algebra ii for dummies

[find (1)]!=1){ thememset (Vis,0,sizeof(VIS)); num=0; the for(intI=2; i1; i++){94 if(Vis[find (i)])Continue; theVis[find (i)]=++num; the } thenum=0;98 for(intI=2; i1; i++) Aboutnum=num*Ten+Vis[find (i)]; -a.mat[id[num]][id[mem[t]]]+=1;101 }102 }103 }104 return; the }106 107 intMain () {108 #ifndef Online_judge109Freopen ("count.in","R", stdin); theFreopen ("Count.out","W", stdout);111 #endif thesc

1. Determinant1.1 Second-order determinant1.2 third-order determinant1.3 Number of reverse order1.4 N-Step determinant2. The nature of the determinantProperty 1 The determinant is equal to its transpose determinant.Property 2 swaps the determinant of two rows (columns), determinant.Property 3 The determinant of a row (column) in which all elements are multiplied by the same multiplier K, equals the number k multiplied by this determinant.Property 4 Determinant If there are two rows (column) elem

transformation.Matrix elimination Element Method:determinant TypeCalculation (0 descending order method)Other properties of the determinant:The law of ClydeMatrixFollow the law1. Linear Properties2. Operational and polynomial of n-order matricesElementary matrix and its role in multiplicationFor the unit matrix, the matrix obtained by making an elementary transformation becomes the elementary matrix.Together there are three primary transformations:The block rule of multiplication:Two frequently

Representation of graphs Adjacency Matrix Correlation matrix (horizontal longitudinal point, direction Graph 1 in-1 The connectivity of graphs Non-Tourienton/non-connected Forward graph strong connectivity/single-sided connectivity/weak connectivity Proof method: Specific analysis of the special case of the Reach matrix (Eulerian graph has the necessary and sufficient conditions, Hamilton full/essential Euler

1. Determinant 1.1 Second and third-order determinantSecond-order determinant = A * D-b * C　　　　Third-order determinant = A11*a22*a33+a12*a23*a31+a13*a21*a32-a31*a22*a13-a32*a23*a11-a33*a21*a12Three lines minus three lines1.2 full rank and reverse order numberAll-in-all arrangement: 1,3,2 2,1,3 2,3,1 3,1,2 3,2,1The total number of permutations is: 3*2*1 = 3!reverse order Number: Two elements of the size and position of the relationship does not match and there is an inverse (the general default f

] ofRec; V:Array[0..2002000] ofBoolean; N,m,i,j,k,l,st,ed,ww,top,tar,ans,x:longint;functionmin (aa,bb:longint): Longint;begin ifAa Thenexit (AA); exit (BB);End;procedureAdd (st,ed,ww:longint);beginInc (top); A[TOP].S:=St; A[TOP].E:=Ed; A[TOP].W:=ww; A[top].next:=B[st]; B[ST]:=top;End;functionBfs:boolean;varHead,tail,x,u:longint; Y:rec;beginFillchar (v,sizeof (v), false); Tail:=1; head:=0; d[st]:=1; V[ST]:=true; q[1]:=St; whileHead Do beginInc (head); x:=Q[head]; U:=B[x]; whileU>0 Do begi

1.The calculate the slope:the covariance of X and Y divided by the variance of X　　From NumPy import CoVslope_density = CoV (wine_quality["quality"],wine_quality["density"]) [0,1]/wine_quality["Density"].var () #cov ( X, y) is the function from NumPy, which returns a 2*2 Metric,.var () is Pandas function.2.To get the INTERCEPT:B = Y-ax (x and Y is the mean value of each column)Intercept_density = wine_quality["Quality"].mean ()-wine_quality["Density"].mean () * (Calc_slope (wine_quality[) Density

Topic Link: PortalTest instructionsGiven your three numbers, P,q,n, p stands for a + B, Q for a*b;Then ask A^n + b^nset F[i] = A^i +b^i; f[0]=2,f[1]=p;f[i]* (a+b) = a^ (i+1) + b^ (i+1) +a^i*b + b^i*a;F[i]*p = f[i+1] + a*b*[ a^ (i-1) + b^ (i-1) ]F[i+1] = f[i]*p + q*f[i-1];And then use the matrix to speed up a bit (PS. The input of this problem is very pit .... ）The code is as follows:#include UVA 10655 contemplation! Algebra (Matrix fast Power)

[Problem 2015s02] set \ (a,b,c\) is plural and \ (bc\neq 0\), proving that the following \ (n\) Order matrix \ (a\) can be diagonalization:\[a=\begin{pmatrix} A B \ C A B \ C A am P B \ \ddots \ddots \ddots \ C A b\\ C A \end{pmatrix}.\][Question 2015S02] Fudan Advanced Algebra II (Level 14) weekly (second teaching week)

[Linear algebra] matrix addition 1 # Include 2 Using Namespace STD; 3 4 5 Int Main () 6 { 7 Int Matrixa [ 100 ] [ 100 ]; // Matrixa 8 Int Matrixb [ 100 ] [ 100 ]; // Matrixb 9 Int Plusresult [ 100 ] [ 100 ]; // Matrixa + matrixb = plusresult (this is a maxtrix) 10 Int M, N; 11 Cout " Enter the required and number of rows and columns in the matrix. " 12 Cin> m> N; 13 Cout " Enter matrix " 14

{ - mat ans; toans.v[0][0]=ans.v[1][1]=1; + - while(k)///after the second division to be determined * the { * if(k1) $ans=ans*x;Panax Notoginsengx=x*x; -k>>=1; the } + A returnans; the } + - Long LongSolve () $ { $ if(n==0) - return 2; - if(n==1) the returnp; - if(n==2)Wuyi returnp*p-2*Q; the - Mat A; Wua.v[0][0]=p; -a.v[0][1]=-Q; Abouta.v[1][0]=1; $A=pow_mod (a,n-2); - - returna.v[0][1]*p+a.v[0][0]* (p*p-2*q); - } A + int

structMatrix9 {Ten intr,c; OneLL m[3][3]; AMatrixintArrint_c): R (_r), C (_c) {} - }; - voidInit (Matrix m,intAintb) the { -m.m[1][1]=a,m.m[1][2]=-b; -m.m[2][1]=1, m.m[2][2]=0; - } + matrix Multiply (matrix A,matrix b) - { + Matrix M (A.R,B.C); A for(intI=1; ii) { at for(intj=1; jj) { -m.m[i][j]=0; - for(intk=1; kk) -m.m[i][j]+=a.m[i][k]*B.m[k][j]; - } - } in returnm; - } toMatrix Matrix_pow (Matrix M,intN) + { - if(n==0){ them.m[1][1]=m.m[2][2

1, 110501/10035 Primary arithmetic (elementary school arithmetic)Note the output format#include #includestring.h>#include#include#includeusing namespaceStd;typedefLong LongLld;lld A, b;intMain () { while(SCANF ("%lld", a)! =EOF) {scanf ("%lld",b); if(a==0b==0) Break; intCnt=0; inttot=0; while(a!=0|| b!=0) { if(a%Ten+b%Ten+cnt>=Ten) {CNT=1; Tot++; } Else{cnt=0; } A/=Ten; b/=Ten; } if(tot==0) printf ("No"); Elseprintf"%d", tot); if(tot1) printf ("Ca

^nf (x) \rd x}$.6. For integer $a, b$, define $a \equiv b\ (\mod m) $ when and only if $m \mid (A-B) $ (that is, $m $ divisible $a-b$). When a positive integer $m $, what is the solution to a linear equation group? $$\bex \sedd{\ba{rrrrrrl} x+2y-z\equiv1\ (\mod m) \ \ 2x-3y+z\ Equiv4\ (\mod m) \ 4x+y-z\equiv9\ (\mod m) \ea} \eex$$7. Set $\tt$ is real, $n $ is the natural number, ask $$\bex \sex{\ba{cc} e^{-i\tt}2i\sin \tt\\ 0e^{i\tt} \ea}^n. \eex$$8. Set $A, B\in m_n (\BBC) $ ($n $ order complex

1.0 and 1 (duality:0--1,--+)X + 0 = x, x 1 = XX + 1 = 1, x 0 = 02. idempotentx + x = x, x x = X3. involution(x ') ' = X4. Complementarityx + x ' = 1, x X ' = 05. Commutativex + y = y + X, xY = y X6. Associative(x + y) + z = x + (y + z) = x + y + Z(xY) ·Z = x(YZ) = xY Z7. DistributiveT(Y + Z) = xY + x ZX + (yZ) = (X + Y) · (X +z)8. SimplificationTY + xY ' = x, (x + Y) · (x + Y ') = XX + xY = X, x (x + Y) = X9. Multiplying and factoring(X + Y) ·(X ' + Z) = xZ + X ' · YTY + X ' ·z = (X + z) · (X '

\ \end{pmatrix}\to \begin{pmatrix}0 0 \ 0 A + B \ \ \end{pmatrix}.$$ Because the rank of the matrix does not change under the elementary transformation of the block, the basic formula of rank can be obtained $ $r (A) +r (b) =r (a^2) +r (b) =r\begin{pmatrix} a^2 0 \ 0 B \ \ \end{pmatrix}=r\begin{pmatrix}0 0 \ 0 a+b \ \end{pmatrix}=r (a+b). $$Certificate Act Two (Geometric method-using linear transformation theory) refer to [question 2014a12] and its solution.Certificate Law III (Algebraic

[Problem 2015s01] set \ (M_n (\mathbb{r}) \) is the real linear space of the entire formation of the \ (n\) Order matrix, \ (\varphi\) is a linear transformation on \ (M_n (\mathbb{r}) \), so that for a given \ (A, B\in m_n (\mathbb{r}), or \ (\varphi (AB) =\varphi (a) \varphi (b) \), or \ (\varphi (AB) =\varphi (b) \varphi (a) \) is established. Proof: either \ (\varphi (AB) =\varphi (a) \varphi (b) \) is true for any of the \ (A,b\in m_n (\mathbb{r}) \), or \ (\varphi (AB) =\varphi (b) \varphi

\), and the conclusion of the proof is the formula (13).\[r/\cap i_k\cong r/i_1\times r/i_2\times\cdots\times r/i_n\tag{13}\]First, easy authentication \ (r\to r/i_1\times r/i_2\times\cdots\times r/i_n\) is the homomorphism mapping, if it can be proved that it is full-shot, by the homomorphism fundamental theorem can be concluded. The proof method is the same as the essence in the elementary number theory, we need to construct for each dimension \ (r_k= (\cdots,0,a_k,0\cdots) \). This condition

arithmetic comparer (>,≥,Select * from R,s where r.a=s.c;except (division)Given the relationship R (x, y) and S (y) where x, Y is a set of properties (which can also be a single attribute), y in Y and s in R are attributes (sets) with the same name and can have different property names.But must originate from the same domain set. When solving R÷s, group R by the value of X, and then examine each group, such as Y in a group that contains all Y in S, then take the value of x in that group as a tu

necessary and sufficient condition to prove its existence is \ (| g_k|\) coprime, the full use of the cycle group has just been discussed proof \ (a\) decomposition of each factor is its generation of the elements of the group, the necessity is through the construction of two \ (p-\) Order (refer to the next article) of the product to export contradictions. In addition, if \ (G=g_1\times g_2\) and \ (G_1\leqslant h\), it is easy to prove that there is \ (H=g_1\times (G_2\cap H) \).\[h= (H\cap g