Find the expected number of carries when adding two numbers, each at most N digits long.
Explanation:
let E[i] = probability that addition at ith place results in carry.
Then E[1] = 45/100
and E[i] = 45/100 + 10/100 * E[i-1] for i > 1
(Justification in next section)
By linearity of expectation,
Answer[N] = E[1] + E[2] … E[N]
Therefore, one could first pre-compute all E[i]'s and then Answer[i]'s in O(N) time overall.
This gives complexity of O(N+T).
However, it is also possible to have O(1) time per testcase.
This can be done by rewriting the above equation as: (0.5 - E[i]) = 10/100 * (0.5 - E[i-1])
Therefore, (0.5 - E[i]) = (0.5 - E[1]) / 10i-1 = 0.00…(i times 0)…05
So, Answer[N] = E[1] + E[2] … E[N] = 0.5 * n - 0.0555…(n times 5)…5
Justifications:
Adding the digits at ith place results in a carry if
The sum of those two digits is more than 9. All such pairs of digits are (1, 9), (2, 9), (3, 9) …, (2, 8), (3, 8), …, (3, 7), (4, 7), …, (9, 1). There are 45 such pairs, therefore probability of this happening is 45/100.
The sum of those two digits is exactly 9, and there was a carry from i-1th position. The sum of digits is 9 for the pairs (0, 9), (1, 8), (2, 7), … (9, 0). The probability of this happening is 10 / 100 * E[i-1].
why we are considering each pair two times like when we are calculating pairs with 9 we’ve already considerd 9+8…but again when calculating pairs for 8 we included 8+9…aren’t they same ???
I used the formula E[i]=0.45*i+E[i-1]/10 and got it accepted . I don’t understand how your formula and mine can give the same results. Also your justification is quite unclear. Could you please elaborate it properly.
The sum of those two digits is exactly 9, and there was a carry from i-1th position. The sum of digits is 9 for the pairs (0, 9), (1, 8), (2, 7), … (9, 0). The probability of this happening is 10 / 100 * E[i-1].
can someone plz explain why we r multiplying E[i-1]*10/100;
No, they are not same. Adding 1 and 9 is different from adding 9 and 1. This is because both numbers are chosen independently.
If you consider 1+9 same as 9+1, then you are giving more “weight” to the case when both digits are same(there will be 10/55 such cases as compared to 10/100 in the original setting).
To see this quickly, make your base 2, and consider 1 digit numbers. Carry is there only when you add 1+1. Since you consider 0+1 same as 1+0, you will give answer 1/3. whereas, it is clearly 1/4(probability that both numbers are 1).