CSCI 150 Discrete Mathematics Homework 5 Solution

Transcription

1 CSCI 150 Discrete Mathematics Homework 5 Solution Saad Mneimneh Computer Science Hunter College of CUNY Problem 1: Happy Birthday (if it applies to you)! Based on the size of the class, there is approximately a 45% chance that someone s birthday is today. Eight candles are placed on the perimeter of a rectangular cake. For the sake of the math, each candle can be thought of as a point on the actual perimeter of the rectangle. Here s a possible example (just an example): This problem involves cutting cake, so we assume that cuts are straight and non-trivial, i.e. no piece should be empty or equal to the entire cake. (a) I ask you to cut the cake to make two pieces, and I let you take the piece with the most candles (I am nice). How many candles can you guarantee? Hint: Some simple logic. Solution: You can guarantee 7. Cut out the tiniest corner and take the rest. The corner can contain at most one candle (right at the corner). Observe that you can t do better, e.g. when all four corners have a candle. (b) If I cut the cake instead, but I still let you take the piece with the most candles, how many candles can you guarantee? Hint: Pigeon-hole. Solution: No matter how I cut the cake, one of the pieces will have at least 8/ = 4 candles by piegonhole (8 objects, two boxes). So you can t get less. (c) If I let you cut the cake, and give you the piece with the most candles, but require that the two pieces must have the same amount of cake (same area). How many candles can you guarantee? Hint: pigeon-hole with a twist. Solution: Pick one of the 8 points on the perimeter, say point p, and construct the line that goes through p and the center of the rectangle. Due to symmetry,

2 this line cuts the rectangle in two pieces of equal area (we will skip the formal argument). Observe that p is on the perimeter of both pieces. Aside from p, 7 points remain on the perimeter of the rectangle. By pigeon hole principle, one of the pieces must have at least 7/ = 4 among these on its perimeter. Together with p, that s five. Problem : Grid Consider the following grid: Experiment with this activity: Color five of the points black. Observe that at least two black points must satisfy the following property: The line segment that joins the two black points passes through a third one (either black or white). If you attempt to prove this property, what technique will you use and why? Try to formally prove this property. Hint: Think of the coordinates of each point as being either even or odd. Solution: The statement of the problem suggests the use of pigeonhole, since no matter how we choose the 5 points, they must satisfy the condition. A point can be categorized as (even, even), (even, odd), (odd, even), or (odd, odd) based on the parity of its two coordinates. If we imagine 4 boxes for these categories, placing the points in 4 boxes means that one of the boxes must contain at least two points. Let p = (x, y) and q = (a, b) be two such points. Then x and a have the same parity, and y and b have the same parity. This means x + a is even, and y + b is even. Therefore, the mid point z = ( x + a, y + b ) has integer coordinates, and therefore, must be another point on the grid. Problem 3: Languages In a certain class there are 5 students: 14 speak Spanish, 1 speak French, 6 speak French and Spanish, 5 speak German and Spanish, and speak all three. The 6 that speak German all speak another language. How many speak no foreign language? Solution: S = 14 F = 1 F S = 6 G S = 5

3 F G S = G = 6 G F is missing but we can determine it from the information that says all German speaking students also speak another language. Therefore, Now, G = (G S) (G F ) G = G F + G S G F S 6 = G F + 5 G F = 3 G F S = = 0 So 0 students speak a foreign language. Therefore, 5 students don t. Note: Another way to solve the problem is to say the following: since G is entirely covered by other sets, i.e. no element in G that is not in another set, then G F S is the same as F S, so we can do inclusion-exclusion on F and S alone to obtain the result. Problem 4: A 3-digit number Consider this question: How many 3-digit numbers are there if the first digit cannot be 1, the second digit cannot be, and the third digit cannot be 3. This can be easily answered using the product rule. There are 8 choices for the first digit (it can t be 0 or 1), there are 9 choices for the second digit (it can t be ), and there are 9 choices for the third digit (it can t be 3). Therefore, there are such numbers. (a) Reproduce the same result using inclusion-exclusion. Let P i be the set of three digit numbers that have i in the i th digit, and count P 1 P P 3. Solution: If the first digit is 1, then the second and third digit have possibilities. If the second digit is, then the first digit can be anything except 0, and the third digit has 10 possibilities, giving 90 combinations. Using similar arguments, we have: P 1 = P = 9 10 P 3 = 9 10 P 1 P = 10 P 1 P 3 = 10 P P 3 = 9 P 1 P P 3 = 1 Using the inclusion-exclusion formula, we get P 1 P P 3 = = 5

4 Therefore, the 3-digit numbers that satisfy the condition that digit i th cannot be i, must be the total minus the above, which is = 648. (b) What if in addition, the three digits must be different? Solution: We can solve this using the same strategy but in addition require that the 3 digits are different. If the first digit is 1, then this leaves 9 choices for the second digit, and 8 choices for the third. So P 1 = 9 8. If the second digit is, then the first digit can be anything by 0 or, and that s 8 possibilities. The third digit has also 8 possibilities because it can t be the same as the first or the second. Using similar arguments, we get: P 1 = 9 8 P = 8 8 P 3 = 8 8 P 1 P = 8 P 1 P 3 = 8 P P 3 = 7 P 1 P P 3 = 1 Using the inclusion-exclusion formula, we get P 1 P P 3 = = 178 Subtracting from the total, which is now (first digit can be anything by 0, second digit can be anything but the first, and third digit can be anything but the first and second), we get 470. Problem 5: Integer solutions How many non-negative integer solutions are there of the equation x+y+z = 0 satisfying x 10, y 5, and z 15. Hint: we have seen how to solve this when the constraints are lower bounds, i.e. using > instead of. Let S x be the set of all solutions in which x > 10. Define S y and S z similarly, and find S x S y S z and subtract this number from all possible solutions (with only x, y, z 0). Solution: ( Let S be the) set of ( all possible ) non-negative solutions. We know that S = =. Let P be the property that x 11, P be the property that y 6, and P 3 be the property that z 16. Then we want the solutions in S possessing none of the properties P 1, P, and P 3. Let N(i) be the number of solutions possessing property P i, and N(i, j) be the number of solutions possessing properties P i and P j, and finally N(1,, 3) be the number of solutions possessing all three properties. Then the number of solutions possessing none of these properties is (by inclusion-exclusion): ( ) [ N(i) ] N(i, j) + N(1,, 3) i<j i Consider N(1). If x 11, then x = 11 + x where x 0, ( and ) the equation 11 becomes x + y + z = 9. The number of solutions is N(1) =. Similarly,

5 ( ) ( ) 16 6 N() =, and N(3) =. Next, N(1, ) is the number of solutions satisfying ( x ) 11 and y 6, resulting in an equation x + y + z = 3 with 5 N(1, ) =. Now N(1, 3), N(, 3), and N(1,, 3) are all zero, so we have: ( ) ( 11 ) ( 16 ) ( 6 ) ( 5 + ) = 51