4
$\begingroup$

enter image description here

  • This is a modified Futoshiki
  • Place the numbers 1 to 5 such that each row, and column contains each of the digits 1 to 5.
  • Squares with same color, contains each of the digits 1 to 5 too.
$\endgroup$
4
$\begingroup$

Here is my solution, it was kinda easy, waiting for more:

enter image description here

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ you are right! wait till tomorrow for the green check. $\endgroup$ – Jamal Senjaya Oct 12 '17 at 10:08
  • 1
    $\begingroup$ ^vote with a note: This solution would probably get more approval if it were to show a couple of intermediate stages, especially those that were not straightforward. $\endgroup$ – humn Oct 12 '17 at 17:15
2
$\begingroup$

I'm posting this to expand upon the logical contradiction I spotted which forces the unique solution.

futoshiki-1

These are simple giveways.

The green 5 because 5 can't be on a less than, and the middle green is less than by proxy. The same proxy forces the 1 on the top row.

The 1 in the blue is now bottom row, and cannot be left because of the equals. This completes the 1's.

The 5 in the blue is on the left, and so with the equals (row 4 column 3= row 3 column 4), the 5 in the bottom row is in the middle, and this forces the 5 on column 5.

Next

futoshiki-2

We notice X is either 2 or 3. If Y is the other one, then 4 is bottom right, and row 4 column 4 by the equals. so row 3 ends with either 2 3 or 3 2.

The blue of the top row is also 4, as is the remaining green, and so the middle blue is 5.

However, this means the remaining bottom left blue, Z, cannot be 1 2 3 4 or 5.

Therefore Y is 4.

| improve this answer | |
$\endgroup$

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.