Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
5th → 3
6th → a, 0th → b, ab=24
2nd → 0|5|6
3rd → a, 0th → b, a+b=0+6n
⛔Avoid
5th → a, 1st → b, a+b=8
⟦0,6⟧ ∋ 1,2,5
Jump(0,5) = 1
2nd → a, 1st → b, a+b=6
#125034_v2.2
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 3 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 4 │ 3 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 4 │ 3 │ │ │ │ │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 4 │ 3 │ │ 0 │ │ │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │ 3 │ │ 0 │ 5 │ │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 3 │ 1 │ 0 │ 5 │ │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 4 │ 3 │ 1 │ 0 │ 5 │ 2 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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Proof of 2023-12-15 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
Plainly, our first step follows from ✅「5th → 3」. Next, according to ✅「6th → a, 0th → b, ab=24」, we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ X │ 3 │ │ │ │ │ Y │
└───┴───┴───┴───┴───┴───┴───┘
where {X,Y} = {4,6}.
(1) We proceed to show that (X,Y) = (4,6).
------------------------------
If on the contrary (1) does not hold, then (X,Y) = (6,4), and in view of ✅「3rd → a, 0th → b, a+b=0+6n」, we get
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│4th│ 3▲│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 6 │ 3 │ │ 2 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, to avoid ⛔「Jump(0,5) = 1」, we need [1st] = 0 or 5. We need to avoid ⛔「5th → a, 1st → b, a+b=8」 too, so [1st] = 0.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 6 │ 3 │ │ 2 │ │ 0 │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
But then we would match ⛔「⟦0,6⟧ ∋ 1,2,5」, which is a contradiction.
------------------------------
We have verified (1). Accordingly, we have:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 3 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 4 │ 3 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 4 │ 3 │ │ │ │ │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
It now follows from ✅「3rd → a, 0th → b, a+b=0+6n」 that [3rd] = 0:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 4 │ 3 │ │ │ │ │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 4 │ 3 │ │ 0 │ │ │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
To match ✅「2nd → 0|5|6」, we need [2nd] = 5.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 4 │ 3 │ │ 0 │ │ │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 4 │ 3 │ │ 0 │ 5 │ │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「2nd → a, 1st → b, a+b=6」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 4 │ 3 │ │ 0 │ 5 │ │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 3 │ 1 │ 0 │ 5 │ │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 4 │ 3 │ 1 │ 0 │ 5 │ 2 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.2