Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟦2,6⟧ ∋ 0,1,3
4th → a, 1st → b, a+b=1+6n
⟨? ⋯ 3 ⋯ (?+4) ⋯⟩
⟨ ⁴ᵗʰc ²ⁿᵈa ⁰ᵗʰb ⟩, a > b > c
3rd → 3|4|6
#125034_v2.5
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
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Step 1 │ 2 │ │ │ │ │ │ │▒
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Step 2 │ 2 │ │ │ │ │ │ 4 │▒
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Step 3 │ 2 │ │ │ 3 │ │ │ 4 │▒
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Step 4 │ 2 │ │ 1 │ 3 │ │ │ 4 │▒
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Step 5 │ 2 │ │ 1 │ 3 │ │ 6 │ 4 │▒
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Step 6 │ 2 │ │ 1 │ 3 │ 5 │ 6 │ 4 │▒
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Step 7 │ 2 │ 0 │ 1 │ 3 │ 5 │ 6 │ 4 │▒
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Proof of 2024-05-07 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「⟨? ⋯ 3 ⋯ (?+4) ⋯⟩」, we have [6th] = 0|1|2. Since ✅「⟦2,6⟧ ∋ 0,1,3」 implies 0 and 1 are not in corners, we have [6th] = 2.
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 2 │ │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, we consider the value of [0th]. As ✅「⟦2,6⟧ ∋ 0,1,3」 implies that 0,1,3 are not in corners, we have
(1) [0th] = 6|5|4.
(2) In view of ✅「⟨ ⁴ᵗʰc ²ⁿᵈa ⁰ᵗʰb ⟩, a > b > c」, we have [0th] != 6. We claim that [0th] = 4 actually.
------------------------------
For, suppose on the contrary [0th] = 5. Then, using ✅「⟨ ⁴ᵗʰc ²ⁿᵈa ⁰ᵗʰb ⟩, a > b > c」, we get [2nd] = 6:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2▲│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ │ │ 6 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 3 │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
It then follows from ✅「⟦2,6⟧ ∋ 0,1,3」 that { [5th], [4th], [3rd] } = {0,1,3}, and that [1st] = 4:
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│6th│*5 │*4 │*3 │2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ ▬ │ ▬ │ ▬ │ 6 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Using ✅「3rd → 3|4|6」, we have [3rd] = 3:
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│6th│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ ▬ │ ▬ │ 3 │ 6 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
But then we cannot not match ✅「4th → a, 1st → b, a+b=1+6n」, which is a contradiction.
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We have verified our claim in (2). Accordingly, we get
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│6th│5th│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ │ │ │ │ │▒
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Step 2 │ 2 │ │ │ │ │ │ 4 │▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, note that to match ✅「⟦2,6⟧ ∋ 0,1,3」, we cannot place 6 at 3rd. So, in view of ✅「3rd → 3|4|6」, we get [3rd] = 3:
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│6th│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ │ │ │ │ 4 │▒
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Step 3 │ 2 │ │ │ 3 │ │ │ 4 │▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
We proceed to consider how to match ✅「4th → a, 1st → b, a+b=1+6n」. Observe that there are only two ways to do so:
(3) {[4th], [1st]} = {0,1} | {1,6}.
We need to use 1 in both cases, so 1 = [4th] | [1st]. To match ✅「⟦2,6⟧ ∋ 0,1,3」, we cannot have 1 = [1st]. It follows that 1 = [4th]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│1st│0th│▒
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│ 2 │ │ │ 3 │ │ │ 4 │▒
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Step 4 │ 2 │ │ 1 │ 3 │ │ │ 4 │▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, by (3), we have [1st] = 0|6. Again using ✅「⟦2,6⟧ ∋ 0,1,3」, we see that [1st] != 0. So, [1st] = 6:
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│6th│5th│4th│3rd│2nd│ 1■│0th│▒
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│ 2 │ │ 1 │ 3 │ │ │ 4 │▒
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Step 5 │ 2 │ │ 1 │ 3 │ │ 6 │ 4 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Finally, to match ✅「⟨ ⁴ᵗʰc ²ⁿᵈa ⁰ᵗʰb ⟩, a > b > c」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ 1 │ 3 │ │ 6 │ 4 │▒
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Step 6 │ 2 │ │ 1 │ 3 │ 5 │ 6 │ 4 │▒
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Step 7 │ 2 │ 0 │ 1 │ 3 │ 5 │ 6 │ 4 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.5