Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨? ⋯ 1 (?+2) ⋯ 2 ⋯⟩ (?≠1)
Jump(4,5) = 2
⟨ ⁶ᵗʰa ¹ˢᵗb ⟩, (ab)₁₀ ≥ 13
6th|5th|4th|2nd|0th → 6
⛔Avoid
⟦2,4⟧ ∋ 1
5th → a, 2nd → b, a+b=2+4n
2nd → a, 1st → b, a+b=10
⟨? ⋯ 2 ⋯ (?−3) ⋯⟩ (?≠5)
#125034_v2.2
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 3 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 3 │ 1 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ 1 │ 5 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 1 │ 5 │ │ │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 1 │ 5 │ │ │ 4 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 1 │ 5 │ 0 │ │ 4 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 3 │ 1 │ 5 │ 0 │ 2 │ 4 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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Proof of 2023-12-10 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
We begin with determining the value of [6th]. By ✅「⟨? ⋯ 1 (?+2) ⋯ 2 ⋯⟩ (?≠1)」, we have [6th] = 0|3|4. If [6th] = 4, then the preceding pattern becomes:
⟨4 ⋯ 16 ⋯ 2 ⋯⟩
Matching it would be a contradiction, however, as we need to avoid ⛔「⟦2,4⟧ ∋ 1」. Hence, [6th] = 0|3. To match ✅「⟨ ⁶ᵗʰa ¹ˢᵗb ⟩, (ab)₁₀ ≥ 13」, we nedd [6th] >= 1, thus [6th] = 3.
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 3 │ │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Now, note that to avoid ⛔「⟨? ⋯ 2 ⋯ (?−3) ⋯⟩ (?≠5)」, we need to match:
(1) ⟨3 ⋯ 0 ⋯ 2 ⋯⟩
Also, note that ✅「⟨? ⋯ 1 (?+2) ⋯ 2 ⋯⟩ (?≠1)」 becomes:
(2) ⟨3 ⋯ 15 ⋯ 2 ⋯⟩
In particular, 1,5 are adjacent to each other in this order. To match (2), there are four possibilities:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(2.1) │ 3 │ 1 │ 5 │ │ │ │ │
├───┼───┼───┼───┼───┼───┼───┤
(2.2) │ 3 │ │ 1 │ 5 │ │ │ │
├───┼───┼───┼───┼───┼───┼───┤
(2.3) │ 3 │ │ │ 1 │ 5 │ │ │
├───┼───┼───┼───┼───┼───┼───┤
(2.4) │ 3 │ │ │ │ 1 │ 5 │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
We proceed to show that (2.1) holds actually.
------------------------------
Case (2.2):
If (2.2) holds, then by ✅「Jump(4,5) = 2」, we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│*4 │*3 │2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ │ 1 │ 5 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
If 0 is at 5th or 2nd, then we would match ⛔「5th → a, 2nd → b, a+b=2+4n」. Hence, 0 = [1st].
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ │ 1 │ 5 │ │ 0 │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
This results in a contradiction, however, as we cannot match (1) now.
Case (2.3):
If (2.3) holds, then by ✅「Jump(4,5) = 2」, we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│4th│*3 │*2 │1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ 4 │ │ 1 │ 5 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
But then to match (2) we would match ⛔「⟦2,4⟧ ∋ 1」 too, which is a contradiction.
Case (2.4):
If (2.4) holds:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│*2 │*1 │*0 │
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ │ │ │ 1 │ 5 │ 2 │
└───┴───┴───┴───┴───┴───┴───┘
then unavoidably we would match ⛔「⟦2,4⟧ ∋ 1」.
------------------------------
We have verified that (2.1) holds indeed. Combining it with ✅「Jump(4,5) = 2」, we obtain
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│ 4■│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 3 │ 1 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ 1 │ 5 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 1 │ 5 │ │ │ 4 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Now we consider where to place 6. By ✅「6th|5th|4th|2nd|0th → 6」, we see that 6 = [2nd] or [0th]. To avoid ⛔「2nd → a, 1st → b, a+b=10」, we need 6 = [0th].
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 1 │ 5 │ │ │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 1 │ 5 │ │ │ 4 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, recall that we need to match (1). We finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 1 │ 5 │ │ │ 4 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 1 │ 5 │ 0 │ │ 4 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 3 │ 1 │ 5 │ 0 │ 2 │ 4 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.2