100 Prisoners and a Light Bulb

There are 100 prisoners in solitary cells. There's a central living room with one light bulb; this bulb is initially off. No prisoner can see the light bulb from his or her own cell. Everyday, the warden picks a prisoner equally at random, and that prisoner visits the living room. While there, the prisoner can toggle the bulb if he or she wishes. Also, the prisoner has the option of asserting that all 100 prisoners have been to the living room by now. If this assertion is false, all 100 prisoners are shot. However, if it is indeed true, all prisoners are set free and inducted into MENSA, since the world could always use more smart people. Thus, the assertion should only be made if the prisoner is 100% certain of its validity. The prisoners are allowed to get together one night in the courtyard, to discuss a plan. What plan should they agree on, so that eventually, someone will make a correct assertion?

Firing Squad Synchronization

Consider a finite but arbitrary number of identical finite state machines (soldiers) arranged in a line.
At time t = 0, each soldier is initialized to the quiescent (idle) state, except for the soldier on the far left (the general). The state of each soldier at each discrete time-step t > 0 is dependent on its state and the state of its two neighbors at time t - 1 (except for the two soldiers at either end, each of whose state depends only on itself and its sole neighbor). In addition, if a soldier and its neighbors are in the quiescent state, then the soldier will remain quiescent at the next time-step. The problem is to define a finite set of states and state transition rules for the soldiers such that all soldiers enter a distinguished state (fire) at the same time and for the very first time.

Four Points, Two Distinct Distances

Find all configurations of 4 points in a plane with only 2 distinct values for distances between pairs of points.

Tumblers on a Rotating Table

A blind gnome and an evil goblin take turns to play a game. Four tumblers are placed at the corners of a square table. The initial configuration of the tumblers (facing up or facing down) is chosen by the evil goblin. When the blind gnome gets his turn, he is allowed to specify a subset of the four tumblers and flip them simultaneously. To be precise, he may choose "one tumbler", "two diagonally opposites", "two adjacent", "three tumblers" or "four tumblers" lying in front of him, and flip them simultaneously. After flipping, if all four tumblers are upright, he wins the game! Otherwise, the game continues and the evil goblin is allowed to rotate the table by an amount of his choice. Can the blind gnome win the game with a *deterministic* strategy?

Geometry With Only a Compass

Using only a compass (and without a straight edge or a ruler), is it possible to identify (a) the midpoint of two points? (b) the center of a circle? (c) all four corners of a square, given two of them?

Dijkstra's Self-Stabilization Protocol

There are n+1 processors named 0, 1, ..., n. Processor i has a counter C(i) that takes values in the range [0, n]. Its initial value is arbitrarily chosen from [0, n]. Processor 0 is said to be privileged if C(0) = C(n). Processor i, where i > 0, is said to be privileged if C(i) ≠ C(i-1). At successive clock ticks, exactly one out of possibly several privileged processors is arbitrarily chosen and its counter is updated as follows: If processor 0 is chosen, we set C(0) ← (C(0) + 1) mod (n+1). Otherwise, we set C(i) ← C(i-1). Prove that after a bounded number of clock ticks, exactly one processor will be privileged. And that this will continue to hold forever.

Three NOT Gates from Two NOT Gates

Design a 3-input 3-output logic circuit that negates the 3 signals. You have an infinite supply of AND and OR gates but only two NOT gates.

© Copyright 2008—2017, Gurmeet Manku.