Given integers  ,  we can easily and always uniquely recover integers    such that

       

It is enough to define:

       

where    is defined below, and the existence is subjected to the conditions:

•       is an integer
•    
•    

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It’s Sunday, h21:19. Starbucks. I’ll have to leave in less than 40m. The time is passing by awfully fast.

---

I am looking for the 2-colorings of triangles, which avoid unicolor tetrahedra. I’ll try the colorings    which are split into a composition of

   

and

   

where

   

and

   

Then for every tetrahedron  ,  and ,  we obtain the following -values of its walls

   

namely:

   

Assume  .  Then

•       (is divisible by 3)
•    
•    

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Let

   

Let’s define a coloring    as follows:

   
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Then it’s easy to show that there is no unicolor tetrahedron.

(This is a very weak result but still a nice exercise :-)).

The following trivial observation (OK, it’s a version of Axiom of Choice :-)) is still useful:

every family admits a pinning of cardinality not higher than that of the family itself.

It follows easily from my earlier intro how small pinnings lead to cliques. But here let me write down my first steps. Consider 2-coloring:

   

What can we say about the pinnings when there is no unicolor tetrahedron?

• When then it is necessary and sufficient that there is at least one triangle of each color; and it’s easily possible, hence there may be no unicolor tetrahedron (there is one altogether anyway :-)).
• Now let  .  Any two triangles have at least one vertex in common (because  )).&nbsp Thus if no more than 2 triangles are painted in one of the colors then they admit a 1-element pinning, hence the tetrahedron of the remainng 4 vertices is unicolored (has all its triangular faces colored by the other color). Thus the assumption about the absence of the unicolored tetrahedrons implies that there are at least 3 different triangles painted in each of the colors. And indeed, such coloring, say in  ,  free of unicolored tetrahedrons, is possible:

    blue 

    red 

  And the remaining 4 triangles (for a total of 10) can be painted arbitrarily.

  This is already a result of interest. When    has any finite number of elements then any 5-element subset houses at least 3 triangles of each color which is a significant restriction.

Now let  ,  say  .  Then  ,  contains    different triangles colored blue. If    then there is a vertex, say vertex  ,  which pins at least 2 blue triangles, thus leaving at the most one blue triangle (inside  ).  Thus there are at least 2 more blue triangles within  ,  for a total of at least 5 different blue triangles in the whole  .

Next, let  ,  contain    different triangles colored blue. Then there are 4*3 = 12 incident pairs blue_triangle-vertex, and  ,  which means that there is a vertex, say vertex  ,  which pins at least 3 blue triangles, thus leaving at the most one blue triangle (inside  ).  Thus there are at least 2 more blue triangles within  ,  for a total of at least 6 different blue triangles in the whole  .

In the remaining case of    there are at least 5 different blue triangles in    right away.

Thus there are at least 5 different blue triangles always (under the assumption of no unicolored tetrahedron).

Everything that was said about blue holds also for red. Thus if  ,  and there is no unicolor tetrahedron, then there are at least 5 different triangles of each color.

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Yesterday (Friday) a friend of mine had an operation. I was too timid to ask a few days earlier about its exact time and location. But I was worried, there was a possibility that he might need someone. Thus early Friday morning (and I am not a morning person) I called the default hospital. I was told that they don’t have any notices in advance in the case of out-patient operations. I wondered if it was true, or was it just the receptionist. After two calls I just got on the bus and went to the hospital. There definitely they had no record of his operation scheduled at any time, be it in- or out-patient. Now I was more worried because he didn’t answer my phone made just after h.17 (5pm). Thus I got on the bus, and then walked fast some distance to his house. There were cars parked on the street, and they looked all the same to me more or less. One of them could have been his. All doors were locked, and the house looked dead but there was a chance of him being unconscious inside. He doesn’t know it yet but I actually called police (after many calls–someone in the hospital also advised me to call 911). The conversation with the police department took a few minutes. Some questions were supposed to check if I make sense, I guess (What is your FIRST name? Spell it please. Was it your LAST name?). A police car came some 25 minutes after the conversation. They were very good (the ones in Sunnyvale on a completely different occasions were not at all). They checked everything possible and decided that most likely my friend is not inside, that breaking in is not justified (I got this feeling too, because his car was not there after all). And they were right. But I made a few more calls to my friend. Finally he answered, somewhat annoyed, quite justifiably. Oooph. And his surgery, according to him, went fine but for a certain complication.

He had his surgery at a different hospital, on an out-patient basis–no wonder I (or rather police, after I told them about the possibility of that other hospital) didn’t get any information from them.

All that my worrying and good will amounted to was unnecessarily bothering a poor fellow, weak after his surgery. I am not making too much sense.

---

I should and want to think intensively about applying pinnings to Ramsey Theory. And I should continue to write about it despite having only and barely some initial observations and computations. Perhaps I should still focus on the triangles-tetraedra case (the 2-color case).

---

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I’ve written to MK about my willingness to write for Delta (in Polish) on unicolor triangles and on the minusops-cyclands-Inteland theme. We are old friends from student years, in the same group, and student military service (almost an oxymoron :-)). He passed my case to his young Delta coworkers.

---

Here at Starbucks I talked to many people about their activities. One of them happened to be molecular biologist (neurons, etc.) B.Y. from UM. After the initial improvised meeting we made an appointment for today, and had here a working session. B.Y. has presented some of his research topics and open questions. It gave me also an occasion–during the second part of the meeting–to get acquainted with his small daughter. She was quite bored by our conversation though :-). I hope for an interesting development involving applications of mathematics and parallel processing (or computers in general) to biology. A lot of it was done already over many years but it is still as interesting and exciting as always.

---

There were today, and there always are, many more silly things to mention but I already have written way too much.

I hope my pinning will allow me to get more results related to Ramsey Theory. Consider coloring (just a function)

   

and  .  A c-pinning is a set such that
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Let me call    the pinfree set. Set    is free of color  .

I got an elegant way (efficient, algebraic!) to introduce the multiplication of integers. Actually more than that. The common approach works not only for Inteland (integers) but also for all cyclands. Thus the conceptual sequence is:

minusop   ⇒   cycland   Inteland

It’s 2012, and to spend time on foundations of integers may seem silly. Nevertheless it was worth it! Technically, everything is easy, virtually trivial (especially after it’s done). But my Inteland (no more “InteGland”, no more “G”) has conceptual value, it allows regular mathematicians to have a good feeling about it, it has value for education. Inteland means a mild but nice breakthrough since the times of Peano and Hilbert. The topic in the past was so tedious that only logicians cared about it. No textbook on number theory would bother to introduce integers (or natural numbers) thoroughly. They would say this and that about induction and perhaps a few words about commutativity and associativity (?), and otherwise they would rely on the reader’s good intuition of natural numbers.
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Enough of this shameless self-praising. It’s time to write all of it again from scratch. Should I go to Starbucks at Arborland. I am at the Glenco Crossing shopping center, as usually. The one at Arborland is open 24h/day all year except for the X’mass.

The coming Tuesday (in 2 days) will be messed up for me.

OK, plans, or rather vague possibilities:

• Mathematics (not related directly to computers):
• Topology–go back, and seriously, to my universal functions. I had better proofs than those which I had published soon afterwards. It bothers me. But I’d like also to obtain new results. Of course!
• Finish Integland; and start an exposition of number theory (what for? also, which direction to choose–I have a few)?
• Learn or obtain number-theoretic results which lead to a more efficient (computer) search of baroque numbers.
• publish one way or another my old result about combinatorial circles (ECC of equal weight)
• Edit anew, improve and republish my several old results.
• Just write-write-write-… on many of the attractive themes which I have touched upon in the past (OK, that simplified my task of listing my mathematical plans :-)).

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Computer related plans:

• learn more than just a bit about HMTL+CSS+JavaScript.
• Get an ability of working with the bit images. I have more than one possible image processing project in mind.
• Go back to programs (on mathematical games, number-theoretical, combinatorial) which I have written in the past.
• Etc.

Etc. (it’s hopeless, it’s impossible to list so much, and it does not make too much sense in the first place, or does it?).

Hm, it was a disappointing, unsatisfactory exercise, so incomplete.

There is a regular Starbucks patron who talks loud and in a voice which cannot be ignored. When he does so I can’t concentrate on anything. Hm, just as I wrote these words he started to talk in a softer voice, less loud, for the first time. We at our online pharmacy store here cialis 5 mg pharmacy provide the FDA approved this drug after analyzing its clinical efficacy and safety. Kamagra comes in many versions like Generic kamagra, canada pharmacy tadalafil , Edegra, Silagra, Penegra etc. A detailed surgical valve analysis is performed to determine the functional type and the super cialis cheap segmental localization. What Changes while Lovemaking Session? Such complications india viagra for sale of aging can ultimately effect on how one performs in bed. But it’ll take me a while before I can get going, especially that from time to time he raises his voice again (not too bad–what happened/ :-)). (So far there is only one such regular around here).

Let  V  be a  v-element set. Let’s call every set of two different triangles which share an edge–a  folder;  and every set of three different triangles which together have only four vertices–a  tent.

Let

c : (V ## 3) → {0 1}

be an arbitrary 2-coloring of triangles. Let

• Δ_e  :=  |c^-1(e)|  be the number of triangles colored  e = 0 or 1;
• integer  B  be the number of tetrahedra which have two faces painted in one, and the other two faces in the other color;
• C  :=  (v ## 4) – B;
• S  be the number of unicolor tents;
• Q  be the number of the remaining tents, i.e.;

  Q  :=  4*(v ## 4) – S

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Let’s assume that (for the given coloring) there are no unicolored tetrahedra, i.e. that

B + C   =   (v ## 4)

Then

• S  =  C
• Q  =  4*B + 3*C

---

Each folder is contained in exactly 2 different tents, and each tent contains exactly 3 different folders. Thus, in particular, the number of folders is

(v ## 2) * ((v-2) ## 2)   =   6 * (v ## 4)

---

(I’m not doing anything, I’m just typing; I am trying to extend my edge coloring approach–in the past I have not obtained any sharp result for the triangles-tetrahedra case, and it’s tantalizing). Each unicolor tent contains 3 unicolor folders (and no other), and each bicolor tent contains 1 unicolor folder, and 2 bicolor folders. Let  F_e  be the number of e-colored folders,  e=1 or 2. Then

• F₁  =  (1/2) * (3*S + Q)
• F₂  =  Q

(My past analysis indicates that if I get enough of the 2-2 balanced colored tetrahedra (on the assumption that there are no unicolored tetrahedra) then I will get the desired contradiction, i.e. the existence of at least one unicolor tetrahedron, hopefully for a low value of  v.

One kind of inequalities comes from the distribution of triangle colors

e₀ + e₁  =  ((v-2) ## 2)

which share edge  e. The more even split the fewer unicolored folders, and more bicolored folders (thus we get a lower bound on the number of the unicolored folders, and an upper bound on the bicolored folders). More is needed.