Is the picture a
Which one is real Qubic: the 3d 4x4x4 Tic tac toe in which you can leave your token on the highest plattform anytime, or the one in that your token falls down? Which one did Patashnik and Victor Allis solve in their work (references)? --anon1
Patashnik and Allis independently solved qubic (3d 4x4x4 tic tac toe). From the original paper (his masters thesis) by Allis, it seemed that he only weakly solved qubic. However I have not read his "Qubic solved again" paper mentioned in this page, but I think it is the same paper that was in his thesis. --anon2
The image was of Score Four although it is misnamed qubic.jpg. That is why I moved the image to the Score Four page. -- IanOsgood 16:03, 17 February 2007 (UTC)
I actually own the Parker Brothers game. I'm wondering if I should take a photo and include it here, and if so, how best to do that. Do I put the photo on Picasa or some such and link to it, or upload it directly? kogorman ( talk) 17:03, 9 February 2016 (UTC)
I put an unclear tag on this.
...so how does it get down to just five?
Also, the whole section needs a reference. Where did this come from?
Jeh (
talk)
19:39, 11 November 2015 (UTC)
Rewrite
The cube structure makes the 8 corner-points and 8 centre-points extremely important; each of these is a member of 6 planes (1×flat, 2×vertical, 2×diagonal-vertical, 1×cross-vertical) of 16 points. The flat and 2×vertical planes only contain 4 important points, while the 2×diagonal-vertical and cross-vertical contain 8 powerpoints.
O begins and places his first peg A on any one of the 16 powerpoints. Even if X places his first peg on a plane involving A there will be other planes involving A where O can place his second peg B; and on any such plane there will be 3 or 7 available powerpoints.
Even if X places his second peg on a plane including A & B there still remains a plane where O can place a third peg C onto one of the 2 or 6 available powerpoints. Here X must be foolish and allow a fourth O ‘1’ to be placed in that plane.
Once A, B, C & 1 are placed then there is a forced win with a further 5 pegs by O since X must respond with x1; then 2, x2; 3, x3 and so on.
I have had to redo our summary as the Salisbury Games Group notes are now unavailable. I am sure I recollect that somehow we identified 5 points but typos happen, memory fades and I have rewritten the description. I seem to recollect that our tame mathematician argued that mathematically or rather topologically, the 4 x 4 x 4 structure can be transformed by symmetries and reflections so that many powerpoint choices are ‘identical’ - this may have reduced the choice to '5' ?! JK — Preceding unsigned comment added by Nojoking ( talk • contribs) 11:58, 6 December 2015 (UTC)
The last version of the "Fool's Mate strategy" section is completely unreferenced and has been so for over a year despite the add of a CN tag. It must therefore be regarded as pure original research. Accordingly it has been removed from the article and is preserved here for reference. Click the "Show" link at the right end of this box to view it. Please do not modify it. Please do not restore any form of this to the article without adding reliable sources for each claim of fact. | ||||
---|---|---|---|---|
The following discussion has been closed. Please do not modify it. | ||||
The cube structure makes the 8 corner-points and 8 centre-points extremely important; each of these is a member of 6 planes (1×flat, 2×vertical, 2×diagonal-vertical, 1×cross-vertical) of 16 points. The flat and 2×vertical planes only contain 4 important points, while the 2×diagonal-vertical and cross-vertical contain 8 powerpoints. O begins and places his first peg A on any one of the 16 powerpoints. Even if X places his first peg on a plane involving A there will be other planes involving A where O can place his second peg B; and on any such plane there will be 3 or 7 available powerpoints. Even if X places his second peg on a plane including A & B there still remains a plane where O can place a third peg C onto one of the 2 or 6 available powerpoints. Here X must be foolish and allow a fourth O ‘1’ to be placed in that plane. Once A, B, C & 1 are placed then there is a forced win with a further 5 pegs by O since X must respond with x1; then 2, x2; 3, x3 and so on. .A.x3..3..B .1..5..2..w x1.x2..4... .C....x4..w |
At least a third of the article 3-D Tic-Tac-Toe (Atari console game) (article formerly called 3-D Tic-Tac-Toe, while this one was called Qubic replicates this article. The rest is about a program for the Atari 2600. Does that program for a long-obsolete console really deserve its own article? It's not as if its existence is particularly notable - there have been many Qubic programs before and since. Jeh ( talk) 02:35, 12 November 2015 (UTC)
Concur. It would make sense for this article to (briefly) list known implementations, and perhaps a redirect from any known implementations. On that topic, should the main article be named Qubic? My understanding is that it is a trade name owned by Parker Brothers. It seems better to move this to 3-D Tic-Tac-Toe, and redirect this name to that article. 47.32.142.141 ( talk) 21:22, 7 February 2016 (UTC) Oops. sorry. The above was by me when I had not signed in. kogorman ( talk) 16:02, 9 February 2016 (UTC)
It gets worse. I was concerned that Qubic might need disambiguation, and looked it up on Google. Sure enough, there is currently an app named Qubic available on Amazon, which I have just installed on my Kindle Fire. It is a puzzle game having nothing to do with tic-tac-toe. Moreover, there are other items and businesses that use the name. So I'm thinking the Qubic entry should be disambiguation. kogorman ( talk) 16:51, 9 February 2016 (UTC)
I just made a version of this game that allows for physically playing against algorithms on an electronic board.
You can see it here: https://www.kickstarter.com/projects/xno/4play
I was wondering if I could mention this addition somewhere in the article, since I do think it represents a new type of play that hasn’t been possible before. However, I understand if this just seems like advertising. Any advice/direction would be appreciated. 5^.5*.5+.5=phi ( talk) 17:45, 2 October 2019 (UTC) — Preceding unsigned comment added by 5^.5*.5+.5=phi ( talk • contribs)
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Is the picture a
Which one is real Qubic: the 3d 4x4x4 Tic tac toe in which you can leave your token on the highest plattform anytime, or the one in that your token falls down? Which one did Patashnik and Victor Allis solve in their work (references)? --anon1
Patashnik and Allis independently solved qubic (3d 4x4x4 tic tac toe). From the original paper (his masters thesis) by Allis, it seemed that he only weakly solved qubic. However I have not read his "Qubic solved again" paper mentioned in this page, but I think it is the same paper that was in his thesis. --anon2
The image was of Score Four although it is misnamed qubic.jpg. That is why I moved the image to the Score Four page. -- IanOsgood 16:03, 17 February 2007 (UTC)
I actually own the Parker Brothers game. I'm wondering if I should take a photo and include it here, and if so, how best to do that. Do I put the photo on Picasa or some such and link to it, or upload it directly? kogorman ( talk) 17:03, 9 February 2016 (UTC)
I put an unclear tag on this.
...so how does it get down to just five?
Also, the whole section needs a reference. Where did this come from?
Jeh (
talk)
19:39, 11 November 2015 (UTC)
Rewrite
The cube structure makes the 8 corner-points and 8 centre-points extremely important; each of these is a member of 6 planes (1×flat, 2×vertical, 2×diagonal-vertical, 1×cross-vertical) of 16 points. The flat and 2×vertical planes only contain 4 important points, while the 2×diagonal-vertical and cross-vertical contain 8 powerpoints.
O begins and places his first peg A on any one of the 16 powerpoints. Even if X places his first peg on a plane involving A there will be other planes involving A where O can place his second peg B; and on any such plane there will be 3 or 7 available powerpoints.
Even if X places his second peg on a plane including A & B there still remains a plane where O can place a third peg C onto one of the 2 or 6 available powerpoints. Here X must be foolish and allow a fourth O ‘1’ to be placed in that plane.
Once A, B, C & 1 are placed then there is a forced win with a further 5 pegs by O since X must respond with x1; then 2, x2; 3, x3 and so on.
I have had to redo our summary as the Salisbury Games Group notes are now unavailable. I am sure I recollect that somehow we identified 5 points but typos happen, memory fades and I have rewritten the description. I seem to recollect that our tame mathematician argued that mathematically or rather topologically, the 4 x 4 x 4 structure can be transformed by symmetries and reflections so that many powerpoint choices are ‘identical’ - this may have reduced the choice to '5' ?! JK — Preceding unsigned comment added by Nojoking ( talk • contribs) 11:58, 6 December 2015 (UTC)
The last version of the "Fool's Mate strategy" section is completely unreferenced and has been so for over a year despite the add of a CN tag. It must therefore be regarded as pure original research. Accordingly it has been removed from the article and is preserved here for reference. Click the "Show" link at the right end of this box to view it. Please do not modify it. Please do not restore any form of this to the article without adding reliable sources for each claim of fact. | ||||
---|---|---|---|---|
The following discussion has been closed. Please do not modify it. | ||||
The cube structure makes the 8 corner-points and 8 centre-points extremely important; each of these is a member of 6 planes (1×flat, 2×vertical, 2×diagonal-vertical, 1×cross-vertical) of 16 points. The flat and 2×vertical planes only contain 4 important points, while the 2×diagonal-vertical and cross-vertical contain 8 powerpoints. O begins and places his first peg A on any one of the 16 powerpoints. Even if X places his first peg on a plane involving A there will be other planes involving A where O can place his second peg B; and on any such plane there will be 3 or 7 available powerpoints. Even if X places his second peg on a plane including A & B there still remains a plane where O can place a third peg C onto one of the 2 or 6 available powerpoints. Here X must be foolish and allow a fourth O ‘1’ to be placed in that plane. Once A, B, C & 1 are placed then there is a forced win with a further 5 pegs by O since X must respond with x1; then 2, x2; 3, x3 and so on. .A.x3..3..B .1..5..2..w x1.x2..4... .C....x4..w |
At least a third of the article 3-D Tic-Tac-Toe (Atari console game) (article formerly called 3-D Tic-Tac-Toe, while this one was called Qubic replicates this article. The rest is about a program for the Atari 2600. Does that program for a long-obsolete console really deserve its own article? It's not as if its existence is particularly notable - there have been many Qubic programs before and since. Jeh ( talk) 02:35, 12 November 2015 (UTC)
Concur. It would make sense for this article to (briefly) list known implementations, and perhaps a redirect from any known implementations. On that topic, should the main article be named Qubic? My understanding is that it is a trade name owned by Parker Brothers. It seems better to move this to 3-D Tic-Tac-Toe, and redirect this name to that article. 47.32.142.141 ( talk) 21:22, 7 February 2016 (UTC) Oops. sorry. The above was by me when I had not signed in. kogorman ( talk) 16:02, 9 February 2016 (UTC)
It gets worse. I was concerned that Qubic might need disambiguation, and looked it up on Google. Sure enough, there is currently an app named Qubic available on Amazon, which I have just installed on my Kindle Fire. It is a puzzle game having nothing to do with tic-tac-toe. Moreover, there are other items and businesses that use the name. So I'm thinking the Qubic entry should be disambiguation. kogorman ( talk) 16:51, 9 February 2016 (UTC)
I just made a version of this game that allows for physically playing against algorithms on an electronic board.
You can see it here: https://www.kickstarter.com/projects/xno/4play
I was wondering if I could mention this addition somewhere in the article, since I do think it represents a new type of play that hasn’t been possible before. However, I understand if this just seems like advertising. Any advice/direction would be appreciated. 5^.5*.5+.5=phi ( talk) 17:45, 2 October 2019 (UTC) — Preceding unsigned comment added by 5^.5*.5+.5=phi ( talk • contribs)