A binary counter of unspecified length supports two operations: increment (which increases its value by one) and reset (which sets its value back to zero). Show that, starting from an initially zero counter, any sequence of n increment and reset operations takes time $\mathbf{O}\mathbf{\left(}\mathbf{n}\mathbf{\right)}$; that is, the amortized time per operation is O(1) .

Be using the taxation technique of amortised analysis to establish approximatelybinary increase & resets operations on a binary counter.

This taxation approach applies taxes solely to specific operations, ensuring that the overall cost of those operations does not exceed the tax paid. The tax paid by those operations is the amortised cost.

Description:

Always one bit gets coded as 1 when a counter is incremented.

• When change any bit of something like the binary counters to 1 , set amortised cost to$\$2$.

• Switch it to 1 using $\$1$and preserve some other$\$1$for future reset operations.

• As a result, whenever a bit is set to 1 , a$\$1$credit is provided.

• Assume the counter is reset after

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• As a result, the reset procedure always has enough money to return the counter to zero.

• As a result, O(n) time will be required for n increment and reset operations.

As a result, the average cost per operation will be O(1) .

1 bit is get coded as 1 when computer counting ingormation is increase. And binary counter is 1 when set amortised cost is $2 . So, as a result the reset procedure always has enough money to way back the counter till zero.