Medium | Discrete Maths |

Assume 100 zombies are walking on a straight line, all moving with the same speed. Some are moving towards left, and some towards right. If a collision occurs between two zombies, they both reverse their direction. Initially all zombies are standing at 1 unit intervals. For every zombie, you can see whether it moves left or right, can you predict the number of collisions?

Hint

On every collision, assume that the two zombies don't reverse direction but simply cross each other.

Solution

Since we can assume that zombies can pass through each other, for a zombie moving right, count the number of zombies to its right moving left. Add this number for every right moving zombie. That is the number of collisions.

Source: CSEblog

Enable Like and Comment Medium | Discrete Maths |

An infection spreads among the squares of an nXn checkerboard in the following manner. If a square has two or more infected neighbors, it becomes infected itself. (Each square has 4 neighbors only!). Prove that you cannot infect the whole board if you begin with fewer than n infected squares.

Hint

Invariance

Solution

Perimeter of infected area can't increase. It stays constant or decreases. Initially maximum perimeter is 4*k if k blocks are infected. But to infect all blocks, the perimeter must increase to 4*n, k<n. This is not possible

Source: P. Winkler

Enable Like and Comment Hard | Discrete Maths |

A group of 5 people want to keep their secret document in a safe. They want to make sure that in future, only a majority (>=3) can open the safe. So they want to put some locks on the safe, each of the locks have to be opened to access the safe. Each lock can have multiple keys; but each key only opens one lock. How many locks are required at the minimum? How many keys will each member carry?

Hint

For each group of 2 ppl, there must be a lock which none of them have a key to.

Answer

10 locks, 6 Keys.

Solution

For each group of 2 ppl, there must be a lock which none of them have a key to. But the key of such a lock will be given to the remaining 3 ppl of group. Thus, we must have atleast 5C2 = 10 Locks. Each lock has 3 keys, which is given to unique 3-member subgroup. So each member should have 10*3/5 = 6 keys.

Hard | Discrete Maths |

We have a beam balance (with two pans to compare weights) and a positive integer N. How do we select fewest number of pebbles to weigh all possible integers from 1 to N

Solution

We will require the set (1,3,9.....3^x )

where x is lowest integer with 3^x > N. This is true because each number now has exactly one ternary representation. Any 2*3^i can always be represented as 3^(i+1) - 3^i. So, there is a unique way of representing a number in the form of sigma s_i*3^i where s_i belongs to {0, 1, -1}. So, this is optimal

Verify that this can be used to weigh all integers from 1 to N

Number of pebbles = log_{base:3} (2N+1)

Solution by Palak Bhushan

If each weight w_i has k copies, then 2k+1 combinations can be weighed using them (from -k*w_i to 0 to k*w_i). So, if we choose w_i = (2k+1)^{i-1}, i=1...p, then everything till k*((2k+1)^p - 1)/((2k+1)-1) = ((2k+1)^p-1)/2 can be weighted, thus requiring p=log_{2k+1} (2N+1) number of distinct weights.

where x is lowest integer with 3^x > N. This is true because each number now has exactly one ternary representation. Any 2*3^i can always be represented as 3^(i+1) - 3^i. So, there is a unique way of representing a number in the form of sigma s_i*3^i where s_i belongs to {0, 1, -1}. So, this is optimal

Verify that this can be used to weigh all integers from 1 to N

Number of pebbles = log_{base:3} (2N+1)

Solution by Palak Bhushan

If each weight w_i has k copies, then 2k+1 combinations can be weighed using them (from -k*w_i to 0 to k*w_i). So, if we choose w_i = (2k+1)^{i-1}, i=1...p, then everything till k*((2k+1)^p - 1)/((2k+1)-1) = ((2k+1)^p-1)/2 can be weighted, thus requiring p=log_{2k+1} (2N+1) number of distinct weights.

Source: leino

Enable Like and Comment Hard | Discrete Maths |

Several gas stations on a circular trek have between them just enough gas for one car to make a complete round trip. Prove that if you start at the right station with an empty tank you shall be able to make it all the way around.

Hint

Induction

Solution

Take N=2 gas stations. Assume their capacities are x1 and x2 (in terms of distance). Now, x1+x2 = 1. Assume y1 and y2 be clockwise distances from 1 to 2 and 2 to 1 (clockwise), y1+y2 = 1. Hence we cant have x1<y1 and x2<y2 both. Thus, let x1 >= y1. In this case we start with station 1 and complete the journey. Assume the path is possible for N=K stations, we want to prove for N = K+1 stations, using same argument as above, xi>=yi for some i, and hence reaching xi with zero fuel ensures reaching x(i+1). Thus we merge xi and x(i+1), and the problem is reduced to N=K stations, which we assumed!

Another method is to imagine having a big tank with enough gas for a round trip and enough room for going through the motions of emptying every gas station on your way. Start at any station and mind to record the amount of gasoline on reaching gas stations on your way around. At the end of the trip, when you pull into the station of departure with the original amount of gas, check your list. The station marked with the least number is the one where you want to start on an empty tank.

Another method is to imagine having a big tank with enough gas for a round trip and enough room for going through the motions of emptying every gas station on your way. Start at any station and mind to record the amount of gasoline on reaching gas stations on your way around. At the end of the trip, when you pull into the station of departure with the original amount of gas, check your list. The station marked with the least number is the one where you want to start on an empty tank.

Source: P. Winkler

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