simplying of the statements

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Hege
Hege on 30 Jul 2020
Commented: Hege on 30 Jul 2020
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)

Accepted Answer

Sriram Tadavarty
Sriram Tadavarty on 30 Jul 2020
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 Comment
Hege
Hege on 30 Jul 2020
Hi Sriram, Thank you so much for th explanation. This is working for my script.Thank you.

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More Answers (1)

Walter Roberson
Walter Roberson on 30 Jul 2020
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 Comments
Walter Roberson
Walter Roberson on 30 Jul 2020
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 on 30 Jul 2020
Hi Walter. Thank you so much for the explanation. This works very well. Thank you.

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