algebra ii workbook

,len), add (V,u,0);} Queueint>QintD[n],s,t;BOOLBFS () { while(!q.empty ()) Q.pop ();memset(d,0,sizeofD);intI,u,v; Q.push (s), d[s]=1; while(!q.empty ()) {U=q.front (), Q.pop (); for(I=head[u];i;i=e[i].next) {if(!d[v=e[i].v]e[i].len) {d[v]=d[u]+1;if(v==t)return 1; Q.push (v); } } }return 0;}intDinic (intXintFlow) {if(x==t)returnFlowintI,u,v,k,remain=flow; for(I=head[x];iremain;i=e[i].next) {if(d[v=e[i].v]==d[x]+1e[i].len) {k=dinic (V,min (E[i].len,remain));if(!k) d[v]=0; e[i]

the relationship s. Compare to discover:The X1 image set is only Y1 and cannot contain all the values of the attribute y in the relationship s, so the X1 is ruled out;And X2 's image set contains all the values of attribute y in relation s, so the r÷s end result is X2,Maybe now you know a little bit about how the division operation is done, so let's take a look at what the division can solve.Take a look at the following small example:With relationship R,s and RS, for rs÷s resultsIt's easy to ge

[Problem 2014a13] set \ (v\) is a \ (n\) dimensional linear space on a number of domains \ (k\), \ (\varphi\) is a power 0 linear transformation on \ (v\) and satisfies \ (\mathrm{r} (\varphi) =n-1 \), verification: \ (v\) is about linear transformation \ (\varphi\) of the loop space, that is, the existence of vector \ (\alpha\in v\), so that \[v=l (\alpha,\varphi (\alpha), \cdots,\varphi^{n-1} (\ Alpha), \varphi^n (\alpha), \cdots). \][Question 2014a13] Fudan Advanced

assume that the line expression is as follows:Then calculate the error function:The error function e is obtained by the deviation of the coefficient, a, a, and the 0:The value of the coefficients, a, a, a, and a linear expression is obtained by the above formula:The curves obtained by the least squares are as follows:Linear algebraic approach to least squares: Similarly, the expression of a fitted line is set to:The purpose of fitting is to make the data points satisfy the above function expres

[Problem 2015s03] set \ (g (x) =x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n\) is an irreducible polynomial on the Number field \ (\mathbb{k}\). Set \ (v\) is the \ (n\) dimension linear space on the number domain \ (\mathbb{k}\), \ (\varphi\) is a linear transformation on \ (v\), \ (\alpha_1\neq 0,\alpha_2,\cdots,\alpha_n\) is \ (v\) Vector in, meet \[\varphi (\alpha_1) =\alpha_2,\,\varphi (\alpha_2) =\alpha_3,\,\cdots,\,\varphi (\alpha_{n-1}) =\alpha_n,\,\ Varphi (\alpha_n) =-a_n\alpha_1-a_{n-1}\alpha_2

Label: SP strong on 2014 problem BS nbsp space mat[Question 2014a07]Set \ (A \) to the fourth-order square matrix on the rational number field \ (\ mathbb {q} \), \ (\ alpha_1, \ alpha_2, \ alpha_3, \ alpha_4 \) is the four-dimensional column vector on \ (\ mathbb {q} \), satisfying the following conditions: \ [A \ alpha_1 = \ alpha_2, \, \, A \ alpha_2 = \ alpha_3 ,\, \, A \ alpha_3 = \ alpha_4, \, \, A \ alpha_4 =-\ alpha_1-\ alpha_2-\ alpha_3-\ alpha_4. \] proof: If \ (\ alpha_1 \ NEQ 0 \), \

", stdin); -Freopen ("Bzoj_1013.out","W", stdout); the #endif * intN; $scanf"%d",n);Panax Notoginseng for(intI=1; i1; i++) - for(intj=1; j) thescanf"%LF",a[i][j]); + A for(intI=1; i){ the for(intj=1; j){ +a[i][n+1]-=a[i][j]*a[i][j]-a[i+1][j]*a[i+1][j]; -a[i][j]=2* (a[i+1][j]-a[i][j]); $ } $ } - - Solve (n); the - for(intI=1; i)Wuyiprintf"%.3LF", a[i][n+1]); theprintf"\ n"); - return 0; Wu}Linear

Ultraviolet A 10655 Contemplation! Algebra (matrix fast power) Given the valueA + BAndABYou will have to find the valueAn + bn Input The input file contains several lines of inputs. Each line does t the last line contains3Non-negative integersP,QAndN. HerePDenotes the valueA + BAndQDenotes the valueAB. Input is terminated by a line containing only two zeroes. This line shoshould not be processed. Each number in the input file fits in a signed32-Bit in

Least squares problem:Before the combination of orthogonal, subspace W, orthogonal projection, orthogonal decomposition theorem and best approximation principle in vector space, the least squares problem can be solved satisfactorily.First of all, we have to explain the problem itself, that is, in the process of production, for the giant linear systems ax=b, may be no solution, but we are urgently need a solution, satisfies the solution is the most approximate solution of the equation.Below we sy

matrix A and e just can reflect these 3 elementary transformations.The proof that the determinant is equal to the original determinant:This problem is very simple, but we should be able to realize the meaning of this theorem, it makes the row transformation and column transformation has the equivalence, that is, the application of the transformation of the row is applicable to the column.A brief proving process: defining determinant A and writing out its transpose matrix a^t.The A determinant i

Matrix equation:We have previously introduced the linear combination of vectors, the form of X1a1+x2a2+xnan, that we can use to express them with [] formulas. (This expression is sought for convenience and unity of computation), and we give the following definition to give another form of the linear combination of vectors.It can be seen that the right side of the equation, the form of a vector combination, we use the algebraic nature of the vector to sum it, we will eventually get a vector B, th

Based on the previous chapters, we can easily draw the concept of eigenvectors and eigenvalues.First we know that the product of a and n dimensional vector v of n x n matrices will get an n-dimensional vector, then we now find that, after calculating u=av, the resulting vector u is collinear with V, that is, vector v is multiplied by matrix A to get the vector u "stretched" with respect to vector V, which satisfies the following equation:Av =λv=uSo here we call λ the eigenvalues of matrix A, and

Principle Analysis:This section describes the transpose of the Matrix. The transpose of the matrix will change the row and column elements of the matrix, that is, the first column of the second row (with C21, after the same) and the first row of the second column (C12) element swap position, the original C31 and C13 Exchange. namely CIJ and cji Exchange.(Fill in the illustration here)C + + language:The first thing we think about is to remove the column J of row I from Row J, which is simple enou

is divided into two steps: Judging the legality of two matrices; The K line of the A matrix is extracted and multiplied by the column I of the b Matrix, and the first column I of the target matrix is obtained. The following two kinds of writing are the above ideas, the first one faster, occupy less memory, the second closer to People's thinking (the second folding please expand).Template Matrixoperator* (matrix//operator Overloading * Overloading for point multiplication{ /

Reference: "Linear algebra" Tongji University, fourth edition 1. Elementary transformations of matrices 1) definition 2) equivalence relationship between matrices: definition, Nature (3) 3) row ladder-shaped matrix; the simplest form; standard type; equivalence class 2. Elementary matrices 1) definition 2) Theorem 1 (Relationship between elementary transformation and elementary matrix) 3) theorem 2 (the necessary and sufficient conditions for

Reference: Hiraoka and luckily Programmer's Math 3: linear algebra. 1. Vector-- What is the space 1.1 vector? Vectors, which can be seen as a pile of permutations.In space, a vector can represent a point, for example (2,3) that represents a point with a horizontal axis of 2 on a two-dimensional plane, an ordinate of 3, or a directed segment that points to it from the origin.When emphasizing the concept of "permutation number", it is generally indicate

The MIT Challenge My friend Scott Young recently finished a astounding Feat:he completed all the courses in MIT's fabled computer science C Urriculum, from Linear Algebra to Theory of computation, in less than one year. More importantly, he does it all in his own, watching the lectures online and evaluating himself using the actual exams. (see Scott's FAQ page for the details about how he ran this challenge.) That works out to around 1 course every 1.

1, Proof: The third class of block elementary transformation is a compound of several third class elementary transformations. In particular, the third class of block elementary transformations does not change the value of the determinant. 2, set $n \, (N\geq 2) $ order Phalanx $A = (A_{ij} (x)) $, where each element $a _{ij} (x) $ is a polynomial about the undecided meta $x $. If $k $ is a positive integer, all algebra cofactor $x that satisfy $A ^k

This paper turns from the public number---meets the mathematical---graphic mathematical---linear algebra partThank you for meeting the Math Working Group to explain the obscure and acting knowledgeable mathematical knowledge of university textbooks in an easy-to-understand and lively and interesting way.This time we mainly do a review, and then further the determinant of the geometric meaning of the animation to show the explanation. We say that matri

The following list does not contain errors in the Chinese publishing format, and contains only errors in the mathematical dimension and irregularities in the narrative or discourse. Please note that this errata is not updated regularly. Welcome you to the Fudan University "Advanced Generation of Mathematics (third edition)" Textbook Error and improper place to correct! 43rd page, Exercise 1: $a _{1n}a_{2,n-1}a_{3,n-2}\cdots a_{n1}$. The 198th page, Theorem 4.4.1, proves the 9th line: $0