Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ⁵ᵗʰa ²ⁿᵈb ⟩, (ab)₁₀ ≤ 32
⟨⋯ 3 ⋯ 0 ⋯⟩
⛔Avoid
max ⊢1⊣ ≥ 3
⟦1,4⟧ ∋ 3,5
⟨ ³ʳᵈa ²ⁿᵈb ¹ˢᵗc ⟩, a > b > c
Jump(2,4) ≥ 1
⟨⋯ Perm(0,1,3) ⋯⟩
5th → a, 2nd → b, a+b=4
#125034_v2.8
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 1 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │ 2 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ 2 │ 4 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 2 │ 4 │ │ 3 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 2 │ 4 │ │ 3 │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 2 │ 4 │ 5 │ 3 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2025-02-04 WR
══════════════════════
(1) We first show that 1 is in corners (5th or 0th).
------------------------------
If this is not the case, then to avoid ⛔「max ⊢1⊣ ≥ 3」, we need 0,2 to be next to 1. Combining this with ⛔「Jump(2,4) ≥ 1」, we see that one of the following holds:
⟨⋯ 4210 ⋯⟩ or ⟨⋯ 0124 ⋯⟩.
Given that 1 is not in corners, we have the following possible cases:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(2) │ 0 │ 1 │ 2 │ 4 │ │ │
├───┼───┼───┼───┼───┼───┤
(3) │ │ 0 │ 1 │ 2 │ 4 │ │
├───┼───┼───┼───┼───┼───┤
(4) │ 4 │ 2 │ 1 │ 0 │ │ │
├───┼───┼───┼───┼───┼───┤
(5) │ │ │ 0 │ 1 │ 2 │ 4 │
├───┼───┼───┼───┼───┼───┤
(6) │ │ 4 │ 2 │ 1 │ 0 │ │
├───┼───┼───┼───┼───┼───┤
(7) │ │ │ 4 │ 2 │ 1 │ 0 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
However, all of them are contradictions:
If (2)⟨0124--⟩ holds, then we cannot match ✅「⟨⋯ 3 ⋯ 0 ⋯⟩」;
If (3)⟨-0124-⟩ holds, then to match ✅「⟨⋯ 3 ⋯ 0 ⋯⟩」 we would finish by ⟨301245⟩, which matches ⛔「⟨⋯ Perm(0,1,3) ⋯⟩」 nevertheless;
If (4)⟨4210--⟩ holds, then we cannot match ✅「⟨ ⁵ᵗʰa ²ⁿᵈb ⟩, (ab)₁₀ ≤ 32」;
If (5)⟨--0124⟩ holds, then to match ✅「⟨ ⁵ᵗʰa ²ⁿᵈb ⟩, (ab)₁₀ ≤ 32」 we would finish by ⟨350124⟩, which matches ⛔「5th → a, 2nd → b, a+b=4」 however;
If (6)⟨-4210-⟩ or (7)⟨--4210⟩ hold, then we match ⛔「⟨ ³ʳᵈa ²ⁿᵈb ¹ˢᵗc ⟩, a > b > c」.
------------------------------
We have verified (1). Now, let us consider what [5th] is. By ✅「⟨ ⁵ᵗʰa ²ⁿᵈb ⟩, (ab)₁₀ ≤ 32」, it is 0|1|2|3. In view of ✅「⟨⋯ 3 ⋯ 0 ⋯⟩」, it is not 0. Therefore
(8) [5th] = 1|2|3, and we show that [5th] = 1 actually.
------------------------------
If on the contrary [5th] != 1, then by (1), we have [0th] = 1:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ │ │ 1 │
└───┴───┴───┴───┴───┴───┘
and [5th] = 2|3:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(9) │ 2 │ │ │ │ │ 1 │
├───┼───┼───┼───┼───┼───┤
(10) │ 3 │ │ │ │ │ 1 │
└───┴───┴───┴───┴───┴───┘
Case (9) is not possible, for combining it with ⛔「Jump(2,4) ≥ 1」 we get
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 4 │ │ │ │ 1 │
└───┴───┴───┴───┴───┴───┘
and we would match ⛔「⟦1,4⟧ ∋ 3,5」. Therefore, case (10) holds:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ │ │ │ │ 1 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, to match ✅「⟨ ⁵ᵗʰa ²ⁿᵈb ⟩, (ab)₁₀ ≤ 32」 and avoid ⛔「max ⊢1⊣ ≥ 3」 at the same time, we need
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ │ │ ▬ │ ▬ │ 1 │
└───┴───┴───┴───┴───┴───┘
where ▬ are 0 or 2. Combining this with ⛔「Jump(2,4) ≥ 1」, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ │ 4 │ 2 │ 0 │ 1 │
└───┴───┴───┴───┴───┴───┘
but it matches ⛔「⟨ ³ʳᵈa ²ⁿᵈb ¹ˢᵗc ⟩, a > b > c」, which is a contradiction.
------------------------------
We have verified (8). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 1 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, to avoid ⛔「max ⊢1⊣ ≥ 3」, we need [4th] = 0|2. In view of ✅「⟨⋯ 3 ⋯ 0 ⋯⟩」, it is 2 indeed.
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │ 2 │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Plainly, to avoid ⛔「Jump(2,4) ≥ 1」, we then obtain
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 2 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ 2 │ 4 │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
By ⛔「5th → a, 2nd → b, a+b=4」, we cannot place 3 at 2nd. So, in view of ✅「⟨⋯ 3 ⋯ 0 ⋯⟩」, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 2 │ 4 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 2 │ 4 │ │ 3 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 2 │ 4 │ │ 3 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ │ │ │
└───┴───┴───┴───┴───┴───┘
As a result, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 2 │ 4 │ │ 3 │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 2 │ 4 │ 5 │ 3 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.8