 Practice

Contest

Easy

# Pre-requisites:

Linearity of Expectation, basic probability

# Problem:

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 = 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 + E … 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) / 10i-1 = 0.00…(i times 0)…05

So,             Answer[N] = E + E … E[N] = 0.5 * n - 0.0555…(n times 5)…5

# Justifications:

Adding the digits at ith place results in a carry if

1. 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.

2. 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].

Hence E[i] = 45/100 + 10 * E[i-1] / 100

# Setter’s Solution:

Can be found here

# Tester’s Solution:

Can be found here

13 Likes

can anyone give a link to tutorial on Linearity of expectation

1 Like

http://courses.csail.mit.edu/6.042/fall05/ln14.pdf

5 Likes

I started to see a pattern after I divided each of the test cases by 0.45.

I got 1, 2.1, 3.21 … as you can notice each one is the previous divided by 10 + next integer.

so I pre calculated each E[i] as follows;

d = 1;
E = 0.45;

for each i from 2 to 100000
d /= 10; d += i; E[i] = 0.45 * d;

Then for each n I read I write E[n].

still dnt able to get the concept can anyone please clarify that how there are 100 cases and 45 cases that gives carry one …

u can see
for each 1 digit number like 0 u can add(0-9)
so for 10 1-digit numbers cases are 10*10=100

now for carries
for number 9 u can get carry by adding (1-9 any number)
lly for number 8 u can get carry by adding(2-9 any number)

so total case are
9+8+7+6+5+4+3+2+1=45

hence E[x]=45/100

1 Like

Can someone explain setter’s appraoch?
The one that uses DP.

1 Like

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.

4 Likes

Yes I also have the same doubt. There seems to be a redundancy both while considering and calculating answer.

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).

1 Like

Actually, your formula is same as mine. What you meant was:

Ans[i] = 0.45*i + Ans[i-1]/10

=> Ans[i] - Ans[i-1] = 0.45 + (Ans[i-1] - Ans[i-2]) / 10

=> E[i] = 0.45 + E[i-1] / 10

Just the reccurence above. Please let me know exactly what is unclear in the editorial.

arr=0.45

``````    for i in range(1,100001):
if(i%2==1):
p =  ((l[i-1]/10.0) + 0.9)
else:
p = ((l[i-2]/100.0) + .44)
l.append(p)
n=input()
print l[n-1]+(n-1)/2

``````

Can anyone explain dp approach to the question

1 Like

In the last line : E[i] = 45/100 + 9 * E[i-1] / 100 ?? I think it should be E[i] = 45/100 + 10 * E[i-1] / 100

You should have got over this doubt by seeing the answer 0.45 for single digit numbers. If you don’t include repetition then it should have been 0.30 as there would be 30 such cases. But as the sum of 1,2,3…,10 is 55; it takes less than a millisecond to observe that number of cases is 45 (1+2+3+…+9).

Adding (21 and 19) and (29 and 11) are considered to be different although in both cases the carry is due to 9+1, for the reason cum explanation mentioned by @utkarsh_lath

I figured out the recurrence and used the fact that the answer has to be correct for six decimal digits only.
Have a look at my solution. link

Though I don’t understand why this code didn’t work. I thought long double would take care of precision.

``````int t,i,j,n;
long double arr,d,num=0.045;
arr=0.0;
d=0.45;
for(i=1;i<100001;++i)
{
arr[i]=arr[i-1]+d;
d+=num;
num/=10;
}
cin>>t;
for(i=0;i<t;++i)
{
cin>>n;
cout<<arr[n]<<endl;
}
``````

You got it right, it gives WA due to precision issues.

But you seem to not to know where precision is coming into picture.

It is coming when you print your result !

cout prints upto 6 significant digits.

This fact often gives wrong answer and people remain clueless about their error.

@bb_1: How do u get this E[i]=0.45*i+E[i-1]/10 recurrence relation?
plz explain ur approach bcz I didn’t understand from editorial.

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