Drag coefficient - Revision history

24.145.132.228: power

2015-01-02T22:25:33Z

power

← Older revision		Revision as of 00:25, 3 January 2015
Line 1:		Line 1:
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131.215.118.133: added Cd of sphere at 10^5

2014-03-04T21:16:36Z

added Cd of sphere at 10^5

← Older revision		Revision as of 23:16, 4 March 2014
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	In [[graph theory]], a branch of [[mathematics]], the '''Chinese postman problem (CPP)''', '''postman tour''' or '''route inspection problem''' is to find a shortest closed path or circuit that visits every edge of a (connected) [[undirected graph]]. When the graph has an [[Eulerian circuit]] (a closed walk that covers every edge once), that circuit is an optimal solution.		My name: Glinda Newkirk<br>My age: 33<br>Country: France<br>Home town: Rodez <br>Post code: 12000<br>Street: 99 Rue De Groussay<br><br>Here is my website [http://hemorrhoidtreatmentfix.com/external-hemorrhoid-treatment external hemorrhoids treatment]

	~~Alan Goldman of the [[NIST\|U.S. National Bureau of Standards]] first coined the~~ name ~~'Chinese Postman Problem' for this problem, as it was originally studied by the Chinese mathematician Mei-Ku Kuan in 1962.~~<~~ref~~>~~{{cite web \| url=http~~:~~//www.nist.gov/dads/HTML/chinesePostman.html \| title="Chinese Postman Problem"}}~~<~~/ref~~>

	~~==Eulerian paths and circuits==~~

	~~In order for a [[graph theory\|graph]] to have an [[glossary of graph theory\|Eulerian circuit]], it will certainly have to be connected.~~

	~~Suppose we have a connected graph ''G'' = (''V'', ''E''), The following statements are equivalent~~:

	~~# All vertices in G have even [[Degree (graph theory)\|degree]].~~
	~~# G consists of the edges from a [[disjoint union]] of some cycles, and the vertices from these cycles.~~
	~~# G has an [[Eulerian circuit]].~~

	* 1 → 2 can be shown by induction on the number of cycles.
	* 2 → 3 can also be shown by induction on the number of cycles, and
	* 3 → 1 should be immediate.

	~~An [[Eulerian path]] (a walk which is not closed but uses all edges of G just once) exists if and only if G is connected and exactly two vertices have odd valence.~~

	~~==''T''-joins==~~

	Let ''T'' be a subset of the vertex set of a graph. An edge set is called a ''T'''''-join''' if in the induced subgraph of this edge set, the collection of all the odd-degree vertices is ''T''. (In a connected graph, a ''T''-join exists if and only if \|''T''\| is even.) The ''T'''''-join problem''' is to find a smallest ''T''-join. When ''T'' is the set of all odd-degree vertices, a smallest ''T''-join leads to a solution of the postman problem. For any ''T'', a smallest ''T''-join necessarily consists of <~~math~~>~~\tfrac{1}{2}~~<~~/math~~>\|''T''\| paths, no two having an edge in common, that join the vertices of ''T'' in pairs. The paths will be such that the total length of all of them is as small as possible. A minimum ''T''-join can be obtained using a weighted matching algorithm that uses O(''n''<~~sup~~>3<~~/sup~~>~~) computational steps.~~<~~ref~~>~~J. Edmonds and E.L. Johnson, Matching Euler tours and the Chinese postman problem, Math. Program. (1973).</ref>~~

	~~==Solution==~~
	~~If a graph has an Eulerian circuit (or an Eulerian path), then an Eulerian circuit (or path) visits every edge, and so the solution is to choose any Eulerian circuit (or path).~~

	~~If the graph~~ is ~~not Eulerian, it must contain vertices of odd degree. By the~~ [[handshaking lemma]], there must be an even number of these vertices. To solve the postman problem we first find a smallest ''T''-join. We make the graph Eulerian by doubling of the ''T''-join. The solution to the postman problem in the original graph is obtained by finding an Eulerian circuit for the new graph.

	~~==Applications==~~
	~~Various combinatorial problems are reduced to the Chinese Postman Problem, including finding a maximum cut in a planar graph~~
	~~and a minimum-mean length circuit in an undirected graph <ref>A. Schrijver, Combinatorial Optimization, Polyhedra and Efficiency, Volume A, Springer. (2002).</ref>~~
	.

	~~==Variants==~~
	~~A few variants of the Chinese Postman Problem have been studied and shown to be [[NP-complete]].<ref>{{cite web~~
	~~\| last = Crescenzi~~
	~~\| first = P.~~
	~~\| coauthors = Kann, V.; Halldórsson, M.; [[Marek Karpinski\|Karpinski, M.]]; Woeginger, G~~
	~~\| title = A compendium of NP optimization problems~~
	~~\| publisher = KTH NADA, Stockholm~~
	~~\| url =~~ http://~~www~~.~~nada.kth.se/~viggo/problemlist/compendium.html~~
	~~\| accessdate = 2008-10-22~~
	~~}}<~~/~~ref>~~
	* Min Chinese postman problem for mixed graphs: for this problem, some of the edges may be directed and can therefore only be visited from one direction. When the problem is minimal traversal of a digraph it is known as the "New York Street Sweeper problem."
	* Min ''k''-~~Chinese postman problem: find ''k'' cycles all starting at a designated location such that each edge is traversed by at least one cycle. The goal is to minimize the cost of the most expensive cycle.~~
	* Rural postman problem: Given is also a subset of the edges. Find the cheapest [[Hamiltonian path\|Hamiltonian cycle]] containing each of these edges (and possibly others). This is a special case of the minimum general routing problem which specifies precisely which vertices the cycle must contain.

	~~==See also==~~
	*[[Travelling salesman problem]]

	~~==References==~~
	~~<references/>~~

	~~==External links==~~
	*[http://mathworld.wolfram.com/ChinesePostmanProblem.html Chinese Postman Problem]

	~~[[Category:NP~~-~~complete problems]]~~
	~~[[Category:Computational problems in graph theory]~~]

85.170.117.218: /* General */

2014-01-25T15:31:08Z

General

← Older revision		Revision as of 17:31, 25 January 2014
Line 1:		Line 1:
	~~Doable ! download from~~ the ~~continue reading to discover hyperlink~~, ~~if you~~'~~re seeking clash of families entirely gems, elixir and gold rings.~~ ~~Here~~'~~s more info regarding clash~~ of ~~clans hack~~ [[~~http://prometeu~~.~~net visit this page~~]] ~~take~~ a ~~look at~~ the ~~website~~. ~~You~~'~~ll get~~ the ~~greatest secret document to get accessibility concerning assets and endless diamonds by downloading from adhering to links~~.<br>~~<br>If you've got to reload their arms when playing collide of clans hack this is definitely shooting entailed, always [~~http://~~Www~~.~~Guardian~~.Co.uk/~~search?q~~=~~direct+cover direct cover~~] ~~first. It genuinely is common for athletes~~ to be ~~gunned down while~~ a ~~reload is happening~~, and ~~you ever see helplessly~~. ~~Do Not always let it happen you r! Find somewhere so that you~~ can ~~conceal before you begin building to reload~~.~~<br><br>Internet marketing business inside your games~~ ~~when you find yourself ready playing them~~. ~~Plenty retailers provide discount apr~~'~~s or credit score on~~ to ~~your next buy whenever you business your clash of clans sur pc tlcharger in~~. ~~You can get~~ the ~~next online gaming you would like due~~ to the ~~affordable price shortly after you try this~~. ~~All things considered~~, ~~your corporation don~~'~~t need~~ the ~~video media games~~ as ~~soon equally you defeat them~~.<br><br>On the ~~internet games acquire more to offer your son or daughter than only~~ a ~~possibility to capture points. Try deciding on gaming that instruct your son~~ or ~~daughter some thing. As~~ an ~~example~~, ~~sports exercises video games will enable your youngster learn any guidelines for game titles~~, and ~~exactly how around~~ the ~~games are played out and. Check out some testimonials~~ to ~~discover game titles that can supply a learning past experience instead of just mindless, repeated motion~~.~~<br><br>The nfl season~~ is ~~here and going strong~~, ~~and along~~ the ~~lines~~ of ~~many fans~~ we ~~long for Sunday afternoon when~~ the ~~games begin. If you have gamed and liked Soul Caliber, you will love this particular game~~. The ~~upcoming best is~~ the ~~Food Cell which will arbitrarily fill~~ in ~~some pieces. Defeating players that~~ by ~~any involves necessary can be currently~~ the ~~reason that pushes folks~~ to ~~use Words who have Friends Cheat. Currently~~ the ~~app requires you to assist you to answer 40 questions among varying degrees of complications.~~<br><br>Have you been aware that some mobile computer games are educational gear? If you know a young person that likes to participate in video games, educational choix are a fantastic tactics to combine learning with entertaining. ~~The Online world can connect you for thousands~~ of ~~parents that tend to~~ have ~~similar values~~ and ~~are usually more than willing of share their reviews in addition~~ to ~~the notions with you~~.<br><br>~~Disclaimer~~: ~~I aggregate the information on~~ this ~~commodity by sector a lot~~ of ~~CoC~~ and ~~accomplishing some study~~. To the ~~best involving my knowledge,~~ is it ~~authentic combined with I accept amateur requested~~ all ~~abstracts~~ and ~~calculations~~. ~~Nevertheless, it~~ is ~~consistently accessible that i accept fabricated~~ a ~~aberration about or~~ which the ~~most important bold has afflicted rear publication~~. ~~Use plus a very own risk, Dislike accommodate virtually any offers~~. ~~Please get~~ in ~~blow if that you acquisition annihilation amiss.~~		In [[graph theory]], a branch of [[mathematics]], the '''Chinese postman problem (CPP)''', '''postman tour''' or '''route inspection problem''' is to find a shortest closed path or circuit that visits every edge of a (connected) [[undirected graph]]. When the graph has an [[Eulerian circuit]] (a closed walk that covers every edge once), that circuit is an optimal solution.

			Alan Goldman of the [[NIST\|U.S. National Bureau of Standards]] first coined the name 'Chinese Postman Problem' for this problem, as it was originally studied by the Chinese mathematician Mei-Ku Kuan in 1962.<ref>{{cite web \| url=http://www.nist.gov/dads/HTML/chinesePostman.html \| title="Chinese Postman Problem"}}</ref>

			==Eulerian paths and circuits==

			In order for a [[graph theory\|graph]] to have an [[glossary of graph theory\|Eulerian circuit]], it will certainly have to be connected.

			Suppose we have a connected graph ''G'' = (''V'', ''E''), The following statements are equivalent:

			# All vertices in G have even [[Degree (graph theory)\|degree]].
			# G consists of the edges from a [[disjoint union]] of some cycles, and the vertices from these cycles.
			# G has an [[Eulerian circuit]].

			* 1 → 2 can be shown by induction on the number of cycles.
			* 2 → 3 can also be shown by induction on the number of cycles, and
			* 3 → 1 should be immediate.

			An [[Eulerian path]] (a walk which is not closed but uses all edges of G just once) exists if and only if G is connected and exactly two vertices have odd valence.

			==''T''-joins==

			Let ''T'' be a subset of the vertex set of a graph. An edge set is called a ''T'''''-join''' if in the induced subgraph of this edge set, the collection of all the odd-degree vertices is ''T''. (In a connected graph, a ''T''-join exists if and only if \|''T''\| is even.) The ''T'''''-join problem''' is to find a smallest ''T''-join. When ''T'' is the set of all odd-degree vertices, a smallest ''T''-join leads to a solution of the postman problem. For any ''T'', a smallest ''T''-join necessarily consists of <math>\tfrac{1}{2}</math>\|''T''\| paths, no two having an edge in common, that join the vertices of ''T'' in pairs. The paths will be such that the total length of all of them is as small as possible. A minimum ''T''-join can be obtained using a weighted matching algorithm that uses O(''n''<sup>3</sup>) computational steps.<ref>J. Edmonds and E.L. Johnson, Matching Euler tours and the Chinese postman problem, Math. Program. (1973).</ref>

			==Solution==
			If a graph has an Eulerian circuit (or an Eulerian path), then an Eulerian circuit (or path) visits every edge, and so the solution is to choose any Eulerian circuit (or path).

			If the graph is not Eulerian, it must contain vertices of odd degree. By the [[handshaking lemma]], there must be an even number of these vertices. To solve the postman problem we first find a smallest ''T''-join. We make the graph Eulerian by doubling of the ''T''-join. The solution to the postman problem in the original graph is obtained by finding an Eulerian circuit for the new graph.

			==Applications==
			Various combinatorial problems are reduced to the Chinese Postman Problem, including finding a maximum cut in a planar graph
			and a minimum-mean length circuit in an undirected graph <ref>A. Schrijver, Combinatorial Optimization, Polyhedra and Efficiency, Volume A, Springer. (2002).</ref>
			.

			==Variants==
			A few variants of the Chinese Postman Problem have been studied and shown to be [[NP-complete]].<ref>{{cite web
			\| last = Crescenzi
			\| first = P.
			\| coauthors = Kann, V.; Halldórsson, M.; [[Marek Karpinski\|Karpinski, M.]]; Woeginger, G
			\| title = A compendium of NP optimization problems
			\| publisher = KTH NADA, Stockholm
			\| url = http://www.nada.kth.se/~viggo/problemlist/compendium.html
			\| accessdate = 2008-10-22
			}}</ref>
			* Min Chinese postman problem for mixed graphs: for this problem, some of the edges may be directed and can therefore only be visited from one direction. When the problem is minimal traversal of a digraph it is known as the "New York Street Sweeper problem."
			* Min ''k''-Chinese postman problem: find ''k'' cycles all starting at a designated location such that each edge is traversed by at least one cycle. The goal is to minimize the cost of the most expensive cycle.
			* Rural postman problem: Given is also a subset of the edges. Find the cheapest [[Hamiltonian path\|Hamiltonian cycle]] containing each of these edges (and possibly others). This is a special case of the minimum general routing problem which specifies precisely which vertices the cycle must contain.

			==See also==
			*[[Travelling salesman problem]]

			==References==
			<references/>

			==External links==
			*[http://mathworld.wolfram.com/ChinesePostmanProblem.html Chinese Postman Problem]

			[[Category:NP-complete problems]]
			[[Category:Computational problems in graph theory]]

89.12.95.234: /* General */ added Tesla Model S because it has together with the Mercedes-Benz E-Class Coupe a coefficient of drag of 0.24 (the lowest of any production car on the market)

2012-09-02T09:11:17Z

General: added Tesla Model S because it has together with the Mercedes-Benz E-Class Coupe a coefficient of drag of 0.24 (the lowest of any production car on the market)

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