Suppose that five ones and four zeros are arranged around a circle. Between any two equal bits you insert a 0 and between any two unequal bits you insert a 1 to produce nine new bits. Then you erase the nine original bits. Show that when you iterate this procedure, you can never get nine zeros.
Answers
Step-by-step explanation:
original" (and any subsequent words) was ignored because we limit queries to 32 words.
Answer:
Using backward reasoning we want to show that "We can never get nine 0's". Step-by-step explanation: Basically in order to create nine O's, the previous step had to have all O's or all 1's. There is no other way possible, because between any two equal bits you insert a If we consider two cases for the second- to-last step: There were 9 0's: We obtain nine O's if all bits in the previous step were the same, thus all bit were O's or all bits were 1's. If the previous step contained all O's, then we have the same case as the current iteration step. Since initially the circle did not contain only O's, the circle had contain something else than only O's some point and thus there exists a point the circle contained only 1's.