This puzzle consists of a 6×6 square board and 36 arrow cards with 6 arrows in 6 colors. The goal is to place all arrow cards on the board such that

- the entire board is covered with cards;
- each arrow points to exactly one arrow of the same color in the same row or column.

The example above shows *incorrect* placements of arrows: The yellow arrow cannot point to another arrow, the left blue arrow points to two different blue arrows, and it is not possible to add to the four purple arrows, while correctly placed, two more purple arrows without violating the rules.

Indeed, if the six arrows of one color are correctly placed, the arrows can be connected to a closed circuit. The figure above shows two closed circuits of six arrow cards. Note that the circuit paths may well cross. The following puzzles have a few arrows already placed, and need to be completed. Below to the left is simple example that shows how to find the solution.

First consider the green arrows. The arrow at e4 points to the right and therefore we must have another green arrow at f4, either pointing up or down. As there is already a green arrow at f6, the new arrow at f4 must point up. This leaves us with two more green arrows to place. Because we already have three horizontal arrows, the remaining arrows must be both vertical, and point to existing horizontal arrows. The only possibility is to place the new green arrows at c6 and e2, as shown above to the right.

Now let’s consider the red arrows. The one at e3 points down, leaving no choice but to place a red arrow pointing left at e1. There are two more horizontal arrows to place, and the only possibilities are b5 and f3, see below to the left.

Turning to blue, there clearly needs to be a left pointing arrow at f1, and the two remaining vertical arrows need to go to a1 and b6.

For yellow we have only two arrows given, but there are not many free spots available. We first are forced to put a left pointing arrow at e6 and then a down pointing arrow at d6. The remaining horizontal arrows go to c5 and d4.

The purple arrow at d4 can only be reached by a right pointing arrow at a5, and the one at c3 only by an up pointing arrow at c1. Then the arrow at c1 requires a left pointing arrow at d1, and the final purple arrow goes to a3.

Filling the remaining spots with cyan arrows is now easy, giving the solution.

Below is a new puzzle to warm up:

And here is a more difficult one. In both cases, there is only one solution.

### Like this:

Like Loading...

I want to understand why the purple arrows won’t work in the incorrect example. Can we add two purple arrows next to each other and pointing to each other?

LikeLike

You are right. I forgot to say that all arrows of one color need to form a chain. In other words, the circuits shown in the second image must always have length 6.

LikeLike