Rearrange the digits in ⟨125034⟩ to meet the rules below.
⟨5th 4th 3rd 2nd 1st 0th⟩
✅Match
median {p5, p4, p0} = 2
{p4, p3, p2} = ? + {0,2,3}
Sim⟨ ⁵ᵗʰ3 ⁴ᵗʰ5 ³ʳᵈ2 ²ⁿᵈ4 ¹ˢᵗ1 ⁰ᵗʰ0 ⟩ ≥ 2
⛔Avoid
min ⊢4⊣ = 1
max {p5, p2, p1} = 5
----- Information -----
🔲 「median {p5, p4, p0} = 2」
The median of { [5th], [4th], [0th] } is 2.
There are 216 permutations matching this pattern.
Examples: ⟨523140⟩, ⟨420531⟩, ⟨410352⟩.
🔲 「{p4, p3, p2} = ? + {0,2,3}」
{ [4th], [3rd], [2nd] } = { ?, ?+2, ?+3 }.
There are 108 permutations matching this pattern.
Examples: ⟨354210⟩, ⟨052431⟩, ⟨213405⟩.
🔲 「Sim⟨ ⁵ᵗʰ3 ⁴ᵗʰ5 ³ʳᵈ2 ²ⁿᵈ4 ¹ˢᵗ1 ⁰ᵗʰ0 ⟩ ≥ 2」
⟨pᵢ⟩ matches this pattern if and only if #{ i: pᵢ = qᵢ } ≥ 2, where ⟨qᵢ⟩ ∶= ⟨352410⟩.
There are 191 permutations matching this pattern.
Examples: ⟨054213⟩, ⟨350241⟩, ⟨243510⟩.
🔳 「min ⊢4⊣ = 1」
The least element of ⊢4⊣ is 1. Also written as B(x=4, r=1), ⊢4⊣ denotes the neighborhood of 4 with radius one.
There are 528 permutations avoiding this pattern.
Examples: ⟨452130⟩, ⟨325401⟩, ⟨105432⟩.
🔳 「max {p5, p2, p1} = 5」
The greatest element of { [5th], [2nd], [1st] } is 5.
There are 360 permutations avoiding this pattern.
Examples: ⟨025134⟩, ⟨153402⟩, ⟨140235⟩.
#125034_v2.7
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 2 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 2 │ 5 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 2 │ 5 │ 4 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 2 │ 5 │ 4 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 2 │ 5 │ 4 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 2 │ 5 │ 4 │ 0 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
Proof of 2024-10-11 Q3
══════════════════════
By ✅「median {p5, p4, p0} = 2」, we have
(1) 2 ∈ {p5, p4, p0}.
On the other hand, by ✅「{p4, p3, p2} = ? + {0,2,3}」, there are three possibilities for {p4, p3, p2}:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ ▬ │ ▬ │ ▬ │ │ │
└───┴───┴───┴───┴───┴───┘
(A) {0,2,3};
(B) {1,3,4};
(C) {2,4,5}.
(2) We show that case (C) holds actually.
------------------------------
(2.1) If case (A) holds, then by (1), we have 2 = p4:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │ 2 │0 3│0 3│ │ │
└───┴───┴───┴───┴───┴───┘
Comparing with ⟨352410⟩, there will be at most one agreed positional digit (at 1st). So, we cannot match ✅「Sim⟨ ⁵ᵗʰ3 ⁴ᵗʰ5 ³ʳᵈ2 ²ⁿᵈ4 ¹ˢᵗ1 ⁰ᵗʰ0 ⟩ ≥ 2」, which is a contradiction.
(2.2) Else, suppose case (B) holds:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │134│134│134│ │ │
└───┴───┴───┴───┴───┴───┘
Let us consider where to place 5. As it is the maximum digit, we need 5 ∉ {p5, p2, p1} in order to avoid ⛔「max {p5, p2, p1} = 5」. It implies 5 = p0:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ │134│134│134│ │ 5 │
└───┴───┴───┴───┴───┴───┘
Then, using (1), we get 2 = p5:
┌───┬───┬───┬───┬───┬───┐
│ 5▲│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │134│134│134│ │ 5 │
└───┴───┴───┴───┴───┴───┘
and p1 = 0 follows:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1▲│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │134│134│134│ 0 │ 5 │
└───┴───┴───┴───┴───┴───┘
Again we cannot match ✅「Sim⟨ ⁵ᵗʰ3 ⁴ᵗʰ5 ³ʳᵈ2 ²ⁿᵈ4 ¹ˢᵗ1 ⁰ᵗʰ0 ⟩ ≥ 2」, which is a contradiction.
------------------------------
We have verified (2). Now
(3) {p4, p3, p2} = {2,4,5}.
By (1), we get 2 = p4, which is our first step:
┌───┬───┬───┬───┬───┬───┐
│5th│ 4■│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ │ 2 │4 5│4 5│ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
As 5 != p2 by ⛔「max {p5, p2, p1} = 5」, we get 5 = p3 and 4 = p2 as well:
┌───┬───┬───┬───┬───┬───┐
│5th│4th│ 3■│ 2■│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 2 │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 2 │ │ 2 │ 5 │ │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 3 │ │ 2 │ 5 │ 4 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ 3 │ │
└───┴───┴───┴───┴───┴───┘
Next, note that to match ✅「Sim⟨ ⁵ᵗʰ3 ⁴ᵗʰ5 ³ʳᵈ2 ²ⁿᵈ4 ¹ˢᵗ1 ⁰ᵗʰ0 ⟩ ≥ 2」, we need one more agreed positional digit. Since p1 != 1 by ⛔「min ⊢4⊣ = 1」, we need
(4) 3 = p5 or 0 = p0.
On the other hand, to match ✅「median {p5, p4, p0} = 2」, we need a digit greater than 2 in {p5, p0}. The only idle digit satisfying this is 3, so
(5) 3 = p5|p0.
It follows that 3 = p5, for otherwise 3 = p0 and we cannot match (4). Accordingly, we get
┌───┬───┬───┬───┬───┬───┐
│ 5■│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ │ 2 │ 5 │ 4 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 4 │ 3 │ 2 │ 5 │ 4 │ │ │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ 1 │ │ │ 0 │ │ │
└───┴───┴───┴───┴───┴───┘
Finally, to avoid ⛔「min ⊢4⊣ = 1」, we finish by
┌───┬───┬───┬───┬───┬───┐
│5th│4th│3rd│2nd│ 1■│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 3 │ 2 │ 5 │ 4 │ │ │▒
├───┼───┼───┼───┼───┼───┤▒
Step 5 │ 3 │ 2 │ 5 │ 4 │ │ 1 │▒
├───┼───┼───┼───┼───┼───┤▒
Step 6 │ 3 │ 2 │ 5 │ 4 │ 0 │ 1 │▒
└───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.7