Algebraic space - Revision history

en>TakuyaMurata at 04:10, 9 January 2015

2015-01-09T04:10:10Z

← Older revision		Revision as of 06:10, 9 January 2015
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	~~In [[combinatorial game theory]]~~, ~~the~~ '''~~mex~~'~~'', or "''m''inimum ''ex''cludant", of a [[Set (mathematics)\|set]] of [[ordinal number\|ordinal]]s denotes the smallest ordinal not contained~~ in ~~the set~~.		Hello, my title is Andrew and my spouse doesn't like it at all. What I love doing is soccer but I don't have the time recently. Distributing manufacturing is where my primary earnings arrives from and it's something I truly appreciate. I've usually cherished residing in Alaska.<br><br>my web-site: authentic psychic readings, [http://www.Khuplaza.com/dent/14869889 http://www.Khuplaza.com/dent/14869889],

	~~Some examples:~~
	:<~~math~~>~~\mbox{mex}(\left \{1, 2, 3\right \}) = 0~~<~~/math~~>
	:~~<math>\mbox{mex}(\left \{0~~, ~~1, 4, 7, 12\right \}) = 2<~~/~~math>~~
	~~:<math>\mbox{mex}(\left \{0, 1, 2, 3, \ldots\right \}) = \omega<~~/~~math>~~
	~~:<math>\mbox{mex}(\left \{0, 2, 4, 6, \ldots\right \}) = 1<~~/~~math>~~
	~~:<math>\mbox{mex}(\left \{0, 1, 2, 3, \ldots, \omega\right \}) = \omega+1<~~/~~math>~~

	~~<!-- COMMENTED OUT EXAMPLES~~
	:~~<math>\mbox{mex}(\left \{0, 1, 2, 3, 50\right \}) = 4<~~/~~math>~~
	~~:<math>\mbox{mex}(\left \{4, 50, \omega, \omega+29\right \}) = 0<~~/~~math>~~
	~~-->~~
	~~where ω is the [[limit ordinal]] for the natural numbers~~.

	In the [[Sprague-Grundy theorem\|Sprague-Grundy theory]] the minimum excluded ordinal is used to determine the [[nimber]] of a [[Normal play convention\|normal-play]] impartial game, which is a game in which either player has the same moves in each position and the last player to move wins. ~~The nimber is equal to 0 for a game that is lost immediately by the first player, and is equal to the mex of the nimbers of all possible next positions for any other game.~~

	~~For example, in a one-pile version of [[Nim~~]], the game starts with a pile of ''n'' stones, and the player to move may take any positive number of stones. If ''n'' is zero, the nimber is 0. If ''n'' is 1, the player to move will leave 0 stones, and the mex of { 0 }, 1, gives the nimber for this case. If ''n'' is 2, the player to move can leave 0 or 1 stones, giving 2 as the mex of { 0, 1 }. In general, the player to move with a pile of ''n'' stones can leave anywhere from 0 to n-1 stones; the mex of the numbers { 0, 1, ..., ''n''-1 } is always ''n''. From this we can conclude that the first player wins if ''n'' is not zero (by taking all the stones).

	If we change the game so that the player to move can only take up to 3 stones, then when ''n''=4, the successor states have nimbers { 1, 2, 3 }, giving a mex of 0. Now the first player loses; the second player's strategy is to respond to whatever move the first player makes by taking the rest of the stones. For ''n''=5, the nimbers of the successor states 2, 3, and 4 are 2, 3, and 0 (as we just calculated); the mex of { 0, 2, 3 } is 1, so starting with 5 stones in this game is a win for the first player.

	~~(See [[nimber]]s for more details on the meaning of nimber values.)~~

	~~==References==~~
	* {{cite book \| last=Conway \| first=John H. \| authorlink=John Horton Conway \| title=[[On Numbers and Games]] \| edition=2nd \| publisher=A.K. Peters \| year=2001 \| isbn=1-56881-127-6 \| page=124 }}

	~~[[Category:Combinatorial game theory]]~~

en>Mark viking: Added wl

2014-01-07T01:50:52Z

Added wl

← Older revision		Revision as of 03:50, 7 January 2014
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	~~She~~ is ~~recognized~~ by the ~~name~~ of ~~Myrtle Shryock~~. ~~The factor she adores most~~ is ~~physique building~~ and ~~now she~~ is ~~attempting~~ to ~~earn money~~ with it. ~~California~~ is ~~our birth location~~. ~~I am~~ a ~~meter reader but I~~ strategy on ~~changing it~~.~~<br><br>Look into my weblog ::~~ [~~http://www~~.~~pornextras~~.~~info/blog/187567 at home std testing~~]		In [[combinatorial game theory]], the '''mex''', or "''m''inimum ''ex''cludant", of a [[Set (mathematics)\|set]] of [[ordinal number\|ordinal]]s denotes the smallest ordinal not contained in the set.

			Some examples:
			:<math>\mbox{mex}(\left \{1, 2, 3\right \}) = 0</math>
			:<math>\mbox{mex}(\left \{0, 1, 4, 7, 12\right \}) = 2</math>
			:<math>\mbox{mex}(\left \{0, 1, 2, 3, \ldots\right \}) = \omega</math>
			:<math>\mbox{mex}(\left \{0, 2, 4, 6, \ldots\right \}) = 1</math>
			:<math>\mbox{mex}(\left \{0, 1, 2, 3, \ldots, \omega\right \}) = \omega+1</math>

			<!-- COMMENTED OUT EXAMPLES
			:<math>\mbox{mex}(\left \{0, 1, 2, 3, 50\right \}) = 4</math>
			:<math>\mbox{mex}(\left \{4, 50, \omega, \omega+29\right \}) = 0</math>
			-->
			where ω is the [[limit ordinal]] for the natural numbers.

			In the [[Sprague-Grundy theorem\|Sprague-Grundy theory]] the minimum excluded ordinal is used to determine the [[nimber]] of a [[Normal play convention\|normal-play]] impartial game, which is a game in which either player has the same moves in each position and the last player to move wins. The nimber is equal to 0 for a game that is lost immediately by the first player, and is equal to the mex of the nimbers of all possible next positions for any other game.

			For example, in a one-pile version of [[Nim]], the game starts with a pile of ''n'' stones, and the player to move may take any positive number of stones. If ''n'' is zero, the nimber is 0. If ''n'' is 1, the player to move will leave 0 stones, and the mex of { 0 }, 1, gives the nimber for this case. If ''n'' is 2, the player to move can leave 0 or 1 stones, giving 2 as the mex of { 0, 1 }. In general, the player to move with a pile of ''n'' stones can leave anywhere from 0 to n-1 stones; the mex of the numbers { 0, 1, ..., ''n''-1 } is always ''n''. From this we can conclude that the first player wins if ''n'' is not zero (by taking all the stones).

			If we change the game so that the player to move can only take up to 3 stones, then when ''n''=4, the successor states have nimbers { 1, 2, 3 }, giving a mex of 0. Now the first player loses; the second player's strategy is to respond to whatever move the first player makes by taking the rest of the stones. For ''n''=5, the nimbers of the successor states 2, 3, and 4 are 2, 3, and 0 (as we just calculated); the mex of { 0, 2, 3 } is 1, so starting with 5 stones in this game is a win for the first player.

			(See [[nimber]]s for more details on the meaning of nimber values.)

			==References==
			* {{cite book \| last=Conway \| first=John H. \| authorlink=John Horton Conway \| title=[[On Numbers and Games]] \| edition=2nd \| publisher=A.K. Peters \| year=2001 \| isbn=1-56881-127-6 \| page=124 }}

			[[Category:Combinatorial game theory]]

en>Headbomb: Various citation cleanup + AWB fixes using AWB (8062)

2012-07-30T19:45:27Z

Various citation cleanup + AWB fixes using AWB (8062)

New page

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