Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
{p6, p3, p1} = ? + {0,3,4}
⟦1,3⟧ ∋ 5
Sim⟨ ⁶ᵗʰ4 ⁵ᵗʰ2 ⁴ᵗʰ3 ³ʳᵈ5 ²ⁿᵈ0 ¹ˢᵗ1 ⁰ᵗʰ6 ⟩ ≥ 2
Jump(3,6) ≥ 2
⛔Avoid
⟨ ⁵ᵗʰb ²ⁿᵈa ⟩, a > b
⟨ ⁰ᵗʰ↓ ⟩ after 2×⟨→⟩
#125034_v2.6
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ 3 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ 3 │ 5 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ 3 │ 5 │ │ 6 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ │ 3 │ 5 │ │ 6 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 4 │ 3 │ 5 │ │ 6 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 4 │ 3 │ 5 │ 1 │ 6 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 4 │ 3 │ 5 │ 1 │ 6 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-07-30 Q1(m=6)
═══════════════════════════
Notation: if Nth -> a, then we write pN = a.
To begin with, observe that by combining ⛔「⟨ ⁵ᵗʰb ²ⁿᵈa ⟩, a > b」 with ⛔「⟨ ⁰ᵗʰ↓ ⟩ after 2×⟨→⟩」, we have the following required pattern:
(1) p5 > p2 > p0.
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ► │ │ │ ► │ │ ► │
└───┴───┴───┴───┴───┴───┴───┘
For later reference, we denote the set {p5, p2, p0} by S.
On the other hand, let T := {p6, p3, p1}. By ✅「{p6, p3, p1} = ? + {0,3,4}」, one of the following holds:
(i) T = {0,3,4};
(ii) T = {1,4,5};
(iii) T = {2,5,6}.
(2) We claim that both case (i) and (ii) give contradictions. Therefore, case (iii) holds actually.
------------------------------
(2.1) Suppose case (i) holds:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ / │ ► │ │ / │ ► │ / │ ► │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
where "/" are occupied by 0,3,4. Then, T would provide at most one agreed positional digit for ✅「Sim⟨ ⁶ᵗʰ4 ⁵ᵗʰ2 ⁴ᵗʰ3 ³ʳᵈ5 ²ⁿᵈ0 ¹ˢᵗ1 ⁰ᵗʰ6 ⟩ ≥ 2」. We need to match (1) as well, so S would not provide any agreed positional digit. This implies that we need p4 to be an agreed positional digit in order to match the preceding pattern:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ / │ ► │ 3 │ / │ ► │ / │ ► │
└───┴───┴───┴───┴───┴───┴───┘
But this is a contradiction because 3 ∈ T by our initial assumption.
------------------------------
(2.2) Else, suppose case (ii) holds:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ / │ ► │ │ / │ ► │ / │ ► │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ 3 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
where "/" are occupied by 1,4,5. Let us consider where to place 3. By (1) and ✅「Jump(3,6) ≥ 2」, we have 3 = p2, and 6 = p5 follows:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ / │ 6 │ │ / │ 3 │ / │ ► │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
As a consequence, S and { p4 } provide no agreed positional digit for ✅「Sim⟨ ⁶ᵗʰ4 ⁵ᵗʰ2 ⁴ᵗʰ3 ³ʳᵈ5 ²ⁿᵈ0 ¹ˢᵗ1 ⁰ᵗʰ6 ⟩ ≥ 2」. So, T has to provide at least two agreed positional digits. It implies that
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 4 │ 6 │ │ 5 │ 3 │ 1 │ ► │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
In view of ✅「⟦1,3⟧ ∋ 5」, this is a contradiction.
------------------------------
We have verified our claim in (2). As a result, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ / │ ► │ │ / │ ► │ / │ ► │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 3 │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
where "/" are occupied by 2,5,6. Now, T would provide at most one agreed positional digit for ✅「Sim⟨ ⁶ᵗʰ4 ⁵ᵗʰ2 ⁴ᵗʰ3 ³ʳᵈ5 ²ⁿᵈ0 ¹ˢᵗ1 ⁰ᵗʰ6 ⟩ ≥ 2」. while by (1), S would not provide any agreed positional digit. Therefore, we need p4 to be an agreed positional digit, and T provide one agreed positional digit. This gives our first and second steps:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ 3 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ │ 3 │ 5 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ 6 │ │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Next, to match T = {2,5,6} and ✅「Jump(3,6) ≥ 2」, we get
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ 3 │ 5 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ │ 3 │ 5 │ │ 6 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 2 │ │ 3 │ 5 │ │ 6 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, in view of (1), we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│ 2■│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ │ 3 │ 5 │ │ 6 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 4 │ 3 │ 5 │ │ 6 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 4 │ 3 │ 5 │ 1 │ 6 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 4 │ 3 │ 5 │ 1 │ 6 │ 0 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.6