The puzzle of water jugs

Transcription

1 The puzzle of water jugs Umesh P N May 4, 2012 Contents 1 The general problem 2 2 Analysis 2 3 Solution 2 4 Examples Example 1: p = 5, q = 3, r = With an extra container Without an extra container Example 2: p = 15, q = 25, r = Example 3: p = 15, q = 25, r = A more general solution 6 6 Further problems 7 1

2 1 The general problem Given two jars of volumes a and b litres, measure c litres. a, b and c are integers. 2 Analysis We need to arrive at c by adding and/or subtracting a and b multiple times. So, a general solution is obtained by solving whwre x and y are integers (positive, negative or zero). ax + by = c (1) Since gcd(a, b) divides both a and b, it should divide c as well. So, if a and b are not realatively prime, the solution reduces to a x + b y = c (2a) where p = gcd(a, b) a = a/p b = b/p c = c/p (2b) (2c) (2d) (2e) It is often easier to solve the equation where p, q and r are non-negative integers. px qy = r (3) 3 Solution There are many methods to solve (3). The general method is to expand m n or n m as a continued fraction, solve the equation px qy = 1 and extend the solution to (3). Greatest Common Divisor. The largest number that divides both a and b. For example, gcd(100, 80) = 20. 2

3 An easier method to solve particular problems is to solve it as a modular equation, i.e. px r (mod q) (4) (4) is guaranteed to have one (and only one) solution in the range x = 0, 1 (q 1), so by choosing the smaller number as q, this problem is easily solved. 4 Examples 4.1 Example 1: p = 5, q = 3, r = 4. This is the question Kuttyedathi asked in Google+. It can be stated as You have two jugs that can measure 3 liters and 5 liters. How can you measure exactly 4 liters? For this problem, (3) becomes or which are reduced to 5x 3y = 4 3x 5y = 4 (5a) (5b) and 5x 4 (mod 3) (6a) 3x 4 (mod 5) (6b) Putting x = 0, 1, 2 in (6a), we find x = 2 is a solution. So, x = 2, y = 2 is a solution. That means Similarly = 4 (7a) is a solution to (6b). These are the two basic solutions of the problem = 4 (7b) 3

4 4.1.1 With an extra container From (7a), we get the following solution: Solution 1 Fill the 5-liter jar twice and empty that in a separate container. From that container, fill the 3-liter jug twice. The container will have 4 liter water left. Similarly, (7b) will lead to the following solution: Solution 2 Fill the 3-liter jar three times and empty that in a separate container. From that container, fill the 5-liter jug once. The container will have 4 liter water left Without an extra container What if we don t have an extra container to hold the intermediate measures of water? Extra container corresponds to the temporary values in the equations (7a) and (7b). We need to rewrite it in such a way that each intermediate value is between -(volume of the last jug used) and +(volume of the next jug used). (7a) is Rewriting, = 4 This leads to the following solution: 5 (3 (5 3)) = 4 Solution 3 This can be done in the following steps: Fill the 5-liter jar and pour it to the 3-liter jar. 2 liter will remain in the 5-liter jug. Empty the 3-liter jug and pour the remaining 2 liter water from the 5-liter jug into it. (Now, it can accommodate 1 more liter.) Fill the 5-liter jug and pour it into the 3-liter jug. After 1 liter is poured, 3-liter jug will be full, and the 5-liter jug will have 4 liter of water. Use the 4 liter water in the 5-liter jug as you wish. 4

5 Similarly, rewriting (7b), = 4 Or, leading to the following solution. ( ) + 3 = 4 Solution 4 This can be done in the following steps: Fill the 3-liter jug and pour it in the 5-liter jug. Now the 5-liter jug can accommodate 2 liter more. Fill the 3-liter jug again and pour it in the 5-liter jug. After the 5-liter jug is full, the 3-liter jug will have 1 liter left. Empty the 5-liter jug. Pour the 1 liter water remaining in the 3-liter jug into the 5-liter jug. Fill the 3-liter jug again and pour it into the 5-liter jug. Now the 5-liter jug will have = 4 liter water. Use the 4 liter water in the 5-liter jug as you wish. 4.2 Example 2: p = 15, q = 25, r = 20 Here is another question: You have two jugs that can measure 15 liters and 25 liters. How can you measure exactly 20 liters? We need to solve 15x + 25y = 20 Since gcd(15, 25) = 5, using (2), this reduces to 3x + 5y = 4 This is same as the previous problem, so the solution is the same. 5

6 4.3 Example 3: p = 15, q = 25, r = 14 Here is another question: You have two jugs that can measure 15 liters and 25 liters. How can you measure exactly 14 liters? We need to solve 15x + 25y = 14 Since gcd(15, 25) = 5 doesn t divide 14, this is not solveable. 5 A more general solution See (1), (2a) or (3). There is one equation and two variables. That means there are more than one solutions. In fact, there are infinite number of solutions to these equations. If (x 0, y 0 ) is a solution to (3), then all values of (x, y) satisfying x = x 0 + kq y = y 0 + kp where k is an integer, are also solutions. Now, for (5a), we see that x 0 = 2, y 0 = 2 is a solution. So, x = 2+3k, y = 2+5k should be the general solution. That means, all of the following are solutions: x = 2, y = 2 x = 5, y = 7 x = 8, y = 12 will be solutions to (7a). If the number of equations is less than/equal to/greater than the number of variables, the system has zero/one/more than one solutions. It is easy to verify: p(x 0 + kq) q(y 0 + kp) = (px 0 qy 0 ) + (kpq kpq) = r. 6

7 Similarly, for (5b), we see that x 0 = 3, y = 1 is a solution. So, x = 3 + 5k, y = 1 + 3k the general solution. That means, all of the following are solutions: will be solutions to (7b). x = 3, y = 1 x = 8, y = 4 x = 13, y = 7 These are solutions with extra container. By suitably arranging the terms, we ll be able to find solutions without extra container also. 6 Further problems There are many problems that follow: 1. How can a solution without containers be obtained from a solution with containers (rather than trial and error)? 2. How can we find a solution that requires minimum number of steps? 3. If there is another requirement that no water should be wasted (without an extra container), is the problem solveable? Solutions to these are left to the reader as exercises :) Can we? I am not certain now. Does anybody care to prove this or provide a counterexample? 7