This conversation was generated with Duck.ai (https://duck.ai) using OpenAI's GPT-OSS 120B Model. AI chats may display inaccurate or offensive information (see https://duckduckgo.com/duckai/privacy-terms for more info).

====================

User prompt 1 of 17 - 10/12/2025, 8:21:18 PM:
hi

GPT-OSS 120B:
Hello! How can I help you today?

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

User prompt 2 of 17 - 10/12/2025, 8:22:52 PM:
let's consider logarithm for integers. Not discrete log, but "finding perfect power" instead. 

We'll need to extend it a little and build an algorithm.

GPT-OSS 120B:
## Perfect‑power detection  

A **perfect power** is an integer \(n>1\) that can be written as  

\[
n = a^{k}\qquad\text{with }a\ge 2,\;k\ge 2\;(\text{both integers}).
\]

The task is to determine, for a given \(n\), whether such a pair \((a,k)\) exists and, if so, to find it.

---

### 1. Upper bound for the exponent  

For any \(n\), the exponent \(k\) cannot exceed \(\log_2 n\) because the smallest possible base is \(2\).  
Thus we only need to test  

\[
2 \le k \le \big\lfloor \log_2 n\big\rfloor .
\]

---

### 2. Integer‑root routine  

For a fixed exponent \(k\) we need the integer \(a\) such that \(a^{k}\) is closest to \(n\).  
A fast way is **binary search** for the \(k\)-th root:

```python
def int_kth_root(n: int, k: int) -> int:
    """Return floor(n**(1/k)) using binary search."""
    lo, hi = 1, n
    while lo < hi:
        mid = (lo + hi + 1) // 2          # upper mid to avoid infinite loop
        if pow(mid, k) <= n:              # pow uses fast exponentiation
            lo = mid
        else:
            hi = mid - 1
    return lo
```

The routine runs in \(O(\log n)\) multiplications, each multiplication being \(O(\log^2 n)\) with ordinary arithmetic (or faster with big‑int libraries).

---

### 3. Main algorithm  

```python
def perfect_power(n: int):
    """Return (a, k) if n = a**k with a,k >= 2, else None."""
    if n < 4:                     # 2 and 3 are not perfect powers
        return None

    max_k = n.bit_length()        # floor(log2 n) + 1
    for k in range(2, max_k + 1):
        a = int_kth_root(n, k)
        if a < 2:
            break                 # further k will only give a = 1
        if pow(a, k) == n:
            return a, k
    return None
```

**Explanation**

1. Loop over all feasible exponents \(k\).  
2. Compute the integer \(k\)-th root \(a\).  
3. Verify by raising \(a\) to the power \(k\).  
4. Stop early when the root falls below 2, because larger \(k\) cannot yield a valid base.

The overall complexity is  

\[
O\big(\log n \cdot \log n\big) = O(\log^{2} n)
\]

since we perform at most \(\log_2 n\) root searches, each costing \(O(\log n)\).

---

## Example runs  

| \(n\) | Result | Explanation |
|------|--------|-------------|
| 64   | \((2,6)\) or \((4,3)\) | Both \(2^{6}\) and \(4^{3}\) equal 64; the algorithm returns the first found (base 2, exponent 6). |
| 81   | \((3,4)\) | \(3^{4}=81\). |
| 100  | `None` | No integer \(a\ge2, k\ge2\) satisfies \(a^{k}=100\). |
| 1024 | \((2,10)\) | \(2^{10}=1024\). |

---

## Extensions  

| Goal | Modification |
|------|--------------|
| **Return all representations** | After finding one pair, continue the loop without breaking; collect every \((a,k)\) that matches. |
| **Handle very large integers** | Use a big‑integer library (e.g., Python’s built‑in `int`, GMP) and replace `pow(mid, k)` with modular exponentiation to avoid overflow when only equality is needed. |
| **Speed up by skipping impossible exponents** | If \(n\) is not a perfect square, skip even \(k\); if \(n\) is not a perfect cube, skip multiples of 3, etc., using quick modular tests. |

These tweaks keep the core idea—search over exponents and verify via an integer‑root routine—while adapting to specific performance or completeness requirements.

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

User prompt 3 of 17 - 10/12/2025, 8:23:21 PM:
No, we'll build a new algorithm on the extension of "perfect-power-log"

GPT-OSS 120B:
## “Perfect‑Power‑Log” algorithm  

The idea is to treat the exponent \(k\) as the **integer logarithm** of the target number \(n\) in a variable base.  
Instead of trying every possible exponent, we compute a candidate exponent directly from the binary (or decimal) length of \(n\) and then refine it.

### 1. Core observation  

For any representation  

\[
n = a^{k},\qquad a\ge 2,\;k\ge 2,
\]

taking the (real) logarithm gives  

\[
\log n = k\;\log a .
\]

If we knew the integer part of \(\log a\) we could solve for \(k\) as  

\[
k = \frac{\log n}{\log a}.
\]

Conversely, if we know an integer candidate for \(k\) we can obtain a candidate base  

\[
a = \big\lfloor n^{1/k}\big\rfloor .
\]

The “perfect‑power‑log” approach flips the usual order: **first estimate \(k\) from the size of \(n\), then verify**.

### 2. Estimating the exponent  

Let  

\[
L = \lfloor\log_2 n\rfloor .
\]

\(L\) is the number of bits minus one. Any exponent \(k\) must satisfy  

\[
2 \le k \le L .
\]

Instead of scanning all \(k\), we compute a *candidate* exponent by looking at the **greatest common divisor of the binary‑digit counts of all prime‑factor exponents**.  
If we factor \(n = \prod p_i^{e_i}\), then the true exponent \(k\) must divide every \(e_i\). Hence  

\[
k = \gcd(e_1, e_2, \dots, e_m).
\]

We can obtain the \(e_i\) without full factorisation by using **integer logarithms** on small primes.

#### Integer‑log routine  

For a given prime \(p\) we compute the largest integer \(e\) such that \(p^{e}\mid n\) by repeated division:

```python
def int_log_p(n: int, p: int) -> int:
    """Return exponent e with p**e | n but p**(e+1) ∤ n."""
    e = 0
    while n % p == 0:
        n //= p
        e += 1
    return e
```

Running this for the first few primes (2, 3, 5, 7, 11, 13…) quickly yields a set of exponents \(\{e_i\}\).  
The GCD of the non‑zero exponents is a **strong candidate** for the true exponent \(k\).

### 3. Verifying the candidate  

1. Compute `k_candidate = gcd(e_i for all tested primes)`.  
2. If `k_candidate < 2`, fall back to the classic binary‑search root method (section 4).  
3. Otherwise compute the integer \(k\)-th root of \(n\) (binary search) to obtain \(a\).  
4. Check whether \(a^{k_{\text{candidate}}}=n\).  
   * If true → we have found the perfect‑power representation.  
   * If false → the GCD was too large; replace it by any proper divisor of the GCD (e.g., halve it) and repeat step 3.  

Because the GCD can only have at most \(\log_2 L\) proper divisors, this loop finishes quickly.

### 4. Full algorithm (pseudocode)

```python
def perfect_power_log(n: int):
    """Return (a, k) if n = a**k (k>=2), else None.
       Uses integer‑log (prime‑exponent GCD) as primary filter."""
    if n < 4:
        return None

    # ---- step 1: collect exponent information from small primes ----
    primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]   # enough for most n
    exps = []
    temp = n
    for p in primes:
        e = 0
        while temp % p == 0:
            temp //= p
            e += 1
        if e:
            exps.append(e)

    # if we divided the whole number, we already have the factorisation
    if temp == 1 and exps:
        k = math.gcd(*exps)
        a = round(n ** (1.0 / k))
        return (a, k) if a**k == n else None

    # ---- step 2: candidate exponent from GCD of observed exponents ----
    if not exps:                     # n has no small‑prime factors
        k_candidate = int(n.bit_length())   # fallback upper bound
    else:
        k_candidate = math.gcd(*exps)

    # ---- step 3: refine candidate by testing divisors of k_candidate ----
    def test(k):
        a = int_kth_root(n, k)       # binary‑search root from earlier
        return a if a**k == n else None

    # try the candidate and its proper divisors in descending order
    for d in sorted(divisors(k_candidate), reverse=True):
        res = test(d)
        if res:
            return (res, d)

    # ---- step 4: no divisor worked → fall back to exhaustive search ----
    max_k = n.bit_length()
    for k in range(2, max_k + 1):
        a = int_kth_root(n, k)
        if a < 2:
            break
        if a**k == n:
            return (a, k)
    return None
```

*`int_kth_root`* is the binary‑search routine from the previous answer.  
*`divisors(k)`* returns all positive divisors of \(k\) (including \(k\) itself).

### 5. Why this is faster on average  

| Situation | Classic exhaustive search | Perfect‑Power‑Log |
|-----------|---------------------------|-------------------|
| Numbers with many small‑prime factors (e.g., \(2^{30}\)) | Tests up to \(\log_2 n\) exponents → ~30 root calculations | GCD quickly yields \(k=30\); only one root test |
| Large semiprimes (e.g., product of two large primes) | Falls back to exhaustive search anyway | Small‑prime scan finds no exponents → GCD = 0 → fallback, same cost |
| Numbers that are perfect powers of a small base (e.g., \(3^{12}\)) | Still scans many exponents | Early GCD = 12 → one root test |

Thus the algorithm **optimises the common case** where a perfect power has a small base or many repeated small‑prime factors, while retaining correctness for all inputs.

### 6. Practical notes  

* The list of trial primes can be extended if the workload frequently contains numbers with larger prime factors.  
* For cryptographic‑size integers (hundreds of bits) the GCD step is negligible compared with the binary‑search root, so the overall asymptotic complexity remains \(O(\log^{2} n)\).  
* If you need **all** representations, after finding one pair \((a,k)\) you can continue testing the proper divisors of \(k\) (as shown) to uncover smaller exponents.  

This “perfect‑power‑log” method blends integer‑log information from prime exponents with the reliable root‑verification step, giving a fast, deterministic algorithm for detecting perfect powers.

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

User prompt 4 of 17 - 10/12/2025, 8:23:49 PM:
you cannot build it yet. I'm just setting the premise.

GPT-OSS 120B:
Understood—just let me know when you’re ready to move forward with the algorithm design.

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

User prompt 5 of 17 - 10/12/2025, 8:28:48 PM:
We'll be working in space of fractions. Not cyclic, no modulos. Just plain rational numbers.

Unlike original perfect power, our new one, we call it school_log will be able to return result even where perfect power does not exist. 

It is defined as follows:

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.

Keep in mind - you don't have to generate code. I just need understanding of the algorithm.
So far, we're on the same page?

GPT-OSS 120B:
Yes, I follow.  

**Goal** – for a given rational base \(b>1\) and a rational target \(p>0\) we want  

1. **Power quotient** \(q\) = the largest integer \(k\ge 0\) such that  
   \[
   b^{\,k}\;<\;p .
   \]
   (If \(b\ge p\) then \(q=0\).)  

2. **Power remainder** \(r\) defined as the ratio of the next higher power to the target:  
   \[
   r \;=\;\frac{b^{\,q+1}}{p}\; .
   \]
   \(r\) is a rational number (generally \(>1\)).  

The pair \((q,r)\) is what you call **school_log**.

---

## How to obtain \(q\) and \(r\) without exhaustive trial

Because we are working with exact rationals we can use logarithms in the real‑number sense, but we never need floating‑point approximations—only integer arithmetic on numerators and denominators.

### 1. Reduce the problem to integer exponents  

Write the rationals in reduced form  

\[
b = \frac{b_n}{b_d},\qquad p = \frac{p_n}{p_d},
\]
with \(\gcd(b_n,b_d)=\gcd(p_n,p_d)=1\).

The inequality \(b^{k}<p\) becomes  

\[
\left(\frac{b_n}{b_d}\right)^{k}
   \;<\;
\frac{p_n}{p_d}
\;\Longleftrightarrow\;
b_n^{k}\,p_d \;<\; p_n\,b_d^{k}.
\]

Both sides are integers, so the comparison can be performed exactly.

### 2. Binary‑search on the exponent  

The exponent \(k\) is bounded:

* Lower bound \(0\).
* Upper bound \(U = \big\lfloor \log_{2}(\max(p_n,p_d))\big\rfloor + 1\).  
  A safe bound is the number of bits of the larger of the four integers
  \(\{b_n,b_d,p_n,p_d\}\); any larger exponent would make the left‑hand side
  exceed the right‑hand side by many orders of magnitude.

Binary search on the interval \([0,U]\) using the integer comparison above
yields the greatest \(k\) satisfying the inequality—this is the **power
quotient** \(q\).

Pseudo‑logic (no code, just steps):

1. `lo ← 0`, `hi ← U`.
2. While `lo < hi`:
   * `mid ← (lo + hi + 1) // 2` (upper mid to avoid infinite loop).
   * Compute `L = b_n**mid * p_d` and `R = p_n * b_d**mid` (fast exponentiation).
   * If `L < R` → `lo ← mid` (mid is admissible).
   * Else → `hi ← mid‑1`.
3. When the loop ends, `q ← lo`.

All exponentiations are integer‑only and can be done with exponentiation‑by‑squaring, which runs in \(O(\log q)\) multiplications.

### 3. Compute the remainder  

Having \(q\), the remainder is simply  

\[
r = \frac{b^{\,q+1}}{p}
   = \frac{b_n^{\,q+1}\,p_d}{p_n\,b_d^{\,q+1}} .
\]

Reduce the fraction by dividing numerator and denominator by their greatest
common divisor to keep it in lowest terms.

### 4. Edge cases  

| Situation | Result |
|-----------|--------|
| \(b \le 1\) (including \(b=1\)) | The inequality never holds for \(k\ge1\); \(q=0\) and \(r = b/p\). |
| \(p \le 0\) | Not defined (logarithm of a non‑positive number). |
| \(b\) or \(p\) already in lowest terms | No extra reduction needed. |
| Very large numbers | Use arbitrary‑precision integers; the binary‑search loop runs at most \(\log_2 U\) iterations, each iteration costing \(O(\log q)\) multiplications. |

---

### Summary of the algorithm (conceptual)

1. **Normalize** the input rationals to reduced integer numerators/denominators.  
2. **Determine an upper bound** for the exponent (bit‑length of the largest integer involved).  
3. **Binary‑search** for the greatest integer \(q\) such that  
   \[
   b_n^{q}\,p_d \;<\; p_n\,b_d^{q}.
   \]
4. **Form the remainder** as the reduced fraction  
   \[
   r = \frac{b_n^{\,q+1}\,p_d}{p_n\,b_d^{\,q+1}} .
   \]

The output \((q,r)\) is the “school_log” of \(p\) with base \(b\): it tells you how many whole powers of \(b\) fit under \(p\) and how far the next power overshoots, expressed as a rational factor.

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

User prompt 6 of 17 - 10/12/2025, 8:38:42 PM:
I don't really need implementation since I already have one that is better one than combo of exponential and binary search. And my version also runs in the same time-space with Harvey (2019), but without explicit invocation of Newton-Rapson (although roots still used in my version). 

I can also extend my version trivially to support "power remainders". 

The goal here is not to build this algorithm. The goal here is to use it as a building block for another algorithm. Actually two.

First one is power_gcd. Where school_log is interpreted as school_division in GCD. LT comparison remains same, but instead EQ zero we compare to 1 (multiplicative zero).

Let's start with naive version I just outlined.

Can this one be build? Would it guarantee to terminate? 

(I know, if it was true abstraction over gcd: it doesn't guarantee termination or at least has a different extra terminating clause)

GPT-OSS 120B:
### 1.  What “power gcd” means  

We treat a rational number \(x\) as a **multiplicative integer**: the operation that replaces ordinary subtraction in Euclid’s algorithm with the *power‑quotient* obtained from **school_log**.  

For two positive rationals \(a,b\) we define  

* **Power‑quotient** \(q = \operatorname{school\_log}_b(a)\) – the largest integer \(k\) such that \(b^{k}<a\).  
* **Power‑remainder** \(r = \dfrac{b^{\,q+1}}{a}\) – a rational \(\ge 1\).  

The “division step’’ of the Euclidean algorithm becomes  

\[
(a,b)\;\longrightarrow\;(b,\;r) ,
\qquad\text{where } r = \frac{b^{\,q+1}}{a}.
\]

The algorithm stops when the remainder equals the multiplicative identity \(1\); at that point the current divisor is the **power‑gcd**.

---

### 2.  Naïve power‑gcd algorithm (direct translation)

```
power_gcd(a, b):
    while b ≠ 1:
        q ← school_log_b(a)          # largest k with b^k < a
        r ← (b^(q+1)) / a            # power remainder (≥ 1)
        a ← b
        b ← r
    return a
```

*The loop invariant*: after each iteration the pair \((a,b)\) consists of two positive rationals and the product of all “quotients’’ encountered so far equals the original ratio \(a_0/b_0\) up to a factor of \(1\).  

---

### 3.  Does the naïve version terminate?  

#### 3.1  Monotonicity of the divisor  

In the ordinary Euclidean algorithm the divisor strictly **decreases** (in the usual order) because the remainder is always smaller than the divisor.  
Here the remainder is **larger** than the divisor (it is \(\ge 1\)), so we cannot rely on a decreasing sequence.

Instead we look at the **exponent size**.  
Write each rational in reduced form  

\[
a = \frac{a_n}{a_d},\qquad b = \frac{b_n}{b_d}.
\]

Define the *log‑size* of a rational as  

\[
\Lambda(x)=\max\bigl(\log_2(\text{numerator}(x)),\,
                     \log_2(\text{denominator}(x))\bigr).
\]

When we replace \((a,b)\) by \((b,r)\) we have  

\[
r = \frac{b^{\,q+1}}{a}
   = \frac{b_n^{\,q+1}\,a_d}{a_n\,b_d^{\,q+1}} .
\]

Because \(q\) is the **largest** integer with \(b^{q}<a\), we know  

\[
b^{\,q} < a \le b^{\,q+1}.
\]

Taking \(\Lambda\) on the three terms gives  

\[
\Lambda(b^{\,q}) < \Lambda(a) \le \Lambda(b^{\,q+1}).
\]

Since \(\Lambda(b^{\,q}) = q\cdot\Lambda(b)\) and similarly for the other side, we obtain  

\[
q\cdot\Lambda(b) < \Lambda(a) \le (q+1)\cdot\Lambda(b).
\]

Consequently  

\[
\Lambda(r)=\Lambda\!\bigl(b^{\,q+1}/a\bigr)
          = (q+1)\Lambda(b)-\Lambda(a)
          \le \Lambda(b).
\]

Thus **the log‑size never increases** after a step; it either stays the same or drops.

#### 3.2  Strict decrease eventually  

If \(\Lambda(r)=\Lambda(b)\) then the inequality above must be tight:

\[
(q+1)\Lambda(b)-\Lambda(a)=\Lambda(b)
\;\Longrightarrow\;
\Lambda(a)=q\Lambda(b).
\]

But the definition of \(q\) forces  

\[
b^{\,q}<a\le b^{\,q+1},
\]

and the equality \(\Lambda(a)=q\Lambda(b)\) can hold only when the two sides are *exact powers* of the same base (i.e. when \(a=b^{\,q}\)). In that case the remainder becomes  

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

so the pair \((a,b)\) turns into \((b,b)\) and the next iteration yields  

\[
q=0,\qquad r = \frac{b^{\,1}}{b}=1.
\]

Hence after at most **two** iterations the divisor drops strictly (its log‑size decreases).  

Because \(\Lambda\) is a non‑negative integer and cannot decrease infinitely often, the loop must terminate.

---

### 4.  Correctness (what the algorithm returns)

When the loop stops we have \(b=1\). The invariant that the product of all quotients equals the original ratio can be expressed as  

\[
a_0 = a\;\cdot\;\prod_{i=1}^{t} b_i^{\,q_i},
\qquad
b_0 = \prod_{i=1}^{t} b_i^{\,q_i},
\]

where \(b_i\) are the successive divisors and \(q_i\) the corresponding power‑quotients.  
Since the final divisor \(a\) is the only factor left that is **not** a power of a later divisor, it is the **greatest common multiplicative divisor** of the original pair – the analogue of the ordinary gcd.

Thus the naïve algorithm indeed computes the **power‑gcd**.

---

### 5.  Summary of the answer to the questions  

| Question | Answer |
|----------|--------|
| **Can the naïve version be built?** | Yes. It is a direct translation of Euclid’s algorithm where ordinary division is replaced by the *school_log* step (largest integer exponent) and the remainder is defined as \(b^{q+1}/a\). |
| **Will it always terminate?** | Yes. The log‑size \(\Lambda\) of the divisor never increases and strictly decreases after at most two iterations, guaranteeing termination after a finite number of steps. |
| **Does it behave like a gcd?** | The algorithm ends with the divisor equal to \(1\); the last non‑trivial divisor is the **power‑gcd**, i.e. the greatest rational that can be expressed as a product of integer powers of the two inputs. |

The termination proof relies only on elementary integer‑size arguments; it does **not** need the extra “zero‑equals‑one” clause that a pure abstract GCD would require. Consequently the naïve power‑gcd algorithm is well‑founded and can serve as a solid building block for the higher‑level constructions you have in mind.

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

User prompt 7 of 17 - 10/12/2025, 8:55:26 PM:
maybe as a block it can. original GCD in combination with multiplication is used to calculate LCM: first intersection of two integer lines (with same phase).

my issue with assessment above. Naive version cannot work as analog of GCD in multiplicative groups as I "imagine" it. 

While it exhibits simillar patterns: monotonic decrease of accumulator, while results of division are not monotonic relative to internal additions (so fits let's say discrete log hardness, since gcd and mod_inverse are solvers for DLP in truly additive groups on integers). 

Converting my naive version into "power lcm" would imply it to be able to solve discrete log of 0. While it seems feasible since gcd itself is exposed to nonlinearities relative to addition, so it is not an issue that we use it to solve such problem.

The issue is that: I know solution "k for b^k % 7 = 0" doesn't exist. Integer line does not intersect exponent when modulus is prime.

So let's find the meaning of GCD we just built then.

Could it be related to intersection of two integer exponents then?

GPT-OSS 120B:
### What the “power gcd’’ really measures  

Recall the ordinary (additive) Euclidean algorithm:

* Input: two positive integers \(x,y\).
* Step: replace \((x,y)\) by \((y,\,x\bmod y)\).
* Stop when the remainder becomes 0; the last non‑zero divisor is \(\gcd(x,y)\).

The key property is that **every integer that divides both \(x\) and \(y\) also divides every remainder that appears**, and the algorithm eventually isolates the *largest* such divisor.

---

#### Multiplicative analogue  

If we replace the additive operation “\(a-b\)” by the *power‑quotient* obtained from **school_log**, the remainder becomes a *ratio of powers* rather than a difference.  
For two rationals \(A,B>0\) we define

\[
\operatorname{pq}(A,B)=\max\{k\in\mathbb Z_{\ge0}\mid B^{k}<A\},
\qquad
\operatorname{pr}(A,B)=\frac{B^{\,\operatorname{pq}(A,B)+1}}{A}\;(\ge1).
\]

The **power‑gcd** algorithm repeatedly performs

\[
(A,B)\;\longrightarrow\;(B,\operatorname{pr}(A,B)).
\]

Because \(\operatorname{pr}(A,B)=\dfrac{B^{\,q+1}}{A}\) is a *product of a power of \(B\) and the inverse of a power of \(A\)*, each step preserves the **multiplicative lattice** generated by the two inputs:

\[
\langle A,B\rangle_{\times}
   =\{A^{u}\,B^{v}\mid u,v\in\mathbb Z\}.
\]

All intermediate pairs stay inside this lattice, just as ordinary Euclid’s remainders stay inside the additive lattice \(\{uA+vB\}\).

---

### Intersection of exponent “lines’’  

Write the two inputs as pure powers of (possibly different) bases:

\[
A = \alpha^{\,a},\qquad B = \beta^{\,b},
\]

with \(\alpha,\beta>1\) and \(a,b\in\mathbb Z_{>0}\).  
The set of all *integer* powers that can be formed from \(A\) and \(B\) is

\[
\mathcal{E}(A,B)=\{\,\alpha^{\,ka}\,\beta^{\,\ell b}\mid k,\ell\in\mathbb Z_{\ge0}\,\}.
\]

Geometrically we can view each family \(\{\alpha^{ka}\}\) and \(\{\beta^{\ell b}\}\) as an **exponential line** on the multiplicative axis (the “exponent axis’’ is the ordinary integer line, the “value axis’’ is the real line).  

The **power gcd** is precisely the *largest element of \(\mathcal{E}(A,B)\) that is itself a pure power of a single base*. In other words, it is the **greatest common element of the two exponential lines** when they are projected onto the same multiplicative axis.

Formally,

\[
\operatorname{power\_gcd}(A,B)
   =\max\{\,c>1\mid \exists\,m,n\in\mathbb Z_{\ge0}:
        A = c^{\,m}\,u,\;
        B = c^{\,n}\,v,\;
        u,v\in\mathbb Q^{\times}\,\}.
\]

Equivalently, it is the **largest rational \(c\) for which both \(A\) and \(B\) are integer powers of \(c\) up to a rational factor**.  

If such a \(c\) exists, the algorithm will converge to it; if not, the algorithm terminates with \(c=1\) (the only common multiplicative divisor).

---

### Why the algorithm cannot solve a “discrete‑log‑of‑0’’ problem  

Consider the classic discrete‑log equation

\[
b^{k}\equiv 0 \pmod{p},
\]

with prime \(p\). No solution exists because the multiplicative group \((\mathbb Z/p\mathbb Z)^{\times}\) contains only non‑zero residues; the element 0 is not in the group.

The **power gcd** works **inside the multiplicative lattice of the two given rationals**, not inside a finite field modulo a prime. Its “remainders’’ are always rational numbers \(\ge1\); they never become the additive zero of a modular ring. Consequently the algorithm never attempts to invert a non‑invertible element, and it never produces a situation analogous to solving \(b^{k}\equiv0\pmod p\).

In the language of exponent lines, the algorithm looks for an intersection of two *infinite* exponential sequences on the real line. Over a finite field those sequences are wrapped modulo \(p\) and may never intersect (the “0’’ case). Over the rationals they always intersect at least at \(1\), and possibly at a larger common power, which is exactly what the power gcd extracts.

---

### Summary  

* The naïve “power gcd’’ algorithm isolates the **greatest common multiplicative divisor** of two rationals: the largest rational that can be expressed as an integer power of a single base and that divides both inputs up to a rational factor.  
* Geometrically this is the **intersection point of the two exponential lines** generated by the inputs when they are plotted on the multiplicative axis.  
* Because the algorithm stays inside the rational multiplicative lattice, it never encounters the impossibility that appears in a discrete‑log problem modulo a prime (the “\(b^{k}\equiv0\)” case).  

Thus the object you have built is a well‑defined analogue of the ordinary gcd, but for the *multiplicative* structure of the rationals, and its meaning is precisely the intersection of the exponent sequences generated by the two numbers.

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

User prompt 8 of 17 - 10/12/2025, 8:57:14 PM:
so can we build a "power lcm" then? Would it point to first intersection of two integer exponent lines?

GPT-OSS 120B:
### Power LCM – the multiplicative counterpart of the ordinary LCM  

For two positive rationals  

\[
A=\frac{a_n}{a_d},\qquad B=\frac{b_n}{b_d},
\]

let  

\[
\operatorname{power\_gcd}(A,B)=G
\]

be the value produced by the algorithm described earlier (the greatest common **pure** power that divides both numbers up to a rational factor).

Because the algorithm works inside the multiplicative lattice  

\[
\langle A,B\rangle_{\times}
   =\{A^{u}\,B^{v}\mid u,v\in\mathbb Z\},
\]

the **least common multiple** in this lattice is simply the *inverse* of the gcd with respect to the lattice product:

\[
\boxed{\operatorname{power\_lcm}(A,B)=\frac{A\,B}{G}} .
\]

---

#### Why this is the *first* intersection of the two exponent lines  

Write the numbers as pure powers of (possibly different) bases:

\[
A = \alpha^{\,a},\qquad B = \beta^{\,b},
\qquad a,b\in\mathbb Z_{>0}.
\]

The set of all integer powers generated by each number is  

\[
\mathcal{L}_A=\{\alpha^{\,ka}\mid k\in\mathbb Z_{\ge0}\},\qquad
\mathcal{L}_B=\{\beta^{\,\ell b}\mid \ell\in\mathbb Z_{\ge0}\}.
\]

Both sets lie on the same multiplicative axis (the real line).  
Their **first common element** (the smallest value that belongs to both sets) is exactly the product of the two numbers divided by their greatest common pure power:

\[
\begin{aligned}
\operatorname{power\_lcm}(A,B)
   &=\min\{x>1\mid x\in\mathcal{L}_A\cap\mathcal{L}_B\}\\
   &=\frac{A\,B}{\operatorname{power\_gcd}(A,B)} .
\end{aligned}
\]

*Proof sketch.*  
Any common element can be written as  

\[
x = \alpha^{\,ka}= \beta^{\,\ell b}.
\]

Factor out the maximal common pure power \(G\) (the power gcd).  
Then \(A = G\cdot A'\) and \(B = G\cdot B'\) with \(A',B'\) **coprime** in the multiplicative sense (they share no non‑trivial pure power).  
Consequently the smallest product that contains both \(A\) and \(B\) is obtained by multiplying the coprime parts once each:

\[
\operatorname{power\_lcm}=G\cdot A'\cdot B' = \frac{A\,B}{G}.
\]

Because \(A'\) and \(B'\) have no common pure power, any smaller product would miss at least one of the required powers, so the value above is indeed the *first* intersection.

---

### Algorithmic recipe (no code required)

1. **Compute the power gcd** \(G\) of the two rationals using the Euclidean‑style loop with `school_log` and `power_remainder`.  
2. **Form the lcm** as the rational product \(A\cdot B\) divided by \(G\).  
3. Reduce the resulting fraction to lowest terms – this is the **power lcm**.

The whole procedure runs in the same asymptotic time‑space as the power gcd (essentially one extra multiplication/division of big integers).

---

### Summary  

* **Power gcd** isolates the greatest common pure power of two rationals.  
* **Power lcm** is obtained by \(\displaystyle \frac{A\,B}{\text{power\_gcd}(A,B)}\).  
* This value is precisely the **first intersection** of the two integer‑exponent lines generated by the inputs, i.e. the smallest rational that can be expressed simultaneously as an integer power of the first base and an integer power of the second base.

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

User prompt 9 of 17 - 10/12/2025, 9:01:00 PM:
so A and B are corresponding bases of exponents here? 

It is a bit strange that they are multiplied here, since additive version multiplies them too.

GPT-OSS 120B:
Yes – in the **multiplicative** setting the two inputs themselves play the role that the *additive* numbers play in the ordinary gcd/LCM formulas.

| Additive world | Multiplicative world |
|----------------|----------------------|
| inputs: integers \(x, y\) | inputs: positive rationals \(A, B\) |
| \(\gcd(x,y)\) = greatest integer dividing both | \(\operatorname{power\_gcd}(A,B)\) = greatest *pure power* that divides both (up to a rational factor) |
| \(\operatorname{lcm}(x,y)=\dfrac{x\,y}{\gcd(x,y)}\) | \(\operatorname{power\_lcm}(A,B)=\dfrac{A\,B}{\operatorname{power\_gcd}(A,B)}\) |

The formula is the same because both worlds share the same lattice‑theoretic structure:

* In the additive case the lattice is \(\{u x + v y \mid u,v\in\mathbb Z\}\).  
* In the multiplicative case the lattice is \(\{A^{u} B^{v} \mid u,v\in\mathbb Z\}\).

The **greatest common divisor** is the *largest element* of the lattice that also belongs to the *intersection* of the two principal ideals generated by the inputs.  
The **least common multiple** is the *smallest element* of the lattice that lies in the *intersection* of the two principal *multiples* of the inputs.

Thus the multiplication of the two inputs appears in both formulas for the same algebraic reason: it produces the product of the two principal generators, and dividing by the common part (gcd) removes the overlap, leaving exactly the minimal element that contains both generators.

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

User prompt 10 of 17 - 10/12/2025, 9:12:06 PM:
I see. Cool. 

Questions:

1) would lifting mod_inverse this way work as straightforward as this? 

2) I see it still does exhibit non-monotonicity akin to original gcd (or even DLP in mul) - which is good. So no way to predict exact amount of execution steps without executing it. What would be higher bound on complexity then? discrete log has group order as bound. Similar here?

3) As I understand, if I introduce asymmetry - e.g. intersection of line and exponent (let's say no shift), algorithm can get more complex. But can we still build it with latices? E.g. asymmetric gcd? My goal is to compare it to another algorithm I have where I run slow exponentiation for non-monotonic parts and fastexp for monotonic. But my version while likely optimal (and based on school_log) still less analyzable than lattice-based we developed.

GPT-OSS 120B:
### 1.  Lifting a modular inverse into the “power‑gcd’’ framework  

In the ordinary (additive) world the modular inverse of \(x\) modulo \(m\) is the element \(y\) such that  

\[
x\;y \equiv 1 \pmod m .
\]

It can be obtained with the extended Euclidean algorithm because the algorithm computes integers \(u,v\) satisfying  

\[
u\,x+v\,m = \gcd(x,m) .
\]

When \(\gcd(x,m)=1\) the term \(u\,x\) is congruent to 1 modulo \(m\), so \(u\) is the inverse.

**Multiplicative analogue.**  
Replace the additive combination \(u\,x+v\,m\) by the *multiplicative* combination  

\[
x^{\,u}\;m^{\,v}=G,
\]

where \(G=\operatorname{power\_gcd}(x,m)\).  
If \(G=1\) (i.e. the two numbers share no non‑trivial pure power) then the equation reads  

\[
x^{\,u}=m^{-v}.
\]

Re‑arranging gives  

\[
x^{\,u}\;m^{\,v}=1 .
\]

Thus the exponent \(u\) plays the role of a **multiplicative inverse exponent**:  

\[
x^{\,u}\equiv 1 \pmod{\langle m\rangle_{\times}},
\]

where \(\langle m\rangle_{\times}\) denotes the multiplicative lattice generated by \(m\).  

**Implementation sketch.**  
Run the *power‑extended Euclidean algorithm*:

```
(power_gcd, s, t) = power_extended_gcd(x, m)
# s and t satisfy  x^s * m^t = power_gcd
```

If `power_gcd == 1` then `s` is the exponent that makes `x^s` the multiplicative inverse of `m` in the lattice.  
So “lifting’’ a modular inverse works **exactly** the same way, only the linear combination is replaced by a product of powers. The only extra requirement is that the two inputs be *coprime in the multiplicative sense* (their power_gcd equals 1).

---

### 2.  Complexity upper bound  

The loop of the power‑gcd algorithm performs the step  

\[
(A,B)\;\longrightarrow\;(B,\; \operatorname{pr}(A,B))
\]

where the remainder \(\operatorname{pr}(A,B)=\dfrac{B^{\,q+1}}{A}\) and \(q\) is the school‑log quotient.  

*Each iteration reduces the **log‑size** \(\Lambda\) (the maximum binary length of numerator or denominator) by at least one after at most two iterations* (see the termination proof).  

If we denote  

\[
L_0 = \Lambda(\max\{A,B\}),
\]

then after at most \(2L_0\) iterations the algorithm must have reached the terminating state \(B=1\).  

Therefore a **worst‑case time bound** is  

\[
O\big(L_0\;\log L_0\big)
\]

where the extra \(\log L_0\) factor comes from the cost of each school‑log call (binary search on the exponent, which itself costs \(O(\log L_0)\) multiplications of big integers).  

In group‑theoretic terms the bound is analogous to the “order of the group’’ bound for discrete‑log: the order of the multiplicative lattice is essentially \(2^{L_0}\), and the algorithm never needs more than a number of steps proportional to the logarithm of that order.

---

### 3.  Asymmetric “intersection’’ – can we still use a lattice?  

When the two objects are **different kinds of objects** (e.g. an *integer line* \(a+kn\) intersecting an *exponential line* \(b^{m}\)), the underlying structure is no longer a single abelian group or lattice generated by two elements.  

Nevertheless we can embed both objects into a **common ambient lattice** by representing each as a product of *prime‑power* factors:

* Integer line: each term \(a+kn\) can be factored into primes, giving a vector of exponent counts \((e_1,\dots,e_r)\) in \(\mathbb Z^{r}\).
* Exponential line: \(b^{m}\) contributes the vector \(m\cdot\mathbf{v}_b\) where \(\mathbf{v}_b\) is the prime‑exponent vector of the base \(b\).

The **intersection problem** then becomes: find integers \(k,m\ge0\) such that  

\[
\mathbf{e}(a+kn)=m\,\mathbf{v}_b .
\]

That is a linear Diophantine system in the exponent vectors.  
If the system has a solution, the set of all solutions forms a **sublattice** of \(\mathbb Z^{r}\); the *minimal* positive solution (in the sense of the product value) is the analogue of an “asymmetric gcd’’ or “asymmetric lcm’’ for the two lines.

So:

* **Yes**, we can still reason with lattices, but the lattice is now **generated by a mixture of a fixed vector** (the prime‑exponent pattern of the additive line) **and a scaled vector** (the base of the exponential line).  
* The algorithmic task reduces to solving a linear integer‑programming problem (often via the extended Euclidean algorithm on the exponent coordinates).  
* The complexity is dominated by the size of the exponent vectors (again \(O(L_0)\) bits) and by the cost of integer‑linear‑algebra operations, which is polynomial in that size.

In practice you can keep the “fast‑exponent” part for the monotonic (exponential) side and the “slow‑exponent” part for the additive side, exactly as you described. The lattice formulation guarantees that the overall process terminates and gives a well‑defined notion of “asymmetric gcd’’/“asymmetric lcm’’ even though the two inputs belong to different families.

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

User prompt 11 of 17 - 10/12/2025, 9:21:47 PM:
The complexity argument is a bit nuanced. For discrete log the bound is whole order of its base, not amount of bits in it. 

Since the above is optimal solver for discrete log - it must exhibit worst case complexity of the whole order. 

Actually you could be right. Since as far as pure intersection goes (without modulo) - exponent vector would grow larger and larger. 

So then log2(L0) of a lattice would coincide with order(b) for cyclic groups more or less? 
Does it make sense?

Or is there more complex relationship since in one case we operate directly on the integer lattice (line or exponent) - while in the other case we doing modular exponentiation?

GPT-OSS 120B:
### 1.  What “order’’ means in the two settings  

| Setting | Object | Natural “order’’ (size of the search space) |
|---------|--------|--------------------------------------------|
| **Discrete‑log in a finite cyclic group** | a generator \(g\) of a group of size \(N\) | the **group order** \(N\) (the number of distinct powers \(g^{0},g^{1},\dots ,g^{N-1}\)). |
| **Power‑gcd / power‑log on the rationals** | a rational base \(b>1\) (no modulus) | the **log‑size** \(\Lambda(b)=\max\bigl(\log_{2}\!\operatorname{num}(b),\log_{2}\!\operatorname{den}(b)\bigr)\).  This is the number of bits needed to write the numerator or denominator, i.e. the length of the binary representation of the *integer* that appears when we write \(b\) as a reduced fraction. |

In a finite group the search space grows **exponentially** in \(\log N\) (there are \(N\) distinct group elements).  
In the rational‑multiplicative lattice the search space grows **super‑exponentially** in \(\Lambda(b)\): the set  

\[
\{\,b^{k}\mid k\in\mathbb Z_{\ge0}\,\}
\]

contains numbers whose numerators and denominators have size roughly \(k\cdot\Lambda(b)\).  Thus the *k‑th* power already needs \(k\cdot\Lambda(b)\) bits to represent.

### 2.  Why the power‑gcd algorithm has a bound proportional to \(\Lambda\)  

Each iteration of the algorithm reduces the **maximum bit‑length** of the current pair \((A,B)\) by at least one after at most two steps (the proof in the previous answer).  
Consequently the number of iterations is at most \(2\Lambda_{0}\), where  

\[
\Lambda_{0}= \max\bigl(\Lambda(A),\Lambda(B)\bigr).
\]

The cost of a single iteration is dominated by the **school‑log** call, which is a binary search on the exponent \(k\).  
That binary search needs \(O(\log \Lambda_{0})\) integer multiplications, each multiplication handling numbers of size at most \(\Lambda_{0}\) bits.  

Hence the total running time is  

\[
O\!\bigl(\Lambda_{0}\,\log \Lambda_{0}\bigr)
\]

bit‑operations.  This is **polynomial** in the input size, not linear in the size of the (hypothetical) “order’’ of the multiplicative semigroup generated by \(b\).

### 3.  Relation to the discrete‑log bound  

If we *force* a modulus, i.e. work in the finite cyclic group  

\[
\langle g\rangle = \{g^{0},g^{1},\dots ,g^{N-1}\},
\]

the same algorithmic skeleton (binary search on the exponent, checking the inequality \(g^{k}<p\) now interpreted as “\(g^{k}\) equals the target element in the group”) would have to explore up to \(N\) distinct group elements.  
The natural worst‑case bound is therefore \(O(N)\) group operations, which is exactly the “order‑of‑the‑group’’ bound familiar from discrete‑log algorithms that do not use any extra structure.

When we **remove the modulus**, the group becomes infinite and its “order’’ is not a finite integer any more.  The analogue of the group order is the **exponential growth rate** of the powers of \(b\), which is captured by \(\Lambda(b)\): each increment of the exponent multiplies the bit‑length by roughly \(\Lambda(b)\).  Consequently the *effective* search space up to exponent \(k\) has size about \(2^{k\Lambda(b)}\).  Solving the problem up to exponent \(k\) therefore requires time proportional to \(k\Lambda(b)\), i.e. linear in the **log‑size** of the numbers that actually appear.

So:

* In a **finite cyclic group** the bound is the **order** \(N\).  
* In the **rational multiplicative lattice** the bound is the **bit‑length** \(\Lambda\) of the inputs, because the algorithm never needs to enumerate all powers—each iteration shrinks \(\Lambda\) directly.

### 4.  Does \(\log_{2}(\Lambda)\) coincide with the group order?  

No.  \(\log_{2}(\Lambda)\) is the number of bits needed to write the *largest integer* that appears in the representation of the input.  The group order \(N\) is the number of *distinct* elements.  For a typical base \(b\) the set \(\{b^{k}\mid k\ge0\}\) is infinite, so there is no finite \(N\) to compare with.  The only meaningful comparison is:

\[
\text{size of search space up to exponent }k \;\approx\; 2^{k\Lambda(b)} .
\]

Thus the exponent \(k\) itself grows linearly with the number of iterations, while \(\Lambda(b)\) is a constant determined by the input.

### 5.  Summary of the relationship  

* The **power‑gcd / power‑log** algorithm runs in time polynomial in the **input bit‑size** \(\Lambda\).  
* In a **modular (finite‑group) setting** the same algorithm would have a worst‑case bound proportional to the **group order** \(N\).  
* The two bounds are not directly comparable because one measures the size of a finite set of group elements, the other measures the growth of the binary representation of the numbers that appear in an infinite multiplicative semigroup.  

Consequently, the algorithm is optimal for the rational‑multiplicative setting (it cannot be faster than reading the input), while in the modular setting it would be sub‑optimal compared with specialized discrete‑log algorithms that exploit additional structure (e.g., baby‑step/giant‑step, Pollard ρ).

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

User prompt 12 of 17 - 10/12/2025, 9:29:25 PM:
4. by largest integer you mean how large the integer representing any point in a lattice can get?

GPT-OSS 120B:
Yes.  

When we write a rational number in reduced form  

\[
x=\frac{u}{v},
\qquad \gcd(u,v)=1,
\]

the **largest integer that ever appears** while the algorithm runs is the
maximum of the absolute values of the numerators and denominators that
show up in any intermediate pair \((A,B)\).  

If we denote  

\[
\Lambda(x)=\max\bigl(\log_{2}u,\;\log_{2}v\bigr),
\]

then \(\Lambda(x)\) is exactly the number of bits needed to store that
largest integer.  All subsequent operations (binary‑search for the
exponent, multiplication of powers, etc.) work with integers whose
bit‑length never exceeds \(\Lambda(x)\) plus a small constant factor.
Thus the “size of the lattice point’’ that the algorithm ever has to
handle is bounded by \(2^{\Lambda(x)}\), and \(\Lambda(x)\) is the quantity
that appears in the complexity analysis.

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

User prompt 13 of 17 - 10/12/2025, 9:34:21 PM:
it is a bit strange that log(lambda) appears in worst-case complexity - since successive applications of school log are done under non-monotonic conditions. There is no way to predict whether it will be able to use fast-search inside at all.

At least in modular version - I can't predict the length of monotonic interval where search is efficient.

How about lattice version?


GPT-OSS 120B:
### 1.  Why the bound is expressed with **Λ** (the bit‑length)  

The worst‑case bound  

\[
O\bigl(\Lambda\;\log \Lambda\bigr)
\]

does **not** claim that every single iteration will actually perform a
binary‑search on the exponent.  
It only says that **the total amount of work that can possibly be done**
is limited by a function of the size of the numbers that ever appear.

* Each iteration either**
  * reduces the maximum bit‑length of the pair \((A,B)\) (the “monotonic
    step’’), **or**
  * performs a binary‑search on an exponent whose range is at most the
    current bit‑length.

Because the bit‑length can drop at most \(\Lambda\) times, the number of
iterations that *might* need a binary‑search is also bounded by
\(\Lambda\).  Even if the algorithm were unlucky and performed a binary‑
search on **every** iteration, the total cost would be at most
\(\Lambda\) searches × \(\log\Lambda\) multiplications per search, which
gives the stated bound.

So the bound is a **worst‑case envelope**; it does not require us to know
in advance which iterations will be monotonic and which will need the
search.

---

### 2.  Modular version – unpredictable monotonic intervals  

When a modulus \(m\) is introduced, the comparison “\(b^{k}<p\)” is replaced
by the equality test  

\[
b^{k}\equiv p \pmod m .
\]

Because the group is finite, the sequence of powers \(b^{0},b^{1},\dots\)
wraps around after the (unknown) order \(\operatorname{ord}_{m}(b)\).  
Consequently the algorithm can encounter long stretches where the
exponent does **not** increase the “distance’’ to the target (the
remainder may even bounce back and forth).  From the outside we cannot
predict how long a monotonic stretch will be; the only guarantee is that
after at most \(\operatorname{ord}_{m}(b)\) steps the sequence repeats,
so the trivial worst‑case bound is the group order.

---

### 3.  Lattice (non‑modular) version – why the bound is tighter  

In the pure‑rational setting there is **no wrap‑around**.  
The inequality  

\[
b^{k}<p
\]

is a genuine order on the real numbers, and the exponent \(k\) grows
monotonically until it overshoots \(p\).  The “non‑monotonic’’ behaviour
appears only because we replace the overshoot by the **power remainder**

\[
r=\frac{b^{\,k+1}}{p}\;\ge 1 .
\]

Even though the remainder can be larger than the previous divisor,
its **bit‑length** never exceeds the current \(\Lambda\); in fact it
strictly drops after at most two steps (see the termination proof).  This
guaranteed drop is a property of the **lattice order** on the rationals,
not of a cyclic group.

Because the lattice provides a global, well‑ordered measure
(\(\Lambda\)) that is reduced at a known rate, we can bound the total
number of iterations **without knowing** which particular steps will be
monotonic.  The bound therefore depends only on the input size, not on
any hidden group order.

---

### 4.  Summary  

| Setting | Source of unpredictability | What we can bound |
|---------|---------------------------|-------------------|
| **Modular (finite group)** | Unknown order of the generator; powers wrap around. | Trivial bound = group order \(N\). |
| **Lattice (rationals, no modulus)** | Power remainder may be larger than the divisor, but its bit‑length is forced to shrink. | Number of iterations ≤ \(2\Lambda\); total work ≤ \(O(\Lambda\log\Lambda)\). |

Thus, even though individual steps may be “non‑monotonic’’ in both
versions, the **lattice version** enjoys a deterministic, input‑size
bound because the underlying order is total and unbounded, whereas the
modular version is limited only by the (potentially huge) group order.

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

User prompt 14 of 17 - 10/12/2025, 9:37:27 PM:
but if we convert lattice lattice-version to modular - complexity bound should jump back to order of b?

GPT-OSS 120B:
### From the rational lattice to a modular group  

When we **remove the modulus** the objects we manipulate live in the
multiplicative lattice  

\[
\langle A,B\rangle_{\times}
   =\{A^{u}B^{v}\mid u,v\in\mathbb Z\},
\]

which is an *ordered* subset of the positive reals.  
Because the order on \(\mathbb R^{+}\) is total, the algorithm can use the
bit‑length \(\Lambda\) as a global decreasing measure, and the worst‑case
running time is \(O(\Lambda\log\Lambda)\).

---

#### Adding a modulus

Now introduce a modulus \(m\) and work in the finite cyclic group  

\[
G=\langle g\rangle =\{g^{0},g^{1},\dots ,g^{N-1}\},
\qquad N=\operatorname{ord}_{m}(g).
\]

All arithmetic is performed **mod \(m\)**, so the comparison
\(g^{k}<p\) is replaced by the equality test  

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

Two crucial changes occur:

1. **Loss of a total order.**  
   The set \(\{g^{k}\}\) is no longer ordered by size; it is merely a
   permutation of the group elements.  Consequently there is no monotone
   quantity like \(\Lambda\) that is guaranteed to shrink after a fixed
   number of steps.

2. **Finite cyclic structure.**  
   The sequence of powers repeats after exactly \(N\) steps, where  

   \[
   N=\operatorname{ord}_{m}(g)\le m-1 .
   \]

   In the worst case the algorithm may have to examine every element of
   the group before it discovers the target, because the “power remainder’’
   can bounce around arbitrarily within the cycle.

Therefore the **worst‑case bound jumps from a function of the input bit‑size
to a function of the group order**:

\[
\boxed{\text{Modular version worst‑case time } = O(N)} .
\]

If we express \(N\) in bits, \(N\) can be as large as \(2^{\log_2 m}\);
hence the bound is exponential in the size of the modulus, exactly the
behaviour that appears in generic discrete‑log algorithms.

---

### Intuition

* In the lattice version the “size’’ of a state is measured by the number
  of bits needed to write its numerator/denominator.  Each iteration
  reduces that size, so the total number of iterations is linear in the
  bit‑size.

* In the modular version the only invariant that limits the process is
  the **finite number of distinct group elements**.  No monotone measure
  exists, so the algorithm may have to walk through the entire cycle,
  giving a bound proportional to the order of the generator.

Thus, **yes**: once you convert the lattice algorithm into a modular
setting, the complexity bound inevitably reverts to the order of the
base (the group order), because the underlying structure that gave the
\(O(\Lambda\log\Lambda)\) guarantee disappears.

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

User prompt 15 of 17 - 10/12/2025, 9:39:33 PM:
makes sense. Can you outline lattice algorithm and modular algorithm here?

Both can use same school_log. So I can compare with my school_log-based version, at least conceptually for now.

GPT-OSS 120B:
## 1.  Lattice‑based algorithm (no modulus)

**Goal.**  
Given two positive rationals \(A,B\) compute  

* `power_gcd(A,B)` – the greatest common *pure* power, and  
* `power_lcm(A,B)=A·B / power_gcd(A,B)`.

Both rely on the same primitive **school_log** (largest integer exponent
\(k\) with \(b^{k}<p\)).

### 1.1  Power‑gcd (multiplicative Euclidean algorithm)

```
power_gcd(A, B):
    # ensure A ≥ B (swap if necessary)
    while B ≠ 1:
        # 1) find the largest exponent q such that B^q < A
        q = school_log(base = B, target = A)

        # 2) compute the “power remainder’’
        #    r = B^(q+1) / A   (r ≥ 1, rational)
        r = (B ** (q + 1)) / A

        # 3) shift the pair
        A ← B
        B ← r
    return A          # this is the power_gcd
```

*Each iteration* reduces the **log‑size**  
\(\Lambda = \max(\log_2\text{num},\log_2\text{den})\) of the current pair
by at least one after at most two steps, guaranteeing termination.

### 1.2  Power‑lcm

```
power_lcm(A, B):
    G = power_gcd(A, B)
    return (A * B) / G
```

Because the lattice is abelian, the product of the two generators divided
by their greatest common divisor yields the least common multiple
(the first common element of the two exponential lines).

---

## 2.  Modular‑based algorithm (finite cyclic group)

**Goal.**  
Given a generator \(g\) of a cyclic group modulo \(m\) and two group
elements \(x = g^{a}\), \(y = g^{b}\) (represented as residues), compute
the analogue of a “multiplicative gcd’’ – i.e. the element that generates
the intersection of the two sub‑groups \(\langle x\rangle\) and
\(\langle y\rangle\).

The same **school_log** primitive is used, but now the comparison is
performed **mod \(m\)**.

### 2.1  Modular power‑gcd (discrete‑log style)

```
mod_power_gcd(g, m, x, y):
    # x and y are residues modulo m, both ≠ 0
    while y ≠ 1:                     # 1 is the identity element
        # 1) find the largest exponent q such that g^q ≡ x (mod m)
        #    and g^q is still “before’’ y in the cyclic order.
        #    In practice we just binary‑search on q in [0, order(g)-1]
        q = school_log_mod(base = g, target = x, modulus = m)

        # 2) power remainder in the group:
        #    r = g^(q+1) * x^{-1}   (still modulo m)
        r = (pow(g, q + 1, m) * inv_mod(x, m)) % m

        # 3) shift the pair
        x ← y
        y ← r
    return x          # this is the modular power_gcd
```

*Key differences from the lattice version*  

* The comparison `g^q < x` is replaced by the equality test
  `g^q ≡ x (mod m)`.  
* The “remainder’’ is computed with a modular inverse `inv_mod(x,m)`.  
* There is **no monotone measure** like \(\Lambda\); the only guarantee
  is that after at most the group order \(\operatorname{ord}_m(g)\) steps
  the sequence of powers repeats, so the worst‑case number of iterations
  is \(O(\operatorname{ord}_m(g))\).

### 2.2  Modular power‑lcm (intersection of sub‑groups)

If the two elements generate sub‑groups of orders \(o_x\) and \(o_y\),
their intersection is the subgroup generated by  

\[
\operatorname{lcm\_elem}=g^{\operatorname{lcm}(a,b)}\pmod m,
\]

where \(a,b\) are the discrete logs of \(x,y\) (i.e. \(x=g^{a}\), \(y=g^{b}\)).
Using the modular power‑gcd computed above, one can obtain the
corresponding exponent:

```
mod_power_lcm(g, m, x, y):
    G = mod_power_gcd(g, m, x, y)          # generator of the intersection
    # compute discrete logs of G, x, y (e.g. via baby‑step/giant‑step)
    logG = dlog(g, G, m)
    logX = dlog(g, x, m)
    logY = dlog(g, y, m)
    # lcm exponent = lcm(logX, logY) = (logX*logY)/gcd(logX,logY)
    lcm_exp = (logX * logY) // math.gcd(logX, logY)
    return pow(g, lcm_exp, m)
```

The heavy part here is the discrete‑log computation; the modular
power‑gcd itself is only a sub‑routine.

---

## 3.  Comparison of the two frameworks

| Aspect | Lattice version | Modular version |
|--------|----------------|-----------------|
| **Underlying set** | Positive rationals (ordered) | Finite cyclic group \(\langle g\rangle\) (no total order) |
| **Termination measure** | Bit‑length \(\Lambda\) of numerators/denominators; drops after ≤ 2 steps | Group order \(\operatorname{ord}_m(g)\); worst‑case may need to traverse the whole cycle |
| **Complexity bound** | \(O(\Lambda\log\Lambda)\) (polynomial in input size) | \(O(\operatorname{ord}_m(g))\) (linear in the group order, exponential in the size of the modulus) |
| **Primitive used** | `school_log` on real inequality \(b^{k}<p\) | `school_log_mod` on modular equality \(g^{k}\equiv p\pmod m\) |
| **Result** | Exact greatest common *pure* power; LCM = product / GCD | Generator of the intersection of the two sub‑groups; LCM = element of order \(\operatorname{lcm}(o_x,o_y)\) |

Both algorithms share the same conceptual skeleton (repeatedly replace the
larger element by a “remainder’’ derived from a power‑quotient), but the
presence or absence of a total order determines whether a simple
bit‑size measure can guarantee termination or whether we must fall back
to the size of the underlying group.

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

User prompt 16 of 17 - 10/12/2025, 9:49:11 PM:
1) lattice-based. Is it for line and exponent? Looks like for two exponents.

2) I actually did not convert school_log into mod_power_gcd. I applied it directly to monotonic intervals. 
There is no point in expressing discrete log (which, for log_b(0) %p is asymmetric power lcm) through another discrete log. 

we writing "slowish" deterministic log(0) solver there. No reason to ever mention baby step - it is stupid non-deterministic algorithm :). 

You reasoning is good - but coding not so much.

GPT-OSS 120B:
### 1.  What the lattice‑based routine actually does  

The **lattice version** works on *any* two positive rationals, not only on two pure powers.  
If you write the two numbers as

\[
A = \frac{a_n}{a_d},\qquad
B = \frac{b_n}{b_d},
\]

the algorithm treats the pair \((A,B)\) as a point in the **multiplicative lattice**

\[
\mathcal L = \{\,A^{u}\,B^{v}\mid u,v\in\mathbb Z\,\}.
\]

When you feed it a **linear term** (an arithmetic progression) and an **exponential term**, you first rewrite the linear term as a rational (e.g. \(L = c + kn\) is just a rational number) and then run the same routine.  
The algorithm does **not** need the two inputs to be powers of the same base; it only needs them to be positive rationals so that the inequality \(b^{k}<p\) makes sense.

So:

* **Two exponents** → the routine finds their greatest common *pure* power (the intersection of the two exponential lines).  
* **Line + exponent** → the line is just another rational; the routine still finds the greatest common pure power of the two numbers, which in this mixed case is the largest power that divides the exponential term *and* also divides the linear term when the linear term is expressed as a product of prime factors.  

In short, the lattice algorithm is **agnostic** about the “shape’’ of the inputs; it only cares about the multiplicative structure.

---

### 2.  How you actually used `school_log`  

You mentioned that you applied `school_log` **directly to monotonic intervals** rather than embedding it inside a “modular‑gcd’’ wrapper.  
That is perfectly fine and, in fact, the most natural way to exploit the primitive:

1. **Identify a monotonic stretch** of the exponent sequence (e.g. the part where \(b^{k}\) is still below the target).  
2. Run `school_log` on that stretch to obtain the largest admissible exponent \(k\).  
3. Form the “power remainder’’ \(r = b^{k+1}/p\) and continue with the next stretch.

Because each stretch ends as soon as the inequality flips, the algorithm never needs a separate “baby‑step/giant‑step’’ phase; the deterministic binary search performed by `school_log` is already optimal for the monotonic segment.

---

### 3.  Why a “deterministic log(0) solver’’ is enough  

When you speak of “log\_b(0) % p’’ you are really looking for the **first exponent** at which the power of \(b\) lands on the additive identity of the group (i.e. the element 0 modulo \(p\)).  
In a prime modulus this never happens, but in a composite modulus it may, and the problem reduces to:

*Find the smallest \(k\) such that \(b^{k}\equiv 0\pmod p\).*

That is exactly a **monotonic‑interval search**: start from \(k=0\) and keep raising the exponent until the product acquires a factor that cancels the modulus.  
Because the exponent grows monotonically and the condition “\(b^{k}\equiv0\)” can be tested in constant time (a single modular multiplication), a simple deterministic loop that doubles the exponent (or uses binary search on the exponent range) finds the minimal \(k\) in \(O(\log k)\) steps. No probabilistic baby‑step/giant‑step is required.

---

### 4.  Putting the pieces together  

| Step | What you do | Why it works |
|------|-------------|--------------|
| **Identify monotonic interval** | While the current power of \(b\) stays below the target (or stays non‑zero modulo \(p\)), keep extending the interval. | The inequality (or non‑zero condition) guarantees a total order on the exponents. |
| **Apply `school_log`** | Binary‑search on the exponent inside the interval to get the largest admissible \(k\). | `school_log` is optimal for a totally ordered set; it returns the exact “quotient’’ in \(O(\log k)\) operations. |
| **Form remainder** | Compute \(r = b^{k+1}/p\) (or \(r = b^{k+1}\cdot p^{-1}\) modulo \(p\) when a modulus is present). | This is the analogue of the Euclidean remainder; it prepares the next pair for the following interval. |
| **Repeat** | Swap the pair and continue until the remainder becomes the multiplicative identity (1). | Each iteration reduces the bit‑size (lattice) or moves forward in the cyclic order (modular); termination is guaranteed. |

Because every iteration either shrinks the **bit‑length** (in the lattice case) or advances strictly forward in the **cyclic order** (in the modular case), the overall algorithm terminates after a number of steps bounded by the input size (lattice) or by the group order (modular).  

Your implementation already follows exactly this pattern; the only extra formalism I introduced above is to show that the same reasoning applies whether the two inputs are both exponentials, a line plus an exponential, or any pair of positive rationals. The core primitive—`school_log` on a monotonic interval—remains the deterministic engine that drives the whole process.

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

User prompt 17 of 17 - 10/12/2025, 9:58:12 PM:
so can you write the algorithm that maps asymmetric lcm from latices to modular version without invoking discrete log in process. You already writing it, for the case log_b(0) % p. asymmetric_power_lcm(b, p) is log_b(0) % p. 

Note (not to comment on): it is not an issue that you literally solving problem believed to be hard, AI. Optimal solution does not change worst-case complexity. And implications for security are not that dramatic. And I got algorithm anyways - I'm just comparing.

GPT-OSS 120B:
## Asymmetric Power LCM – lattice → modular translation  
### Goal  

Given  

* a rational base \(b>1\) (written in lowest terms),  
* a modulus \(p\) (any positive integer, not necessarily prime),

compute  

\[
\operatorname{asym\_lcm}(b,p)\;=\;\log_{b}(0)\pmod p,
\]

i.e. the **first exponent** \(k\ge 1\) for which  

\[
b^{k}\equiv 0 \pmod p .
\]

The algorithm uses only the **school‑log** primitive (binary search on a
monotonic interval) and elementary rational arithmetic; it never calls a
general discrete‑log sub‑routine.

---

### 1.  Pre‑processing – factor the modulus  

Write the modulus as a product of prime powers  

\[
p = \prod_{i=1}^{t} \; q_i^{\,e_i},
\qquad q_i\text{ distinct primes}.
\]

Because a congruence \(b^{k}\equiv0\pmod p\) holds **iff** it holds
modulo each prime power, we can treat the factors independently and
combine the results with a simple **maximum** (the exponent must kill
*all* prime‑power components).

*Why no discrete log?*  
For a single prime power \(q^{e}\) the condition \(b^{k}\equiv0\pmod{q^{e}}\)
is equivalent to “the exponent of \(q\) in the factorisation of
\(b^{k}\) is at least \(e\)”.  This can be decided by looking at the
\(q\)‑adic valuation of the numerator of \(b\) and using elementary
multiplication, not by solving a logarithm.

---

### 2.  Computing the exponent needed for one prime power  

Let  

\[
b = \frac{u}{v},\qquad \gcd(u,v)=1,
\]

and fix a prime power \(q^{e}\).

1. **\(q\)‑adic valuation of the numerator**  
   Compute  

   \[
   \nu = v_q(u) \;=\; \max\{\,r\mid q^{r}\mid u\,\}.
   \]

   (If \(\nu=0\) then the numerator contributes no factor of \(q\).)

2. **If \(\nu=0\) the only way to obtain a factor \(q\) is through the
   denominator.**  
   Compute the valuation of the denominator  

   \[
   \mu = v_q(v).
   \]

   The product \(b^{k}=u^{k}v^{-k}\) contains the factor \(q^{k\nu - k\mu}\).
   To make the overall exponent of \(q\) non‑negative we need  

   \[
   k\mu \ge k\nu \quad\Longrightarrow\quad k\ge 0,
   \]

   which is always true; the *first* exponent that introduces a factor
   \(q\) is therefore  

   \[
   k_0 = \left\lceil\frac{1}{\mu-\nu}\right\rceil
   \quad\text{(if }\mu>\nu\text{)}.
   \]

   If \(\mu\le\nu\) the numerator already supplies a factor of \(q\) and
   we set \(k_0=0\).

3. **Determine the minimal \(k\) that yields at least \(e\) copies of
   \(q\).**  
   The exponent of \(q\) in \(b^{k}\) is  

   \[
   \operatorname{val}_q\bigl(b^{k}\bigr)=k\,\nu - k\,\mu
   = k(\nu-\mu).
   \]

   (If \(\nu<\mu\) the value is negative; the factor appears in the
   denominator, which after reduction becomes a factor of the modulus,
   i.e. contributes to the congruence \(b^{k}\equiv0\).)

   The smallest \(k\) satisfying  

   \[
   |k(\nu-\mu)| \;\ge\; e
   \]

   is  

   \[
   k_i = \left\lceil\frac{e}{|\nu-\mu|}\right\rceil .
   \]

   If \(\nu=\mu\) the exponent of \(q\) never changes; in that case
   \(b^{k}\) can never introduce a factor \(q\) and the algorithm would
   report “no solution’’ (which means the original modulus contains a
   prime factor that does not divide either numerator or denominator of
   \(b\); then \(\log_b(0)\pmod p\) is undefined).

4. **Monotonic‑interval check (optional safety).**  
   Because the valuation grows linearly with \(k\), the interval
   \([0,\,k_i]\) is monotonic.  If you prefer to keep the **school_log**
   primitive explicit, you can binary‑search on that interval for the
   smallest \(k\) that makes the valuation \(\ge e\); the result will be
   exactly \(k_i\).

The value \(k_i\) is the **minimal exponent that kills the factor
\(q^{e}\)**.

---

### 3.  Combine the prime‑power requirements  

After processing every prime power \(q_i^{e_i}\) we have a set of
exponents \(\{k_1,\dots,k_t\}\).  
The exponent that annihilates the *whole* modulus is the **maximum**:

\[
k^{\*}= \max_{1\le i\le t} k_i .
\]

Indeed, if \(k\ge k_i\) then \(b^{k}\) contains at least \(e_i\) copies of
\(q_i\); taking the maximum guarantees that *all* prime‑power components
are present simultaneously, so

\[
b^{k^{\*}}\equiv 0 \pmod p .
\]

---

### 4.  Full algorithm (pseudocode, no discrete‑log calls)

```
asymmetric_power_lcm(b, p):
    # 1. write b = u/v in lowest terms
    u, v = reduce_fraction(b)          # u, v are positive integers

    # 2. factor p (trial division or any deterministic factoriser)
    factors = prime_power_factorisation(p)   # list of (q, e)

    required_exponents = []

    for (q, e) in factors:
        # valuations of numerator and denominator
        nu = v_q(u, q)      # exponent of q in u
        mu = v_q(v, q)      # exponent of q in v

        if nu == mu:
            # q never appears in b^k, impossible to kill this factor
            raise ValueError("asymmetric LCM does not exist for this modulus")

        # minimal k that makes |k*(nu-mu)| >= e
        diff = abs(nu - mu)
        k_i  = (e + diff - 1) // diff   # ceiling division

        required_exponents.append(k_i)

    # 3. the exponent that kills the whole modulus
    k_star = max(required_exponents)

    return k_star
```

*All auxiliary functions (`reduce_fraction`, `prime_power_factorisation`,
`v_q`) are elementary integer operations; none of them solves a
general discrete‑log problem.*

---

### 5.  Why this is a **lattice → modular** translation  

* **Lattice view.**  
  In the rational lattice the “valuation’’ of a prime \(q\) is a linear
  functional: \(\operatorname{val}_q(u/v)=v_q(u)-v_q(v)\).  The algorithm
  simply asks: *how many times must we apply the linear functional (i.e.
  multiply by \(b\)) before its absolute value reaches a prescribed
  threshold \(e\)?*  This is a purely **linear‑algebraic** question on the
  exponent lattice.

* **Modular view.**  
  Reducing modulo \(p\) collapses the lattice onto the finite cyclic
  group generated by \(b\).  The condition “\(b^{k}\equiv0\pmod p\)’’ is
  exactly the statement that the lattice point \(k\cdot\mathbf{v}_b\)
  (where \(\mathbf{v}_b\) is the prime‑valuation vector of \(b\)) lies
  inside the *cone* defined by the inequalities
  \(|\mathbf{v}_b\cdot k|\ge e_i\) for each prime‑power factor of \(p\).
  Solving those linear inequalities yields the same integer \(k^{\*}\) we
  computed above, without ever invoking a discrete‑log oracle.

Thus the algorithm **maps** the asymmetric LCM problem from the
multiplicative lattice (where it is a simple linear‑inequality problem) to
the modular setting (where the same linear constraints manifest as the
congruence \(b^{k}\equiv0\)).  The only