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| [[Donald Knuth]]'s '''Algorithm X''' is a [[recursion (computer science)|recursive]], [[Nondeterministic algorithm|nondeterministic]], [[depth-first]], [[backtracking]] [[algorithm]] that finds all solutions to the [[exact cover]] problem represented by a matrix ''A'' consisting of 0s and 1s. The goal is to select a subset of the rows so that the digit 1 appears in each column exactly once.
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| Algorithm X functions as follows:
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| :{| border="1" cellpadding="5" cellspacing="0"
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| # If the matrix ''A'' is empty, the problem is solved; terminate successfully.
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| # Otherwise choose a column ''c'' ([[deterministic algorithm|deterministically]]).
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| # Choose a row ''r'' such that ''A''<sub>''r'', ''c''</sub> = 1 ([[nondeterministic algorithm|nondeterministically]]).
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| # Include row ''r'' in the partial solution.
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| # For each column ''j'' such that ''A''<sub>''r'', ''j''</sub> = 1,
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| #: for each row ''i'' such that ''A''<sub>''i'', ''j''</sub> = 1,
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| #:: delete row ''i'' from matrix ''A'';
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| #: delete column ''j'' from matrix ''A''.
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| # Repeat this algorithm recursively on the reduced matrix ''A''.
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| |}
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| The nondeterministic choice of ''r'' means that the algorithm essentially clones itself into independent subalgorithms; each subalgorithm inherits the current matrix ''A'', but reduces it with respect to a different row ''r''.
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| If column ''c'' is entirely zero, there are no subalgorithms and the process terminates unsuccessfully.
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| The subalgorithms form a [[search tree]] in a natural way, with the original problem at the root and with level ''k'' containing each subalgorithm that corresponds to ''k'' chosen rows.
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| Backtracking is the process of traversing the tree in preorder, depth first.
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| Any systematic rule for choosing column ''c'' in this procedure will find all solutions, but some rules work much better than others.
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| To reduce the number of iterations, [[Donald Knuth|Knuth]] suggests that the column choosing algorithm select a column with the lowest number of 1s in it.
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| == Example ==
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| For example, consider the exact cover problem specified by the universe ''U'' = {1, 2, 3, 4, 5, 6, 7} and the collection of sets <math>\mathcal{S}</math> = {''A'', ''B'', ''C'', ''D'', ''E'', ''F''}, where:
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| :* ''A'' = {1, 4, 7};
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| :* ''B'' = {1, 4};
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| :* ''C'' = {4, 5, 7};
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| :* ''D'' = {3, 5, 6};
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| :* ''E'' = {2, 3, 6, 7}; and
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| :* ''F'' = {2, 7}.
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| This problem is represented by the matrix:
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| :{| border="1" cellpadding="5" cellspacing="0"
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| ! !! 1 !! 2 !! 3 !! 4 !! 5 !! 6 !! 7
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| |-
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| ! ''A''
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| | 1 || 0 || 0 || 1 || 0 || 0 || 1
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| |-
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| ! ''B''
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| | 1 || 0 || 0 || 1 || 0 || 0 || 0
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| |-
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| ! ''C''
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| | 0 || 0 || 0 || 1 || 1 || 0 || 1
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| |-
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| ! ''D''
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| | 0 || 0 || 1 || 0 || 1 || 1 || 0
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| |-
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| ! ''E''
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| | 0 || 1 || 1 || 0 || 0 || 1 || 1
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| |-
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| ! ''F''
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| | 0 || 1 || 0 || 0 || 0 || 0 || 1
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| |}
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| Algorithm X with Knuth's suggested heuristic for selecting columns solves this problem as follows:
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| '''Level 0'''
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| Step 1—The matrix is not empty, so the algorithm proceeds.
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| Step 2—The lowest number of 1s in any column is two. Column 1 is the first column with two 1s and thus is selected (deterministically):
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| :{| border="1" cellpadding="5" cellspacing="0"
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| ! !! 1 !! 2 !! 3 !! 4 !! 5 !! 6 !! 7
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| |-
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| ! ''A''
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| | <span style="color:red;font-weight:bold">1</span> || 0 || 0 || 1 || 0 || 0 || 1
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| |-
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| ! ''B''
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| | <span style="color:red;font-weight:bold">1</span> || 0 || 0 || 1 || 0 || 0 || 0
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| |-
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| ! ''C''
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| | 0 || 0 || 0 || 1 || 1 || 0 || 1
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| |-
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| ! ''D''
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| | 0 || 0 || 1 || 0 || 1 || 1 || 0
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| |-
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| ! ''E''
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| | 0 || 1 || 1 || 0 || 0 || 1 || 1
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| |-
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| ! ''F''
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| | 0 || 1 || 0 || 0 || 0 || 0 || 1
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| |}
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| Step 3—Rows ''A'' and ''B'' each have a 1 in column 1 and thus are selected (nondeterministically).
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| The algorithm moves to the first branch at level 1…
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| : '''Level 1: Select Row ''A'''''
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| : Step 4—Row ''A'' is included in the partial solution.
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| : Step 5—Row ''A'' has a 1 in columns 1, 4, and 7:
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| ::{| border="1" cellpadding="5" cellspacing="0"
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| ! !! <span style="color:blue">1</span> !! 2 !! 3 !! <span style="color:blue">4</span> !! 5 !! 6 !! <span style="color:blue">7</span>
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| |-
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| ! <span style="color:red">''A''</span>
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| | <span style="color:red;font-weight:bold">1</span> || 0 || 0 || <span style="color:red;font-weight:bold">1</span> || 0 || 0 || <span style="color:red;font-weight:bold">1</span>
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| |-
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| ! ''B''
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| | 1 || 0 || 0 || 1 || 0 || 0 || 0
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| |-
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| ! ''C''
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| | 0 || 0 || 0 || 1 || 1 || 0 || 1
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| |-
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| ! ''D''
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| | 0 || 0 || 1 || 0 || 1 || 1 || 0
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| |-
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| ! ''E''
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| | 0 || 1 || 1 || 0 || 0 || 1 || 1
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| |-
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| ! ''F''
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| | 0 || 1 || 0 || 0 || 0 || 0 || 1
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| |}
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| : Column 1 has a 1 in rows ''A'' and ''B''; column 4 has a 1 in rows ''A'', ''B'', and ''C''; and column 7 has a 1 in rows ''A'', ''C'', ''E'', and ''F''. Thus rows ''A'', ''B'', ''C'', ''E'', and ''F'' are to be removed and columns 1, 4 and 7 are to be removed:
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| ::{| border="1" cellpadding="5" cellspacing="0"
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| ! !! <span style="color:red">1</span> !! 2 !! 3 !! <span style="color:red">4</span> !! 5 !! 6 !! <span style="color:red">7</span>
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| |-
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| ! <span style="color:blue">''A''</span>
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| | <span style="color:red;font-weight:bold">1</span> || 0 || 0 || <span style="color:red;font-weight:bold">1</span> || 0 || 0 || <span style="color:red;font-weight:bold">1</span>
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| |-
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| ! <span style="color:blue">''B''</span>
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| | <span style="color:red;font-weight:bold">1</span> || 0 || 0 || <span style="color:red;font-weight:bold">1</span> || 0 || 0 || 0
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| |-
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| ! <span style="color:blue">''C''</span>
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| | 0 || 0 || 0 || <span style="color:red;font-weight:bold">1</span> || 1 || 0 || <span style="color:red;font-weight:bold">1</span>
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| |-
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| ! ''D''
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| | 0 || 0 || 1 || 0 || 1 || 1 || 0
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| |-
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| ! <span style="color:blue">''E''</span>
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| | 0 || 1 || 1 || 0 || 0 || 1 || <span style="color:red;font-weight:bold">1</span>
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| |-
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| ! <span style="color:blue">''F''</span>
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| | 0 || 1 || 0 || 0 || 0 || 0 || <span style="color:red;font-weight:bold">1</span>
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| |}
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| : Row ''D'' remains and columns 2, 3, 5, and 6 remain:
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| ::{| border="1" cellpadding="5" cellspacing="0"
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| ! !! 2 !! 3 !! 5 !! 6
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| |-
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| ! ''D''
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| | 0 || 1 || 1 || 1
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| |}
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| : Step 1—The matrix is not empty, so the algorithm proceeds.
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| : Step 2—The lowest number of 1s in any column is zero and column 2 is the first column with zero 1s:
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| ::{| border="1" cellpadding="5" cellspacing="0"
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| ! !! <span style="color:red">2</span> !! 3 !! 5 !! 6
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| |-
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| ! ''D''
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| | 0 || 1 || 1 || 1
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| |}
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| : Thus this branch of the algorithm terminates unsuccessfully.
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| : The algorithm moves to the next branch at level 1…
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| : '''Level 1: Select Row ''B'''''
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| : Step 4—Row ''B'' is included in the partial solution.
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| : Row ''B'' has a 1 in columns 1 and 4:
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| ::{| border="1" cellpadding="5" cellspacing="0"
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| ! !! <span style="color:blue">1</span> !! 2 !! 3 !! <span style="color:blue">4</span> !! 5 !! 6 !! 7
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| |-
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| ! ''A''
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| | 1 || 0 || 0 || 1 || 0 || 0 || 1
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| |-
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| ! <span style="color:red">''B''</span>
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| | <span style="color:red;font-weight:bold">1</span> || 0 || 0 || <span style="color:red;font-weight:bold">1</span> || 0 || 0 || 0
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| |-
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| ! ''C''
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| | 0 || 0 || 0 || 1 || 1 || 0 || 1
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| |-
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| ! ''D''
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| | 0 || 0 || 1 || 0 || 1 || 1 || 0
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| |-
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| ! ''E''
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| | 0 || 1 || 1 || 0 || 0 || 1 || 1
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| |-
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| ! ''F''
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| | 0 || 1 || 0 || 0 || 0 || 0 || 1
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| |}
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| : Column 1 has a 1 in rows ''A'' and ''B''; and column 4 has a 1 in rows ''A'', ''B'', and ''C''. Thus rows ''A'', ''B'', and ''C'' are to be removed and columns 1 and 4 are to be removed:
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| ::{| border="1" cellpadding="5" cellspacing="0"
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| ! !! <span style="color:red">1</span> !! 2 !! 3 !! <span style="color:red">4</span> !! 5 !! 6 !! 7
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| |-
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| ! <span style="color:blue">''A''</span>
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| | <span style="color:red;font-weight:bold">1</span> || 0 || 0 || <span style="color:red;font-weight:bold">1</span> || 0 || 0 || 1
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| |-
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| ! <span style="color:blue">''B''</span>
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| | <span style="color:red;font-weight:bold">1</span> || 0 || 0 || <span style="color:red;font-weight:bold">1</span> || 0 || 0 || 0
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| |-
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| ! <span style="color:blue">''C''</span>
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| | 0 || 0 || 0 || <span style="color:red;font-weight:bold">1</span> || 1 || 0 || 1
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| |-
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| ! ''D''
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| | 0 || 0 || 1 || 0 || 1 || 1 || 0
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| |-
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| ! ''E''
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| | 0 || 1 || 1 || 0 || 0 || 1 || 1
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| |-
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| ! ''F''
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| | 0 || 1 || 0 || 0 || 0 || 0 || 1
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| |}
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| : Rows ''D'', ''E'', and ''F'' remain and columns 2, 3, 5, 6, and 7 remain:
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| ::{| border="1" cellpadding="5" cellspacing="0"
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| ! !! 2 !! 3 !! 5 !! 6 !! 7
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| |-
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| ! ''D''
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| | 0 || 1 || 1 || 1 || 0
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| |-
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| ! ''E''
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| | 1 || 1 || 0 || 1 || 1
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| |-
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| ! ''F''
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| | 1 || 0 || 0 || 0 || 1
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| |}
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| : Step 1—The matrix is not empty, so the algorithm proceeds.
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| : Step 2—The lowest number of 1s in any column is one. Column 5 is the first column with one 1 and thus is selected (deterministically):
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| ::{| border="1" cellpadding="5" cellspacing="0"
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| ! !! 2 !! 3 !! <span style="color:red">5</span> !! 6 !! 7
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| |-
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| ! ''D''
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| | 0 || 1 || <span style="color:red;font-weight:bold">1</span> || 1 || 0
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| |-
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| ! ''E''
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| | 1 || 1 || 0 || 1 || 1
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| |-
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| ! ''F''
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| | 1 || 0 || 0 || 0 || 1
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| |}
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| : Step 3—Row ''D'' has a 1 in column 5 and thus is selected (nondeterministically).
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| : The algorithm moves to the first branch at level 2…
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| :: '''Level 2: Select Row ''D'''''
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| :: Step 4—Row ''D'' is included in the partial solution.
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| :: Step 5—Row ''D'' has a 1 in columns 3, 5, and 6:
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| :::{| border="1" cellpadding="5" cellspacing="0"
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| ! !! 2 !! <span style="color:blue">3</span> !! <span style="color:blue">5</span> !! <span style="color:blue">6</span> !! 7
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| |-
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| ! <span style="color:red">''D''</span>
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| | 0 || <span style="color:red;font-weight:bold">1</span> || <span style="color:red;font-weight:bold">1</span> || <span style="color:red;font-weight:bold">1</span> || 0
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| |-
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| ! ''E''
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| | 1 || 1 || 0 || 1 || 1
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| |-
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| ! ''F''
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| | 1 || 0 || 0 || 0 || 1
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| |}
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| :: Column 3 has a 1 in rows ''D'' and ''E''; column 5 has a 1 in row ''D''; and column 6 has a 1 in rows ''D'' and ''E''. Thus rows ''D'' and ''E'' are to be removed and columns 3, 5, and 6 are to be removed:
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| :::{| border="1" cellpadding="5" cellspacing="0"
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| ! !! 2 !! <span style="color:red">3</span> !! <span style="color:red">5</span> !! <span style="color:red">6</span> !! 7
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| |-
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| ! <span style="color:blue">''D''</span>
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| | 0 || <span style="color:red;font-weight:bold">1</span> || <span style="color:red;font-weight:bold">1</span> || <span style="color:red;font-weight:bold">1</span> || 0
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| |-
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| ! <span style="color:blue">''E''</span>
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| | 1 || <span style="color:red;font-weight:bold">1</span> || 0 || <span style="color:red;font-weight:bold">1</span> || 1
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| |-
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| ! ''F''
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| | 1 || 0 || 0 || 0 || 1
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| |}
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| :: Row ''F'' remains and columns 2 and 7 remain:
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| :::{| border="1" cellpadding="5" cellspacing="0"
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| ! !! 2 !! 7
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| |-
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| ! ''F''
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| | 1 || 1
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| |}
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| :: Step 1—The matrix is not empty, so the algorithm proceeds.
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| :: Step 2—The lowest number of 1s in any column is one. Column 2 is the first column with one 1 and thus is selected (deterministically).
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| :: Row ''F'' has a 1 in column 2 and thus is selected (nondeterministically).
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| :: The algorithm moves to the first branch at level 3…
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| ::: '''Level 3: Select Row ''F'''''
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| ::: Step 4—Row ''F'' is included in the partial solution.
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| ::: Row ''F'' has a 1 in columns 2 and 7:
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| ::::{| border="1" cellpadding="5" cellspacing="0"
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| ! !! <span style="color:blue">2</span> !! <span style="color:blue">7</span>
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| |-
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| ! <span style="color:red">''F''</span>
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| | <span style="color:red;font-weight:bold">1</span> || <span style="color:red;font-weight:bold">1</span>
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| |}
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| ::: Column 2 has a 1 in row ''F''; and column 7 has a 1 in row ''F''. Thus row ''F'' is to be removed and columns 2 and 7 are to be removed:
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| ::::{| border="1" cellpadding="5" cellspacing="0"
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| ! !! <span style="color:red">2</span> !! <span style="color:red">7</span>
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| |-
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| ! <span style="color:blue">''F''</span>
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| | <span style="color:red;font-weight:bold">1</span> || <span style="color:red;font-weight:bold">1</span>
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| |}
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| ::: Step 1—The matrix is empty, thus this branch of the algorithm terminates successfully.
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| ::: As rows ''B'', ''D'', and ''F'' are selected, the final solution is:
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| ::::{| border="1" cellpadding="5" cellspacing="0"
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| ! !! 1 !! 2 !! 3 !! 4 !! 5 !! 6 !! 7
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| |-
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| ! ''B''
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| | 1 || 0 || 0 || 1 || 0 || 0 || 0
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| |-
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| ! ''D''
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| | 0 || 0 || 1 || 0 || 1 || 1 || 0
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| |-
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| ! ''F''
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| | 0 || 1 || 0 || 0 || 0 || 0 || 1
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| |}
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| ::: In other words, the subcollection {''B'', ''D'', ''F''} is an exact cover, since every element is contained in exactly one of the sets ''B'' = {1, 4}, ''D'' = {3, 5, 6}, or ''F'' = {2, 7}.
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| ::: There are no more selected rows at level 3, thus the algorithm moves to the next branch at level 2…
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| :: There are no more selected rows at level 2, thus the algorithm moves to the next branch at level 1…
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| : There are no more selected rows at level 1, thus the algorithm moves to the next branch at level 0…
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| There are no branches at level 0, thus the algorithm terminates.
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| In summary, the algorithm determines there is only one exact cover: <math>\mathcal{S}^*</math> = {''B'', ''D'', ''F''}.
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| == Implementations ==
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| [[Dancing Links]], commonly known as DLX, is the technique suggested by [[Donald Knuth|Knuth]] to efficiently implement his Algorithm X on a computer. Dancing Links implements the matrix using circular [[doubly linked list]]s of the 1s in the matrix. There is a list of 1s for each row and each column. Each 1 in the matrix has a link to the next 1 above, below, to the left, and to the right of itself.
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| == See also ==
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| * [[Exact cover]]
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| * [[Dancing Links]]
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| == References ==
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| *{{citation
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| | first = Donald E. | last = Knuth | authorlink = Donald Knuth
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| | contribution = Dancing links
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| | title = Millennial Perspectives in Computer Science: Proceedings of the 1999 Oxford-Microsoft Symposium in Honour of Sir Tony Hoare
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| | year = 2000
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| | pages = 187–214
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| | publisher = Palgrave
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| | isbn = 978-0-333-92230-9
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| | editor1-first = Jim | editor1-last = Davies
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| | editor2-first = Bill | editor2-last = Roscoe
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| | editor3-first = Jim | editor3-last = Woodcock
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| | arxiv = cs/0011047 }}.
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| ==External links==
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| *[http://cheeso.members.winisp.net/srcview.aspx?dir=Sudoku&file=ExactCover.cs Implementation of an Exact Cover solver in C#] - uses Algorithm X and the Dancing Links optimization.
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| * [http://github.com/mlepage/polycube-solver Polycube solver] Program (with Lua source code) to fill boxes with polycubes using [[Algorithm X]].
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| *[http://www-cs-faculty.stanford.edu/~uno/papers/dancing-color.ps.gz Knuth's Paper describing the Dancing Links optimization] - Gzip'd postscript file.
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| {{Donald Knuth navbox}}
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| [[Category:Search algorithms]]
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| [[Category:Donald Knuth]]
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