Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
5th → 1|2|5|6
⟨? ⋯ 1 ⋯ (?+2) ⋯⟩ (?≠1)
⟨⋯ ? ⋯ 6 ⋯ (?+4)⟩ (?≠2)
6th → a, 5th → b, a+b=5+6n
⟨⋯ ? ⋯ 6 ⋯ (?+3)⟩ (?≠3)
Jump(1,4) = 1
3rd → 1|2|6
⛔Avoid
⟨ ²ⁿᵈa ¹ˢᵗb ⟩, (ab)₁₀ ≥ 46
#125034_v2.2
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ 6 │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 2 │ │ 6 │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 2 │ 1 │ 6 │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 2 │ 1 │ 6 │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 2 │ 1 │ 6 │ 4 │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 3 │ 2 │ 1 │ 6 │ 4 │ 0 │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2023-12-21 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
We begin with two observations. Firstly, we have
(1) [5th] = 2|5|6.
It is because of ✅「5th → 1|2|5|6」, and that if [5th] = 1, then ✅「Jump(1,4) = 1」 implies [3rd] = 4:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ │ 4 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
which contradicts ✅「3rd → 1|2|6」.
Secondly, by ✅「⟨⋯ ? ⋯ 6 ⋯ (?+4)⟩ (?≠2)」, exactly one of the following holds:
(2) ⟨⋯ 0 ⋯ 6 ⋯ 4⟩ or ⟨⋯ 1 ⋯ 6 ⋯ 5⟩.
Now, let us determine where to place 6. By ✅「⟨⋯ ? ⋯ 6 ⋯ (?+4)⟩ (?≠2)」, 6 is not in corners; and to avoid ⛔「⟨ ²ⁿᵈa ¹ˢᵗb ⟩, (ab)₁₀ ≥ 46」, we need 6 ≠ [2nd]. Hence
(3) 6 = [5th] or [4th] or [3rd] or [1st]. We claim that 6 = [3rd] actually.
------------------------------
(3.1) Case "6 = [5th] or [4th]":
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ * │ * │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Suppose 6 = [5th] or [4th]. By ✅「⟨? ⋯ 1 ⋯ (?+2) ⋯⟩ (?≠1)」 and (1), we have [6th] ≠ 1, [5th] ≠ 1, and [5th] ≠ 0. Therefore, it follows from (2) that:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ * │ * │ │ │ │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
Then, by ✅「Jump(1,4) = 1」 and ✅「3rd → 1|2|6」, we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 0 │ * │ * │ 2 │ 1 │ │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
It is a contradiction, however, as we fail to match ✅「⟨? ⋯ 1 ⋯ (?+2) ⋯⟩ (?≠1)」.
(3.2) Case "6 = [1st]":
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ │ │ 6 │ │
└───┴───┴───┴───┴───┴───┴───┘
Suppose 6 = [1st]. By ✅「3rd → 1|2|6」, we have [3rd] = 1|2. It is not 1, for otherwise by ✅「Jump(1,4) = 1」 we have [5th] = 4:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│*5 │4th│*3 │2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 4 │ │ 1 │ │ 6 │ │
└───┴───┴───┴───┴───┴───┴───┘
which contradicts (1). Accordingly, [3rd] = 2.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ 2 │ │ 6 │ │
└───┴───┴───┴───┴───┴───┴───┘
By (1), we have [5th] = 2|5|6. So, we see that [5th] = 5.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 5 │ │ 2 │ │ 6 │ │
└───┴───┴───┴───┴───┴───┴───┘
Then, matching (2) becomes matching ⟨⋯ 0 ⋯ 6 ⋯ 4⟩. It implies [0th] = 4, and using ✅「Jump(1,4) = 1」 we have [2nd] = 1 as well:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2▲│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 5 │ │ 2 │ 1 │ 6 │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
It is a contradiction because we cannot match ✅「⟨? ⋯ 1 ⋯ (?+2) ⋯⟩ (?≠1)」 anyway.
------------------------------
By (3.1) and (3.2), our claim in (3) is verified. Accordingly, we get our first step:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ 6 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
We proceed to determine the value of [0th]. By (2), we have [0th] = 4|5. Actually it is not 4, for otherwise it follows from ✅「Jump(1,4) = 1」 that
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│*2 │1st│*0 │
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ │ 6 │ 1 │ │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
which contradicts ✅「⟨⋯ ? ⋯ 6 ⋯ (?+3)⟩ (?≠3)」.
As a result, we have [0th] = 5. Recalling (1), we get [5th] = 2 as well.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ 6 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ │ 6 │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 2 │ │ 6 │ │ │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 3 │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
By ✅「⟨⋯ ? ⋯ 6 ⋯ (?+4)⟩ (?≠2)」, 1 is at the left of 6, so 1 = [6th] or [4th]. It is not in corners by ✅「⟨? ⋯ 1 ⋯ (?+2) ⋯⟩ (?≠1)」, so 1 = [4th].
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 2 │ │ 6 │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 2 │ 1 │ 6 │ │ │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ 3 │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Now there is only one way to match ✅「6th → a, 5th → b, a+b=5+6n」:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 2 │ 1 │ 6 │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 2 │ 1 │ 6 │ │ │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
In view of ✅「Jump(1,4) = 1」, we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 2 │ 1 │ 6 │ │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 2 │ 1 │ 6 │ 4 │ │ 5 │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 3 │ 2 │ 1 │ 6 │ 4 │ 0 │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.2
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