Most Sudoku puzzles published have only one solution. If there is more than one solution, it is probably a mistake. That said, puzzles with incomplete clues can have multiple solutions. In the extreme case, a puzzle with no clues has 6,670,903,752,021,072,936,960 solutions according to Wikipedia.
I don't know if it's possible to have exactly 3 solutions, ...
This is a difficult question to answer, but it's important to clarify one point in advance: what do we mean by "algorithm"?
An algorithm, in this context, refers to a sequence of steps that you do when solving the Rubik's cube to achieve a certain outcome. It refers to any repeated sequence. For instance, there's an algorithm for moving a corner to the ...
I'm not sure if you don't understand the example puzzle, or if you don't understand how to find the solution. I'll explain both, just in case.
The example takes a word/phrase that is 13 letters long with no repeated letters. That means it contains exactly half of the letters in the alphabet (26 total). Then the 13 remaining letters are written below, in ...
I have seen two dissenting opinions on this subject (and in my opinion, the first option is right):
By definition, a Sudoku has only one solution. Anything else is just a grid of numbers. Sometimes, there are errors in a publication, and a starting grid has multiple solutions, but, then the starting grid was not a Sudoku!
From Wikipedia: The number of ...
This puzzle is called a "Baguenaudier" which is French for "Time Waster".
This solution assumes that you are holding the handle end on the left. The rings are numbered from right to left starting at 1.
The solution to the puzzle involves a couple possible moves.
1) The first ring is always available to put on or take off of the bar. You take it off the ...
There are several types of clues embedded in these puzzles.
Cryptic clues where the words have no meaning. One is meant to break words apart into constituent letters and then reconstruct.
The clues are literal, however they are veiled in metaphor and use of hyperbole and exploitation of synonyms.
The clues are totally descriptive (easiest, most obvious and ...
I can sympathize. This is from The Guardian, 2015:
From left to right, starting with image 2, we have:
not a square
leaving image 1 because it is the only image that is not an odd one out!
Which I think qualifies it as an odd one out.
So the answer given might not be unique, which makes some of the puzzles unsolvable.
The only way you can have a solve-all sequence is if you have a sequence of moves that goes through all 43 quintillion configurations of the Rubik's Cube. In order to do this, you need to draw a transition graph between all the states of the Rubik's Cube and find a Hamiltonian cycle through them.
This sequence of moves doesn't necessarily have to be 43 ...
This can be solved in 16 steps. Lets use binary here, 0 is off, 1 is on the handle
You start with 11111
1 - Remove the first ring to get 01111
2 - Remove the third ring to get 01011
3 - Put back the first ring to get 11011
4 - Remove the first two rings to get 00011
5 - Remove the fifth ring to get 00010
6 - Put back the first ...
A random configuration of gears is solvable 1/3 of the time*
For details on what kind of configuration is solvable, refer to the claim after the image below.
Your hunch is right, we can use modular arithmetic here.
Let's mark each gear with a letter, starting with the top one as A, going from top to bottom, left to right, as in this figure:
Let's denote ...
In the classic book "Winning Ways (for your mathematical plays)" by Berlekamp, Conway and Guy there is a small section devoted to wire puzzles (like the Tavern puzzles) in the second-to-last chapter.
They use a technique they call a "magic mirror". Imagine a fun-house mirror that distorts your view of the puzzle. Some bits of the body of the puzzle get ...
The key is that the four center cubules on each face of a Rubik's Revenge are indistinguishable.
When you do the move sequence to swap those two edges in the "3x3x3" phase of solving a 4x4x4 cube (or to flip a single edge or swap two corners), a bunch of the center cubules get reoriented, but you don't notice because they're all the same colour and ...
The angels should play the game, as long as the formula for the computable function must be fixed in advance and finite.
Proof: let F be the shortest encoding of the formula (length l(F)) using a set of symbols S of size n(S). Then the guessing angel can simply count up from 0 in base n(S), and some time before n(S)l(F) they will encounter the formula.
The easiest explanation would be that in a 3x3 cube, only one cube is out of position, but in a 4x4 cube two cubes are out of position.
In a 15 puzzle (the sliding puzzle where you try to put the numbers in order) half of all possible initial positions are unsolvable. They call the solvable positions "even" and the unsolvable positions "odd". The "odd" ...
The canonical Einstein's Puzzle can be solved using a grid of possibilities, and the easiest way to do so is to track each of the possible states.
Each letter in this grid stands for the associated entity:
Nationality stands for Enlgishman, Swede, Dane, Norwegian, and German
Colors (house color) stand for Blue, Green, Red, White, Yellow
Animals stand for ...
Some simple strategies I've found are:
Look for zeroes. Any time you have $X + Y = X$ in the ones column, you know that $Y = 0$.
Look for doubles. Any time you have numbers of the format $A + A = B$ in the ones column, you know that $B$ is even, and you know that either $0 < A < 5$ OR that there is a carrying one to the next column.
Look for ...
Does an algorithm exist?
Yes. Consider every valid state of the Rubik's cube. It can be brought to the solved state in 20 moves or less. For each state, apply the sequence of moves followed by its inverse. This giant algorithm is guaranteed to solve any cube.
Now, does a reasonable-length algorithm exist?
No. I will show that any such algorithm should ...
The bin-packing problem is NP-complete, and any instance of the bin-packing problem can be turned into a 2D packing problem. (Represent each integer $k$ with a $1\times k$ rectangle, and make the bins of size $n$ into several $1\times n$ holes with tiny "channels" connecting them into one full shape.) Therefore the 2D packing problem is at ...
By definition, all valid Sudoku puzzles should have only one solution. In point of fact, many of the techniques used for solving puzzles depend on there being only one solution. All of the Unique Rectangle techniques for example, only work if there is one, and only one, solution.
The optimal way to solve this is to create linear equations of the problem.
I'll take the following problem:
K Y O T O
O S A K A +
T O K Y O
What you need to do to solve this is to set up a sequence of linear equations that represent the carryovers and equations. I'll denote $x_n$ as the single-digit carryover, and $C_n$ as the double-digit ...
The Situation that you are talking about is a lot more common than you know. And it's not only limited to odd-one-out puzzles. There are riddles which are basically "guess what I am thinking" puzzles. Odd-one-out puzzles usually land in this basket.
There is nothing wrong with your thinking or your reasoning. The fault, almost always, lies with the question....
This puzzle is indeed broken, and has no solutions. (I ran into the same contradiction as Arnaud.)
The organizer, Serkan Yürekli, made a post on the Logic Masters Deutschland forum around that time, confirming this. The 1,1,3 clue at the top was the problem, and because of this they ignored the top arrow when checking solution keys.
(In addition, the ...
What you seem to be asking (assuming the squares are completely filled), is to find an Hamiltonian path on a connected subset of the square grid graph.
Defintions (lifted from Wikipedia)
A Hamiltonian path (or traceable path) is a path in an undirected or directed graph that visits each vertex exactly once.
A square grid graph is the graph whose ...
The answer is no for "iterating"/repeating a single algorithm because the maximum order (of a permutation) of an algorithm is 1260, which is not 43 quintillion. In other words, no one move sequence can be repeated more than 1260 times before the cube is restored to the original state it was in before iterating that algorithm.
However, the answer is yes for ...
This is how you can get started -
Check out this article on Wikipedia which provides an insight. It also mentions some techniques used in Steganography.
So, when you see a puzzle with steganography tag added you can start by looking at the following things -
First word/letters of each line/paragraph
Last word/letters of each line/paragraph
Lights Out puzzles have a polynomial time algorithm by linear algebra.
"Draw this shape without picking up your pencil" puzzles have a polytime algorithm which follows from the proof of which graphs have Eulerian circuits.
Sliding puzzles like the 15-puzzle are easy (though hard to solve in a minimal number of moves).
IMO these examples kind of prove your ...