Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨⋯ Perm(0,1,4) ⋯⟩
⟨⋯ 2 ⋯ 5 ⋯ 1 ⋯⟩
6th → a, 5th → b, a+b=5
⟨⋯ Perm(1,6) ⋯⟩
⛔Avoid
⟨⋯ 4 ⋯ a ⋯⟩, a = 0|5
6th → a, 0th → b, a+b=8
2nd → 1
#125034_v2.4
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ │ 1 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ │ │ 1 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 2 │ │ │ │ 1 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 2 │ 5 │ │ │ 1 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 2 │ 5 │ │ │ 1 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 2 │ 5 │ 0 │ │ 1 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 3 │ 2 │ 5 │ 0 │ 4 │ 1 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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Proof of 2024-04-23 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
To begin with, note that by ✅「⟨⋯ 2 ⋯ 5 ⋯ 1 ⋯⟩」, ✅「⟨⋯ Perm(0,1,4) ⋯⟩」, and ✅「⟨⋯ Perm(1,6) ⋯⟩」, we need to match
(1) ⟨⋯ 2 ⋯ 5 ⋯ Perm(0,1,4,6) ⋯⟩.
Four consecutive boxes are needed for Perm(0,1,4,6). In view of (1), there are two possibilities:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(2) │ │ │ ▬ │ ▬ │ ▬ │ ▬ │ │
├───┼───┼───┼───┼───┼───┼───┤
(3) │ │ │ │ ▬ │ ▬ │ ▬ │ ▬ │
└───┴───┴───┴───┴───┴───┴───┘
Actually (2) is not possible, for otherwise it follows from (1) that
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 5 │ ▬ │ ▬ │ ▬ │ ▬ │ │
└───┴───┴───┴───┴───┴───┴───┘
and we cannot match ✅「6th → a, 5th → b, a+b=5」, which is a contradiction.
Therefore, (3) holds. A fortiori,
(4) 1 = [3rd] | [2nd] | [1st] | [0th].
By ⛔「2nd → 1」, plainly 1 != [2nd]. On the other hand, if 1 is at the boundary of the four consecutive boxes (i.e. 3rd or 0th), then by ✅「⟨⋯ Perm(1,6) ⋯⟩」 and (3), we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ 1 │ 6 │ ▬ │ ▬ │
├───┼───┼───┼───┼───┼───┼───┤
│ │ │ │ ▬ │ ▬ │ 6 │ 1 │
└───┴───┴───┴───┴───┴───┴───┘
It follows that we cannot match ✅「⟨⋯ Perm(0,1,4) ⋯⟩」, which is a contradiction. Therefore, 1 = [1st]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ ▬ │ ▬ │ 1 │ ▬ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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Then, to match both ✅「⟨⋯ Perm(1,6) ⋯⟩」 and ✅「⟨⋯ Perm(0,1,4) ⋯⟩」, we have to place 6 at 0th.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 2 │ │ │ │ ▬ │ ▬ │ 1 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ 3 │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we determine where to place 2. By (1), we have
(5) 2 = [6th] | [5th].
Since we have to avoid ⛔「6th → a, 0th → b, a+b=8」, so 2 = [5th]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 3 │ │ 2 │ │ ▬ │ ▬ │ 1 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, by (1), we get
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 4 │ │ 2 │ 5 │ ▬ │ ▬ │ 1 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 2 │ 5 │ ▬ │ ▬ │ 1 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Finally, in view of ⛔「⟨⋯ 4 ⋯ a ⋯⟩, a = 0|5」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 6 │ 3 │ 2 │ 5 │ 0 │ ▬ │ 1 │ 6 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 3 │ 2 │ 5 │ 0 │ 4 │ 1 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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Q.E.D.
#125034_v2.4