The "Who caught the fish" logical riddle:

  Given:

   1) If Tom didn't, then Dick did.
   2) "Tom did" and "Jim didn't" are not both true statements.
   3) If Dick did, then Tom did and Jim did.

  Is it true that Jim did?

I will convert each statement to DCI program settings that
change any machine state which conflicts with it to one that
doesn't. If the riddle allows, with care/luck the resulting
program won't have crashes were it tries to simultaneously set
and reset the same flip-flop.
I will then determine which machine states the completed
program leaves unchanged and see if they provide a consistent
answer to the riddle. 

Assign flip-flop A to Tom, B to Dick, and C to Jim; let 1/true
signify "did catch" and 0/false signify "didn't catch".

In strict Boolean logic terms, the first statement says only
that having both A and B "didn't" is inconsistent -- setting
either or both to "did" yields states consistent with it.
The statement's colloquial meaning is to set B to "did" while
leaving A as "didn't", and the corresponding program is:

 . __1_ __2_ __3_ __4_ __5_ __6_
A: ..|. .F|. ..|. ..|. ..|. ..|.
B: ..|. .F|S ..|. ..|. ..|. ..|.
C: ..|. ..|. ..|. ..|. ..|. ..|.


The second statement says states with both A true and C false
are not valid, but it doesn't provide guidance as to how they
should be changed. The program could toggle either or both A
and C flip-flops. I think toggling both is most aesthetic, so
I'll add programming for two sets of rods:

 . __1_ __2_ __3_ __4_ __5_ __6_
A: ..|. .F|. T.|R T.|. ..|. ..|.
B: ..|. .F|S ..|. ..|. ..|. ..|.
C: ..|. ..|. .F|. .F|S ..|. ..|.


The third statement may be split into two sub-statements (this
can help avoid a crash):

   3a) if B is true and A is false, set A true
   3b) if B is true and C is false, set C true

I'll assign 3a to logic rod 6, exhausting the available "Set"
clock rods. Since clock rod 4 sets only C, I can use logic rod
5 for 3b's condition and use an OR-slider to join logic rods 4
and 5 to clock rod 4:

. . . . . . . . . |--o----|
 . __1_ __2_ __3_ __4_ __5_ __6_
A: ..|. .F|. T.|R T.|. ..|. .F|S
B: ..|. .F|S ..|. ..|. T.|. T.|.
C: ..|. ..|. .F|. .F|S .F|. ..|.

The resulting state transition table is:

abc
---
000 -> 010
001 -> 011
010 -> 111
011 -> 111
100 -> 001
101 -> 101
110 -> 011
111 -> 111

Note we can start from any state and cycle the clock until we
reach one of two stable states: 101 and 111. We can answer the
riddle that yes, Jim did catch fish, as did Tom, but the boys
didn't tell enough to determine whether or not Dick did.

Also note that if we had programmed only one of the transitions
for statement 2, the stable states would have remained the same
and reachable from any starting state.

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