Simple Pro ofs of Rectangle Tiling Theorem Da vid J
94K - views

Simple Pro ofs of Rectangle Tiling Theorem Da vid J

C MacKa Ca endish Lab oratory Madingley Road Cam bridge CB3 OHE mackaymraoca m ac u Ma 28 2003 Draft 20 Abstract If 57356nite um er of rectangles ev ery one of whic has at least one in teger side erfectly tile big rectangle then the big rectangle als

Tags : MacKa endish Lab
Download Pdf

Simple Pro ofs of Rectangle Tiling Theorem Da vid J




Download Pdf - The PPT/PDF document "Simple Pro ofs of Rectangle Tiling Theor..." is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.



Presentation on theme: "Simple Pro ofs of Rectangle Tiling Theorem Da vid J"— Presentation transcript:


Page 1
Simple Pro ofs of Rectangle Tiling Theorem Da vid J.C. MacKa Ca endish Lab oratory Madingley Road Cam bridge CB3 OHE mackay@mrao.ca m. ac .u Ma 28, 2003 Draft 2.0. Abstract If nite um er of rectangles, ev ery one of whic has at least one in teger side, erfectly tile big rectangle, then the big rectangle also has at least one in teger side. presen pro ofs of this theorem, oth accessible to ten-y ear-old. The pro ofs generalize to other situations. In tro duction Here are three theorems. In all three, large rectangle is partitioned in to smaller rectangles, with sides

parallel to those of the large rectangle. Theorem If nite numb er of ctangles, every one of which has at le ast one inte ger side, erfe ctly tile big ctangle, then the big ctangle also has at le ast one inte ger side. Theorem If nite numb er of ctangles, every one of which has at le ast one rational side, erfe ctly tile big ctangle, then the big ctangle also has at le ast one rational side. Theorem If nite numb er of ctangles, every one of which has at le ast one algebraic side, erfe ctly tile big ctangle, then the big ctangle also has at le ast one algebraic side. Do

these theorems ha simple solutions? Opinions ary When as in tro duced to theorem 2, heard that it as celebrated for ha ving an elemen tary solution that as hard to nd. Ho ev er, according to the IBM ‘P onder This ebsite (IBM 1999a), Although this problem (theorem 1) ma app ear simple at rst, the solution is actually fairly complex. In fact, all three theorems ha pro ofs that ten-y ear-old can understand. [F ourteen pro ofs of theorem ere published agon (1987), but as not are of this when wrote the presen pap er.] STOP maximize our enjo ymen of this puzzle, ou ma wish to think

ab out these theorems efore turning the page. Pro ofs ahead
Page 2
(a) (b) Figure Illustration of the hec erb oard pro of. Dashed/blue lines are an in teger grid. The hec erb oard squares are of size 2. In the lefthand example (a), the cen tral small rectangle violates the constrain of the theorem: it has oth sides non-in teger. In the righ thand example (b), ev ery small rectangle has one side of in teger length. Ev ery small rectangle co ers equal amoun ts of blac and white, so the large rectangle ust do the same. In order to pro the second theorem it suces to pro the

rst, since, taking an instance of theorem 2, if the rational sides ha lengths =n can scale up all the lengths factor of the lo est common ultiple of th us con erting all the rational sides to in teger sides. So, ho should pro theorem 1? And what ab out theorem 3? Pr oof by complex integra tion standard pro of of theorem (IBM 1999b) in olv es in tegrating the complex function er the o-dimensional plane. The in tegral er an rectangle is zero if and only if at least one side is an in teger; so the in tegral er eac small rectangle is zero; the in tegral er the big rectangle is the sum of

the small rectangles in tegrals, so it equals zero to o; therefore the big rectangle has an in teger side. This pro of is probably not accessible to most ten-y ear olds; nor is the second solution giv en at (IBM 1999b), whic in olv es large prime um ers. And neither of these pro ofs helps pro theorem 3. no presen pro ofs of theorem 1, oth accessible to ten-y ear-olds. The rst pro of is similar in haracter to the complex in tegral pro of sk etc hed ab e. The second pro of uses completely dieren approac h, whic can applied immediately to theorem 3. Both pro ofs ha een published

efore (W agon 1987). Chec erb oard pro of of theorem ak the big rectangle and align its ottom left corner with half-inte ger che ckerb ar that is, hec erb oard whose squares ha side (gure 1). Let the ottom left corner blac k. If the big rectangle has oth sides non-in teger, then the blac area co ered will exceed the white area co ered; and if the blac and white areas co ered are equal then the big rectangle ust ha an in teger side (gure 2).
Page 3
x; x; x; Figure non-in teger rectangle whose ottom-left corner is aligned with the in teger grid alw ys co ers more blac

than white. (The origin of the in teger grid is at the ottom left of blac square.) Here zo om in on the upp er righ corner of the rectangle in gure 1, whic ust lo ok lik one of the three pictures in the top ro w. or the purp ose of nding the excess blac area, can transform the rst situation in to the second and the second in to the third the constructions sho wn in the ottom ro w, whic lea unc hanged the dierence et een blac and white co erage. In the third situation, only blac area is co ered, therefore in all three situations, the blac area exceeds the white

area. or those who prefer form ulae to sym olic pro ofs, the excess blac area for rectangle with upp er corner at x; is where denotes the in teger nearest to
Page 4
Figure Mak graph in whic h, for ev ery small rectangle, there are edges corresp onding to parallel ‘sp ecial edges. Ho ev er, ev ery small rectangle, since it has an in teger side, ust co er equal areas of blac and white. So the big rectangle ust co er equal areas of blac and white and so ust ha an in teger side. Hex pro of of all three theorems Radford Neal oin ted out the similarit of these puzzles to the task of pro

ving that the game of Hex cannot end in dra w. In the three rectangle puzzles, at le ast one of the big rectangle’s sides ust ha sp ecial prop ert Similarly in Hex, when all pieces ha een pla ed, at le ast one pla er ust ha connected his sides together (Gale and an Rijswijc 2002). (In fact, in Hex, exactly one pla er ac hiev es this result.) Ho can apply this analogy? Dep ending whether wish to pro theorem 1, 2, or 3, will call um er sp ecial if it is inte ger ational or algebr aic resp ectiv ely The sum and dierence of sp ecial um ers are also sp ecial. dene oin x; in the

plane to sp ecial if oth co ordinates and are sp ecial. ak the set of small rectangles, and asso ciate with eac small rectangle four ertices and two edges (gure 3). The four ertices are the four corners of the rectangle. The edges are parallel sides of the rectangle that are oth sp cial in length. (Ev ery small rectangle has suc sp ecial edges, the statemen of the problem.) Lemma If vertex is sp cial, then its neighb ouring vertex is also sp cial. no dene graph iden tifying all the ertices that ha iden tical co ordinates (gure 4). In this graph there ma double edges

connecting ertices. Suc double edges are not merged in to single edge. Lemma Ap art fr om the four vertic es at the orners of the big ctangle, which have de gr 1, al other vertic es have de gr two or four.
Page 5
Figure ertices iden tied. Let’s align our co ordinate system suc that the ottom left ertex of the big rectangle is at the origin. This ertex is th us sp ecial. No w, mak alk through the graph starting from this ertex. eac step of the alk, adv ance along an edge that has not een tra ersed efore to another ertex. Since ev ery ertex (other than the four corner ertices)

has ev en degree, there will alw ys another edge to tra erse, unless mo in to one of the corner ertices. There are nite um er of edges, so ust ev en tually mo to corner ertex. It’s imp ossible that this corner ertex is the one started from (since it had degree one, and nev er tra erse an edge wice). Th us our alk ust ha tak en us to another corner. By lemma 1, our alk establishes that the big rectangle has sp ecial corners, th us it has at least one side that is sp ecial in length. Commen Both pro ofs can applied to the analogous problem in higher dimensions where eac small yp

errectangle has one ‘sp ecial dimension. In the hec erb oard pro of, use the half-in teger hec erb oard in dimensions. yp errectangle that has one corner at the origin and one at co ers an excess blac olume equal to =1 where is the in teger closest to or the hex pro of, assign to eac yp errectangle ertices and edges, all asso- ciated with parallel sp ecial sides. ckno wledgments Thanks to Graeme Mitc hison, Geo Hin ton, Radford Neal, and adashi okieda. References Gale, D., and an Rijswijc k, J., (2002) Hex cannot end in dra Pro of. Electronic publication: http://www.cs.ualberta.ca /~ja vhar

/hex /he x-ga lepr oof. htm
Page 6
Figure The red/highligh ted alk starts from the ottom-left corner and pro ceeds through the graph along sp ecial edges. Ev en tually the alk emerges at another corner (the ottom- righ corner, as it happ ens) and th us ecomes witness to the fact that those corners dier sp ecial distance. IBM, (1999a) onder This. http://www.research.ibm.co m/p onde r/ Ma 1999 hallenge. IBM, (1999b) onder This, Ma 1999 solution. http://domino.research.ibm .com /Co mm/w wwr_ pond er. nsf/ solu tion s/M ay19 99.h tml agon, S. (1987) ourteen pro ofs of result ab

out tiling rectangle. The meric an Mathematic al Monthly 94 (7): 601{617.