Talk:Hitori

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Missing rule[edit]

Is there some rule that I'm not aware of? For example, take the following puzzle (yoinked from http://www.mathsnet.net/puzzles/hitori/index.html):

4 4 4 3
1 3 4 2
3 1 1 1
1 2 3 1

When I tell the applet to check, it indicates there a mistake in the solution. Based on the rules on the article alone, it looks perfect. --Sigma 7 04:00, 26 May 2007 (UTC)[reply]

Have you rechecked row 3 against "there are no duplicate numbers in any row or column"? —C.Fred (talk) 04:06, 26 May 2007 (UTC)[reply]
Okay, now I see the problem. I was thinking that the black cells seperated each row/column. --Sigma 7 04:08, 26 May 2007 (UTC)[reply]


Complexity[edit]

Does anybody know if Hitori is NP-complete? If so, is there some article on the web that we could link to from this page? 68.150.245.6 15:09, 27 July 2007 (UTC)[reply]

Yes, it is. The proof will appear in the upcoming book "Games, Puzzles, and Computation", from AK Peters. I'll try to get a version of the result onto arxiv as well. At the moment there is nothing online to link to. Bobhearn (talk) 02:40, 14 January 2009 (UTC)[reply]

Multiple solutions?[edit]

The article doesn't say whether a (properly formulated) Hitori can have multiple solutions. I assume it can't, but this should probably be said explicitly one way or the other. —RuakhTALK 01:52, 22 March 2008 (UTC)[reply]

Isolated Diagonally[edit]

The part: "The remaining un-filled cells must not be isolated from the rest of the grid by filled-in cells either vertically, horizontally or diagonally". The diagonally part is wrong according to Nikoli website (http://www.nikoli.co.jp/en/puzzles/hitori/). Plus I cannot think of anyway to be isolated diagonally as well and not violate the first rule.Ajorians (talk) 00:03, 19 October 2008 (UTC)[reply]

I rephrased it to "The remaining un-filled cells must form a single component connected horizontally and vertically" -- is that clear to the target audience, or have I gone over into math-speak? Joule36e5 (talk) 00:59, 13 November 2008 (UTC)[reply]

Game rule incomplete?[edit]

The rule of the game is supposedly as follows:

Two cells are called adjacent, if they touch at a full cell side (i.e. horizontally or vertically).
Find the single existing complete coloring of the game board in black and white, such that
  1. in every row and column, exactly one occurrence of each number contained is colored white, the others , if any, are all colored black
  2. no two adjacent cells are colored black
  3. the cells colored white make up a sole connected component

In my opinion, it's preferable to have a rule formulation which isn't procedural but declarative. One should never mix rules and solving strategies.

Currently missing in the rule definition is just the point not to blacken all occurrences of a number in a row or column. By the way, blackening hides information; one should consequently demonstrate why this doesn't change the game. In fact, this game is solved incrementally, so that blacking cells doesn't ceals essential information matters most.

Generally, randomly choosen cell number schemes aren't valid puzzles. They may be insolvable, e.g.

1 1
1 1

or they may have multiple solutions, e.g.

1 2
2 1
1 2

Puzzle solvers don't like either of these situations. The first one is outright wicked ( “I tried very long and finally thought I must have messed it up, and now you say that it doesn't solve at all!?” ), the second at least annoying ( “This is our own solution, if you should have found another, go and check it for yourself.” ) It's the puzzle makers duty to offer only puzzles with a single valid solution. So the article should also note that there is some work done in preparing and selecting appropriate puzzles.

-- 79.217.206.194 (talk) 08:32, 9 December 2009 (UTC)[reply]

Observation[edit]

I have observed in all Hitori's that I have seen so far that a number that is unique in both the row and the column it is in, is always white. In fact, when I first started with Hitori's this is exactly how I would start solving them. Since there is no evident reason why this should be the case given the four simple rules:

  • Single value at most once in the column
  • Single value at most once in the row
  • No neighboring black squares (hor/ver)
  • Connectivity: All white cells must be continuous

Can any one proof that I am right or show me a Hitori in which a unique value (in both row and column) is removed? Pluisjenijn (talk) 19:48, 30 April 2011 (UTC)[reply]

If a cell such as you describe were black, then turning that cell white would never create a contradiction, and so it would be a second solution to the puzzle. So, the property you're describing is a side effect of well-formed puzzles having a single solution. I don't think this is important enough to mention in the text. Joule36e5 (talk) 05:11, 30 October 2015 (UTC)[reply]

Equivalent bullet points[edit]

The following two bullet points under 'Solving Techniques' appear to be describing the same thing:

  • "In a sequence of three identical, adjacent numbers; the centre number must be white and cells on either side must be black. If one of the end numbers were white this would result in either two adjacent filled in cells or two white cells in the same row/column, neither of which are allowed."
  • "Any number that has two identical numbers on opposite sides of itself must be white, because one of the two identical numbers must be black, and it cannot be adjacent to another black cell."

Sbj42 (talk) 13:21, 22 August 2016 (UTC)[reply]

The first point is specific to THREE identical, adjacent numbers. The other rule is for two identical numbers on each side of cell with a different number. Dhugot (talk) 04:48, 17 February 2017 (UTC)[reply]