2
$\begingroup$

Here is a square filled with numbers and holes.

enter image description here

The square is divided in multiple invisible areas which all have their own specific logical sequence.
Although every numbers in an area is found with the same logic, the numbers used to find the next number can come from another area.
Can you find every positive integers missing by finding out the logical sequences?

NOTE
This is an new experimental puzzle for me, so let me know how it is in the comments. Hopefully it will go well.

NOTE2
I used the word "sequence" quite loosely here.
Some numbers might not be found by using the numbers "immediately" preceding them

EXAMPLE

enter image description here

Rule of area 1 is +1 the previous number
Rule of area 2 is 5 times the upper number
Rule of area 3 is 2 times the upper number

HINT
The correct solution will be found with a total of 4 areas.
The shapes of the areas are quite basic. No weird awkward shapes.

HINT2
All areas are of equal sizes

$\endgroup$
  • 3
    $\begingroup$ This seems way too broad. The left column is fine, but the rest... I'm not so sure about. It seems dangerously close to "Guess what I'm thinking". $\endgroup$ – Deusovi Nov 1 '16 at 4:39
  • $\begingroup$ @Deusovi Should I make the areas visible and see how it goes? $\endgroup$ – stack reader Nov 1 '16 at 4:48
  • $\begingroup$ ^vote with a note: A small-scale example fully solved would be quite helpful. Could be just a few cells with only 2 areas, woudn't even need to be well-clued. $\endgroup$ – humn Nov 1 '16 at 6:01
  • $\begingroup$ @humn I added a small example, let me know what you think. Since I made the puzzle it seems very easy to me but, perhaps not knowing where the areas are is too hard? $\endgroup$ – stack reader Nov 1 '16 at 6:27
  • $\begingroup$ Excellent example, stack reader, completely clarifies the puzzle for me. Would be worth formatting. (The original puzzle still looks tough, but what else is new . . . a good hint might be the total number of areas.) $\endgroup$ – humn Nov 1 '16 at 6:31
2
$\begingroup$

Equally divide the square into 4 3x3 quadrants. Top left quadrant - sum of the cell above and to the left. Bottom left - sum of all the cells above. Top right - sum of the cell 2 steps to the left and of the cell to the left. Bottom right- sum of the cell 3 steps to the left and of the cell 3 steps above. The numbers (as we fill the rows) are 0,1,2,3,5,8,1,2,4,6,10,16,2,4,8,12,20,32,3,7,14,6,12,22,6,14,28,12,24,44,12,28,56,24,48,88.
$$\require{action}\require{enclose}\toggle{\enclose{roundedbox}{\text{ Click here to toggle grid }}}{\enclose{roundedbox}{\text{ Click here to toggle grid }}\begin{array}{|c|c|}\hline\begin{array}{c}\bbox[pink]{\begin{array}{c|c|c}{\textbf0} & \textbf1 & 2 \\\hline\textbf1 & 2 & \textbf4 \\\hline\ 2\ &\ 4\ &\ \textbf8\ \end{array}}\\\hline\bbox[cyan]{\begin{array}{c|c|c}\textbf3 & \textbf7 & 14 \\\hline 6 & 14 & 28 \\\hline 12 & \textbf{28} & 56 \end{array}} \end{array} & \begin{array}{c} \bbox[yellow]{ \begin{array}{c|c|c} 3 & \textbf5 & 8 \\\hline 6 & 10 & \textbf{16} \\\hline 12 & \textbf{20} & 32 \end{array}}\\\hline \begin{array}{c|c|c} \textbf6 & 12 & 22 \\\hline 12 & \textbf{24} & 44 \\\hline 24 & 48 & \textbf{88} \end{array} \end{array} \\\hline \end{array}}\endtoggle$$

Got a lot of help for this from Angzuril and mactro's answers.

$\endgroup$
  • $\begingroup$ Welcome to puzzling and good job on solving this puzzle. This is exactly the answer I was looking for. $\endgroup$ – stack reader Nov 2 '16 at 6:08
4
$\begingroup$

My solution (partially matching that of Angzuril):

enter image description here

Yellow: sum of the cell directly above, and directly to the left (in case one cell is missing, treat it as 1)

Blue: sum of two cells directly to the left

Red: 4 times the cell 2 above

Green: Sum of cells 3 and 4 to the left. In case there is no cell 4 to the left, it's two times the 3rd cell.

$\endgroup$
  • $\begingroup$ Very nice try! but the logic for the "green" area seems a little bit more like 2 rules than one. Here is a hint, the numbers in the top half of the square are all good, and the blue area's logic is correct. The logic for the yellow area is also correct, but the size was not. $\endgroup$ – stack reader Nov 1 '16 at 23:19
  • $\begingroup$ ^vote of approval for being the only solution with reasons for the edge cells near the top left corner $\endgroup$ – humn Nov 3 '16 at 7:46
3
$\begingroup$

My solution, probably not intended solution.

enter image description here

Yellow: Sum of the cell directly above, and directly to the left

Blue: Double cell directly above

Green: Sum of the 2 cells to the left

Red: Form of $2 \times (a - b^2)$ , where a is 3 cells to the left, and b is 3 cells above a. The complexity of this rule leads me to believe it is not intended.

$\endgroup$
  • 1
    $\begingroup$ You could cheat a little and say Red is $\frac12$ of the cell directly below... :) $\endgroup$ – GentlePurpleRain Nov 1 '16 at 17:50
  • $\begingroup$ @GentlePurpleRain Thanks for the edits, I would like to avoid circular references however :P $\endgroup$ – Angzuril Nov 1 '16 at 17:52
  • $\begingroup$ @Angzuril Very good job! But the red area logic is indeed a little bit too complex. Look at my comment to the other answer for some hints. $\endgroup$ – stack reader Nov 1 '16 at 23:28
2
$\begingroup$

my answer is,

$$\require{action}\require{enclose}\toggle{\enclose{roundedbox}{\text{ Click here to toggle grid }}}{\enclose{roundedbox}{\text{ Click here to toggle grid }}\begin{array}{|c|c|}\hline\begin{array}{c}\bbox[pink]{\begin{array}{c|c|c}{\textbf0} & \textbf1 & 2 \\\hline\textbf1 & 2 & \textbf4 \\\hline 2 & 4& \textbf8 \\\hline \ \textbf3\ & \ \textbf7\ & 15 \end{array}}\\\hline\bbox[cyan]{\begin{array}{c|c|c}\text{-4}\ & 12\ & 28 \\\hline 12 & \textbf{28} & 44 \end{array}} \end{array} & \begin{array}{c} \bbox[yellow]{ \begin{array}{c|c|c} 3 & \textbf5 & 8 \\\hline 6 & 10 & \textbf{16} \\\hline 12 & \textbf{20} & 32 \end{array}}\\\hline \begin{array}{c|c|c} \textbf6 & 14 & 30 \\\hline \text{-8} & \textbf{24} & 56 \\\hline 24 & 56 & \textbf{88} \end{array} \end{array} \\\hline \end{array}}\endtoggle$$

Top left (pink) is sum of top and left
Top right (yellow) is sum of two left ones
Bottom left (cyan) is left +16
Bottom right (white) is three to the left doubled

original format:
I am not on a computer so sorry for formatting. _ separate cells and ' separates groups, and bold is the last row in a group,

0_1_2'3_5_8
1_2_4'6_10_16.
2_4_8'12_20_32
3_7_15'6_14_30
-4_12_28'-8_24_56
12_28_44'24_56_88

$\endgroup$
  • $\begingroup$ I'm not sure your bottom left logic works that well. $\endgroup$ – stack reader Nov 2 '16 at 1:09
  • $\begingroup$ Oops -6 should have been -4.. $\endgroup$ – ev3commander Nov 2 '16 at 1:10
  • $\begingroup$ Otherwise I think its OK.... $\endgroup$ – ev3commander Nov 2 '16 at 1:11
  • $\begingroup$ Sorry if it was not clearly mentioned in the question but only positive integers are expected. I will add it to the question now, along with a few more hints.(+1 for the nice try) $\endgroup$ – stack reader Nov 2 '16 at 1:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.