Logic Puzzles in C#.NET

Printing QR Code in C#.NET Logic Puzzles

To solve the version where N s additional piece of information is that one of his kids was born after he bought the house, change the last predicate in the outer query s lter to age1 < age2.

Let n be the zero-based position of the number in the sequence (0, 1, 2, 3, ). The given numbers are what you get if you begin with the number n, then take the factorial n times in sequence that is, 0, 1!, 2!!, 3!!!, and so on. The lengthy last number in the sequence is 3!!!, so the next number in the sequence is 4!!!!. It is probably prudent not to include the actual number in this space because it wouldn t leave space for anything else.

The puzzle doesn t restrict you to placing the item you re weighing on one side of the scale and the weights on the other. Therefore, you can place weights on both sides. To simplify the solution s explanation, rst assume that there was a restriction to place the item you re weighing on one side of the scale and the weights on the other. Given a set of weights, to measure some item s weight (call it w), you need to use a subset of the weights you have that is, each weight from your set of weights will be either used or not used to weigh the item. So any w in the range 1 through n must be representable with a binary system, where each bit represents a different weight from your set of weights, and only the bits of the participating weights will be turned on. The best strategy is to use the positional values of binary representation. For example, to represent any integer in the range 1 through 100, you need 7 bits (1, 2, 4, 8, 16, 32, and 64). Notice that you get a geometric sequence (also known as a geometric progression) with a common ratio 2 (1 20, 1 21, 1 22, 1 23, and so on). To use any set of weights, their total weight must be at least the largest weight you need. The simpli ed formula for the sum of the geometric sequence in our case is 2num_weights 1, and this sum must be greater than or equal to n. Hence, the minimum number of weights required is ceiling(log2(n+1)). Next, remove the restriction to place weights only on one side of the scale. Now each weight from your set of weights can assume one of three roles: rst, placed on the same side of the scale as the item you re weighing (a negative value); second, placed on the other side of the scale (a positive value); and third, not used (a 0 value). If you think about it, just like you can represent numbers using a binary system where each bit represents a different power of 2, you can represent numbers using a ternary system where each trit (ternary digit) represents a different power of 3. A ternary system where each trit can be 1, 0, and +1 is known as a balanced ternary system. As an example, in this system the number 150 is represented as 1 (0) + 3 ( 1) + 9 ( 1) + 27 (0) + 81 ( 1) + 243 (+1). Though very cumbersome,

such a system provides the optimal solution in terms of the number of weights required to weigh any object. With a set of weights that are consecutive powers of 3 starting with 1 and on (1, 3, 9, 27, 81, . . .) whose sum is s, you can express any number in the range s through s. In our case, only the positive numbers are relevant. So in order to be able to weigh any w in the range 1 through n, you need the sum of the values represented by the trits to be greater than or equal to n. This time, the common ratio of our geometric sequence is 3. The simpli ed sum of the geometric sequence is (3num_weights 1) 2. To represent any integer in the range 1 through n, the minimum number of weights required is ceiling(log3(2 n+1)).