Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨? ⋯ 3 ⋯ (?+2) ⋯⟩ (?≠3,1)
4th → a, 1st → b, a+b=4+5n
6th → a, 2nd → b, a+b=8
4th → a, 0th → b, |a-b|=2
#125034_v2.1
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
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Step 1 │ 2 │ │ │ │ │ │ │
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Step 2 │ 2 │ │ │ │ 6 │ │ │
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Step 3 │ 2 │ │ 3 │ │ 6 │ │ │
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Step 4 │ 2 │ │ 3 │ │ 6 │ 1 │ │
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Step 5 │ 2 │ │ 3 │ │ 6 │ 1 │ 5 │
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Step 6 │ 2 │ │ 3 │ 4 │ 6 │ 1 │ 5 │
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Step 7 │ 2 │ 0 │ 3 │ 4 │ 6 │ 1 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Proof of 2023-11-21 Q1(m=6)
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Notation: if nth -> a, then we write [nth] = a.
We begin with the left corner. In view of ✅【⟨? ⋯ 3 ⋯ (?+2) ⋯⟩ (?≠3,1)】, we need [6th] = 0 or 2 or 4 in order that (?+2) <= 6. We need to match ✅【6th → a, 2nd → b, a+b=8】 too. When [6th] = 0 or 4, there is no solution to "[6th] + [2nd] = 8". Therefore, we have [6th] = 2, and thus [2nd] = 6.
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│ 2■│1st│0th│
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Step 1 │ 2 │ │ │ │ │ │ │
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Step 2 │ 2 │ │ │ │ 6 │ │ │
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--- Idle ---
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│ 1 │ │ │ 3 │ 0 │ 4 │ 5 │
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(1) Next, we determine the value of [4th]. Claim that indeed [4th] = 3.
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(2) If [4th] = 5 or 1, then by ✅【4th → a, 0th → b, |a-b|=2】 we have [0th] = 3:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│3rd│2nd│1st│ 0▲│
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│ 2 │ │ * │ │ 6 │ │ 3 │
└───┴───┴───┴───┴───┴───┴───┘
Then 3 is in corners, which contradictis ✅【⟨? ⋯ 3 ⋯ (?+2) ⋯⟩ (?≠3,1)】.
(3) Else, if [4th] = 4, then ✅【4th → a, 0th → b, |a-b|=2】 cannot be matched:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ 4 │ │ 6 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 3 │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
(4) Else, if [4th] = 0:
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│6th│5th│ 4▲│3rd│2nd│1st│0th│
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│ 2 │ │ 0 │ │ 6 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
then we cannot match ✅【4th → a, 0th → b, |a-b|=2】.
------------------------------
Combining (1), (2), (3), (4), we have verified that [4th] = 3.
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│6th│5th│ 4■│3rd│2nd│1st│0th│
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│ 2 │ │ │ │ 6 │ │ │
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Step 3 │ 2 │ │ 3 │ │ 6 │ │ │
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Now there is only one way to match ✅【4th → a, 1st → b, a+b=4+5n】:
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│6th│5th│4th│3rd│2nd│ 1■│0th│
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│ 2 │ │ 3 │ │ 6 │ │ │
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Step 4 │ 2 │ │ 3 │ │ 6 │ 1 │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
and only one way to match ✅【4th → a, 0th → b, |a-b|=2】:
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│6th│5th│4th│3rd│2nd│1st│ 0■│
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│ 2 │ │ 3 │ │ 6 │ 1 │ │
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Step 5 │ 2 │ │ 3 │ │ 6 │ 1 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, as ✅【⟨? ⋯ 3 ⋯ (?+2) ⋯⟩ (?≠3,1)】 implies that 3 is at the left of 4, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│ 3■│2nd│1st│0th│
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│ 2 │ │ 3 │ │ 6 │ 1 │ 5 │
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Step 6 │ 2 │ │ 3 │ 4 │ 6 │ 1 │ 5 │
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Step 7 │ 2 │ 0 │ 3 │ 4 │ 6 │ 1 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.1