Step 1 — Apply Clue (2)

Pump	P1	P2	P3	P4	P5	P6	P7	P8	P9	P10	P11	P12	P13	P14	P15	P16	P17	P18	P19	P20
Value	H/M	H/M	H/M	H/M	H/M	L	H/M	H/M	H/M	H/M	?	?	?	?	?	?	?	?	?	?

P6 is the only Low among P1–P10.

So P6 = L and every other in P1–P10 is in {H, M}.

Step 2 — Apply Clues (1) and (3) on P1–P5

Pump	P1	P2	P3	P4	P5	P6
Value	H	M	H	M	H	L

• Clue (1): Among P1–P5, exactly three H.
• Clue (3): The only equal consecutive pair in the entire row is (P7, P8).

⇒ Therefore, no equal neighbors among P1–P5; they must alternate H/M.

The only alternating length-5 pattern with exactly three H is H M H M H.

(So far fixed: P1=H, P2=M, P3=H, P4=M, P5=H, P6=L.)

Step 3 — Pin the special pair (Clue 3)

Case	P1	P2	P3	P4	P5	P6	P7	P8
A	H	M	H	M	H	L	H	H
B	H	M	H	M	H	L	M	M

Clue (3): P7 and P8 are the only consecutive equal pumps.

So P7 = P8 ∈ {H, M} and no other neighbors anywhere can be equal.

We consider two cases:

Case A: P7=P8=H

Case B: P7=P8=M

Step 4 — Propagate to P9–P10 (neighbors must differ)

Case	P7	P8	P9	P10
A	H	H	M	H
B	M	M	H	M

Because only (P7,P8) can match, every other adjacent pair must differ.

• If P7=P8=H ⇒ P8≠P9 ⇒ P9=M ⇒ P10≠P9 ⇒ P10=H.
• If P7=P8=M ⇒ P9=H ⇒ P10=M.

Combine with Steps 1–2:

• Case A (P1–P10): H M H M H L H H M H
• Case B (P1–P10): H M H M H L M M H M

Step 5 — Eliminate Case B (contradiction with totals/placement)

We now use:

• Clue (4): No H at P16–P20.
• Clue (5): #H = 2 × #L overall.

Count in Case B after P1–P10

• H = 4 (P1,3,5,9), M = 5 (P2,4,7,8,10), L = 1 (P6).

Let total L = k ⇒ total H = 2k and total M = 20 − 3k.

• If k=4 ⇒ total H=8 ⇒ remaining H (P11–P20) = 4.

But P16–P20 can’t be H, so all 4 H must fit in P11–P15 with no equal neighbors and also P10= M ⇒ P11 ≠ M. It’s impossible to place 4 non-adjacent H into 5 slots while also respecting the boundary with P10—contradiction.

• If k=3 ⇒ total H=6 ⇒ remaining H = 2 and remaining L = 2.

Since P16–P20 can’t be H, we must place the two H in P11–P15. To avoid any equal neighbors from P10 onward (P10=M), P11 cannot be M, but with only H and M available in P11–P15 (because both L must fall in P16–P20 to hit L total), you’re forced into a same-level clash (two M’s consecutively) or violate counts—contradiction.

Therefore Case B is impossible.

Hence P7=P8=H is forced.

Step 6 — Lock P1–P10 (the front half is unique)

Pump	P1	P2	P3	P4	P5	P6	P7	P8	P9	P10
Value	H	M	H	M	H	L	H	H	M	H

Counts so far: H=6, M=3, L=1.

Step 7 — Work out remaining totals (Clues 4 & 5)

Let total L = k. Clue (5): total H = 2k. Since we already have H=6 and L=1:

• P16–P20 have no H (Clue 4), so any remaining H must be in P11–P15.
• P10=H, so P11 ≠ H (can’t create another equal neighbor).
• In P11–P15, the H’s must be non-adjacent (still only one equal pair allowed overall, and that’s P7–P8).

The only value of k that fits all limits is k = 4, giving totals:

• L = 4, H = 8, M = 8.

So remaining to place across P11–P20:

• H left = 8 − 6 = 2 (both must be in P12–P15),
• L left = 4 − 1 = 3,
• M left = 8 − 3 = 5.

Step 8 — Where can the last two H go?

Because P11 ≠ H and H’s can’t be adjacent, the two H in P11–P15 must be a non-adjacent pair chosen from {P12, P13, P14, P15}. The only valid location sets are:

• {P12, P14}
• {P12, P15}
• {P13, P15}

(That’s 3 patterns.)

Step 9 — Finish each pattern (no H in P16–P20; no other equal neighbors)

For each H-pattern, fill the rest with M/L so that:

• there are exactly three L total in P11–P20,
• no equal neighbors appear anywhere (remember only P7–P8 match),
• P16–P20 contain no H.

Each H-pattern yields two valid M/L alternation completions ⇒ 3 × 2 = 6 total solutions.

Final 6 Solutions (full table)

#	P1	P2	P3	P4	P5	P6	P7	P8	P9	P10	P11	P12	P13	P14	P15	P16	P17	P18	P19	P20	H-slots among P12–P15
1	H	M	H	M	H	L	H	H	M	H	M	H	M	H	M	L	M	L	M	L	{P12,P14}
2	H	M	H	M	H	L	H	H	M	H	M	H	M	H	L	M	L	M	L	M	{P12,P14}
3	H	M	H	M	H	L	H	H	M	H	M	H	M	L	H	M	L	M	L	M	{P12,P15}
4	H	M	H	M	H	L	H	H	M	H	M	H	L	M	H	M	L	M	L	M	{P12,P15}
5	H	M	H	M	H	L	H	H	M	H	M	L	H	M	H	M	L	M	L	M	{P13,P15}
6	H	M	H	M	H	L	H	H	M	H	L	M	H	M	H	M	L	M	L	M	{P13,P15}

✅ Checks (true for every row):

• P1–P5 have exactly three H (Clue 1).
• Only P6 is Low among P1–P10 (Clue 2).
• The only equal consecutive pair is (P7, P8) (Clue 3).
• P16–P20 contain no H (Clue 4).
• Totals are H=8, M=8, L=4 (Clue 5).

Exactly 8.

Brief why:

• Let total L = $k$. By (5), total H $=2k$, so total M $=20-3k$.
• From the forced first half (by clues 1–3):

$P1\!-\!P10 = \text{H M H M H L H H M H}$ ⇒ so far H=6, M=3, L=1.

• Then in $P11\!-\!P20$:

H left $=2k-6$, M left $=17-3k$, L left $=k-1$.

• By (4) no H in $P16\!-\!P20$, and by (3) no equal consecutive levels anywhere except $P7,P8$.

Because $P10=$H, at most 2 H can be placed non-adjacently in $P11\!-\!P15$.

Hence $2k-6 \le 2 \Rightarrow k \le 4$, and since $k\ge 3$ (to have $2k-6\ge 0$), try $k=3$ and $k=4$.

• $k=3$ ⇒ no H left; $P11\!-\!P20$ must use only M and L with no equal neighbors, forcing strict alternation → would give 5 L (not 2). Impossible.
• Thus $k=4$ is the only feasible total: H $=8$, L $=4$, so M $=20-8-4=8$.