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 Post subject:
PostPosted: Sep.10.03 7:42 pm 
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Location: Henti aimgiin Galshar sumiin uyach Bazarwaaniin gert
goyo bodolt bainaa.

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 Post subject:
PostPosted: Sep.10.03 8:45 pm 
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Эзэрхийлэн Эзлэгч Гишvvн
Эзэрхийлэн Эзлэгч Гишvvн
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Posts: 985
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GAVANA,er ni gavana ch geltgui humuus goy bodlogo taviach
joohon zavtai baigaa deeree jool oroldii.olon hunees yanz buriin sonin bodolt garch magad shuudee

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 Post subject:
PostPosted: Sep.15.03 11:36 pm 
1. Graph-n songomol theorm-s shuud garah baih.
Oroi - Irmeg + Tal = 2 - 2*nuh biluu? Tegeheer n-Irmeg+1=2 gedgees Irmeg = n - 1

2. 1, 2, ... n -n yamar neg selgemliig a1, a2, ...aN geye. Graf-n oroinuudiig 1,2...n gej dugaarlaad, holboosiig ni a1->a2->a3 geh met holbogddog gevel buh modnii too, yalgaatai selgemliin tootei tentsuu bolno. Ooroo helbel n!

Anduursan bol zasch ogoorei.


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 Post subject:
PostPosted: Sep.16.03 2:12 am 
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Эзэрхийлэн Эзлэгч Гишvvн
Эзэрхийлэн Эзлэгч Гишvvн
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Joined: Apr.16.03 2:49 pm
Posts: 985
Location: DAMBADARJAA
Detective wrote:
Та нар графын онолын талаар дажгуй мэддэг юм шиг байна! Дахиад нэг хялбар бодлого тавья, залхуутай санагдсан бол хэлээрэй! Би нэг тодорхойлолтыг графын талаар мэдэхгуй хумууст зориулж бичлээ шуу, учир нь доорх бодлогыг бодоход заавал графын талаар уншсан байх шаардлагагуй! Мэддэг нэг нь tree(мод)-ийг хэлээд байна гэж ойлгоорой!

Тодорхойлолт: Мод
Хэрвээ хавтгай дээрх дурын байрлалтай цэгуудийг дараах маягаар хооронд нь холбовол мод ууснэ. Уунд: 1) Дурын 2 цэгийг холбосон холбоос эсвэл цорын ганц байна, эсвэл тэр 2 цэг ямарч холбоосоор холбогдохгуй. 2) Эдгээр цэгуудийг холбосон холбоосууд дотор цикл уусгэдэг дэд олонлог байхгуй. Жишээлбэл, (1,2), (2,3), (3,1) нь 1-ээс эхлээд 1 дээр дуусаж байгаа 3 урттай цикл уусгэж байна, уунд 1,2,3 нь цэгууд, (1,2),(2,3),(3,1) нь цэгуудийг холбосон холбоосууд болно. 3) Дээрх цэгуудээс дурын 2-ыг нь авахад энэ 2-ыг холбосон нэг зам заавал оршин байна. Жишээлбэл, (1,2,3) нь 1 ба 3 гэсэн цэгуудийг холбож байгаа зам болно, гэхдээ мэдээж (1,2), (2,3) гэсэн холбоосууд байж байх ёстой!

Бодлого 1) Хэрвээ ямарваа нэгэн модны цэгуудийн тоо n бол тууний цэгуудийг холбосон холбооснуудын тоо n-1 гэж батла!

Бодлого 2) n цэгууд огогдсон байг. Эдгээр цэгуудийг холбож хэдэн янзын мод уусгэж болох вэ?


uuruur helbel modnii irmegiig chiglelt gej uzehgui gesen ug uu:-) :-)

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 Post subject:
PostPosted: Sep.16.03 4:06 pm 
bi chiglelgui bodson shuu.
gypsy, 2ulangaar ni bodchihoo...

yer ni todorhoiloltoo bodohoor, chiglelt gevel yamar ch 2 tsegiin hoorond zam olddog geheer, zaaval hos chigleltei boloh geed baih yum bish uu? esvel cycle uusne. cycle uusvel yamar yumnii mod boloh vee dee.
ps: iim yum bodoogui udsan baina. (hayaagaa manah)


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 Post subject:
PostPosted: Sep.17.03 3:44 pm 
haha, shuud sanaand orsno bichsen chin shaal hudlaa yum hiichij. hariug ni harahad indukts ashiglah tovogtei bololtoi...
bi ch too toolohgoo muu l doo (dahiad hayagaa manav)


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 Post subject:
PostPosted: Sep.17.03 4:23 pm 
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Эзэрхийлэн Эзлэгч Гишvvн
Эзэрхийлэн Эзлэгч Гишvvн
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Joined: Apr.16.03 2:49 pm
Posts: 985
Location: DAMBADARJAA
end joohon bulterchiheed zaschihlaa shuu :D

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 Post subject:
PostPosted: Sep.20.03 5:18 am 
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Жинхэнэ Гишvvн
Жинхэнэ Гишvvн

Joined: Aug.21.03 10:26 pm
Posts: 26
Location: khuduu barag l govid...
1+1=3

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 Post subject:
PostPosted: Sep.20.03 2:55 pm 
Detective wrote:
Bodlogo 1)-iig yaaj bodsonoo helj o'gooch!

Minii bodoltiig tootsohgui baigaa yum uu?
GYPSY zovshoorno biz dee?

Indukts esvel invariant ashiglaj bolno. Neg oroi nemegdehed 1 tal nemegdene gedgiig helehed boloh baih.


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 Post subject:
PostPosted: Sep.20.03 4:58 pm 
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Эзэрхийлэн Эзлэгч Гишvvн
Эзэрхийлэн Эзлэгч Гишvvн
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Joined: Apr.16.03 2:49 pm
Posts: 985
Location: DAMBADARJAA
bi yag naadah shig chini setgesen bolovch hudlaa bainaa...

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 Post subject:
PostPosted: Sep.22.03 8:39 pm 
Detective wrote:
To Ai:

Chi <tal> gej irmeg, esvel holboosiig heleed baigaa yum uu?

Graph chin ooroo baigaa havtgaig neg tal gej tootsno. Deer minii bichsen tomiyog harval ih zugeer baina. Herev zovshoorohgui bol, havtgai gedgee ch hangalttai tom radiustai bombortsog gej uzeerei.


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 Post subject:
PostPosted: Sep.23.03 4:58 pm 
Toonii onoliin bish tomiyonii bodlogo.
http://mongolduu.com/forum/viewtopic.ph ... 461#456461


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 Post subject:
PostPosted: Sep.23.03 7:04 pm 
unsh l daa, :(

Herev chi Mongold baigaa bol Battsengel bagshiin bichsen nomiig MUIS-n havias olj unshaarai. Tend euler-n theorem-g invairant ashiglaad bodtson baigaa.


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 Post subject:
PostPosted: Sep.25.03 11:57 pm 
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Эзэрхийлэн Эзлэгч Гишvvн
Эзэрхийлэн Эзлэгч Гишvvн
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Joined: Apr.16.03 2:49 pm
Posts: 985
Location: DAMBADARJAA
Detective wrote:
Ai wrote:
Detective wrote:
To Ai:
Chi <tal> gej irmeg, esvel holboosiig heleed baigaa yum uu?

Graph chin ooroo baigaa havtgaig neg tal gej tootsno. Deer minii bichsen tomiyog harval ih zugeer baina. Herev zovshoorohgui bol, havtgai gedgee ch hangalttai tom radiustai bombortsog gej uzeerei.

Ai, ter euler-iin tomyog chin' zo'vhon planar graf deer heregledeg yum baina. Bi say nom so'hoj uzlee. Bidnii yariad baigaa modond tal baij bolohgui shuu dee, hervee 1 tal baidag l yum bol ter taliig burduulj baigaa irmeguud n' cikl uusgechihne.


euleriin tomiyogoor gargah ni zuv shuu dee. planar graph gej yu bolohoo sanatsgaay.

irmeguud ni ogtloltsohguigeer bairluulj boldog graph iig planar graph geh busuu.

mod maani ingehed planar graphiin negen jishee shuudee. tiim bolohoor euleriin tomiyog aih zuilgui ashiglaj bolno gej bodoj baina. harin ter bodlogiin 2r hesgiig bodohod minii bachuu uhaan daanch hureltsehgui ni shig baina:-D

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 Post subject:
PostPosted: Oct.07.03 9:25 pm 
1. Natural too n tei hariltsan anhnii toonuudiin toog ol. (Euler-n theorem-g batal) Uuniig enuuhendee Euler-n too geed todorhoiloyo.
2. Natural too n -g huvaagchuudiin Euler-n toonuudiin niilber-g ol. (n-tei tentsuu bolohiig batal)
3. zaa za boliloo. Odoohondoo ene 2-g bodoh hun baina uu?


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 Post subject:
PostPosted: Oct.07.03 9:30 pm 
Oor neg amarhan bodlogo taviatii.
{Ai} (bi bish shuu :razz:) ni olonloguudiin daraala gej uzie. Daraalliin element ni omnoh element ba tuunii power set-n negdelees togtono. ( A[i+1] = A[i] U P(A[i]) )

A[0] iin chadal ni 1 bol, A[2003]-g ol. Mon A[2004] ni yamar negen toonii kwadrat bish bolohiig batal.


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 Post subject:
PostPosted: Oct.09.03 12:42 am 
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Эзэрхийлэн Эзлэгч Гишvvн
Эзэрхийлэн Эзлэгч Гишvvн
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Joined: Apr.16.03 2:49 pm
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Location: DAMBADARJAA
AIchan ingej shuuraad yahav . umnu ni tavigdsan bodlogoo buren shiidchii.tegeed daraagiinh ruu orii

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 Post subject:
PostPosted: Oct.09.03 12:50 am 
bi ali deer buugaad ogtson shdee. hehe.
Yer ni chadahaasaa hiihgui bol onoo tuuhgui shdee. Suuld ni olimpiad duustsan hoino oilgoson shdee. :razz:


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 Post subject: --------------------
PostPosted: Oct.30.03 7:55 pm 
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Эрхэм Гишvvн
Эрхэм Гишvvн

Joined: Sep.08.03 7:30 pm
Posts: 32
Ai wrote:
1. Natural too n tei hariltsan anhnii toonuudiin toog ol. (Euler-n theorem-g batal) Uuniig enuuhendee Euler-n too geed todorhoiloyo.
2. Natural too n -g huvaagchuudiin Euler-n toonuudiin niilber-g ol. (n-tei tentsuu bolohiig batal)
3. zaa za boliloo. Odoohondoo ene 2-g bodoh hun baina uu?



1. phi(n) -r Euler function -g temdegleye.

(m1,m2)=1 baih m1 ba m2 toonii huvid phi(m1*m2)=phi(m1)*phi(m2) gedgiig

uldegdliin tuhai hyatadiin theorem-g ashiglan hyalbarhan harj bolno.
n = p1^a1 * ... * pk^ak baig. uund pi -uud ni prime numbers.

tegvel phi(n)=phi(p1^a1)*...*phi(pk^ak),

herev p ni prime bol phi(p^k)=p^k-p^(k-1) gedeg ni ilt.

iimees phi(n)=n*urjver(1-1/p) for p | n and p prime.

2. n = sum(phi(d)) for d | n.

proof. Herev x ni 0<=x<n baih too bol (x,n)=n/d baih n-iin huvaagch d oldoh ni

medeej. Iimees n = sum(Nd) for d | n, end Nd ni (x,n)=n/d baih 0<=x<n nuhtsliig

hangah x toonuudiin too (uuruur helbel 0,1,...,n-1 gesen n shirheg toog ul

ogtloltsoh angiudad huvaalaa). Odoo Nd = phi(d) gedgiig haruulbal bodlogo

shiidegdlee gesen ug. d ni n-iin yamar neg huvaagch baig. a = n/d geye. a -d

huvaagdah 0<=x<n baih duriin x toonii huvid 0<=y<d ba x=a*y baih y too oldono.

Iimees (x,n)=a * b helbertei baina. uunees gadna b ni y ba d-g hoyulang ni

huvaadag uyed l a*b ni x ba n -g hoyulang ni huvaana. iimees (x,n) =a * (y,d)

boloh ba (x,n)=a baih 0<=x<n -g hangah x ni a*y helbert baina. uund y ni 0<=y<d

ba (y,d)=1 baina. iimees 0<=x<n -g hangah (x,n)=n/d baih x too yag phi(d)

shirheg orshin baina. uuruur helbel Nd = phi(d) ==> n = sum(phi(d)) for d | n.


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 Post subject: ----------------
PostPosted: Nov.01.03 6:13 pm 
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Эрхэм Гишvvн
Эрхэм Гишvvн

Joined: Sep.08.03 7:30 pm
Posts: 32
Detective wrote:
Та нар графын онолын талаар дажгуй мэддэг юм шиг байна! Дахиад нэг хялбар бодлого тавья, залхуутай санагдсан бол хэлээрэй! Би нэг тодорхойлолтыг графын талаар мэдэхгуй хумууст зориулж бичлээ шуу, учир нь доорх бодлогыг бодоход заавал графын талаар уншсан байх шаардлагагуй! Мэддэг нэг нь tree(мод)-ийг хэлээд байна гэж ойлгоорой!

Тодорхойлолт: Мод
Хэрвээ хавтгай дээрх дурын байрлалтай цэгуудийг дараах маягаар хооронд нь холбовол мод ууснэ. Уунд: 1) Дурын 2 цэгийг холбосон холбоос эсвэл цорын ганц байна, эсвэл тэр 2 цэг ямарч холбоосоор холбогдохгуй. 2) Эдгээр цэгуудийг холбосон холбоосууд дотор цикл уусгэдэг дэд олонлог байхгуй. Жишээлбэл, (1,2), (2,3), (3,1) нь 1-ээс эхлээд 1 дээр дуусаж байгаа 3 урттай цикл уусгэж байна, уунд 1,2,3 нь цэгууд, (1,2),(2,3),(3,1) нь цэгуудийг холбосон холбоосууд болно. 3) Дээрх цэгуудээс дурын 2-ыг нь авахад энэ 2-ыг холбосон нэг зам заавал оршин байна. Жишээлбэл, (1,2,3) нь 1 ба 3 гэсэн цэгуудийг холбож байгаа зам болно, гэхдээ мэдээж (1,2), (2,3) гэсэн холбоосууд байж байх ёстой!

Бодлого 1) Хэрвээ ямарваа нэгэн модны цэгуудийн тоо n бол тууний цэгуудийг холбосон холбооснуудын тоо n-1 гэж батла!

Бодлого 2) n цэгууд огогдсон байг. Эдгээр цэгуудийг холбож хэдэн янзын мод уусгэж болох вэ?





1. Tree gedeg chin graph -n tuhain tohioldol yum l daa, acyclic, connnected graph-g tree gedeg yum.

Theorem: Let T be a tree on n vertices. Then T has n-1 edges.

proof: (proof by strong induction on |E(T)| - irmewgiin too )

Base case - |E(T)| =0. Since T is connected |V(T)|=1 (oroin too) and the theorem hold.

Inductive case -

Assume: If T is acyclic and connected then T is acyclic and has |V(T)|-1 edges, where |E(T)|<k.

Show: If T' is acyclic and connected then T' is acyclic and has |V(T)|-1 edges, where |E(T)|=k.

Remove some edge e from T'. This divides T' into two components, R and S. Each of these components is connected and acyclic, so they each are acyclic and have n-1 edges by the inductive hypothesis. Let nR and nS be the respective number of vertices in each component. Then n=nR+nS and |E(T)|=nR-1+nS-1+1=n-1 as required.

2. How many trees are there on n vertices?

Hypothesis: n^(n-2)
Proof: What then does n^(n-2) count?
Suppose we have n objects, O1, … On, and we pick one. There are n ways to do it. If we throw it back and pick again, independently, a total of n-2 times, there will be a total of n^(n-2) ways to do all this.
There is a neat way to describe this. We can give a name to each choice, say xj to choosing the j-th object. If we define addition for the x’s, we can describe each potential choice as
(x1 + x2 + , . . ., + xn),
where each term xj represents an individual choice, and the plus signs each represent the word “or”. Then repeating this choice n-2 times can be represented as
(x1 + x2 + , . . ., + xn)^(n-2),
where each monomial obtained by multiplying this out will represent a sequence of n-2 choices. This expression loses some of the information about the choices, namely the ordering in which they are made, but it is quite useful in letting us keep track of how many times a given set of choices can be made, as we shall see. Notice that if we replace all the xj ‘s by 1’s, our expression counts the number of our possible choices, which is n^(n-2). We will now describe a given sequence of choices graphically:
Let f(j) = xk indicate that we chose our k-th object on our j-th choice.
Then we can draw a directed graph, and put in a directed edge (j,k) from vertex j to vertex k, for each such choice. Remember that we make n-2 choices and each can be any object.
The resulting graph will have the following properties

1. There will be exactly one edge from each vertex with index up to n-2, and none from the last two vertices.
2. It can have directed cycles or even loops.

Our plan is to make each such graph into a tree in a reversible way.
Now a tree is different from one of our graphs in the following respects.

First it is an undirected graph.
We can change this by introducing a direction to each edge, namely toward the last vertex. If we do so every vertex other than the last will have an edge from it.

The difference between our graphs and trees is then the following:

Our graphs have no edge coming from vertex n-1 while a directed tree has one.

Our graphs can have loops and directed cycles, trees cannot.

There may be no edge coming into vertex n in one of our graphs, but there must be at least one in every directed tree.
And our graphs have n-2 edges while trees have n-1 of them.
We will convert one of our graphs into a tree by adding to it a directed path from vertex n-1 to vertex n that passes through and destroys every cycle in our graph.
This leaves us with three questions: how do we order the cycles on this path? How do we pass through a cycle to destroy it? And, how do we reverse this process to regain our graph from the tree it creates?
We label each cycle in one of our graphs by the smallest index of the vertices in it. Then we order the cycles by their labels and traverse them in that order. We destroy a cycle by entering it immediately after the label vertex and exiting at the label vertex, omitting the edge from the label vertex to its successor in the cycle.
So we will have a path that goes from vertex n-1 to the first cycle then the next, etc. and finally ends at vertex n.
Do we really get a tree after introducing this path? Well, there are no cycles left and there are n-1 vertices, and that defines a tree.
Notice that the smallest vertex index on the path from n-1 to n in the resulting tree will mark the end of the first cycle destroyed, the smallest index beyond it marks the end of the next tree, and so on.
This means that given a tree, we can examine the path in it from vertex n-1 to n, find the smallest vertex in it and make it the end of the first cycle, and so on until we have reconstructed our original graph. And that concludes our proof.


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 Post subject: Re: ----------------
PostPosted: Nov.03.03 10:17 pm 
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Эзэрхийлэн Эзлэгч Гишvvн
Эзэрхийлэн Эзлэгч Гишvvн
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Joined: Apr.16.03 2:49 pm
Posts: 985
Location: DAMBADARJAA
ideal ha ee yatsu dane :-)

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 Post subject:
PostPosted: Nov.03.03 10:25 pm 
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Эзэрхийлэн Эзлэгч Гишvvн
Эзэрхийлэн Эзлэгч Гишvvн
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Joined: Apr.16.03 2:49 pm
Posts: 985
Location: DAMBADARJAA
graph iin bas neg bodlog nemeed tavichihay
uul ni ch tugeemel bodlogo l doo.

n oroitoi buren chiglelt graph dahi 3 urttai cycle iin too hamgiin ihdee hed baih ve?

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PostPosted: Nov.07.03 4:09 am 
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Гялалзах Одны Гялбаа Гишvvн
Гялалзах Одны Гялбаа Гишvvн
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Joined: Aug.12.03 5:45 pm
Posts: 272
Location: usa wa
kusta wrote:
2toonii onoliin bodlogo:
1.81026521 ni haritsangui oirhon 2 sondgoi anhnii toonii urjver gedeg ni medegdej baigaa bol urjigdehuund zadal


2.a ba n ni hariltsan anhnii bol (x-a)*n =x*n -a(mod n) baih gartsaagui buguud hureltseetei nuhtsul ni n anhnii too baina gej batal. x*n eer x iin n zergiig temdeglev. :D


suuliin bodlogiin tuhaid tsaash sonirhson hun baival
www.cse.iitk.ac.in/news/primality.html iig uzeerei.hiisen ajluudiiig ni tataj avj bas bolno.sayhan hiigdsen, batlagdsan ajluud baigaa yum baina.(M.Agrawal,N.Kayal and N.Saxena:PRIMES is in P,2002) :-)


ankhnii too-g angliar iu geh ve? tsus gej buh ium angli deer uzseer baigaad eh hele martakhn
mun gartsaagui buguud hureltseetei nuhtsul--iig bas angliar iu geh ve


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 Post subject:
PostPosted: Nov.07.03 5:13 am 
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Гялалзах Одны Гялбаа Гишvvн
Гялалзах Одны Гялбаа Гишvvн
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Joined: Aug.12.03 5:45 pm
Posts: 272
Location: usa wa
Ai wrote:
C deer neg iimerhuu l bolloo. gomen na compuer ashiglaad. ;)
Code:
  unsigned int p = (unsigned int)sqrt(n);
  unsigned int q;
  if (p & 1 == 0)
    p--;
  while (p>1) {
    q = n/p;
    if (p*q == n && isPrimeNumber(q)) {
      printf("%u*%u=%u\n", p, q, n);
      break;
    }
    p-=2;
  }
  if (p <= 1)
    printf("none\n");


java deer neg iimerhuu bollo. ai dundaas n iavsan shig biana. bi zahaas n iavlaa hehe

c:\autosave\SondgoiDrv.java
public class SondgoiDrv
{
....public static void main(String[] args)
....{
..........int LargeNum = 81026521;
......... int num1 =0 ;
..........int num2 = 0;

..........for (num1 = 1; num1 < Math.sqrt(LargeNum); num1 = num1 + 2)
..........{
...............if (LargeNum % num1 == 0)
...............num2 = LargeNum / num1;
...........}
..........num1 = LargeNum / num2;
..........System.out.println("num1 is "+ num1 + "\nnum2 is "+ num2+"\n"+num1 +" * "+ num2 + " = " + num1*num2);
....}
}

output is:

C:\AUTOSAVE>java SondgoiDrv
num1 is 8753
num2 is 9257
8753 * 9257 = 81026521


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PostPosted: Nov.07.03 11:20 am 
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Эзэрхийлэн Эзлэгч Гишvvн
Эзэрхийлэн Эзлэгч Гишvvн
User avatar

Joined: Apr.16.03 2:49 pm
Posts: 985
Location: DAMBADARJAA
Bishrelt wrote:
kusta wrote:
2toonii onoliin bodlogo:
1.81026521 ni haritsangui oirhon 2 sondgoi anhnii toonii urjver gedeg ni medegdej baigaa bol urjigdehuund zadal


2.a ba n ni hariltsan anhnii bol (x-a)*n =x*n -a(mod n) baih gartsaagui buguud hureltseetei nuhtsul ni n anhnii too baina gej batal. x*n eer x iin n zergiig temdeglev. :D


suuliin bodlogiin tuhaid tsaash sonirhson hun baival
www.cse.iitk.ac.in/news/primality.html iig uzeerei.hiisen ajluudiiig ni tataj avj bas bolno.sayhan hiigdsen, batlagdsan ajluud baigaa yum baina.(M.Agrawal,N.Kayal and N.Saxena:PRIMES is in P,2002) :-)


ankhnii too-g angliar iu geh ve? tsus gej buh ium angli deer uzseer baigaad eh hele martakhn
mun gartsaagui buguud hureltseetei nuhtsul--iig bas angliar iu geh ve


anhnii toog prime number
zailshgui buguud hureltseetei nuhtsul gehiig yag onooj medehgui baina. yurunhiiduu if and only if gej baival zailshgui buguud hureltseetei nuhtsul geed oilgood yavchihna.
necessary and sufficient condition gedeg ch biluu

_________________
Yтрэмийн талбайд очиж...


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 Post subject:
PostPosted: Nov.19.03 1:19 pm 
Offline
Хvндэт Гишvvн
Хvндэт Гишvvн
User avatar

Joined: Aug.15.03 7:49 pm
Posts: 161
mergen olon zaluuchuudad ahiad neg bodlogo.
G ni zereg ni 4 baih togsgolog buleg.teguul G g Abeliin buleg bolohiig bas,Z4 ba Z2xZ2 toi izomorphic bolohiig batal gej.(end Z2 gej Z iin 2moduliarh uldegdliin angi.Z4 mon adilhan) :oh:


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