[
   \boxed{\text{#steps}_{\max}\le \operatorname{ord}_{p}(g)} .
   \]

   This is the exact analogue of the Euclidean bound “\(\le c\log_2 b\)’’: a *static* quantity derived from the algebraic structure of the problem, not from the particular target \(t\).

---

### 6.  Summary  

* The Euclidean bound \(O(\log b)\) comes from the binary length of the **smaller** input.  
* In the power‑log setting the analogous static quantity is the **order of the base** \(\operatorname{ord}_{p}(g)\), i.e. the size of the cyclic subgroup generated by the base.  
* Both bounds are obtained by exhibiting a monotone potential that can decrease only a limited number of times.  
* The asymmetry of the power‑log problem (fixed base, variable exponent) means the potential is *single‑valued* rather than a function of a pair, but the logical inference—*“the algorithm cannot visit more distinct states than the size of the underlying group’’*—is identical.  

Thus, yes: the \(O(\log b)\) bound for Euclid’s algorithm is directly analogous to the “order of the base’’ bound for your monoid‑based power‑log algorithm, even though the concrete objects (bit‑length vs. group order) differ.