LOK is a new original puzzle book, created by the musical and visual artist Blaž Gracar. It is his first published work in this domain and considering the attention to detail put into puzzle design, solutions system, and illustrations, I would say that’s an impressive start, and I am looking forward to his future endeavors.
The puzzle mechanics of LOK are quite innovative, and despite my long-time puzzle history, I cannot relate them to anything I have seen before. Each puzzle consists of a grid partially filled with letters and the goal is to black out all cells by marking certain words and then triggering their effects. For example, if one has the letters L, O, K in consecutive cells and marks the resulting word LOK, then they must black out the cells of all three letters plus an additional cell by choice.
There are 5 more trigger words that appear in the book, but since it is the solver’s challenge to figure out what their effects are, I will not spoil them for you. All I can say is that each of them works uniquely and sometimes solving the puzzles gives the feeling of planning ahead the moves of a chess game.
In addition to the funny-sounding trigger words, new chapters introduce various other mechanics, building upon the previous ones and making the puzzles progressively more complex. You will encounter ”conductors“ that let you connect separated letters, “clouds” that let you black out a given configuration of cells, and other interesting concepts.
The book is split into 12 parts: 8 chapters with a total of 80 puzzles, 2 expansions with 10 extra puzzles, Rules that explain all the concepts in the book that have been left to the solver to discover, and Solutions that provide the answers to all puzzles. The parts are separated by beautifully illustrated pages of cute worm-like creatures, called ”Loks”, performing various activities, such as flying in space or working in a factory, on 10 additional solvable grids. Of course, the author could assemble the book just with all the 100 puzzles inside, and it still would have been great, but that extra touch of art is what makes LOK even more special for me.
While the market is currently flooded with all kinds of derivative logic books filled with computer generated Sudokus and Kakuros, LOK differentiates itself as an ingenious passion project of someone who simply wants to contribute to the world of puzzling. You can get the digital version of LOK for free from the author’s website by clicking the button below. If you like, there you can also purchase the physical edition and solve the puzzles the way they are intended to be solved, with pencil and eraser on paper. For those who prefer to keep their puzzle books in immaculate condition, a transparent draw board is included in the package. Thanks!
- appropriate for all ages
- more than 90 carefully hand-crafted puzzles
- surprising mechanics taught through practical solving
- explained rules and solutions included
- free e-book version available to download
We have written the numbers from 1 to 12 on the faces of a regular dodecahedron. Then, we have written on each vertex the sum of the five numbers on the faces incident with it. Is it possible that 16 of these 20 sums are the same?
We color the vertices in five colors as shown in the image below, such that each face of the dodecahedron has 1 vertex of each color. Then, the sum of the four numbers from each color must be equal to the sum of all numbers written on the faces: 1 + 2 + … + 12 = 78.
If 16 of the vertices have the same number written on them, then by the pigeonhole principle there will be 4 vertices with identical colors and identical numbers. Since 78 is not divisible by 4, we conclude that this is impossible.
Your goal is to switch the positions of the three white knights with the positions of the three black knights. What is the least number of moves required to do this?
The least number of moves required to switch the positions of the knights is 16. An example is shown below.
Next, we prove that it is impossible to switch the positions of the knights with less than 16 moves. Since the white knights occupy 2 black and 1 white squares, they need to end up on 2 white and 1 black squares, and each knight must make at least 2 moves in order to get to the opposite side, the total number of moves for the white knights should be an odd number, larger or equal to 2+2+2=6. The same applies to the total number of moves for the black knights. Therefore, the only possible way to get a total number of moves less than 16 is if both the white knights and the black knights move exactly 7 times.
We assume it is possible to switch the positions with 7+7=14 moves in total. Then, the white knight on A2 and one of the white knights on A1, A3 should make 2 moves each, and the third white knight should make 3 moves and land on a white square, either D1 or D3. Without loss of generality, we assume the knight on A1 makes 3 moves: A1-B3-C1-D3. Then, the knight on A2 should make 2 moves: A2-C3-D1, and the knight on A3 should make 2 moves: A3-B1-D2.
We make the same argument for the black knights. Since it is impossible that the white knight on A1 moves along the trajectory A1-B3-C1-D3 and also the black knight on D3 moves along the trajectory D3-C1-B3-A1, we conclude that the black knight on D1 should make 3 moves: D1-C3-B1-A3, the black knight on D2 should make 2 moves: D2-B3-A1, and the black knight on D3 should make 2 moves: D3-C1-A2.
This is only possible if:
- the knight on A1 moves to C1 after the knight on D3 moves to A2
- the knight on D3 moves to A2 after the knight on A2 moves to C3
- the knight on A2 moves to C3 after the knight on D1 moves to B1
- the knight on D1 moves to B1 after the knight on A3 moves to D2
- the knight on A3 moves to D2 after the knight on D2 moves to B3
- the knight on D2 moves to B3 after the knight on A1 moves to C1
We get a contradiction which means that the least number of moves is 16.
Recognize the phrases depicted by the “terrible rebuses” below.
- Blood is thicker than water.
- Home is where the heart is.
- It takes two to tango.
- Haste makes waste.