Consider the following code, and write the sequence of statements (Si) that will be executed for the values of (x,y) = (0,0), (0,1), (1,0), (1,1), (-1,-1)
S1 if x==0 and y==0:
S2 elif x>0 and y>0:
S3 elif x>0 or y>0:
S4
else:
S5
S6
If (x,y) = (0,0): ----------------- S1, S2, S6 If (x,y) = (0,1): ----------------- S1, S4, S6 If (x,y) = (1,0): ----------------- S1, S4, S6 If (x,y) = (1,1): ----------------- S1, S3, S6 If (x,y) = (-1,-1): ----------------- S1, S5, S6
Consider the following code, and write the sequence of statements (Si) that will be executed for...
A sequential circuit has one input (X), a clock input (CLK), and
two outputs (S and V). X, S and V are all one-bit signals. X
represents a 4-bit binary number N, which is input least
significant bit first. S represents a 4-bit binary number equal to
N + 3, which is output least significant bit first. At the time the
fourth input occurs, V = 1 if N + 3 is too large to be represented
by 4 bits;...
For Python 3.7+.
TEST THE CODE WITH THE FOLLOWING GLOBAL CODE AFTER
YOU'RE DONE:
print('\nStart of A2 Student class demo ')
s1 = Student('David Miller', major = 'Hist',enrolled = 'y',
credits = 0, qpoints = 0)
s2 = Student('Sonia Fillmore', major = 'Math',enrolled = 'y',
credits = 90, qpoints = 315)
s3 = Student('A. Einstein', major = 'Phys',enrolled = 'y', credits
= 0, qpoints =
0)
s4 = Student('W. A. Mozart', major = 'Mus',enrolled = 'n', credits
= 29, qpoints...
Draw a stack diagram to show how the following code is executed and write the generated output. def sequence(n, m): if n < 0: return 1 elif n + m > 5: return sequence(n, m - 1) else: return n * sequence(m - 1, n + 1) print(sequence(5, 3))
Consider the following state diagram, which items on the state table is correct for the switch between states and output values. (Fig. 31) So S1 YO S Sz %0 Ss S2 S4 0 So Next state Z2Z1 Current state A. B. C. D. S4 S5 S6 S7 X=0 54 S5 S5 S5 X=1 X=0 S1 01 S6 00 S7 01 S4 10 Fig. 31 X=1 10 00 00 00 A. Line A on the table OB Line B on the...
Draw a stack diagram to show how the following code is executed and write the generated output. def sequence(n, m): if n < 0: return 1 elif n + m > 5: return sequence(n, m - 1) else: return n * sequence(m - 1, n + 1) print(sequence(5, 3))
Show how the following four instructions will be executed within the MIPS pipeline. Also, show the forwarding paths needed. Use the graphical notation showing all stages of MIPS pipeline. Indicate all data dependencies. Which dependencies are data hazards that will be resolved via forwarding? Which dependencies are data hazards that will cause a stall? add $s3, $s4, $s6 sub $s5, $s5, $s2 lw $s7, 100 ($s5) add $s8, $s7, $s2
Please solve step by step.
Find the following:
a. lambda for s3
b. lambda for s4
c. lambda for s5
d. lambda for s6
-1 A,-60 /hr S1 S4 C3 -3 0.7 μ,-70 /hr 70 /hr 4 S3 ,3 0.8 35 / hr S5 -30 /hr S2 Hs- 20 /hr 0.2 2-45/hr S6 10 / hr
-1 A,-60 /hr S1 S4 C3 -3 0.7 μ,-70 /hr 70 /hr 4 S3 ,3 0.8 35 / hr S5 -30 /hr S2 Hs-...
Translate each of the following pseudo-instructions into MIPS instructions. You should Produce a minimal sequence of MIPS instructions to accomplish the required computation. (8 Points) 1) bgt $t1, 100, Label # bgt means branch if greater than 2) ble $s2, 10, Next # ble means branch if less than or equal 3) ror $s0, $s4, 7 # ror means rotate right $s4 by 7 bits and store the result in $s0 4) neg $s5, $s4 # $s5 will have the...
Consider the following code to be executed on a pipelined processor lw $s1, 40(Ss6) add $s6, $s2, $s2 sw Ss6, 48(Ss1) a. Include stalls/nops in the code so it executes correctly in the cases of (i) No forwarding (ii) ALU-ALU for warding, (iii) Full forwarding b. In each case calculate the number of clock cycles required to execute the code c. Assume further that the clock cycle time is 110 ps with no forwarding, 120 ps with ALU-ALU forwarding and...
java
a. What is printed by the following code? Explain your answer by showing how you computed the values of pos, s1, s2 and 53. String str = "2900 Bedford Ave Brooklyn NY 12230-6843"; int pos = str.indexOf(" ", str.indexOf(" ", 'B') + 1); String si = str.substring(o, pos); System.out.println("pos: + post sl: + sl); pos - str.indexOf("B", pos+1); pos = str.indexOf("B", pos+1); String s2 = str.substring(pos, str. indexOf(" ", pos)): System.out.println("pos: + pos + + s2); String s3 =...