Mavzu: Matrisalarning asosiy xarakteristikalarini xisoblash Masalaning qo’yilishi
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MUHAMMAD AL-XORAZMIY NOMIDAGI TOSHKENT AXBOROT TEXNOLOGIYALARI UNIVERSITETI “Mashinali o`qitishga kirish” fanidan AMALIY ISH – 1 Bajardi: Turkmanov Sh. Tekshirdi: Kubayev S. SAMARQAND-2023 Mavzu: Matrisalarning asosiy xarakteristikalarini xisoblash Masalaning qo’yilishi: Matlabda matritsalarni chapdan o’ngga burishda fliplr komandasidan foydalanish: A=[-1 0 1; 0 -1 0; 1 -1 1] 1-rasm. A matritsa yaratdik 2-rasm. fliplr(A) buyrug’i orqali matrit chapdan o’ngga burishda sonlarni ekranga chiqardik Endi shu komandani qo’lda bajarib chiqamiz: >> for i=1:3; for j=1:3; C(i,j)=A(3-i+1,j);end; end; C 3-rasm. Bu matrisani qo’lda bajardik va >> for i=1:3; for j=1:3; C(i,j)=A(3-i+1,j);end; end; C shu kodni ekranga chiqard 4-rasm. Buda esa yuqoridan pastga burishda flipud(A) komandasidan foydalandik. Endi shu amalni algoritmi bilan tanishib chiqamiz: >> for i=1:3; for j=1:3; C(j,i)=A(j,3-i+1); end; end; C 5-rasm. Bu rasmda A matrisani algoritm ko’rinishi 6-rasm. Bu rasmda matritsani soat strelkasiga qarshi 90 0 ga burish uchun ishlatiladigan rot90(A) komandasidan foydalandik. Endi shu amalning bajarilish tartibi ya’ni algoritmi haqida ko’rib chiqamiz: >> for i=1:3; for j=1:3; C(i,j)=A(j,3-i+1); end; end; C 7-rasm. Matrisani 90 ga burgandagi algortim ko’rinishida ekranga chiqardik. Dasturning kodi: A=[-1 0 1; 0 -1 0; 1 -1 1] A =
-1 0 1 0 -1 0 1 -1 1 fliplr(A) ans = 1 0 -1
1 -1 1 for i=1:3; for j=1:3; C(i,j)=A(3-i+1,j);end; end; C C = 1 -1 1
-1 0 1 flipud(A) ans = 1 -1 1
-1 0 1 for i=1:3; for j=1:3; C(j,i)=A(j,3-i+1); end; end; C C = 1 0 -1
1 -1 1 rot90
Error in rot90 (line 22)
Error in dot (line 21) if isinteger(a) || isinteger(b) rot90(A) ans =
1 0 1 0 -1 -1 -1 0 1 for i=1:3; for j=1:3; C(i,j)=A(j,3-i+1); end; end; C C = 1 0 1
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