Design (draw a datapath) for a block that will help your processor decide whether to take a branch for different conditions. As
input, your block will be given an opcode for the type of branch and signals from the ALU which perform the actual comparisons. Your block will produce a single signal branch as output.
Inputs:
• zero - 1 if R[rs1] - R[rs2] = 0, 0 otherwise
• lt - 1 if R[rs1] < R[rs2], 0 otherwise
• opcode (2b) - 00: beq (==), 01: bne (!=), 10: blt (<), 11: bge (>=)
Output:
• branch - 1 if branch should be taken, 0 if not
Inputs:
• zero - 1 if R[rs1] - R[rs2] = 0, 0 otherwise
• lt - 1 if R[rs1] < R[rs2], 0 otherwise
• opcode (2b) - 00: beq (==), 01: bne (!=), 10: blt (<), 11: bge (>=)
Output:
• branch - 1 if branch should be taken, 0 if not
1 answer:
Answer:
See explaination
Explanation:
Here are the steps taken to draw the diagram!
opcode0 and the opcode1 signals are inverted/not to determine the case whether it is 00, 01, 10, 11.
Then they are 'and'ed with the corresponding Zero/lt signal to confirm if the case is true, for eg. zero = 1 if opcode = 00 (beq).
Then all the cases are 'or'ed to get the Branch signal!
See attachment for the diagram
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Answer:
p(x,n)
1. if(n==0) [if power is 0]
2. then result =1.
3.else
4. { result=1.
5. for i=1 to n.
6. { result = result * x. } [each time we multiply x once]
7. return result.
8. }
Let's count p(3,3)
30, so come to else part.
i=1: result = result *3 = 3
i=2: result = result *3 = 9
i=2: result = result *3 = 27
Explanation:
here the for loop at step 4 takes O(n) time and other steps take constant time. So overall time complexity = O(n)