10:00 12:30. Do not open this problem booklet until the start of the examination is announced.

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1 0 I :00 1:30 (1),. Do not open this problem booklet until the start of the examination is announced. () 4.. Answer the following 4 problems. Use the designated answer sheet for each problem. (3). Do not take the answer sheets and the problem booklet out of the examination room.. Fill the following blank with your examinee s number. No.

2 1 3 3 z 3 O z (0, 0,z 0 ) z 0 3 P 1, P, P 3 (x 1,y 1,z 1 ), (x,y,z ), (x 3,y 3,z 3 ) P1 0, P 0, P 3 0 (x0,y 0,z 0 ) (x 0,y 0 ) (1) P 0 1, P 0, P 0 3 (x0 1,y0 1 ), (x0,y0 ), (x0 3,y0 3 ) () P 0 (x 0,y 0 ) 4P1 0P 0P 3 0 P 0 (3) 3 4P 1 P P 3 4P 1 P P 3 (4) P 0 4P1 0P 0P 3 0 P 0 3 4P 1 P P 3 P P 0 P (5) P P 1 P P P 0 P1 0 P 0 P 4P 1 P P 3 P P 0 4P1 0P 0P 3 0 4P 1 P P 3 O

3 Problem 1 Answer the following questions on the projection of objects in a three dimensional space onto a two dimensional screen. For all triangles in the following, assume that each has three vertices that are not on a line, and that each vertex has a positive z coordinate. Suppose that the projection center (viewpoint) is at the origin O of the three dimensional coordinate system, and the projection plane (screen) is perpendicular to the z-axis and intersects with the z-axis at the coordinates (0, 0,z 0 ), where z 0 is a positive constant. Let P 1, P,andP 3 be the three vertices of a triangle and have the coordinates (x 1,y 1,z 1 ), (x,y,z ), and (x 3,y 3,z 3 ), respectively. The points P 0 1, P 0,andP 0 3 are de ned to be the projections of P 1, P,andP 3, respectively, onto the projection plane. To a projected point (x 0,y 0,z 0 ), coordinates (x 0,y 0 )of the two dimensional projection plane coordinate system are associated. (See the gure.) (1) Calculate the projection plane coordinates P 0 1 (x0 1,y0 1 ), P 0 (x0,y0 ), and P 0 3 (x0 3,y0 3 ). () Assume that a point P 0 (x 0,y 0 ) on the projection plane is given. Find a method to determine whetherornotthepointp 0 lies inside the triangle 4P 0 1 P 0 P 0 3. Assume that if P 0 coincides with one of the vertices or on an edge of the triangle, then it is not regarded as inside the triangle. (3) Calculate a normal vector of the triangle 4P 1 P P 3 in the three dimensional coordinate system, and give an equation of the plane that contains the triangle 4P 1 P P 3. (4) Suppose that the point P 0 on 4P 0 1 P 0 P 0 3 is obtained by projecting a point P on 4P 1P P 3 onto the screen. Write the coordinates of P using the coordinates of P 0. (5) Next assume that P isthemiddlepointofp 1 and P. Derive the condition for P 0,which is the projection of P, being the middle point of P1 0 and P 0, and explain its geometric interpretation. Also derive the condition for P 0 being the center of mass of 4P1 0P 0P 3 0 when P is the center of mass of 4P 1 P P 3. Assume that 4P 1 P P 3 and O are not on a plane. 3

4 N 1 L N L N = {a 1 a n {0, 1} n N, a n N+1 =1} (1) L 3 4 () L 3 8 (3) L N N 4

5 Problem For an integer N 1, consider the following language L N L N = {a 1 a n {0, 1} n N, a n N+1 =1}. Answer the following questions. (1) Construct a nondeterministic nite automaton with 4 states which accepts L 3. () Construct a deterministic nite automaton with 8 states which accepts L 3. (3) Prove that any deterministic nite automaton which accepts L N has at least N states. 5

6 3 n X = {x 1,x,...,x n } k X X N 1: Select(k, X) { : if ( X <N) { 3: return X k ; 4: } 5: X 5 m = d X 5 e X 1,X,...,X m ; 6: /* i<m X i =5 X m 5 */ 7: for (i =1tom) { 8: y i Select(d X i e, X i); 9: } 10: a Select(d m e, {y 1,y,...,y m }); 11: X small X a ; 1: X large X a ; 13: if ( X small = k 1) { 14: return a; 15: } 16: if ( X small k) { 17: return Select(k, X small ); 18: } 19: else { 0: return Select(k X small 1, X large ); 1: } : } (1) 3 () N X X small 3 4 X (3) T (n) T (n) n = X (4) T (n) O(n) 6

7 Problem 3 For a sequence of n di erent integers X = {x 1,x,...,x n }, consider the following algorithm that computes the k-th smallest integer in the sequence. X denotes the number of elements in X. N is a positive constant. 1: Select(k, X) { : if ( X <N) { 3: return the k-th smallest integer in X computed by using a heap; 4: } 5: Divide X into m = d X 5 e sequences of 5 or less integers, and let them be X 1,X,...,X m ; 6: /* We let X i =5fori<m. X m can be smaller than 5. */ 7: for (i =1tom) { 8: y i Select(d X i e, X i); 9: } 10: a Select(d m e, {y 1,y,...,y m }); 11: X small the sequence of the integers in X that are smaller than a; 1: X large the sequence of the integers in X that are larger than a; 13: if ( X small = k 1) { 14: return a; 15: } 16: if ( X small k) { 17: return Select(k, X small ); 18: } 19: else { 0: return Select(k X small 1, X large ); 1: } : } Answer the following questions on the above pseudo code. (1) Describe what a heap mentioned in line 3 is. () Prove that 1 4 X X small 3 4 X after line 11 when the constant N is large enough. (3) Let the worst case computational complexity of the above algorithm be T (n). Show a recurrence equation of T (n). (Note that n = X.) (4) Show that T (n) is O(n). 7

8 4 10 Mbps (1) D- AND OR NOT () 10 Mbps 3 (3) () 8

9 Problem 4 Manchester code is used as an encoding method of data in 10 Mbps Ethernet. As shown in the gure below, data 1 is expressed by a 0 1 transition, and data 0 is expressed by a 1 0 transition. Transitions that signify data are generated with regular time intervals as shown in the arrow marks of the gure. (1) Design a Manchester encoder, using D ip- ops, AND, OR, and NOT gates, whose input data is synchronized with a clock. () Explain at least two reasons to use Manchester code in 10 Mbps Ethernet. The answer should be within three lines. (3) Manchester code requires the network bandwidth twice as wide as the data transfer bandwidth. Discuss a coding method that utilizes the network bandwidth better, without much degrading the advantages of Manchester code answered in question (). 9

10 0 II :30 16:00 (1),. Do not open this problem booklet until the start of the examination is announced. () 4.. Answer the following 4 problems. Use the designated answer sheet for each problem. (3). Do not take the answer sheets and the problem booklet out of the examination room.. Fill the following blank with your examinee s number. No.

11 1 N 4 =exp μ i N 1 N i (z 0,z 1,...,z N 1 ) N (Z 0,Z 1,...,Z N 1 ) Z k = N 1 X j=0 z j jk (k =0, 1,..., N 1) (1) z j 0 Z 0 Z N k =1,,..., N 1 Z k = Z N k z z () z j g j = z j + iz j+1 ³j =0, 1,..., N 1 (g 0,g 1,...,gN 1) N (G 0,G 1,...,GN 1) N X 1 G k = g j ³k jk N =0, 1,..., 1 Z 0 Z N j=0 G 0 ( ) (3) () k = 1,,..., N 4 1 Z k Z N k G k G N k

12 Problem 1 Let N be a positive integer that is a multiple of 4, and μ i =exp N be the primitive Nth root of 1 (i is the imaginary unit). For a sequence of complex numbers (z 0,z 1,...,z N 1 ), its N-point discrete Fourier transform (Z 0,Z 1,...,Z N 1 )isde nedas Z k = N 1 X j=0 Answer the following questions. z j jk (k =0, 1,..., N 1). (1) Assume that all z j are real (that is, their imaginary parts are zeros). Show that Z 0 and Z N are real numbers. Also show that Z k = Z N k for k = 1,,..., N 1. ( z is the complex conjugate of z.) () Assume that all z j are real numbers. Let g j = z j + iz j+1 ³j =0, 1,..., N 1 and (G 0,G 1,...,GN 1) bethe N -point discrete Fourier transform of (g 0,g 1,...,gN 1), that is, Express Z 0 and Z N be used.) G k = N 1 X g j ³k jk =0, 1,..., j=0 N 1. in terms of G 0. (Complex conjugates or the real/imaginary parts may (3) Express Z k and Z N k in terms of G k and G N k for k =1,,..., N 4 assumptions as question (). 1, under the same 3

13 c + E w E w c c E 1 w 1 E w E 1 + E w 1 w + E 1 w 1 E w E 1 E w 1 w E eval(e) w s (w, s) (w 1,s 1 ) (w,s ) (w 1,s 1 ) (w,s ) c v c (cw, s) (w, v c s) v = eval(v 1 + v ) (+w, v v 1 s) (w, vs) v = eval(v 1 v ) ( w, v v 1 s) (w, vs) eval(e) =v E w (w, ²) (², v) ² (1) () (3) 4

14 Problem Consider translating an expression made of integer constants (denoted by c below) and binary operators + and to a string in post x notation as follows, where E w means that expression E is translated to string w. c c E 1 w 1 and E w imply E 1 + E w 1 w + E 1 w 1 and E w imply E 1 E w 1 w Let us write eval(e) for the value resulting from evaluating an expression E by usual integer arithmetics. We then introduce the following rules for rewriting a pair (w, s) ofastringw in post x notation and a sequence (stack) s of values, where (w 1,s 1 ) (w,s ) means that pair (w 1,s 1 ) is rewritten into pair (w,s ). if the value of c is v c then (cw, s) (w, v c s) v = eval(v 1 + v )implies(+w, v v 1 s) (w, vs) v = eval(v 1 v )implies( w, v v 1 s) (w, vs) The transitive closure of is denoted by. We now want to prove a proposition that if eval(e) =v and E w hold, then (w, ²) (², v) holds, where ² denotes an empty sequence or an empty string. Answer the following questions. (1) To prove the proposition, should we use induction on what? () In order for induction to go through, we should generalize the proposition. How should the proposition be generalized? (3) Prove the generalized proposition by induction. 5

15 3 U (0 ) U U =(v 1,v,v 3,v 4,v 5 ) (v 1,v 3,v 4 ) U U i U[i] U U f(u, V ) U = V U, V U X f(u, V )= {dist(u[i],v[i]) c} i=1 dist(v,w) v w c U, V U 0 = V 0 U U 0 V V 0 f(u 0,V 0 ) min(u, V ) Umin 0, V min 0 min(u, V ) = min{f(u 0,V 0 ) U 0 U, V 0 V, U 0 = V 0 } = f(umin,v 0 min) 0 U 0 U U 0 U k = Umin 0 = V min 0 1 ` k U 0 min[`] =U[i`], V 0 min[`] =V [j`] i`, j` min(u, V ) = f (U 0 min[1],u 0 min[],...,u 0 min[k]), (V 0 min[1],v 0 min[],...,v 0 min[k]) = f((u[i 1 ],U[i ],...,U[i k ]), (V [j 1 ],V[j ],...,V[j k ])) i 1 <i < <i k j 1 <j < <j k (1) U, V, c min(u, V ) (i 1,i,...,i k ) (j 1,j,...,j k ) U = ((0, ), (1, ), (, ), (3, ), (3, 3), (4, 3), (4, )) V = ((1, 1), (, 1), (, 0), (3, 0), (3, 1), (4, 1), (5, 1)) c = 1. 1 ` k dist(umin 0 [`],V0 min [`]) c () U i j U[i..j] 1 ` k ` min(u[1..i`],v[1..j`]) = f ((U[i 1 ],U[i ],...,U[i`]), (V [j 1 ],V[j ],...,V[j`])) (3) min(u, V ) 6

16 Problem 3 Let U be a sequence of points in two dimensional space, and let us call subsequence of U a sequence of points that can be constructed from U by removing some (more than or equal to 0) points. For example, (v 1,v 3,v 4 ) is a subsequence of U, whereu =(v 1,v,v 3,v 4,v 5 ). Let U[i] denote the i-th point of the sequence U, andlet U denote the number of points in U. For two point sequences U and V such that U = V, we consider the following function. U X f(u, V )= {dist(u[i],v[i]) c}, i=1 where dist(v, w) denotes the Euclidean distance between v and w, and c is a real positive constant. For given U and V (possibly U 6= V ), let min(u, V ) be the minimum of f(u 0,V 0 )whereu 0 and V 0 are subsequences of U and V, respectively, such that U 0 = V 0.LetU 0 min and V 0 min be the subsequences that give the minimum. That is, min(u, V ) = min{f(u 0,V 0 ) U 0 U, V 0 V, U 0 = V 0 } = f(umin,v 0 min), 0 where U 0 U means that U 0 is a subsequence of U. Let k = U 0 min = V 0 min and determine i` and j` for 1 ` k as U 0 min[`] =U[i`], V 0 min[`] =V [j`]. Thus we have min(u, V ) = f (U 0 min[1],u 0 min[],...,u 0 min[k]), (V 0 min[1],v 0 min[],...,v 0 min[k]) = f((u[i 1 ],U[i ],...,U[i k ]), (V [j 1 ],V[j ],...,V[j k ])). Here assume that i 1 <i < <i k and j 1 <j < <j k. Answer the following questions. (1) Compute min(u, V ), (i 1,i,...,i k ), and (j 1,j,...,j k )when U = ((0, ), (1, ), (, ), (3, ), (3, 3), (4, 3), (4, )), V = ((1, 1), (, 1), (, 0), (3, 0), (3, 1), (4, 1), (5, 1)), c = 1.. Note that dist(umin 0 [`],V0 min [`]) c holds for 1 ` k. () Let U[i..j] denote the subsequence from the i-th point to the j-th point of U. Show that the following equation holds for an arbitrary integer ` such that 1 ` k: min(u[1..i`],v[1..j`]) = f ((U[i 1 ],U[i ],...,U[i`]), (V [j 1 ],V[j ],...,V[j`])). (3) Describe an algorithm for computing min(u, V ) and its computational complexity. 7

17 byte/sec 10 msec Sequence Number Packet Type Data Size data (3 Bytes) (1 Byte) ( Bytes) Packet Type: 1 DATA 0 ACK DATA ACK ACK Sequence Number Sequence Number Sequence Number DATA ACK DATA Sequence Number (modulo 4 )1 (1) () (3) (4) (a) (b) 8

18 Problem 4 We assume that two computers are connected by a network of data-link layer whose packet size is xed as 1000 bytes. We also assume that the bandwidth and the latency of this data-link layer network are 10 6 byte/sec and 10 msec, respectively. No packet ow controlling mechanism is supported, and packet arrival is not guaranteed. The following is an un nished design of a network protocol on top of this data-link layer. Packet Format Sequence Number Packet Type Data Size data (3 Bytes) (1 Byte) ( Bytes) Packet Type: 1 DATA 0 ACK Processing in the receiver When a DATA packet is received, an ACK packet, whose Sequence Number is the same as the received DATA packet, is sent back to the sender. Processing in the sender The initial Sequence Number is randomly chosen. After sending a DATA packet, the sender waits for the ACK packet of the DATA packet, and the sender sends the next DATA packet after the receipt of the ACK packet. The Sequence Number is incremented by 1 (modulo 4 )ateachpacketsending. Answer the following questions. (1) Calculate the maximum throughput when no packet is lost. () Modify the protocol so as to consider packet loss. The protocol may be written in a pseudo code. (3) Modify the protocol so as to utilize the data-link network bandwidth better. (4) Assume that 64 computers are connected with a network of 8 8 mesh topology using the same data-link layer as above. Design a new network protocol based on the above protocol so that all the computers may communicate each other. Explain the new protocol from the following points of view. (a) Packet format. (b) Sending, receiving, and routing on each computer. 9