Medium | Discrete Maths |

In a circle are light bulbs numbered 1 through n, all initially on. At time t, you examine bulb number t, and if it’s on, you change the state of bulb t + 1 (modulo n); i.e., you turn it off if it’s on, and on if it’s off. If bulb t is off, you do nothing. Prove that if you continue around and around the ring in this manner, eventually all the bulbs will again be on.

Hint

It doesn't matter what the rules are, only thing that matters is that all bulbs should not go off!

Solution

Suppose every orientation of light bulbs & position of current pointer (t modulo n) forms a different state. Since all bulbs don’t go off, we must repeat a state after finite number of steps. Also notice that every state has a unique pre-state. Suppose the two states are at time T1 & T2. Notice that at time 0 all bulbs are on. Thus moving backwards from both states, we arrive at T2-T1, where all bulbs are on again.

Source: P. Winkler

Enable Like and Comment Medium | Discrete Maths |

There is a 6x8 rectangular chocolate bar made up of small 1x1 bits. We want to break it into the 48 bits. We can break one piece of chocolate horizontally or vertically, but cannot break two pieces together! What is the minimum number of breaks required?

Answer

47

Solution

For a chocolate of size mxn, we need mn - 1 steps. By breaking an existing piece horizontally or vertically, we merely increase the total number of pieces by one. Starting from 1 piece, we need mn - 1 steps to get to mn pieces.

Another way to reach the same conclusion is to focus on "bottom left corners of squares": Keep the chocolate rectangle in front of you and start drawing lines corresponding to cuts. Each cut "exposes" one new bottom left corner of some square. Initially, only one square's bottom left corner is exposed. In the end, all mn squares have their bottom left corners exposed.

Another way to reach the same conclusion is to focus on "bottom left corners of squares": Keep the chocolate rectangle in front of you and start drawing lines corresponding to cuts. Each cut "exposes" one new bottom left corner of some square. Initially, only one square's bottom left corner is exposed. In the end, all mn squares have their bottom left corners exposed.

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.

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