Water Sort Solver Guide: Par Table, Spare Math and the Stuck Scan (2026)

Last updated: October 2026 · Tested on Playnesta's Water Sort Puzzle (Playnesta-original, MIT: js/main.js, js/levels.js, stylesheet, analytics bridge and touch shim) · 11 levels · capacity 4 units per glass · par per level 6 / 9 / 9 / 12 / 9 / 14 / 14 / 14 · 1 spare glass (easy/medium) · 2 spare glasses (hard) · Undo / Restart / prev-next

Method: every number below is a constant in this site's shipped rules module js/main.js and js/levels.js (Playnesta-original, MIT), read and verified this session — not quoted from a third party. Verified constants: capacity is 4 units per glass; a legal pour src→dst requires src non-empty, dst with free space, and dst empty or top(dst) == top(src); a pour moves the whole top run of the source, capped by free space in the target (min of the two); a level is solved when every glass is empty or full (4) and single-colour; the library holds 11 levels with frozen tube layouts and min_steps (par) 6, 6, 6, 6, 9, 9, 12, 9, 14, 14, 14 — each BFS-verified offline as solvable and non-trivial (min 3 pours) before release. This guide is the solver layer: how the minimum is computed and how to use it. The game page covers operation; the companion Water Sort strategy guide covers playing technique — read that first if pours, spare discipline and top-run reading are new to you.

1. The par table: what the solver proved

Par is not a guess. Each level's tube layout is frozen in js/levels.js, and an offline breadth-first search explored every legal pour sequence from that start until the solve condition held. The first time the search reaches solved, that depth is the minimum — no shorter route exists. The shipped par values are those minima:

Easy (3 colours, 1 spare): level 1 par 6 · level 2 par 6 · level 3 par 6 · level 4 par 6
Medium (4 colours, 1 spare): level 5 par 9 · level 6 par 9 · level 7 par 12 · level 8 par 9
Hard (5 colours, 2 spares): level 9 par 14 · level 10 par 14 · level 11 par 14

Level 7 is the medium outlier (12 versus 9 around it) — expect one longer forced chain there. Hard levels are uniformly 14: the second spare does not shorten the optimum, it only makes the search survivable. Use the table as a diagnostic: if your move counter is already past par with no glass close to done, you have taken a branch the solver would never take — undo to the last point where the spare was free.

2. Spare-glass math: why one buffer is enough until hard

A pour needs a destination, so the number of spare glasses caps how many colours you can hold "in flight" at once. Count the glasses: levels 1–8 have 4 tubes plus one empty ([]) — 1 spare; levels 9–11 have 5 tubes plus two empties — 2 spares. The rule of thumb falls out of the layouts:

With 1 spare and 3 colours (easy), exactly one colour can be parked while the other two settle. Fill the spare with a second colour and the position is dead — there is nowhere left to split a run.
With 1 spare and 4 colours (medium), the same spare must serve three unsettled colours in rotation. That is why medium punishes an early wrong pour: the buffer has no slack for a second mistake.
With 2 spares and 5 colours (hard), one spare is the working glass and the other stays the true buffer. If both spares hold mixed colours at once, you are back to the medium trap with one extra colour — restart is usually cheaper than digging out.

Never start a level by pouring into the spare "to see what happens". The spare's job is to stay empty until a planned sequence needs it, then to be emptied again immediately.

3. The stuck scan: find the forced glass

When no pour looks good, run this scan in order — it mirrors what the solver checks at every node:

A glass whose top run exactly fills a same-colour glass. That pour is never wrong: it completes or extends a home glass without consuming the spare.
A glass whose top colour appears nowhere else on top. Its run has exactly one legal destination family — pour it there before that destination gets claimed.
A full mixed glass (4 units, more than one colour). Nothing can land on it until its top run leaves, so its top run is forced to move first. Pick the destination that splits the smallest run.
Two colours competing for one glass. One of them must yield. Undo one pour and give the glass to the colour with fewer alternative landings.

If none of the four fires, the position needs a temporary split: use the spare for exactly one pour, then re-run the scan. A second consecutive spare pour without progress is the signal to undo, not to keep exploring.

4. Three tiers, three search shapes

The rules never change — capacity 4, same pour rule, same solve condition — only the search space grows:

Easy (levels 1–4, par 6): the whole tree fits in your head. Plan the full 6 pours before touching anything; if you cannot see 6, your first pour is wrong.
Medium (levels 5–8, par 9–12): one colour more, same single spare. Level 7's par 12 means a forced chain roughly twice as long as easy — protect the spare through the middle third, where most branches die.
Hard (levels 9–11, par 14): five colours, two spares, and every optimum is 14. The extra spare is not a shortcut; it is what lets a 14-step chain survive one mid-sequence split. Expect to use Undo more, not less — the solver's tree here is orders of magnitude wider, and human search needs the same backtracking.

Move up only when you finish the current tier at or near par without restarting every level.

5. Undo versus restart, with the counter as judge

Undo steps back one pour and restores the move counter, so a short detour is free — use it the moment the counter passes par with nothing close to done. Restart reloads the frozen start instantly; prefer it when the spare is full of a mixed colour and the scan in section 3 finds no forced glass — that position is outside every optimal path and digging out costs more than the 6–14 pours a fresh attempt needs. Neither is failure: the solver itself explores by backtracking, and your Undo button is the same operation.

FAQ

How is this different from the Water Sort strategy guide?

The strategy guide teaches playing technique: reading top runs, protecting the spare, pouring whole runs, freeing buried colours, and choosing which colour to sort first. This page teaches the solver layer above it: the frozen par table, the spare-glass counting that makes a position alive or dead, the four-step stuck scan, and how to read the move counter against par. Technique first, solver second.

What does par actually guarantee?

Par is the length of the shortest legal pour sequence from the frozen start, proven by exhaustive search — no shorter solution exists, and at least one solution at exactly par exists. It does not promise your line is short; it promises a short line exists. Missing par by a few moves is normal; doubling it means a wrong early branch.

Can every level really be finished?

Yes. All 11 frozen layouts were BFS-verified solvable and non-trivial (minimum 3 pours) before release. There are no guessing levels: every start has a route to all-empty-or-full-single-colour under the pour rule. Being stuck means an earlier pour closed an option, not that the level is broken — undo to where the spare was last free.

When should I restart instead of undoing?

Undo when the last one or two pours caused the jam — the counter barely moved and the spare was recently free. Restart when the spare holds a mixed colour, the scan finds no forced glass, and the counter is already past par: that position is off every optimal path, and a fresh 6–14 pours is cheaper than excavation.

Why are all three hard levels par 14?

Because the frozen hard layouts each need the same minimum depth under identical rules (capacity 4, 5 colours, 2 spares). The uniform 14 is a property of those three starts, not a cap — and it tells you the difficulty is chain length, not a hidden trick. The mechanics are the same as easy; only the search is longer.