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Log inNot sure, but willing to bet $5...
Not sure, but willing to bet $5...
Frank.Schwartz on Sat Nov 22, 2008 10:54 am
Hi all,
I’ve been working on a four color proof (yeah, I know one was already found, but I want one that won’t outweigh a Manhattan phone book….)
I’ve checked it over a few times and I think it’s valid, but I’m not sure that I haven’t missed something, so I’ll give five bucks and a hearty congratulations to the first poster who points out a logical mistake in my attempt. (Typos and grammatical errors don’t count, although I think I’ve gotten rid of them.)
The illustrations are on my "Tagged" web site, and I’ve provided links to them where they are referenced, but for some reason they seem not to work on my computer (they all take me to Figure 7). If you have the same problem, try copying the link addresses into the URL field.(I think I've fixed this problem, but I'm leaving the link addresses visible in case the editor changes it back to the wrong link again...)
Here it is:
Let us assume there is some map M containing N territories such that five colors are necessary to draw the map, and N is the minimum number of territories that such a map can possibly contain.
(Trivial notes:
For the sake of this proof, "drawing" or "adding" a territory are equal in meaning. Imagine that a territory is drawn by exclusively tracing its perimeter. Thus, even if all of the outer territories of M are drawn first, it will still take the drawing of N territories to create M; in such a case, the final territory will have to be "drawn" even after it is fully visible on the map. This stipulation removes an ambiguity that might otherwise arise.
Also for the sake of this proof, a map’s identity is defined solely by the relationship of bordering vs. non-bordering between the territories that make it up. See Figure 1 for an example of what this means.
http://www.tagged.com/photo_view.html?photoId=86043360&uid=5379893398#state=7_0
Also, a reference to the color of an outer border in this proof is obviously not referring to the border line color itself, but the color of the territory that it borders.)
First let’s examine what we know about M (There is an unstated assumption being made in this list regarding some of the territories drawn, but it will be shown to be a valid one at the end of the proof.)
http://www.tagged.com/photo_view.html?photoId=86043360&uid=5379893398#state=6_0
Map M can be drawn as a contiguous map that does not have any territories that touch each other at a single point without bordering each other. (If M contains two such territories, extend the point to become a border between the two territories, and the resulting map will be no less restrictive; M will still require at least the same number of colors. See Figure 3.) Let us refer to such a map hereinafter as a contiguous restrictive map (because a border is restrictive while a point is not).
http://www.tagged.com/photo_view.html?photoId=86043360&uid=5379893398#state=5_0
M may be drawn as a contiguous restrictive map in stages, adding a single territory with each stage, such that stagei is a map containing i territories. Let us refer to this process hereinafter as drawing the map incrementally, and refer to the the last territory added at stagei as territoryi.
If M is drawn incrementally, then immediately before the final territory is drawn, only four colors are necessary, so the final territory must border at least four other territories (otherwise it could be assigned one of the already used colors that is not assigned to any of the three or less territories that it borders). Since the territories of M could be drawn in any order, we know that each territory in it could be drawn as the final territory, which means that each territory in it must border at least 4 other territories.
M can be drawn incrementally in such a manner that it is contiguous at each stage. Let us refer hereinafter to the process of drawing the map in this manner as expanding the map (because without resizing, each stage will cover a larger contiguous area than the previous stage).
If M is expanded, then at every stage prior to stageN-1, at most three colors are forced to the outer borders of the map when the map is restricted to four colors.
If M is expanded, then at each stage, any territory with an outside border may be considered to be the most recently drawn territory.
If M is expanded, then no two territories in stageN-1 may share more than one border, because any two such territories would surround a set of territories that could have no effect on the remainder of the map.
If M is expanded, then no two territories with outer borders in stageN-1 may share any border that does not reach the outer border of the map, because any two such territories would separate two submaps thereby preventing them from influencing each other. (See Figure 4.)
http://www.tagged.com/photo_view.html?photoId=86043360&uid=5379893398#state=4_0
If M is expanded, then the map at any stage previous to stageN may not contain any entirely surrounded submap bordering less than four territories, as such a submap could have no effect on the remainder of the map at that stage. (See Figure 5.) Because of this condition, if M is expanded, then at every stage prior to stageN, every territory in the map can be drawn as a rectangle. (See Figure 6 for an example of how to do this. Drawing only rectangular territories until stageN is not necessary for the proof, but it can make it easier to visualize what’s going on.)
http://www.tagged.com/photo_view.html?photoId=86043360&uid=5379893398#state=3_0
http://www.tagged.com/photo_view.html?photoId=86043360&uid=5379893398#state=2_0
If M is expanded, then at stageN-1, if the map is restricted to four colors, all four colors must be forced the outside borders of the map.
If M is expanded, then at stageN-2, the map’s outer borders can be divided into two contiguous sections: sectionN1, which will border territoryN-1, and sectionN1′ which will not. Note these sections must share exactly two territories, which means that at least one territory will exist in each that does not exist in the other. Also note that since sectionN1′ will be a part of the outer border of stageN-1, the territories that make it up must be a simple strip such that if they are the first territories drawn, then when the last of them is drawn, the map at that stage can be still drawn with two colors. In other words, in that strip, no two territories may border each other on any border that does not meet sectionN1′. (See Figure 7.)
http://www.tagged.com/photo_view.html?photoId=86043360&uid=5379893398#state=1_0
If M is expanded, then at stageN-3, the map’s outer borders can be divided into sectionN2, which will border territoryN-2, and sectionN2′ (a part of which is the unaltered sectionN1′), which will not.
If M is expanded, then at stageN-4, the map’s outer borders can be divided into sectionN3, which will border territoryN-3, and sectionN3′ (a part of which is the unaltered sectionN1′), which will not.
If M is expanded, then at stageN-5, the map’s outer borders can be divided into sectionN4, which will border territoryN-4, and sectionN4′ (a part of which is the unaltered sectionN1′), which will not.
If the map at stageN-2 is restricted to four colors, sectionN1 and sectionN1′ must each be forced to the same three colors. That is, neither section may have less than three colors, and whichever three colors are chosen for sectionN1′ must also be used for sectionN1. If this were not the case, territoryN-1 could simply be assigned the third color used in sectionN1′ and this would make a fifth color unnecessary at stageN.
If the map at stageN-3 is restricted to four colors, sectionN2′ must still be forced to three colors, and sectionN2 must be forced to contain at least two of those colors. If this were not the case, territoryN-2 could simply be assigned a color already used in sectionN1′ and the immediately preceding condition would remain unfulfilled.
If the map at stageN-4 is restricted to four colors, sectionN3′ must still be forced to three colors, and sectionN3 must be forced to contain at least one of those colors. If this were not the case, territoryN-3 could simply be assigned a color already used in sectionN1′ and the immediately preceding condition would remain unfulfilled.
If the map at stageN-5 is restricted to four colors, sectionN4′ must still be forced to three colors, and sectionN4 must be forced to contain the fourth color. If this were not the case, territoryN-4 could simply be assigned the fourth color and the immediately preceding condition would remain unfulfilled. This would precipitate the condition required for stageN-1, Since there is no N = N-5, stageN-5 is an impossible situation.
Therefore, M can’t possibly exist unless the unstated assumption mentioned above is unprovable.
The assumption is as follows: The last four territories drawn before territoryN-1 have no borders in sectionN1′. In order to be sure that the final given conditions apply, this (now stated) assumption must be true.
It is fairly easily seen from Figure 7 that in the case of territoryN-2, there will be at least one territory to choose that fits the condition, due to the condition stated in (9) above.
However, at stageN-3, it is necessary to consider the possibility that stageN-4 will not contain all of sectionN1′. Therefore we must analyze what the situation is at stageN-3. First, sectionN1′ must obviously contain at least three territories, as it is still forced to contain at least three colors. Second, stageN-3 cannot be a simple strip of territories, for such a strip would be able to drawn with only two colors. Third, sectionN1′ cannot include two territories that border each other on a border that does not touch the outside border of sectionN1′, because that would violate the condition stated in (9) above- i.e. sectionN1′ must be the edge of a simple strip of territories. As a result of this, stageN-3 must contain at least one territory that has no border in sectionN1′, and that territory can be chosen as territoryN-4, allowing us to continue the logic as above to the next stage back.
At this point, we are helped by a sort of Catch-22. As long as sectionN1′ is still intact, the above logic can be applied at each stage, which keeps sectionN1′ intact for one more stage.
At stageN-4, the same problem crops up, but again it is solved by the same reasoning, which leads us back another stage, where we can apply the same reasoning yet once more to finally solve the problem.
I’ve been working on a four color proof (yeah, I know one was already found, but I want one that won’t outweigh a Manhattan phone book….)
I’ve checked it over a few times and I think it’s valid, but I’m not sure that I haven’t missed something, so I’ll give five bucks and a hearty congratulations to the first poster who points out a logical mistake in my attempt. (Typos and grammatical errors don’t count, although I think I’ve gotten rid of them.)
The illustrations are on my "Tagged" web site, and I’ve provided links to them where they are referenced, but for some reason they seem not to work on my computer (they all take me to Figure 7). If you have the same problem, try copying the link addresses into the URL field.(I think I've fixed this problem, but I'm leaving the link addresses visible in case the editor changes it back to the wrong link again...)
Here it is:
Let us assume there is some map M containing N territories such that five colors are necessary to draw the map, and N is the minimum number of territories that such a map can possibly contain.
(Trivial notes:
For the sake of this proof, "drawing" or "adding" a territory are equal in meaning. Imagine that a territory is drawn by exclusively tracing its perimeter. Thus, even if all of the outer territories of M are drawn first, it will still take the drawing of N territories to create M; in such a case, the final territory will have to be "drawn" even after it is fully visible on the map. This stipulation removes an ambiguity that might otherwise arise.
Also for the sake of this proof, a map’s identity is defined solely by the relationship of bordering vs. non-bordering between the territories that make it up. See Figure 1 for an example of what this means.
http://www.tagged.com/photo_view.html?photoId=86043360&uid=5379893398#state=7_0
Also, a reference to the color of an outer border in this proof is obviously not referring to the border line color itself, but the color of the territory that it borders.)
First let’s examine what we know about M (There is an unstated assumption being made in this list regarding some of the territories drawn, but it will be shown to be a valid one at the end of the proof.)
- M can be drawn as a contiguous map; i.e. as a map that can be drawn to fill an entire rectangle (such as a standard 8.5 X 11 page) in such a manner that no part of the rectangle does not contain at least part of one or more territories. (If map M contains "holes", we could eliminate them by extending the territories around them until they are gone. If M has blank areas on the edge of the rectangle, these areas could be filled by expanding territories. See Figure 2.)
http://www.tagged.com/photo_view.html?photoId=86043360&uid=5379893398#state=6_0
http://www.tagged.com/photo_view.html?photoId=86043360&uid=5379893398#state=5_0
http://www.tagged.com/photo_view.html?photoId=86043360&uid=5379893398#state=4_0
http://www.tagged.com/photo_view.html?photoId=86043360&uid=5379893398#state=3_0
http://www.tagged.com/photo_view.html?photoId=86043360&uid=5379893398#state=2_0
http://www.tagged.com/photo_view.html?photoId=86043360&uid=5379893398#state=1_0
Therefore, M can’t possibly exist unless the unstated assumption mentioned above is unprovable.
The assumption is as follows: The last four territories drawn before territoryN-1 have no borders in sectionN1′. In order to be sure that the final given conditions apply, this (now stated) assumption must be true.
It is fairly easily seen from Figure 7 that in the case of territoryN-2, there will be at least one territory to choose that fits the condition, due to the condition stated in (9) above.
However, at stageN-3, it is necessary to consider the possibility that stageN-4 will not contain all of sectionN1′. Therefore we must analyze what the situation is at stageN-3. First, sectionN1′ must obviously contain at least three territories, as it is still forced to contain at least three colors. Second, stageN-3 cannot be a simple strip of territories, for such a strip would be able to drawn with only two colors. Third, sectionN1′ cannot include two territories that border each other on a border that does not touch the outside border of sectionN1′, because that would violate the condition stated in (9) above- i.e. sectionN1′ must be the edge of a simple strip of territories. As a result of this, stageN-3 must contain at least one territory that has no border in sectionN1′, and that territory can be chosen as territoryN-4, allowing us to continue the logic as above to the next stage back.
At this point, we are helped by a sort of Catch-22. As long as sectionN1′ is still intact, the above logic can be applied at each stage, which keeps sectionN1′ intact for one more stage.
At stageN-4, the same problem crops up, but again it is solved by the same reasoning, which leads us back another stage, where we can apply the same reasoning yet once more to finally solve the problem.
Last edited by Frank.Schwartz on Sat Nov 22, 2008 10:11 pm; edited 1 time in total
Frank.Schwartz-
Number of posts: 7
Age: 49
Location: Brooklyn
Registration date: 2008-11-22 
Re: Not sure, but willing to bet $5...
Frank.Schwartz on Sat Nov 22, 2008 1:40 pm
Both this note and the proof itself have been cross-posted from the Mensa Science forum:
http://www.community.us.mensa.org/forums/t/10282.aspx
First: Thanks to Candace Harvey for telling me about this forum.
Just a short note:
By the time I posted thiis here, I had already edited the proof above from when I first posted it in the above forum, to correct a typo. Item (1) on the list previously said:
"M can be drawn as a contiguous map; i.e. as a map that can be drawn to fill an entire rectangle (such as a standard 8.5 X 11 page) in such a manner that no part of the rectangle does contain a territory."
Obviously this is not what I meant to say, as can be seen by several of the illustrations. However, the sentence was grammatically correct, and resulted in logical incoherency, so a few questions regarding this fact occurred to me, and I think I should add a couple of details to my $5 offer.
(If there is a winner, I will give the prize, re-examine the proof in light of the discovered error, and decide whether to try to correct it and offer another prize, or to admit defeat -which I don't do easily... so it might be a good idea if I set some ground rules...)
First, that I am not allowed to collect the prize myself, though if I find a logical error before anybody else does and post the correction to at least one forum that it has been posted to, that particular error shall no longer entitle anyone to the prize.
Second, a clarification of what a logical error in the proof is. In the event of a typo or grammatical error that results in a logical error, the prize is still claimable if the error is consistent with the rest of the reasoning; i.e. if I ultimately based a conclusion of the proof on the error, because that would imply that after I made the typo or grammatical error, I accepted its premise, a logical error the discovery of which would entitle the finder to the prize. By the same reasoning, however, even a genuine logical error would not entitle the finder to the prize if it is not used in the proof.
(For example, if my claim that all stages before the last two are representable with a contiguous map made purely of rectangular territories was incorrect, this would not entitle the finder to the prize unless it was presented with a proof that the claim is in fact essential to my proof, which I don't believe it is. The proof is purely in the text; the illustrations are just provided as an aid in understanding the text.)
Third, considering the fact that this is posted to two forums already I should clarify that since there can only be one prize per offer, the winner will be the first to post the error from among all forums and/or web sites that the proof is posted to. In the event of a tie (very unlikely unless someone is having a bit of fun by simultaneous posting under different names), the prize will be split.
Fourth, I will not be held legally liable for any of these offers, for obvious reasons. (If you think you deserve the prize, and I disagree, you don't get the prize. If you're right in such a disagreement, I lose a lot of valuable input- not only from you but from all those who would agree that I've compromised my integrity, and not only on this issue but possibly in future issues as well. The monetary prize is only five dollars, definitely not worth going to court over, and the other part of the prize certainly would not be worth much either if you had to sue me for it.)
Finally, each offer may be withdrawn at any time and for any reason so long as I withdraw it before it is rightfully claimed by someone who has already found the error, posted what it is, and was the first to do so.
http://www.community.us.mensa.org/forums/t/10282.aspx
First: Thanks to Candace Harvey for telling me about this forum.
Just a short note:
By the time I posted thiis here, I had already edited the proof above from when I first posted it in the above forum, to correct a typo. Item (1) on the list previously said:
"M can be drawn as a contiguous map; i.e. as a map that can be drawn to fill an entire rectangle (such as a standard 8.5 X 11 page) in such a manner that no part of the rectangle does contain a territory."
Obviously this is not what I meant to say, as can be seen by several of the illustrations. However, the sentence was grammatically correct, and resulted in logical incoherency, so a few questions regarding this fact occurred to me, and I think I should add a couple of details to my $5 offer.
(If there is a winner, I will give the prize, re-examine the proof in light of the discovered error, and decide whether to try to correct it and offer another prize, or to admit defeat -which I don't do easily... so it might be a good idea if I set some ground rules...)
First, that I am not allowed to collect the prize myself, though if I find a logical error before anybody else does and post the correction to at least one forum that it has been posted to, that particular error shall no longer entitle anyone to the prize.
Second, a clarification of what a logical error in the proof is. In the event of a typo or grammatical error that results in a logical error, the prize is still claimable if the error is consistent with the rest of the reasoning; i.e. if I ultimately based a conclusion of the proof on the error, because that would imply that after I made the typo or grammatical error, I accepted its premise, a logical error the discovery of which would entitle the finder to the prize. By the same reasoning, however, even a genuine logical error would not entitle the finder to the prize if it is not used in the proof.
(For example, if my claim that all stages before the last two are representable with a contiguous map made purely of rectangular territories was incorrect, this would not entitle the finder to the prize unless it was presented with a proof that the claim is in fact essential to my proof, which I don't believe it is. The proof is purely in the text; the illustrations are just provided as an aid in understanding the text.)
Third, considering the fact that this is posted to two forums already I should clarify that since there can only be one prize per offer, the winner will be the first to post the error from among all forums and/or web sites that the proof is posted to. In the event of a tie (very unlikely unless someone is having a bit of fun by simultaneous posting under different names), the prize will be split.
Fourth, I will not be held legally liable for any of these offers, for obvious reasons. (If you think you deserve the prize, and I disagree, you don't get the prize. If you're right in such a disagreement, I lose a lot of valuable input- not only from you but from all those who would agree that I've compromised my integrity, and not only on this issue but possibly in future issues as well. The monetary prize is only five dollars, definitely not worth going to court over, and the other part of the prize certainly would not be worth much either if you had to sue me for it.)
Finally, each offer may be withdrawn at any time and for any reason so long as I withdraw it before it is rightfully claimed by someone who has already found the error, posted what it is, and was the first to do so.
Frank.Schwartz- Number of posts: 7
Age: 49
Location: Brooklyn
Registration date: 2008-11-22 -
Re: Not sure, but willing to bet $5...
Frank.Schwartz on Sat Nov 22, 2008 10:14 pm
I just edited the post to fix the links; they seem to work now on my computer.
Last edited by Frank.Schwartz on Sun Nov 23, 2008 11:26 pm; edited 2 times in total (Reason for editing : Changed subject line per Candace's suggestion...)
Frank.Schwartz- Number of posts: 7
Age: 49
Location: Brooklyn
Registration date: 2008-11-22 -
Re: Not sure, but willing to bet $5...
Frank.Schwartz on Mon Nov 24, 2008 12:42 pm
I found an error on one of the seven diagrams...it doesn't affect the proof, so I left it alone to see if anyone else might notice it, because I'm figuring that most of the people looking at this post like a good puzzle..
It is independent of anything else in the proof (including the other 6 diagrams), so you only need to look at that one diagram to spot it, and a failure to notice it will not affect your ability to solve the larger puzzle of whether or not the proof is valid, nor your ability to find an error in the proof itself if one exists. (So showing the corrected diagram would not make the $5 prize any easier to obtain, other than in that the current diagram might provide a very short-lived dead end route to the person who finds the error in it.)
I've already corrected the diagram on my computer, and after someone points out the diagram error in a post on one of the forums this is posted in, I'll post the corrected version- but if if nobody spots it before this Wednesday evening, I probably won't be able to post it until sometime in December.
If you want to take credit for finding the error in the diagram, but want to leave the puzzle up for others, send me a private message.
Happy hunting, folks!
It is independent of anything else in the proof (including the other 6 diagrams), so you only need to look at that one diagram to spot it, and a failure to notice it will not affect your ability to solve the larger puzzle of whether or not the proof is valid, nor your ability to find an error in the proof itself if one exists. (So showing the corrected diagram would not make the $5 prize any easier to obtain, other than in that the current diagram might provide a very short-lived dead end route to the person who finds the error in it.)
I've already corrected the diagram on my computer, and after someone points out the diagram error in a post on one of the forums this is posted in, I'll post the corrected version- but if if nobody spots it before this Wednesday evening, I probably won't be able to post it until sometime in December.
If you want to take credit for finding the error in the diagram, but want to leave the puzzle up for others, send me a private message.
Happy hunting, folks!
_________________
________________
|==E___|>o______
|)>N___|>o______
|>>+__|>o_______
|===B_|>o_______
|===K+_______|>o
|>>+__|>o_______
|)>N___|>o_______
|==E___|>o_______
Your move...
Frank.Schwartz- Number of posts: 7
Age: 49
Location: Brooklyn
Registration date: 2008-11-22 -
Re: Not sure, but willing to bet $5...
Frank.Schwartz on Tue Nov 25, 2008 11:50 pm
I found a another typo:
In (19) The last sentence should obviously read:
"Since there is no N-1 = N-5, stageN-5 is an impossible situation."
I left out "-1".
I hope it's the last typo, but I suspect there might be a couple more....
It is obviously a trivial error, but I looked over my previous stipulation regarding typos, and it seems that had I not been the first to post this, I'd be out $5 on a technicality...
So (*sigh*)... I'm going to add to the rules regarding typos, grammatical errors and such to include that trivial errors don't win the prize, with the definition of a trivial error as follows:
A trivial error is either one that differs from its correction in a manner such that the conclusion based on it's premise remains valid after correction due to a common underlying assertion that can be derived by what came before, or a statement that is inconsistent with it's stated or implied premise and inconsistent with the conclusion that is immediately derived from it, such that when the statement is corrected it becomes consistent with both of these.
In short, a trivial error before correction has consistency with both sides of itself or with neither, and when corrected with a minor adjustment, respectively remains or becomes consistent on both sides.
In the case above, the error is trivial because the underlying assertion that two unequal numbers would have to be equal for that stage to exist is implied by what preceded the statement, and is what makes the corrected statement a valid premise for the conclusion.
Also, in the event of an inconsistent statement, what follows the misstatement but is consistent with the corrected version is not considered an error, even though it is inconsistent with it's apparent premise before the correction, although I would consider spotting that inconsistency equivalent to spotting the typo or trivial error itself.
(For example, consider a line of reasoning went: "We know that 2r = k, therefore 1/2 of k and r are not equal to each other. Because of this, 1/4 of k is equal to 1/2 of r." The last statement in this argument appears to be an erroneous conclusion, but makes perfect sense as soon as the word "not" is removed from the preceding clause. Thus, the error is in the clause "1/2 of k and r are not equal to each other", and is trivial because simply removing the word "not" brings consistence.)
The prize is only offered for the discovery of a non-trivial logical error essential to the proof.
A non-trivial logical error essential to the proof is one that makes an assumption (or draws a conclusion) that is unsupported by both obvious evidence and logic, and is used as a premise or conclusion. An example of something supported by obvious evidence is the fact that four colors cannot be forced to a four territory map without surrounding one of the territories.
Given that I'm narrowing my requirements a second time, I guess it's fair to give an example of a logical error that would be non-trivial and essential to the proof, especially because I might have to narrow my requirements even further in the future. Regardless, no requirements are applied until after I've posted them.
An example of a non-trivial logical error essential to the proof would be if I'd based my proof on an erroneous underlying premise, such as that a single center territory could replace any sub-map that contains no outer borders at any stage without affecting the available choices for the outer border's color configuration. This might be a very useful premise, but when found to be unjustified (which it is; in fact there is a simple counter-example), the whole proof would likely fail. I could not correct it simply by adding the word "not" or rephrasing a sentence, and leaving the rest of the proof intact, because the conclusion based on the premise would become inconsistent with the corrected version.
I do want to get rid of typos and grammatical errors, so If you find a typo, grammatical error, or a logical error that might be one, post it immediately, and it will be exempt from future restrictions, even if you don't prove in the same post that it fulfilled the restrictions thus far. Even if it doesn't, you might alert me to another loophole, which I will promptly close with another restriction.
If your posting of a typo or trivial error inspires another restriction, you get the prize, and another will be immediately offered.
If you later show that the typo you found is one that constittutes a non-trivial error essential to the proof (since it's conceivable that such a thing might not be immediately evident), you get the prize, and I'll have to look over the proof and see if it can be successfully repaired.
If someone else shows that the typo you found constitutes a non-trivial error essential to the proof, you share the prize with that person 60/40 (you get 60% for being the first poster).
(If you find a non-trivial error essential to the proof, you can also send me a private message if you want to claim the prize and leave it as a puzzle for others until I'm ready to offer another prize. In such a case, typos & such in the text will always be corrected, but the cause of non-triviality will be the entertaining mystery.)
BTW, I have heard some evidence that lightly favors this proof's validity. Because this proof could conceivably be applied to a map of any surface passing through any number of dimensions, this proof predicts that for any map of any given surface of any given number of dimensions, the maximum number of colors needed is also the minimum number of territories needed to force that many colors. As it turns out, this is in fact already proven to be the case (or so I've heard).
Also, I found some more errors in the same diagram- it was diagram 6...as mentioned in the proof, diagram 6 is not important to the proof. However, because the additional errors interfered with the clarity of the diagram's purpose, I have posted a corrected version:
http://www.servimg.com/image_preview.php?i=8&u=13290863
Unfortunately, I can't edit the original post anymore, so it will still point to the old diagram...
In (19) The last sentence should obviously read:
"Since there is no N-1 = N-5, stageN-5 is an impossible situation."
I left out "-1".
I hope it's the last typo, but I suspect there might be a couple more....
It is obviously a trivial error, but I looked over my previous stipulation regarding typos, and it seems that had I not been the first to post this, I'd be out $5 on a technicality...
So (*sigh*)... I'm going to add to the rules regarding typos, grammatical errors and such to include that trivial errors don't win the prize, with the definition of a trivial error as follows:
A trivial error is either one that differs from its correction in a manner such that the conclusion based on it's premise remains valid after correction due to a common underlying assertion that can be derived by what came before, or a statement that is inconsistent with it's stated or implied premise and inconsistent with the conclusion that is immediately derived from it, such that when the statement is corrected it becomes consistent with both of these.
In short, a trivial error before correction has consistency with both sides of itself or with neither, and when corrected with a minor adjustment, respectively remains or becomes consistent on both sides.
In the case above, the error is trivial because the underlying assertion that two unequal numbers would have to be equal for that stage to exist is implied by what preceded the statement, and is what makes the corrected statement a valid premise for the conclusion.
Also, in the event of an inconsistent statement, what follows the misstatement but is consistent with the corrected version is not considered an error, even though it is inconsistent with it's apparent premise before the correction, although I would consider spotting that inconsistency equivalent to spotting the typo or trivial error itself.
(For example, consider a line of reasoning went: "We know that 2r = k, therefore 1/2 of k and r are not equal to each other. Because of this, 1/4 of k is equal to 1/2 of r." The last statement in this argument appears to be an erroneous conclusion, but makes perfect sense as soon as the word "not" is removed from the preceding clause. Thus, the error is in the clause "1/2 of k and r are not equal to each other", and is trivial because simply removing the word "not" brings consistence.)
The prize is only offered for the discovery of a non-trivial logical error essential to the proof.
A non-trivial logical error essential to the proof is one that makes an assumption (or draws a conclusion) that is unsupported by both obvious evidence and logic, and is used as a premise or conclusion. An example of something supported by obvious evidence is the fact that four colors cannot be forced to a four territory map without surrounding one of the territories.
Given that I'm narrowing my requirements a second time, I guess it's fair to give an example of a logical error that would be non-trivial and essential to the proof, especially because I might have to narrow my requirements even further in the future. Regardless, no requirements are applied until after I've posted them.
An example of a non-trivial logical error essential to the proof would be if I'd based my proof on an erroneous underlying premise, such as that a single center territory could replace any sub-map that contains no outer borders at any stage without affecting the available choices for the outer border's color configuration. This might be a very useful premise, but when found to be unjustified (which it is; in fact there is a simple counter-example), the whole proof would likely fail. I could not correct it simply by adding the word "not" or rephrasing a sentence, and leaving the rest of the proof intact, because the conclusion based on the premise would become inconsistent with the corrected version.
I do want to get rid of typos and grammatical errors, so If you find a typo, grammatical error, or a logical error that might be one, post it immediately, and it will be exempt from future restrictions, even if you don't prove in the same post that it fulfilled the restrictions thus far. Even if it doesn't, you might alert me to another loophole, which I will promptly close with another restriction.
If your posting of a typo or trivial error inspires another restriction, you get the prize, and another will be immediately offered.
If you later show that the typo you found is one that constittutes a non-trivial error essential to the proof (since it's conceivable that such a thing might not be immediately evident), you get the prize, and I'll have to look over the proof and see if it can be successfully repaired.
If someone else shows that the typo you found constitutes a non-trivial error essential to the proof, you share the prize with that person 60/40 (you get 60% for being the first poster).
(If you find a non-trivial error essential to the proof, you can also send me a private message if you want to claim the prize and leave it as a puzzle for others until I'm ready to offer another prize. In such a case, typos & such in the text will always be corrected, but the cause of non-triviality will be the entertaining mystery.)
BTW, I have heard some evidence that lightly favors this proof's validity. Because this proof could conceivably be applied to a map of any surface passing through any number of dimensions, this proof predicts that for any map of any given surface of any given number of dimensions, the maximum number of colors needed is also the minimum number of territories needed to force that many colors. As it turns out, this is in fact already proven to be the case (or so I've heard).
Also, I found some more errors in the same diagram- it was diagram 6...as mentioned in the proof, diagram 6 is not important to the proof. However, because the additional errors interfered with the clarity of the diagram's purpose, I have posted a corrected version:
http://www.servimg.com/image_preview.php?i=8&u=13290863
Unfortunately, I can't edit the original post anymore, so it will still point to the old diagram...
Last edited by Frank.Schwartz on Wed Nov 26, 2008 1:41 pm; edited 2 times in total (Reason for editing : (spelling correction: conceivably), added new figure 6...)
Frank.Schwartz- Number of posts: 7
Age: 49
Location: Brooklyn
Registration date: 2008-11-22 -
Clarification for figure 5...
Frank.Schwartz on Tue Dec 16, 2008 12:52 am
This is not so much a correction as a clarification; the old Figure 5 was valid in that it supported the text, but this one is a little more specific, removing an unnecessary and contradictory extraneous case that existed in the original diagram....
http://www.servimg.com/image_preview.php?i=9&u=13290863
(The original had the second part of the diagram as a concievable option of the first.)
Anyway, on Thanksgiving, I showed the proof to my sister-in-law (who, although not a Mensan, is probably qualified; very good in math and logic...), and she didn't find any errors, although to be fair, I didn't have a copy of the proof with me at the time; I think I showed it to her on the back of a paper plate, drawing the diagrams and explaining them as I went along...)
So- anyone see any flaws in the proof yet?
http://www.servimg.com/image_preview.php?i=9&u=13290863
(The original had the second part of the diagram as a concievable option of the first.)
Anyway, on Thanksgiving, I showed the proof to my sister-in-law (who, although not a Mensan, is probably qualified; very good in math and logic...), and she didn't find any errors, although to be fair, I didn't have a copy of the proof with me at the time; I think I showed it to her on the back of a paper plate, drawing the diagrams and explaining them as I went along...)
So- anyone see any flaws in the proof yet?
Frank.Schwartz- Number of posts: 7
Age: 49
Location: Brooklyn
Registration date: 2008-11-22 -
Tao's "Editorial policy on submissions concerning famous problems"
ratboy on Tue Dec 30, 2008 7:29 pm
You might find some of these guidelines useful if you seriously want someone to invest time and energy in verifying your proof: Editorial policy on submissions concerning famous problems.
ratboy- Number of posts: 8
Registration date: 2008-04-06 -
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