Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
2nd → 0|1|4|5
⟨ ⁵ᵗʰ↓ ⁴ᵗʰ↓ ³ʳᵈ↑ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ↑ ⟩ after 5−⟨⋯⟩
⟨ − ▧ ▧ ▢ ▢ ▢ ⟩, Σ▧ ≥ Σ▢
⛔Avoid
⟨⋯ ? ⋯ 5 ⋯ (?−2)⟩ (?≠5)
⟨ ⁵ᵗʰ= ⁴ᵗʰ↓ ³ʳᵈ↑ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ= ⟩ after 5−⟨⇌⟩
4th → a, 2nd → b, |a-b|=4
#125034_v2.9
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 5 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 5 │ 3 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 5 │ 3 │ │ │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ 3 │ 2 │ │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ 3 │ 2 │ 0 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 3 │ 2 │ 0 │ 4 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2025-07-15 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「⟨ ⁵ᵗʰ↓ ⁴ᵗʰ↓ ³ʳᵈ↑ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ↑ ⟩ after 5−⟨⋯⟩」, we have
(1)
{[5th], [4th], [1st]} = {3,4,5} and
{[3rd], [2nd], [0th]} = {0,1,2}.
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ ▬ │ ? │ ? │ ▬ │ ? │
└───┴───┴───┴───┴───┴───┘
(2) A fortiori, 5 ∈ {[5th], [4th], [1st]}.
(2.1) We show that 5 = [5th] indeed.
------------------------------
(2.2) If 5 is at 1st:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ ▬ │ ? │ ? │ 5 │ ? │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
then the sum [2nd] + [1st] + [0th] is at least 5 + 1 + 0 = 6, while the sum [4th] + [3rd] is at most 4 + 2 = 6.
So, to match ✅「⟨ − ▧ ▧ ▢ ▢ ▢ ⟩, Σ▧ ≥ Σ▢」, we need [4th] = 4, [3rd] = 2, {[2nd], [0th]} = {0,1}, and [5th] = 3:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│ 4▲│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ 4 │ 2 │ │ 5 │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
But then we would fail to avoid ⛔「⟨⋯ ? ⋯ 5 ⋯ (?−2)⟩ (?≠5)」, which is a contradiction.
------------------------------
(2.3) On the other hand, suppose 5 = [4th]:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ 5 │ ? │ ? │ ▬ │ ? │
└───┴───┴───┴───┴───┴───┘
By ✅「2nd → 0|1|4|5」 and (1), we have [2nd] = 0 or 1, and by ⛔「4th → a, 2nd → b, |a-b|=4」 we have [2nd] != 1. So [2nd] = 0:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ 5 │ ? │ 0 │ ▬ │ ? │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Then, using (1), we have 2 possibilities:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ 5 │ ? │ 0 │ 4 │ ? │
├───┼───┼───┼───┼───┼───┤
│ 4 │ 5 │ ? │ 0 │ 3 │ ? │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
To avoid ⛔「⟨⋯ ? ⋯ 5 ⋯ (?−2)⟩ (?≠5)」, we get
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ 5 │ 1 │ 0 │ 4 │ 2 │
├───┼───┼───┼───┼───┼───┤
│ 4 │ 5 │ 2 │ 0 │ 3 │ 1 │
└───┴───┴───┴───┴───┴───┘
In both cases, we match ⛔「⟨ ⁵ᵗʰ= ⁴ᵗʰ↓ ³ʳᵈ↑ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ= ⟩ after 5−⟨⇌⟩」, which is a contradiction.
------------------------------
Combining (2), (2.2) and (2.3), we have verified (2.1). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 5 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 0 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Next, we consider what [4th] and [1st] are. By (1) we have two possibilities:
([4th], [1st]) = (4,3) or (3,4).
(3) We claim that it is (3,4) actually.
------------------------------
For, if instead it is (4,3):
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ 4 │ │ │ 3 │ │
└───┴───┴───┴───┴───┴───┘
then, combining ✅「2nd → 0|1|4|5」 with ⛔「4th → a, 2nd → b, |a-b|=4」, we get [2nd] = 1:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ 4 │ │ 1 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
To avoid ⛔「⟨ ⁵ᵗʰ= ⁴ᵗʰ↓ ³ʳᵈ↑ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ= ⟩ after 5−⟨⇌⟩」 we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 5 │ 4 │ 0 │ 1 │ 3 │ 2 │
└───┴───┴───┴───┴───┴───┘
However, note that it fails to match ✅「⟨ − ▧ ▧ ▢ ▢ ▢ ⟩, Σ▧ ≥ Σ▢」.
------------------------------
We have verified (3) and get
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ 5 │ 3 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 5 │ 3 │ │ │ 4 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Then, observe that to match ✅「⟨ − ▧ ▧ ▢ ▢ ▢ ⟩, Σ▧ ≥ Σ▢」, 2 cannot be placed at 2nd or 0th. Consequently, 2 = [3rd]:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ 3 │ │ │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 5 │ 3 │ 2 │ │ 4 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨ ⁵ᵗʰ= ⁴ᵗʰ↓ ³ʳᵈ↑ ²ⁿᵈ↑ ¹ˢᵗ↓ ⁰ᵗʰ= ⟩ after 5−⟨⇌⟩」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 5 │ 3 │ 2 │ │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 5 │ 3 │ 2 │ 0 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 5 │ 3 │ 2 │ 0 │ 4 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.9