The Homeless Traveling Salesman: Investigating Quadrilaterals with a Variable Home Point

Transcription

1 Te Homeless Traveling Salesman: Invesigaing Quadrilaerals wi a Variable Home Poin Lawrence Garcia Dana Tompson Meropolian Sae College of Denver Nor Park Universiy Te radiional raveling salesman problem involves minimizing e oal disance raveled for a salesman wo needs o visi a given number of ciies (visiing eac ciy only once) and en reurn ome. We examined an exension of e problem by allowing e posiion of e ome poin o vary. We fixed four poins, represening ciies, on a Caresian grid and allowed e coordinaes of e ome poin o cange. For eac posiion, as e ome poin moves rougou e plane, e salesman as several possible pas from wic o coose. Te pa wi e sores leng is deermined and called e bes pa for a ome poin. We assigned eac possible bes pa a color and en colored eac ome poin according o is bes pa, creaing colored regions on e grap. A Malab program (see appendix) was used o creae colored graps for differen arrangemens of e four ciies. We cose o examine cases a involved placing e four poins on e grid o form cerain quadrilaerals including a rombus, a kie, and a rapezoid. Melissa Dejarlais proved in er unpublised 999 SUMSRI paper a ere are four, five, or six disinc bes pas for any configuraion of four disinc ciies. For eac of our quadrilaerals, we examined e condiions necessary for a fif or six color o be presen in e grap. Rombus Consider a rombus ABCD in wic e verices ave e following coordinaes: (0, ), (-, 0), (0, -), and (, 0), were > 0 and > 0. Wen = and =, e rombus is a square and e colored regions are elemenary, eac occupying one quadran of e plane and all meeing a e origin. A B D C Figure : =, =

2 Varying or sligly wile keeping e oer consan also yields suc fourcolor figures. However, once is sufficienly large, a fif color (green) comes in on bo sides of e rombus, as can be seen in Figure wen = 6, four imes e value of. Similarly, wen is sufficienly small, a fif color (black) comes in above and below e rombus. In Figure 3 wen =, one-four e value of, e black region is simply a 90 roaion and scaling of e green region. Figure : =, = 6 Figure 3: =, = Te grap below indicaes e number of colors presen for values of and ranging from 0. o 0. A red poin indicaes e presence of five colors for a given and, and a blue poin indicaes e presence of four colors Figure : versus As can be seen, for any given value of, if is sufficienly large or small, ere will be a fif color. Te lineariy and symmery (abou e line = ) of e grap lead us o believe a ere is a criical raio of o (and vice versa) a deermines weer or no a fif color is presen. Once we know is criical raio for a specific case, e following lemma proves a e raio is criical for all rombi. 5

3 Lemma : Similar figures wi n fixed poins in e Caresian plane will ave similar colored regions for e raveling salesman problem. Proof: Given n fixed poins in e Caresian plane and any given ome ciy, scaling e enire figure by a facor of k will also scale e lengs of e possible bes pas by a facor of k. Tus e pa wi e minimum leng for e given ome poin prior o scaling will sill be e bes pa afer scaling. Terefore, scaling e figure by a facor of k will lead o a scaling of e colored regions by a facor of k as well, wic means a similar figures ave similar colored regions for e raveling salesman problem. QED For our rombus, scaling and by e same consan scales e enire figure. Tus if we find e criical value of above wic a fif color is presen for a given, e raio of o for a paricular case will be e criical raio for all rombi. Before we find e criical raio, le us examine some oer aspecs of is figure. Te six possible bes pas for any given se of four fixed poins are as follows. p = H A D C B H or H A C D B H p = H A B D C H or H A D B C H p3 = H A B C D H or H A C B D H p = H B A D C H or H B D A C H p5 = H B A C D H or H B C A D H p6 = H C B A D H or H C A B D H Dejarlais proved a for any given figure, regardless of e posiion of e ome poin, eier e wo opions lised above for eac pa will always be of equal leng or one of e wo opions will always be sorer an e oer. For our rombus, we cose = and. Tus e coordinaes of e verices of rombus ABCD are (0, ), (-,0), (0, -), and (,0), respecively. In is case, e firs opion for eac pa lised above will always be sorer an or equal o e second opion lised. In p, e firs opion is sorer because e second one crosses iself. In p, e wo opions are e same leng because eac side of e rombus is e same leng. By similar reasoning, e firs opions for e oer four pas are all sorer an or equal o e second opions. Tus we can simply look a e firs opion for eac pa, ignoring e second. Te following car liss e color a corresponds o eac of e pas. Color Red Green Blue Cyan Black Yellow Pa p = H A D C B H p = H A B D C H p3 = H A B C D H p = H B A D C H p5 = H B A C D H p6 = H C B A D H 6

4 We can find e equaions for e boundaries beween any wo colored regions by seing e lengs for e wo pas equal o eac oer and simplifying e expressions. We will le d(pi) denoe e leng of e i pa. Some of e equaions for e boundaries are elemenary. Looking a e boundary beween red and blue, we se d(p) = d(p3) and obain HA AD DC CB BH = HA AB BC CD DH. Canceling erms, we obain AD BH = AB DH. Since e figure is a rombus, all e sides ave e same leng. Tus e equaion for is boundary simplifies o BH = DH, wic is simply e y-axis. Te same is rue for e boundary beween cyan and yellow. Similar reasoning sows a e red and cyan regions and e blue and yellow regions mee along e x-axis. Te equaions for e boundaries beween blue and cyan and beween red and yellow simplify o HA HD = HB HC and HA HB = HC HD, respecively. By inspecion, one can see a e origin saisfies eac of ese equaions. Tus e four colors, red, blue, cyan, and yellow, are eac limied o one quadran of e plane and all four mee in e cener of e figure a e origin. Of greaer ineres are e boundaries were e fif color mees e oer four. We ope o find e criical value of greaer an above wic e fif color, green, is presen on bo sides of e rombus. Tus we will look a e boundaries were green mees red, blue, cyan, and yellow, in oer words e equaions were d(p) equals d(p), d(p3), d(p), and d(p6). Teorem : If e raio of o (or vice versa) for a rombus wi verices a (0, ), (-, 0), (0, -), and (, 0), were > 0 and > 0, is greaer an.337 /, en ere will be a fif color presen in e raveling salesman grap. Tis is e criical raio for a rombus. Proof: Seing d(p) equal o d(p), d(p3), d(p), and d(p6), canceling erms as above, and rearranging yields four boundary equaions for e green region. Colors Boundary Equaion b Red/Green HC HB = BC BD b Blue/Green HC HD = BC BD b3 Cyan/Green HA HB = AD BD b Yellow/Green HA HD = AD BD Since BC = AD in a rombus, ese equaions are all yperbolas were e difference of e disances o wo fixed poins are equal o e same consan, BC BD, wic is denoed by a. Figure 5 sows e relevan brances of ese yperbolas for = and = 5. Te algebraic compuaions and Malab program used o produce e yperbolas in is figure are included in e appendix. 7

5 Figure 5: =, = 5 Due o e symmery of e figure and e yperbolas abou e y-axis, we can concenrae on wen e green is presen on e lef side, knowing a i will simulaneously be presen on e rig side as well. Te following graps sow e relevan brances of b and b3, e yperbolas governing e lef side of e figure, for =, 5, 30, and 50, wen =. Figure 6: =, = Figure 7: =, = 5

6 Figure : =, = 30 Figure 9: =, = 50 Poin D saisfies e equaions for bo of ese yperbolas, wic means a b and b3 will always inersec a D. Te fif color, green, is only presen wen b and b3 inersec a second ime o e lef of poin B. Due o e symmery of e yperbolas abou e x-axis, is inersecion mus occur on e x-axis. Tus e criical -value is e maximum -value a wic ese yperbolas fail o inersec e second ime along e x-axis. We know a yperbolas inersec only wen eir asympoes inersec; erefore, e criical value is e one a wic e asympoes exending o e lef for eac yperbola are parallel. Te asympoes of e yperbolas will be parallel o eac oer wen ey are bo parallel o e x-axis, and wen one asympoe is parallel o e x-axis e oer will be as well. Tus, we will narrow our focus o finding e a wic e asympoe of b is parallel o e x-axis. Figure 0 sows e asympoe of b P b c T V a Figure 0 9

7 I suffices o sow a wo disinc poins on e asympoe ave e same y- coordinae. Poin T on e asympoe is e midpoin of BC; us is coordinaes are (-, - / ). Te coordinaes for P on e asympoe can be found by saring a poin T, moving a leng a along e uni vecor from T o V, and en moving a leng b along e uni vecor from V o P, wic is perpendicular o e firs uni vecor. Tus e coordinaes for P are as follows: (, ) (, ) P = (, ) a b 6 6. Te geomery of yperbolas enables us o calculae e following values for a and b: a = c = b = c (BC BD) = BC = a = 6 ( (6 6, and ) ( ), 6 ), wic simplifies o b = 6 Tus e y-coordinae for P is Seing e y-coordinaes for T and P equal o eac oer, we obain Simplifying is equaion yields ( 6 ) = 6. ( 6 ) ( 6 ) 6 6 = Using e fzero command in Malab, wic uses e bisecion meod o approximae e roos of an equaion o any given olerance level, we found a.337 wi a olerance of Tus e criical -value is.337. Terefore, wen =, if e raio of o is greaer an.337 /, ere will be a fif color presen in e raveling salesman grap. However, according o e previous lemma, we know a is is e criical raio for all rombi. Terefore, if e raio of o (or vice versa) for a given rombus is greaer an.337 /, en ere will be a fif color presen. Tis is e criical raio for all rombi. QED Te following grap sows e yperbolas wen =.337 and =, e criical values a wic e yperbolas fail o inersec a second ime

8 Figure : =, =.337 Corollary : Le σ be e obuse angle of a rombus. If σ is greaer an.0575, en ere will be a fif color presen in e raveling salesman grap. Tis is e criical angle for a rombus. Proof: Te criical raio of o (and vice versa) for a rombus is.337 /. Using is raio, we can find e criical angle, σ, using an(σ / ) =.337 /. Solving is equaion gives us σ QED Rig Kie A figure a is a slig variaion of e rombus is a kie ABCD wi verices a (0, ), (-, 0), (0, -), and (, 0), were > 0 and > 0. In is figure, angle A is always 90 ; us we will refer o is figure as a rig kie. Te six possible bes pas for e rig kie are e same as ose for e rombus, for e same reasoning. Te color corresponding o eac possible pa is mainained as well for consisency. Color Red Green Blue Cyan Black Yellow Pa p = H A D C B H p = H A B D C H p3 = H A B C D H p = H B A D C H p5 = H B A C D H p6 = H C B A D H We will begin by examining cases were. Te following graps are for = 5, 50, and 500 wen =. 3

9 Figure : =, = 5 Figure 3: =, = 50 Figure : =, = 500 From ese graps, e fif color, green, seems o appear only for -values a are muc larger an e criical -value for e rombus. Te green regions appear on bo sides of e figure well above e figure. Wile e green regions sill appear symmeric abou e y-axis, ey are no longer symmeric abou e x-axis. Te symmery of e colored regions corresponds o e symmery of e kie. Jus as we did wi e rombus, we will examine e boundaries beween colored regions by seing e lengs of wo pas equal o eac oer and finding e boundary equaion. Te boundaries beween red and blue and beween cyan and yellow, b and b, are bo e y-axis, jus as ey were for e rombus. Te equaions for e boundaries beween red and cyan and beween blue and yellow simplify o b3: HC HA = CB AB and b: HC HA = CD AD, respecively. In e rig kie, CB = CD and AB = AD; erefore, b3 = b. By inspecion, B and D saisfy ese equaions; us b3 (b) is a yperbola wi foci a A and C, passing roug poins B and D. Noe a e four-color poin inside e figure is no a e origin, as i was for e rombus. Insead, i occurs a e poin were e yperbola crosses e y- axis. We wis o deermine e criical raio of o for above wic a fif color, in our case green, is presen. Noe a is raio will no also be e criical raio of o for a deermines wen a fif color, in our case black, appears above or below e figure. For e rombus, ese raios were e same, because e figures were similar, simply roaed 90. For e rig kie, allowing o be less an creaes a new siuaion. Teorem : If e raio of o for a rig kie wi verices a (0, ), (-, 0), (0, -), and 3

10 (, 0), were > 0 and > 0, is greaer an 7.90 /, en ere will be a fif color presen in e raveling salesman grap. Proof: Te boundary equaions for e fif color, were green mees red, blue, cyan, and yellow, respecively, simplify o e same boundary equaions for green a governed e rombus. Colors Boundary Equaion b5 Red/Green HC HB = BC BD b5* Blue/Green HC HD = BC BD b6 Cyan/Green HA HB = AD BD b6* Yellow/Green HA HD = AD BD Clearly ese boundary equaions are symmeric abou e y-axis; us we need only examine e presence of green on one side of e figure. Terefore, we will focus on b5 and b6, knowing a b5* and b6* are simply reflecions of b5 and b6 across e y- axis. (We use e bi* noaion o indicae a yperbola a is simply a reflecion of bi.) Noe a e equaions for b5 and b6 are bo saisfied by poin D. Te following graps sow e relevan brances of b3, b5, and b6 for = 75, 5, 50, and 500 wen =. Te algebraic compuaions and Malab programs used o plo ese yperbolas are similar o e ones sown in e appendix for e yperbola in e case of e rombus. Figure 5: =, = 75 Figure 6: =, = 5 33

11 Figure 7: =, = 50 Figure : =, = 500 I is ineresing o noe a wen b5 and b6 inersec, b3 also passes roug e inersecion poin. Tis ricolor poin occurs because green mees red along b5, and green mees cyan along b6; us red and cyan mus mee wen b5 and b6 inersec. Terefore, beween e figure and e ricolor inersecion poin, b3 is e governing boundary separaing e red and cyan regions, and beyond e inersecion poin, b5 and b6 are e governing boundaries separaing red from green and green from cyan. Jus as in e case of e rombus, we need o find e criical -value above wic b5 and b6 inersec e second ime. Once again, our ask comes down o finding e -value a wic e asympoes of e yperbolas are parallel. Looking a b6, we see a for a fixed, b6 will also be fixed. By our lemma, we know a e criical raio for a fixed will be e criical raio for all, since e rig kies will be similar; us we will le =. To find e for wic e asympoes are parallel, we will find wo poins on eac asympoe, find e slope of eac asympoe, se e slopes equal o eac oer, and solve for. Following e same meods used for e rombus, we can find e coordinaes for wo poins on eac of e asympoes. Te asympoe of b6 conains e poins Te slope of e asympoe of b6 is wic simplifies o ( 3 ) 3 (, ) and (-, ) (,) 3 3 ( 3 ) -( 3 ( 3 ) --( ) 36 3 ), 6 (,). 3 ) 3 )

12 Te asympoe of b5 conains e poins ( 6 6 6, - ) and (, - ) (, ) (, ). 6 Tus e slope of e asympoe of b5 is 6 - -(- - ( ) ), wic simplifies o ( ( 6 6 ) ) Seing ese wo slopes equal o eac oer and using e fzero command in Malab, as described previously, we find a Hence, for a rig kie wi =, if e raio of o is greaer an 7.90 /, ere will be a fif color presen in e raveling salesman grap. According o our lemma, we know a is raio is a criical raio for all rig kies. Terefore, if e raio of o for a rig kie is greaer an 7.90 /, en ere will be a fif color presen. QED Te following grap sows b3, b5, and b6 wen = 7.90 and =, e criical values a wic e yperbolas fail o inersec a second ime. Figure 9: =, = 7.90 In e rig kie ABCD, angles B and D are congruen, and we will refer o em as e side angles. We will refer o C as e boom angle. 35

13 Corollary : Le σ be a side angle of a rig kie ABCD. If σ is greaer an 33.33, en ere will be a fif color presen in e raveling salesman grap. Tis is a criical angle for e rig kie. Proof: Since e side angles are congruen, we can le σ = B wiou any loss of generaliy. Le O represen e origin. Tus B = ABO OBC. Since ABO is a rig, isosceles riangle, ABO mus be 5. Using e criical raio of o for a rig kie, we can find e criical measure of OBC, using an ( OBC) = 7.90 /. Solving is equaion gives us OBC.33. Combining ese seps gives us σ = B = ABO OBC 5.33 = 33.33, wic means σ QED We will now look a cases of e rig kie were 0 <. In e following graps, = and = 0.5, 0.5, and. Figure 0: =, = 0.5 Figure : =, = 0.5 Figure : =, = From ese graps, i appears a we can find a criical value for below wic a fif color, black, appears below e rig kie. Teorem 3: If e raio of o for a rig kie wi verices a (0, ), (-, 0), (0, -), and (, 0), were > 0 and > 0, is less an 0.30 /, en ere will be a fif color presen beyond e boom angle in e raveling salesman grap. Proof: Te boundary equaions for e fif color appearing below e rig kie are e equaions were black mees cyan and yellow, or d(p5) = d(p) and d(p5) = d(p6). Colors Boundary Equaion b7 Cyan/Black HD HC = AD AC b7* Yellow/Black HB HC = AD AC Tese boundaries are yperbolas wi foci a C and D and a C and B wi e same value for a. Te wo yperbolas inersec once a A. As in e previous cases, e fif color is presen beyond e second inersecion of ese wo yperbolas. We wis o find e criical value for below wic ese yperbolas inersec a second ime. Following e argumen used for e rombus and because e boundaries are symmeric 36

14 abou e y-axis, we can simply find e a wic e asympoe of one of e yperbolas is parallel o e y-axis. As before, i will be sufficien o find e a wic wo disinc poins on e asympoe ave e same x-coordinae. Once again, we can find e coordinaes for wo poins on one of e asympoes. Te asympoe of b7 conains e poins 3 6 ( 6 6, ) and (, ) (, ) (, ). Seing e x-coordinaes for ese wo poins equal o eac oer, we obain = Simplifying is expression gives us = 0. Using Malab s fzero command, we find a Tus wen e =, e criical -value below wic a fif color will appear for e rig kie is Once more, we can use our lemma o conclude a is raio olds for all rig kies. Terefore, if e raio of o for a rig kie is less an 0.30 /, ere will be a fif color presen in e raveling salesman grap. QED Corollary 3: Le σ be e boom angle of a rig kie ABCD. If σ is greaer an , en ere will be a fif color presen in e raveling salesman grap. Tis is anoer criical angle for e rig kie. Proof: If e raio of o for a rig kie is less an 0.30 /, en ere will be a fif color presen in e raveling salesman grap. Using is raio, we can find e criical boom angle, using an(σ / ) = / Solving is equaion gives us σ Since e fif color occurs wen e raio of o is less an 0.30 /, e fif color will occur wen e boom angle is greaer an is criical angle. QED In all of e raveling salesman graps for e rig kie, we failed o see any black appear above e rig angle of e figure. We can explain is fac by examining e boundaries separaing black from red and blue. Teorem : For a rig kie wi verices a (0, ), (-, 0), (0, -), and (, 0), were > 0 and > 0, a fif color can never appear direcly beyond e rig angle in a raveling salesman grap. Proof: Te boundary equaions for e fif color appearing above e rig kie are e equaions were black mees red and blue, in oer words d(p5) = d(p) and d(p5) = d(p3). 37

15 Colors Boundary Equaion b Red/Black HA HD = AC BC b* Blue/Black HA HB = AC BC Tese boundaries are yperbolas wi foci a A and D and a A and B a bo pass roug C and are symmeric abou e y-axis. As before, e fif color, black, will be presen only if ese yperbolas inersec a second ime, somewere above poin A. We know a e yperbolas will inersec wen eir asympoes inersec. Te following figure sows an asympoe of b and demonsraes a ere are cases wen e asympoes will no inersec beyond poin A. Figure 3: =, = Before e asympoes inersec, ey mus reac a poin wen ey are parallel. Tus, we can sow a e yperbolas will no inersec beyond poin A by proving a ere are no values for a wic e asympoes become parallel. Due o e symmery of e yperbolas abou e y-axis, e problem is reduced o proving a eier one of e asympoes can never be parallel o e y-axis. As before, we will le = and en use our lemma o generalize e resuls. Using e meod discussed earlier, e coordinaes for wo poins on b are (,) and (,) ( 6 ) ( 6 ) (,) 3 3 (,). Seing e x-coordinaes of ese poins equal o eac oer gives us = ( 6 ) ( 6 ) 3 3, wic reduces o ( 6 ) ( 6 ) = 0. We now need o prove a is equaion as no posiive soluions. Firs we will make e following subsiuion: a = 6. 3

16 Tus our equaion becomes Solving for a, we find a = ±. Tus we ave wo cases o examine. Case: a = a a = 0. 0 = 6 Tus ere are no soluions for Case. Case : a = - a = = = 6 = 6 a = = 6 = = 3 = = 6 Tus ere are no posiive soluions for Case. 6 Terefore wen =, ere are no posiive values for a wic e asympoes will be parallel o e y-axis, wic means a e yperbolas will never cross beyond poin A. According o our lemma for similar figures, we can generalize is saemen o say a e yperbolas governing e presence of a fif color beyond e rig angle of a rig kie will never inersec. Terefore, for all posiive values of and, a fif color can never appear direcly beyond e rig angle of a rig kie in e raveling salesman grap. QED Trapezoid Consider an isosceles rapezoid ABCD in wic e verices ave e following coordinaes: (0,0), (-,), (,), (,0), were 0 or 0. If eier or is equal o zero, en e figure ceases o be a rapezoid. We will label AD as b and BC as b and 39

17 we will keep consan a = wile we vary. Since b = and b =, b / = and b / = / =. Le us firs examine values of were 0 < < and ence, b > b. If we begin wi =.5, en ere are four colors presen in e grap. A value of =, owever, will produce a grap wi six colored regions. Similarly, a value of =.5 will produce a six color grap wile a value of = 3.5 will produce a grap wi only four colors. Figure : (op) lef =.5, =, rig =, = ; (boom) lef =.5, =, rig = 3.5, = Wen =, b = b and ABCD is a recangle, wic produces a grap wi four colors. Figure 5: =, = 0

18 Wen >, b < b. If we coose =.5, en ere are four colors presen in e grap, wile a value of = 5.5 will produce a grap wi six colors Figure 6: lef =.5, =, rig = 5.5, = Te grap below indicaes e number of colors presen for values of and ranging from 0 o 0. A blue poin represens e presence of four colors for a given and, and a green poin indicaes e presence of six colors Figure 7: versus As can be seen, for =, ere are wo poins were e number of colors presen sifs from four o six wen 0 < <. In e case of =, i appears a ere will always be four colors in e grap regardless of e eig. For values of wen >, ere is only

19 one criical value were e number of colors presen sifs from four o six. Te lineariy of e grap leads us o believe a ere are criical raios of o a will deermine weer or no six colors are presen. Red, yellow, blue and cyan are e four colors a are always presen in our grap. Te following able sows e boundary equaions for green and black, e fif and six colors, wi e oer four colors. Colors Boundary Equaion Boundary Equaion Colors Red/Green Blue/Green HC-HB = BCAD-AB-BD HD-HC = BD-BC HC-HB =-BC-ADCDAC HA-HB = AC-BC Yellow/Black Blue/Black Cyan/Green HA-HB = AD-BD HD-HC = AD-AC Cyan/Black Black/Green HDHB-HA-HC = BD-AC HDHB-HA-HC = BD-AC Green/Black Yellow/Green HA-HD = -BD-CDBCAD HA-HD = ACAB-BC-AD Red/Black Lemma : Isosceles rapezoids will generae eier four or six color regions. Proof: Since ABCD is a quadrilaeral, ere is e possibiliy of four, five, or six colors for e raveling salesman grap. Because ABCD is an isosceles rapezoid, AB = CD (e legs of an isosceles rapezoid are congruen) and AC = BD (e diagonals of an isosceles rapezoid are congruen). By subsiuing ese values ino e boundary equaions above, e following equaions are obained. Colors Boundary Equaion Boundary Equaion Colors Red/Green Blue/Green HC-HB = BCAD-AB-BD HD-HC = BD-BC HC-HB =-BC-ADABBD HA-HB = BD-BC Yellow/Black Blue/Black Cyan/Green HA-HB = AD-BD HD-HC = AD-BD Cyan/Black Black/Green HDHB-HA-HC = BD-AC HDHB-HA-HC = BD-AC Green/Black Yellow/Green HA-HD = -AC-ABBCAD HA-HD = -BC-ADACAB Red/Black Te Red/Green and Yellow/Black boundary equaions are e equaions of opposie brances of e same yperbola. Since e foci are symmeric abou e line x =, e yperbola brances will also be symmeric abou e line x =. By e same argumen, is is also rue of e Yellow/Green and e Red/Black equaions. Te Black/Green and e Green/Black equaions are, of course, equal as well as symmeric abou x =. Since AB = CD, e Blue/Green and e Blue/Black equaions as well as e Cyan/Green and e Cyan/Black equaions are equaions of congruen yperbolas. Also, since e foci poins A and B of e Blue/Black and e Cyan/Green yperbolas are symmeric o e foci poins D and C of e Blue/Green and e Cyan/Black yperbolas abou e line x =, respecively, e congruen yperbolas are also symmeric abou e line x =. Since eac of e boundary equaions a govern e green and black regions are e same, equal, or congruen yperbolas a are all symmeric abou e line x =, e green and black regions will only appear under idenical condiions. Tus five colors will never be presen, only four or six. Q.E.D. We will now invesigae e criical values of a cause e sif beween four and six colors. Even oug ere are five equaions a govern eac e green and black regions, e enire region can be discovered by only examining ree of e boundary

20 equaions for eac of e wo colors. In e graps above were 0 < <, e green and yellow and e green and black regions do no inersec. Te boundary equaions sill exis bu ey fall wiin e region of anoer color. In is region, e oer color will always be e bes pa and ence e boundary colors do no inersec in e grap. Since red, blue and cyan all inersec green in ese graps, e inersecion of e green sides of e Red/Green, Blue/Green, and Cyan/Green boundary equaions will deermine e green region in e grap. Wen >, e green and red regions as well as e green and black regions do no inersec. By e same reasoning, e green region will be deermined by e inersecion of e green sides of e Yellow/Green, Blue/Green, and e Cyan/Green boundary. Since = b /, we will find criical values of and use em wi e raio b /. Teorem 5: For any isosceles rapezoid ABCD were b / = : i) if.766 < b / < 3.003, ere are six colors presen in e raveling salesman grap; ii) if b / >.7, ere are six colors presen in e raveling salesman grap; iii) for all oer posiive values of b /, e raveling salesman grap will ave four colors. Proof: i) Te lower criical -value occurs wen e ree relevan boundary equaions ave a common soluion. green green = green region green = green side of boundary = Red/Green boundary = Cyan/Green boundary = Blue/Green boundary Figure : color boundaries green wen is equal o e criical value Using a Malab program (see appendix) a uses Newon s Meod, we are able o find e inersecion of e Blue/Green and Cyan/Green boundary equaions as well as e inersecion of e Red/Green and e Cyan/Green boundary equaions. Te program en minimizes e disance beween e wo wiin a given olerance. Using e Malab program, we find a =.766 is a criical value. Wen <.766, ere is no common soluion o e ree equaions and ence no green region. 3

21 green green green = green region green = green side of boundary = Red/Green boundary = Cyan/Green boundary = Blue/Green boundary Figure 9: color boundaries wen is less an e criical value Wen >.766, e equaions bound a region of e grap wic produces e green color. Tus, wen b / >.766, e color green will appear on e grap. green green = green region green = green side of boundary = Red/Green boundary = Cyan/Green boundary = Blue/Green boundary Figure 30: color boundaries wen is greaer an e criical value Using Lemma, e raio b / >.766 will also be wen e black region will appear on e grap. Te upper -value for e criical raio occurs wen a common soluion sops exising for e ree equaions. More specifically, e yperbolas formed from e Red/Green and e Cyan/Green boundary equaions ave no poin of inersecion wiin e green side of e Blue/Green boundary equaion and us eir asympoes are parallel in is region.

22 green green green = green region green = green side of boundary = Red/Green boundary = Cyan/Green boundary = Blue/Green boundary Figure 3: color boundaries wen is less an e criical value green green green = green region green = green side of boundary = Red/Green boundary = Cyan/Green boundary = Blue/Green boundary Figure 3: color boundaries wen is equal o e criical value green green Figure 33: color boundaries wen is greaer an e criical value To find e for wic e asympoes are parallel, as in e case of e kie, we will find wo poins on eac asympoe, find e slope of eac asympoe, se em equal o eac 5

23 6 oer, and solve for. Following e same meods used for e rombus and e kie, we can find e coordinaes for wo poins on eac of e asympoes. Te asympoe of e Red/Green boundary equaion conains e poins ( ) ( ) ( )( ) ( ) ( ) slope and as, 0, 0 0,0 0 0, and, ( ) ( ) Te asympoe of e Cyan/Green boundary equaion conains e poins ( ) ( ) ( ) slope and as,, 0 0 0, 0 0, and, ( ) ( ) ( ) ( ) ( ) ( ) By seing e slopes equal o eac oer, we produce e following equaion: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = By using e fzero command in Malab o solve for, we obain a criical -value of Wen < e green region is presen in e grap. Since green and black occur under e same condiions, e black region is also presen. Tus, wen b / < 3.003, ere are six colors presen in e grap. Combining e wo resuls wi Lemma gives us e open inerval (.766,3.003) for b / were six colors will be seen in e raveling salesman grap for all isosceles rapezoids wi b / =.

24 ii) For values of, were >, e Yellow/Green, Blue/Green, and Cyan/Green boundary equaions deermine e green region. Below e criical value, e Yellow/Green and Blue/Green boundary equaions ave no common soluion on e green side of e Cyan/Green boundary causing ere o be no green region. green green green = green region green = green side of boundary = Yellow/Green boundary = Cyan/Green boundary = Blue/Green boundary Figure 3: Yellow/Green, Cyan/Green, and Blue/Green boundaries wen is less an e criical value A e criical value, e asympoes of e yperbolas formed by e Yellow/Green and Blue/Green boundary equaions are parallel. green green green = green region green = green side of boundary = Yellow/Green boundary = Cyan/Green boundary = Blue/Green boundary Figure 35: Yellow/Green, Cyan/Green, and Blue/Green boundaries wen is equal o e criical value 7

25 For values larger an e criical value, e wo yperbolas inersec and e green region is creaed. green green = green region green = green side of boundary = Yellow/Green Figure 3: Yellow/Green, Cyan/Green, and Blue/Green boundaries wen is greaer an e criical value Te asympoe of e yperbola formed by e Yellow/Green boundary equaion conains e poins (,0) and (,0) (,0) ( 0 0) ( 0, ), and as slope ( 0 0) ( 0 0) Te asympoe of e yperbola formed by e Blue/Green boundary equaion conains e poins. green

26 9 ( ) ( ) ( ) slope and as,, , and, 6 ( ) ( ) ( ) ( ) ( ) Since e criical value of is found wen e asympoes are parallel, we will se e wo slopes equal o eac oer o obain e following equaion. ( ) ( ) ( ) ( ) ( ) ( ) ( ) = As before, we use e fzero command in Malab o find a.7 is a roo of e equaion and ence e criical value of. Wen and ence b / >.7, e green region, as well as e black by Lemma, is presen in e grap. Using Lemma, we can now sae a for all isosceles rapezoids wi b / =, ere will be six colors presen if b / >.7. iii) We sowed earlier in Lemma a ere can only be four or six colors presen for an isosceles rapezoid in a raveling salesman grap. Since we ave found e posiive b / values were six colors are presen in e grap, all oer posiive values will ave four colors in eir respecive raveling salesman graps. Q.E.D. We can use our meods o find criical b / values wen b / by using b / = n, were n is any real number excep zero because b can never be zero. Ten, given any isosceles rapezoid, we can calculae n and wrie e coordinaes of e verices of a similar isosceles rapezoid as (0,0), (-, /n), (, /n), (,0). Firs, we subsiue ese new coordinaes ino our Malab program a uses Newon s meod o find e firs criical value. We en find e criical values wen e asympoes are parallel, as we did before. Ideally some condiions could be found for all isosceles rapezoids, using a sep funcion.

27 Acknowledgemens Firs and foremos, we would like o ank Angela Gran for all er elp and never-ending suppor. We would also like o ank Sidelia Reyna, Dr. Dennis Davenpor, Bonia Porer, and Dugald Hucings. Addiionally, we would like o ank e NSA, e WEDGE Co, and all oer conribuors o SUMSRI. Finally, we would like o ank Dr Earl Barnes, our linear programming advisor. 50

28 Appendix Malab Program for Coloring e Traveling Salesman Graps %Tis program produces e regions corresponding o e six possible %pas for e raveling salesman problem wi four ciies. %Te following need o be defined before running e program: %P: wic conains e coordinaes of e four poins %,k: amoun grid goes beyond e grap of P( sides,k verical) %sep: disance beween grid poins d=0*p'; %creae e disance array e same size as P' for i=: %calculae all disances (some no needed) for j=: d(i,j)=norm(p(i,:)-p(j,:)); end end xlef = min(p(:,))-; xrig = max(p(:,)); yboom = min(p(:,))-k; yop = max(p(:,))k; axis([xlef, xrig,yboom, yop]) old on P=[P;P(,:)]; %cange P in order o plo grap plo(p(:,)',p(:,)') P=P(:,:); %cange P back o original form for furer use x=xlef:sep:xrig; y=yboom:sep:yop; clear f for i = :leng(x) for j = :leng(y); X=[x(i),y(j)]; %deermine e size of e mes %calculae e six pas' lengs a=d(,3)d(3,)d(,); b=d(,)d(,3)d(3,); f() = norm(x-p(,:))min(a,b)norm(x-p(,:)); a=d(,)d(,)d(,3); b=d(,)d(,)d(,3); f() = norm(x-p(,:))min(a,b)norm(x-p(3,:)); 5

29 end end a=d(,)d(,3)d(3,); b=d(,3)d(3,)d(,); f(3) = norm(x-p(,:))min(a,b)norm(x-p(,:)); a=d(,)d(,)d(,3); b=d(,)d(,)d(,3); f() = norm(x-p(,:))min(a,b)norm(x-p(3,:)); a=d(,3)d(3,)d(,); b=d(,)d(,3)d(3,); f(5) = norm(x-p(,:))min(a,b)norm(x-p(,:)); a=d(3,)d(,)d(,); b=d(3,)d(,)d(,); f(6) = norm(x-p(3,:))min(a,b)norm(x-p(,:)); [m,q]=min(f); if q== plo(x(i),y(j),'.r') end if q== plo(x(i),y(j),'.g') end if q==3 plo(x(i),y(j),'.b') end if q== plo(x(i),y(j),'.c') end if q==5 plo(x(i),y(j),'.k') end if q==6 plo(x(i),y(j),'.y') end Algebraic Compuaions and Malab Program for e Hyperbolas Governing e Fif Color in a Rombus Due o e symmery of e figure, we need only find e equaion for any one of e yperbolas and en we can reflec a yperbola across e x-axis and y-axis. We will find e equaion for e firs of e yperbolas lised, HC HB = BC BD. Leing =, e coordinaes of B, C, and D are (-, 0), (0, -), and (,0) respecively. Le e coordinaes of e movable ome poin be (x, y). Tus e equaion becomes 5

30 x ( y ) ( x ) y = 6. Rearranging and en squaring bo sides yields x ( y ) = ( 6 ) ( 6 ) ( x ) y ( x ) y wic simplifies o y 6 6 x 96 = ( 6 ) ( x ) y Squaring again, we obain., y y(6 6 x 96) (6 6 x 96) = ( 6 ) (( x ) y ). Disribuing and regrouping, we arrive a a quadraic equaion, (( 6 ) ) y ((6 6 x 96)) y (( 6 ) ( x ) ) (6 6 x 96) = 0. We use is equaion and e quadraic formula in e following Malab program o grap e relevan brances of e yperbolas. %Tis program draws e relevan brances of e yperbolas for e %rombus case of e raveling salesman problem, were e wid of e %rombus is and e eig,, is variable and mus be enered prior %o running e program. X=[0 0-0]; Y=[ 0-0 ]; plo(x,y) old on x=-0:0.:0; L=leng(x); S=sqr(6^); a=*(s-)^-*^; for i=:l b=-**(6*s-*x(i)-96); c=*((s-)^)*(x(i))^-(6*s-*x(i)-96)^; y(i)=(-b-sqr(b^-*a*c))/(*a); y(i)=-y(i); end plo(x,y) plo(x,y) plo(-x,y) plo(-x,y) axis('equal') 53

31 Twoypina Program %woypina %Tis program uses Newon's Meod o %find e inersecion of wo yperbolas x=origx; y=origy; dab=norm(p(,:)-p(,:)); dac=norm(p(,:)-p(3,:)); dad=norm(p(,:)-p(,:)); dbc=norm(p(,:)-p(3,:)); dbd=norm(p(,:)-p(,:)); dcd=norm(p(3,:)-p(,:)); es=; wile es>.0000 q=[x,y]; da=norm(q-p(,:)); db=norm(q-p(,:)); dc=norm(q-p(3,:)); dd=norm(q-p(,:)); f=dc-dbdbddab-dbc-dad;%red/green boundary g=da-dbdbd-dad;%cyan/green boundary fx=(x-p(3,))/dc-(x-p(,))/db; fy=(y-p(3,))/dc-(y-p(,))/db; gx=(x-p(,))/da-(x-p(,))/db; gy=(y-p(,))/da-(y-p(,))/db; H=[fx,fy;gx,gy]; b=[f;g]; del=h\b; es=norm(del); x=x-del(); y=y-del(); end Twoypinb Program %woypinb %Tis program uses Newon's Meod o 5

32 %find e inersecion of wo yperbolas x=origx; y=origy; dab=norm(p(,:)-p(,:)); dac=norm(p(,:)-p(3,:)); dad=norm(p(,:)-p(,:)); dbc=norm(p(,:)-p(3,:)); dbd=norm(p(,:)-p(,:)); dcd=norm(p(3,:)-p(,:)); es=; wile es>.0000 r=[x,y]; da=norm(r-p(,:)); db=norm(r-p(,:)); dc=norm(r-p(3,:)); dd=norm(r-p(,:)); f=da-dbdbd-dad;%cyan/green boundary g=dd-dcdbc-dbd;%blue/green boundary fx=(x-p(,))/da-(x-p(,))/db; fy=(y-p(,))/da-(y-p(,))/db; gx=(x-p(,))/dd-(x-p(3,))/dc; gy=(y-p(,))/dd-(y-p(3,))/dc; H=[fx,fy;gx,gy]; b=[f;g]; del=h\b; es=norm(del); x=x-del(); y=y-del(); end Newme Program %Newme Program %Minimizes e disance beween wo poins %Calls woypina and woypinb dab=norm(p(,:)-p(,:)); dac=norm(p(,:)-p(3,:)); dad=norm(p(,:)-p(,:)); dbc=norm(p(,:)-p(3,:)); dbd=norm(p(,:)-p(,:)); 55

33 dcd=norm(p(3,:)-p(,:)); es=; wile es>.0000 z=[x,y]; da=norm(z-p(,:)); db=norm(z-p(,:)); dc=norm(z-p(3,:)); dd=norm(z-p(,:)); f=dc-dbdbddab-dbc-dad; g=da-dbdbd-dad; =dcdb-da-dddad-dbc fx=(x-p(,))/d-(x-p(,))/d; fy=(y-p(,))/d-(y-p(,))/d; gx=(x-p(,))/d-(x-p(3,))/d; gy=(y-p(,))/d-(y-p(3,))/d; H=[fx,fy;gx,gy]; b=[f;g]; del=h\b; es=norm(del); x=x-del(); y=y-del(); end 56