That was good! I used induction to proof the problem. First define some terms, (1) A rectangular is a rectangular of any size filled with black square(s), a single black square is a 1*1 rectangular. (2) A L shape is something likewww.ddhw.com ******* * * (* stands for black square) (3) Two rectangulars are contiguous if and only if they intersect or share a (or part of ) edge. Let's introduce several lemmas. (1) Rectangular can not fill more black squares by itself (Proof omiited) (2) L shape can generate a whole rectanguar (Proof omitted)www.ddhw.com (3) If two rectangulars are not contiguous, they can not fill more black squares by themself Proof, since two rectangulars are not contiguous, No such thing like * ** Can be found in them, so we can not use the rule (4) When the process of black square growing can not continue, only rectangular left.www.ddhw.com Proof, (i) at the beginning, only 1*1 rectangular (ii) If two rectangulars are contiguous, they form 0,1 or 2 L shape(s), filling these L shape (lemma 2) gets a new rectangular which contain both 2 original rectangulars. Now we prove the generalized form of the original problem To generate a rectangular of m*n , at least m+n-1 original squares are needed. Proof by induction (1) 1*1 rectangular needs 1+1-1=1 original black square, (note 1*1 rectangular is not generated, must be original) (2) By lemma (1), we need two contiguous (lemma 3) rectangulars to fill more black squares.www.ddhw.com Suppose the two rectangulars are m1*n1, m2*n2, they have m1+n1-1, m2+n2-1 original black squares respectively. Since the two rectangular must overlap at least one row or column so the new rectangular (suppose m*n) must be less than (not equal to) (m1+m2)*(n1+n2), that is m The generalized form is correct www.ddhw.com |