2024-02-20 Q1(m=6)

Rearrange the digits in ⟨1263045⟩ to meet the rules below.

⟨6th 5th 4th 3rd 2nd 1st 0th⟩

✅Match

⟨ ⁶ᵗʰa ⁵ᵗʰb     ²ⁿᵈc     ⟩, (abc)₁₀ ≤ 203

Sim⟨ ⁶ᵗʰ2 ⁵ᵗʰ4 ⁴ᵗʰ6 ³ʳᵈ5 ²ⁿᵈ0 ¹ˢᵗ1 ⁰ᵗʰ3 ⟩ ≥ 2

Jump(3,5) = 2

⟨ ⁶ᵗʰa ⁵ᵗʰb     ²ⁿᵈc     ⟩, (abc)₁₀ ≥ 132

⛔Avoid

⟨       ³ʳᵈa   ¹ˢᵗb   ⟩, (ab)₁₀ ≥ 24

#125034_v2.3

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│5th│4th│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 1 │   │   │   │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │   │   │ 2 │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │   │ 5 │ 2 │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │   │ 5 │ 2 │   │ 3 │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 4 │ 5 │ 2 │   │ 3 │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 4 │ 5 │ 2 │ 0 │ 3 │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 4 │ 5 │ 2 │ 0 │ 3 │ 6 │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

Proof of 2024-02-20 Q1(m=6)
═══════════════════════════

Notation: if nth -> a, then we write [nth] = a.

By ✅「⟨ ⁶ᵗʰa ⁵ᵗʰb     ²ⁿᵈc     ⟩, (abc)₁₀ ≤ 203」 and ✅「⟨ ⁶ᵗʰa ⁵ᵗʰb     ²ⁿᵈc     ⟩, (abc)₁₀ ≥ 132」, we see that

(1) [6th] = 1|2.

(1.1) We claim that [6th] = 1 actually.

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

If on the contrary [6th] = 2, then by ✅「⟨ ⁶ᵗʰa ⁵ᵗʰb     ²ⁿᵈc     ⟩, (abc)₁₀ ≤ 203」, we have [5th] = 0 and X := [2nd] = 1|3:

┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│ 5▲│4th│3rd│*2 │1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 0 │   │   │ X │   │   │
└───┴───┴───┴───┴───┴───┴───┘

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │   │ 6 │ 3 │   │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

In view of ✅「Jump(3,5) = 2」, we have X != 3. Therefore, X = 1:

┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 0 │   │   │ 1 │   │   │
└───┴───┴───┴───┴───┴───┴───┘

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │   │ 6 │ 3 │   │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

But then we would match ⛔「⟨       ³ʳᵈa   ¹ˢᵗb   ⟩, (ab)₁₀ ≥ 24」, which is a contradiction.

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

Our claim in (1.1) is thus verified. Accordingly, we have

       ┌───┬───┬───┬───┬───┬───┬───┐
       │ 6■│5th│4th│3rd│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 1 │   │   │   │   │   │   │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

Next, note that to avoid ⛔「⟨       ³ʳᵈa   ¹ˢᵗb   ⟩, (ab)₁₀ ≥ 24」, we need

(2) [3rd]  = 0|2.

Combining this with ✅「Jump(3,5) = 2」, we have

(3) 3 != [0th],

for otherwise [3rd] would be 5, contradicting (2):

┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │   │   │ 5 │   │   │ 3 │
└───┴───┴───┴───┴───┴───┴───┘

As a result, to match ✅「Sim⟨ ⁶ᵗʰ2 ⁵ᵗʰ4 ⁴ᵗʰ6 ³ʳᵈ5 ²ⁿᵈ0 ¹ˢᵗ1 ⁰ᵗʰ3 ⟩ ≥ 2」, [6th] = 1 (step 1), (2), and (3), we need 

(4) at least two of the following hold:

(i) [5th] = 4;
(ii) [4th] = 6;
(iii) [2nd] = 0.

(5) We are ready to show that [3rd] = 2.

If this is not the case, then by (2), we have [3rd] = 0:

┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │   │   │ 0 │   │   │   │
└───┴───┴───┴───┴───┴───┴───┘

Then, to match (4), we need 4(i) and 4(ii) hold:

┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ 4 │ 6 │ 0 │   │   │   │
└───┴───┴───┴───┴───┴───┴───┘

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │ 2 │   │ 3 │   │   │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

However, it follows that we could not match ✅「Jump(3,5) = 2」, which is a contradiction.

Our claim in (5) is verified. Thus, we get

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│5th│4th│ 3■│2nd│1st│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
       │ 1 │   │   │   │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │   │   │ 2 │   │   │   │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │   │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘

We proceed by considering how to match ✅「Jump(3,5) = 2」. There are two ways to do so:

    ┌───┬───┬───┬───┬───┬───┬───┐
    │6th│5th│4th│3rd│2nd│1st│0th│
    ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(6) │ 1 │ * │   │ 2 │ * │   │   │
    ├───┼───┼───┼───┼───┼───┼───┤
(7) │ 1 │   │ * │ 2 │   │ * │   │
    └───┴───┴───┴───┴───┴───┴───┘

Actually (6) is not possible, as we need to match (4) as well. Therefore, (7) holds and we have

{ [4th], [1st] } = {3,5}.

Since ⛔「⟨       ³ʳᵈa   ¹ˢᵗb   ⟩, (ab)₁₀ ≥ 24」 forbids [1st] = 5, we get

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│5th│ 4■│3rd│2nd│ 1■│0th│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
       │ 1 │   │   │ 2 │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │   │ 5 │ 2 │   │   │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │   │ 5 │ 2 │   │ 3 │   │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │   │ 6 │   │ 0 │ 4 │   │
└───┴───┴───┴───┴───┴───┴───┘

Finally, by applying (4), we finish by

       ┌───┬───┬───┬───┬───┬───┬───┐
       │6th│ 5■│4th│3rd│ 2■│1st│ 0■│▒
       ╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
       │ 1 │   │ 5 │ 2 │   │ 3 │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 4 │ 5 │ 2 │   │ 3 │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 4 │ 5 │ 2 │ 0 │ 3 │   │▒
       ├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 4 │ 5 │ 2 │ 0 │ 3 │ 6 │▒
       └───┴───┴───┴───┴───┴───┴───┘▒
        ▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒

--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│   │   │   │   │   │   │   │
└───┴───┴───┴───┴───┴───┴───┘

Q.E.D.

#125034_v2.3