Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
6th|5th|2nd|0th → 4
⟨ ⁶ᵗʰ↓ ⁵ᵗʰ= ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↑ ¹ˢᵗ= ⁰ᵗʰ↓ ⟩ after 6−⟨⇌⟩
⟨⋯ 0 ⋯ ? 5 ⋯ (?+2)⟩ (?≠3)
⛔Avoid
⟨ ⁶ᵗʰa ³ʳᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≤ 346
#125034_v2.7
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 3 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 1 │ │ 3 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 1 │ │ 3 │ │ 5 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 1 │ 0 │ 3 │ │ 5 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ │ 1 │ 0 │ 3 │ 2 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ │ 1 │ 0 │ 3 │ 2 │ 5 │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 6 │ 1 │ 0 │ 3 │ 2 │ 5 │ 4 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-11-19 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
Consider the "=" positions of ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ= ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↑ ¹ˢᵗ= ⁰ᵗʰ↓ ⟩ after 6−⟨⇌⟩」. Plainly, we need [3rd] = 3.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 3 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
The preceding pattern also implies
(1) {[5th], [1st]} = {0,6} | {1,5} | {2,4}.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ │ 3 │ │ ▬ │ │
└───┴───┴───┴───┴───┴───┴───┘
On the other hand, by ✅「⟨⋯ 0 ⋯ ? 5 ⋯ (?+2)⟩ (?≠3)」, one of the following holds:
(2) ⟨⋯ 05 ⋯ 2⟩;
(3) ⟨⋯ 25 ⋯ 4⟩;
(4) ⟨⋯ 45 ⋯ 6⟩.
(5) Based on (1), we proceed to show that {[5th], [1st]} = {1,5}.
------------------------------
(5.1) If {[5th], [1st]} = {0,6}, then to match ✅「⟨⋯ 0 ⋯ ? 5 ⋯ (?+2)⟩ (?≠3)」, we need
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ │ 3 │ │ 6 │ │
└───┴───┴───┴───┴───┴───┴───┘
Then, noting that cases (3) and (4) are not possible, case (2) holds and we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ 5 │ 3 │ │ 6 │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
But then it cannot match the 4th position of ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ= ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↑ ¹ˢᵗ= ⁰ᵗʰ↓ ⟩ after 6−⟨⇌⟩」, which is a contradiction.
(5.2) Else, suppose {[5th], [1st]} = {2,4}:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ │ 3 │ │ ▬ │ │
└───┴───┴───┴───┴───┴───┴───┘
then (2) and (3) are not possible, so (4) holds and we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│ 4▲│3rd│2nd│ 1▲│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 4 │ 5 │ 3 │ │ 2 │ 6 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
In view of ✅「⟨⋯ 0 ⋯ ? 5 ⋯ (?+2)⟩ (?≠3)」, 0 is to the left of 5, so we reach
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 4 │ 5 │ 3 │ 1 │ 2 │ 6 │
└───┴───┴───┴───┴───┴───┴───┘
It does not match the 6th position of ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ= ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↑ ¹ˢᵗ= ⁰ᵗʰ↓ ⟩ after 6−⟨⇌⟩」, however.
------------------------------
We have verified (5). Consequently, one of the following holds:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(6) │ │ 1 │ │ 3 │ │ 5 │ │
├───┼───┼───┼───┼───┼───┼───┤
(7) │ │ 5 │ │ 3 │ │ 1 │ │
└───┴───┴───┴───┴───┴───┴───┘
Case (6) holds actually, for otherwise by ✅「⟨⋯ 0 ⋯ ? 5 ⋯ (?+2)⟩ (?≠3)」, we have
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 5 │ │ 3 │ │ 1 │ │
└───┴───┴───┴───┴───┴───┴───┘
and it does not match the 6th position of ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ= ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↑ ¹ˢᵗ= ⁰ᵗʰ↓ ⟩ after 6−⟨⇌⟩」.
Therefore, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ 3 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 1 │ │ 3 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 1 │ │ 3 │ │ 5 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Next, we consider where to place 0. It cannot be at a "↓" position of ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ= ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↑ ¹ˢᵗ= ⁰ᵗʰ↓ ⟩ after 6−⟨⇌⟩」, so
(8) 0 = [4th] | [2nd].
(9) We show that 0 = [4th] indeed.
------------------------------
For, suppose on the contrary 0 = [2nd]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ │ 3 │ 0 │ 5 │ │
└───┴───┴───┴───┴───┴───┴───┘
Then ✅「⟨⋯ 0 ⋯ ? 5 ⋯ (?+2)⟩ (?≠3)」 gives [0th] = 2:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ │ 3 │ 0 │ 5 │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
and ✅「6th|5th|2nd|0th → 4」 gives [6th] = 4:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ 1 │ │ 3 │ 0 │ 5 │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
This is a contradiction because it does not match the 6th position of ✅「⟨ ⁶ᵗʰ↓ ⁵ᵗʰ= ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↑ ¹ˢᵗ= ⁰ᵗʰ↓ ⟩ after 6−⟨⇌⟩」.
------------------------------
We have verified (9) and get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 1 │ │ 3 │ │ 5 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 1 │ 0 │ 3 │ │ 5 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Now, recall to one of (2), (3), (4) holds. Plainly, case (2) does not hold. If case (4) holds, then we cannot avoid ⛔「⟨ ⁶ᵗʰa ³ʳᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≤ 346」. Therefore, case (3) holds. We finish by
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│ 2■│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 1 │ 0 │ 3 │ │ 5 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ │ 1 │ 0 │ 3 │ 2 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ │ 1 │ 0 │ 3 │ 2 │ 5 │ 4 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 6 │ 1 │ 0 │ 3 │ 2 │ 5 │ 4 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.7