ANSWERS TO THE EXERCISES 9/18 



9/5. 1: Once under a pebble it would stay there. 2: B must be the paper 

 (where the fly sticks), and D the stove (where it never stops). 3: From 

 protocol to matrix; the protocol gives a unique matrix, but the matrix 

 can give only a set of protocols. Or, if the matrix is lost it can 

 be restored from the protocol, but a lost protocol cannot be restored 

 from the matrix. 



9/6. 1: (100,0,0), (25,75,0), (62,19,19), (32,61,7), etc., if taken to the nearest 

 unit. 3: Face 3 tends to come up, face 4 is tending to go down; there- 

 fore X = 4. 4: Consider 100 molecules, and let x of the 100 /4's be 

 dissociated. Ignore the 5's. Each A has two possible states, dissociated 

 or not, and in each interval of time has the probabilities of staying in its 

 state or changing: 



j Dissociated Not Dissociated 



Dissociated 

 Not Dissociated 



0999 0-01 



0001 0-99 



5: If X and y are the numbers dissociated and not, respectively, then for 

 equilibrium: 



X = 0-999X + OOlj 

 100 = X + y; 



Therefore, x = 90it. 7: Each insect can be only in one of 3; if there 

 are n insects, the number of distinct populations is \{n +2)(« + 1). 



9/7. 1: 



After D: 



Thus, transitions from C are markedly affected by what preceded the C. 



9/10. \:t'i = -.ti. 2: No. 3: Yes. 



9/11. 1: The probabilities are (so far as the evidence goes) 0175 and 0-825; 

 so the entropy is 0-67 bits. 2: The probabilities are -5-2, \, \%; so the 

 entropy is 0-94 bits. 3: 2-6 bits. 4: 5-2 bits. 5: 2-6 x // bits. 6: 0. 



9/12. 2: It always falls below 1 bit. 



9/13. 1: The terminal equilibrium is with all in B; and any sequence must 

 eventually become . . . B B B B . . . . This has no variety, so the 

 entropy must be zero. 2: Entropy is calculated when the whole is at 

 terminal equilibrium and in this case the terminal equilibrium does 

 not permit the assumption "when it is at /I". 



9/16. 1: Yes, for 62 is fewer than 3i4; 34 is gi^ so four sugars might be sufficient 

 if suitably chosen. 2: Every number in it is the same, e.g. that at the 

 endof S.9/10. 



9/17. 2: It must be at least 2000 bits per min., if the assumptions are correct. 

 3: Each finger has variety of log2 3 in -j^o min, and 300 log2 3 in 1 

 minute; so all 10, being independent, have 3000 log2 3 bits in one 

 minute; so the bound is 4800 bits/min. 4: 5540 symbols/hour, 



9/18. 1: b can follow only a or b, not d; so Xb must be ab; similarly Xc must 

 be ac ; and XX must be dX. 



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