Triangle-Free Planar Graphs as Segment Intersection Graphs

Triangle-Free Planar Graphs as Segment

Intersection Graphs

Natalia de Castro Francisco Javier Cobos Juan Carlos Dana

Alberto M′a rquez

Departamento de Matem′a tica Aplicada I

Universidad de Sevilla

https://www.doczj.com/doc/087913704.html,.es/

{natalia,cobos,dana,almar}@us.es

Marc Noy?

Departamento de Matem′a tica Aplicada II

Universitat Polit`e cnica de Catalunya

http://www.upc.es/

noy@ma2.upc.es

Abstract

We prove that every triangle-free planar graph is the intersection graph of a set of segments in the plane.Moreover,the segments can be chosen in

only three directions(horizontal,vertical and oblique)and in such a way

that no two segments cross,i.e.,intersect in a common interior point.This

particular class of intersection graphs is also known as contact graphs.

Communicated by H.de Fraysseix and and J.Kratochv′?l:

submitted October1999;revised February2001and July2001.

N.de Castro et al.,Segment Intersection Graphs ,JGAA ,6(1)7–26(2002)81Introduction

Given a set S of segments in the plane,its intersection graph has a vertex for every segment and two vertices are adjacent if the corresponding segments intersect.Intersection graphs of segments and other geometrical objects have been widely studied in the past.For instance,if the segments are contained in a straight line then we have the interval graphs [5],a well-known family of perfect graphs.If the segments are chords of a circle then the intersection graph is called a circle graph ,see for instance [8,10].

An interesting result,due to de Fraysseix,Osona de Mendez and Pach [4],and independently to Ben-Arroyo Hartman,Newman and Ziv [2],says that every planar bipartite graph is the contact graph (a particular case of intersection graph,where no two segments cross)of a set of horizontal and vertical segments and,in a di?erent formulation,Tamassia and Tollis proved in [11]almost the same result.On the other hand,it is known that the recognition of such graphs is an NP-complete problem (see [6],[7]and recently [3]).

This result provides a partial answer to a question of Scheinerman [9]:Is every planar graph the intersection graph of a set of segments in the plane?

The main result in this paper,which is a signi?cant extension of [4],is that every triangle-free planar graph is the intersection graph of a family of segments.Moreover,the segments can be drawn in only three directions and in such a way that they do not cross,i.e.the segments do not have any interior point in common.This particular class of intersection graphs is also known as contact graphs.We call such a representation a segment representation of the

graph.

h1h2h2h5

h6

h3h4v1

v1v2v3v4v5v6v7o1o2o3Figure 1:A segment representation of a planar graph.

Our technique is similar in spirit to the one used in [2]for bipartite planar graphs with some transformations.A key point in our proof is Gr¨o tzsch’s The-orem [12],which guarantees that every planar triangle-free graph is 3-colorable.

N.de Castro et al.,Segment Intersection Graphs,JGAA,6(1)7–26(2002)9 The sketch of the proof is as follows.Fixing an embedding of a given triangle-free planar graph G,adding new vertices and(induced)paths between the vertices of the graph we can obtain a new triangle-free graph embedded in the plane which is a subdivision of a3-connected graph.Starting with a3-coloring,a segment representation in three directions is obtained for this new graph using several technical lemmas.Finally,removing the dummy vertices and paths,we obtain a segment representation of the graph G.The three directions considered are horizontal,vertical and oblique(parallel to the bisector of the second quadrant of the plane).

2Pseudo-convex faces with three directions

In this section we present some preliminaries and we indicate the basic tech-niques that are used.

Let G be a planar graph,and consider a?xed embedding of G in the plane. The embedding divides the plane into a number of regions that we call faces of the graph.Let I G be a segment representation of G.We assume throughout the paper that no segment continues beyond its last intersection with a segment in another direction.The segment representation also partitions the plane into regions which we call faces of the representation,whose are simple n-polygons (if n≥4,there is no pair of non-consecutive edges sharing a point).

The proof of our main result is based on adapting the concept of“convex polygon”for segments in three directions.

A segment path of length n is a sequence of horizontal,vertical and oblique segments P={s1,...,s n}such that segment s i intersects s i?1and s i+1,for 1

We consider monotone paths of the segment representation with respect to the following two lines:the bisector of the second quadrant of the plane (l1={x+y=0})and the line(l2={x?2y=0}).

Since we can draw a monotone path rightward or leftward,we distinguish four di?erent kinds of monotone paths:

?A path P is an increasing monotone path to the right(IMPR)if its hor-izontal segments are drawn rightward,its vertical segments are drawn upward,and its oblique segments are drawn rightward and downward.

?A path P is a decreasing monotone path to the right(DMPR)if its hor-izontal segments are drawn rightward,its vertical segments are drawn downward,and its oblique segments are drawn rightward and downward.

?A path P is a decreasing monotone path to the left(DMPL)if its horizontal segments are drawn leftward,its vertical segments are drawn downward, and its oblique segments are drawn leftward and upward.

N.de Castro et al.,Segment Intersection Graphs,JGAA,6(1)7–26(2002)10?A path P is an increasing monotone path to the left(IMPL)if its horizontal segments are drawn leftward,its vertical segments are drawn upward,and its oblique segments are drawn leftward and

upward.

n

Figure2:Monotone paths.

Let I G be a segment representation of a plane triangle-free graph G.Recall that the face boundaries of the representation correspond to closed walks in G.

A face F of a segment representation is pseudo-convex if its boundary can be divided into an IMPR,a DMPR,a DMPL and an IMPL(clockwise).It can be seen that the partition of a pseudo-convex face into monotone paths is not unique,but there are only a few possibilities and this does not modify the proofs of the theorems.From now on,we consider the boundary of a pseudo-convex face clockwise.

Pseudo-convex Face Non Pseudo-convex Face

Figure3:The four paths of a pseudo-convex face.

Given a pseudo-convex face,we de?ne the upper subpath as the union of its IMPR and its DMPR and,the lower subpath as the union of its DMPL and IMPL subpath.Analogously,we de?ne the right subpath as the union of the DMPR and DMPL,and the left subpath as the union of the IMPL and IMPR. In this way the boundary of a pseudo-convex face can be considered as the

N.de Castro et al.,Segment Intersection Graphs,JGAA,6(1)7–26(2002)11 union of the upper and lower subpaths,and also as the union of the right and left subpaths.We say that the upper and lower subpaths are opposite subpaths to each other(analogously with the right and the left subpaths).

Figure4:The upper,lower,right and left subpaths of a pseudo-convex face.

Note that if F is a pseudo-convex face,and P is a monotone path inside F connecting two segments in its boundary,then P partitions F into two new pseudo-convex faces.Moreover,if two segments on the boundary of F can be extended inside F until they cross,those extended segments also would split F into two new pseudo-convex faces.

In order to build a segment representation of a plane graph G,we proceed by representing the boundary of the outerface of G by contact segments,using a pseudo-convex representation.If two vertices of the outerface are joined in G by a path,we represent this path as a monotone segment path inside the representation of the outerface,obtaining two pseudo-convex faces.Recursively we represent monotone paths of G until the representation of the graph is com-pleted.

We see then that in order to represent a graph by contact segments it is necessary to join segments by segment paths inside the faces of the represen-tation.We need to extend segments of the boundary of the faces inside them to contact the segments.However it is not always possible to extend segments inside a pseudo-convex face.A segment of a pseudo-convex is extensible along one of its ends if and only if its angle inside the face is concave.Thus,a segment of a pseudo-convex face is extensible by one of its ends,or by both or by none, depending on the adjacent segments(see Figure5).

N.de Castro et al.,Segment Intersection Graphs ,JGAA ,6(1)7–26(2002)12Non-extensible Extensible by one end Extensible by both ends

Figure 5:Extensible segments.

It is obvious that in order to join two segments inside a pseudo-convex face,one of them must be extended,but this is not su?cient in general as we can see in Figure 6:segment r 2can be extended inside F ,but we can not join it with r 1because r 2intersects other segments of the face.The segment r 3can be also extended inside the face,but we cannot join r 1with r 3because they never contact.

r 1

2

3r r Figure 6:Extension of segments.

According to the above remark,we have two di?erent problems when we try to join segments inside a pseudo-convex face;when the segments can be extended inside the face,but they ?nd some “obstacle”before the intersection (other segments of the face),and on the other hand when the segments cannot be extended or they never contact.

To solve these problems we de?ne two basic operations in the segment repre-sentation.The ?rst operation is enlarging a segment of the representation,and the second one is changing the sense of drawing of some segments.Although we are interested in some speci?c segments,these operations will transform other segments of the representation.

N.de Castro et al.,Segment Intersection Graphs,JGAA,6(1)7–26(2002)13 Enlarging a segment means to increase its length,and obviously we also need to increase the length of other segments of the representation and make a translation of other segments without modifying their length.With this op-eration we can keep away the obstacles and to overlap parallel segments,if we want to join them by a segment path.The other operation allows us to extend a segment by one of its ends,if it was not possible before.It changes the sense of drawing of its adjacent segment and the other segments of its monotone path, to preserve the pseudo-convexity of the face.

Figure7:Transforming the segment representation.

The following technical lemmas allow us to construct a topological equivalent representation where we modify some segments,allowing to join the segments by a path of any length.

Lemma1Let I G be a segment representation of a subdivision of a3-connected plane graph G such that all its faces are pseudo-convex,let k be a positive real number and let s be a segment on the boundary of a face F of I G.Then I G can be transformed into another segment representation

I G is topologically equivalent to I G(that is,the faces of I G and

I G are pseudo-convex;

2.it is possible to enlarge s(and at most two segments on the opposite subpath

of F),such that the length of I G is k+l.

N.de Castro et al.,Segment Intersection Graphs ,JGAA ,6(1)7–26(2002)14Proof:

The proof is by induction on the number of faces of the graph G .Firstly,we suppose that F is just the boundary of I G ,and let s be the segment of F that we want to enlarge.

We can distinguish three cases depending on the direction of s .

Case 1Let s be a horizontal segment.

If s is a horizontal segment in the upper subpath of F (analogously if s is in the lower subpath)we look for a horizontal segment s ′in the lower subpath of F .If there exists such segment we enlarge s and s ′by the same amount and we make a translation of all the segments between s and s ′,transforming F into another pseudo-convex face (see Figure 8(a)).

If there does not exist a horizontal segment s ′in the lower subpath of F ,there must exist a vertical segment s ′and an oblique segment s ′′which are adjacent in the lower subpath of F ,because the boundary of F is a closed path.In this case,we increase the length of s ,s ′and s ′′and we make a translation of the rest of the segments (see Figure 8(b)).

s s's s's s

s'

s''s'

s''

(a)(b)Figure 8:How to enlarge a segment of a pseudo-convex face by any amount,using parallel segments (a),using three segments (b).

Case 2If s is a vertical segment,the proof is as Case 1,considering the bound-

ary of F as the union of the right and left subpaths.

Case 3If s is an oblique segment,the proof is similar to the other cases,con-

sidering the boundary of F as the union of the upper and lower subpaths.

N.de Castro et al.,Segment Intersection Graphs ,JGAA ,6(1)7–26(2002)15

Suppose that G has two bounded faces and let us denote by u 1,...,u n the vertices of G which are in both faces.Let {s 1,...,s n }be the segments of I G corresponding to u 1,...,u n ,respectively,and let us denote by F 1and F 2the boundary of the bounded faces in I G .F F s s s s s'

s

1

23

4

12Figure 9:The adjacent segments are s 1,...,s 4.

Let s be the segment of F 1we want to enlarge.By considerations given above,this transformation will enlarge other segments of F 1,but if none of these segments belong to F 2,we can reduce to Case 1because the transformations does not modify F 2.

If we have selected two (or three)segments in F 1to enlarge them,they will modify F 2if and only if F 1and F 2share more than one point of those segments.Thus,we can suppose that F 1∩F 2∩s 1is not just one point,and neither is F 1∩F 2∩s n .Otherwise,we do not consider the segments s 1and s n in the subpath {s 1,...,s n }.According to that,the path {s 1,...,s n }is monotone because the faces of I G are pseudo-convex.

Suppose that s is in F 1.As in the case where G had only one bounded face,in the opposite subpath of F 1we look for a parallel segment or two contiguous segments non-parallel to s .If these segments are not in F 2we only modify F 1according to the ?rst case.

Let us suppose that s belongs to F 1and we have found a parallel segment s ′in the opposite part of F 1.Since F 1∩F 2is a monotone path and the segments are in opposite parts of F 1,just one of them belongs to F 2.Now,the task is to ?nd a parallel segment in the opposite part of F 2,and we are sure that this new segment does not belongs to F 1,so enlarging all selected segments,a new representation I G ′equivalent to I G is obtained.

The same proof remains valid if we choose two contiguous segments s ′and s ′′in F 1and they belong also to F 2.

Suppose now that I G has n bounded pseudo-convex faces with boundaries F 1,...,F n ,and let s be a segment in the boundary of a face F .Since G is a subdivision of a 3-connected graph,there exists a face (for example F n )adjacent

N.de Castro et al.,Segment Intersection Graphs ,JGAA ,6(1)7–26(2002)16to the unbounded face (i.e.,the boundary of the face in G corresponding to F n has an edge in the unbounded face H of G )and such that s is not in F n ∩H .Thus,if we construct the graph G 1as the graph obtained from G by deleting all the edges of the face of G corresponding to F n which are contained in the outerface,then we can assure that s belongs to G 1.

G

2

v

G 1

I I G G 1

s 1

23

45s s s s

Figure 10:Induction.

Let v 1,...,v k be the vertices of G ,corresponding to the segments s 1,...,s k of F n ,which are also on the outerface,where v 2,...,v k ?1are of degree two.By removing the segments of I G corresponding to the vertices v 2,...,v k ?1,we obtain a segment representation I G 1of G 1.

Since I G 1has n ?1faces,enlarging in any amount some of the segments,we can transform its segment representation into another one which has the same convex faces.To obtain this new representation,we have to enlarge other segments in I G 1besides s ,but we have preserved the structure of the segments in the representation.So,we are able to represent again the segments corre-sponding to v 2,...,v k ?1enlarging the length of one (or two)of them,if it is necessary,to obtain a segment representation of G .2

We can see,using Lemma 1,that by enlarging some segments,a segment representation with pseudo-convex faces can be transformed into another one preserving the topology of the embedding.But this is not su?cient to join segments inside a face,so we need another kind of transformation.

N.de Castro et al.,Segment Intersection Graphs ,JGAA ,6(1)7–26(2002)17

Changing the sense of drawing of the segments of a monotone path of a pseudo-convex face F (for instance,the segments which were drawn to the left are drawn to the right)it is possible to draw a new pseudo-convex face F only change in the length of some of the segments (see Figure 11).123424F

F s 1

s s s s 3

s s s Figure 11:How to restructure a face through a path.

In the segment representation,the change of the face F to

I G satisfying:

1.

I G

have the same facial walks),and the faces of

F corresponding to F has been restructured through a monotone

path.

Proof:We ?rst suppose that F is just the boundary of I G .It is easily seen that we can change the drawing of the segments so that they belong to the previous or the following monotone path.

Suppose now that I G has n bounded pseudo-convex faces.Let F be the face that we want to restructure through the monotone path s 1,...,s n .It is neces-sary to change the sense of drawing of all the segments of the restructured path,thus the other faces that contain in their boundaries the segments s 1,...,s n will be also restructured,but they will preserve their pseudo-convexity.

The main idea is that no other face of the representation will be modi?ed,except for the length of some segments,but by Lemma 1it is possible to increase the length of the segments of the representation preserving the pseudo-convexity.

N.de Castro et al.,Segment Intersection Graphs,JGAA,6(1)7–26(2002)18 In fact,we can consider the faces adjacent to F.The union of these faces determines a region of the plane,R,whose boundary is also a pseudo-convex face.Since restructuring F only modi?es the segments in the interior of R but never in its boundary,the faces of the representation that are not adjacent to F will keep their structure.

For instance,in Figure12,we have restructured the face F and this trans-formation forces restructuring the shaded face.Other faces of the segment representation have been transformed in the length of some segments of their boundaries,but by Lemma1they preserve their pseudo-convexity.

Figure12:How to restructure the faces adjacent to the restructured face F.

2 3Triangle-free graphs

We need the following result proved by Barnette[1]:

Theorem1If G is a subdivision of a3-connected graph,there exists a family Q1,...,Q n?1of paths in G such that the family of subgraphs de?ned as G1=G, G2=G1?Q1,...,G n=G n?1?Q n?1satis?es the properties

1.Each G k is a subdivision of a3-connected graph.

2.G n is a subdivision of K4.

Moreover,if we?x a subgraph of G which is a subdivision of K4,G can be reduced,by deleting edges and paths,to this subgraph.

An easy consequence is that every planar triangle-free graph which is a sub-division of a3-connected graph can be reduced,by deleting edges and paths with internal vertices of degree two,to a subdivision of a?xed complete subgraph K4 in such a way that in each step we have a subdivision of a3-connected graph.

N.de Castro et al.,Segment Intersection Graphs ,JGAA ,6(1)7–26(2002)19This result will allow us to build a segment representation in three directions of subdivisions of 3-connected graphs as follows.Firstly we build a segment rep-resentation of a subdivision of K 4and then insert,in reverse order,the edges and paths that were removed to obtain from G the subdivision of K 4.

In order to do this,we need two basic operations:(1)insert a segment path between two segments of the same face and (2)join two segments of the same face.The second operation is,at least,as di?cult as the ?rst one,so we will concentrate only in the second operation.

As we see in Figure 6,we can ?nd some problems in the segment repre-sentation to join two segments inside a face.Most of the cases can be solved by changing the drawing of the segments of the representation with the basic operations “enlarge segments”and “restructure faces”,but there are four rare cases that need an special operation.

We need at this point some new de?nitions.Given a 3-colored plane graph G with colors {h,o,v },a path P ={u 1,...,u n }of G is rare if it veri?es one of the following conditions:

Case 1.u 1is colored as h ,u 2as o ,and u n as v .

Case 2.u 1is colored as v ,u n ?1as h ,and u n as o .

Case 3.u 1is colored as h ,u 2as o ,u n ?1as h ,and u n as o .

Case 4.u 1is colored as o ,u 2as v ,u n ?1as v ,and u n as h .

s s v v 12

n n-1v 1w 1

r 1w 2r 2

w n r n s n

w n-1

Figure 13:Rare face (four rare paths).

A face of a graph is called rare if there exists a monotone path containing a rare subpath (see Figure 13).We say that a rare face is repaired if we subdivide its rare path with a new vertex colored as v between u 1and u 2in the ?rst and the third case,and between u n ?1and u n in the second case,or a new vertex colored as h between u 1and u 2in the fourth case.

N.de Castro et al.,Segment Intersection Graphs,JGAA,6(1)7–26(2002)20 In the following lemma,when we say that two segments s and t can be joined by a segment path of length k,we mean a path{s,s1,s2,...,s k,t}.

Lemma3Let I G be a segment representation of a plane graph G and let s1and s2be two distinct non adjacent segments in a pseudo-convex non-rare face F. Then I G can be transformed into another segment representation

I G have the same faces and face boundaries as G,and the faces of

F,corresponding to F.

Proof:The proof falls naturally into two cases:the segments have the same direction,or they do not.

Figure14:Joining segments

Case1:s1and s2do not have the same direction.

Suppose the segments cannot be joined inside F directly.The lines containing

N.de Castro et al.,Segment Intersection Graphs ,JGAA ,6(1)7–26(2002)21s 1and s 2indicate us which end has to be extended for the segments to intersect.Let us mark these ends.

1.At least one of the segments is extensible by the marked end but it in-tersects other segments of F (Figure 14(b)).Then we can enlarge some segments of F using Lemma 1and save the obstacle.

2.No segment is extensible by the marked end (Figure 14(c)).Then we have to restructure the face through the path containing s 1(or s 2)to make it extensible.Note that there exists exactly four con?gurations where the segments are not extensible and restructuring the face does not solve the problem,but these are the rare paths which are excluded.

S S 1

2

Non pseudo-convex face Non pseudo-convex face

Figure 15:Segments s 1and s 2cannot be adjacent.

We need that s 1and s 2are non adjacent segments because we cannot join them by a segment path inside F if their angle inside F is concave.Note that in this case F will split into two faces,but one of them will not be pseudo-convex (see Figure 15).

Case 2:s 1and s 2have the same direction.

In this case,it su?ces to join the segments by a segment path of length one,because this path can be substituted by any path of length greater than one.

Again there are two subcases:

1.If the segments belong to opposite subpaths (they are drawn in opposite sense).By Lemma 1,we can enlarge s 1and s 2until there exists a perpen-dicular line that crosses both segments,and a segment path of any length can be drawn inside the face.

N.de Castro et al.,Segment Intersection Graphs ,JGAA ,6(1)7–26(2002)22

(a)

(b)

s s 1

2

s s 12s s 12s s 12

Figure 16:Joining segments.

2.If the segments are drawn in the same sense and they cannot be extended,we have to restructure the face through the path containing s 1(or s 2)to make it extensible.

2

The interest of these Lemmas is that they allow us to proof the following.Lemma 4Any triangle-free plane graph which is a subdivision of a 3-connected graph can be represented by segments in three directions.

Proof:

As G is a triangle-free plane graph,in virtue of Gr¨o tzsch’s Theorem [12],G admits a 3-coloring with the colors h ,v and o (horizontal,vertical and oblique,respectively).Since G is a subdivision of a 3-connected graph,we can ?nd a vertex,not in the outerface of G ,connected by three disjoint paths to three vertices in the outerface.These paths and the outerface determine a graph K ,which is a subdivision of K 4.

Using Theorem 1[1]it is possible to build a sequence of graphs G 1,...,G n and a sequence of paths Q 1,...,Q n ?1such that G 1=G ,G i is a subdivision of a 3-connected plane graph,G i is obtained from G i ?1by deleting the path Q i ?1,and the graph obtained from G n ?1deleting Q n ?1is K .

When the path Q i is deleted,a new face F appears in G i +1as the union of two faces in G i .The boundary of F can be divided into two paths;on the one hand the path beginning in the ?rst vertex of Q i and ending in the last one and,on the other hand,the rest of the boundary.If one of them is rare,we must

N.de Castro et al.,Segment Intersection Graphs ,JGAA ,6(1)7–26(2002)23repair F by subdividing an edge with a new vertex (that we label as a repaired vertex),to make the face F non-rare.So,we can construct another sequence G ′1,...,G ′n where G ′i is G i with a new repaired vertex,if necessary.

u u u u Rare face (four rare paths)Repaired face

s s v v 1

2n n-1v 1w 1

r 1w 2r 2

w n r n

s n

w n-1Figure 17:Repairing rare paths.

It is easy to give a segment representation of K ′with all its faces pseudo-convex.It su?ces to represent pseudo-convexly the outerface of K ′and proceed according to the above https://www.doczj.com/doc/087913704.html,ing Lemma 3it is possible to add the path Q i to the segment representation of G ′i +1to obtain a pseudo-convex face repre-sentation of G ′i .Since every subgraph is a subdivision of a 3-connected graph,the path Q i cannot join adjacent vertices and so we can apply Lemma 3.

In order to obtain a segment representation of the graph G ,we must remove the repaired vertices.These segments cannot be removed directly,because we must change the drawing of one of the segments adjacent to the repaired vertex,so it is necessary to modify the representation.Since this is the last operation of the representation,and the repaired vertex is not adjacent to any other segment,it is possible to remove it,as illustrated in Figure 18.

Notice that we could not remove the repaired vertices if the rare path would have exactly three segments u 1,u 2and u 3,but this case is not possible because joining u 1with u 3it will be form a triangle,and G was a triangle-free graph.2

We can now formulate our main result as follows.

Theorem 2Every triangle-free planar graph is the contact graph of a set of segments in three directions.

Proof:Let G be a triangle-free plane graph.Since G has no triangles,we can obtain a 3-coloring of G using Gr¨o tzsch’s Theorem [12].The colors will be labeled as h ,v and o (horizontal,vertical and oblique,respectively).We can build a new triangle-free plane graph G 1,subdivision of a 3-connected graph,which contains G as a subgraph,by adding new vertices and edges joining the

N.de Castro et al.,Segment Intersection Graphs,JGAA,6(1)7–26(2002)24

Figure18:How to remove the repaired vertex u in the four cases. blocks of G.If this is the case,these vertices are labeled as dummy vertices. When an added edge produces a triangle,we subdivide it,and when a new edge joins two vertices with the same color,we subdivide it too.We call these new vertices and edges virtual.All the new vertices(dummy and virtual)can be colored so that the3-coloring is preserved.

By Lemma4,G1admits a segment representation.Out of this segment representation we must remove all the vertices and edges added.

A virtual edge(or a path built with virtual edges and vertices)in the segment representation of G1is an adjacency between two segments(or a virtual path joining two segments).It su?ces to break this adjacency and to shorten these segments(note that the segments do not cross,they only contact).The dummy vertices are removed as the repaired vertices in Lemma4(see Figure18).2 4Concluding remarks

The hypothesis that the graph has no triangles can be relaxed in some cases.

No doubt there exist planar graphs with triangles that admit segment rep-

N.de Castro et al.,Segment Intersection Graphs ,JGAA ,6(1)7–26(2002)25resentations (see Figure 1).The problem lies in the fact that we have followed a constructive proof to obtain a segment representation using the pseudo-convex faces representation of a subdivision of a 3-connected graph.This construction would not be possible in general if the graph contains triangles,because if we ?x the three directions of the segments we can observe that the graph in Figure 19cannot be represented by non-crossing

segments.

h v v o 11

v 1h 1222o 2v 2

h 2Figure 19:Two segments must cross.

In this example we see the intersection graph of a family of segments,but this representation by segments of the graph contains regions that do not correspond to faces of the embedding of the graph.So,the topology of the embedding of the plane graph cannot be preserved and inductive arguments are no longer possible.This problem admits a partial solution as follows.A 3-coloring of a plane graph G with colors {h,v,o }is good if all the triangles of G are colored (clockwise),as h ?v ?o .

The proof of Theorem 2can be adapted yielding the following result:

Theorem 3Let G be a 3-colored plane graph.The coloring is good if and only if there exists a segment representation I G verifying that the faces of G correspond to faces of I G ,and the boundaries of the faces are preserved.

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Comb.Theory,Serie B,62:289–315,1994.

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of graphs.Ph D.thesis,,Princeton University,1984.

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数学图形知识点

数学图形知识点 数学图形知识点 一、图形的变换 1、轴对称图形：把一个图形沿着某一条直线对折，两边能够完 全重合，这样的图形叫做轴对称图形，这条直线叫做对称轴。 2、成轴对称图形的特征和性质：①对称点到对称轴的距离相 等;②对称点的连线与对称轴垂直;③对称轴两边的图形大小形状完 全相同。 3、物体旋转时应抓住三点：①旋转中心;②旋转方向;③旋转角度。旋转只改变物体的位置，不改变物体的形状、大小。 二、因数与倍数 1、因数和倍数：如果整数a能被b整除，那么a就是b的倍数，b就是a的因数。 2、一个数的因数的求法：一个数的因数的个数是有限的，最小 的是1，最大的是它本身，方法是成对地按顺序找。 3、一个数的倍数的求法：一个数的倍数的个数是无限的，最小 的是它本身，没有最大的，方法时依次乘以自然数。 4、2、 5、3的倍数的特征：个位上是0、2、4、 6、8的数，都 是2的倍数。个位上是0或5的数，是5的倍数。一个数各位上的 数的和是3的倍数，这个数就是3的`倍数。 5、偶数与奇数：是2倍数的数叫做偶数(0也是偶数)，不是2 的倍数的数叫做奇数。

6、质数和和合数：一个数，如果只有1和它本身两个因数的数叫做质数(或素数)，最小的质数是2。一个数，如果除了1和它本身还有别的因数的数叫做合数，最小的合数是4。 三、长方体和正方体 1、长方体和正方体的特征：长方体有6个面，每个面都是长方形(特殊的有一组对面是正方形)，相对的面完全相同;有12条棱，相对的棱平行且相等;有8个顶点。正方形有6个面，每个面都是正方形，所有的面都完全相同;有12条棱，所有的棱都相等;有8个顶点。 2、长、宽、高：相交于一个顶点的三条棱的长度分别叫做长方体的长、宽、高。 3、长方体的棱长总和=(长+宽+高)×4正方体的棱长总和=棱长×12 4、表面积：长方体或正方体6个面的总面积叫做它的表面积。 5、长方体的表面积=(长×宽+长×高+宽×高)×2 S=(ab+ah+bh)×2 正方体的表面积=棱长×棱长×6 6、表面积单位：平方厘米、平方分米、平方米相邻单位的进率为100 7、体积：物体所占空间的大小叫做物体的体积。 8、长方体的体积=长×宽×高用字母表示：V=abh长=体积÷(宽×高)宽=体积÷(长×高) 高=体积÷(长×宽) 正方体的体积=棱长×棱长×棱长用字母表示：V=a×a×a 9、体积单位：立方厘米、立方分米和立方米相邻单位的进率为1000

小学数学 图形与几何 知识点归纳汇总

小学数学图形与几何知识点归纳汇总 图形与几何 一线和角 （1）线 *直线 直线没有端点；长度无限；过一点可以画无数条，过两点只能画一条直线。 *射线 射线只有一个端点；长度无限。 *线段 线段有两个端点，它是直线的一部分；长度有限；两点的连线中，线段为最短。 *平行线 在同一平面内，不相交的两条直线叫做平行线。 两条平行线之间的垂线长度都相等。 *垂线 两条直线相交成直角时，这两条直线叫做互相垂直，其中一条直线叫做另一条直线的垂线,相交的点叫做垂足。 从直线外一点到这条直线所画的垂线的长叫做这点到直线的距离。 （2）角 （1）从一点引出两条射线，所组成的图形叫做角。这个点叫做角的顶点，这两条射线叫做角的边。 （2）角的分类 锐角：小于90°的角叫做锐角。 直角：等于90°的角叫做直角。 钝角：大于90°而小于180°的角叫做钝角。 平角：角的两边成一条直线，这时所组成的角叫做平角。平角180°。 周角：角的一边旋转一周，与另一边重合。周角是360°。

1长方形 （1）特征 对边相等，4个角都是直角的四边形。有两条对称轴。 （2）计算公式 c=2(a+b)s=ab 2正方形 （1）特征： 四条边都相等，四个角都是直角的四边形。有4条对称轴。 （2）计算公式 c=4a s=a2 3三角形 （1）特征 由三条线段围成的图形。内角和是180度。三角形具有稳定性。三角形有三条高。（2）计算公式 s=ah/2 （3）分类 按角分 锐角三角形：三个角都是锐角。 直角三角形：有一个角是直角。等腰三角形的两个锐角各为45度，它有一条对称轴。钝角三角形：有一个角是钝角。 按边分 不等边三角形：三条边长度不相等。 等腰三角形：有两条边长度相等；两个底角相等；有一条对称轴。 等边三角形：三条边长度都相等；三个内角都是60度；有三条对称轴。

初中数学知识点总结汇总结构图

有理数数轴：数轴是规定了原点、正方向、单位长度的一条直线。 有理数 概念：凡能写成形式的数，都是有理数。（正整数、0、负整数统称整数；正分数、负分数统称分数；整数和分数统称有理数.注意：0即不是正数， 也不是负数；-a不一定是负数，+a也不一定是正数；π不是有理数。） 有理数的分类：①有理数 正有理数 零 负有理数 正整数 正分数 负整数 负分数 ②有理数 整数 分数 正整数 零 负整数 正分数 负分数 相反数 (1)只有符号不同的两个数，我们说其中一个是另一个的相反数；0的相反数还是0； (2)相反数的和为0 ? a+b=0 ? a、b互为相反数。 绝对值：正数的绝对值是其本身，0的绝对值是0，负数的绝对值是它的相反数；注意：绝对值的意义是数轴上表示某数的点离开原点的距离； 有理数比大小 （1）正数的绝对值越大，这个数越大； （2）正数永远比0大，负数永远比0小； （3）正数大于一切负数； （4）两个负数比大小，绝对值大的反而小； （5）数轴上的两个数，右边的数总比左边的数大； （6）大数-小数＞0，小数-大数＜0。 互为倒数：乘积为1的两个数互为倒数；注意：0没有倒数；若a≠0，那么的倒数是； 若ab=1? a、b互为倒数；若ab=-1? a、b互为负倒数。 有理数乘方的法则 （1）正数的任何次幂都是正数； （2）负数的奇次幂是负数；负数的偶次幂是正数；注意：当n为正奇数时: (-a)n=-a n 或(a -b)n=-(b-a)n , 当n为正偶数时: (-a)n =a n 或(a-b)n=(b-a)n . 科学记数法：把一个大于10的数记成a×10n的形式，其中a是整数数位只有一位的数，这种记数法叫科学记数法. 近似数的精确位：一个近似数，四舍五入到那一位，就说这个近似数的精确到那一位。 有效数字：从左边第一个不为零的数字起，到精确的位数止，所有数字，都叫这个近似数的有效数字。 举几个例子：3一共有1个有效数字，0.0003有一个有效数字，0.1500有4个有效数字， 1.9*10^3有两个有效数字（不要被10^3迷惑，只需要看1.9的有效数字就可以了，10^n 看作是一个单位）。 整式的加单项式：在代数式中，若只含有乘法（包括乘方）运算。或虽含有除法运算，但除式中不含字母的一类代数式叫单项式。 单项式的系数与次数：单项式中不为零的数字因数，叫单项式的数字系数，简称单项式的系 数；系数不为零时，单项式中所有字母指数的和，叫单项式的次数。 多项式：几个单项式的和叫多项式。

图形的初步认识知识点

? ? ? ? ? ?图形的初步认识 一、本章的知识结构图 一、立体图形与平面图形 立体图形：棱柱、棱锥、圆柱、圆锥、球等。 1、几何图形 平面图形：三角形、四边形、圆等。 主（正）视图---------从正面看 2、几何体的三视图侧（左、右）视图-----从左（右）边看 俯视图---------------从上面看 （1）会判断简单物体（直棱柱、圆柱、圆锥、球）的三视图。 （2）能根据三视图描述基本几何体或实物原型。 3、立体图形的平面展开图 （1）同一个立体图形按不同的方式展开，得到的平现图形不一样的。 （2）了解直棱柱、圆柱、圆锥、的平面展开图，能根据展开图判断和制作立体模型。 4、点、线、面、体 （1）几何图形的组成 点：线和线相交的地方是点，它是几何图形最基本的图形。 线：面和面相交的地方是线，分为直线和曲线。 面：包围着体的是面，分为平面和曲面。 体：几何体也简称体。 （2）点动成线，线动成面，面动成体。 例1 （1）如图1所示，上面是一些具体的物体，下面是一些立体图形，试找出与下面立体图形相类似的物体。 （2）如图2所示，写出图中各立体图形的名称。 图 1 图2 解：（1）①与d类似，②与c类似，③与a类似，④与b类似。 （2）①圆柱，②五棱柱，③四棱锥，④长方体，⑤五棱锥。 例2 如图3所示，讲台上放着一本书，书上放着一个粉笔盒，指出右边三个平面图形分别是左边立体图形的哪个视图。 图3 解：（1）左视图，（2）俯视图，（3）正视图 练习 1．下图是一个由小立方体搭成的几何体由上而看得到的视图，小正方形中的数字表示该位置小立方块的个数，则从正面看它的视图为（）

人教版七年级上册数学知识结构

一：有理数 知识网络： 正分数 负分数 正整数 负整数 概念、定义： 1、 大于0的数叫做正数（positive number ）。 2、 在正数前面加上负号“-”的数叫做负数（negative number ）。 3、 整数和分数统称为有理数（rational number ）。 4、 人们通常用一条直线上的点表示数，这条直线叫做数轴（number axis ）。 5、 在直线上任取一个点表示数0，这个点叫做原点（origin ）。 6、 一般的，数轴上表示数a 的点与原点的距离叫做数a 的绝对值（absolute value ）。 7、 由绝对值的定义可知：一个正数的绝对值是它本身；一个负数的绝对值是它 的相反数；0的绝对值是0。 8、 正数大于0，0大于负数，正数大于负数。 9、 两个负数，绝对值大的反而小。 10、 有理数加法法则

（1）同号两数相加，取相同的符号，并把绝对值相加。 （2）绝对值不相等的异号两数相加，取绝对值较大的加数的负号，并用较大的绝对值减去较小的绝对值，互为相反数的两个数相加得0。 （3）一个数同0相加，仍得这个数。 11、有理数的加法中，两个数相加，交换交换加数的位置，和不变。 12、有理数的加法中，三个数相加，先把前两个数相加，或者先把后两个数相加， 和不变。 13、有理数减法法则 减去一个数，等于加上这个数的相反数。 14、有理数乘法法则 两数相乘，同号得正，异号得负，并把绝对值向乘。 任何数同0相乘，都得0。 15、有理数中仍然有：乘积是1的两个数互为倒数。 16、一般的，有理数乘法中，两个数相乘，交换因数的位置，积相等。 17、三个数相乘，先把前两个数相乘，或者先把后两个数相乘，积相等。 18、一般地，一个数同两个数的和相乘，等于把这个数分别同这两个数相乘，再 把积相加。 19、有理数除法法则 除以一个不等于0的数，等于乘这个数的倒数。 20、两数相除，同号得正，异号得负，并把绝对值相除。0除以任何一个不等于 0的数，都得0。 21、求n个相同因数的积的运算，叫做乘方，乘方的结果叫做幂（power）。在 a n 中，a叫做底数（base number），n叫做指数（exponeht）

立体图形的知识点整理

立体图形的知识点整理 一、长方体、正方体都有6个面，12条棱，8个顶点。正方体是特殊的长方体。 二、圆柱的特征：一个侧面、两个底面、无数条高。 三、圆锥的特征：一个侧面、一个底面、一个顶点、一条高。 四、表面积：立体图形所有面的面积的和，叫做这个立体图形的表面积。 五、体积：物体所占空间的大小叫做物体的体积。容器所能容纳其它物体的体积叫做容器的容积。 六、圆柱和圆锥三种关系： ①等底等高：体积1︰3 ②等底等体积：高1︰3 ③等高等体积：底面积1︰3 七、等底等高的圆柱和圆锥： ①圆锥体积是圆柱的1/3， ②圆柱体积是圆锥的3倍， ③圆锥体积比圆柱少2/3， ④圆柱体积比圆锥多2倍。 八、等底等高的圆柱和圆锥：锥1、差2、柱3、和4。 九、立体图形公式推导： 【1】圆柱的侧面展开后得到一个什么图形?这个图形的各部分与圆柱有何关系?(圆柱侧面积公式的推导过程)

①圆柱的侧面展开后一般得到一个长方形。 ②长方形的长相当于圆柱的底面周长，长方形的宽相当于圆柱的高。 ③因为：长方形面积=长×宽，所以：圆柱侧面积=底面周长×高。 ④圆柱的侧面展开后还可能得到一个正方形。 正方形的边长=圆柱的底面周长=圆柱的高。 【2】我们在学习圆柱体积的计算公式时，是把圆柱转化成以前学过的一种立体图形(近似的)进行推导的，请你说出这种立体图形的名称以及它与圆柱体有关部分之间的关系? ①把圆柱分成若干等份，切开后拼成了一个近似的长方体。 ②长方体的底面积等于圆柱的底面积，长方体的高等于圆柱的高。 ③因为：长方体体积=底面积×高，所以：圆柱体积=底面积×高。即：V=Sh。 【3】请画图说明圆锥体积公式的推导过程? ①找来等底等高的空圆锥和空圆柱各一只。

初中数学图形与几何知识点

小学数学总复习基础知识 第二单元 空间与图形 （一）图形的认识、测量 1、长度单位是用来测量物体的长度的。常用的长度单位有：千米、米、分米、厘米、毫米。 2、长度单位：（进率10） 3、面积单位是用来测量物体的表面或平面图形的大小的。常用的面积单位有：平方千米、公顷、平方米、平方分米、平方厘米。 4、测量和计算土地面积，通常用公顷作单位。边长100米的正方形土地，面积是1公顷。 5、测量和计算大面积的土地，通常用平方千米作单位。边长1000米的正方形土地，面积是1平方千米。 6、面积单位：（100） 7、体积单位是用来测量物体所占空间的大小的。常用的体积单位有：立方米、立方分米（升）、立方厘米（毫升）。 8、体积单位：（1000） 9、常用的质量单位有：吨、千克、克。 10、质量单位： 11、常用的时间单位有：世纪、年、季度、月、旬、日、时、分、秒。 12、时间单位：（60） 13 、高级单位的名数改写成低级单位的名数应该乘以进率； 低级单位的名数改写成高级单位的名数应该除以进率。 14 1、用直尺把两点连接起来，就得到一条线段；把线段的一端无限延长，可以得到一条射线；把线段的两端无限延长，可以得到一条直线。线段、射线都是直线上的一部分。线段有两个端点，长度是有限的；射线只有一个端点，直线没有端点，射线和直线都是无限长的。 2、从一点引出两条射线，就组成了一个角 。角的大小与两边叉开的大小有关，与边的长短无关。角的大小的计量单位是（°）。 3、角的分类：小于90度的角是锐角；等于90度的角是直角；大于90度小于180度的角是钝角；等于180度的角是平角；等于360度的角是周角。 4、相交成直角的两条直线互相垂直；在同一平面不相交的两条直线互相平行。 5、三角形是由三条线段围成的图形。围成三角形的每条线段叫做三角形的边，每两条线段的交点叫做三角形的顶点。 6、三角形按角分，可以分为锐角三角形、直角三角形和钝角三角形。按边分，可以分为等边三角形、等腰三角形和任意三角形。 7、三角形的内角和等于180度。 8、在一个三角形中，任意两边之和大于第三边。

新人教版小学数学六年级图形与几何知识点

图形与几何 一线和角 （1）线 * 直线 直线没有端点；长度无限；过一点可以画无数条，过两点只能画一条直线。 * 射线 射线只有一个端点；长度无限。 * 线段 线段有两个端点，它是直线的一部分；长度有限；两点的连线中，线段为最短。 * 平行线 在同一平面内，不相交的两条直线叫做平行线。 两条平行线之间的垂线长度都相等。 * 垂线 两条直线相交成直角时，这两条直线叫做互相垂直，其中一条直线叫做另一条直线的垂线,相交的点叫做垂足。 从直线外一点到这条直线所画的垂线的长叫做这点到直线的距离。 （2）角 （1）从一点引出两条射线，所组成的图形叫做角。这个点叫做角的顶点，这两条射线叫做角的边。 （2）角的分类 锐角：小于90°的角叫做锐角。 直角：等于90°的角叫做直角。 钝角：大于90°而小于180°的角叫做钝角。 平角：角的两边成一条直线，这时所组成的角叫做平角。平角180°。 周角：角的一边旋转一周，与另一边重合。周角是360°。 二平面图形 1长方形 （1）特征 对边相等，4个角都是直角的四边形。有两条对称轴。 （2）计算公式 c=2(a+b) s=ab 2正方形 （1）特征：

四条边都相等，四个角都是直角的四边形。有4条对称轴。 （2）计算公式 c= 4a s=a2 3三角形 （1）特征 由三条线段围成的图形。内角和是180度。三角形具有稳定性。三角形有三条高。 （2）计算公式 s=ah/2 （3）分类 按角分 锐角三角形：三个角都是锐角。 直角三角形：有一个角是直角。等腰三角形的两个锐角各为45度，它有一条对称轴。 钝角三角形：有一个角是钝角。 按边分 不等边三角形：三条边长度不相等。 等腰三角形：有两条边长度相等；两个底角相等；有一条对称轴。 等边三角形：三条边长度都相等；三个内角都是60度；有三条对称轴。 4平行四边形 （1）特征 两组对边分别平行的四边形。 相对的边平行且相等。对角相等，相邻的两个角的度数之和为180度。平行四边形容易变形。 （2）计算公式 s=ah 5 梯形 （1）特征 只有一组对边平行的四边形。 中位线等于上下底和的一半。 等腰梯形有一条对称轴。 （2）公式 s=(a+b)h/2=mh

初中数学丰富的图形世界知识点归纳

初中数学丰富的图形世界知识点归纳 集团文件发布号：（9816-UATWW-MWUB-WUNN-INNUL-DQQTY-

第一章丰富多彩的图形总结济宁附中李涛 1、几何图形 从实物中抽象出来的各种图形，包括立体图形和平面图形。 立体图形：有些几何图形的各个部分不都在同一平面内，它们是立体图形。 平面图形：有些几何图形的各个部分都在同一平面内，它们是平面图形。 2、点、线、面、体 （1）几何图形的组成 点：线和线相交的地方是点，它是几何图形中最基本的图形。 线：面和面相交的地方是线，分为直线和曲线。 面：包围着体的是面，分为平面和曲面。 体：几何体也简称体。 （2）点动成线，线动成面，面动成体。点、线、面、体都是几何图形。 任何一个几何体都由点、线、面构成，点无大小，线有曲直而无粗细，平面是无限延伸的，面有平面和曲面，面面相交得线，线线相交得点。 本节拓展习题：将一个平面按一定方式旋转得到什么样的几何体 3、生活中的立体图形 圆柱（圆柱的侧面是曲面，底面是圆） 柱体 生活中的立体图形棱柱：三棱柱、四棱柱（长方体、正方体）、五棱柱、…… （棱柱的侧面是若干个小长方形构成，底面是多边形） (按名称分)锥体圆锥（圆锥的侧面是曲面，底面的圆）

棱锥（棱锥的侧面是若干个三角形构成，底面是多边形） 球体 还有一种分类看是否有曲面：曲面体和多面体。 棱柱与圆柱的异同点相同点：圆柱、棱柱都有（相同的）个底面 不同点：a.圆柱的底面是（圆）形，棱柱的底面是（多边形）形。 b.圆柱的侧面是一个（曲）面，棱柱的侧面是（平行四边形）形 4、棱柱及其有关概念： 棱：在棱柱中，任何相邻两个面的交线，都叫做棱。侧棱：相邻两个侧面的交线叫做侧棱。 n棱柱有两个底面，n个侧面，共（n+2）个面；3n条棱，n条侧棱；2n个顶点。 1.性质：棱柱的所有侧棱长都相等，棱柱的上下两个底面是相同的多边形， 2.分类：1.根据侧棱是否与底面垂直分为直棱柱和斜棱柱。直棱柱的侧面是长方形。斜棱柱的侧面是平行四边形。2.棱柱还可以根据底面多边形的边数（或侧棱的条数）分类的，如：五棱柱说明它有五条侧棱而不是五条棱，它的底面为五边形。 3.将一个图形折叠后能否变成棱柱，一要看有无两个底面，二要看底面的形状，（底面边数要与侧面数相同），三要看两个底面的位置。 补充：（棱柱）的顶点数V、面数F及棱数E间有关系V+F-E=2 这个公式叫欧拉公式 5、正方体的平面展开图：11种 一个正方体的表面沿某些棱剪开，可得到十一种不同的平面图形，这些平面图形经过折叠后又能围成一个正方体，圆柱和圆锥的侧面展开图分别是长方形和扇形。任何一个立体图形的表面沿某些棱剪开都可以得到不同的平面图形，必须提高自己的空间想象力。

初一数学第四章[几何图形初步]知识点汇总

方向教育《几何图形初步》 1

一、知识结构框图 二、具体知识点梳理 （一）几何图形(是多姿多彩的) 平面图形：三角形、四边形、圆等. 1、几何图形立体图形：棱柱、棱锥、圆柱、圆锥、球等. 主（正）视图---------从正面看； 2、几何体的三视图侧（左、右）视图-----从左（右）边看； 俯视图---------------从上面看. （1）会判断简单物体（直棱柱、圆柱、圆锥、球）的三视图. （2）能根据三视图描述基本几何体或实物原型. 3、立体图形的平面展开图 （1）同一个立体图形按不同的方式展开，得到的平面图形不一样的. （2）了解直棱柱、圆柱、圆锥的平面展开图，能根据展开图判断和制作立体模型. 4、点、线、面、体 （1）几何图形的组成 点：线和线相交的地方是点，它是几何图形最基本的图形. 线：面和面相交的地方是线，分为直线和曲线. 面：包围着体的是面，分为平面和曲面. 体：几何体也简称体. （2）点动成线，线动成面，面动成体.

（二）直线、射线、线段 1、基本概念 2、直线的性质 经过两点有一条直线，并且只有一条直线. 简称：两点确定一条直线. 3、画一条线段等于已知线段（1）度量法（2）用尺规作图法 4、线段的大小比较方法（1）度量法（2）叠合法 5、线段的中点（二等分点）、三等分点、四等分点等 5、 定义：把一条线段平均分成两条相等线段的点叫做线段的中点. 图形： 符号：若点M是线段AB的中点，则AM=1/2BM=AB，AB=2AM=2BM. 6、线段的性质：两点的所有连线中，线段最短.简称：两点之间，线段最短. 7、两点的距离：连接两点的线段长度叫做这两点的距离. 8、点与直线的位置关系（1）点在直线上；（2）点在直线外. (三）角 1、角：有公共端点的两条射线组成的图形叫做角，两条射线的公共端点叫做这个角的顶点，这两条射线叫做这个角的边。或：角也可以看成是一条射线绕着它的端点旋转而成的。 2、平角和周角：一条射线绕着它的端点旋转，当终边和始边成一条直线时，所形成的角叫做平角。终边继续旋转，当它又和始边重合时，所形成的角叫做周角。 角的表示： ①用数字表示单独的角，如∠1，∠2，∠3等。 ②用小写的希腊字母表示单独的一个角，如∠α，∠β，∠γ，∠θ等。 ③用一个大写英文字母表示一个独立（在一个顶点处只有一个角）的角，如∠B，∠C 等。

- 青岛版小学数学知识结构脉络图

青岛版小学数学知识结构脉络图 同和小学 魏建 6.常见的量 （1）认识长度、面积、体积、容积、质量、时间等单位和单位间的进率 （2）不同单位的改写 数与运算 数与 代数 比与例比 式与方程 常见的量 1. 数的认识 （1）整数、小数、分数、百分数和负数的意义、读写，认识数的组成、数位和计算单位。 （2）整数、小数、分数、百分数和负数的大小比较。 （3）大数的改写，分数、小数、百分数的互化。 （4）因数和倍数的认识，知道奇数、偶数、合数、质数的概念，会求最小公倍数合作大公因数。 2.数的运算 （1）整数、小数、分数、百分数的四则混合运算算理和计算方法 （2）四则混合运算的顺序和简便计算 （3）用四则混合运算解决问题 3.运算定律和基本性质 （1）认识加法运算定律、乘法运算定律 （2）减法和除法的性质 （3）积、商的变化规律 （4）分数、小数、比和比例的基本性质 4.比与比例 （1）比和比例的认识 （2）比例的基本性质，利用比例的基本性质解比例 （3）正比例和反比例的意义和判断，用正、反比例解决实际问题 （4）比例尺=图上距离：实际距离，比例尺的分类 5.式与方程 （1）用字母表示数、数量关系和公式 （2）方程和等式的意义 （3）等式的基本性质，以及用等式的基本性质解方程 （4）列方程解决问题

平面图形 图形与变换 图形与位置 1.线 （1）认识直线、射线和线段 （2）认识平行与垂直 （3） 图 形 与 几 何 立 体 图 形 2.角 （1）认识角 （2）角的大小和分类 （3）量角和画角 3.多边形的认识 （1）认识三角形，知道三角形的特性、三角形的分类和内角和 （2）认识正方形、长方形 （3）认识平行四边形和梯形的特征 （4）认识圆的各部分组成及相互关系 4.求平面图形的周长和面积 （1）求长方形、正方形、三角形和圆的周长 （2）求三角形、正方形、长方形、平行四边形、梯形和圆的面积 5.立体图形 （1）认识长方体、正方体、圆柱、圆锥的特征 （2）求长方体、正方体、圆柱的表面积 （3）求长方体、正方体、圆柱、圆锥的体积或容积 （8） 6.图形变换 （1）轴对称图形和轴对称变换 （2）平移和旋转现象及作图 （3）图形按比例放大或缩小 （9） 7.位置 （1）认识8个方向 （2）用方向和距离确定物体的位置 （3）用数对确定物体的位置 （10）