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Arrows

cinelli
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Arrows

#244830

Postby cinelli » August 16th, 2019, 4:26 pm

Consider a set of sixteen arrows and a four-by-four chessboard. One arrow has one spot, eight have two spots and seven have three spots.

When an arrow is placed on a square of the board pointing north, south, east or west, the single spot means it points to the immediately adjacent cell. The 2-spot arrow points to the cell two away and the 3-spot arrow points to the cell three away. Just seven arrows of the set can be placed to map a closed circuit, as follows:

.----- ----- ----- -----
| | | | |
| 3 | 2 | | <3 |
| v | v | | |
----- ----- ----- -----
| | | | |
| | | | |
| | | | |
----- ----- ----- -----
| | | | ^ |
| | 2> | | 2 |
| | | | |
----- ----- ----- -----
| | ^ | | |
| 1> | 3 | | |
| | | | |
----- ----- ----- -----

This puzzle is to use all sixteen arrows to make a sixteen-long closed circuit. All sixteen squares must be used.

Why would this challenge be impossible if the set consisted of one 1-spot, seven 2-spot and eight 3-spot arrows?

Cinelli

jfgw
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Re: Arrows

#244839

Postby jfgw » August 16th, 2019, 4:56 pm

cinelli wrote:Why would this challenge be impossible if the set consisted of one 1-spot, seven 2-spot and eight 3-spot arrows?


Spoiler to second part:
Consider that the chess board would have black and white squares. Arrows with one spot or three spots would point to a different colour square. In order to finish on the same square that you started, there must be an even number of moves to a different colour, therefore, the the total number of 1-spot and 3-spot arrows must be even for a solution to be possible.

Julian F. G. W.

jfgw
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Re: Arrows

#244846

Postby jfgw » August 16th, 2019, 5:35 pm

Spoiler,

.----- ----- ----- -----
| | | | |
| 3 | 2 | 2 | <3 |
| v | v | v | |
----- ----- ----- -----
| | | | |
| 2> | 2> | 2 | <3 |
| | | v | |
----- ----- ----- -----
| | ^ | | ^ |
| 3> | 1 | <2 | 2 |
| | | | |
----- ----- ----- -----
| | ^ | ^ | |
| 3> | 3 | 3 | <2 |
| | | | |
----- ----- ----- -----


Julian F. G. W.

jfgw
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Re: Arrows

#244866

Postby jfgw » August 16th, 2019, 6:43 pm

Method:

Rather than visualise this with arrows, I drew lines. One advantage is that a line can be in either direction so I could determine this later. It was just a case of drawing lines between dots on a 4 x 4 grid.

Since there are seven 7-spot arrows, seven of the eight possible lines of length 3 must exist. The missing one must be an edge one otherwise there would be a closed loop around the edge which would not pass through the inner four dots. I decided to miss out the one on the right, the choice being arbitary due to symmetry.

There is only one line of length 1 so at least one line (possibly both) on the right edge must be of length 2 so I drew one in (top to lower-middle, symmetry again).

The rest was mostly, "make the next one a line of length 1, fill in the rest and see if it works". There were a few different branches to explore but not many.

Drawing dots and lines in Photoshop makes undoing changes easy!

The solution is unique except for rotations, reflections and reversals (16 discrete solutions).


Julian F. G. W.

Gengulphus
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Re: Arrows

#244934

Postby Gengulphus » August 17th, 2019, 12:32 am

cinelli wrote:Consider a set of sixteen arrows and a four-by-four chessboard. One arrow has one spot, eight have two spots and seven have three spots.

When an arrow is placed on a square of the board pointing north, south, east or west, the single spot means it points to the immediately adjacent cell. The 2-spot arrow points to the cell two away and the 3-spot arrow points to the cell three away. Just seven arrows of the set can be placed to map a closed circuit, as follows:

.----- ----- ----- -----
| | | | |
| 3 | 2 | | <3 |
| v | v | | |
----- ----- ----- -----
| | | | |
| | | | |
| | | | |
----- ----- ----- -----
| | | | ^ |
| | 2> | | 2 |
| | | | |
----- ----- ----- -----
| | ^ | | |
| 1> | 3 | | |
| | | | |
----- ----- ----- -----

This puzzle is to use all sixteen arrows to make a sixteen-long closed circuit. All sixteen squares must be used.

Why would this challenge be impossible if the set consisted of one 1-spot, seven 2-spot and eight 3-spot arrows?

Spoiler...

Consider the seven 3-spot arrows. Each one must point all the way from one side of the board to the corresponding cell on the opposite side. Furthermore, the corresponding cell on the opposite side cannot have a 3-spot arrow pointing back the opposite way, since that would make a two-long closed circuit. So each of the four rows and each of the four columns can only have one 3-spot arrow pointing along it, and since there are seven 3-spot arrows, precisely one row or column doesn't have a 3-spot arrow pointing along it. By rotating the board appropriately, we can ensure that it is the first row or the second row.

We have four vertical and three horizontal 3-spot arrows, so the 3-spot arrows give us four transitions between even and odd rows and three transitions between even and odd columns. The 2-spot arrows don't give us any transitions of either type, and the 1-spot arrow gives us one transition between even and odd rows if it points vertically or one transition between even and odd columns if it points horizontally. Since the total number of transitions of each type in a closed circuit must be even, the 1-spot arrow must point horizontally.

There must be a 3-spot arrow pointing along each of the last row, the first column and the last column. Each of them points from one corner on that row or column to the other, and since a closed circuit is formed, no corner can have two arrows in it or two arrows pointing to it. So those arrows must be positioned as in the following diagram or its left/right-reflected mirror image:

+-----+-----+-----+-----+
| | | | |
| 3 | | | |
| v | | | |
+-----+-----+-----+-----+
| | | | |
| | | | |
| | | | |
+-----+-----+-----+-----+
| | | | |
| | | | |
| | | | |
+-----+-----+-----+-----+
| | | | ^ |
| 3> | | | 3 |
| | | | |
+-----+-----+-----+-----+

There cannot be a 3-spot arrow pointing along the first row, as it would have to point from right to left to avoid having two arrows in the top-left corner cell, but that produces a four-long closed circuit. So the row that doesn't have a 3-spot arrow pointing along it is the first row, not the second.

At this point, a quick detour to answer the question about one 1-spot, seven 2-spot and eight 3-spot arrows: essentially the same arguments as above say that every row and every column must have a 3-spot arrow pointing along it, but that produces a four-long closed circuit between the four corner cells. Not saying that's superior to Julian's checkerboard-colouring argument - it's inferior, because that argument also says e.g. that there's no solution for three 1-spot, seven 2-spot and six 3-spot arrows and this one is highly specific to the case of one 1-spot, seven 2-spot and eight 3-spot arrows. But it does drop out very conveniently at this point in my solution to the puzzle!

Now consider the following colouring of the board with the four colours Red, Yellow, Green and Blue (I've not used coloured text to make it visually more obvious because coloured text conflicts with spoiler protection):

+-----+-----+-----+-----+ 
| | | | |
| R | Y | R | Y |
| | | | |
+-----+-----+-----+-----+
| | | | |
| G | B | G | B |
| | | | |
+-----+-----+-----+-----+
| | | | |
| R | Y | R | Y |
| | | | |
+-----+-----+-----+-----+
| | | | |
| G | B | G | B |
| | | | |
+-----+-----+-----+-----+

The three 3-spot arrows we have placed produce a Red->Green->Blue->Yellow transition in the sixteen-long closed circuit, i.e. an overall Red->Yellow transition, and the 2-spot arrows don't produce any colour transition no matter where they're placed. The other five arrows produce colour transitions as follows:

3-spot arrow pointing along second row: between Green and Blue in either direction, which I'll abbreviate to Green<->Blue.
3-spot arrow pointing along third row: Red<->Yellow
3-spot arrow along second column: Yellow<->Blue
3-spot arrow along third column: Red<->Green
Horizontally-pointing 1-spot arrow: Red<->Yellow or Green<->Blue

So if we travel once around the closed circuit, ending up where we started and only looking at the overall effect of the three 3-spot arrows we have already placed, not their individual effects, and count the colours at both the start and end of each of the six colour transitions, we will see Red 4 times, Yellow 4 times, Green 2 times and Blue 2 times if the 1-spot arrow does a Red<->Yellow transition, or each colour 3 times if it does a Green<->Blue transition. But the number of times we see each colour in that count must be twice the number of times we encountered the colour along that circuit (not counting the encounters with Green in the bottom left corner and Blue in the bottom right corner since they're ignored when we look at the overall effect of the three 3-spot arrows we have already placed), so:

* The 1-spot arrow must do a Red<->Yellow transition and so must be along the first row or the third row.
* The three Green cells other than the one in the bottom left corner must all be in the only encounter with Green, so must be connected together by 2-spot arrows.
* The three Blue cells other than the one in the bottom right corner must all be in the only encounter with Blue, so must be connected together by 2-spot arrows.

The leftmost and rightmost cells on the second row must be connected together in one direction or the other by a 3-spot arrow, so we get a five-long chain between the two middle cells on the last row going through those three Green cells and then the three Blue cells, or the same thing traversed in the reverse order. Furthermore, each of those middle cells on the last row must be connected to the corresponding cell on the first row by a 3-spot arrow, so that extends to a seven-long chain between the two middle cells on the top row, either consisting of a 3-spot arrow going from the Yellow one to the three Green cells (with two 2-spot arrows between them), then a 3-spot arrow to the three Blue cells (again with two 2-spot arrows between them), then via a 3-spot arrow to the Red middle cell on the top row, or the same thing in the opposite order. So we have two possibilities:

+-----+-----+-----+-----+          +-----+-----+-----+-----+
| | | | | | | | | |
| 3 | 3 | | | | 3 | | 3 | |
| v | v | | | | v | | v | |
+-----+-----+-----+-----+ +-----+-----+-----+-----+
| | | | | | | | | |
| 2> | 2> | 2 | <3 | | 3> | 2 | <2 | <2 |
| | | v | | | | v | | |
+-----+-----+-----+-----+ +-----+-----+-----+-----+
| | | | | | | | | |
| | | | | | | | | |
| | | | | | | | | |
+-----+-----+-----+-----+ +-----+-----+-----+-----+
| | ^ | ^ | ^ | | | ^ | ^ | ^ |
| 3> | 2 | 3 | 3 | | 3> | 3 | 2 | 3 |
| | | | | | | | | |
+-----+-----+-----+-----+ +-----+-----+-----+-----+

where in each case the three-long path starting in the top left corner ends in the top right corner, and the seven-long path starting in the occupied middle-of-top-row cell ends in the unoccupied middle-of-top-row cell. To make it a closed circuit, therefore, the six remaining arrows must connect the unoccupied middle-of-top-row cell to the top left corner and the top right corner to the occupied middle-of-top-row cell. There are only two colour transitions that are done outside those two paths, for the 1-spot arrow and for the 3-spot arrow pointing along the third row, both of which are Red<->Yellow transitions. Now split according to which of the two possibilities it is:

The first possibility

The three-long path does a Red->Yellow transition overall and the seven-long path a Yellow->Red transition. Each of the connections must produce no overall colour transition, so clearly one of the connections must involve only 2-spot arrows and the other both the 1-spot arrow and the third-row 3-spot arrow, doing opposite colour transitions to each other. It's easy to see that the connection that involves only 2-spot arrows must either involve just one arrow pointing along the first row, or three arrows pointing down to the third row, then along the third row, then back up to the first row. If it involves three 2-spot arrows, we end up having visited every cell of the colour being connected up, so the cells available for the other connection are all the same colour, which makes it impossible to use either the 1-spot arrow or the 3-spot arrow on it. However, if it involves just one 2-spot arrow along the top row, there isn't that problem, and so the first possibility expands into the following two cases:

+-----+-----+-----+-----+          +-----+-----+-----+-----+
| | * | | * | | * | | * | |
| 3 | 3 | <2 | | | 3 | 3 | | <2 |
| v | v | | | | v | v | | |
+-----+-----+-----+-----+ +-----+-----+-----+-----+
| | | | | | | | | |
| 2> | 2> | 2 | <3 | | 2> | 2> | 2 | <3 |
| | | v | | | | | v | |
+-----+-----+-----+-----+ +-----+-----+-----+-----+
| | | | | | | | | |
| | | | | | | | | |
| | | | | | | | | |
+-----+-----+-----+-----+ +-----+-----+-----+-----+
| | ^ | ^ | ^ | | | ^ | ^ | ^ |
| 3> | 2 | 3 | 3 | | 3> | 2 | 3 | 3 |
| | | | | | | | | |
+-----+-----+-----+-----+ +-----+-----+-----+-----+

where in each case I've marked the start and end of the now eleven-long path with asterisks. It's now easy to see that in each case, those two endpoints must each be connected to the third row by a 2-spot arrow, then that the two ends of the third row must be connected to each other by the 3-spot arrow, and then that there's only one way to use the 1-spot arrow and the remaining 2-spot arrow to finish the connection. So we get the following two solutions, both of which can be checked to be valid:

+-----+-----+-----+-----+          +-----+-----+-----+-----+
| | | | | | | | | |
| 3 | 3 | <2 | 2 | | 3 | 3 | 2 | <2 |
| v | v | | v | | v | v | v | |
+-----+-----+-----+-----+ +-----+-----+-----+-----+
| | | | | | | | | |
| 2> | 2> | 2 | <3 | | 2> | 2> | 2 | <3 |
| | | v | | | | | v | |
+-----+-----+-----+-----+ +-----+-----+-----+-----+
| | ^ | | | | ^ | | | |
| 2> | 2 | <1 | <3 | | 2 | 2> | <1 | <3 |
| | | | | | | | | |
+-----+-----+-----+-----+ +-----+-----+-----+-----+
| | ^ | ^ | ^ | | | ^ | ^ | ^ |
| 3> | 2 | 3 | 3 | | 3> | 2 | 3 | 3 |
| | | | | | | | | |
+-----+-----+-----+-----+ +-----+-----+-----+-----+

The second possibility

Both of the paths do Red->Yellow colour transitions overall, so each of the connections must produce an overall Yellow->Red colour transition. So each of the connections must use exactly one of the 1-spot arrow and the third-row 3-spot arrow, and each of those two arrows must do a Yellow->Red colour transition. That forces the third-row 3-spot arrow to go from the right-hand end of the row to the left-hand end. Now note that if the 1-spot arrow is also on the third row, the two unoccupied cells on the first row cannot be connected to each other by a single 2-spot arrow along the first row, since that wouldn't be in accoordance with the required connection pattern. So the paths can only be validly extended with 2-arrow connections between all four first-row cells and the corresponding third-row cells, but that doesn't work because it completes a six-long closed circuit involving all four corner cells and the two end cells of the third row.

So the the 1-spot arrow must be on the first row. It cannot point from the second cell to the third, since that would complete an eight-long closed circuit, so it must make one of the two connections we want, and so the second possibility expands into the following two cases:

+-----+-----+-----+-----+          +-----+-----+-----+-----+
| | | * | * | | * | * | | |
| 3 | <1 | 3 | | | 3 | | 3 | <1 |
| v | | v | | | v | | v | |
+-----+-----+-----+-----+ +-----+-----+-----+-----+
| | | | | | | | | |
| 3> | 2 | <2 | <2 | | 3> | 2 | <2 | <2 |
| | v | | | | | v | | |
+-----+-----+-----+-----+ +-----+-----+-----+-----+
| | | | | | | | | |
| | | | | | | | | |
| | | | | | | | | |
+-----+-----+-----+-----+ +-----+-----+-----+-----+
| | ^ | ^ | ^ | | | ^ | ^ | ^ |
| 3> | 3 | 2 | 3 | | 3> | 3 | 2 | 3 |
| | | | | | | | | |
+-----+-----+-----+-----+ +-----+-----+-----+-----+

where in each case the eleven-long path starting and ending in the asterisked cells needs to be closed off with a 3-spot arrow going from right to left along the third row and four 2-spot arrows. This is easily checked not to be possible in either case by trying to extend the path further using those arrows.

So there are the two solutions I've arrived at for the first possibility, which can be checked to be reversals of each other, and obviously all of their rotations and reflections, and no others.

Julian's method is also valid, of course, but the backtrackings by a different distance make his lines rather hard to depict in detail in ASCII diagrams!

Gengulphus

UncleEbenezer
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Re: Arrows

#245050

Postby UncleEbenezer » August 17th, 2019, 3:43 pm

For minimal graphics, I'll use l/r/u/d for left/right/up/down instead of arrows.
It's small enough for informed trial-and error, without too much error. I used literal back-of-envelope. Likely my solution matches what others have posted, modulo reflection/rotation [edit, no, it seems we're all different, though Julian's is similar]!

Guiding principle 1: there are 7 legs of length 3. They can't include all four edges, as that would make a closed circuit of just four legs. And since we need to visit all squares, we can't have any two legs sharing both vertices. Thus we need three edges of length 3, and all four inner side-to-side lines.
This also gives us why it's impossible to deploy 8 legs of length 3.

Guiding principle 2: we need to cover ground efficiently and avoid leaving inaccessible orphan squares (e.g. at a diagonal). Use the length 2 legs to accomplish this, and leave the length 1 leg for where it's needed in the endgame.

So starting clockwise from top left,

3r  2d  2l  3d
3r 2d 2d 2l
2r 2r 1u 3l
2u 3u 3u 3l

UncleEbenezer
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Re: Arrows

#245115

Postby UncleEbenezer » August 17th, 2019, 7:53 pm

Hmm, on checking Gengulphus' full post, I think my solution is the same (modulo reflection) as the second variant of his first possibility.

cinelli
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Re: Arrows

#245303

Postby cinelli » August 18th, 2019, 9:05 pm

Well done all. I am glad you enjoyed solving this puzzle. I too have just two different solutions, up to rotations and reflections, and these correspond to Gengulphus's. Julian's is the same as his first and Uncle's his second.

Cinelli


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