Cody

# Rohiit Priyadarshan Muralidharan

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 on 30 Mar 2020 on 30 Mar 2020 I solved this question with an if condition and the size of my solution was 39. I found the size of the leading solution to be 14. This is the leading solution: function ans = in_prod(x,y) "no"; try x*y; end How is this even correct? I tried to paste the same solution instead of mine and the assertion failed. z is not mentioned anywhere in this code. Can someone enlighten me? on 30 Mar 2020 on 30 Mar 2020 on 30 Mar 2020 on 30 Mar 2020 on 30 Mar 2020 on 30 Mar 2020 on 30 Mar 2020 on 30 Mar 2020 on 30 Mar 2020 on 30 Mar 2020 on 29 Mar 2020 on 29 Mar 2020 Can someone help me to solve this question? I did not learn image processing. So I tried to solve this question assuming that the pixels either have a value of 1 or 0. I am a beginner. This is my second day into coding. I would really love some help. This is my code: [Ix,Iy]=size(I); for n = 1:Ix %for Rmin if I(n,:)==0 n = n + 1; else n = n; break end end Rmin = n for m = 1:Iy %for Cmin if I(:,m)==0 m = m + 1; else m = m; break end end Cmin = m for o = Rmin:Ix %for Rmax if I(o,Cmin) == 1 o = o+1; elseif I(o,Cmin) == 0 o = o - 1; break; end end Rmax = o for p = Cmin:Iy %for Cmax if I(Rmin,p) == 1 p = p+1; elseif I(Rmin,p) == 0 p = p - 1; break; end end Cmax = p Rpix = Rmax - Rmin + 1 Cpix = Cmax - Cmin + 1 on 29 Mar 2020 on 29 Mar 2020 on 29 Mar 2020 on 29 Mar 2020 on 29 Mar 2020 on 29 Mar 2020 on 29 Mar 2020 on 29 Mar 2020 on 29 Mar 2020 on 29 Mar 2020 on 29 Mar 2020 Rohiit Priyadarshan Muralidharan received Solver badge for Solution 2180668 on 29 Mar 2020 on 29 Mar 2020
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