simplying of the statements

1 次查看（过去 30 天）
Hege
Hege 2020-7-30
评论： Hege 2020-7-30
I have attached two parts(part 1 and part 2) from my script and there are few parts like this and I want to simplify these hard coded statements. For easyness, I have attached two different statements only in here which means I just want understand my weakness once somebody guided me. All these i,j,k values are in 1:10 range. just expecting two different simplified statements for part(1) and part(2).
if ((j==1)&&(k==1)||(j==2)&&(k==2)||(j==3)&&(k==3)||(j==4)&&(k==4)||(j==5)&&(k==5)||(j==6)&&(k==6)||(j==7)&&(k==7)||(j==8)&&(k==8)||(j==9)&&(k==9)||(j==10)&&(k==10))&&((i==1)||(i==10));
B(i,j,k)=1;
else B(i,j,k)=0;%----------part (1)
if ((i==10)&&(j==1)||(i==9)&&(j==2)||(i==8)&&(j==3)||(i==7)&&(j==4)||(i==6)&&(j==5)||(i==5)&&(j==6)||(i==4)&&(j==7)||(i==3)&&(j==8)||(i==2)&&(j==9)||(i==1)&&(j==10))&&((k==1)||(k==10));
B(i,j,k)=1;
else B(i, j,k)=0;%----------part(2)

请先登录，再回答此问题。

采纳的回答

Sriram Tadavarty
Sriram Tadavarty 2020-7-30
Hi Yasasween,
Two things to make a simplied equation:
  1. Assigning B with zeros, as all the dimensions are known at the first place. Implies, one can get away with else condition (i.e. B = zeros(10,10,10)
  2. Now, placing the if conditions for the value of 1
On the Simplication part, lets start with the first condition placed
% First condition
((j==1)&&(k==1)||(j==2)&&(k==2)||(j==3)&&(k==3)||(j==4)&&(k==4)||(j==5)&&(k==5)||(j==6)&&(k==6)||(j==7)&&(k==7)||(j==8)&&(k==8)||(j==9)&&(k==9)||(j==10)&&(k==10))&&((i==1)||(i==10));
% As can be seen in the above condition, the first part has j and k are equal, and then it is valid for i equals 1 or 10
% So, this can be the update or simplied statement for it
((j == k)) && (any(i == [1 10]))
% Second condition
((i==10)&&(j==1)||(i==9)&&(j==2)||(i==8)&&(j==3)||(i==7)&&(j==4)||(i==6)&&(j==5)||(i==5)&&(j==6)||(i==4)&&(j==7)||(i==3)&&(j==8)||(i==2)&&(j==9)||(i==1)&&(j==10))&&((k==1)||(k==10))
% As can be seen in the above condition, other than the last condiiton, all other conditions lead to a value of 11 when added
((i+j == 11)) || (any(k == [1 10]))
% In a similar fashion, based on the pattern of the conditions, the simplications can be made
Hope this helps.
Regards,
Sriram
  1 个评论
Hege
Hege 2020-7-30
Hi Sriram, Thank you so much for th explanation. This is working for my script.Thank you.

请先登录，再进行评论。

更多回答（1 个）

Walter Roberson
Walter Roberson 2020-7-30
B(i,j,k) = ismember(i,[1 10]) && ismember(j, 1:10) && j == k; %part 1
B(i,j,k) = ismember(k,[1 10]) && ismember(j, 1:10) && i+j == 11; %part 2
Depending on the permitted range of j values, there would be some cases where these could be simplified to
B(i,j,k) = ismember(i,[1 10]) && j == k; %part 1
B(i,j,k) = ismember(k,[1 10]) && i+j == 11; %part 2
  3 个评论
显示 1更早的评论
Walter Roberson
Walter Roberson 2020-7-30
if ((j==1)&&(k==1)||(j==2)&&(k==2)||(j==3)&&(k==3)||(j==4)&&(k==4)||(j==5)&&(k==5)||(j==6)&&(k==6)||(j==7)&&(k==7)||(j==8)&&(k==8)||(j==9)&&(k==9)||(j==10)&&(k==10))&&((i==1)||(i==10));
(j==1)&&(k==1) -- okay that is starting conditions that do not necessarily mean much in themselves but we will keep them in mind
(j==2)&&(k==2) -- oh look, j is the same as k, and that was true for (j==1)&&(k==1) -- maybe that is the pattern ?
(j==3)&&(k==3) -- yup, still j == k. And that pattern carries through to (j==10)&&(k==10))&&((i==1)||(i==10))
So the part in () up to but excluding the &&((i==1)||(i==10)) has the rule that j and k must be equal and that they must be in the range 1 to 10. Because they must be equal, you can skip testing that they are both in the range 1 through 10 if you have already tested that they are equal.
The ((i==1)||(i==10)) part can be written multiple ways. Since I already used ismember(j, 1:10) to enforce that j and k are 1, 2, 3, ... 10 (and not, for example, -2 or 7.5), then it makes sense to use the same style and test ismember(i, [1 10])
if ((i==10)&&(j==1)||(i==9)&&(j==2)||(i==8)&&(j==3)||(i==7)&&(j==4)||(i==6)&&(j==5)||(i==5)&&(j==6)||(i==4)&&(j==7)||(i==3)&&(j==8)||(i==2)&&(j==9)||(i==1)&&(j==10))&&((k==1)||(k==10));
That starts with (i==10)&&(j==1) and after seeing the previous part, our thought will be along the lines of whether j will increase at the same rate that i increases, or whether one of them will decrease when the other one increases. And (i==9)&&(j==2) immediately focuses us on i decreasing at the same rate that j increases. We can see quickly that the same pattern holds up to (i==1)&&(j==10) . So the sum is constant and we can use that as the test.
Hege
Hege 2020-7-30
Hi Walter. Thank you so much for the explanation. This works very well. Thank you.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Programming 的更多信息

标签

产品


版本

R2017a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by