New answers tagged grid-deduction
4
If I've interpreted the rules correctly, then this should be the solution:
Explanation
We start off by
Next, we
Now, we can note that
Building off of this, we can now determine
The next step involves a bit of "what if?" reasoning - if there's a more straightforward logical way to proceed, then let me know:
We're at the home stretch now! Next, ...
14
Answer to the "more difficult" question:
I claim that
and here's why:
Easier analogue:
I claim that
and here's why:
Normal question:
0
Here’s a solution that has only some non-unique pipes. My strategy is having that sort of outer parity loop, and filling in the middle in any way. This can be made for any odd square, and I believe the inner fill for a 5x5 is impossible using this strategy. Wouldn’t be shocked if you could do something similar with a 6x6, and I will look into it some more....
6
There is one solution for $G$, which is as follows
Notice first that
Now
Suppose instead
3
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
Step 6:
Step 7:
Step 8:
Step 9:
Step 10:
7
I have the same final answer as the others, but using only one repeated method, namely
This gives the following first step
1
I came up with the same solution as the other answers, but I did it a different way
The solution:
Steps:
Edit: I realized that this method is not guaranteed to work.
5
Step 1:
Step 2:
Step 3:
Step 4/solution:
6
Solution:
Deduction process:
3
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
Step 6:
Step 7:
Step 8:
Step 9:
Step 10:
Step 11:
Step 12:
Step 13:
Step 14/solution:
2
To expand Lukas Rotter's comment into an answer,
We can get this far by logical deduction ("escape" to make room for a polyomino, draw walls to prevent two polyominoes of the same type from touching, fill all room left, etc.):
However, the 6 and 13 can fill the remaining squares near them either way. Also, the empty 3-area on the bottom row can be ...
2
Completed Sudoku:
This is confirmed by the FUN FACT, which refers to
Step-by-step solution:
4
Pull up a chair, this is going to take a while. Especially since I keep using far too many words. There are 19 steps, I kid you not.
The coordinate system used in this answer:
Note that there are four ways to make 10 (1-9, 2-8, 3-7, 4-6) and two ways to make 5 (1-4, 2-3)
Step 1:
Interim Observation 1:
Step 2:
Step 3:
Step 4:
Step 5:
Interim ...
11
The solution is (drumroll please...)
Explanation:
We start by
Next,
And here's the coup de grâce:
4
Just noticed that this puzzle - although previously solved - has no step-by-step solution in place. So to assist other puzzlers in understanding the process involved in solving it, here's a write-up of the logic...
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
0
I confirmed via constraint programming that the following solution is unique:
You have narrowed the (6,5) entry to two choices: 1 or 3. The unique solution has a 1, so choose 3 instead. You will quickly get a contradiction, which therefore implies that (6,5) must instead be 1.
Alternatively, suppose the (4,6) entry is 2 (rather than 4). Then (2,6) must ...
0
My interpretation for the two pairs connected with / and - is slightly different: I assume that the calculation is top cell by lower cell, or top cell minus lower cell.
So the conditions are that row 5, column 2 = 5 and row 6, column 2 = 1. And row 5, column 6 minus row 6, column 6 = 3. Since row 5, column 5 ≠ 5 and row 5, column 6 ≠ 1, these two cells are 6 ...
6
Bass has shown that there are "obviously" multiple solutions. It appears that there are in fact exactly 24.
from pysmt.shortcuts import Symbol, LE, GE, And, Int, Equals, NotEquals, Plus, Minus, Times
from pysmt.typing import INT
grid = [[Symbol(f"g{row}{col}", INT) for col in range(6)] for row in range(6)]
inrange = And(And(GE(grid[row][...
20
COMPLETED GRID
The first step:
Next:
An important side note:
Moving on:
The top shaded region:
Hopefully, finishing up:
17
I am officially an idiot.
I spent several hours figuring out brilliant deductions and got really good progress with many actual numbers on the grid, and even though I got stuck at places, there was always some clever bit that got me just that much forward.
In the end, I was just about to fill the grid in two different ways to show that the puzzle must be ...
4
Here's a solution with a fully worked-out logical path. Blue means "no circle here"
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
Step 6:
Step 7:
Step 8/solution:
The rest is trivial deduction.
6
Here's the finished grid:
You can see some of the original letter clues got covered by the stars. I'm not sure how I feel about that; on one hand, it's just a puzzle rule that allows for a unique solution. On the other hand, it feels like having two words intersect in a crossword puzzle even though they have different letters at the intersection square. ...
2
First, let's list all the possibilities for the 3x3 box in the middle of the left edge:
Now, for the deductions:
That gives us:
It should now be clear that:
Hopefully that's enough for a "next move"
Disclaimer: I did not figure this out myself, but used this sudoku solver to point me in the right direction. Images made with Penpa+
0
Nice to see it describes this way. The column C5 will only allow one of the 3,5,6 values in the available cells. R4 has the 3 and 5, which removes the 3 and 5 from the candidates. This forces the 6 onto that position. Refered to as the naked value.
10
R6C5 doesn't have anything ruling out it being a 6... but R4C5 could only be a 6. There are no other options for R4C5: placing a 1, 2, 3, 4, 5, 7, 8, or 9 would break the rules. You know you have to fill a box with some number, and that is the only one left.
This is one of two basic Sudoku techniques, the "naked single" -- when a cell only has one ...
1
The only possible number that could go in cell (R4, C5) is 6, as shown by the the grayed out numbers: 3 and 5 are eliminated by the row, and the rest are eliminated by the column.
1
This is one solution, which has not one but two closed circuits.
Here is the solved puzzle:
2
First, let's make some general observations.
Because of how the dominos connect, as we follow the loop, we will encounter every digit in exactly two runs. Those runs will have lengths 2 and 4.
The loop can go through a corner in only one way.
So we start by noticing that the bottom zeroes must be part of the same run. (The other run of zeroes is at the ...
5
This was a surprisingly difficult puzzle! To start, some basic single-clue deductions:
Next,
Next, some deductions involving:
Some steps flow nicely from that:
Next, check
And now I had to use an advanced connectivity deduction:
And that breaks in to the rest of the puzzle:
And with that, the puzzle is solved!
6
Solution:
Explanation:
7
I think I got it:
Sorry about covering up some of the clues, they are pretty easy to figure out though, because they follow the 3/16 theme.
EDIT: Full write-up below, with different colours this time, since hindsight allows for more efficient deductions :-)
There are couple of easy deductions we can often make from the connectivity:
never close off an ...
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