Martin's Logic Puzzles

Graph Colouring #3

2009-08-28T11:43:00.004-05:00

An easy colouring problem. (1-3)

Hexagonal Masyu #5-8

2009-08-28T11:29:00.005-05:00

I made 4 more recently... The instructions are pretty simple, I think. Not really play tested these puzzles, so there may be flaws.

#5

#6

#7

#8

Graph Colouring #2

2007-06-20T18:11:00.001-05:00

I was bored at a conference and drew this. I think there isn't too many tricks that can be done with this kind of a puzzle, or maybe I'm just not imaginative enough.

Graph Colouring #1

2007-04-09T10:36:00.000-05:00

I hastily produced one puzzle, which should be pretty simple. There's only one small trick to it...

Graph Colouring Instructions

2007-04-09T10:30:00.000-05:00

Here is a classic graph theory problem: Put numbers (or "colours") 1, 2, or 3 into each circle so that adjacent circles receive different numbers. I guess some puzzles may use more colours, and that will be indicated for each puzzle.

Here is a simple example:

Notice that the circle on the top is adjacent to circles already coloured 1 and 2. So that circle must be coloured with 3. Then the circle on the right must be coloured with 1. So the solution for this puzzle is
.

MinMax Path #1-3

2007-04-06T18:44:00.000-05:00

I'll start off with three small instances of this puzzle. I'm not sure if it will still be interesting when the graph gets larger...

Puzzle #1

Puzzle #2

Puzzle #3

MinMax Path Instruction

2007-04-06T17:59:00.000-05:00

Given a graph, assign to the edges of the graph weights 1, 2, ..., m where m is the total number of edges. Assign a unique weight to each edge. Then find a shortest path joining the two black vertices, where the "length" of the path is the sum of the weights of all the edges in the path. The goal is to find an assignment of the edges so that the length of the shortest path is as large as possible. Note that there may be several optimal ways of assigning the weights, but the optimal value is unique.

I guess this may be a bit confusing, so here is a simple example: There are 4 edges in this graph, so we need to put weights 1, 2, 3, 4 on the edges.

There are essentially three ways to put the weights...

The shortest path between the two black vertices in the first assignment is 3. This is 4 in the second assignment, and 5 in the third assignment. Since 5 is the largest of them all, this is the answer for this puzzle.

Hexagonal Masyu #4

2007-02-21T18:35:00.000-05:00

This is one where I was trying for the "artistic" side of making a puzzle... I marked 19 cells in the middle in some kind of a symmetric pattern for putting in circles. Surprisingly, it was kind of tough to even make a correct loop that passes through these cells, but eventually I managed to do it. I had to put 3 circles along the rim of the puzzle in order to make it easier (for me) and unique, hence breaking the symmetric look that I was going for...oops. Anyway, I would have liked to have more black circles in there, but alas, that doesn't work. So...after all that useless rambling, here is the puzzle...

Hexagonal Masyu #3

2007-02-17T22:09:00.000-05:00

So far, my favourite instance of the original Masyu puzzle is #20 from Nikoli's Puzzle Box 7. That was a good one. Here's a hexagonal version that pays a bit of homage to that, though my version may be easier... (Note: I didn't spend enough time play-testing the puzzles that I post, so they may have some errors in there, or they may be a lot easier than I intended...)

Hexagonal Masyu #2

2007-02-15T10:12:00.000-05:00

My second attempt at making this puzzle. I was aiming for a hard puzzle, and was hoping that the centre portion would take some effort to figure out, but perhaps the cells on the rim of the puzzle give it away...

Hexagonal Masyu #1

2007-02-15T10:10:00.000-05:00

Here is my first attempt at making this puzzle. I hope it has a unique solution, and should be pretty simple...

Hexagonal Masyu Instruction

2007-02-15T10:00:00.000-05:00

Masyu (and I still don't know how to properly pronounce that) is one of my favourite Nikoli puzzles. I find it to be quite easy to do, though sometimes too easy... I've created an extension of this puzzle, tiling the plane with hexagons instead of squares.

Instruction: Find a simple closed loop that travels between adjacent hexagons (cells) via the centres of the hexagons. At each cell containing a white circle, the loop must go straight through the cell, and then make a turn in the very next cell on at least one side of the white cell. At each cell containing a black circle, the loop must make a turn at the cell, and then continue straight for 2 more cells in both directions. The loop can only make 120/240-degree turns, i.e. no sharp 60-degree turns.

Here is an example:

Breaking the Loop #1

2007-02-14T23:37:00.000-05:00

I first encountered "Breaking the Loop" in WPC15 Bulgaria. The puzzle type is invented (I believe) by Vladimir Portugalov, who is quite pleasant to meet while I was in the WPC. Initially I thought this puzzle is impossibly hard to solve, but after I've solved the WPC instance of the puzzle with the help of my officemate, I got quite interested in it and made one myself.

Instruction: Find a loop that visits all grid nodes, and locate 16 breakpoints (some of which are marked by "x" in the diagram). There are two breakpoints in each row and each column of nodes. The 16 breakpoints break the loop into 16 segments, and the midpoints of all 16 segments are shown as dots. For an example, see this page.

I have two versions of the same puzzle in here. The hard version is shown here, and I think this might be just a bit too hard. An easier version is hidden behind this link. It's the hard version with a few more breakpoints added, making it (what I believe) a more reasonable puzzle.