- input a
- input b
- internal c = a and b
- internal d = b and e
- output e = c or d
The space to the left is available for any
logic diagrams, truth tables or other tools
you might use in answering the following:
(0.5 points each)
| a | b | c | d | e | ||
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | ||
| 0 | 1 | 0 | 0/1 | 0/1 | ||
| 1 | 0 | 0 | 0 | 0 | ||
| 1 | 1 | 1 | 1 | 1 | ||
A = ____0____ B = ____1____
b)
How do you set this flipflop?
A = ____1____ B = ____1____
c)
How do you reset this flipflop?
A = ____0____ B = ____0____
7 did the problem perfectly, 5 more got part a) right, 3 more got part b) right, and 2 more got part c) right.
9 students gave essentially the logic diagram above, 5 gave variant logic diagram with more "flipfloppy" layout. 13 gave truth tables. Only one student earned no credit on this problem.
_ ___ _ s - 1 bit, the sign of the signed-magnitude mantissa
|_|___|_| e - 2 bits, the exponent, biased so 10 = 0
|s| e |m| note: e = 00 means the mantissa is not normalized.
m - 1 bit plus 1 hidden bit, the magnitude of the mantissa,
the point is between the hidden bit and the m bit.
Give decimal equivalents of all 16 floating point values below.
Feel free to use the space to the right as scratch space. (2.5 points)
0000 = __0.0____
1000 = _-0.0_____
9 had trouble with zero
0001 = __0.25___
1001 = _-0.25____
16 had trouble with this non-normalized value
0010 = __0.5____
1010 = _-0.5_____
0011 = __0.75___
1011 = _-0.75____
0100 = __1.0____
1100 = _-1.0_____
0101 = __1.5____
1101 = _-1.5_____
0110 = __2.0____
1110 = _-2.0_____
0111 = __3.0____
1111 = _-3.0_____
2 did perfectly, 4 earned no credit at all. 2 seriously misinterpreted the sign bit.
; best solution, based on 165 = 5 x 33 -- base score 3.0
ADDSL R3,R3,2 ; R3 times 5
ADDSL R3,R3,5 ; R3 times 33
; next best solution, based on 165 = 10100101 -- base score 2.5
MOVE R1,R3
ADDSL R3,R1,2 ; 101
ADDSL R3,R1,3 ; 101001
ADDSL R3,R1,2 ; 10100101
; brute force solution, based on 165 = 3 x 5 x 11 -- base score 2.0
ADDSL R3,R3,1 ; R3 times 3
ADDSL R3,R3,2 ; R3 times 5
MOVE R1,R3 ; \
ADDSL R3,R1,2 ; | R3 times 11 = 1011
ADDSL R3,R1,1 ; /
6 gave the optimal solution, one more tried but made an error. 1 gave the next best solution, anf 4 more tried but made errors. 1 gave the brute-force solutioin and 7 made errors in trying this. There were several ad-hoc solutions tried, and a few tried to give general purpose multiply algorithms to multiply by a constant. Most efforts in this direction were peppered with errors. Of the 4 who earned no credit at all, 3 wrote considerable meaningless code.
FIRST: TBIT R4,0 ; 1
BCR A ; 2
SL R3,1
A: TBIT R4,1 ; 3
BCR B ; 4
SL R3,2
B: TBIT R4,2 ; 5
BCR C ; 6
SL R3,4
C: TBIT R4,3 ; 7
BCR D ; 8
SL R3,8
D:
SECOND: NEG R4,R4 ; 1
ADDI R4,R4,15 ; 2
BTRUNC R4,2 ; 3
SL R3,1
SL R3,1
SL R3,1
SR R4,2 ; 4
BTRUNC R4,2 ; 5
SL R3,4
SL R3,4
SL R3,4
a)
For random initial values of R4 from 0 to 15, how many instructions
would you expect to be executed in an average run through each?
(1 point)
First: __10_______________________
Second: ___8_______________________
5 did perfectly, itis difficult to classify wrong answers, but several earned partial credit for off-by one errors and several more gave the minimum number of instructions executed for the first solution while giving the maximum number of instructions for the second solution. 3 earned no credit at all.
The method that works here is to count the minimum number of instructions that could be executed in a pass through the code (the comments in the code give this count), and then count the total number of instructions. Take the average of these two numbers to get the average number of instructions.
b)
The first fragment given above uses base 2, while the second fragment
uses base 4. For random initial values of R4 from 0 to 15, how many
instructions would you expect to be executed in an average run through
the base 16 solution?
(1 point)
__10.5____________________________
Just 1 got full credit here, 2 more had small penalties for off-by-one errors, while a few got a bit of credit for off by 1 errors on the minimum or maximum. 14 earned no credit at all. Some gave instruction counts far higher than the number of instructions in the code, impossible answers in this context, since the code contains no loops.
The base 16 solution is shown below, with minimum instruction counts.
BASE16: NEG R4,R4 ; 1
ADDI R4,R4,15 ; 2
BTRUNC R4,4 ; 3
SL R3,1
SL R3,1
SL R3,1
SL R3,1
SL R3,1
SL R3,1
SL R3,1
SL R3,1
SL R3,1
SL R3,1
SL R3,1
SL R3,1
SL R3,1
SL R3,1
SL R3,1
c)
What does it do? In a language such as C or Java, you can express the
correct answer in one very simple assignment statement.
(1 point)
___R3=R3<<R4_____________________
3 did well. 5 more guessed that it was a left shift but were confused about how far it shifts. 3 got partial credit for indicating the right operands with a multiply or exponentiate operation, since those change the operand in the right way but by the wrong amount. 8 got no credit at all.