Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨⋯ 2 ⋯ 5 ⋯⟩
⟨ ³ʳᵈa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≤ 243
4th → a, 0th → b, |a-b|=4
5th → a, 1st → b, |a-b|=2
⛔Avoid
⟨⋯ Perm(0,1,5) ⋯⟩
⟨ ▧ ▧ ▧ − ▢ ▢ ⟩, Σ▧ ≤ Σ▢
3rd|1st|0th → 4
#125034_v2.9
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 4 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 4 │ │ │ │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 4 │ │ │ 5 │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 4 │ │ │ 5 │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 4 │ │ 2 │ 5 │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 4 │ 1 │ 2 │ 5 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2025-06-24 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「4th → a, 0th → b, |a-b|=4」, we have
(1) {[4th], [0th]} = {0,4} or {1,5}.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ │ │ │ ▬ │
└───┴───┴───┴───┴───┴───┘
(1.1) We show that indeed {[4th], [0th]} = {0,4}.
------------------------------
Suppose on the contrary {[4th], [0th]} = {1,5}. There are two possibilities:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(2) │ │ 5 │ │ │ │ 1 │
├───┼───┼───┼───┼───┼───┤
(3) │ │ 1 │ │ │ │ 5 │
└───┴───┴───┴───┴───┴───┘
(2.1) If (2) holds, then using ✅「⟨⋯ 2 ⋯ 5 ⋯⟩」, we get
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 5 │ │ │ │ 1 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
To match ✅「⟨ ³ʳᵈa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≤ 243」, we need [3rd] = 0:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 5 │ 0 │ │ │ 1 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
It follows that we cannot match ✅「5th → a, 1st → b, |a-b|=2」 and avoid ⛔「3rd|1st|0th → 4」 at the same time. This shows a contradiction.
(3.1) On the other hand, if (3) holds:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ │ │ │ 5 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, to match ✅「5th → a, 1st → b, |a-b|=2」, we need
(3.2) {[5th], [1st]} = {0,2} or {2,4}.
A fortiori, 2 ∈ {[5th], [1st]}, whence 2!=[3rd]. Therefore, to match ✅「⟨ ³ʳᵈa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≤ 243」, we need [3rd] = 0:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ 0 │ │ │ 5 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, using (3.2), we have
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ - │ 1 │ 0 │ │ - │ 5 │
└───┴───┴───┴───┴───┴───┘
where "-" are occupied by 2,4. Using ⛔「3rd|1st|0th → 4」, we get 4 = [5th] and 2 = [1st]:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ 1 │ 0 │ │ 2 │ 5 │
└───┴───┴───┴───┴───┴───┘
Note that we fail to avoid ⛔「⟨ ▧ ▧ ▧ − ▢ ▢ ⟩, Σ▧ ≤ Σ▢」. This shows a contradiction.
------------------------------
We have shown that (2) and (3) give contradictions. Therefore, (1.1) is verified. Combining (1.1) with ⛔「3rd|1st|0th → 4」, we get 4 = [4th] and 0 = [0th]:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 4 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 4 │ │ │ │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Then, we consider where to place 5. By ✅「⟨⋯ 2 ⋯ 5 ⋯⟩」, it is not at 5th. By ✅「⟨ ³ʳᵈa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≤ 243」, it is not at 3rd either. Therefore,
(4) 5 = [2nd] or [1st].
(4.1) We show that 5 = [1st].
------------------------------
If on the contrary 5 = [2nd]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 4 │ │ 5 │ │ 0 │
└───┴───┴───┴───┴───┴───┘
then in view of ✅「⟨⋯ 2 ⋯ 5 ⋯⟩」, we have 2 = [5th] or [3rd]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ - │ 4 │ - │ 5 │ │ 0 │
└───┴───┴───┴───┴───┴───┘
Noting that if 2 = [3rd] then we cannot match ✅「⟨ ³ʳᵈa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≤ 243」, we get 2 = [5th]:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 4 │ │ 5 │ │ 0 │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
We are unable to match ✅「5th → a, 1st → b, |a-b|=2」 now. This shows a contradiction.
------------------------------
We have verified (4.1). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 4 │ │ │ │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 4 │ │ │ 5 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Given that 5 = [1st], ✅「5th → a, 1st → b, |a-b|=2」 implies [5th] = 3:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 4 │ │ │ 5 │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 4 │ │ │ 5 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨⋯ Perm(0,1,5) ⋯⟩」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 4 │ │ │ 5 │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 4 │ │ 2 │ 5 │ 0 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 4 │ 1 │ 2 │ 5 │ 0 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.9