数学建模实验报告机械工程及自动化75班丁鑫四人追击问题问题：在一个边长为1的正方形跑道的四个顶点上各站有一人，他们同时开始以等速顺时针追逐下一人，在追逐过程中，每个人时刻对准目标，试模拟追击路线。

并讨论：（1）四个人能否追到一起？（2）若能追到一起，则每个人跑过多少路程？（3）追到一起所需要的时间（设速率为1）？（4）如果四个人追逐的速度不一样，情况又如何呢分析：先建立坐标系，设计程序使从A,B,C,D四个点同时出发，画出图形并判断。

程序设计流程：四个人追击的速度相等，则有。

针对这种情形，可有以下的程序。

hold onaxis([0 2 0 2]);gridA=[0,0];B=[0,1];C=[1,1];D=[1,0];k=0;s1=0;s2=0;s3=0;s4=0; %四个人分别走过的路程t=0;v=1;dt=0.002;while k<10000k=k+1;plot(A(1),A(2),'r.','markersize',15);plot(B(1),B(2),'b.','markersize',15);plot(C(1),C(2),'m.','markersize',15);plot(D(1),D(2),'k.','markersize',15);e1=B-A;d1=norm(e1);e2=C-B;d2=norm(e2);e3=D-C;d3=norm(e3);e4=A-D;d4=norm(e4);fprintf('k=%.0f ',k)fprintf('A(%.2f,%.2f) d1=%.2f ',A(1),A(2),d1)fprintf('B(%.2f,%.2f) d2=%.2f ',B(1),B(2),d2)fprintf('C(%.2f,%.2f) d3=%.2f ',C(1),C(2),d3)fprintf('D(%.2f,%.2f) d4=%.2f\n',D(1),D(2),d4) A=A+v*dt*e1/d1;B=B+v*dt*e2/d2;C=C+v*dt*e3/d3;D=D+v*dt*e4/d4;t=t+dt;s1=s1+v*dt;s2=s2+v*dt;s3=s3+v*dt;s4=s4+v*dt;if norm(A-C)<=5.0e-3&norm(B-D)<=5.0e-3 breakendendts1s2s3s4部分运行结果：k=481 A(0.52,0.52) d1=0.04 B(0.52,0.48) d2=0.04 C(0.48,0.48) d3=0.04 D(0.48,0.52) d4=0.04 k=482 A(0.52,0.52) d1=0.04 B(0.52,0.48) d2=0.04 C(0.48,0.48) d3=0.04 D(0.48,0.52) d4=0.04 k=483 A(0.52,0.52) d1=0.04 B(0.52,0.48) d2=0.04 C(0.48,0.48) d3=0.04 D(0.48,0.52) d4=0.04k=484 A(0.52,0.51) d1=0.04 B(0.51,0.48) d2=0.04 C(0.48,0.49) d3=0.04 D(0.49,0.52) d4=0.04 k=485 A(0.52,0.51) d1=0.04 B(0.51,0.48) d2=0.04 C(0.48,0.49) d3=0.04 D(0.49,0.52) d4=0.04 k=486 A(0.52,0.51) d1=0.03 B(0.51,0.48) d2=0.03 C(0.48,0.49) d3=0.03 D(0.49,0.52) d4=0.03 k=487 A(0.52,0.51) d1=0.03 B(0.51,0.48) d2=0.03 C(0.48,0.49) d3=0.03 D(0.49,0.52) d4=0.03 k=488 A(0.52,0.51) d1=0.03 B(0.51,0.48) d2=0.03 C(0.48,0.49) d3=0.03 D(0.49,0.52) d4=0.03 k=489 A(0.52,0.51) d1=0.03 B(0.51,0.48) d2=0.03 C(0.48,0.49) d3=0.03 D(0.49,0.52) d4=0.03 k=490 A(0.52,0.50) d1=0.03 B(0.50,0.48) d2=0.03 C(0.48,0.50) d3=0.03 D(0.50,0.52) d4=0.03 k=491 A(0.52,0.50) d1=0.02 B(0.50,0.48) d2=0.02 C(0.48,0.50) d3=0.02 D(0.50,0.52) d4=0.02 k=492 A(0.52,0.50) d1=0.02 B(0.50,0.48) d2=0.02 C(0.48,0.50) d3=0.02 D(0.50,0.52) d4=0.02 k=493 A(0.51,0.50) d1=0.02 B(0.50,0.49) d2=0.02 C(0.49,0.50) d3=0.02 D(0.50,0.51) d4=0.02 k=494 A(0.51,0.50) d1=0.02 B(0.50,0.49) d2=0.02 C(0.49,0.50) d3=0.02 D(0.50,0.51) d4=0.02 k=495 A(0.51,0.50) d1=0.02 B(0.50,0.49) d2=0.02 C(0.49,0.50) d3=0.02 D(0.50,0.51) d4=0.02 k=496 A(0.51,0.50) d1=0.01 B(0.50,0.49) d2=0.01 C(0.49,0.50) d3=0.01 D(0.50,0.51) d4=0.01 k=497 A(0.51,0.49) d1=0.01 B(0.49,0.49) d2=0.01 C(0.49,0.51) d3=0.01 D(0.51,0.51) d4=0.01 k=498 A(0.51,0.49) d1=0.01 B(0.49,0.49) d2=0.01 C(0.49,0.51) d3=0.01 D(0.51,0.51) d4=0.01 k=499 A(0.50,0.49) d1=0.01 B(0.49,0.50) d2=0.01 C(0.50,0.51) d3=0.01 D(0.51,0.50) d4=0.01 k=500 A(0.50,0.50) d1=0.01 B(0.50,0.50) d2=0.01 C(0.50,0.50) d3=0.01 D(0.50,0.50) d4=0.01 k=501 A(0.50,0.50) d1=0.01 B(0.50,0.50) d2=0.01 C(0.50,0.50) d3=0.01 D(0.50,0.50) d4=0.01 k=502 A(0.50,0.50) d1=0.00 B(0.50,0.50) d2=0.00 C(0.50,0.50) d3=0.00 D(0.50,0.50) d4=0.00 t =1.0040s1 =1.0040s2 =1.0040s3 =1.0040s4 =1.0040从运行的结果来看，如果四个人的追击速度相同，均为1，可有以下的结果：（1）四人最后可以追到一起。

（2）每个人跑过相等的路程，均为1.0040.（3）追到一起的时间为1.0040秒。

如果四个人追击的速度不一样，可取，运行程序。

运行程序：hold onaxis([0 2 0 2]);gridA=[0,0];B=[0,1];C=[1,1];D=[1,0];k=0;s1=0;s2=0;s3=0;s4=0; %四个人分别走过的路程t=0;v=1;dt=0.001;while k<10000k=k+1;plot(A(1),A(2),'r.','markersize',15);plot(B(1),B(2),'b.','markersize',15);plot(C(1),C(2),'m.','markersize',15);plot(D(1),D(2),'k.','markersize',15);e1=B-A;d1=norm(e1);e2=C-B;d2=norm(e2);e3=D-C;d3=norm(e3);e4=A-D;d4=norm(e4);fprintf('k=%.0f ',k)fprintf('A(%.2f,%.2f) d1=%.2f ',A(1),A(2),d1)fprintf('B(%.2f,%.2f) d2=%.2f ',B(1),B(2),d2)fprintf('C(%.2f,%.2f) d3=%.2f ',C(1),C(2),d3)fprintf('D(%.2f,%.2f) d4=%.2f\n',D(1),D(2),d4)A=A+v*dt*e1/d1;B=B+2*v*dt*e2/d2;C=C+2*v*dt*e3/d3;D=D+v*dt*e4/d4;t=t+dt;s1=s1+v*dt;s2=s2+v*dt;s3=s3+v*dt;s4=s4+v*dt;if norm(A-C)<=1.0e-3&norm(B-D)<=1.0e-3breakendendts1s2s3s4部分运行结果：k=662 A(0.41,0.32) d1=0.06 B(0.43,0.27) d2=0.00 C(0.43,0.27) d3=0.00 D(0.43,0.27) d4=0.05 k=663 A(0.41,0.32) d1=0.05 B(0.43,0.27) d2=0.00 C(0.43,0.27) d3=0.00 D(0.43,0.27) d4=0.05 k=664 A(0.41,0.32) d1=0.05 B(0.43,0.27) d2=0.00 C(0.43,0.27) d3=0.00 D(0.43,0.27) d4=0.05 k=665 A(0.41,0.32) d1=0.05 B(0.43,0.27) d2=0.00 C(0.43,0.27) d3=0.00 D(0.43,0.28) d4=0.05 k=666 A(0.41,0.31) d1=0.05 B(0.43,0.28) d2=0.00 C(0.43,0.28) d3=0.00 D(0.43,0.28) d4=0.04 k=667 A(0.41,0.31) d1=0.04 B(0.43,0.28) d2=0.00 C(0.43,0.28) d3=0.00 D(0.43,0.28) d4=0.04 k=668 A(0.41,0.31) d1=0.04 B(0.43,0.28) d2=0.00 C(0.43,0.28) d3=0.00 D(0.43,0.28) d4=0.04 k=669 A(0.41,0.31) d1=0.04 B(0.43,0.28) d2=0.00 C(0.43,0.28) d3=0.00 D(0.43,0.28) d4=0.04 k=670 A(0.41,0.31) d1=0.04 B(0.43,0.28) d2=0.00 C(0.43,0.28) d3=0.00 D(0.43,0.28) d4=0.04 k=671 A(0.41,0.31) d1=0.03 B(0.43,0.28) d2=0.00 C(0.43,0.28) d3=0.00 D(0.43,0.28) d4=0.03k=672 A(0.41,0.31) d1=0.03 B(0.43,0.28) d2=0.00 C(0.43,0.28) d3=0.00 D(0.43,0.28) d4=0.03 k=673 A(0.41,0.31) d1=0.03 B(0.43,0.28) d2=0.00 C(0.43,0.28) d3=0.00 D(0.43,0.28) d4=0.03 k=674 A(0.41,0.31) d1=0.03 B(0.43,0.28) d2=0.00 C(0.43,0.28) d3=0.00 D(0.43,0.28) d4=0.03 k=675 A(0.41,0.31) d1=0.03 B(0.43,0.28) d2=0.00 C(0.43,0.29) d3=0.00 D(0.43,0.28) d4=0.03 k=676 A(0.41,0.31) d1=0.02 B(0.43,0.29) d2=0.00 C(0.43,0.28) d3=0.00 D(0.43,0.29) d4=0.02 k=677 A(0.41,0.31) d1=0.02 B(0.43,0.28) d2=0.00 C(0.43,0.29) d3=0.00 D(0.43,0.29) d4=0.02 k=678 A(0.42,0.30) d1=0.02 B(0.43,0.29) d2=0.00 C(0.42,0.29) d3=0.00 D(0.42,0.29) d4=0.02 k=679 A(0.42,0.30) d1=0.02 B(0.42,0.29) d2=0.00 C(0.42,0.29) d3=0.00 D(0.42,0.29) d4=0.02 k=680 A(0.42,0.30) d1=0.02 B(0.42,0.29) d2=0.00 C(0.42,0.29) d3=0.00 D(0.42,0.29) d4=0.02 k=681 A(0.42,0.30) d1=0.02 B(0.42,0.29) d2=0.00 C(0.42,0.29) d3=0.00 D(0.42,0.29) d4=0.01 k=682 A(0.42,0.30) d1=0.01 B(0.42,0.29) d2=0.00 C(0.42,0.29) d3=0.00 D(0.42,0.29) d4=0.01 k=683 A(0.42,0.30) d1=0.01 B(0.42,0.29) d2=0.00 C(0.42,0.29) d3=0.00 D(0.42,0.29) d4=0.01 k=684 A(0.42,0.30) d1=0.01 B(0.42,0.29) d2=0.00 C(0.42,0.29) d3=0.00 D(0.42,0.29) d4=0.01 k=685 A(0.42,0.30) d1=0.01 B(0.42,0.29) d2=0.00 C(0.42,0.29) d3=0.00 D(0.42,0.29) d4=0.01 k=686 A(0.42,0.30) d1=0.01 B(0.42,0.29) d2=0.00 C(0.42,0.29) d3=0.00 D(0.42,0.29) d4=0.00t =0.6860s1 =0.6860s2 =1.3720s3 =1.3720s4 =0.6860从运行的结果来看，如果四个人的追击速度不同，可有以下的结果：（1）四人最后可以追到一起。