矩阵中水平-1和垂直-1的广义数学方程

时间：2018-08-14 10:15:23

标签： math wolfram-mathematica symbolic-math discrete-mathematics

矩阵表示。 图片2。我这样写的水平组-1。但我不确定是否正确？ enter image description here

我需要一个矩阵中的一组水平-1和一组垂直-1的广义数学方程。如果一行中有-1的组并且列中有1的组。附图中突出显示的红色元素是必填组。

2 个答案:

答案 0 :(得分：1)

编辑：

在评论中与OP交谈后，听起来他可能想要让计算机通过并搜索矩阵的行或列中-1的连续块的方法。我在这里提供一个示例，您可以在python中使用numpy来搜索行：

import numpy as np;

V = np.array([
    [0,-1,-1,-1,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,-1,0,0,0,0,-1,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,-1,-1,0,0,0,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,-1,-1,1,0,0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,-1,-1,0,0,0,-1,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,-1,-1,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,-1,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,-1,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,-1,0,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1,0],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1],
    [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]])

def group_finder(A, num):
    print("Searching for groups of %d's in the rows of this matrix:"%num)
    print(A)
    print

    m = M.shape[0]
    n = M.shape[1]

    #find groups in rows
    for i in range(m):

        last_matched = False;
        index_of_group_start = None;

        #go through the entries in each row
        for j in range(n) :

            this_matched = (M[i][j] == num)

            if(this_matched):

                if(last_matched):
                    continue #keep processing the matching numbers

                else:
                    index_of_group_start = j

            else:
                if(last_matched and index_of_group_start+1 != j):

                    print("There is a group of %d's in row %d."
                            %(num, i+1))#add one because numpy is zero-indexed.

                    print("It starts at column %d and ends at column %d"
                                %(index_of_group_start+1, j))
                    print

            #update values for next iteration
            last_matched = this_matched


group_finder(V, -1)

当我在计算机上运行该代码时，这就是我得到的输出：

Searching for groups of -1's in the rows of this matrix:
[[ 0 -1 -1 -1  0  0  0  0  0  0  0  0  0  0  0  0  0  0]
 [ 0  0  0  0 -1  0  0  0  0 -1  0  0  0  0  0  0  0  0]
 [ 0  0  0  0  0 -1 -1  0  0  0  0  0  0  0  0  0  0  0]
 [ 0  0  0  0  0  0  0 -1 -1  1  0  0  0  0  0  0  0  0]
 [ 0  0  0  0  0  0  0  0  0  0 -1 -1  0  0  0 -1  0  0]
 [ 0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0]
 [ 0  0  0  0  0  0  0  0  0  0  0  0 -1 -1  0  0  0  0]
 [ 0  0  0  0  0  0  0  0  0  0  0  0  0 -1  0  0  0  0]
 [ 0  0  0  0  0  0  0  0  0  0  0  0  0 -1  0  0  0  0]
 [ 0  0  0  0  0  0  0  0  0  0  0  0  0 -1  0  0  0  0]
 [ 0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0]
 [ 0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0]
 [ 0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 -1  0]
 [ 0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 -1  0]
 [ 0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 -1]
 [ 0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 -1]
 [ 0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 -1]
 [ 0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0]]

There is a group of -1's in row 1.
It starts at column 2 and ends at column 4

There is a group of -1's in row 3.
It starts at column 6 and ends at column 7

There is a group of -1's in row 4.
It starts at column 8 and ends at column 9

There is a group of -1's in row 5.
It starts at column 11 and ends at column 12

There is a group of -1's in row 7.
It starts at column 13 and ends at column 14

旧答案：

我认为您很难描述您的要求。可以理解的我想我可以帮助您描述您的问题。

似乎您需要在电子表格程序（例如Microsoft Excel或LibreOffice）中找到-1的连续单元格。您需要可以在矩阵中找到一组连续的负数，然后对其进行操作的东西。您正在寻找类似COUNTIF函数的方法，但不是要 counting 具有特定属性的单元格，而是要尝试 highlight 或 select < / em>具有特定属性的单元格。具体来说，您要突出显示包含-1的单元格，并具有也包含-1的非对角相邻单元格。您的问题是如何使电子表格程序做到这一点。

在这种情况下，您为问题使用了错误的标签。仅仅因为有人知道数学并不意味着他们知道电子表格软件。我建议更改您问题上的标签。一些不错的标签可能是calc-libreoffice，excel或spreadsheet。实际上，similar question with the excel tag有一个答案，尽管它不处理行。

您有机会尝试理解数学家如何在矩阵中描述一组负数。在那种情况下，我认为您的问题最好用“我应该使用什么数学的表达式来描述一个完全由-1组成且限于一行或一部分的子矩阵？只有一列。”

该问题的一个答案是使用一种矩阵下标表示法。我会将您的第一组标记为V 5-11-12 ，将您的第二组标记为V 7-10、15 。然后我只想说V 5，11-12 中的条目都是-1。

答案 1 :(得分：0)

<?php 

print_r($_POST["innerArray"]);

?>

在第一行中添加了两个额外的<!DOCTYPE html> <html> <head> <script src="jquery-3.3.1.min.js"></script> <script src="scripts.js"></script> </head> <body> </body> </html> ，以演示代码能够找到多个array = { {0,-1,-1,-1, 0, 0, 0, 0, 0,-1,-1, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0,-1, 0, 0, 0, 0,-1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0,-1,-1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0,-1,-1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1,-1, 0, 0, 0,-1, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1,-1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,-1}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}}; 跨度的功能。

函数定义

-1

水平扫描

-1
fromTo[y_] := #[[{1, -1}]] & /@ SplitBy[MapIndexed[
     If[# == -1, Last[#2]] &, y], Head] /. {a_, a_} -> Nothing

wherearethey[a_, dir_] := Grid[Prepend[DeleteCases[MapIndexed[
     {Last[#2], fromTo[#1]} &, If[dir == "Row", a, Transpose[a]]],
    {__, {}}], {dir, "{{From, To}}"}], Alignment -> Left]

因此，在第一行中，位置2至4和位置10至11有wherearethey[array, "Row"] 。

垂直扫描

Row  {{From, To}}
1    {{2,4},{10,11}}
3    {{6,7}}
4    {{8,9}}
5    {{11,12}}
7    {{13,14}}

-1
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