Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨⋯ Perm(0,2,4) ⋯⟩
5th → a, 2nd → b, |a-b|=1
⟨ ⁵ᵗʰa ²ⁿᵈb ⟩, min⟦a,b⟧ = 0
⛔Avoid
⟨⋯ Perm(0,3) ⋯⟩
{p4, p2, p1} = ? + {0,2,3}
Jump(1,2) ≥ 2
5th|2nd|0th → 0
max ⊢3⊣ = 3
#125034_v2.8
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 2 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 3 │ │ │ 2 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ │ 0 │ 2 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ │ 0 │ 2 │ │ 5 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 4 │ 0 │ 2 │ │ 5 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 4 │ 0 │ 2 │ 1 │ 5 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2025-03-25 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「⟨ ⁵ᵗʰa ²ⁿᵈb ⟩, min⟦a,b⟧ = 0」 and ⛔「5th|2nd|0th → 0」, we have
(1) 0 = [4th] | [3rd].
(2) We show that 2 = [2nd]. It is proved by contradiction.
Suppose on the contrary that 2 != [2nd]. Then
2 = [5th] | [4th] | [3rd] | [1st] | [0th].
------------------------------
(2.1) If 2 = [5th]:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
then ✅「⟨⋯ Perm(0,2,4) ⋯⟩」 implies
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ ▬ │ ▬ │ │ │ │
└───┴───┴───┴───┴───┴───┘
where ▬ are occupied by 0,4. But then we cannot avoid ⛔「Jump(1,2) ≥ 2」, which is a contradiction.
------------------------------
(2.2) Else if 2 = [4th]:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 2 │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
then by (1), we have
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 2 │ 0 │ │ │ │
└───┴───┴───┴───┴───┴───┘
In view of ✅「⟨⋯ Perm(0,2,4) ⋯⟩」, one of the following holds:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ 2 │ 0 │ │ │ │
├───┼───┼───┼───┼───┼───┤
│ │ 2 │ 0 │ 4 │ │ │
└───┴───┴───┴───┴───┴───┘
No matter which happens, we use ⛔「Jump(1,2) ≥ 2」 and get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ 2 │ 0 │ 1 │ │ │
├───┼───┼───┼───┼───┼───┤
│ 1 │ 2 │ 0 │ 4 │ │ │
└───┴───┴───┴───┴───┴───┘
Note that we fail to match ✅「5th → a, 2nd → b, |a-b|=1」, which is a contradiction.
------------------------------
(2.3) Else if 2 = [3rd]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 2 │ │ │ │
└───┴───┴───┴───┴───┴───┘
then (1) implies
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ 2 │ │ │ │
└───┴───┴───┴───┴───┴───┘
In view of ✅「⟨⋯ Perm(0,2,4) ⋯⟩」, one of the following holds:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(A) │ 4 │ 0 │ 2 │ │ │ │
├───┼───┼───┼───┼───┼───┤
(B) │ │ 0 │ 2 │ 4 │ │ │
└───┴───┴───┴───┴───┴───┘
Consider case (A) first:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ 0 │ 2 │ │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
To avoid ⛔「max ⊢3⊣ = 3」, we need 3,5 to be adjacent. The possible cases are:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ 0 │ 2 │ ▬ │ ▬ │ 1 │
├───┼───┼───┼───┼───┼───┤
│ 4 │ 0 │ 2 │ 1 │ ▬ │ ▬ │
└───┴───┴───┴───┴───┴───┘
However, the first disagrees with ⛔「Jump(1,2) ≥ 2」, while the second disagrees with ✅「5th → a, 2nd → b, |a-b|=1」.
So, case (A) does not hold, and case (B) holds instead:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ 2 │ 4 │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
To match ✅「5th → a, 2nd → b, |a-b|=1」 and avoid ⛔「⟨⋯ Perm(0,3) ⋯⟩」 at the same time, we have
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ 0 │ 2 │ 4 │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
But then we cannot avoid ⛔「Jump(1,2) ≥ 2」 and ⛔「max ⊢3⊣ = 3」 at the same time. It shows a contradiction.
------------------------------
(2.4) Else if 2 = [1st]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ │ 2 │ │
└───┴───┴───┴───┴───┴───┘
Then, to match ✅「⟨⋯ Perm(0,2,4) ⋯⟩」 and (1) at the same time, we need
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 0 │ 4 │ 2 │ │
└───┴───┴───┴───┴───┴───┘
By ⛔「Jump(1,2) ≥ 2」 and ⛔「⟨⋯ Perm(0,3) ⋯⟩」, we get
┌───┬───┬───┬───┬───┬───┐
│ 5▲│ 4▲│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 0 │ 4 │ 2 │ 1 │
├───┼───┼───┼───┼───┼───┤
│ 3 │ 5 │ 0 │ 4 │ 2 │ 1 │
└───┴───┴───┴───┴───┴───┘
It matches ⛔「{p4, p2, p1} = ? + {0,2,3}」 however, which is a contradiction.
------------------------------
(2.5) Lastly, suppose 2 = [0th]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ │ │ 2 │
└───┴───┴───┴───┴───┴───┘
Note that we cannot match ✅「⟨⋯ Perm(0,2,4) ⋯⟩」 and (1) at the same time. It shows a contradiction.
------------------------------
By (2.1)-(2.5), we have verified (2). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 2 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, by ✅「5th → a, 2nd → b, |a-b|=1」, we have [5th] = 1|3. In view of ⛔「Jump(1,2) ≥ 2」, it is 3 indeed:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ 2 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 3 │ │ │ 2 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
By (1), we have 0 = [4th] | [3rd]. To avoid ⛔「⟨⋯ Perm(0,3) ⋯⟩」, we get 0 = [3rd]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ │ │ 2 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ │ 0 │ 2 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Next, we consider the value of [0th]. To match ✅「⟨⋯ Perm(0,2,4) ⋯⟩」, it is not 4; on the other hand, if it is 1, then
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ ▬ │ 0 │ 2 │ ▬ │ 1 │
└───┴───┴───┴───┴───┴───┘
where ▬ are occupied by 4,5. This however matches ⛔「{p4, p2, p1} = ? + {0,2,3}」, which is a contradiction. Therefore, we need [0th] != 1 as well.
Consequently, [0th] = 5:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ │ 0 │ 2 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ │ 0 │ 2 │ │ 5 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「max ⊢3⊣ = 3」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ │ 0 │ 2 │ │ 5 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 4 │ 0 │ 2 │ │ 5 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 4 │ 0 │ 2 │ 1 │ 5 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.8