## Discussion to One More #290 (regarding symmetries in str8ts)

### Discussion to One More #290 (regarding symmetries in str8ts)

Hi everyone,

this week provided two nice extra puzzles:

is.gd/290_one_more_more_str8t

and

is.gd/290_one_more_more_str8t

After still not having found THE simple way for that one, this week's puzzle with its symmetric grid is a good opportunity to talk about symmetries in str8ts as mentioned some days earlier.

In general:

There's the flip symmetry in vertical and horizontal direction: If you flip both grid and solution, you will also get a valid str8ts solution. However, there's no non-trivial (white cells in row E only) solution that is symmetric to itself owing to the rules of the games. Given the hints, you can be sure that the flipped version violates. Given a symmetric grid, there needs to be a hint somewhere off-central for uniqueness.

There's a symmetry in the numbers ('global hi/low'): If you flip all numbers (1->9, 2->8,...) of a solution, you will also get a valid str8ts. Again, the hints will only allow for one of the two. Follows that there's no unique solution with just digit 5 as hints.

There's a point reflection symmetry: With symmetric grid (as here), you also get a valid solution by turning your solution 180 degree. Again, there's no all symmetric solution itself (except row E and col 5 empty).

There's also reflection symmetry in diagonal direction: Same arguments. But this time there might be an all symmetric solution.

The symmetries and the uniqueness considerations give rise to a (my oppinon new) str8ts rule: Given a symmetric grid and all symmetric hints (in the current puzzle e.g. B2=H8=2, C7=G3=1) there cannot be a unique solution (nothing to determine the asymetric part). In turn, given a symmetric grid and hints being symmetrical but one (e.g. only D6=1 or D6=1 and B2=H8=2), you can delete the unsymmetric hint digit directly from the corresponding cell as candidate (here F4<>1). Very seldomly applicable, but might be a very powerful tool for solving. If in both versions the corresponding symmetric hints were given in addition, solving becomes extremely neat whenapplying that rule (maybe a plea to Klaus...).

In the special case of this puzzle:

As mentined before, the grid is a very restrictive one, see the 4-cell hi/low interactions and many more. A lot of inner strings that reduce the number of hints needed. For example, there's a lot of conclusion even w/o any hints given. HJ3 and AB7 automatically become 23 (78) and AB4 HJ6 78(23). Furthermore, you can easily show (disproving via solver with deciding on hi/low before) that E5=3(7), setti 3,4,6,7, no 5 in A2-8 and J2-8, the well-known 7-fish, maybe also more (like no 5 in the outer ring at all).

If you see the solutions, there's a lot of symmetric fields (for both versions): A123478,B1479,D9,E5+corresponding lower half. Would guess that these are general to the geometry, not specific to the hints given. Guess also the non-symmetric part is more or less fixed, too. Would guess (??) that this puzzle has only one possible basic solution plus the variants of point reflection and global hi/low (number symmetry), so 4 in total.

For the only black cell hint versions: You will at least need two hints: One to decide on the asymmetric part and one for the high low decision. C7 or G3 determine hi/low, but nothing about (a)symmetry. Conversely, D5 or F5 decide (a)symmetry but not hi/low. BH28 don't help for neither, so only needed if there's more variants. So possibly redundant...

For the white cell hint versions: 2 hint versions are, as presented by Klaus, possible. In general, 1 number could determine both binary variants (hi/low + symmetry). Maybe there's also an 1 hint unique solutions starting from one of the non-symmetric, non-5 fields...

@Klaus:

a)In the current version, does the puzzle also solve omitting the B2=2? Even if very 'chainy' then...Does neither influence the hi/low nor the (a)symmetric question.

b)If there existed a 1 hint solution, it must be a hint answering both options. How about e.g. F7=9, D6=1 or C3=9, does your solver provide a solution for any of those cases? Would be keen to know...

@Leren: Asymmetry is induced by D5 or F5 and the range of the corresponding rows, the set of numbers for col 5 is fixed by the grid. Hope this answers your previous findings why the cols have the same range but not the rows...

CL

this week provided two nice extra puzzles:

is.gd/290_one_more_more_str8t

and

is.gd/290_one_more_more_str8t

After still not having found THE simple way for that one, this week's puzzle with its symmetric grid is a good opportunity to talk about symmetries in str8ts as mentioned some days earlier.

In general:

There's the flip symmetry in vertical and horizontal direction: If you flip both grid and solution, you will also get a valid str8ts solution. However, there's no non-trivial (white cells in row E only) solution that is symmetric to itself owing to the rules of the games. Given the hints, you can be sure that the flipped version violates. Given a symmetric grid, there needs to be a hint somewhere off-central for uniqueness.

There's a symmetry in the numbers ('global hi/low'): If you flip all numbers (1->9, 2->8,...) of a solution, you will also get a valid str8ts. Again, the hints will only allow for one of the two. Follows that there's no unique solution with just digit 5 as hints.

There's a point reflection symmetry: With symmetric grid (as here), you also get a valid solution by turning your solution 180 degree. Again, there's no all symmetric solution itself (except row E and col 5 empty).

There's also reflection symmetry in diagonal direction: Same arguments. But this time there might be an all symmetric solution.

The symmetries and the uniqueness considerations give rise to a (my oppinon new) str8ts rule: Given a symmetric grid and all symmetric hints (in the current puzzle e.g. B2=H8=2, C7=G3=1) there cannot be a unique solution (nothing to determine the asymetric part). In turn, given a symmetric grid and hints being symmetrical but one (e.g. only D6=1 or D6=1 and B2=H8=2), you can delete the unsymmetric hint digit directly from the corresponding cell as candidate (here F4<>1). Very seldomly applicable, but might be a very powerful tool for solving. If in both versions the corresponding symmetric hints were given in addition, solving becomes extremely neat whenapplying that rule (maybe a plea to Klaus...).

In the special case of this puzzle:

As mentined before, the grid is a very restrictive one, see the 4-cell hi/low interactions and many more. A lot of inner strings that reduce the number of hints needed. For example, there's a lot of conclusion even w/o any hints given. HJ3 and AB7 automatically become 23 (78) and AB4 HJ6 78(23). Furthermore, you can easily show (disproving via solver with deciding on hi/low before) that E5=3(7), setti 3,4,6,7, no 5 in A2-8 and J2-8, the well-known 7-fish, maybe also more (like no 5 in the outer ring at all).

If you see the solutions, there's a lot of symmetric fields (for both versions): A123478,B1479,D9,E5+corresponding lower half. Would guess that these are general to the geometry, not specific to the hints given. Guess also the non-symmetric part is more or less fixed, too. Would guess (??) that this puzzle has only one possible basic solution plus the variants of point reflection and global hi/low (number symmetry), so 4 in total.

For the only black cell hint versions: You will at least need two hints: One to decide on the asymmetric part and one for the high low decision. C7 or G3 determine hi/low, but nothing about (a)symmetry. Conversely, D5 or F5 decide (a)symmetry but not hi/low. BH28 don't help for neither, so only needed if there's more variants. So possibly redundant...

For the white cell hint versions: 2 hint versions are, as presented by Klaus, possible. In general, 1 number could determine both binary variants (hi/low + symmetry). Maybe there's also an 1 hint unique solutions starting from one of the non-symmetric, non-5 fields...

@Klaus:

a)In the current version, does the puzzle also solve omitting the B2=2? Even if very 'chainy' then...Does neither influence the hi/low nor the (a)symmetric question.

b)If there existed a 1 hint solution, it must be a hint answering both options. How about e.g. F7=9, D6=1 or C3=9, does your solver provide a solution for any of those cases? Would be keen to know...

@Leren: Asymmetry is induced by D5 or F5 and the range of the corresponding rows, the set of numbers for col 5 is fixed by the grid. Hope this answers your previous findings why the cols have the same range but not the rows...

CL

### Re: Discussion to One More #290 (regarding symmetries in str

Hi CL,

great post. unfortunately my geometrical vision is rather poor, so i have some trouble to see all the symmetry.

but i can tell what my solver thinks about this special grid. the black cells alone, without any numerical hint, reduce the possibilities this way:

here e.g. F7=1, F7=3, F7=7 or F7=9 all lead to a unique solution.

this means that with this structure indeed one hint is enough to give a valid str8t.

the solutions for (one and four) and (two and three) are number symmetric, but none of them are not point symmetric. in the solution of the first two cases D3=2 and in the other cases D3=8. so i think your guess(??) "that this puzzle has only one possible basic solution" is not true.

in case of F7=2 or F7=8 the solutions possible are not unique anymore, e.g. F7=2 and G2=5 lead to a unique solution, but F7=2 and G2=6 gives a whole bunch of solutions.

all very interesting, thanks for your input.

great post. unfortunately my geometrical vision is rather poor, so i have some trouble to see all the symmetry.

but i can tell what my solver thinks about this special grid. the black cells alone, without any numerical hint, reduce the possibilities this way:

here e.g. F7=1, F7=3, F7=7 or F7=9 all lead to a unique solution.

this means that with this structure indeed one hint is enough to give a valid str8t.

the solutions for (one and four) and (two and three) are number symmetric, but none of them are not point symmetric. in the solution of the first two cases D3=2 and in the other cases D3=8. so i think your guess(??) "that this puzzle has only one possible basic solution" is not true.

in case of F7=2 or F7=8 the solutions possible are not unique anymore, e.g. F7=2 and G2=5 lead to a unique solution, but F7=2 and G2=6 gives a whole bunch of solutions.

all very interesting, thanks for your input.

### Re: Discussion to One More #290 (regarding symmetries in str

Hi Klaus,

thanks for the answer with a lot of insights!

Had just clicked around a while with the online solver to disprove a few things and speculated. But yours is a little more helpful in that case to prove things

I have had another look on #14 solution, so I realized that there is more than 4 solutions in total. In #14 there's a UR in FG79 which cannot be resolved by putting a 2 there (you cannot decide if it is the other variant of the UR or the flipped version of the same solution (D3=2)). So F7=2 is not unique.

Symmetries are not that complicated, easier than finding a jellyfish (to me ): Only number symmetry and point symmetry apply here owing to the shape of the grid. Non of the solutions is symmetric (BC5=56 and GH5=56 cannot be the same time...). But the symmetry makes flipping one solution another solution. E.g. flipping F7=1 gives one of the F7=2 solution, flipping F7=3 gives a different F7=2 solution, and so on for all others. So if there's any solution, there's 4 (or more) solutions: Another one by flipping, third one by changing numbers, and fourth by doing both. If you count all, you'll get a multiple of 4.

So symmetries cannot tell you if the hints provide a unique solution, but they may tell you that certain hints, all 5s or everything symmetric, will not provide uniqueness. See the proposed 'strategy'.

Btw, have there been a lot of single hint puzzles so far?

Best,

CL

thanks for the answer with a lot of insights!

Had just clicked around a while with the online solver to disprove a few things and speculated. But yours is a little more helpful in that case to prove things

I have had another look on #14 solution, so I realized that there is more than 4 solutions in total. In #14 there's a UR in FG79 which cannot be resolved by putting a 2 there (you cannot decide if it is the other variant of the UR or the flipped version of the same solution (D3=2)). So F7=2 is not unique.

Symmetries are not that complicated, easier than finding a jellyfish (to me ): Only number symmetry and point symmetry apply here owing to the shape of the grid. Non of the solutions is symmetric (BC5=56 and GH5=56 cannot be the same time...). But the symmetry makes flipping one solution another solution. E.g. flipping F7=1 gives one of the F7=2 solution, flipping F7=3 gives a different F7=2 solution, and so on for all others. So if there's any solution, there's 4 (or more) solutions: Another one by flipping, third one by changing numbers, and fourth by doing both. If you count all, you'll get a multiple of 4.

So symmetries cannot tell you if the hints provide a unique solution, but they may tell you that certain hints, all 5s or everything symmetric, will not provide uniqueness. See the proposed 'strategy'.

Btw, have there been a lot of single hint puzzles so far?

Best,

CL

### Re: Discussion to One More #290 (regarding symmetries in str

Hi CL,

i dont know of any other 1-hint puzzle.

i've been wondering for a while whats the minimum number of hints and still get an interesting puzzle. indeed before seeing Ulrichs pattern of #14, and playing a bit around with it, my bet would have been something like 3, maybe 2. but 1, i was a bit surprised. still wondering if there is a 1- or 2-black hint(s)-puzzle!?

arent you a Sudoku-guy? do you know whats the minimum number there?

and i'm wondering if the set of solutions to the #14-pattern forms some kind of mathematical structure, like a group or something!?

greets

i dont know of any other 1-hint puzzle.

i've been wondering for a while whats the minimum number of hints and still get an interesting puzzle. indeed before seeing Ulrichs pattern of #14, and playing a bit around with it, my bet would have been something like 3, maybe 2. but 1, i was a bit surprised. still wondering if there is a 1- or 2-black hint(s)-puzzle!?

arent you a Sudoku-guy? do you know whats the minimum number there?

and i'm wondering if the set of solutions to the #14-pattern forms some kind of mathematical structure, like a group or something!?

greets

### Re: Discussion to One More #290 (regarding symmetries in str

to your sudoku question: not a sudoku guy at all...never tackled s/th more challenging than SZ newspaper versions...but wiki states 16/17 plus is the minimum of hints. Very obvious that at least 8 different numbers have to appear.

to your group structure question: think kind of. If you think of solutions as 'vectors' and number/geometric flipping as operators/'matrices' maybe the operators form some kind of very simple group structure. But assume much more interesting -if at all- for sudoku where there's no neighbor relations to be considered.

to your hint assumption: Most interesting point Interesting to know I guess. As pointed out in the initial post, don't think it possible with 1 black cell for that grid. Even 2 is hard. And black hints are typically much less restrictive than white hints. Especially when considering grids with medium/mixed compartment sizes. Any combination of 2 black cells works here? So I believe 2 hints could be possible. So keep on trying out!!

to your group structure question: think kind of. If you think of solutions as 'vectors' and number/geometric flipping as operators/'matrices' maybe the operators form some kind of very simple group structure. But assume much more interesting -if at all- for sudoku where there's no neighbor relations to be considered.

to your hint assumption: Most interesting point Interesting to know I guess. As pointed out in the initial post, don't think it possible with 1 black cell for that grid. Even 2 is hard. And black hints are typically much less restrictive than white hints. Especially when considering grids with medium/mixed compartment sizes. Any combination of 2 black cells works here? So I believe 2 hints could be possible. So keep on trying out!!

### Re: Discussion to One More #290 (regarding symmetries in str

Hi all,

when I let my solver run over this grid it produced the following result:

There are exactly 4 solutions.

Then I wondered why the solution with F7=7 is excluded and searched for bugs in my solver. But after rereading the posts it was clear that the difference was that my solver runs with UR checks.

Starting with F7 preset with 2, there are 2 UR checked solutions, one has 2 and 3 in FG79 but the UR restriction is lifted by F7 being preset.

Staring with F7=7 yields a unique solution, again with a lifted UR condition in FG79.

When running the solver with UR restrictions switched off, the result is:

This is very similar to Klaus' solution but with further restrictions in B6, E5 and H4.

Using UR checks or not boils down to the question if you want to solve or generate a grid.

BTW, in the original grid one can safely omit the black 2; the remaining two black values of 1 and 9 determine one of the 4 solutions of the UR picture.

Greetings

when I let my solver run over this grid it produced the following result:

There are exactly 4 solutions.

Then I wondered why the solution with F7=7 is excluded and searched for bugs in my solver. But after rereading the posts it was clear that the difference was that my solver runs with UR checks.

Starting with F7 preset with 2, there are 2 UR checked solutions, one has 2 and 3 in FG79 but the UR restriction is lifted by F7 being preset.

Staring with F7=7 yields a unique solution, again with a lifted UR condition in FG79.

When running the solver with UR restrictions switched off, the result is:

This is very similar to Klaus' solution but with further restrictions in B6, E5 and H4.

Using UR checks or not boils down to the question if you want to solve or generate a grid.

BTW, in the original grid one can safely omit the black 2; the remaining two black values of 1 and 9 determine one of the 4 solutions of the UR picture.

Greetings

### Re: Discussion to One More #290 (regarding symmetries in str

Hi hone,

of course you only can use UR-arguments when its safe that there is only one solution, thats the basis of UR-arguments. we here discussed the grid alone, without any clues, where its clear that there is no unique solution. so UR based elimination is not valid.

creating a puzzle UR's are of course not valid too. you cannot try to find a unique solution by assuming there is a unique solution.

and i dont understand your "BTW, in the original grid one can safely omit the black 2". it seems to me that without the black 2, F7=7 as well as F7=9 gives at least one possible solution.

of course you only can use UR-arguments when its safe that there is only one solution, thats the basis of UR-arguments. we here discussed the grid alone, without any clues, where its clear that there is no unique solution. so UR based elimination is not valid.

creating a puzzle UR's are of course not valid too. you cannot try to find a unique solution by assuming there is a unique solution.

and i dont understand your "BTW, in the original grid one can safely omit the black 2". it seems to me that without the black 2, F7=7 as well as F7=9 gives at least one possible solution.

### Re: Discussion to One More #290 (regarding symmetries in str

Hi Klaus,

the remark regarding the black 2 was again under the assumption, that the grid has to be solved as it is and UR restrictions can be applied like if the grid would appear as a weekly extreme. At least I had the opinion that those published puzzles are UR proofed.

The only solution then is the one with F7=9.

the remark regarding the black 2 was again under the assumption, that the grid has to be solved as it is and UR restrictions can be applied like if the grid would appear as a weekly extreme. At least I had the opinion that those published puzzles are UR proofed.

The only solution then is the one with F7=9.

### Re: Discussion to One More #290 (regarding symmetries in str

Hi hone,

i dont exactly understand what you mean.

for extremes you can clearly assume a unique solution as given. when you talk about ommiting clues, here the black 2, then this new variation clearly has at least one solution (the original one will still hold here) , but that this is still the only one is not safe anymore. you have to proof this, and within this proof you cannot use UR-arguments. in your case of leaving the black 2 i think uniqueness is not given anymore. so your "in the original grid one can safely omit the black 2" doesnt make sense to me.

secondly i dont understand how you came to your first picture.

if i am not mistaken in the pure grid F7=1, F7=3, F7=7 or F7=9 all lead to a unique solution, F7=2 or F7=8 to multiple solutions. but your F7 consists of 1289. why so?

i dont exactly understand what you mean.

for extremes you can clearly assume a unique solution as given. when you talk about ommiting clues, here the black 2, then this new variation clearly has at least one solution (the original one will still hold here) , but that this is still the only one is not safe anymore. you have to proof this, and within this proof you cannot use UR-arguments. in your case of leaving the black 2 i think uniqueness is not given anymore. so your "in the original grid one can safely omit the black 2" doesnt make sense to me.

secondly i dont understand how you came to your first picture.

if i am not mistaken in the pure grid F7=1, F7=3, F7=7 or F7=9 all lead to a unique solution, F7=2 or F7=8 to multiple solutions. but your F7 consists of 1289. why so?

### Re: Discussion to One More #290 (regarding symmetries in str

Hi Klaus,

Regarding the first picture:

This is the result as if this grid was published as weekly extreme and UR restrictions could be applied.

So with UR restrictions there are only 4 solutions. For cells with 4 possible values each value leads to a unique solution under the premise that UR restrictions are further applied.

Now setting the black 1 and 9 reduces the solution to the one with F7=9.

The black 2 is not needed, as one can see in the first picture, that 2 can neither appear in row B nor column 2.

So if the grid with only black 1 and 9 would be published, and one can assume it UR proofed, you would only find the solution with F7=9 (which is the same as the original solution).

The F7=7 and F7=3 possibilities were obviously dropped during the solver run because of the UR in FG79.

If F7 is preset with 7 than the situation is different because the UR condition in FG79 is lifted (like in #289 the 56-UR condition in HJ26 by presetting a 6 into H2).

Back to the basic grid without the Black 2 the result with no UR checks is:

Choosing F7=7 leads immediately to UR FG79.

Choosing F7=8 leaves the choice of 7 or 9 in G7. The first is the UR FG79, the other leads to a UR in CD13.

So this grid is an example for a unique UR proofed solution (F7=9) with only two black cell hints.

Regarding the first picture:

This is the result as if this grid was published as weekly extreme and UR restrictions could be applied.

So with UR restrictions there are only 4 solutions. For cells with 4 possible values each value leads to a unique solution under the premise that UR restrictions are further applied.

Now setting the black 1 and 9 reduces the solution to the one with F7=9.

The black 2 is not needed, as one can see in the first picture, that 2 can neither appear in row B nor column 2.

So if the grid with only black 1 and 9 would be published, and one can assume it UR proofed, you would only find the solution with F7=9 (which is the same as the original solution).

The F7=7 and F7=3 possibilities were obviously dropped during the solver run because of the UR in FG79.

If F7 is preset with 7 than the situation is different because the UR condition in FG79 is lifted (like in #289 the 56-UR condition in HJ26 by presetting a 6 into H2).

Back to the basic grid without the Black 2 the result with no UR checks is:

Choosing F7=7 leads immediately to UR FG79.

Choosing F7=8 leaves the choice of 7 or 9 in G7. The first is the UR FG79, the other leads to a UR in CD13.

So this grid is an example for a unique UR proofed solution (F7=9) with only two black cell hints.