Puzzle challenges!

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Puzzle challenges!

Postby Ephraim » August 8th, 2017, 7:41 pm

Hello everyone!

This topic is a recopilation of puzzles that will make you think a lot.

Rules:

The person who solved the last puzzle can post another puzzle. If he/she doesn't know any other one, he/she can just say «I don't know any other one», and then, any person can give one, but only the first posted after the solution of the last puzzle will be the official one to solve.

List of puzzles:

The nine braziers
The holy water
The 100 green-eyed dragons
The two sages and the evil king
The four races
The four witness
The poisoned wine
The impossible math riddle
The 100 tigers and the sheep
The two paths (variation: The two gates)
The old hermit
The hanging one
The elephant
The spy
The 100 coins
The 101 dwarves (variation: The hats again)
The balancing scale
The 20 bags of coins
The spider and the ant
The 20 pirates
The quest of Delfador
The 7 races of Wesnoth
The 4 injured men
The nested riddle
The other impossible math riddle
The footpad, the mage and the lord

Have fun!


The nine braziers:

You enter a room with 9 braziers arranged in a square (3 x 3), all of which are off except the center one, that it is on. In order to advance your path, you must light them all. But they are not normal braziers, since ...

- If you turn on one, the one in the center (every time I talk about «the one in the center» I mean the one that is in the center of the square, NOT in the center of the line) and the opposite one will turn on.

- If you turn one off the center and its opposite turn off too.

- If one that is already on has to be turned back on then it turns off.

- If one that is already turned off has to be turned off, it will turn on.

- If you turn on the center one, it turns on the right line and the top center one too.

- If you turn off the center one, those on the left line and the bottom center turn on.

As I said before, it is a complicated puzzle. If you found the answer, post it here using this method:

Code: Select all
I turn on the center brazier,
I turn on the middle-left brazier,
I turn off the center brazier,
I turn on the bottom right brazier,
...
(Note: This is just an example)


Solution and Winner:
Last edited by Ephraim on November 26th, 2017, 11:41 am, edited 29 times in total.
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Re: A very difficult puzzle

Postby Velensk » August 8th, 2017, 9:06 pm

Just to confirm, the rules are as follows:
-You can toggle any of the switches.
-If you toggle a switch that isn't in the center you also toggle the center and the opposite switch.
-The center is conditional, if you toggle it while it's on it toggles one set of 4 switches and if you toggle it while it's off it toggles those four switches opposite switch.

Assuming my understand is correctly, I'm fairly confident it's impossible. I sketched the matrix and I don't think there's any way to go from all 0s to all 1s or vice versa. I haven't spent a lot of time trying to solve it but that'd be my guess just looking at the math.
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Re: A very difficult puzzle

Postby Ephraim » August 8th, 2017, 9:09 pm

Yeah, you are right with the rules.
Last edited by Ephraim on August 9th, 2017, 9:35 am, edited 1 time in total.
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Re: A very difficult puzzle

Postby Samonella » August 9th, 2017, 2:33 am

I puzzled over it for a bit and didn't come up with anything. Had a good time though. :)

Velensk wrote:-If you toggle a switch that isn't in the center you also toggle the center and the opposite switch.
Just to be clear, "opposite" means this:
Code: Select all
1 0 0
0 1 0
0 0 1
as opposed to this:
Code: Select all
0 0 1
0 1 0
0 0 1
right?

Also, it's not something silly like walking to a different side of the square so "left" is different, is it?
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Re: A very difficult puzzle

Postby Inky » August 9th, 2017, 3:14 am

Samonella wrote:Also, it's not something silly like walking to a different side of the square so "left" is different, is it?

Ohh I like this solution, very creative! :D :eng: Or maybe, you could just pick up the braziers and rearrange them :whistle:

If it's not something creative and "opposite" means the first matrix in Samonella's post, then it doesn't seem possible? Or I'm not interpreting it correctly? :?
Basically, the number of times you turn the center on should have the same parity as the number of times you turn it off, and the number of times you toggle each of the 4 opposite pairs should also have the same parity (opposite to the one above) so then there's an even total of operations but since every operation toggles the center there should be an odd number? :hmm:
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Re: A very difficult puzzle

Postby ElderofZion » August 9th, 2017, 4:19 am

Isn't there a feedback? If switching any brazier outside the center causes the center and the opposite one to switch on, then by the rules switching on the center (and it doesnt work only for the center, in fact if switching on any brazier causes the center and the opposite one to light on, well if we follow this logic it gets rather crazy) causes one side to light up, we can go further than that but the idea itself is silly.

So if we assume there is no feedback, which seems to be the only possible option, then, well if i understood it right then the sulution would be like this, surprisingly simple:

Code: Select all
I turn on middle
111
011
001

I turn off middle
011
101
110

I activate top left one or bottom right one
111
111
111


Did I get it right? This is a fun riddle
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Re: A very difficult puzzle

Postby Inky » August 9th, 2017, 5:15 am

ElderofZion wrote:I turn on middle
111
011
001

Ephraim wrote:- If you turn on the center one, it turns on the right line and the top center one too.

I think turning on the middle would actually be this?
Code: Select all
0 1 1
0 1 1
0 0 1
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Re: A very difficult puzzle

Postby Ephraim » August 9th, 2017, 9:30 am

Inky wrote:
ElderofZion wrote:I turn on middle
111
011
001

Ephraim wrote:- If you turn on the center one, it turns on the right line and the top center one too.

I think turning on the middle would actually be this?
Code: Select all
0 1 1
0 1 1
0 0 1

Yes, the top left brazier doesn't switch on, as Inky wrote. I have to say that... there isn't a solution. It can only be solved if the center brazier is turned on yet, at the start of the game. :whistle:
But I want to make it difficult, because...
Code: Select all
0 0 0
0 1 0
0 0 0
I turn off the center brazier,
1 0 0
1 0 0
1 1 0
I turn on the center brazier.
1 1 1
1 1 1
1 1 1


It's too simple. Maybe, if we change the rules like...
If you turn off the center brazier, the left column and the bottom center ones turn off too.


What do you think about it?
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Re: A very difficult puzzle

Postby Ravana » August 9th, 2017, 10:35 am

As I told you in pm, with your original production rules
Code: Select all
{'101110010', '000001111', '011010110', '100100110', '000110111', '000101000', '
010100101', '001100011', '011110001', '111100000', '110110100', '110101011', '01
1101110', '101010101', '001000100', '111111111', '100011110', '111000111', '1001
11001', '011001001', '110001100', '010111010', '110010011', '001111100', '101001
010', '100000001', '010000010', '001011011', '010011101', '111011000', '00001000
0', '101101101'}
can be changed to 111111111. I just picked '000010000' for you as it has least already on.

I wrote nonrecursive DFS algorithm for this.
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Re: A very difficult puzzle

Postby Ephraim » August 9th, 2017, 11:24 am

Ravana wrote:As I told you in pm, with your original production rules
Code: Select all
{'101110010', '000001111', '011010110', '100100110', '000110111', '000101000', '
010100101', '001100011', '011110001', '111100000', '110110100', '110101011', '01
1101110', '101010101', '001000100', '111111111', '100011110', '111000111', '1001
11001', '011001001', '110001100', '010111010', '110010011', '001111100', '101001
010', '100000001', '010000010', '001011011', '010011101', '111011000', '00001000
0', '101101101'}
can be changed to 111111111. I just picked '000010000' for you as it has least already on.

I wrote nonrecursive DFS algorithm for this.


As I said before, this code says that the center brazier must be alredy on, and it's too easy to solve, then.
The solution is turn off the center one, and then, turn on the center one too.
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Re: A very difficult puzzle

Postby Ravana » August 9th, 2017, 11:31 am

That covers 2 of these 32. There are 30 other states that lead to solution.
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Re: A very difficult puzzle

Postby ElderofZion » August 9th, 2017, 12:18 pm

Well, I thought that by "the top center one" you mean top center line, though, now that I think about that it leads to feedback, but well, it seems that Ravana solved it.
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Re: A very difficult puzzle

Postby Ephraim » August 9th, 2017, 12:22 pm

ElderofZion wrote:Well, I thought that by "the top center one" you mean top center line, though, now that I think about that it leads to feedback, but well, it seems that Ravana solved it.


I want to make this puzzle very difficult to solve, and that it hasn't got 32 possible answer. A puzzle that makes the person think a lot, but that has a solution. Someone knows how (changing the rules or something) make it more difficult?
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Re: A very difficult puzzle

Postby Ravana » August 9th, 2017, 12:41 pm

If it is solvable, then by including symmetrical rule, you have infinite solutions by using that rule twice, however many times.
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Re: A very difficult puzzle

Postby Inky » August 9th, 2017, 12:45 pm

Ephraim wrote:I want to make this puzzle very difficult to solve, and that it hasn't got 32 possible answer.

There are 32 starting configurations for which there exists a solution, not 32 different solutions for a given starting configuration.
Ephraim wrote: I have to say that... there isn't a solution.

Well, I liked my idea of rearranging the braziers yourself :P :whistle:

I think there is a simple way to figure out valid configurations for the original problem:

math
If the center is on we can turn the center off to get this
Code: Select all
1 0 0
1 1 0
1 1 0

and then toggle the four opposite pairs to get
Code: Select all
0 1 1
0 1 1
0 0 1
which is the same thing as what happens when you turn the center from off to on.

So we can replace these two rules
Ephraim wrote:- If you turn on the center one, it turns on the right line and the top center one too.
- If you turn off the center one, those on the left line and the bottom center turn on.
with this single rule:
-Toggling the center (turning it either on or off) turns on the right line and the top center one too.

So a valid starting position would just be any linear combination of these 5 matrices:
Code: Select all
   1 0 0      0 1 0      0 0 1      0 0 0      0 1 1   1 1 1
x1*0 1 0 + x2*0 1 0 + x3*0 1 0 + x4*1 1 1 + x5*0 1 1 + 1 1 1
   0 0 1      0 1 0      1 0 0      0 0 0      0 0 1   1 1 1  (mod 2)

(if anyone can even make sense of this "creative" notation :doh: )
which has 32 solutions (each of the five xi can equal 0 or 1) which hopefully give the same as what Ravana posted above? Maybe? :hmm:
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