H***F
发帖数: 2501 | 1 suppose we have a system with bits coming serially. the output is 1 only if
the number at that cycle is divisible by 7. creat a state diagram. any
ideas?
1--->next bit
10=2--->next bit
101=5--->next bit
1011=11--> | l*****x
发帖数: 3431 | 2 寻找"111"string?
if
【在 H***F 的大作中提到】
: suppose we have a system with bits coming serially. the output is 1 only if
: the number at that cycle is divisible by 7. creat a state diagram. any
: ideas?
: 1--->next bit
: 10=2--->next bit
: 101=5--->next bit
: 1011=11-->
| H***F
发帖数: 2501 | 3 不是。111可以被7除，10101=21也可以被7除。。
寻找"111"string?
if
【在 l*****x 的大作中提到】
: 寻找"111"string?
:
: if
| b*****e
发帖数: 1193 | 4 what's the cycle?
4 bits? | H***F
发帖数: 2501 | 5 不停，一直下去。后面的补上来称为新的一位。
【在 b*****e 的大作中提到】
: what's the cycle?
: 4 bits?
| e***y
发帖数: 4307 | 6 states:
start, s1, s2, s3, s4, s5, s6, s7
state transitions (input, next state)
start: (1, s1), (0, start)
s1: (1, s3), (0, s2)
s2: (1, s3), (0, s6)
s3: (1, s7), (0, s6)
s4: (1, s5), (0, s1)
s5: (1, s4), (0, s3)
s6: (1, s6), (0, s5)
s7: (1, s1), (0, s7) ==> output 1 at s7
basically the idea is to represent the remainders of division by 7 using
states s1-s7. there are only 7 different remainders so 7 states are needed.
at each input you'd determine the next remainder and hence the next state. | H***F
发帖数: 2501 | 7 good, thank you
【在 e***y 的大作中提到】
: states:
: start, s1, s2, s3, s4, s5, s6, s7
: state transitions (input, next state)
: start: (1, s1), (0, start)
: s1: (1, s3), (0, s2)
: s2: (1, s3), (0, s6)
: s3: (1, s7), (0, s6)
: s4: (1, s5), (0, s1)
: s5: (1, s4), (0, s3)
: s6: (1, s6), (0, s5)
| j***j
发帖数: 324 | 8 very good and clear. smart...
【在 e***y 的大作中提到】
: states:
: start, s1, s2, s3, s4, s5, s6, s7
: state transitions (input, next state)
: start: (1, s1), (0, start)
: s1: (1, s3), (0, s2)
: s2: (1, s3), (0, s6)
: s3: (1, s7), (0, s6)
: s4: (1, s5), (0, s1)
: s5: (1, s4), (0, s3)
: s6: (1, s6), (0, s5)
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