Puzzle 13: Tetromino Slitherlink

Tetromino Slitherlink. Follow usual rules of Slitherlink. In addition, the gray squares can be partitioned into tetrominoes such that in each tetromino, each number from 0-3 appears. Pretty hard because it’s first time. Well it’s first time; I can’t estimate its difficulty properly.

Puzzle 13: Tetromino Slitherlink

Yay. Okay the tetrominoes are obvious, but putting the numbers aren’t that obvious.

I can obviously imagine a version where the tetrominoes can be ambiguous. As long as the numbers (and hence the loop) are fixed, it doesn’t matter how the tetrominoes are divided. Or otherwise, without given numbers. The latter would be way more difficult actually, because I couldn’t think of an opening without a number. Time to ask someone good at making difficult puzzles.

Puzzle 12: Hidato

Hidato. Fill in a number in each white cell such that for any two consecutive numbers, they are orthogonally or diagonally adjacent. Probably a little harder than usual; numbers 1-80.

Puzzle 12: Hidato

Yay Hidato. Tried a new design trick here; see if you can find it. See that 1 and 80 are adjacent? We have a loop!

Puzzle 11: Yajilin

Yajilin. Shade in black cells and make a loop visiting all remaining white cells (not including clue cells). Black cells may not be orthogonally adjacent; each clue cell points at the number of black cells in the direction of the arrow, passing over other clue cells. Should be some kind of easy.

Puzzle 11: Yajilin

First, I attempted to make exactly identical clues symmetrically (see R1C4 and R10C7, R4C7 and R7C4). But then I got stuck because it’s just impossible to put a 3-clue, putting a 0-clue kinda gives stuffs for long lines, and I just got bored. Hence the given result. Yes, symmetrical givens because it’s quite easy making that in Yajilin.

To keep you entertained, there will be two more puzzles, each appearing in 12 hours after the latest one. If I manage to construct another before all puzzles are released, then there will be more this weekend.

Puzzle 10: Slitherlink

Slitherlink; common logic puzzle I’d say. Google for rules. 17×17, easy-to-medium.

Puzzle 10: Slitherlink (click to enlarge)

UPDATE (7-Aug-2012, 19:38 UTC+7): Fixed an ambiguity around the G.

Okay, so you’re reminded that it’s the beginning of a new week. At least for me. Notice the symmetry?

That was extremely hard to think of. I started by listing all letters where it can form another letter when turned 180 degrees; I got like 7 pairs, including “A” paired with “U/V” (the middle two letters), and also 7 single letters. Then after some failed tries (like, trying SWAP with its reverse being JUMS…wut?), I noticed the word NOW which can be turned to MON. I tried to think of a Monday date that is also rotationally symmetric, and as you see, 6 August is symmetric. Quite. It’s pretty rare to see U with a line, but whatever.

However, before getting the idea of making a “today’s date”, I started with planning that every 10 puzzles (puzzles with puzzle number ending with 0), I will make a large puzzle (at least 289 cells). That’s also why I tried my best to construct two puzzles (Puzzle 8-9), as I haven’t stocked enough to publish this one. I’m also planning to make an extra large puzzle (at least 1395 cells (31×45) or something extremely fancy to overcome the restriction) every 50 puzzles or 100 puzzles, but it seems like I’m having a long way to go…

Puzzle 9: Five Cells

Five Cells. Mark walls along the gridlines to divide the grid into pentominoes such that each number tells exactly how many walls are among the four sides of the cell containing the number. 10×10. Easy-ish or something.

Puzzle 9: Five Cells

This puzzle has symmetry, unusual for Five Cells. Mostly toying around with one trick used at least twice that I remember, plus another trick used also at least twice. Yay. Puzzle 10 is ready, but it will be posted when it strikes 00.00 on 6 August (Monday) my time, for a reason you’ll see…

Update (4-Aug-2012 20:32 UTC+7): Fixed a multiple solution issue.

Puzzle 8: Surveyors Heyawake

Surveyors Heyawake, 10×10. Should be rather easy.

Puzzle 8: Surveyors Heyawake

Hm things seem pretty easy. That’s what happened when I tried making a “normal” Heyawake with absolutely no symmetry. The 6×6 room is pretty fun to toy with though; think of it as a Minesweeper. Oops spoiler; but you don’t think that 11 is going to be a Minesweeper clue don’t you -_-

Hm, Surveyors Heyawake and Smullyanic Dynasty can make some great hybrid, due to both having Minesweeper clue-style. I’ll see how I can toy with it…but I need to try making a Smullyanic Dynasty first.

Puzzle 7: Number In Order

Second puzzle. After this, I (hopefully) return to the IMO 2012 series.

Number In Order. Rules:

– Enter an integer between 1 and x inclusive to each white square. x is a number that vary between puzzles. If you want to go technical, then x is defined as the longest white square run in the puzzle, but this also might be changed depending on the puzzle. So, just assume it’s given.
– Each “run” of white cells (consecutive white cells in the same row/column) must contain all different numbers, and the numbers must be consecutive (for example, 2,5,3,4 is okay, but 1,2,4,5 isn’t).

Yeah, I think that’s it.

Number In Order, 10×10, maximum number in puzzle is 6. Should be easy-medium.

Puzzle 7: Number In Order

Also do you notice that the black cells are exactly like in Puzzle 6? I wanted to go a little further (given a grid with all-filled white cells and some-filled black cells, solve the Akari using the black cells; the lightbulbs denote the squares which are given in Number In Order) but I failed. Well, these two separate puzzles are pretty good enough too.

Puzzle 6: Antisymmetric Light Bulbs

Akari Put some light bulbs on the cells of the grid. Light bulbs illuminate all squares in the four orthogonal directions (up, down, left, right), up until reaching an edge of the grid or a black square. Illuminate all squares, but no light bulb may illuminate another light bulb. A number on a black square determines the number of light bulbs orthogonally adjacent to it.

Expected difficulty Easy • Answer • Comment/E-mail if you want a solution to be published

Puzzle 6: Antisymmetric Light Bulbs
Akari

We interrupt the IMO 2012 series for this post and the next post.

Yay for puzzles. Guess I’m back at “logicsmithing”.

This one is a pretty easy Akari. However, the aesthetic part of this puzzle is rather high by my standards. Opposing givens add up to 3, just like some Slitherlink I’ve seen…

*went browsing for like 30 minutes*

Oh hey I can’t find that puzzle. Whatever; it means I can make it some time soon and claim it as the only one in the last [insert a small time interval (less than a year)] 😛

Whatever. Akari, 10×10. Would go to an easy or something. Did I repeat myself?

Puzzle 5: Word Puzzle

30-Jan-2014: The original puzzle is broken, so here’s a replacement.

Expected difficulty Medium • Answer and solution follow below the puzzle

On an island, there are two kinds of people: knights who answer questions truthfully and knaves who answer questions falsely. You encounter five people from this island, named Alice, Bob, Charlie, Dave, and Erin. As a bored person, you want to figure out whether they are knights or knaves. When questioned, these are the answers, each answer stated by a different person. Statements in parentheses are statements that you know to be true.

Alice: At least one of us is a knight.
Bob: Exactly two of us are knights.
Charlie: At most three of us are knights.
Dave: The number of knights among us is not four.
Erin: YEAAAAAAAAAAAAAAAAAAAAAY

Which of them are knights?

---

Answer and solution; highlight below:

Alice and Charlie must tell at least one truth. If Bob is a knight, then Dave too; this is impossible as together with Alice and Charlie we have at least three knights. So Bob is a knave.

Since Alice and Charlie must tell at least one truth, there is at least one knight, so Alice is a knight.

If Charlie is a knave, then there are at least four knights. But Bob and Charlie are knaves, so there are at most three knights remaining, impossible. So Charlie is a knight. Thus the number of knights cannot exceed three, and so Dave is a knight, and so Erin is a knave.

Thus, the knights are Alice, Charlie, and Dave.

Puzzle 4: Take One

Surveyors Heyawake (v0.1) Follow regular Heyawake rules: Follow dynasty rules. Additionally, no connected white cells in the same row/column may span over one bold border.

There are numbers in the grid. For each “region” (bold-bordered area), exactly one of the numbers act as a Heyawake clue: the number is equal to the number of black squares in the region. All other numbers are Minesweeper clues: each number is equal to the number of black squares on the number or adjacent to the number.

Expected difficulty Hard • Answer • Comment/E-mail if you want a solution to be published

Puzzle 4: Take One
Surveyors Heyawake

Original text
Yeah bad title maybe. But I think this one has some pretty tricks waiting (besides the tricks of Heyawake)… I’m trying to get a few basic tricks I figured out here. However, this would be rated somewhere like medium, or 4 out of 10. Whoa.

And yes, I miscounted; I didn’t see Puzzle 3 in my book so I reported that I had only two puzzles ready, while I have three. 😛

Additional text on 27 September 2013
Take One of Surveyors Heyawake. When the rules are still rather crude. (The finalized version is actually much more complex x3 )