7/15 ANSWERS TO THE EXERCISES 



7/15. 1 : It says that of all the possible rational numbers (infinite in number), 

 the combining proportions will always be found in a small subset 

 (numbering perhaps a few dozen). 2: Of all geometrically possible 

 trajectories, and all possible heat changes, etc., it allows only a few. 



7/19. 1 : Of the transitions (e.g.) a -> o, a-^ b,a->c, etc. all are excluded but 

 one, for the transition from a must be single-valued; similarly from b, etc. 



7/20. 1:8. 2: 17. 3: 12. 4: (i) 1,048,576; (ii) 21,892. 



7/22. 2: The parasites'; evidently some hosts are food to more than one 

 species of parasite. 3: Fis many-one, and causes a fall. 4: As lacking 

 in discrimination. 5: (i) 6 states, (ii) 2 states. "The bath is out of 

 order". 6: The chance that a particular state Si will be the transform 

 of a particular state Sj is l/«. The chance that Sj will not be the trans- 

 form of Sj is 1 — 1//7. The chance that Si will not be the transform 

 of Sk will also be 1 — 1///. So the chance that St will not be the 

 transform of any state is (1 — l/«)". This gives the fraction of the 

 operands that disappear after the transformation. As n tends to in- 

 finity it tends to l/e. So the fraction that remains, to give the variety, 

 is 1 - lie. 

 7/24. 1: 3 states, = 1-58 bits. 2: By another 1-58 bits. 3: "a and 6" 

 becomes, in succession, 5a, 5a + 1, lOa -1- 14, \0a -f Z> -t- 14. If 14 

 is subtracted, 10a -1- 6 is left. Thus the hundred combinations of a 

 and b (if and is allowed) is transformed one-one, after subtraction 

 of 14, to the hundred numbers from to 99. The variety is 100 states 

 or 6-64 bits. 4: All the two-number combinations that are suggested 

 on such occasions. 5: Zero. 6: 2 states, 1 bit; either various circuits 

 or one circuit at various times. 7: No. They may be going together 

 round the same cycle. Distinguish between (i) equality of state between 

 machine and machine considered at one instant, and (ii) equality of 

 state between time and time considered in one machine. 



8/3. 1 : By no more than one cut can do. 



8/4. 1: Yes; "taking the antilog". 2: No; the same value for x' is given 

 by many values of x. 3: The identical transformation. 4: n' = n — 1. 

 5: x' = X - V, y' = - x + ly. 6: 3 loga 26 bits, i.e. 14-1 bits. 7: 

 263=17576. 8: log2 8 -|- logz 7, i.e. 5-8 bits. 9: Not quite: the 

 variety would be 5-7 bits, which is insufficient (loga 52 < log2 56). 

 10: 1 bit; the messages are "courting" and "not-courting", and there 

 are two of them. The complexities of molecule and ritual are irrelevant 

 here. 



8/5. \\ AACBD DBCBCCB. 2: acdbdcd. 3: b d c d b a d. 4: 

 Yes. 5:10,8,7,10,11,9,8. 6:10,8,4,3,-1,-1,3,0,1,1,-1,... 

 7: X = 2, 1, 2, -11, 11, -2, 16, . . . and y = I, 4, -11, 13, -24, 



-13, -93, 8: X = exp (-4/ - sin /)• 9: x = K^ ' + te-^ 



—cos /). 10: X chases a, and follows it closer and closer. 



8/6. 1: No; in the table of transformations there must be'108 rows, so each 

 column must have 108 elements; as only 100 are available, there must 

 be repetitions. 2: (i) 7, (ii) 512. 3: Fitting some device such as a 

 speedometer or tachometer that emits a number proportional to the 

 time-derivative. 4: No; for if the output is steady at zero (as will 

 happen if it is started at zero) a's values cannot be deduced from x's 

 transitions, which are -> for all a. 



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