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Generating Formulas

In a secret code that uses only the letters A, B, C, and D, a word is formed by applying the following rules:
 
Rule 1: A B C D is the base word.
 
Rule 2: If C immediately follows B, then C can be moved to the front of the word.
 
Rule 3: One letter of the same type can be added immediately after an A, a B, or a C.

Notice that Rules 2 and 3 are permissive because they can be applied but need not be. No diagram can be drawn, nor are there any readily derived conditions, so we turn to the questions.

 
Example-1
Which one of the following is not a code word?
  1. A B C D
  2. D A B C
  3. C A B D
  4. A A B B C C D
  5. C C C A B D
Solution
Let’s use elimination on this question. A B C D is the base word. So (A) is a word—eliminate (A).
 
Applying Rule 2 to the base word gives C A B D. So (C) is a word—eliminate.
 
Applying Rule 3 to the base word three times gives A A B B C C D. So (D) is a word—eliminate.
 
Finally, applying Rule 2 to the base word gives C A B D; then applying Rule 3 to C twice gives C C C A B D. So (E) is a word—eliminate.
 
Hence, by process of elimination, the answer is (B).
 
 
Example-2
Which of the following letters can start a word?
I.   A
II.  B
III. C
  1. I only
  2. II only
  3. III only
  4. I and II only
  5. I and III only
Solution
A can start a word since it starts the base word. This eliminates choices (B) and (C) since they don’t contain I.
 
C can start a word since C A B D is formed from the base word by using Rule 2. This eliminates (A) and (D) since they don’t contain III.
 
Hence, the answer is (E), and there is no need to check Statement II.
 
 
Example-3
The word C A A B C C D can be formed from the base word by applying the rules in which one of the following orders?
  1. 22333
  2. 23232
  3. 32233
  4. 3223
  5. 3233
Solution
This question is hard, because we don’t know to which letter(s) in the base word to apply the rules. Furthermore, there is more than one way to generate the word—but, of course, only one of those ways is listed as an answer-choice. We can, however, narrow the number of answer-choices by analyzing the word C A A B C C D.
 
Notice that C occurs three times and A two times. So Rule 3 must have been applied three times, twice to C and once to A.
 
This eliminates choices (B) and (D) since neither has three 3’s.
 
Next, turning to choice (A), we apply Rule 2 to the base word giving C A B D. Now Rule 2 cannot be applied to this word again, since C is not immediately after B. This eliminates (A).
 
Next, choice (C) seems at first glance to be plausible. It begins the same way as does choice (E). But notice in choice (C) that Rule 2 is applied twice in a row. A little fiddling shows that if this is done, then two C’s in a row would come at the beginning of the word. So eliminate (C).
 
Hence, by process of elimination, the answer is (E). As a matter of test taking strategy this would be sufficient analysis of the question.
 
However, it is instructive to verify that the answer is (E). To that end, apply Rule 3 to C in the base word A B C D which gives A B C C D.
 
Next, apply Rule 2 which gives C A B C D.
 
Finally, apply Rule 3 to A and then to C which gives C A A B C C D.
 
 
Example-4
If a fourth rule is added to the other three rules which states that whenever B or D ends a word the sequence obtained by dropping either B or D is still a word, then which of the following would be true?
I.   Some words could end with A.
II.  Some words could start with C and end with C.
III. A C D would be a word.
  1. I only
  2. II only
  3. I and II only
  4. I and III only
  5. I, II, and III
Solution
Start with the base word A B C D. Applying the new rule gives A B C. Then applying Rule 2 gives C A B. Finally, applying the new rule again gives CA. Hence, some words could end with A. So I is true. This eliminates choice (B).
 
Next, starting again with the base word A B C D, apply Rule 3 to C which gives A B C C D. Then apply Rule 2 which gives C A B C D. Finally, apply the new rule which gives C A B C. So some words could start with C and end with C. This eliminates choices (A) and (D).
 
Unfortunately, we have to check the third condition. From the base word all the other words must be derived.
 
Now in the base word, D must be dropped from the end before B can be dropped. Thus A C D cannot be formed.
 
This eliminates choice (E), and therefore the answer is (C).
 
 
Example-5
If a fourth rule is added to the other three rules which states that a word is created whenever the reversed sequence of a word is added to the end of the word itself, then which one of the following is NOT a word?
  1. A B C D D C B A
  2. A B B C C D
  3. C A B C D D C B A C
  4. C A B D D B A C
  5. C B C D D C B A
Solution
This problem can be solved either by deriving the four words offered, or by finding the word that violates one or more of the rules. We shall solve it both ways.
 
Choice (A) is a word because it can be derived by adding the reversed sequence of the base word to the base word itself. So eliminate (A).
 
Choice (B) is word because it can be derived by applying Rule 3, first to B, then to C. So eliminate (B). (Notice that this new rule is also permissive, so it need not be applied for choice (B) to be a word.)
 
Choice (C) is a word because it can be derived by applying Rule 3 to the base word which gives A B C C D. Then applying Rule 2 gives C A B C D.
 
Finally applying the new rule gives C A B C D D C B A C. So eliminate (C).
 
Finally, choice (D) is a word because it can be derived by applying Rule 2 to the base word which gives C A B D. Then applying the new rule gives C A B D D B A C. So eliminate (D).
 
Thus, by process of elimination, the answer is (E).
Turning to the other method, we now show that (E) violates one of the rules.
 
The only way that A can end a word is if the reversed sequence of a word is added to the word itself.* But D C B A is not the reversed sequence of C B C D, which isn’t even a word.
 

Although the latter method was faster than deriving the four words, it can be deceptively hard to spot the choice that violates one or more of the rules.





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