Rearrange the digits in ⟨1263045⟩ to meet the rules below.
⟨6th 5th 4th 3rd 2nd 1st 0th⟩
✅Match
⟨ ⁶ᵗʰa ⁵ᵗʰb ²ⁿᵈc ⟩, (abc)₁₀ ≤ 203
Sim⟨ ⁶ᵗʰ2 ⁵ᵗʰ4 ⁴ᵗʰ6 ³ʳᵈ5 ²ⁿᵈ0 ¹ˢᵗ1 ⁰ᵗʰ3 ⟩ ≥ 2
Jump(3,5) = 2
⟨ ⁶ᵗʰa ⁵ᵗʰb ²ⁿᵈc ⟩, (abc)₁₀ ≥ 132
⛔Avoid
⟨ ³ʳᵈa ¹ˢᵗb ⟩, (ab)₁₀ ≥ 24
#125034_v2.3
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 1 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │ │ │ 2 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ │ 5 │ 2 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ │ 5 │ 2 │ │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 4 │ 5 │ 2 │ │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 4 │ 5 │ 2 │ 0 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 4 │ 5 │ 2 │ 0 │ 3 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
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Proof of 2024-02-20 Q1(m=6)
═══════════════════════════
Notation: if nth -> a, then we write [nth] = a.
By ✅「⟨ ⁶ᵗʰa ⁵ᵗʰb ²ⁿᵈc ⟩, (abc)₁₀ ≤ 203」 and ✅「⟨ ⁶ᵗʰa ⁵ᵗʰb ²ⁿᵈc ⟩, (abc)₁₀ ≥ 132」, we see that
(1) [6th] = 1|2.
(1.1) We claim that [6th] = 1 actually.
------------------------------
If on the contrary [6th] = 2, then by ✅「⟨ ⁶ᵗʰa ⁵ᵗʰb ²ⁿᵈc ⟩, (abc)₁₀ ≤ 203」, we have [5th] = 0 and X := [2nd] = 1|3:
┌───┬───┬───┬───┬───┬───┬───┐
│ 6▲│ 5▲│4th│3rd│*2 │1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 0 │ │ │ X │ │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ 1 │ │ 6 │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
In view of ✅「Jump(3,5) = 2」, we have X != 3. Therefore, X = 1:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│ 2▲│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 2 │ 0 │ │ │ 1 │ │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ 3 │ │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
But then we would match ⛔「⟨ ³ʳᵈa ¹ˢᵗb ⟩, (ab)₁₀ ≥ 24」, which is a contradiction.
------------------------------
Our claim in (1.1) is thus verified. Accordingly, we have
┌───┬───┬───┬───┬───┬───┬───┐
│ 6■│5th│4th│3rd│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
Step 1 │ 1 │ │ │ │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
Next, note that to avoid ⛔「⟨ ³ʳᵈa ¹ˢᵗb ⟩, (ab)₁₀ ≥ 24」, we need
(2) [3rd] = 0|2.
Combining this with ✅「Jump(3,5) = 2」, we have
(3) 3 != [0th],
for otherwise [3rd] would be 5, contradicting (2):
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│1st│ 0▲│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ │ │ 5 │ │ │ 3 │
└───┴───┴───┴───┴───┴───┴───┘
As a result, to match ✅「Sim⟨ ⁶ᵗʰ2 ⁵ᵗʰ4 ⁴ᵗʰ6 ³ʳᵈ5 ²ⁿᵈ0 ¹ˢᵗ1 ⁰ᵗʰ3 ⟩ ≥ 2」, [6th] = 1 (step 1), (2), and (3), we need
(4) at least two of the following hold:
(i) [5th] = 4;
(ii) [4th] = 6;
(iii) [2nd] = 0.
(5) We are ready to show that [3rd] = 2.
If this is not the case, then by (2), we have [3rd] = 0:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3▲│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ │ │ 0 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Then, to match (4), we need 4(i) and 4(ii) hold:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5▲│ 4▲│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
│ 1 │ 4 │ 6 │ 0 │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ 2 │ │ 3 │ │ │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
However, it follows that we could not match ✅「Jump(3,5) = 2」, which is a contradiction.
Our claim in (5) is verified. Thus, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│ 3■│2nd│1st│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ │ │ │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 2 │ 1 │ │ │ 2 │ │ │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ 3 │ 0 │ 4 │ 5 │
└───┴───┴───┴───┴───┴───┴───┘
We proceed by considering how to match ✅「Jump(3,5) = 2」. There are two ways to do so:
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│4th│3rd│2nd│1st│0th│
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡
(6) │ 1 │ * │ │ 2 │ * │ │ │
├───┼───┼───┼───┼───┼───┼───┤
(7) │ 1 │ │ * │ 2 │ │ * │ │
└───┴───┴───┴───┴───┴───┴───┘
Actually (6) is not possible, as we need to match (4) as well. Therefore, (7) holds and we have
{ [4th], [1st] } = {3,5}.
Since ⛔「⟨ ³ʳᵈa ¹ˢᵗb ⟩, (ab)₁₀ ≥ 24」 forbids [1st] = 5, we get
┌───┬───┬───┬───┬───┬───┬───┐
│6th│5th│ 4■│3rd│2nd│ 1■│0th│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ │ │ 2 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 3 │ 1 │ │ 5 │ 2 │ │ │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 4 │ 1 │ │ 5 │ 2 │ │ 3 │ │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ 6 │ │ 0 │ 4 │ │
└───┴───┴───┴───┴───┴───┴───┘
Finally, by applying (4), we finish by
┌───┬───┬───┬───┬───┬───┬───┐
│6th│ 5■│4th│3rd│ 2■│1st│ 0■│▒
╞═══╪═══╪═══╪═══╪═══╪═══╪═══╡▒
│ 1 │ │ 5 │ 2 │ │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 5 │ 1 │ 4 │ 5 │ 2 │ │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 6 │ 1 │ 4 │ 5 │ 2 │ 0 │ 3 │ │▒
├───┼───┼───┼───┼───┼───┼───┤▒
Step 7 │ 1 │ 4 │ 5 │ 2 │ 0 │ 3 │ 6 │▒
└───┴───┴───┴───┴───┴───┴───┘▒
▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒▒
--- Idle ---
┌───┬───┬───┬───┬───┬───┬───┐
│ │ │ │ │ │ │ │
└───┴───┴───┴───┴───┴───┴───┘
Q.E.D.
#125034_v2.3