Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ⁵ᵗʰd ⁴ᵗʰc ²ⁿᵈb ¹ˢᵗa ⟩, a > b > c > d
4th → a, 0th → b, a+b=3+5n
⛔Avoid
6th|5th|2nd|1st|0th → 5
⟨⋯ 2 ⋯ 3 ⋯⟩
{p6, p5, p4, p0} = ? + {0,1,2,3}
⟨ ⁶ᵗʰa ⁵ᵗʰc ¹ˢᵗb ⟩, a > b > c
#125034_v2.6
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 5 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 0 │ │ 5 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 0 │ 1 │ 5 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 0 │ 1 │ 5 │ │ │ 2 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ │ 0 │ 1 │ 5 │ 3 │ │ 2 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 0 │ 1 │ 5 │ 3 │ │ 2 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 4 │ 0 │ 1 │ 5 │ 3 │ 6 │ 2 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-09-03 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
To begin with, we consider where to place 5. By ⛔「6th|5th|2nd|1st|0th → 5」, we have
(1) 5 = [4th] | [3rd].
In view of ✅「⟨ ⁵ᵗʰd ⁴ᵗʰc ²ⁿᵈb ¹ˢᵗa ⟩, a > b > c > d」, we cannot place 5 at 4th. Therefore, 5 = [3rd]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 5 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ 3 │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Next, we consider the required pattern ✅「4th → a, 0th → b, a+b=3+5n」.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ ▬ │ 5 │ │ │ ▬ │
└───┴───┴───┴───┴───┴───┴───┘
Note that it implies [4th] + [0th] = 3|8. There are three possibilities for {[4th], [0th]}:
(i) {[4th], [0th]} = {2,6};
(ii) {[4th], [0th]} = {0,3};
(iii) {[4th], [0th]} = {1,2}.
(2) We show that case (iii) holds indeed.
------------------------------
(2.1) Suppose case (i) holds. Since ✅「⟨ ⁵ᵗʰd ⁴ᵗʰc ²ⁿᵈb ¹ˢᵗa ⟩, a > b > c > d」 implies [4th] != 6, we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 2 │ 5 │ │ │ 6 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 3 │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
To match ✅「⟨ ⁵ᵗʰd ⁴ᵗʰc ²ⁿᵈb ¹ˢᵗa ⟩, a > b > c > d」, we then need
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2▲│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 2 │ 5 │ 3 │ 4 │ 6 │
└───┴───┴───┴───┴───┴───┴───┘
But then we match ⛔「⟨⋯ 2 ⋯ 3 ⋯⟩」, which is a contradiction. So, case (i) does not hold.
(2.2) Suppose case (ii) holds instead. Since ✅「⟨ ⁵ᵗʰd ⁴ᵗʰc ²ⁿᵈb ¹ˢᵗa ⟩, a > b > c > d」 implies [4th] != 0, we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 3 │ 5 │ │ │ 0 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
To match ✅「⟨ ⁵ᵗʰd ⁴ᵗʰc ²ⁿᵈb ¹ˢᵗa ⟩, a > b > c > d」, we then have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2▲│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 3 │ 5 │ 4 │ 6 │ 0 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Again we would match ⛔「⟨⋯ 2 ⋯ 3 ⋯⟩」. So case (ii) also does not hold.
------------------------------
By (2.1) and (2.2), we have verified our claim in (2). Accordingly, we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ ▬ │ 5 │ │ │ ▬ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ 3 │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
where "▬" are occupied by 1,2.
Therefore, since [5th] < [4th] by ✅「⟨ ⁵ᵗʰd ⁴ᵗʰc ²ⁿᵈb ¹ˢᵗa ⟩, a > b > c > d」, we get [5th] = 0:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ 5 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 0 │ ▬ │ 5 │ │ │ ▬ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ 3 │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
(3) Now, we claim that ([4th], [0th]) = (1,2).
------------------------------
For, if on the contrary ([4th], [0th]) = (2,1):
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ 2 │ 5 │ │ │ 1 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ 3 │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
then, to avoid ⛔「⟨⋯ 2 ⋯ 3 ⋯⟩」, we need
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 3 │ 0 │ 2 │ 5 │ │ │ 1 │
└───┴───┴───┴───┴───┴───┴───┘
But then we would match ⛔「{p6, p5, p4, p0} = ? + {0,1,2,3}」, which is a contradiction.
------------------------------
Hence, by (3), we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 0 │ │ 5 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 0 │ 1 │ 5 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 0 │ 1 │ 5 │ │ │ 2 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ 3 │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Then, we consider how to place 3. To avoid ⛔「{p6, p5, p4, p0} = ? + {0,1,2,3}」, it is not at 6th. On the other hand, since 3 is the smallest idle digit, by ✅「⟨ ⁵ᵗʰd ⁴ᵗʰc ²ⁿᵈb ¹ˢᵗa ⟩, a > b > c > d」, it is not at 1st. Therefore, we have 3 = [2nd]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 0 │ 1 │ 5 │ │ │ 2 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ │ 0 │ 1 │ 5 │ 3 │ │ 2 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨ ⁶ᵗʰa ⁵ᵗʰc ¹ˢᵗb ⟩, a > b > c」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 0 │ 1 │ 5 │ 3 │ │ 2 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 4 │ 0 │ 1 │ 5 │ 3 │ │ 2 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 4 │ 0 │ 1 │ 5 │ 3 │ 6 │ 2 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.6
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