Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
Sim⟨ ⁵ᵗʰ0 ⁴ᵗʰ2 ³ʳᵈ5 ²ⁿᵈ4 ¹ˢᵗ1 ⁰ᵗʰ3 ⟩ ≤ 1
min ⊢4⊣ ≤ 1
⟨⋯ 0 ⋯ a ⋯⟩, a = 2|3
5th → 0|2|4
⟨ ⁴ᵗʰa ¹ˢᵗb ⟩, max⟦a,b⟧ = 5
⛔Avoid
{p4, p1, p0} = ? + {0,1,2}
⟨? 5 ⋯ (?+1) ⋯ 1 ⋯⟩ (?≠4)
1st → a, 0th → b, |a-b|=2
Jump(3,4) ≥ 2
max ⊢4⊣ = 4
#125034_v2.8
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 2 │ │ │ 5 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 2 │ │ │ 5 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ │ 3 │ 5 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 0 │ 3 │ 5 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 0 │ 3 │ 5 │ 4 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2025-04-01 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ⛔「max ⊢4⊣ = 4」, 4 is adjacent to 5. Combining this with ✅「min ⊢4⊣ ≤ 1」, we see that one of the following holds:
(1)
(A) ⟨⋯ 045 ⋯⟩
(B) ⟨⋯ 145 ⋯⟩
(C) ⟨⋯ 540 ⋯⟩
(D) ⟨⋯ 541 ⋯⟩.
A fortiori, 4 is not in the left corner (5th). So, ✅「5th → 0|2|4」 implies [5th] = 0|2.
(2) We claim that [5th] = 2.
------------------------------
For, suppose on the contrary [5th] = 0:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 5 │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
We show that it will result in contradiction no matter what [4th] is.
(2.1) If [4th] = 1:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 1 │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
then by (1), we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 1 │ 4 │ 5 │ │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
It will match ⛔「{p4, p1, p0} = ? + {0,1,2}」, which is a contradiction.
------------------------------
(2.2) Else if [4th] = 2:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 2 │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
then we will fail to match ✅「Sim⟨ ⁵ᵗʰ0 ⁴ᵗʰ2 ³ʳᵈ5 ²ⁿᵈ4 ¹ˢᵗ1 ⁰ᵗʰ3 ⟩ ≤ 1」.
------------------------------
(2.3) Else if [4th] = 3:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 3 │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
then in view of ⛔「Jump(3,4) ≥ 2」, we have 4 = [3rd] | [2nd]:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 3 │ ▬ │ ▬ │ │ │
└───┴───┴───┴───┴───┴───┘
Combining this with (1), we get 4 = [2nd]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 3 │ │ 4 │ │ │
└───┴───┴───┴───┴───┴───┘
But again, we will fail to match ✅「Sim⟨ ⁵ᵗʰ0 ⁴ᵗʰ2 ³ʳᵈ5 ²ⁿᵈ4 ¹ˢᵗ1 ⁰ᵗʰ3 ⟩ ≤ 1」.
------------------------------
(2.4) Else if [4th] = 4:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 4 │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
then (1) implies
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 4 │ 5 │ │ │ │
└───┴───┴───┴───┴───┴───┘
Again we fail to match ✅「Sim⟨ ⁵ᵗʰ0 ⁴ᵗʰ2 ³ʳᵈ5 ²ⁿᵈ4 ¹ˢᵗ1 ⁰ᵗʰ3 ⟩ ≤ 1」.
------------------------------
(2.5) Lastly, if [4th] = 5:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 5 │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
then (1) implies
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 5 │ 4 │ 1 │ │ │
└───┴───┴───┴───┴───┴───┘
In view of ⛔「⟨? 5 ⋯ (?+1) ⋯ 1 ⋯⟩ (?≠4)」, it is a contradiction.
------------------------------
By (2.1)-(2.5), we have verified (2). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ 5 │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Next, we consider where to place 5. By ✅「⟨ ⁴ᵗʰa ¹ˢᵗb ⟩, max⟦a,b⟧ = 5」, we have
5 = [4th] | [3rd] | [2nd] | [1st].
(3) We show by contradiction that 5 = [2nd].
------------------------------
(3.1) If 5 = [4th]:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 5 │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
then (1) implies
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 5 │ 4 │ ▬ │ │ │
└───┴───┴───┴───┴───┴───┘
where ▬ is 0 or 1. It then follows from ⛔「Jump(3,4) ≥ 2」 that
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 5 │ 4 │ ▬ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
To match ✅「⟨⋯ 0 ⋯ a ⋯⟩, a = 2|3」, we end at
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 5 │ 4 │ 0 │ 3 │ 1 │
└───┴───┴───┴───┴───┴───┘
It fails to avoid ⛔「⟨? 5 ⋯ (?+1) ⋯ 1 ⋯⟩ (?≠4)」, which is a contradiction.
------------------------------
(3.2) Else if 5 = [3rd]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ 5 │ │ │ │
└───┴───┴───┴───┴───┴───┘
Then in view of (1), we have
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ 5 │ 4 │ │ │
└───┴───┴───┴───┴───┴───┘
We fail to match ✅「Sim⟨ ⁵ᵗʰ0 ⁴ᵗʰ2 ³ʳᵈ5 ²ⁿᵈ4 ¹ˢᵗ1 ⁰ᵗʰ3 ⟩ ≤ 1」 now, which is a contradiction.
------------------------------
(3.3) Lastly, if 5 = [1st]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ │ │ 5 │ │
└───┴───┴───┴───┴───┴───┘
then (1) implies
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ ▬ │ 4 │ 5 │ │
└───┴───┴───┴───┴───┴───┘
where ▬ is 0 or 1. Then, for the position of 3, in view of ✅「Sim⟨ ⁵ᵗʰ0 ⁴ᵗʰ2 ³ʳᵈ5 ²ⁿᵈ4 ¹ˢᵗ1 ⁰ᵗʰ3 ⟩ ≤ 1」, it has to be 4th:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 3 │ │ 4 │ 5 │ │
└───┴───┴───┴───┴───┴───┘
Note that we will fail to match ✅「⟨⋯ 0 ⋯ a ⋯⟩, a = 2|3」, which is a contradiction.
------------------------------
By (3.1)-(3.3), we have verified (3). So, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 2 │ │ │ 5 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Now, by (1), we have two possibilities:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(4) │ 2 │ ▬ │ 4 │ 5 │ │ │
├───┼───┼───┼───┼───┼───┤
(5) │ 2 │ │ │ 5 │ 4 │ ▬ │
└───┴───┴───┴───┴───┴───┘
where ▬ is 0 or 1. Regardless of which case happens, by ⛔「Jump(3,4) ≥ 2」, we get
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(6) │ 2 │ ▬ │ 4 │ 5 │ 3 │ │
├───┼───┼───┼───┼───┼───┤
(7) │ 2 │ │ 3 │ 5 │ 4 │ ▬ │
└───┴───┴───┴───┴───┴───┘
To match ✅「⟨⋯ 0 ⋯ a ⋯⟩, a = 2|3」, 0 cannot be in the right corner (0th). Therefore, we obtain
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
(8) │ 2 │ 0 │ 4 │ 5 │ 3 │ 1 │
├───┼───┼───┼───┼───┼───┤
(9) │ 2 │ 0 │ 3 │ 5 │ 4 │ 1 │
└───┴───┴───┴───┴───┴───┘
Since (8) matches ⛔「1st → a, 0th → b, |a-b|=2」, the answer is (9).
Q.E.D.
#125034_v2.8