Weighing a Link Against the Rest

The general form of the opening moves, and what to do when the answer is real but no longer obvious. One island, one link, two sums.

The two sums

Pick an island and pick one of its links. Add up the most that all its OTHER links could carry between them. Subtract that from the island’s number, and whatever is left over has to run down the link you set aside — because there is nowhere else for it to go. That is the minimum for that link, and it is very often one when everything looked undecided a moment before.

Then run it the other way. Add up the least that all the other links must carry, subtract that from the number, and what remains is the most this link could possibly take. That is the maximum, and it is how a link gets closed by arithmetic rather than by a crossing.

The showpiece is a 3 with exactly two neighbours. Neither link is settled — either could be one or two — but two links cannot supply three bridge ends if one of them is empty, so each of them carries at least one. You have not decided which is doubled, and you do not need to: one bridge goes down each way and the board moves.

Before — a 2 with one large neighbour and one small one

After — at least one bridge has to go left

The 2 at row 2, column 6 sees two islands: the 2 across to its left, and the 1 far below it in the same column. The link to a 1 can never carry more than a single bridge, so the most that link could contribute is one — and the 2 needs two. The shortfall has to run left, so that link carries at least one bridge even though nobody has decided whether it carries two.

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Why it sits above the cheap moves

Because it subsumes them. A full house is this argument in the case where the shortfall happens to equal the whole link; a spent island is this argument in the case where the leftover is zero. They come first because they are the versions you can see without doing any arithmetic, and this is what you do when the same reasoning applies but the numbers no longer announce themselves.

It is also genuinely more expensive to perform. Every link of every island is two additions and two subtractions, and doing that across a 15×15 board is a couple of minutes of concentrated work. Doing it before a free sweep has been exhausted is the single best way to make this puzzle unpleasant — the sweep would have settled half of those links and made the sums shorter.

When you do commit to it, be systematic. Go island by island rather than to the one that looks promising, because an island that is about to give way and an island with plenty of slack look identical until the sum is done. Guessing which island to weigh is how the technique acquires a reputation for being tedious.

What makes a board hard, precisely

A hard board here is one where the solver, always taking the cheapest available move, was at some point forced into this arithmetic — no island with an obvious lack of choice, none finished with links still open, no crossing left to strike out. That is a strong statement about the board rather than about its size: it contains a genuine bottleneck.

It is also why size and difficulty are separate choices on this site. A 15×15 easy board has forty islands and no bottleneck at all; a 7×7 hard board has eight islands and one. Most players find the second takes longer.

Where this sits on the ladder

Rung 1Full house
An island with one open link, or with only just enough room, has no choice at all.
Rung 2Spent island
An island that already has every bridge it asked for closes all its other links.
Rung 3Blocked crossing
A link that would cut across a bridge already built can never be used.
Rung 4Spare capacity
Take one link away and ask what the rest could carry; the shortfall is forced down it.
Rung 5Isolation
A group cannot seal itself off, and a group with one way out has to use it.
Rung 6Assumption
Draw a bridge, follow it until something breaks, and rub it out again.

Common questions

What is the capacity technique in hashi?

Setting one link aside, working out the most the island’s remaining links could carry, and forcing the difference down the link you set aside. Running the same sum with minimums instead gives that link’s maximum.

When should I start doing this arithmetic?

Only after a full sweep for forced and finished islands, and after every crossing has been struck out, both place nothing. The sweep is free and it shortens every sum this technique has to do.

Why does a 3 with two neighbours always give one bridge each way?

Because two links cap out at two bridges each, so if either carried nothing the other could supply at most two — one short. Neither can be empty, so each carries at least one, without deciding which is doubled.