Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
5th|3rd|2nd|1st → 2
⟨ ⁵ᵗʰ↓ ²ⁿᵈ↓ ⁰ᵗʰ↑ ⟩ after 2×⟨←⟩
⟨⋯ 6 ⋯ 2 ⋯ 4 ⋯ 3 ⋯⟩
⟨ ⁶ᵗʰ↑ ⁵ᵗʰ↓ ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↓ ¹ˢᵗ↑ ⁰ᵗʰ↓ ⟩ after ⟨⇌⟩
⟨ ⁶ᵗʰa ⁴ᵗʰb ²ⁿᵈc ⟩, (abc)₁₀ ≥ 53
⛔Avoid
⟨⋯ ? ⋯ 4 ⋯ (?−2)⟩ (?≠4,6)
#125034_v2.2
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 6 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 6 │ │ 2 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ 6 │ │ 2 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 6 │ 0 │ 2 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 6 │ 0 │ 2 │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 6 │ 0 │ 2 │ 4 │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 6 │ 0 │ 2 │ 4 │ 5 │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-01-23 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
We begin with locating the position of 6. Let [n-th] = 6. By ✅「⟨⋯ 6 ⋯ 2 ⋯ 4 ⋯ 3 ⋯⟩」, we have n >= 3.
┌───┬───┬───┬───┬───┬───┬───┐
│*6 │*5 │*4 │*3 │2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ │ / │ / │ / │
└───┴───┴───┴───┴───┴───┴───┘
Since 6 is the maximum digit, it is at the "↓" or "=" positions of ✅「⟨ ⁶ᵗʰ↑ ⁵ᵗʰ↓ ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↓ ¹ˢᵗ↑ ⁰ᵗʰ↓ ⟩ after ⟨⇌⟩」. Therefore, we have
(1) n = 5 or 3:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│*5 │4th│*3 │2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ / │ │ / │ │ / │ / │ / │
└───┴───┴───┴───┴───┴───┴───┘
If 6 = [3rd], then it follows from ✅「⟨⋯ 6 ⋯ 2 ⋯ 4 ⋯ 3 ⋯⟩」 that
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│ 2▲│ 1▲│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ 6 │ 2 │ 4 │ 3 │
└───┴───┴───┴───┴───┴───┴───┘
and we would match ⛔「⟨⋯ ? ⋯ 4 ⋯ (?−2)⟩ (?≠4,6)」, which is a contradiction. Hence, back to (1), we have 6 = [5th].
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 6 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we consider where to place 2. By ✅「⟨⋯ 6 ⋯ 2 ⋯ 4 ⋯ 3 ⋯⟩」, there are at least two digits at the right of 2. Combining this with ✅「5th|3rd|2nd|1st → 2」, we have
(2) 2 = [3rd] or [2nd].
If 2 = [2nd], then by ✅「⟨⋯ 6 ⋯ 2 ⋯ 4 ⋯ 3 ⋯⟩」 we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│*2 │*1 │*0 │
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 6 │ │ │ 2 │ 4 │ 3 │
└───┴───┴───┴───┴───┴───┴───┘
and we would match ⛔「⟨⋯ ? ⋯ 4 ⋯ (?−2)⟩ (?≠4,6)」, which is a contradiction. Hence, back to (2), we have 2 = [3rd].
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 6 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 6 │ │ 2 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
We proceed to determine the value of [6th]. It is not 4 or 3 by ✅「⟨⋯ 6 ⋯ 2 ⋯ 4 ⋯ 3 ⋯⟩」, so
(3) [6th] = 0|1|5. We claim that [6th] = 1 actually.
------------------------------
(3.1) If [6th] = 0, then to match ✅「⟨ ⁶ᵗʰa ⁴ᵗʰb ²ⁿᵈc ⟩, (abc)₁₀ ≥ 53」, we need [4th] = 5:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ 6 │ │ 2 │ │ │ │
├───┼───┼───┼───┼───┼───┼───┤
│ 0 │ 6 │ 5 │ 2 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 3 │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
However, note that we have put 5 at a "↑" position of ✅「⟨ ⁶ᵗʰ↑ ⁵ᵗʰ↓ ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↓ ¹ˢᵗ↑ ⁰ᵗʰ↓ ⟩ after ⟨⇌⟩」. Since 5 can only be increased by using 6, it is a contradiction.
(3.2) Else if [6th] = 5, then again we have put 5 at a "↑" position of ✅「⟨ ⁶ᵗʰ↑ ⁵ᵗʰ↓ ⁴ᵗʰ↑ ³ʳᵈ= ²ⁿᵈ↓ ¹ˢᵗ↑ ⁰ᵗʰ↓ ⟩ after ⟨⇌⟩」, which is a contradiction.
------------------------------
By (3.1) and (3.2), we have verified our claim in (3). Accordingly,
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 6 │ │ 2 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ 6 │ │ 2 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we determine the value of [4th]. By ✅「⟨⋯ 6 ⋯ 2 ⋯ 4 ⋯ 3 ⋯⟩」, it is not 3 or 4. It is not 5 as well, as argued in (3.1). Therefore, [4th] = 0.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 6 │ │ 2 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 6 │ 0 │ 2 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
We check the value of [0th] now. By ✅「⟨⋯ 6 ⋯ 2 ⋯ 4 ⋯ 3 ⋯⟩」, it is not 4. It is not 5 as well, for otherwise we would have [0th] > [2nd], and ✅「⟨ ⁵ᵗʰ↓ ²ⁿᵈ↓ ⁰ᵗʰ↑ ⟩ after 2×⟨←⟩」 cannot be matched:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ 6 │ 0 │ 2 │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Therefore, we need [0th] = 3.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 6 │ 0 │ 2 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 6 │ 0 │ 2 │ │ │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨⋯ ? ⋯ 4 ⋯ (?−2)⟩ (?≠4,6)」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 6 │ 0 │ 2 │ │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 6 │ 0 │ 2 │ 4 │ │ 3 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 6 │ 0 │ 2 │ 4 │ 5 │ 3 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.2
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