A dual of the rectangle-segmentation problem for binary matrices

Transcription

1 A dul of the rectngle-segmenttion prolem for inry mtrices Thoms Klinowski Astrct We consider the prolem to decompose inry mtrix into smll numer of inry mtrices whose -entries form rectngle. We show tht the liner relxtion of this prolem hs n optiml integrl solution corresponding to well known geometric result on the decomposition of rectiliner polygons. MSC: 9C27, 9C46 Introduction In the context of intensity modulted rdition therpy severl decomposition prolems for nonnegtive integer mtrices hve een considered. One of these is the decomposition into smll numer of inry mtrices whose -entries form rectngle. There is n exmple showing tht in generl the liner relxtion of this prolem hs no optiml integrl solution []. On the other hnd, the sme pper contins n lgorithm sed on the revised simplex method tht uses only very few Gomory cuts. In computtionl experiments, this lgorithm provided exct solutions for mtrices of resonle size in short time. In the present pper we consider the specil cse tht the input mtrix is lredy inry: A {, } m n. In this cse the integer optimiztion prolem is equivlent to well studied geometric prolem: the decomposition of rectiliner polygon into the miniml numer of rectngles. Our min result is tht the miniml numer of rectngles in such decomposition equls the optiml ojective in the relxed mtrix decomposition prolem. In other words, the integrlity gp vnishes, provided the input mtrix is inry. This solves Prolem 2 of []. 2 Nottion nd prolem formultion Let A e inry mtrix of size m n. A rectngle mtrix is n m n mtrix S = (s ij ) such tht for some integers k, k 2, l nd l 2 with k k 2 m nd l l 2 n, we hve { if k i k 2 nd l j l 2, s ij = otherwise. The rectngle segmenttion prolem is the following: RSP. Find decomposition A = S + + S t with rectngle mtrices S,..., S t such tht t is miniml. A liner relxtion of this prolem ist the following. The Electronic Journl of Comintorics 6:R89, 29

2 RSP-Relx. Find liner comintion A = x S + + x t S t with rectngle mtrices S,..., S t nd x i for i =,..., t such tht x + + x t is miniml. The integrl prolem RSP cn e formulted in geometric setup s follows. We ssocite A with rectngulr m n-rry of unit squres in the plne. The set P = {(i, j) : ij = } corresponds to rectiliner polygon whose oundry consists of line segments with integer coordintes. Clerly, solution of the prolem RSP is precisely the decomposition of P into the miniml numer of rectngles. In order to stte the solution to the polygon decomposition prolem we need some nottion. Let N, c nd k e the numer of vertices, connected components nd holes of P, respectively. This nottion hs to e clrified y two remrks (see Figure for illustrtions).. If P cn e decomposed into two or more polygons which intersect pirwise in isolted vertices these vertices re counted twice nd we consider the prts s different connected components. 2. Similrly, if the oundry of hole intersects the outer oundry or nother hole only in isolted vertices these vertices re counted twice. P P 2 Figure : Two polygons. The prmeters re N = 8, c = 2 nd k = for P, nd N =, c =, k = for P 2. We cll vertex of P convex if the interior ngle t this vertex is 9 nd concve if it is 27. A chord of P is line segment tht lies completely inside P, connects two concve vertices nd is prllel to one of the coordinte xes. The chords prllel to the x-xis nd the y-xis re clled horizontl nd verticl, respectively. We ssocite iprtite grph G = (H V, E) with P. The vertex sets H nd V re the sets of horizontl nd verticl chords, respectively, nd two chords re connected y n edge if they intersect (Figure 2). Let α e the mximl crdinlity of n 2 2 c d 3 4 c 3 d 4 Figure 2: The grph ssocited to polygon. independent set in the grph ssocited to P. The following theorem of Lipski et l. (which ws 2

3 reproved y severl uthors), chrcterizes the miniml numer of rectngles in decomposition of P. Theorem ([2, 3, 4]). The miniml numer of rectngles in decomposition of P equls N 2 c + k α. We wnt to show tht there is no integrlity gp in the rectngle segmenttion prolem for inry mtrices. This cn e done y proving tht the optiml ojective vlue for RSP-Relx is t lest N 2 c + k α. In order to do this we will present fesile solution for the dul prolem with exctly this ojective vlue. For inry mtrix A, the set of vriles in RSP-Relx corresponds to the set of rectngles tht re completely contined in P. Indexing these rectngles y the numers,..., T, i.e. S (),..., S (T ) re precisely the rectngle mtrices whose -entries re contined in P, we cn reformulte RSP-Relx s follows. T t= s (t) ij x t = for (i, j) P, x t for t =,..., T, T x t min. t= Dulizing, we otin the prolem RSP-Dul: k 2 i=k l 2 y ij for [k, k 2 ] [l, l 2 ] P, j=l y ij mx. (i,j) P This dul prolem hs nice interprettion. The polygon P is considered s set of unit squres nd we wnt to fill these squres with numers such tht the sum over every rectngle contined in P is ounded y, nd the totl sum is mximized under this constrint. We strt with simple oservtion which llows us to restrict our ttention to dul solutions of specil type. Let R e the set of rectngles into which P is decomposed when the oundry lines t the concve vertices re extended until they meet the opposite oundry of P. For n illustrtion see Figure 3, where R = 6 nd, for instnce, the ig squre in the middle is the element [2, 4] [4, 6] R. We cll the elements of R sic rectngles. The following lemm sserts tht we my ssume tht only in the upper left corner of sic rectngle the entry of y is nonzero. Lemm. Suppose y = (y ij ) (i,j) P is fesile for RSP-Dul, nd define y s follows. For every [i, i 2 ] [j, j 2 ] R put i 2 j 2 y i y ij = j for (i, j) = (i, j ), i =i j =j for (i, j) ([i, i 2 ] [j, j 2 ]) \ {(i, j )}. Then y is lso fesile, nd the ojective vlue for y is the sme s for y. 3

4 Figure 3: Decomposition of polygon P into sic rectngles. Proof. Let R := [k, k 2 ] [l, l 2 ] P, nd let R denote the set of sic rectngles hving their upper left corner in R. More formlly, R = {[i, i 2 ] [j, j 2 ] R : (i, j ) R}. The union of the elements of R is rectngle R = [k, k 2 ] [l, l 2 ] P with k 2 i=k l 2 j=l y ij = k 2 l 2 i=k j=l Figure 4 illustrtes the step from R to R for R = [2, 5] [3, 7] nd R = [3, 6] [3, 8]. Now the y ij Figure 4: The trnsition from R to R. The upper left corners of the elements of R re shded in the left drwing. fesiility of y implies k 2 i=k l 2 j=l y ij, nd consequently, the fesiility of y. The finl sttement out the ojective vlues is ovious. 4

5 Clerly, solution y of the form descried in Lemm cn e identified with the function g : R R, [k, k 2 ] [l, l 2 ] y k,l. This is illustrted in Figure 5, where ll the vlues of g re in {, ±}. It will turn out tht these specil vlues re sufficient to define n optiml solution for RSP-Dul. The sme rgument s in Figure 5: An exmple solution for RSP-Dul. the proof of Lemm shows tht for checking the fesiility of such function g it is sufficient to consider the constrints for rectngles tht re unions of sic rectngles. We cll these rectngles essentil. 3 The cse α = Let us ssume α(g) =, i.e. P hs no chords t ll. We fix n orienttion for the lines tht re used to decompose P into sic rectngles. The orienttion is defined y pointing wy from the concve vertices towrds the interior of P (see Figure 6 for n illustrtion). Figure 6: The orienttion of the oundries of the sic rectngles. Figure 7: The intersection points. 5

6 We will define the vlues g(r) for sic rectngles R depending on the orienttion of the oundry of R. First, we need some dditionl nottion. The vertices of sic rectngles tht re either in the interior of P or concve vertices of P re clled intersection points. The set of ll intersection points is denoted y I. In Figure 7 the intersection points re mrked y dots. Definition. A vertex of n essentil rectngle A R is clled source (with respect to A) if the two line segments on the oundry of A tht strt from re oriented wy from. In ddition, let q(a) {,, 2} e the numer of sources for A. Now we cn define function g : R {, ±} which turns out to e n optiml solution for RSP-Dul: g(r) = q(r) (R R). () In order to show the fesiility of this function we oserve tht the vlue extends to essentil rectngles. Lemm 2. For ny essentil rectngle A R, we hve g(r) = q(a). R A In prticulr, g : R {, ±} defines fesile solution for RSP-Dul. Proof. We proceed y induction on A. For A =, the sttement is precisely the definition of g. For A >, A is the union of two rectngles A A 2 s indicted in Figure 8. By induction, we d e c A A 2 f hve Figure 8: The induction step in the proof of Lemm 2. R A i g(r) = q(a i ) (i {, 2}). The vertices,, c nd d re sources for A iff they re sources for the respective A i. The vertex e is not source for ny of the considered rectngles. Finlly, it is esy to see tht f is source in precisely one of the rectngles A i, ecuse it is n intersection point. This yields q(a) = q(a ) + q(a 2 ), hence g(r) = R A R A g(r) + R A 2 g(r) = ( q(a )) + ( q(a )) = q(a). 6

7 We oserve tht every intersection point is source for exctly one sic rectngle with vertex. Hence y simple doule counting rgument, the ojective vlue for the function g is q(r) = R I. R R g(r) = R R R Our next lemm shows tht g is indeed n optiml solution. Lemm 3. The ojective vlue for g equls the upper ound from the miniml decomposition of P into rectngles. In other words, R I = N 2 c + k. Proof. Clerly, we my ssume c =. We proceed y induction on the ojective vlue h := N 2 +k. The vlue h = is possile only if P is single rectngle, nd in this cse R I = =. If h >, P hs t lest one concve vertex. Let e concve vertex such tht no concve vertex is right of. Along the verticl line through we cut the polygon P into two polygons P nd P 2. The three possile situtions re illustrted in Figure 9. P P 2 P P 2 P 2 P () () (c) Figure 9: The possile cuts (dotted lines) in the proof of Lemm 3. For i {, 2}, let N i nd k i e the numers of vertices nd holes of the respective polygons, R i the sets of sic rectngles, I i the sets of intersection points, nd let h i = N i /2 + k i e the corresponding ojective vlues. Cse (). We hve N = N + N 2 2 nd k = k + k 2, thus h = h + h 2. So h nd h 2 re smller thn h nd we cn pply the induction hypothesis to P nd P 2. Let t e the numer of intersection points for P tht re not in I I 2. Clerly, these points lie on the horizontl or on the verticl line through, s indicted in Figure. Let t e the numer of new intersection points on the verticl line, except itself, nd let t 2 e the numer of new intersection points on the horizontl line (including ), so t = t + t 2. In P, P 2 is divided into t + sic rectngles (which gives t dditionl sic rectngles), nd exctly t 2 sic rectngles of P re cut into two prts. We otin nd the clim follows y induction. R = R + R 2 + t, I = I + I 2 + t, 7

8 Figure : Cse (). The new intersection points in Figure : Cse (c). The new intersection points in Cse (). As in cse (), h = h + h 2, nd we cn pply the induction hypothesis to P nd P 2. Agin, we see tht every new intersection point (ll on the verticl line through ) cretes new sic rectngle, nd the rgument is completed s efore. Cse (c). This time we hve N = N + N 2 4 nd k = k + k 2 +, ut gin h = h + h 2, so the induction hypothesis pplies. Agin the t new intersection points lie on the verticl line through, nd P 2 is divided into t + sic rectngles (see Figure ), nd this concludes the proof. 4 The generl cse In this section we show how the generl cse cn e hndled. In the first susection we descrie the solution g in the generl cse, while the second susection is devoted to the proof of the min theorem. 4. Definition of the solution In order to define the solution s in (), we hve to put n orienttion on the chords. To fix such n orienttion let Q = H V e n independent set of size Q = α(g), where H H nd V V. In ddition, let H = H \ H nd V = V \ V. By mximlity of Q nd Hll s theorem, H cn e mtched into V, nd V cn e mtched into H. Let M (H V ) (H V ) e such mtching, i.e. M = H + V. Let Q Q e the suset of vertices tht re not mtched in M. In prticulr, we hve α = Q + M. We cll the elements of Q isolted chords. Now the verticl nd horizontl isolted chords re oriented from top to ottom nd from left to right, respectively. For n edge xy M, we direct every segment of the chords x nd y towrds the intersection point of x nd y. Figure 2 illustrtes this for the polygon in Figure 2 where the underlying mtching is M = {, 2, c3} nd the isolted chords re 4 nd d (see Figure 2 for the leling of the chords). Now we cn define g : R {, ±} y g(r) = q(r) s efore. 8

9 4.2 Proof of the min theorem Figure 2: The orienttion in the generl cse. When we try to prove the fesiility of g y induction on A s in Lemm 2, prolem rises in the cse tht f is the endpoint of horizontl isolted chord s indicted in the left hnd side of Figure 3. Then f is not source for ny of the rectngles A nd A 2. The right hnd side of the d e c d e c A A 2 A A 2 f f Figure 3: The prolem in the induction step for the fesiility of g (nd its solution). sme figure shows solution for this prolem: we just extend the orienttion to the first segment of the oundry immeditely following the isolted chord. Of course, we hve to do this for every isolted chord. Oserve tht in Figure 2 this is done for oth isolted chords (which re 4 nd d in this exmple). After this modifiction the rgument for Lemm 2 proves the following generl version. Lemm 4. For ny essentil rectngle A R, we hve g(r) = q(a). R A In prticulr, g : R {, ±} defines fesile solution for RSP-Dul. Now it remins to prove the optimlity of g. There re some specil intersection points: the points where two chords meet which re mtched in M. Clerly, these re never source for ny incident sic rectngle. But ech of the remining intersection points is source for precisely one sic rectngle. So we otin the ojective vlue g(r) = R I + M. R R 9

10 Lemm 5. We hve R I + M = N 2 c + k α. Proof. As in the proof of Lemm 3 we my ssume c =. We proceed y induction on α. The cse α = ws treted in Section 3. So ssume α >, let Q e some independent set of chords of size Q = α, nd choose ny chord Q (w.l.o.g. horizontl). Now we cut the polygon P long the chord. We hve to distinguish two different cses: either the cut divides P into two polygons P nd P 2, or P stys connected, ut the numer of holes decreses y (see Figure 4 nd 7). Cse (Figure 4). For i {, 2}, denote the prmeters of P i y N i, k i nd α i. Similrly, the corresponding sets of sic rectngles nd intersection points re denoted y R i nd I i, respectively. Oserve tht P i inherits mximum independent set Q i nd corresponding P P P 2 P 2 Figure 4: A cut dividing P into two prts (Cse ). Mtching M i from P : Q i is the set of chords in P tht re lso chords in P i, nd M i is the set of elements xy M such tht x nd y re oth in Q i. We hve N = N + N 2, k = k + k 2, α = α + α 2 +. Hence, y induction, it is sufficient to prove tht R I + M = ( R I + M ) + ( R 2 I 2 + M 2 ). (2) Let T e the set of intersection points on the chord. This set splits into three susets. For i {, 2}, we put T i := {u T : u is vertex of n element of R i, ut not of n element of R 3 i.}, nd in ddition T 3 = T \ (T T 2 ). Oserve tht T 3 is the set of points where the chord meets other chords. For instnce, in Figure 5, we hve T = {, 3 }, T 2 = { 2, 4 }, T 3 = { 5, 6 }. Next we define function φ : T T 2 N. For u in T, let φ(u) e the numer of intersection points for P on the verticl line through u lying elow u. Similrly, for u in T 2, let φ(u) e the numer of intersection points for P on the verticl line through u lying ove u. In oth

11 Figure 5: The intersection points on the cutting chord. Figure 6: The dditionl intersection points. cses, the point u itself is included. In the exmple from Figure 5, we hve φ( ) = 4, φ( 2 ) = 2 nd φ( 3 ) = φ( 4 ) = 3 (see Figure 6). The function φ counts the intersection points of P tht re not intersection points of P or P 2 : I = I + I 2 + φ(c) + T 3. (3) c T T 2 On the other hnd, φ cn e used to count the dditionl sic rectngles. The verticl line through u T divides φ(u) sic rectngles of P 2 into two prts, nd the verticl line through u T 2 divides φ(u) sic rectngles of P into two prts, hence R = R + R 2 + φ(u). (4) u T T 2 The mtching M cn e written s disjoint union M = M M 2 M 3 s follows. For i {, 2}, M i is the set of edges xy M such tht x H, y V, the chord y does not intersect, nd oth vertices of y re in P i. This is consistent with the ove description of the mtchings M nd M 2. The remining mtching edges xy M such tht x H, y V, nd the chord y intersects, re collected in M 3. Clerly, for every such mtching edge xy M 3, the verticl chord y intersects the chord in some point from T 3. On the other hnd, for every point T 3, the chord y tht intersects in does not elong to our mximl independent set Q, so it is mtched in M. Consequently, there is one-to-one correspondence etween M 3 nd T 3, nd we otin Putting together equtions (3), (4) nd (5), we otin R I + M = R + R 2 + M = M + M 2 + T 3. (5) φ(u) u T T 2 I + I 2 + This is (2), nd thus concludes the proof in this cse. φ(u) + T 3 + ( M + M 2 + T 3 ). u T T 2

12 Cse 2 (Figure 7). Essentilly the proof is the sme s in Cse. For the prmeters of P, we otin N = N, k = k, α = α. P P Figure 7: A cut tht kills hole (Cse 2). So induction pplies to P, nd we hve to show tht R I + M = R I + M. (6) As efore, T is the set of intersection points on the chord. In nlogy to Cse, T is the set of u T such tht u sees concve vertex when it looks upwrds, ut u does not see concve vertex when it looks downwrds. Similrly, T 2 is the set of u T such tht u sees concve vertex when it looks downwrds, ut u does not see concve vertex when it looks upwrds, nd finlly T 3 = T \ (T T 2 ), the set of points where meets other chords. For u T, let φ(u) e the numer of intersection points on the verticl line through u lying elow u, nd for u T 2, let φ(u) e the numer of intersection points on the verticl line through u lying ove u (in oth cses we include u itself). By the sme counting rguments s in Cse we otin I = I + φ(u) + T 3, u T T 2 R = R + φ(u), u T T 2 M = M T 3. These equtions imply (6), nd this concludes the proof. As consequence of Lemms 4 nd 5, we otin tht g solves RSP-Dul. Theorem 2. The function g : R {, ±} with g(a) = q(a) defines n optiml solution for RSP-Dul. Corollry. There is no integrlity gp in the rectngle segmenttion prolem for inry input mtrices. 2

13 References [] K. Engel. Optiml mtrix-segmenttion y rectngles. In: Discr. Appl. Mth (29), pp doi:.6/j.dm [2] L. Ferrri, P.V. Snkr nd J. Sklnsky. Miniml Rectngulr Prtitions of Digitized Blos. In: Computer Vision, Grphics nd Imge Processing 28 (984), pp doi:.6/ X(84) [3] W. Lipski, E. Lodi, F. Luccio, C. Mugni nd L. Pgli. On two dimensionl dt orgniztion II. In: Fund. Informtice 2 (979), pp [4] T. Ohtsuki. Minimum dissection of rectiliner regions. In: Proc. IEEE Symposium on Circuits nd Systems. 982, pp