Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟦3,5⟧ ∋ 0,2
4th → a, 1st → b, a+b=5
⟨⋯ Perm(2,5) ⋯⟩
⛔Avoid
⟨ ⁴ᵗʰa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≥ 152
⟦0,4⟧ ∋ 1,2
----- Information -----
🔲 「⟦3,5⟧ ∋ 0,2」
The closed interval given by 3 and 5 contains 0, 2.
120 permutations match this pattern.
Examples: ⟨340215⟩, ⟨302514⟩, ⟨132405⟩.
🔲 「4th → a, 1st → b, a+b=5」
144 permutations match this pattern.
Examples: ⟨420135⟩, ⟨251403⟩, ⟨134520⟩.
🔲 「⟨⋯ Perm(2,5) ⋯⟩」
2,5 are adjacent.
240 permutations match this pattern.
Examples: ⟨014325⟩, ⟨025143⟩, ⟨345210⟩.
🔳 「⟨ ⁴ᵗʰa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≥ 152」
(abc)₁₀ ∶= 100a + 10b + c.
498 permutations match this pattern.
Examples: ⟨345012⟩, ⟨531024⟩, ⟨520314⟩.
🔳 「⟦0,4⟧ ∋ 1,2」
The closed interval given by 0 and 4 contains 1, 2.
120 permutations match this pattern.
Examples: ⟨052134⟩, ⟨425103⟩, ⟨453120⟩.
#125034_v2.2
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 0 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 0 │ │ │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ 0 │ │ │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 0 │ │ 2 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 0 │ 4 │ 2 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 0 │ 4 │ 2 │ 5 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2023-11-28 WR
══════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ⛔「⟨ ⁴ᵗʰa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≥ 152」, we have [4th] = 1 or 0. Together with ✅「4th → a, 1st → b, a+b=5」, there are only two possibilities:
(i) [4th] = 1, [1st] = 4;
(ii) [4th] = 0, [1st] = 5.
(1) We show that (i) is not possible. Suppose on the contrary (i) holds:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ │ │ 4 │ │
└───┴───┴───┴───┴───┴───┘
By ✅「⟨⋯ Perm(2,5) ⋯⟩」, we see that
┌───┬───┬───┬───┬───┬───┐
│5th│4th│*3 │*2 │1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ x │ y │ 4 │ │
└───┴───┴───┴───┴───┴───┘
where {x,y} = {2,5}. To avoid ⛔「⟨ ⁴ᵗʰa ²ⁿᵈb ¹ˢᵗc ⟩, (abc)₁₀ ≥ 152」, we need (x,y) = (5,2):
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3▲│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 1 │ 5 │ 2 │ 4 │ │
└───┴───┴───┴───┴───┴───┘
But then ✅「⟦3,5⟧ ∋ 0,2」 could not be matched, which is a contradiction.
------------------------------
We have verified our claim in (1). Accordingly, we obtain:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 0 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 0 │ │ │ 5 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ 3 │ 4 │
└───┴───┴───┴───┴───┴───┘
Now, to match ✅「⟦3,5⟧ ∋ 0,2」, we need 0 to be between 5 and 3. Thus
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 0 │ │ │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ 3 │ 0 │ │ │ 5 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ 2 │ │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Next, we consider where to place 2. By ✅「⟨⋯ Perm(2,5) ⋯⟩」, 2,5 are adjacent, and by ✅「⟦3,5⟧ ∋ 0,2」, 2 is between 5 and 3. Therefore, we have
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 0 │ │ │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 0 │ │ 2 │ 5 │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ │ │ 4 │
└───┴───┴───┴───┴───┴───┘
Finally, note that there is only one way to avoid ⛔「⟦0,4⟧ ∋ 1,2」. We reach
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│2nd│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 0 │ │ 2 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 0 │ 4 │ 2 │ 5 │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 0 │ 4 │ 2 │ 5 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.2