Knuth's Algorithm X: Difference between revisions

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{{Short description|Algorithm for exact cover problem}}

[[Donald Knuth|Donald Knuth's]] '''Algorithm X''', first presented in November of 2000, is a [[recursion (computer science)|recursive]], [[Nondeterministic_algorithm|nondeterministic]], [[depth-first]], [[brute-force]] [[algorithm]] that finds all solutions to the [[exact cover]] problem represented by a matrix ''A'' consisting of 0s and 1s.

'''Algorithm X''' is an [[algorithm]] for solving the [[exact cover]] problem. It is a straightforward [[Recursion (computer science)|recursive]], [[Nondeterministic algorithm|nondeterministic]], [[depth-first]], [[backtracking]] algorithm used by [[Donald Knuth]] to demonstrate an efficient implementation called DLX, which uses the [[dancing links]] technique.<ref name="knuth">{{cite arXiv | author = Knuth, Donald | author-link = Donald Knuth | title = Dancing links | year = 2000 | eprint = cs/0011047 }}</ref><ref>{{Cite journal |last=Banerjee |first=Bikramjit |last2=Kraemer |first2=Landon |last3=Lyle |first3=Jeremy |date=2010-07-04 |title=Multi-Agent Plan Recognition: Formalization and Algorithms |url=https://ojs.aaai.org/index.php/AAAI/article/view/7746 |journal=Proceedings of the AAAI Conference on Artificial Intelligence |volume=24 |issue=1 |pages=1059–1064 |doi=10.1609/aaai.v24i1.7746 |issn=2374-3468|doi-access=free }}</ref>

The goal is to select a subset of the rows so that the digit 1 appears in each column exactly once.

==Algorithm X functions as follows:==

The exact cover problem is represented in Algorithm X by an [[incidence matrix]] ''A'' consisting of 0s and 1s. The goal is to select a subset of the rows such that the digit 1 appears in each column exactly once.

Algorithm X works as follows:

:{| border="1" cellpadding="5" cellspacing="0"

|

# If the matrix ''A'' ishas emptyno columns, the problemcurrent partial solution is solveda valid solution; terminate successfully.

# Otherwise choose a column ''c'' ([[deterministic algorithm|deterministically]]).

# Choose a row ''r'' such that ''A''_{''r'', ''c''} = 1 ([[nondeterministic algorithm|nondeterministically]]).

# Include row ''r'' in the partial solution.

# For each column ''j'' such that ''A''_{''r'', ''j''} = 1,

#: delete column ''j'' from matrix ''A'';

#: for each row ''i'' such that ''A''_{''i'', ''j''} = 1,

#:: delete row ''i'' from matrix ''A''.

#: delete column ''j'' from matrix ''A''.

# Repeat this algorithm recursively on the reduced matrix ''A''.

|}

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''.

The nondeterministic choice of ''r'' means that the algorithm recurses over independent subalgorithms; each subalgorithm inherits the current matrix ''A'', but reduces it with respect to a different row ''r''.

If column ''c'' is entirely zero, there are no subalgorithms and the process terminates unsuccessfully.

Line 24 ⟶ 25:

Any systematic rule for choosing column ''c'' in this procedure will find all solutions, but some rules work much better than others.

To reduce the number of iterations, [[Donald Knuth|Knuth]] suggests that the column -choosing algorithm select a column with the lowestsmallest number of 1s in it.

== Example ==

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{{mathcal|S}</math>} = {''A'', ''B'', ''C'', ''D'', ''E'', ''F''}, where:

:* ''A'' = {1, 4, 7};

:* ''B'' = {1, 4};

Line 37 ⟶ 38:

This problem is represented by the matrix:

:{| class="wikitable"

:{| border="1" cellpadding="5" cellspacing="0"

! !! 1 !! 2 !! 3 !! 4 !! 5 !! 6 !! 7

|-

Line 63 ⟶ 64:

'''Level 0'''

Step 1—The matrix is not empty, so the algorithm proceeds.

The lowest number of 1s in any column is two and column 1 is the first column with two 1s, thus column 1 is selected (deterministically).

RowsStep ''A''2—The andlowest ''B''number eachof have1s ain any column is two. Column 1 inis the first column 1with two 1s and thus areis selected (nondeterministicallydeterministically).:

:{| class="wikitable"

The algorithm moves to the first branch at level 1.

! !! 1 !! 2 !! 3 !! 4 !! 5 !! 6 !! 7

|-

! ''A''

| <span style="color:red;font-weight:bold">1</span> || 0 || 0 || 1 || 0 || 0 || 1

|-

! ''B''

| <span style="color:red;font-weight:bold">1</span> || 0 || 0 || 1 || 0 || 0 || 0

|-

! ''C''

| 0 || 0 || 0 || 1 || 1 || 0 || 1

|-

! ''D''

| 0 || 0 || 1 || 0 || 1 || 1 || 0

|-

! ''E''

| 0 || 1 || 1 || 0 || 0 || 1 || 1

|-

! ''F''

| 0 || 1 || 0 || 0 || 0 || 0 || 1

|}

Step 3—Rows ''A'' and ''B'' each have a 1 in column 1 and thus are selected (nondeterministically).

The algorithm moves to the first branch at level 1…

: '''Level 1: Select Row ''A'''''

: Step 4—Row ''A'' is included in the partial solution.

: Row ''A'' has a 1 in columns 1, 4, and 7. 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 removed and columns 1, 4 and 7 are removed. Row ''D'' remains and columns 2, 3, 5, and 6 remain:

: Step 5—Row ''A'' has a 1 in columns 1, 4, and 7:

::{| border="1" cellpadding="5" cellspacing="0"

::{| class="wikitable"

! !! <span style="color:blue">1</span> !! 2 !! 3 !! <span style="color:blue">4</span> !! 5 !! 6 !! <span style="color:blue">7</span>

|-

! <span style="color:red">''A''</span>

| <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>

|-

! ''B''

| 1 || 0 || 0 || 1 || 0 || 0 || 0

|-

! ''C''

| 0 || 0 || 0 || 1 || 1 || 0 || 1

|-

! ''D''

| 0 || 0 || 1 || 0 || 1 || 1 || 0

|-

! ''E''

| 0 || 1 || 1 || 0 || 0 || 1 || 1

|-

! ''F''

| 0 || 1 || 0 || 0 || 0 || 0 || 1

|}

: 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:

::{| class="wikitable"

! !! <span style="color:red">1</span> !! 2 !! 3 !! <span style="color:red">4</span> !! 5 !! 6 !! <span style="color:red">7</span>

|-

! <span style="color:blue">''A''</span>

| <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>

|-

! <span style="color:blue">''B''</span>

| <span style="color:red;font-weight:bold">1</span> || 0 || 0 || <span style="color:red;font-weight:bold">1</span> || 0 || 0 || 0

|-

! <span style="color:blue">''C''</span>

| 0 || 0 || 0 || <span style="color:red;font-weight:bold">1</span> || 1 || 0 || <span style="color:red;font-weight:bold">1</span>

|-

! ''D''

| 0 || 0 || 1 || 0 || 1 || 1 || 0

|-

! <span style="color:blue">''E''</span>

| 0 || 1 || 1 || 0 || 0 || 1 || <span style="color:red;font-weight:bold">1</span>

|-

! <span style="color:blue">''F''</span>

| 0 || 1 || 0 || 0 || 0 || 0 || <span style="color:red;font-weight:bold">1</span>

|}

: Row ''D'' remains and columns 2, 3, 5, and 6 remain:

::{| class="wikitable"

! !! 2 !! 3 !! 5 !! 6

|-

Line 80 ⟶ 155:

|}

: Step 1—The matrix is not empty, so the algorithm proceeds.

: The lowest number of 1s in any column is zero and column 2 is the first column with zero 1s, thus this branch of the algorithm terminates unsuccessfully.

: Step 2—The lowest number of 1s in any column is zero and column 2 is the first column with zero 1s:

::{| class="wikitable"

! !! <span style="color:red">2</span> !! 3 !! 5 !! 6

|-

! ''D''

| 0 || 1 || 1 || 1

|}

: Thus this branch of the algorithm terminates unsuccessfully.

: The algorithm moves to the next branch at level 1.1…

: '''Level 1: Select Row ''B'''''

: Step 4—Row ''B'' is included in the partial solution.

: Row ''B'' has a 1 in columns 1 and 4. 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 removed and columns 1 and 4 are removed. Rows ''D'', ''E'', and ''F'' remain and columns 2, 3, 5, 6, and 7 remain:

: Row ''B'' has a 1 in columns 1 and 4:

::{| border="1" cellpadding="5" cellspacing="0"

::{| class="wikitable"

! !! <span style="color:blue">1</span> !! 2 !! 3 !! <span style="color:blue">4</span> !! 5 !! 6 !! 7

|-

! ''A''

| 1 || 0 || 0 || 1 || 0 || 0 || 1

|-

! <span style="color:red">''B''</span>

| <span style="color:red;font-weight:bold">1</span> || 0 || 0 || <span style="color:red;font-weight:bold">1</span> || 0 || 0 || 0

|-

! ''C''

| 0 || 0 || 0 || 1 || 1 || 0 || 1

|-

! ''D''

| 0 || 0 || 1 || 0 || 1 || 1 || 0

|-

! ''E''

| 0 || 1 || 1 || 0 || 0 || 1 || 1

|-

! ''F''

| 0 || 1 || 0 || 0 || 0 || 0 || 1

|}

: 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:

::{| class="wikitable"

! !! <span style="color:red">1</span> !! 2 !! 3 !! <span style="color:red">4</span> !! 5 !! 6 !! 7

|-

! <span style="color:blue">''A''</span>

| <span style="color:red;font-weight:bold">1</span> || 0 || 0 || <span style="color:red;font-weight:bold">1</span> || 0 || 0 || 1

|-

! <span style="color:blue">''B''</span>

| <span style="color:red;font-weight:bold">1</span> || 0 || 0 || <span style="color:red;font-weight:bold">1</span> || 0 || 0 || 0

|-

! <span style="color:blue">''C''</span>

| 0 || 0 || 0 || <span style="color:red;font-weight:bold">1</span> || 1 || 0 || 1

|-

! ''D''

| 0 || 0 || 1 || 0 || 1 || 1 || 0

|-

! ''E''

| 0 || 1 || 1 || 0 || 0 || 1 || 1

|-

! ''F''

| 0 || 1 || 0 || 0 || 0 || 0 || 1

|}

: Rows ''D'', ''E'', and ''F'' remain and columns 2, 3, 5, 6, and 7 remain:

::{| class="wikitable"

! !! 2 !! 3 !! 5 !! 6 !! 7

|-

Line 101 ⟶ 236:

|}

: Step 1—The matrix is not empty, so the algorithm proceeds.

: The lowest number of 1s in any column is one and column 5 is the first column with one 1, thus column 5 is selected (deterministically).

: RowStep ''D''2—The haslowest anumber 1of 1s in any column is one. Column 5 is the first column with one 1 and thus is selected (nondeterministicallydeterministically).:

::{| class="wikitable"

: The algorithm moves to the first branch at level 2.

! !! 2 !! 3 !! <span style="color:red">5</span> !! 6 !! 7

|-

! ''D''

| 0 || 1 || <span style="color:red;font-weight:bold">1</span> || 1 || 0

|-

! ''E''

| 1 || 1 || 0 || 1 || 1

|-

! ''F''

| 1 || 0 || 0 || 0 || 1

|}

: Step 3—Row ''D'' has a 1 in column 5 and thus is selected (nondeterministically).

: The algorithm moves to the first branch at level 2…

:: '''Level 2: Select Row ''D'''''

:: Step 4—Row ''D'' is included in the partial solution.

:: Row ''D'' has a 1 in columns 3, 5, and 6. 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 removed and columns 3, 5, and 6 are removed. Row ''F'' remains and columns 2 and 7 remain:

:: Step 5—Row ''D'' has a 1 in columns 3, 5, and 6:

:::{| border="1" cellpadding="5" cellspacing="0"

:::{| class="wikitable"

! !! 2 !! <span style="color:blue">3</span> !! <span style="color:blue">5</span> !! <span style="color:blue">6</span> !! 7

|-

! <span style="color:red">''D''</span>

| 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

|-

! ''E''

| 1 || 1 || 0 || 1 || 1

|-

! ''F''

| 1 || 0 || 0 || 0 || 1

|}

:: 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:

:::{| class="wikitable"

! !! 2 !! <span style="color:red">3</span> !! <span style="color:red">5</span> !! <span style="color:red">6</span> !! 7

|-

! <span style="color:blue">''D''</span>

| 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

|-

! <span style="color:blue">''E''</span>

| 1 || <span style="color:red;font-weight:bold">1</span> || 0 || <span style="color:red;font-weight:bold">1</span> || 1

|-

! ''F''

| 1 || 0 || 0 || 0 || 1

|}

:: Row ''F'' remains and columns 2 and 7 remain:

:::{| class="wikitable"

! !! 2 !! 7

|-

Line 118 ⟶ 300:

|}

:: Step 1—The matrix is not empty, so the algorithm proceeds.

:: The lowest number of 1s in any column is one and column 2 is the first column with one 1s, thus column 2 is selected (deterministically).

:: 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):

:::{| class="wikitable"

! !! <span style="color:red">2</span> !! 7

|-

! ''F''

| <span style="color:red;font-weight:bold">1</span> || 1

|}

:: Row ''F'' has a 1 in column 2 and thus is selected (nondeterministically).

:: The algorithm moves to the first branch at level 3.3…

::: '''Level 3: Select Row ''F'''''

::: Step 4—Row ''F'' is included in the partial solution.

::: Row ''F'' has a 1 in columns 2 and 7. Column 2 has a 1 in row ''F''; and column 7 has a 1 in row ''F''. Thus row ''F'' is removed and columns 2 and 7 are removed. The matrix is empty, thus this branch of the algorithm terminates successfully.

::: As rowsRow ''BF'', ''D'',has anda ''F''1 arein selected,columns the2 final solutionand is7:

::::{| borderclass="1" cellpadding="5" cellspacing="0wikitable"

! !! <span style="color:blue">2</span> !! <span style="color:blue">7</span>

|-

! <span style="color:red">''F''</span>

| <span style="color:red;font-weight:bold">1</span> || <span style="color:red;font-weight:bold">1</span>

|}

::: 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:

::::{| class="wikitable"

! !! <span style="color:red">2</span> !! <span style="color:red">7</span>

|-

! <span style="color:blue">''F''</span>

| <span style="color:red;font-weight:bold">1</span> || <span style="color:red;font-weight:bold">1</span>

|}

::: No rows and no columns remain:

::::{| class="wikitable"

! &nbsp;

|}

::: Step 1—The matrix is empty, thus this branch of the algorithm terminates successfully.

::: As rows ''B'', ''D'', and ''F'' have been selected (step 4), the final solution in this branch is:

::::{| class="wikitable"

! !! 1 !! 2 !! 3 !! 4 !! 5 !! 6 !! 7

|-

Line 145 ⟶ 362:

::: 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}.

::: There are no more selected rows at level 3, thus the algorithm moves to the next branch at level 2.2…

:: There are no more selected rows at level 2, thus the algorithm moves to the next branch at level 1.1…

: There are no more selected rows at level 1, thus the algorithm moves to the next branch at level 0.0…

There are no branches at level 0, thus the algorithm terminates.

The algorithm is complete.

In summary, the algorithm determines there is only one exact cover: <math>\mathcal{{mathcal|S}^}{{sup|*</math>}} = {''B'', ''D'', ''F''}.

== Implementations ==

[[DancingKnuth's Links]],main commonlypurpose knownin asdescribing DLX,Algorithm isX thewas techniqueto suggesteddemonstrate bythe utility of [[Donalddancing Knuth|Knuthlinks]]. to efficientlyKnuth implementshowed histhat Algorithm X can be implemented efficiently on a computer. Dancingusing Linksdancing implementslinks in a process Knuth calls ''"DLX"''. DLX uses the matrix usingrepresentation circularof the [[doublyexact linkedcover]] list|problem, implemented as [[doubly- linked list]]s of the 1s inof the matrix. There is a list of 1s for: each row and each column. Each 1 in the matrixelement has a link to the next 1 above, below, to the left, and to the right of itself. (Technically, because the lists are circular, this forms a [[torus]]). Because exact cover problems tend to be sparse, this representation is usually much more efficient in both size and processing time required. DLX then uses dancing links to quickly select permutations of rows as possible solutions and to efficiently backtrack (undo) mistaken guesses.<ref name="knuth" />

== See also ==

* [[Exact cover]]

* [[Dancing Links]]

==References==

== External links ==

{{Reflist}}

*{{citation

| first = Donald E. | last = Knuth | authorlinkauthor-link = Donald Knuth

| contribution = Dancing links

| title = Millennial Perspectives in Computer Science: Proceedings of the 1999 Oxford-Microsoft Symposium in Honour of Sir Tony Hoare

Line 170 ⟶ 388:

| pages = 187–214

| publisher = Palgrave

| isbn = 9780333922309978-0-333-92230-9

| editor1-first = Jim | editor1-last = Davies

| editor2-first = Bill | editor2-last = Roscoe

| editor3-first = Jim | editor3-last = Woodcock

| idarxiv = {{arxiv | cs/0011047}} | bibcode = 2000cs.......11047K}}.

==External links==

*[https://www.ocf.berkeley.edu/~jchu/publicportal/sudoku/0011047.pdf Knuth's paper] - PDF file (also {{ArXiv|cs/0011047}})

*[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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