Discuss Sorcerous Solitaire here

[Sofis]

Puzzle Link.

[Applebyd]

OK just as a start,

36 Pegs
37 Holes.

I've never been much good at solitaire in the past!

Is this bigger or smaller than an average (normal) board?

I'm *guessing* that there are a large number of methods of solving
a normal board but only one to solve this.

I know it's not quite correct as you are removing pieces all the time but the shape and odd vertices indicate there isn't a Eularian path. However my math isn't sharp enough know if this matters as you don't have to move the same piece each time.

Ah well someone had to kick the discussion off.

DaveA

[GettinConfused]

This is slightly bigger than the original which had 31 holes (I think). I believe there is only one way to complete the original (except in reverse). If it's anything like that one then there is a definitive set pattern to complete it.

This doesn't help much but at least it might give someone an idea.

[cha]

I'm guessing that this is impossible for some reason. The only peg game I've seen was triangle shaped and had 10 holes. Perhaps there's a way to show this can't be done with an odd number of holes. Or perhaps the shape has something to do with it - either way, I'm betting it can't be done.

[The Cruciverbalist]

I'm with cha on this one. All the wording suggests it.

[cha]

It's the shape. Look at the six haxaganol shapes on the perimeter. There are three outer pegs in each that will only ever be moved by jumping over the center hole of that particular hexagon. Since there are 18 pegs on the perimeter that have only the one option, at some point 18 other pegs must find themselves in that particular center hole. Interestingly, there are no combinations of moves that will allow any perimeter peg from entering any of the center holes of any outer hexagon. There are 36 pegs total. The first move will remove one peg, leaving 35 total - 18 on the outer perimeter, 17 others - not enough to allow all of the outer perimeter pegs to be moved.

[mtc32]

Damn! I was going to say that. If you added the lines to make a full outer hexagon, it might be possible.

[Quailman]

cha: If you consider the second from the left on the bottom of the diagram, it has two options for jumping out of its position. And if it jump to the upper left, it lands in a spot that gives another outer edge peg a chance to jump. Twelve of the eighteen outer pegs have two options for jumping from the outer row. I suspect that this is possible.

[cha]

True - I was wrong about the center peg being the only option, but there will still be 18 needed and only 17 remaining after the first move.

[azu]

the pegs on the outer hexagons which are on the perimeter and face the center(of the whole shape) can only leave there by jumping over the center peg of that hexagon.but on your first move you must pick up one of these center pegs(of the outside hexagons) for you have no other possible moves,thus some other peg must take its place.but in order to do this you must use the center peg of one of the other peripheral hexagons.so you will always have one of the outer hexagons without a peg at the center,so it is impossible to finish with just one peg.

[Gammadragon]

azu this is wrong, as there are other roundabout ways of replacing those pegs, however, I agree with your idea of the outer hexagon pegs requiring some fancy peg-moving to release them.

[azu]

only two ways to get a peg to the center of an outer hexagon a)from the center of an other outer hexagon or b)from the center of the whole shape.only one way for peg to get to the center of the whole shape from the center of an outer pentagon.since you must pick up a peg from the center of an outer hexagon you will always have an empty center in one of the outer hexagons thus one of the pegs on the perimeter which face the center of the entire shape you wont be able to get rid of.thus you will be left with at least two.

[azu]

i am never wrong

[Beartalon]

azu: eventually, there will be a first time, and this might be it.

[SaberKitty]

maybe it is impossible or whatever- but have any of you actually tried it?
best i've been able to get was 5. If getting down to one is impossible, i'd sure like to know how to get 2.

-SK

[Griffin]

Has anyone else tried working backwards? I did and I got it so there was only one place left without a peg. Unfortuately it was not the center . Still I suspect it is possible.

[Timo_Tuokkola]

The rules do not actually say that the pegs can only move along the lines on the board. Is this a requirement of the puzzle or just something we are assuming?

[referee]

You don't need the final peg to be in the center for this puzzle. I guess it must be one of those perimeter-of-outer-hexagons pegs.

Griffin, care to enlighten us?

[Parkboarder]

I think I know what all your problems are... you are all thinking WAY TO MUCH!!!!! all this is is a simple eliminitation. nothing more!

[Parkboarder]

I just constructed a crude but effective version of this puzzel and after a few times of playing I can only seem to get it down to three. I dont know if this is of any help but its quite frustrating !!!

[®ur Išlo]

If your looking for an easy board to set up on the computer. Just save the puzzle board picture as a bitmap onto your computer. Then stretch the image to 200% for both horizontal and vertical by going to the image menu item, then stretch/skew. Now play. Use the paint can feature to fill in the blanks to play.

I've managed to get the pegs down to 2 left. But a few spaces apart. (Not in the middle either). When I tried to bring them closer together, it left me with three pegs in a row, which naturally left two again.

[This message has been edited by ®ur Išlo (edited 02-20-2003 12:53 AM).]

[Griffin]

Referee - I thought the puzzle might be easier if I worked backwards, i.e., started with one peg somewhere, then jump it over a hole and put another peg in the hole that was jumped, and continue to see if I could fill the whole board. Since the starting configuration of the original puzzle has the peg missing in the center, if I work backwards I need to end with the peg missing in the center.

[ZutAlors!]

Maybe this is a silly comment, Griffin, but isn't working backwards just a binary inverse of working forwards? In other words, when working "forwards" you go from:
hole-peg-peg
to
peg-hole-hole

and "backwards" is from:
peg-hole-hole
to
hole-peg-peg

so they're just inverse binary operations, depending on wether you map "peg"->1 or "hole"->1. So, what that boils down to is:

If you worked "backwards", starting from the middle, and got only one hole left, seems like you should be able to do the same sequence of operations, only substituting "peg" for "hole" and vice-versa, and solve the real problem. Right?

[wierdo203]

To work it out, you just need the first sixteen (Wrong), then rotate/flip the board and do the first fifteen (Wrong) in reverse, with holes, as opposed to pegs. [edit]I was thinking of a different puzzle that was similar, it is actually the first 18, then the board can be rotated so that the next move will make it so that the pegs correspond to holes and holes correspond to pegs on the boards. The 18th move would switch whether anything is on the center circle. This is only for getting it on the center.

[This message has been edited by wierdo203 (edited 02-20-2003 05:03 PM).]

[parkboarder]

what do you mean do the first 16 then turn it and do the first 15 in reverse?

[ralphmerridew]

I think I have an impossibility proof:

Color the dots as follows:

code:

---

     R-G B-R


    G-B-R-G-B


   B-R-G-B-R-G


  R-G-B-R-G-B-R


   B-R-G-B-R-G


    G-B-R-G-B


     R-G B-R

---

Note that every row of three contains on red, on blue, and one green, so every move will change the parity of the number of pegs on each color. (The count for two colors goes down by one, and the count for the third goes up by one.)

At first, there are 12 pegs on red, 12 on blue, 12 on green, an even number of each. After the first move, there will be an odd number of pegs on each color. After the second move, there will be an even number of pegs on each color. After 35 moves, there would be an odd number of pegs on each color, requiring at least 3 pegs (one red, one blue, and one green). But at that stage there would be only one peg left. Contradiction.

By the same reasoning, it is impossible if the hole starts out on any other red hex. If the hole starts on a green hex, then it starts 13 red, 12 blue, 11 green (odd, even, odd), which would leave (even, odd, even) after 35 moves, and so is potentially solvable.


Can anybody spot a hole in this reasoning?

[Griffin]

Zutalors - Your absolutely right . I think it just seemed easier because the center hole is not a very good place to start, as ralphmerridew points out.

Ralphmerridew - Looks good to me. And it is possible starting on at least some non red squares, from where it is actually pretty easy to solve.

[HappyFunBall]

Well this topic seems to have died, but just in case anyone was wondering, ralphmerridew's impossibility proof seems completely valid. And as a bonus, note that it works just as well without the gaps on each edge of the hexagon.

[What if...]

So we have established that, without using negative pegs or other weird antimatter devices, it is impossible to win starting in the center? Any objections?

[mathgrant]

Could the wording for the first sentence in the answer be made a little more family-friendly? Like "evil old guy"?

[joatamos]

Hi yall hate to burst your bubble but for my fathers birthady I made hime one(I copyed the pic to the letter) any way hes been able to get it down to 2 pegs and just today he got it to 1 left though not in the center when I get the time (and if he can rember how he did it lol) I'll make a pic and post it

[CrystyB]

add another if: if your dad's solution is valid. There is no way to drop below 2 pegs.

[What if...]

I suspect he didn't start in the middle, as that seems the most likely rule he could forget...