1/-1 matrix - problem #29 from Problem solving strategies

Here is the solution to one of the problems mentioned in the earlier blog post:

Problem 29: Suppose contrary to the claim a1 + b1 + … + a25 + b25 = 0. Then:

(a1 + a2 + … + a25) = - (b1 + … + b25) ⋯ ①

Now note that ai= -1,if number of elements having value -1 in row i is even, and 1 otherwise.

Thus (a1 + a2 + … + a25) = -x + y,  where x is the number of rows having odd number of -1s and y is the number of rows having even number of -1s ⋯ ②

Similarly, (b1 + … + b25) = -m + n, where m is the number of columns having odd number of -1s and n is the number of columns having even number of -1s ⋯ ③

Combining ①, ② and ③ results in the equation (y – x ) = (m - n) ⋯ ④

But (y + x) = 25, the total number of rows in the matrix, and (m + n) = 25, the total number of columns in the matrix. Thus (y + x) = (m + n) ⋯ ⑤

Combining ④ and ⑤ results in the equality y = m, which implies that the number of rows with even number of -1s equals the number of columns with odd number of -1s.

Define z = number of -1s in the matrix.

Suppose y is odd. Then, x must be even and by original definition of x and y, z must be even (even number of rows with odd number of -1s plus odd number of rows with even number of -1s, resulting in even + even). If y is even, z must be odd. Thus parity(z) ≠ parity(y) ⋯⑥

On the other hand, by definition of m and n, it can be shown that if m is odd, then z must be odd and if m is even then z must be even. Thus parity(z) = parity(m). But this contradicts ⑥ since y = m as per an earlier equation. Thus the assumption a1 + b1 + … + a25 + b25 = 0 is wrong.

Tile colouring problem (#8) from Problem solving strategies

Problem: Prove that an a × b rectangle can be covered by 1 × n rectangles iff n|a or n|b.

Solution: The proof is given by means of the following 2 claims.

Claim 1: Given a rectangular area is covered by tiles of dimension 1xn placed in horizontal or vertical orientation, it is possible to remove the tiles one at a time in such a way that at any point of time, the covered region’s top border consists of a sequence of steps increasing from left to right as shown in diagram 1.

Proof: Define step blocks to be rectangular regions defined by the step-like shape of the covered area. In diagram 1, step blocks are represented by the areas shaded blue. Start from the left most step block and consider the top left tile in that block. This tile must fall entirely within the current block (call it block A) or must extend beyond the right of the current step block (it cannot beyond the bottom of the current block since the first block is also the bottommost). In the former case, removing the tile will leave behind a step shape, thereby proving the claim. In the latter case, inspect the next step block for the possibility of removing a tile. The top left tile of this next block (call it block B) cannot extend beyond the bottom of that block since the block A’s top left tile occupies that portion as per the previous statement. Hence the top left tile of block B too must either lie entirely within B or extend beyond the right of B. As before, in the former case, removing the tile will leave a step shape whereas in the latter case, the search for a tile to remove can be repeated with the block to the right of C. Note that this procedure can be repeated at most till the rightmost block is reached, in which case the top left tile cannot extend beyond the right of the block (and must therefore lie entirely within that block), thereby becoming eligible for removal while retaining the step shape.

Claim 2: When tiles are removed as per the procedure of claim 1 and neither a not b are divisible by n, then at any point of time at least one of the step regions on the top border will have a height and a width (as defined by diagram 2) that are both indivisible by n.

Proof: The proof is by induction on r, the number of tiles removed so far under the procedure of claim 1. When r is 0, the covered region and also the only step region is the original axb rectangle. The height and width, namely a and b, are both indivisible by n as per the assumption. Suppose the claim is true for values of r upto R. Now consider r = (R + 1). Suppose the tile being removed at the current iteration lies in a step region whose width is W and height Is H. If either W or H is divisible by n, then there must be some other step region with width and height indivisible by n and that region will remain after iteration (R + 1)  as well. But if say W and H are both indivisible by n. After removing the tile in this iteration, the 2 new step regions would have dimensions (a) (W, H - 1) and (W - n, H) or (b) (W, H - n) and (W - 1, H), depending on the orientation of the tile; in either case, one of the new step regions would have its width and height indivisible by n. Hence proved.

Diagram 1: Step blocks denoted by shaded blue regions

Diagram 2: Width of step regions are denoted by L1 and L2; heights are denoted by H1 and H2. The rectangle  bounded by blue and red lines is the top left tile of its step block and is eligible for removal as per the procedure of claim 1

Colouring problem #24 - 3-colour space

Problem: Every space point is colored with exactly one of the colors red, green or blue. The sets R, G, B consist of the lengths of those segments in space with both end points red, green and blue, respectively. Show that at least one of these sets contains all non-negative real numbers.

Solution:
(With reference to the diagram) Suppose the 3 colours are C1, C2 and C3. Assume that the minimum distances by which no two points with the same colour are separated (i.e. no line segment with end points of the same colour exist) be x, y and z for C1, C2 and C3 respectively. Suppose x >= y >= z. Then as shown in the diagram, any point coloured C1 would form the centre of a sphere of radius x such that the sphere can have only points coloured C2 or C3. As indicated in the diagram, a circle of diameter x√3 would then exist on the surface of the sphere such that all its points are coloured C3 and the chords of the circle form line segments of all possible lengths upto x√3, contradicting the assumption that no two points coloured C3 exist at a distance of z from each other.

Diagram for problem #24: The green circle is formed by points on surface of the sphere that are at a distance of x from the top point (on spherical surface) coloured C2, where x is the radius of the sphere centered at a point coloured C1. The diameter of the circle is xv3. If neither of the colours C1 and C2 have 2 points at a distance of x, then the circle must be coloured C3 entirely.

Colouring problems from Arthur Engel - problem 14

Diagram p14

Problem 14: Every space point is colored either red or blue. Show that among the squares with side 1 in this space there is at least one with three red vertices or at least one with four blue vertices.

Solution: (With reference to diagram p14)

Lemma 1: If there exist 2 red vertices at a distance of 1 unit and there doesn’t exist any unit square having at least 3 red vertices, then there must exist a unit square with 4 blue vertices.

Proof: With reference to diagram 1 on this page – draw parallel unit circles around the 2 red vertices; the circles must be perpendicular to the line joining the 2 red vertices. No point on these 2 circles can be coloured red since otherwise that point, along with the 2 red center points and the counterpart point on the other circle would form a unit square having 3 red vertices.

Lemma 2: If no 2 red vertices lie at a distance of 1 unit apart, then there must exist a unit square with 4 blue vertices.

Proof: 
(a) If there is no red vertex in the space, then all points must be blue and an all-blue unit square exists obviously. 
(b) If a red vertex does exist, draw a unit sphere around it. All points on the sphere must be blue and it is easy to construct a unit square having its 4 vertices on the spherical surface and hence being an all-blue unit square.

Combining lemmas 1 and 2, it follows that there must be a unit square having either 4 blue vertices or (at least) 3 red vertices.

Reference

1. Arthur Engel's book - reference on "Mathematical Olympiads" Wordpress blog