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Rules :

  • Fill with numbers from 1 to 13, and each number can appear max 4 times.
  • Small numbers between 2 cell shows the difference between the cells.
  • A + D = B + C
    Rule

Example :
(In this example, I fill with numbers from 1 to 8, and each number appear max 2 times)
example

original question :
enter image description here

Edit :
I have add more clues to make the puzzle easier (hope not getting too easy), and human solvable.
enter image description here

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  • 3
    $\begingroup$ Did you test-solve this puzzle yourself? Typically, for grid-deduction puzzles, it's best to testsolve to make sure the solution is unique and that there is an interesting solve path for a human solver. $\endgroup$ – Deusovi Mar 27 '17 at 3:20
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For the purposes of writing a walkthrough, it was easier for me to go ahead and re-render the puzzle:

Re-Render

Partial Answer

The natural place to start looking is where a given has already been placed.

Primary Observations





There are two possibilities for each of the blue shaded cells. This leads to four combinations for the total sum of this quad, and therefore four possibilities for the value of the red shaded cell:

{4,5} = 9, red cell = 1 
{4,11} = 15, red cell = 7
{12,5} = 17, red cell = 9
{12,11} = 23, red cell = 15
We can immediately eliminate {12,11} because the red cell is outside the range of possible values.

Looks like that didn't get us very far, so we'll have to look somewhere else.

Parity

Our next step will be to try to determine the parity (even versus odd value) of as many cells as we can.

Overall we know that there will be:
7*4 = 28 Odd cells
6*4 = 24 Even cells

We can determine the parity of all of the sums where two differences are given, based on the following observation.


If the parity of both differences is the same, then regardless of what value goes between the two differences, the parity of the blue shaded cells will be the same. The sum of the quad will therefore be even.

Likewise, if the parity of both differences is not the same, then the parity of the blue shaded cells will be different, and the sum will therefore be odd.

Let's indicate all known parities of sums to make it easier to deduce parities of individual cells.



Since both differences are even, we know that all four cells of the red shaded quad must have the same parity. We'll start looking with this one so we can bifurcate around the parity of the circled sum. We'll simultaneously bifurcate the parity of the upper left corner, resulting in a quadfurcation.



For the red shaded cells, either the values in the top left, top right, bottom left, or bottom right are valid. This parity quadfurcation can extend to the blue shaded cells as well.



Let's get an odd/even count on all four possibilities:

O/E
8/4|8/4
---+---
4/8|4/8


A couple things become apparent:

If the sum of the quad is odd, it must consist of two odd and two even numbers.
If the sum of the quad is even and both differences are odd, it must also consist of two odd and two even numbers.
If the sum of the quad is even and both differences are even, it must consist of all four numbers being the same parity.

Let's shade all quads where there must be two odd and two even numbers red, and all quads where all four numbers must have the same parity blue.



The red shaded cells consist of 18 odd cells and 18 even cells. This leaves us with 10 odd cells and 6 even cells remaining in the blue and unshaded cells.

It is impossible for there to be more than two blue quads with all odd numbers because then the total number of odd numbers would exceed 28. It is likewise impossible for there to be more than one blue quad with all odd numbers. Therefore we can deduce that one of the blue quads is even and the other two are odd. It follows that the unshaded cells consist of two odd and two even numbers.

The bottom half of our quadfurcation consists of one of the blue quads as having all even numbers, which forces the other two blue quads to be odd. Let's pursue this and see if we can find a contradiction.



From our above observation, we can deduce that every quad must have an even number of odds and evens: either two odd and two even, all odd, or all even. Because of this, we can start to understand parity relationships between some of our undefined numbers. Let x be any undefined parity, and y be its opposite. Let u be another undefined parity, and v likewise be its opposite. Let's fill in all remaining parities for the bottom half of our quadfurcation.



I was hoping we would find a contradiction with the blue shaded cells where it would be impossible for them to consist of two odd and two even numbers, but alas, we find everything works. For any value of x and u, the blue shaded cells will have two odd and two even numbers. Let's do a similar analysis on the top half of our quadfurcation.



Unfortunately, our progress stalls at the red shaded cells. We'll have to do trial and error and set temporary possibilities for x and y, as I don't want to add any more forks to our quadfurcation.

Here is scenario 1, where x is odd. Let u be any parity, and v its opposite. Likewise, let r be any parity, and s its opposite.



Alas, we find another consistency. It is possible for the blue shaded cells to consist of two odds and two evens, so long as u and r are opposite. Let's check scenario 2, where x is even.



Once again, we find consistency. For any value of u and r, the blue shaded cells will have two odd and two even numbers.

I don't think we're going to make any further progress looking at parities across the entire board. Let's look at something else.

Large Differences



The next thing we can look at are differences with conspicuously unique qualities. 7 pops out because it is the largest difference. It eliminates 7 as a candidate from either cell, as 0 and 14 are out of range. Let's see what else we can determine.

Which number is larger in a given difference is a binary determination with limited possibilities in a quad. Let's quadfurcate on the possible combinations in the quad with the 7 and 4 differences.


We find a large, but not impossible difference for the blue shaded cell if X is larger or smaller than both its difference-defined neighbors. Unfortunately, this doesn't give us much traction.

We're left with an incredibly large amount of combinatory exhaustion.

Conclusion

I don't believe the original puzzle is human solvable.

Additional Clues

The addition of 8 clues to this puzzle gives us a lot more information to work with. The most natural starting point is the blue shaded cell, which we determined earlier must be 1, 7, or 9. Let's rewrite our pencil marks so they align in a trifurcation.

13 - New Clues
We're still faced with a rather large number of possibilities for the red shaded cells, so let's look at parities again and see if the new clues give us any more information.

14 - No Solution
Two chains of apparently valid parity deductions, indicated in light red and light blue, lead us to a contradiction - the red cell with an X must be both even and odd.

Conclusion

This puzzle has no solution.

As a note, let me say it's great to see someone else inventing puzzles on non-rectangular tilings.

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  • $\begingroup$ One of your images isn't displaying properly. $\endgroup$ – Rand al'Thor Mar 25 '17 at 20:29
  • $\begingroup$ @randal'thor Which one? Everything looks okay to me. $\endgroup$ – paramesis Mar 25 '17 at 20:30
  • $\begingroup$ Just below the paragraph "Here is scenario 1, where x is odd. Let u be any parity, and v its opposite. Likewise, let r be any parity, and s its opposite." $\endgroup$ – Rand al'Thor Mar 25 '17 at 20:32
  • $\begingroup$ @paramesis Checking your conclusion, I recheck my puzzle and answer, I find a small, but fatal mistake in my puzzle, make the puzzle has no solution. Thanks for long and deep analysis for my puzzle. I learn much from your answer and analysis. $\endgroup$ – Jamal Senjaya Mar 29 '17 at 2:11
  • $\begingroup$ I personally think your blog and your non-rectangular grid deduction puzzles are great, but I'm not sure advertising at the top of the post is strictly the best practice. Maybe move it to the bottom of the post as a footnote, perhaps? $\endgroup$ – boboquack Mar 29 '17 at 4:43
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This is the intended puzzle and answer. The red number is the mistake I made.
enter image description here

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