《信息论与编码理论》(王育民 李晖 梁传甲)课后习题答案 高等教育出版社

信息论与编码理论习题解第二章-信息量和熵2.1解:平均每个符号长为:1544.0312.032秒每个符号的熵为9183.03log3123log32比特/符号所以信息速率为444.34159183.0比特/秒2.2解:同步信号均相同不含信息,其余认为等概,每个码字的信息量为3*2=6比特；所以信息速率为600010006比特/秒2.3解:(a)一对骰子总点数为7的概率是366所以得到的信息量为585.2)366(log2比特(b)一对骰子总点数为12的概率是361所以得到的信息量为17.5361log2比特2.4解:(a)任一特定排列的概率为!521,所以给出的信息量为58.225!521log2比特(b)从中任取13张牌,所给出的点数都不相同的概率为13521313521344!13CA所以得到的信息量为21.134log1313522C比特.2.5解:易证每次出现i点的概率为21i,所以比特比特比特比特比特比特比特398.221log21)(807.1)6(070.2)5(392.2)4(807.2)3(392.3)2(392.4)1(6,5,4,3,2,1,21log)(2612iiXHxIxIxIxIxIxIiiixIi2.6解:可能有的排列总数为27720!5!4!3!12没有两棵梧桐树相邻的排列数可如下图求得，YXYXYXYXYXYXYXY图中X表示白杨或白桦，它有37种排法，Y表示梧桐树可以栽种的位置，它有58种排法，所以共有58*37=1960种排法保证没有两棵梧桐树相邻，因此若告诉你没有两棵梧桐树相邻时，得到关于树排列的信息为1960log27720log22=3.822比特2.7解:X=0表示未录取，X=1表示录取；Y=0表示本市，Y=1表示外地；Z=0表示学过英语，Z=1表示未学过英语，由此得比特比特比特比特6017.02log21412log2141910log1094310log10143)11(log)11()1()10(log)10()1()01(log)01()0()00(log)00()0()(8113.04log4134log43)()(02698.04110435log104354310469log10469)1()01(log)01()0()00(log)00()0;(104352513/41)522121()0(/)1())11()1,10()10()1,00(()01(104692513/43)104109101()0(/)0())01()0,10()00()0,00(()00()(4512.04185log854383log83)1()01(log)01()0()00(log)00()0;(8551/4121)0(/)1()10()01(8351/43101)0(/)0()00()00()(,251225131)1(,2513100405451)10()1()00()0()0(,54511)1(,51101432141)10()1()00()0()0(,41)1(,43)0(222222222222222222xypxypxpxypxypxpxypxypxpxypxypxpXYHXHcxpzxpzxpxpzxpzxpzXIzpxpxypxyzpxypxyzpzxpzpxpxypxyzpxypxyzpzxpbxpyxpyxpxpyxpyxpyXIypxpxypyxpypxpxypyxpazpyzpypyzpypzpypxypxpxypxpypxpxp2.8解:令RFTYBAX,,,,，则比特得令同理03645.0)()(5.0,02.03.0)2.05.0(log2.0)()2.05.0(log)2.05.0()2.03.0(log)2.03.0(5.0log5.03.0log3.0)5log)1(2.02log)1(5.0log)1(3.05log2.0log3.02log5.0(2.0log2.0)2.05.0(log)2.05.0()2.03.0(log)2.03.0()()();()(2.0)(,2.05.0)(2.03.0)1(3.05.0)()()()()(5.0max2'2222223102231022222ppIpIppppIppppppppppppppXYHYHYXIpIRPpFPpppBPBTPAPATPTP2.9&2.12解:令X=X1,Y=X1+X2,Z=X1+X2+X3,H(X1)=H(X2)=H(X3)=6log2比特H(X)=H(X1)=6log2=2.585比特H(Y)=H(X2+X3)=6log61)536log365436log364336log363236log36236log361(2222222=3.2744比特H(Z)=H(X1+X2+X3)=)27216log2162725216log2162521216log2162115216log2161510216log216106216log21663216log2163216log2161(222222222=3.5993比特所以H(Z/Y)=H(X3)=2.585比特H(Z/X)=H(X2+X3)=3.2744比特H(X/Y)=H(X)-H(Y)+H(Y/X)=2.585-3.2744+2.585=1.8955比特H(Z/XY)=H(Z/Y)=2.585比特H(XZ/Y)=H(X/Y)+H(Z/XY)=1.8955+2.585=4.4805比特I(Y;Z)=H(Z)-H(Z/Y)=H(Z)-H(X3)=3.5993-2.585=1.0143比特I(X;Z)=H(Z)-H(Z/X)=3.5993-3.2744=0.3249比特I(XY;Z)=H(Z)-H(Z/XY)=H(Z)-H(Z/Y)=1.0143比特I(Y;Z/X)=H(Z/X)-H(Z/XY)=H(X2+X3)-H(X3)=3.2744-2.585=0.6894比特I(X;Z/Y)=H(Z/Y)-H(Z/XY)=H(Z/Y)-H(Z/Y)=02.10解:设系统输出10个数字X等概,接收数字为Y,显然101)(101)()()(9190ijpijpiQjwiiH(Y)=log10比特奇奇奇奇偶18log81101452log211015)(log)()()(log)()(0)(log),()(log),()(22,2222xypxypxpxxpxxpxpxypyxpxypyxpXYHxyxiyxyx所以I(X;Y)=3219.2110log2比特2.11解:（a）接收前一个数字为0的概率2180)0()()0(iiiupuqwbitsppwupuI)1(log11log)0()0(log)0;(2212121（b）同理4180)00()()00(iiiupuqwbitsppwupuI)1(log22)1(log)00()00(log)00;(24122121（c）同理8180)000()()000(iiiupuqwbitsppwupuI)1(log33)1(log)000()000(log)000;(28132121（d）同理))1(6)1(()0000()()0000(42268180ppppupuqwiiibitsppppppppppwupuI42264242268142121)1(6)1()1(8log))1(6)1(()1(log)0000()0000(log)0000;(2.12解:见2.92.13解:(b))/()/()/(1log)()/(1log)()/()/(1log)()/(1log)()/(XYZHXYHxyzpxyzpxypxyzpxyzpxypxyzpxyzpxyzpXYZHxyzxyzxyzxyz(c))/()/(1log)/()()/(1log)/()()/(XZHxzpxyzpxypxyzpxyzpxypXYZHxyzxyz（由第二基本不等式）或0)1)/()/((log)/()()/()/(log)/()()/(1log)/()()/(1log)/()()/()/(xyzpxzpexyzpxypxyzpxzpxyzpxypxzpxyzpxypxyzpxyzpxypXZHXYZHxyzxyzxyzxyz（由第一基本不等式）所以)/()/(XZHXYZH(a))/()/()/()/()/(XYZHXYZHXYHXZHXYH等号成立的条件为)/()/(xzpxyzp,对所有ZzYyXx,,,即在给定X条件下Y与Z相互独立。2.14解:(a))/()/()/()/()/()/(ZXHZXYHZYHYZXHZYHYXH(b))()/()()/()/()()/()/()/()/()()/()()/(0)(,0)/()/()/()()/()/()/()/()/()()/()/()/()/()()/()/()/()/()()/()/()/()()/()/()()/()/()()/()()/()()/(XZHZXHZHZXHZXHZHZYHYXHZYHYXHYZHZYHXYHYXHZHZXHZYHYXHZHZYHYXHZYHYXHYXHYZHZYHYXHYZHYXHYHZYHYXHYXHYZHYHZYHYZHYXHYHYXHYZHYHZYHYXHYHYXHYZHZYHXYHYXH注：baabaaaabaaababababaa221121221121210,02.15解:(a)0)/()/(),(0)/()/(),(XYHYXHYXdXXHXXHXXd(b)),()/()/()/()/(),(XYdYXHXYHXYHYXHYXd(c)),()/()/(),(),()/()/()/()/()/()/()/()/()/()/()/()/()/(),(),(ZXdXZHZXHZYdYXdXZHXYHYZHZXHZXYHZYHYZXHZYHYXHYZHZYHXYHYXHZYdYXd同理2.16解:(a)1)(),(),()()/()/()()()()()()(),(XYHYXIYXSXYHXYHYXHXYHYHXHXYHYHXHYXI又由互信息的非负性，即0);(YXI有0);(YXS，所以1);(0YXS(b)1)()()()/()()(),(),(XHXHXXHXXHXHXXHXXIXXS(c)当且仅当X和Y独立时，I（X；Y）=0，所以当且仅当X和Y独立时，0)(),(),(XYHYXIYXS。2.23解:(a)其它,011,)(21xxpX比特1log)(211121dxXHC(b)令ydydxxy21,2其它,01,21)(yyypY比特443.0log121log2