Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
4th → a, 1st → b, ab=2+4n
Jump(2,3) = 1
⟨ ⁶ᵗʰc ²ⁿᵈa ¹ˢᵗb ⟩, a > b > c
⛔Avoid
⟨⋯ 3 ⋯ 0 ⋯⟩
⟨⋯ ? ⋯ 6 ⋯ (?+1)⟩ (?≠5)
min ⊢5⊣ ≥ 2
6th → a, 3rd → b, ab=0+6n
#125034_v2.4
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 1 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │ 0 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ 0 │ 5 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 0 │ 5 │ │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 0 │ 5 │ 3 │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 0 │ 5 │ 3 │ │ 2 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 0 │ 5 │ 3 │ 4 │ 2 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-04-02 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
We begin with the left corner, i.e. [6th]. Observe that
(i) By ✅「⟨ ⁶ᵗʰc ²ⁿᵈa ¹ˢᵗb ⟩, a > b > c」, it is not 6 or 5;
(ii) By ⛔「⟨⋯ 3 ⋯ 0 ⋯⟩」, it is not 3;
(iii) By ⛔「6th → a, 3rd → b, ab=0+6n」, it is not 0.
Therefore,
(1) [6th] = 4|2|1. We proceed to show that [6th] = 1 indeed.
------------------------------
(1.1) Assume [6th] = 4. Then, to match ✅「⟨ ⁶ᵗʰc ²ⁿᵈa ¹ˢᵗb ⟩, a > b > c」, we need "a"=6 and "b"=5:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│4th│3rd│ 2▲│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ │ │ │ 6 │ 5 │ │
└───┴───┴───┴───┴───┴───┴───┘
Then, to avoid ⛔「min ⊢5⊣ ≥ 2」, we need [0th] = 1|0:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ │ │ │ 6 │ 5 │ ▬ │
└───┴───┴───┴───┴───┴───┴───┘
But it is a contradiction:
● If [0th] = 1, then we would match ⛔「⟨⋯ ? ⋯ 6 ⋯ (?+1)⟩ (?≠5)」;
● Else if [0th] = 0, then we would match ⛔「⟨⋯ 3 ⋯ 0 ⋯⟩」.
Hence, our assumption "[6th] = 4" is not correct.
(1.2) Else, assume [6th] = 2. By ✅「Jump(2,3) = 1」, we get
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ 3 │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, to match ✅「4th → a, 1st → b, ab=2+4n」, we need [1st] = 6:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ 3 │ │ │ 6 │ │
└───┴───┴───┴───┴───┴───┴───┘
But then ✅「⟨ ⁶ᵗʰc ²ⁿᵈa ¹ˢᵗb ⟩, a > b > c」 cannot be matched, which is a contradiction.
------------------------------
By (1.1) and (1.2), our claim in (1) is verified. Accordingly, we get
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 1 │ │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we decide where to place 0. Note that:
(i) To avoid ⛔「⟨⋯ 3 ⋯ 0 ⋯⟩」, we need [0th] != 0;
(ii) By ✅「⟨ ⁶ᵗʰc ²ⁿᵈa ¹ˢᵗb ⟩, a > b > c」, 0 is not at 1st or 2nd;
(iii) To avoid ⛔「6th → a, 3rd → b, ab=0+6n」, we need [3rd] != 0;
(iv) To match ✅「4th → a, 1st → b, ab=2+4n」, we need [4th] != 0.
Consequently, 0 = [5th]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │ 0 │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Now, note that to avoid ⛔「min ⊢5⊣ ≥ 2」, we need 5 adjacent to 1 or 0. We have only one way to do so:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 0 │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ 0 │ 5 │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ 3 │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Then, to match ✅「4th → a, 1st → b, ab=2+4n」, we have [1st] = 6|2. Actually [1st] != 6, for we have to match ✅「⟨ ⁶ᵗʰc ²ⁿᵈa ¹ˢᵗb ⟩, a > b > c」 as well. Hence, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 0 │ 5 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ 0 │ 5 │ │ │ 2 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ 3 │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Then, by ✅「Jump(2,3) = 1」, plainly we have [3rd] = 3.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 0 │ 5 │ │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 0 │ 5 │ 3 │ │ 2 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, in view of ⛔「⟨⋯ ? ⋯ 6 ⋯ (?+1)⟩ (?≠5)」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ 0 │ 5 │ 3 │ │ 2 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 0 │ 5 │ 3 │ │ 2 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 0 │ 5 │ 3 │ 4 │ 2 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.4