Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
{p6, p4, p3, p2} = ? + {0,1,3,4}
2nd → 1|2|6
⟨? ⋯ 5 (?+4) ⋯ 3 ⋯⟩ (?≠1)
⛔Avoid
⟨ ⁴ᵗʰa ²ⁿᵈc ⁰ᵗʰb ⟩, a > b > c
#125034_v2.7
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 0 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 0 │ │ 4 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 0 │ 5 │ 4 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ 5 │ 4 │ │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ 5 │ 4 │ 3 │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 5 │ 4 │ 3 │ 1 │ 2 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 0 │ 5 │ 4 │ 3 │ 1 │ 2 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-10-29 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「⟨? ⋯ 5 (?+4) ⋯ 3 ⋯⟩ (?≠1)」, we have
(1) [6th] = 0|2.
(1.1) We show that [6th] = 0 actually.
------------------------------
For, suppose on the contrary [6th] = 2:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Then, in view of ✅「{p6, p4, p3, p2} = ? + {0,1,3,4}」, one of the following holds:
(2.1) {[4th], [3rd], [2nd]} = {1,4,5};
(2.2) {[4th], [3rd], [2nd]} = {3,5,6}.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │ ▬ │ ▬ │ ▬ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
If (2.1) holds, then [2nd] = 1|4|5. Using ✅「2nd → 1|2|6」, we get [2nd] = 1:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │4 5│4 5│ 1 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
But then we cannot match ✅「⟨? ⋯ 5 (?+4) ⋯ 3 ⋯⟩ (?≠1)」, which is a contradiction.
Therefore, (2.1) does not hold and (2.2) holds instead. A fortiori [2nd] = 3|5|6. Using ✅「2nd → 1|2|6」, we get [2nd] = 6:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ │3 5│3 5│ 6 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Nevertheless, this again contradicts ✅「⟨? ⋯ 5 (?+4) ⋯ 3 ⋯⟩ (?≠1)」.
------------------------------
We have verified (1.1). Accordingly, we get
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 0 │ │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
So, in view of ✅「{p6, p4, p3, p2} = ? + {0,1,3,4}」, we have
(3) {[4th], [3rd], [2nd]} = {1,3,4}.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ │134│134│134│ │ │
└───┴───┴───┴───┴───┴───┴───┘
On the other hand, by ✅「⟨? ⋯ 5 (?+4) ⋯ 3 ⋯⟩ (?≠1)」 with ?=0, we need to have the block "54". Therefore, we obtain:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ │ ▬ │ ▬ │ ▬ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 0 │ │ 4 │ ▬ │ ▬ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 0 │ 5 │ 4 │ ▬ │ ▬ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ 3 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Combining ✅「2nd → 1|2|6」 with (3), we find [2nd] and [3rd] as well:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ 5 │ 4 │ ▬ │ ▬ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 0 │ 5 │ 4 │ ▬ │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 0 │ 5 │ 4 │ 3 │ 1 │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「⟨ ⁴ᵗʰa ²ⁿᵈc ⁰ᵗʰb ⟩, a > b > c」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1■│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 0 │ 5 │ 4 │ 3 │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 0 │ 5 │ 4 │ 3 │ 1 │ 2 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 0 │ 5 │ 4 │ 3 │ 1 │ 2 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.7