Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
{p4, p2, p0} = ? + {0,1,2}
⟦1,4⟧ ∋ 2
⟨ ³ʳᵈb ¹ˢᵗa ⟩, a > b
{p5, p2, p1} = ? + {0,1,3}
#125034_v2.9
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
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Step 1 │ │ │ │ 2 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ 4 │ 2 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ 4 │ 2 │ 5 │ │▒
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Step 4 │ 3 │ │ 4 │ 2 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ │ 4 │ 2 │ 5 │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 0 │ 4 │ 2 │ 5 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
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Proof of 2025-06-03 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
To match ✅「{p4, p2, p0} = ? + {0,1,2}」, we need
(1) {p4, p2, p0} consists of consecutive integers.
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│ 2▲│1st│ 0▲│
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│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
On the other hand, by ✅「{p5, p2, p1} = ? + {0,1,3}」, there are three possibilities for {p5, p2, p1}:
(A) {0,1,3};
(B) {1,2,4};
(C) {2,3,5}.
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│ 2▲│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ ▬ │ │ │ ▬ │ ▬ │ │
└───┴───┴───┴───┴───┴───┘
(2) We show by contradiction that (C) holds actually.
------------------------------
(2.1) Suppose (A) holds:
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│5th│ 4▲│3rd│ 2▲│1st│ 0▲│
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│013│ │ │013│013│ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
We consider what [1st] is. To match ✅「⟨ ³ʳᵈb ¹ˢᵗa ⟩, a > b」, we need [1st] > [3rd]. Since [3rd] ∈ {2,4,5}, we have [3rd] >= 2. Therefore, [1st] = 3:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│ 2▲│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│0|1│ │ │0|1│ 3 │ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ 2 │ 5 │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
However, it follows that no matter whether [2nd] = 0 or 1, we cannot match (1). This shows a contradiction.
------------------------------
(2.2) On the other hand, suppose (B) holds:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│ 2▲│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│124│ │ │124│124│ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
In view of ✅「⟦1,4⟧ ∋ 2」, we have [2nd] = 2:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│ 2▲│ 1 │ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│1|4│ │ │ 2 │1|4│ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ 5 │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
However, observe that we again cannot match (1).
------------------------------
By (2.1) and (2.2), we have verified (2). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│ 2▲│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│235│ │ │235│235│ │
└───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ │ 4 │
└───┴───┴───┴───┴───┴───┘
We consider what [2nd] is. Observe that if it is 5 or 3, then we still cannot match (1). To match (1), there is only one possibility: [2nd] = 2, and {[4th], [0th]} = {0,1}:
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│5th│4th│3rd│ 2■│1st│0th│▒
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Step 1 │3|5│0|1│ │ 2 │3|5│0|1│▒
└───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Plainly, it follows that [3rd] = 4:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│0th│▒
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│ │ │ │ 2 │ │ │▒
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Step 2 │3|5│0|1│ 4 │ 2 │3|5│0|1│▒
└───┴───┴───┴───┴───┴───┘▒
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Then, to match ✅「⟨ ³ʳᵈb ¹ˢᵗa ⟩, a > b」, we need [1st] = 5, and [5th] = 3 follows:
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ 4 │ 2 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │0|1│ 4 │ 2 │ 5 │0|1│▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │0|1│ 4 │ 2 │ 5 │0|1│▒
└───┴───┴───┴───┴───┴───┘▒
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Finally, to match ✅「⟦1,4⟧ ∋ 2」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│ 0■│▒
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│ 3 │ │ 4 │ 2 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ │ 4 │ 2 │ 5 │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 0 │ 4 │ 2 │ 5 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
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Q.E.D.
#125034_v2.9