StackOverflow2013

Note that there are some explanatory texts on larger screens.

plurals

PO
primarykey
Id
12609427
data
AcceptedAnswerId
0
AnswerCount
0
ClosedDate
CommentCount
4
CommunityOwnedDate
CreationDate
2012-09-26T19:58:23.703
FavoriteCount
0
LastActivityDate
2012-09-26T22:12:09.617
LastEditDate
2017-05-23T12:04:38.427
LastEditorUserId
-1
OwnerUserId
330057
ParentId
12609368
PostTypeId
2
Score
1
ViewCount
0
LastEditorDisplayName
text
Body
Clearly, you need to check horizontal, vertical, and diagonal lines. Let us assume that the board is laid out with n always being the bigger number if the two are not equal, and that it is laying on its side" (lego style, not skyscraper style). So it is <code>n</code> across and <code>m</code> tall. The horizontal lines will be <code>n * (m - l)</code> in number. The vertical lines will be <code>m * (n - l)</code> in number. The diagonal lines will be <code>(m - l) * (n - l)</code>, or <code>m*n - l*m - l*n + l*l</code> If we assume <code>n >= m > l</code> then we can safely say that it is within <code>O(n^2)</code>, as we would expect for a two dimensional board. We know that <code>l > n >= m</code> will have no results. If <code>n = m = l</code> we have a constant number (<code>2*n + 2</code>). If <code>n = l > m</code> we are left with an even better case, as we don't have to check the diagonals or the verticals, and you only have to check <code>m</code> lines. If <code>n > l > m</code>, then we can again exclude the verticals, but must consider the diagonals. In any event, it will be less than doing the diagonals, verticals, and horizontals. There is an optimization that can be made, however, based on the <a href="https://stackoverflow.com/users/383124/phant0m">phant0m's</a> comment. It requires that you know what the last move made was. You can safely assume that if a move was made, it was made at a time that there was no winner on the board. Therefore, if there is a win condition on the board after the move, it must have been formed as a result of the most recent move. Therefore, the worst case scenario given this information is that the winning line is formed with the most recent move on the end. You would therefore need to check the 4 line segments that extend <code>l - 1</code> in each direction (horizontal, vertical, forward diagonal, and backward diagonal). This is a total of <code>4 * (2*l - 1)</code>, putting it safely in <code>O(l)</code>. Considering you only need to store one additional coordinate (most recent move) this is a most wise optimization to make.
Tags
Title
singulars
PostAcceptedAnswerId
1. This table or related slice is empty.
PostParentId
1. POAlgorithmic complexity of examining all possible lines on a game board
 singulars
 PostTypePostTypeId
 PTQuestion
PostTypePostTypeId
1. PTAnswer
UserLastEditorUserId
1. USCommunity
UserOwnerUserId
1. UScorsiKa
plurals
PostLinksPostIdRelatedPostId
1. This table or related slice is empty.
PostLinksRelatedPostIdPostId
1. This table or related slice is empty.
PostsAcceptedAnswerId
1. POAlgorithmic complexity of examining all possible lines on a game board
 singulars
 PostTypePostTypeId
 PTQuestion
PostsParentIdCreationDate
1. This table or related slice is empty.
VotesPostIdCreationDate
1. VO
 singulars
 PostPostId
 PO
 UserUserId
 This table or related slice is empty.
 VoteTypeVoteTypeId
 VTUpMod
2. VO
 singulars
 PostPostId
 PO
 UserUserId
 This table or related slice is empty.
 VoteTypeVoteTypeId
 VTAcceptedByOriginator
CommentsPostId

Querying!

Guidance

A row detail

Detail views are divided into sections. All the information in the data section comes from columns in the selected row. The other sections display data from other, related rows.

Related data can be related in a to-one or a to-many fashion. Captions of data related in a to-many fashion link to a list view showing a filtered view of the table.

Try moving around until you find a non-empty to-many entry and click on the label to get to one. You can move back to the root by clicking on the database name in the header.