Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
Sim⟨ ⁶ᵗʰ5 ⁵ᵗʰ0 ⁴ᵗʰ6 ³ʳᵈ1 ²ⁿᵈ2 ¹ˢᵗ3 ⁰ᵗʰ4 ⟩ ≥ 2
⟨ ⁶ᵗʰc ²ⁿᵈd ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c > d
3rd → a, 1st → b, |a-b|=1
⛔Avoid
6th|4th|3rd|1st → 1
⟨ ³ʳᵈ↓ ¹ˢᵗ↑ ⁰ᵗʰ↓ ⟩ after 2×⟨→⟩
#125034_v2.4
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 0 │ │ │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 0 │ │ │ 1 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 0 │ │ 3 │ 1 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 0 │ │ 3 │ 1 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 0 │ 6 │ 3 │ 1 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 0 │ 6 │ 3 │ 1 │ 4 │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-03-12 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
We begin with an observation: by ✅「⟨ ⁶ᵗʰc ²ⁿᵈd ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c > d」, we have "d" = [2nd] < [0th] = "a". It means that 0th will decrease after 2×⟨→⟩. Therefore, the forbidden pattern ⛔「⟨ ³ʳᵈ↓ ¹ˢᵗ↑ ⁰ᵗʰ↓ ⟩ after 2×⟨→⟩」 can be relaxed to:
(1) ⛔「⟨ ³ʳᵈ↓ ¹ˢᵗ↑ ⟩ after 2×⟨→⟩」.
We consider where to place 1. By ⛔「6th|4th|3rd|1st → 1」, we have
(2) 1 = [5th] | [2nd] | [0th].
To match ✅「⟨ ⁶ᵗʰc ²ⁿᵈd ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c > d」, we cannot have "a"=1, so 1 != [0th].
(3) We claim that 1 = [2nd] actually.
------------------------------
For, suppose on the contrary 1 = [5th]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Note that [3rd] != 0, for otherwise we cannot match ✅「3rd → a, 1st → b, |a-b|=1」. Therefore [3rd] > 1, and so 3rd ↓ after 2×⟨→⟩. It follows that to avoid (1), we need to have 1st ↓ after 2×⟨→⟩, or equivalently [3rd] < [1st]. Combining this with ✅「3rd → a, 1st → b, |a-b|=1」, we have
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│*3 │2nd│*1 │0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ │ X │ │ Y │ │
└───┴───┴───┴───┴───┴───┴───┘
where Y = X+1 and X ∈ {2,3,4,5}. There are four cases:
● (X,Y) = (2,3)
● (X,Y) = (3,4)
● (X,Y) = (4,5)
● (X,Y) = (5,6)
(4) We proceed to verify that all of them give contradictions. Doing so will establish our claim in (3).
------------------------------
(4.1) If (X,Y) = (2,3):
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ │ 2 │ │ 3 │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
then we cannot find "c" and "d" to match ✅「⟨ ⁶ᵗʰc ²ⁿᵈd ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c > d」.
(4.2) Else if (X,Y) = (3,4):
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ │ 3 │ │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ │ 0 │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
then there is only one choice of "c" and "d" of ✅「⟨ ⁶ᵗʰc ²ⁿᵈd ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c > d」:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 1 │ │ 3 │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
but then we cannot match ✅「Sim⟨ ⁶ᵗʰ5 ⁵ᵗʰ0 ⁴ᵗʰ6 ³ʳᵈ1 ²ⁿᵈ2 ¹ˢᵗ3 ⁰ᵗʰ4 ⟩ ≥ 2」.
(4.3) Else if (X,Y) = (4,5):
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ │ 4 │ │ 5 │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ 3 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
then by ✅「⟨ ⁶ᵗʰc ²ⁿᵈd ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c > d」, we have "a" = 6:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ │ 4 │ │ 5 │ 6 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ 3 │ 0 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
but then we cannot match ✅「Sim⟨ ⁶ᵗʰ5 ⁵ᵗʰ0 ⁴ᵗʰ6 ³ʳᵈ1 ²ⁿᵈ2 ¹ˢᵗ3 ⁰ᵗʰ4 ⟩ ≥ 2」.
(4.4) Else if (X,Y) = (5,6):
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ │ 5 │ │ 6 │ │
└───┴───┴───┴───┴───┴───┴───┘
then we have no "a" to match ✅「⟨ ⁶ᵗʰc ²ⁿᵈd ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c > d」.
------------------------------
We have verified (4). As a result, we get our first step:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ │ │ │ 1 │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Now, let us consider where to place 0. It cannot be the "a", "b", or "c" of ✅「⟨ ⁶ᵗʰc ²ⁿᵈd ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c > d」:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ / │ │ │ │ 1 │ / │ / │
└───┴───┴───┴───┴───┴───┴───┘
In view of ✅「3rd → a, 1st → b, |a-b|=1」, it is not at 3rd as well:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ / │ │ │ / │ 1 │ / │ / │
└───┴───┴───┴───┴───┴───┴───┘
Hence, 0 = [5th] | [4th].
(5) We claim that 0 = [5th] actually.
------------------------------
Suppose on the contrary 0 = [4th]:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 0 │ │ 1 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Noting that [6th] != 5 by ✅「⟨ ⁶ᵗʰc ²ⁿᵈd ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c > d」, we see that to match ✅「Sim⟨ ⁶ᵗʰ5 ⁵ᵗʰ0 ⁴ᵗʰ6 ³ʳᵈ1 ²ⁿᵈ2 ¹ˢᵗ3 ⁰ᵗʰ4 ⟩ ≥ 2」, we need [1st] = 3 and [0th] = 4:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│ 1▲│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 0 │ │ 1 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Then, ✅「3rd → a, 1st → b, |a-b|=1」 implies [3rd] = 2:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ │ 0 │ 2 │ 1 │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
But now we cannot match ✅「⟨ ⁶ᵗʰc ²ⁿᵈd ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c > d」, which shows a contradiction.
------------------------------
We have verified our claim in (5). Accordingly, we get:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ │ │ │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 0 │ │ │ 1 │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Having [5th] = 0 means that 3rd↓ after 2×⟨→⟩. So, the forbidden pattern (1) can be relaxed to:
(6) ⛔「⟨ ¹ˢᵗ↑ ⟩ after 2×⟨→⟩」.
Combining this with ✅「3rd → a, 1st → b, |a-b|=1」, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│*3 │2nd│*1 │0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ │ X │ 1 │ Y │ │
└───┴───┴───┴───┴───┴───┴───┘
where Y = X+1 and X ∈ {2,3,4,5}. The four possibilities are:
● (X,Y) = (2,3)
● (X,Y) = (3,4)
● (X,Y) = (4,5)
● (X,Y) = (5,6)
(7) We claim that (X,Y) = (3,4) actually.
------------------------------
(7.1) If (X,Y) = (2,3):
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ │ 2 │ 1 │ 3 │ │
└───┴───┴───┴───┴───┴───┴───┘
then we cannot find "c" to match ✅「⟨ ⁶ᵗʰc ²ⁿᵈd ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c > d」.
(7.2) Else if (X,Y) = (4,5):
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ │ 4 │ 1 │ 5 │ │
└───┴───┴───┴───┴───┴───┴───┘
then by ✅「⟨ ⁶ᵗʰc ²ⁿᵈd ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c > d」 we have [0th] = 6:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ │ 4 │ 1 │ 5 │ 6 │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ 3 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
But then we cannot match ✅「Sim⟨ ⁶ᵗʰ5 ⁵ᵗʰ0 ⁴ᵗʰ6 ³ʳᵈ1 ²ⁿᵈ2 ¹ˢᵗ3 ⁰ᵗʰ4 ⟩ ≥ 2」, which is a contradiction.
(7.3) Else if (X,Y) = (5,6):
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 0 │ │ 5 │ 1 │ 6 │ │
└───┴───┴───┴───┴───┴───┴───┘
then we have no "a" to match ✅「⟨ ⁶ᵗʰc ²ⁿᵈd ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c > d」.
------------------------------
We have verified our claim in (7). Accordingly
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 0 │ │ │ 1 │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 0 │ │ │ 1 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ │ 0 │ │ 3 │ 1 │ 4 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
To match ✅「⟨ ⁶ᵗʰc ²ⁿᵈd ¹ˢᵗb ⁰ᵗʰa ⟩, a > b > c > d」, we need "c" = [6th] < [1st] = 4. The only possible choice is [6th] = 2. So, we get
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 0 │ │ 3 │ 1 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 2 │ 0 │ │ 3 │ 1 │ 4 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Finally, to match ✅「Sim⟨ ⁶ᵗʰ5 ⁵ᵗʰ0 ⁴ᵗʰ6 ³ʳᵈ1 ²ⁿᵈ2 ¹ˢᵗ3 ⁰ᵗʰ4 ⟩ ≥ 2」, we need [4th] = 6. We finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 2 │ 0 │ │ 3 │ 1 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 2 │ 0 │ 6 │ 3 │ 1 │ 4 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 2 │ 0 │ 6 │ 3 │ 1 │ 4 │ 5 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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--- Idle ---
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Q.E.D.
#125034_v2.4
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