8/11 ANSWERS TO THE EXERCISES 



simply not there. 5: The inverter requires speedometers to give .vi 

 and X2 as outputs. Then any machine that forms the functions 



JriA-2 — A-2 — -Vl — .^1 + Jf2->^2 + XlX2 



xi(xh - 1) xh- I 



will emit the original input. If an actual transformation is required, 

 then (the functions of iri, etc. above being represented by A\ and Ai) 

 the transformation a'l = k{A\ — a\), a'z — k{A2 ~ az) will give the 

 required behaviour as closely as is desired if k is made positive and 

 large enough (Ex. 8/5/10). 6: —2 has no particular relation to (7,3), 

 whereas 4 has, as the construction of the table showed. 



8/11. 1: / has 3 states; u has 2. 2: / has 3 states; u cannot have more than 6; 

 II actually has 5. 3: Thas 2 states; so has U. 4: 3 states. They are 

 (0,0,0,0), (0,0,1,0) and (0,1,0,1). 



8/13. 1 : One bit per step, for r has two states only. 2 : The numbers of distinct 

 states occupied, at successive states, were: Q: 9,4,3,3,3; R: 1,2,2,2,2; 

 S: 1,1,2,3,5. 3: Because the jump from 1 to 4 would have implied a 

 gain in variety of 3, whereas R can supply only 2 at most. 



8/14. 2: The number of balancings cannot, whatever the method, be fewer 

 than three. For the variety to be transmitted is log2 27 bits, and the 

 transmitter can carry only logi 3 bits per step. 



8/15. 1: Four; the variety from A takes longest. 2: Four steps; (the answer 

 rtuist be the same as to Ex. 1, for the two questions are really identical). 

 3: Three; that from y takes longest. 4: Two steps. 



8/17. 1: A was at (3,2). (Hint: A" was at (-1,0) and B" was at (1,0).) 



2: Yes; the output enables the sequence of input-vectors to be deduced, 



of which the sequence of first components is the a-message. 3: No; 



y's movement is simply A's with half the amplitude. 4: If the letters 



a, b, etc. indicate the respective movements o{ A, B, etc., to right and 



left from suitable zeros, with common scale, then / = ^(a — h), 



n = i(a + b), y = \(l + «), and z = i( — / + n), from which / and n 



are readily eliminated. "Decoding" corresponds to solving these 



simultaneous equations for a and b, the unknowns, in terms of j and z, 



the knowns. 



9/2. 1: The transformation so obtained is determinate; how it was obtained 



is irrelevant. 2 : Since each state must go to some state, the probabilities, 



and the numbers in each column, must add up to 1. 3: No. 4: 2'", 



i.e. 1024. 5: More than one arrow may leave each point. 



9/4. 1 : The actual transition frequencies are 



\ A B 



'a 6 17 



B 17 10 

 As each column's probabilities must add to 1, the first column must be 

 divided by 23 and the second by 27. The estimated probabilities are 

 thus 



± 



A 



B 



A B 



0-26 0-63 

 0-74 0-37 



(This is the system that was, in fact, used 

 to generate the trajectory in Ex.1.) 



282 



