cau 2
ta can tinh h(y/x)=p(x=1)*h(y/x=1) + p(x=2)*h(y/x=2) + p(x=3)*h(y/x=3) + p(x=4)*h(y/x=4)
h(y/x=1) = 0.3*lg(0.3) + 0.4*lg(0.4) + 0.3*lg(0.3) + 0= 1.57 bit
tuong tu ta tinh duoc ket qua sau:(cac ket qua nay co the sai cac ban tinh lai gium)
h(y/x=2) = 1,57 bit
h(y/x=3) = 1.29 bit
h(y/x=4) = 0.72 bit
h(y/x) = 0.4*1.57 + 0.3*1.57 + 0.1*1.29 + 0.2*0.72=1.37 bit cau 3: ta can tinh I(X/Y)=I(Y/X)=H(X) - H(X/Y) = H(Y) - H(Y/X)
TA TINH THEO CONG THUC: I(X/Y)=H(Y) - H(Y/X)
ta can tinh them h(y)
y1=0.3*0.4 + 0*0.3 + 0*0.1 + 0*0.2=0.12 (cong thuc cac ban xem lai gium)
y2=0.26
y3=0.31
y4=0.31
h(y)=1.9
I(X/Y)=H(Y) - H(Y/X)=1.9 - 1.37 = 0.53 bit
cau 4
do ta truyen co 4 gia tri X nen can tao B1.B2,B3,B4
dua y1 vao:
p(x=1)*p(y=1/x=1)=0.4*0.3=0.12 (max)
p(x=2)*p(y=1/x=2)=p(x=3)*p(y=1/x=3)=p(x=4)*p(y=1/x=4)= 0
vay dua B1={ Y1 };
TUONG TU:
dua y2 vao:
p(x=1)*p(y=2/x=1)=0.3*0.4 + 0.3*0.4 + 0.1*0.3 + 0.2*0.2= 0.16 max
p(x=2)*p(y=1/x=2)=0.09
p(x=3)*p(y=1/x=3)=0.01
p(x=4)*p(y=1/x=4)= 0
=> B1 = { y1, y2 }
dua y3
p(x=1)*p(y=2/x=1)= 0.12 max
p(x=2)*p(y=1/x=2)=0.12 max
p(x=3)*p(y=1/x=3)=0.03
p(x=4)*p(y=1/x=4)= 0.04
trong truong hop nay khi nhan duoc y3 ta co the giai ma ra ca hai x1, x2
nen ta chon dua y3 vao B1, B2 deu duoc
ta chon B2 = { y3 };
dua y4 vao:
p(x=1)*p(y=2/x=1)= 0
p(x=2)*p(y=1/x=2)=0.09
p(x=3)*p(y=1/x=3)=0.06
p(x=4)*p(y=1/x=4)= 0.16 max
=> B4 = { y4 }:
luoc do giai ma toi uu:
y1----------------->x1
y2----------------->x1
y3----------------->x2
y4----------------->x4
cau 5:
de ma hoa cho 1, 2, 3, 4 thi duong nhien ta can 2 bit thong tin
trong bang ma haming ta co r1, r2, r4 la cac bit kiem tra
ta can hai bit thong tin nen cac bit do se la r3, r5
vay ta can r1 ->r5 nhu vay can it nhat la 3 bit kiem tra r1, r2, r4
cau 6:\
theo haming ta co ma tran A=
1 0 1 0 1
0 1 1 0 0
0 0 0 1 1
ta tinh cac bo ma bang cach sinh ma nhanh nhu sau:
k=2 co hai tu ma doc lap tuyen tinh
va 4 tu ma can tim
===>> r3 r5 ---------->r1 r2 r3 r4 r5
-------0--1------------------0----1
-------1--0------------------1----0
r1 = r3 + r5
r2 = r3
r4 = r5
=====>
r3-r5------r1-r2-r3-r4-r5
0--1------1--0--0--1--1
1--0------1--1--1--0--0
w1=10011;
w2=11100
w3=11111 ; w1+w2;
w0=00000
