Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨⋯ ? ⋯ 5 ⋯ (?+1)⟩ (?≠4)
⟨ ⁵ᵗʰa ⁴ᵗʰb ³ʳᵈc ⟩, (abc)₁₀ ≥ 154
⛔Avoid
⟦2,3⟧ ∋ 4
Sim⟨ ⁵ᵗʰ3 ⁴ᵗʰ1 ³ʳᵈ0 ²ⁿᵈ4 ¹ˢᵗ5 ⁰ᵗʰ2 ⟩ ≥ 1
min ⊢5⊣ ≤ 2
#125034_v2.8
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ 5 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 0 │ │ 5 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 0 │ │ 5 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 0 │ 3 │ 5 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 0 │ 3 │ 5 │ 4 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2025-02-11 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
Note that ✅「⟨⋯ ? ⋯ 5 ⋯ (?+1)⟩ (?≠4)」 implies 5 is not in corners (5th or 0th). Combining this with ⛔「min ⊢5⊣ ≤ 2」, we see that one of the following holds:
(1) ⟨⋯ 354 ⋯⟩ or ⟨⋯ 453 ⋯⟩.
(2) Let us consider the value of [0th]. By ✅「⟨⋯ ? ⋯ 5 ⋯ (?+1)⟩ (?≠4)」, it is 1|2|3|4. We claim that it is 1 actually.
------------------------------
For, to avoid ⛔「Sim⟨ ⁵ᵗʰ3 ⁴ᵗʰ1 ³ʳᵈ0 ²ⁿᵈ4 ¹ˢᵗ5 ⁰ᵗʰ2 ⟩ ≥ 1」, we cannot have any agreed positional digit with ⟨310452⟩. Therefore, [0th] != 2. On the other hand, if [0th] = 3|4, then by (1), we have
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ ▬ │ 5 │ ▬ │
└───┴───┴───┴───┴───┴───┘
where "▬" are occupied by 3 and 4. But now we have an agreed positional digit with ⟨310452⟩ at 1st, which is a contradiction.
------------------------------
We have verified the claim in (2). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Next, let us consider how to satisfy (1). There are three possibilities:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(3) │ ▬ │ 5 │ ▬ │ │ │ 1 │
├───┼───┼───┼───┼───┼───┤
(4) │ │ ▬ │ 5 │ ▬ │ │ 1 │
├───┼───┼───┼───┼───┼───┤
(5) │ │ │ ▬ │ 5 │ ▬ │ 1 │
└───┴───┴───┴───┴───┴───┘
where again "▬" are occupied by 3 and 4. Noting that 0 is to the left of 5 according to ✅「⟨⋯ ? ⋯ 5 ⋯ (?+1)⟩ (?≠4)」, we see that case (3) does not hold, and case (4) becomes:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ ▬ │ 5 │ ▬ │ │ 1 │
└───┴───┴───┴───┴───┴───┘
As it does not match ✅「⟨ ⁵ᵗʰa ⁴ᵗʰb ³ʳᵈc ⟩, (abc)₁₀ ≥ 154」, case (4) does not hold as well. Therefore, case (5) holds. We get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ ▬ │ 5 │ ▬ │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Next, we determine where to place 0. To match ✅「⟨ ⁵ᵗʰa ⁴ᵗʰb ³ʳᵈc ⟩, (abc)₁₀ ≥ 154」, we need 0 != [5th]. Therefore, 0 = [4th], and [5th] = 2 follows:
┌───┬───┬───┬───┬───┬───┐
│ 5■│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ ▬ │ 5 │ ▬ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 0 │ ▬ │ 5 │ ▬ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ 0 │ ▬ │ 5 │ ▬ │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟦2,3⟧ ∋ 4」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ 0 │ │ 5 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 0 │ 3 │ 5 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 0 │ 3 │ 5 │ 4 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.8