I am using Hive 1.2.1000.2.4.2.0-258.
There are 4850000+ rows in the table, 14511 rows of A between 73 and 74, and 3 cols- group_id, A and B.
Group_id is actually equal to 0.
Almost all of A and B are integers.
I was using the following scripts to find statistic summaries from a table:
select group_id, --group_id=0 a constant
percentile_approx(A , 0.5) as A_mdn,
percentile_approx(A , 0.25) as A_Q1,
percentile_approx(A , 0.75) as A_Q3,
percentile_approx(A , array(0.2,0.15, 0.1,0.05,0.025,0.001)) as A_i,
min(A) as min_A,
percentile_approx(B , 0.5) as B_mdn,
percentile_approx(B , 0.25) as B_Q1,
percentile_approx(B , 0.75) as B_Q3,
percentile_approx(B , array(0.8,0.85, 0.9, 0.95,0.975)) as B_i
from table
group by group_id;
The result I got is:
0
73.21058033222496
73.21058033222496
462.16968382794516
[73.21058033222496,73.21058033222496,73.21058033222496,73.21058033222496,73.21058033222496,73.21058033222496]
0.0
1.0
1.0
2.0
[2.0,3.0,4.0,8.11278644563614,17.0]
Then I change the code as following:
select group_id, --group_id=0 a constant
percentile(cast(A as bigint), 0.5) as A_mdn,
percentile(cast(A as bigint), 0.25) as A_Q1,
percentile(cast(A as bigint), 0.75) as A_Q3,
percentile(cast(A as bigint), array(0.2,0.15, 0.1,0.05,0.025,0.001)) as A_i,
min(A) as min_A,
percentile(cast(B as bigint), 0.5) as B_mdn,
percentile(cast(B as bigint), 0.25) as B_Q1,
percentile(cast(B as bigint), 0.75) as B_Q3,
percentile(cast(B as bigint), array(0.8,0.85, 0.9, 0.95,0.975)) as B_i
from table
group by group_id
The new result is:
0
72.0
6.0
762.0
[3.0,1.0,1.0,0.0,0.0,0.0]
0.0
1.0
1.0
2.0
[2.0,3.0,4.0,9.0,17.0]
To double check the truth, I also load this table to R. Following is the R-result:
A:
Min 0
Q1: 6
Median: 72
Q3: 762
0.2 quantile: 3
0.15 quantile: 1.5
0.1 quantile: 1
0.05 quantile: 0
0.025 quantile:0
0.001 quantile:0
B
Q1: 1
Median: 1
Q3: 2
0.8 quantile: 2
0.85 quantile: 3
0.9 quantile: 4
0.95 quantile: 9
0.975 quantile:17
Obviously, R result is consistent with percentile function, but percentile_approx gives me the wrong answer.
Yeah, the percentile_approx doesn't have any approximation guarantees, except when you set the accuracy to be greater than or equal to the # of data points.
The source for it is here: https://github.com/apache/hive/blob/master/ql/src/java/org/apache/hadoop/hive/ql/udf/generic/NumericHistogram.java
From a quick reading, the gist is that it creates accuracy buckets, and then when it runs out of buckets it merges buckets by finding the two closest buckets and combining them with a weighted sum.
This will break with various inputs though. In particular, if you have datapoints that are very high/very low and are spaced far away from each other, it will break the algorithm. If you first clip your data to be in a range where there are not many outliers, it should perform better.
You might consider randomly sampling the data and computing non-approx percentile instead if your data is too skewed though.
This function returns a true value if "all" the values are integers. You said that almost all of A and B are integers.
Try to cast the complete column A to int and see if you come close to the answer.
I don't think, you will ever get exactly the same answer as R because R's percentile function most likely takes non-integers also.
One way to get the exact answer would be to write your own UDF and use it instead.
Hope this helps!
For example if you take the following example into consideration.
100.00 - Original Number
33.33 - 1st divided by 3
33.33 - 2nd divided by 3
33.33 - 3rd divided by 3
99.99 - Is the sum of the 3 division outcomes
But i want it to match the original 100.00
One way that i saw it could be done was by taking the original number minus the first two divisions and the result would be my third number. Now if i take those 3 numbers i get my original number.
100.00 - Original Number
33.33 - 1st divided by 3
33.33 - 2nd divided by 3
33.34 - 3rd number
100.00 - Which gives me my original number correctly. (33.33+33.33+33.34 = 100.00)
Is there a formula for this either in Oracle PL/SQL or a function or something that could be implemented?
Thanks in advance!
This version takes precision as a parameter as well:
with q as (select 100 as val, 3 as parts, 2 as prec from dual)
select rownum as no
,case when rownum = parts
then val - round(val / parts, prec) * (parts - 1)
else round(val / parts, prec)
end v
from q
connect by level <= parts
no v
=== =====
1 33.33
2 33.33
3 33.34
For example, if you want to split the value among the number of days in the current month, you can do this:
with q as (select 100 as val
,extract(day from last_day(sysdate)) as parts
,2 as prec from dual)
select rownum as no
,case when rownum = parts
then val - round(val / parts, prec) * (parts - 1)
else round(val / parts, prec)
end v
from q
connect by level <= parts;
1 3.33
2 3.33
3 3.33
4 3.33
...
27 3.33
28 3.33
29 3.33
30 3.43
To apportion the value amongst each month, weighted by the number of days in each month, you could do this instead (change the level <= 3 to change the number of months it is calculated for):
with q as (
select add_months(date '2013-07-01', rownum-1) the_month
,extract(day from last_day(add_months(date '2013-07-01', rownum-1)))
as days_in_month
,100 as val
,2 as prec
from dual
connect by level <= 3)
,q2 as (
select the_month, val, prec
,round(val * days_in_month
/ sum(days_in_month) over (), prec)
as apportioned
,row_number() over (order by the_month desc)
as reverse_rn
from q)
select the_month
,case when reverse_rn = 1
then val - sum(apportioned) over (order by the_month
rows between unbounded preceding and 1 preceding)
else apportioned
end as portion
from q2;
01/JUL/13 33.7
01/AUG/13 33.7
01/SEP/13 32.6
Use rational numbers. You could store the numbers as fractions rather than simple values. That's the only way to assure that the quantity is truly split in 3, and that it adds up to the original number. Sure you can do something hacky with rounding and remainders, as long as you don't care that the portions are not exactly split in 3.
The "algorithm" is simply that
100/3 + 100/3 + 100/3 == 300/3 == 100
Store both the numerator and the denominator in separate fields, then add the numerators. You can always convert to floating point when you display the values.
The Oracle docs even have a nice example of how to implement it:
CREATE TYPE rational_type AS OBJECT
( numerator INTEGER,
denominator INTEGER,
MAP MEMBER FUNCTION rat_to_real RETURN REAL,
MEMBER PROCEDURE normalize,
MEMBER FUNCTION plus (x rational_type)
RETURN rational_type);
Here is a parameterized SQL version
SELECT COUNT (*), grp
FROM (WITH input AS (SELECT 100 p_number, 3 p_buckets FROM DUAL),
data
AS ( SELECT LEVEL id, (p_number / p_buckets) group_size
FROM input
CONNECT BY LEVEL <= p_number)
SELECT id, CEIL (ROW_NUMBER () OVER (ORDER BY id) / group_size) grp
FROM data)
GROUP BY grp
output:
COUNT(*) GRP
33 1
33 2
34 3
If you edit the input parameters (p_number and p_buckets) the SQL essentially distributes p_number as evenly as possible among the # of buckets requested (p_buckets).
I've solved this problem yesterday by subtracting 2 of 3 parts from the starting number, e.g. 100 - 33.33 - 33.33 = 33.34 and the result of summing it up is still 100.
Imagine you sell those metallic digits used to number houses, locker doors, hotel rooms, etc. You need to find how many of each digit to ship when your customer needs to number doors/houses:
1 to 100
51 to 300
1 to 2,000 with zeros to the left
The obvious solution is to do a loop from the first to the last number, convert the counter to a string with or without zeros to the left, extract each digit and use it as an index to increment an array of 10 integers.
I wonder if there is a better way to solve this, without having to loop through the entire integers range.
Solutions in any language or pseudocode are welcome.
Edit:
Answers review
John at CashCommons and Wayne Conrad comment that my current approach is good and fast enough. Let me use a silly analogy: If you were given the task of counting the squares in a chess board in less than 1 minute, you could finish the task by counting the squares one by one, but a better solution is to count the sides and do a multiplication, because you later may be asked to count the tiles in a building.
Alex Reisner points to a very interesting mathematical law that, unfortunately, doesn’t seem to be relevant to this problem.
Andres suggests the same algorithm I’m using, but extracting digits with %10 operations instead of substrings.
John at CashCommons and phord propose pre-calculating the digits required and storing them in a lookup table or, for raw speed, an array. This could be a good solution if we had an absolute, unmovable, set in stone, maximum integer value. I’ve never seen one of those.
High-Performance Mark and strainer computed the needed digits for various ranges. The result for one millon seems to indicate there is a proportion, but the results for other number show different proportions.
strainer found some formulas that may be used to count digit for number which are a power of ten.
Robert Harvey had a very interesting experience posting the question at MathOverflow. One of the math guys wrote a solution using mathematical notation.
Aaronaught developed and tested a solution using mathematics. After posting it he reviewed the formulas originated from Math Overflow and found a flaw in it (point to Stackoverflow :).
noahlavine developed an algorithm and presented it in pseudocode.
A new solution
After reading all the answers, and doing some experiments, I found that for a range of integer from 1 to 10n-1:
For digits 1 to 9, n*10(n-1) pieces are needed
For digit 0, if not using leading zeros, n*10n-1 - ((10n-1) / 9) are needed
For digit 0, if using leading zeros, n*10n-1 - n are needed
The first formula was found by strainer (and probably by others), and I found the other two by trial and error (but they may be included in other answers).
For example, if n = 6, range is 1 to 999,999:
For digits 1 to 9 we need 6*105 = 600,000 of each one
For digit 0, without leading zeros, we need 6*105 – (106-1)/9 = 600,000 - 111,111 = 488,889
For digit 0, with leading zeros, we need 6*105 – 6 = 599,994
These numbers can be checked using High-Performance Mark results.
Using these formulas, I improved the original algorithm. It still loops from the first to the last number in the range of integers, but, if it finds a number which is a power of ten, it uses the formulas to add to the digits count the quantity for a full range of 1 to 9 or 1 to 99 or 1 to 999 etc. Here's the algorithm in pseudocode:
integer First,Last //First and last number in the range
integer Number //Current number in the loop
integer Power //Power is the n in 10^n in the formulas
integer Nines //Nines is the resut of 10^n - 1, 10^5 - 1 = 99999
integer Prefix //First digits in a number. For 14,200, prefix is 142
array 0..9 Digits //Will hold the count for all the digits
FOR Number = First TO Last
CALL TallyDigitsForOneNumber WITH Number,1 //Tally the count of each digit
//in the number, increment by 1
//Start of optimization. Comments are for Number = 1,000 and Last = 8,000.
Power = Zeros at the end of number //For 1,000, Power = 3
IF Power > 0 //The number ends in 0 00 000 etc
Nines = 10^Power-1 //Nines = 10^3 - 1 = 1000 - 1 = 999
IF Number+Nines <= Last //If 1,000+999 < 8,000, add a full set
Digits[0-9] += Power*10^(Power-1) //Add 3*10^(3-1) = 300 to digits 0 to 9
Digits[0] -= -Power //Adjust digit 0 (leading zeros formula)
Prefix = First digits of Number //For 1000, prefix is 1
CALL TallyDigitsForOneNumber WITH Prefix,Nines //Tally the count of each
//digit in prefix,
//increment by 999
Number += Nines //Increment the loop counter 999 cycles
ENDIF
ENDIF
//End of optimization
ENDFOR
SUBROUTINE TallyDigitsForOneNumber PARAMS Number,Count
REPEAT
Digits [ Number % 10 ] += Count
Number = Number / 10
UNTIL Number = 0
For example, for range 786 to 3,021, the counter will be incremented:
By 1 from 786 to 790 (5 cycles)
By 9 from 790 to 799 (1 cycle)
By 1 from 799 to 800
By 99 from 800 to 899
By 1 from 899 to 900
By 99 from 900 to 999
By 1 from 999 to 1000
By 999 from 1000 to 1999
By 1 from 1999 to 2000
By 999 from 2000 to 2999
By 1 from 2999 to 3000
By 1 from 3000 to 3010 (10 cycles)
By 9 from 3010 to 3019 (1 cycle)
By 1 from 3019 to 3021 (2 cycles)
Total: 28 cycles
Without optimization: 2,235 cycles
Note that this algorithm solves the problem without leading zeros. To use it with leading zeros, I used a hack:
If range 700 to 1,000 with leading zeros is needed, use the algorithm for 10,700 to 11,000 and then substract 1,000 - 700 = 300 from the count of digit 1.
Benchmark and Source code
I tested the original approach, the same approach using %10 and the new solution for some large ranges, with these results:
Original 104.78 seconds
With %10 83.66
With Powers of Ten 0.07
A screenshot of the benchmark application:
(source: clarion.sca.mx)
If you would like to see the full source code or run the benchmark, use these links:
Complete Source code (in Clarion): http://sca.mx/ftp/countdigits.txt
Compilable project and win32 exe: http://sca.mx/ftp/countdigits.zip
Accepted answer
noahlavine solution may be correct, but l just couldn’t follow the pseudo code, I think there are some details missing or not completely explained.
Aaronaught solution seems to be correct, but the code is just too complex for my taste.
I accepted strainer’s answer, because his line of thought guided me to develop this new solution.
There's a clear mathematical solution to a problem like this. Let's assume the value is zero-padded to the maximum number of digits (it's not, but we'll compensate for that later), and reason through it:
From 0-9, each digit occurs once
From 0-99, each digit occurs 20 times (10x in position 1 and 10x in position 2)
From 0-999, each digit occurs 300 times (100x in P1, 100x in P2, 100x in P3)
The obvious pattern for any given digit, if the range is from 0 to a power of 10, is N * 10N-1, where N is the power of 10.
What if the range is not a power of 10? Start with the lowest power of 10, then work up. The easiest case to deal with is a maximum like 399. We know that for each multiple of 100, each digit occurs at least 20 times, but we have to compensate for the number of times it appears in the most-significant-digit position, which is going to be exactly 100 for digits 0-3, and exactly zero for all other digits. Specifically, the extra amount to add is 10N for the relevant digits.
Putting this into a formula, for upper bounds that are 1 less than some multiple of a power of 10 (i.e. 399, 6999, etc.) it becomes: M * N * 10N-1 + iif(d <= M, 10N, 0)
Now you just have to deal with the remainder (which we'll call R). Take 445 as an example. This is whatever the result is for 399, plus the range 400-445. In this range, the MSD occurs R more times, and all digits (including the MSD) also occur at the same frequencies they would from range [0 - R].
Now we just have to compensate for the leading zeros. This pattern is easy - it's just:
10N + 10N-1 + 10N-2 + ... + **100
Update: This version correctly takes into account "padding zeros", i.e. the zeros in middle positions when dealing with the remainder ([400, 401, 402, ...]). Figuring out the padding zeros is a bit ugly, but the revised code (C-style pseudocode) handles it:
function countdigits(int d, int low, int high) {
return countdigits(d, low, high, false);
}
function countdigits(int d, int low, int high, bool inner) {
if (high == 0)
return (d == 0) ? 1 : 0;
if (low > 0)
return countdigits(d, 0, high) - countdigits(d, 0, low);
int n = floor(log10(high));
int m = floor((high + 1) / pow(10, n));
int r = high - m * pow(10, n);
return
(max(m, 1) * n * pow(10, n-1)) + // (1)
((d < m) ? pow(10, n) : 0) + // (2)
(((r >= 0) && (n > 0)) ? countdigits(d, 0, r, true) : 0) + // (3)
(((r >= 0) && (d == m)) ? (r + 1) : 0) + // (4)
(((r >= 0) && (d == 0)) ? countpaddingzeros(n, r) : 0) - // (5)
(((d == 0) && !inner) ? countleadingzeros(n) : 0); // (6)
}
function countleadingzeros(int n) {
int tmp= 0;
do{
tmp= pow(10, n)+tmp;
--n;
}while(n>0);
return tmp;
}
function countpaddingzeros(int n, int r) {
return (r + 1) * max(0, n - max(0, floor(log10(r))) - 1);
}
As you can see, it's gotten a bit uglier but it still runs in O(log n) time, so if you need to handle numbers in the billions, this will still give you instant results. :-) And if you run it on the range [0 - 1000000], you get the exact same distribution as the one posted by High-Performance Mark, so I'm almost positive that it's correct.
FYI, the reason for the inner variable is that the leading-zero function is already recursive, so it can only be counted in the first execution of countdigits.
Update 2: In case the code is hard to read, here's a reference for what each line of the countdigits return statement means (I tried inline comments but they made the code even harder to read):
Frequency of any digit up to highest power of 10 (0-99, etc.)
Frequency of MSD above any multiple of highest power of 10 (100-399)
Frequency of any digits in remainder (400-445, R = 45)
Additional frequency of MSD in remainder
Count zeros in middle position for remainder range (404, 405...)
Subtract leading zeros only once (on outermost loop)
I'm assuming you want a solution where the numbers are in a range, and you have the starting and ending number. Imagine starting with the start number and counting up until you reach the end number - it would work, but it would be slow. I think the trick to a fast algorithm is to realize that in order to go up one digit in the 10^x place and keep everything else the same, you need to use all of the digits before it 10^x times plus all digits 0-9 10^(x-1) times. (Except that your counting may have involved a carry past the x-th digit - I correct for this below.)
Here's an example. Say you're counting from 523 to 1004.
First, you count from 523 to 524. This uses the digits 5, 2, and 4 once each.
Second, count from 524 to 604. The rightmost digit does 6 cycles through all of the digits, so you need 6 copies of each digit. The second digit goes through digits 2 through 0, 10 times each. The third digit is 6 5 times and 5 100-24 times.
Third, count from 604 to 1004. The rightmost digit does 40 cycles, so add 40 copies of each digit. The second from right digit doers 4 cycles, so add 4 copies of each digit. The leftmost digit does 100 each of 7, 8, and 9, plus 5 of 0 and 100 - 5 of 6. The last digit is 1 5 times.
To speed up the last bit, look at the part about the rightmost two places. It uses each digit 10 + 1 times. In general, 1 + 10 + ... + 10^n = (10^(n+1) - 1)/9, which we can use to speed up counting even more.
My algorithm is to count up from the start number to the end number (using base-10 counting), but use the fact above to do it quickly. You iterate through the digits of the starting number from least to most significant, and at each place you count up so that that digit is the same as the one in the ending number. At each point, n is the number of up-counts you need to do before you get to a carry, and m the number you need to do afterwards.
Now let's assume pseudocode counts as a language. Here, then, is what I would do:
convert start and end numbers to digit arrays start[] and end[]
create an array counts[] with 10 elements which stores the number of copies of
each digit that you need
iterate through start number from right to left. at the i-th digit,
let d be the number of digits you must count up to get from this digit
to the i-th digit in the ending number. (i.e. subtract the equivalent
digits mod 10)
add d * (10^i - 1)/9 to each entry in count.
let m be the numerical value of all the digits to the right of this digit,
n be 10^i - m.
for each digit e from the left of the starting number up to and including the
i-th digit, add n to the count for that digit.
for j in 1 to d
increment the i-th digit by one, including doing any carries
for each digit e from the left of the starting number up to and including
the i-th digit, add 10^i to the count for that digit
for each digit e from the left of the starting number up to and including the
i-th digit, add m to the count for that digit.
set the i-th digit of the starting number to be the i-th digit of the ending
number.
Oh, and since the value of i increases by one each time, keep track of your old 10^i and just multiply it by 10 to get the new one, instead of exponentiating each time.
To reel of the digits from a number, we'd only ever need to do a costly string conversion if we couldnt do a mod, digits can most quickly be pushed of a number like this:
feed=number;
do
{ digit=feed%10;
feed/=10;
//use digit... eg. digitTally[digit]++;
}
while(feed>0)
that loop should be very fast and can just be placed inside a loop of the start to end numbers for the simplest way to tally the digits.
To go faster, for larger range of numbers, im looking for an optimised method of tallying all digits from 0 to number*10^significance
(from a start to end bazzogles me)
here is a table showing digit tallies of some single significant digits..
these are inclusive of 0, but not the top value itself, -that was an oversight
but its maybe a bit easier to see patterns (having the top values digits absent here)
These tallies dont include trailing zeros,
1 10 100 1000 10000 2 20 30 40 60 90 200 600 2000 6000
0 1 1 10 190 2890 1 2 3 4 6 9 30 110 490 1690
1 0 1 20 300 4000 1 12 13 14 16 19 140 220 1600 2800
2 0 1 20 300 4000 0 2 13 14 16 19 40 220 600 2800
3 0 1 20 300 4000 0 2 3 14 16 19 40 220 600 2800
4 0 1 20 300 4000 0 2 3 4 16 19 40 220 600 2800
5 0 1 20 300 4000 0 2 3 4 16 19 40 220 600 2800
6 0 1 20 300 4000 0 2 3 4 6 19 40 120 600 1800
7 0 1 20 300 4000 0 2 3 4 6 19 40 120 600 1800
8 0 1 20 300 4000 0 2 3 4 6 19 40 120 600 1800
9 0 1 20 300 4000 0 2 3 4 6 9 40 120 600 1800
edit: clearing up my origonal
thoughts:
from the brute force table showing
tallies from 0 (included) to
poweroTen(notinc) it is visible that
a majordigit of tenpower:
increments tally[0 to 9] by md*tp*10^(tp-1)
increments tally[1 to md-1] by 10^tp
decrements tally[0] by (10^tp - 10)
(to remove leading 0s if tp>leadingzeros)
can increment tally[moresignificantdigits] by self(md*10^tp)
(to complete an effect)
if these tally adjustments were applied for each significant digit,
the tally should be modified as though counted from 0 to end-1
the adjustments can be inverted to remove preceeding range (start number)
Thanks Aaronaught for your complete and tested answer.
Here's a very bad answer, I'm ashamed to post it. I asked Mathematica to tally the digits used in all numbers from 1 to 1,000,000, no leading 0s. Here's what I got:
0 488895
1 600001
2 600000
3 600000
4 600000
5 600000
6 600000
7 600000
8 600000
9 600000
Next time you're ordering sticky digits for selling in your hardware store, order in these proportions, you won't be far wrong.
I asked this question on Math Overflow, and got spanked for asking such a simple question. One of the users took pity on me and said if I posted it to The Art of Problem Solving, he would answer it; so I did.
Here is the answer he posted:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1741600#1741600
Embarrassingly, my math-fu is inadequate to understand what he posted (the guy is 19 years old...that is so depressing). I really need to take some math classes.
On the bright side, the equation is recursive, so it should be a simple matter to turn it into a recursive function with a few lines of code, by someone who understands the math.
I know this question has an accepted answer but I was tasked with writing this code for a job interview and I think I came up with an alternative solution that is fast, requires no loops and can use or discard leading zeroes as required.
It is in fact quite simple but not easy to explain.
If you list out the first n numbers
1
2
3
.
.
.
9
10
11
It is usual to start counting the digits required from the start room number to the end room number in a left to right fashion, so for the above we have one 1, one 2, one 3 ... one 9, two 1's one zero, four 1's etc. Most solutions I have seen used this approach with some optimisation to speed it up.
What I did was to count vertically in columns, as in hundreds, tens, and units. You know the highest room number so we can calculate how many of each digit there are in the hundreds column via a single division, then recurse and calculate how many in the tens column etc. Then we can subtract the leading zeros if we like.
Easier to visualize if you use Excel to write out the numbers but use a separate column for each digit of the number
A B C
- - -
0 0 1 (assuming room numbers do not start at zero)
0 0 2
0 0 3
.
.
.
3 6 4
3 6 5
.
.
.
6 6 9
6 7 0
6 7 1
^
sum in columns not rows
So if the highest room number is 671 the hundreds column will have 100 zeroes vertically, followed by 100 ones and so on up to 71 sixes, ignore 100 of the zeroes if required as we know these are all leading.
Then recurse down to the tens and perform the same operation, we know there will be 10 zeroes followed by 10 ones etc, repeated six times, then the final time down to 2 sevens. Again can ignore the first 10 zeroes as we know they are leading. Finally of course do the units, ignoring the first zero as required.
So there are no loops everything is calculated with division. I use recursion for travelling "up" the columns until the max one is reached (in this case hundreds) and then back down totalling as it goes.
I wrote this in C# and can post code if anyone interested, haven't done any benchmark timings but it is essentially instant for values up to 10^18 rooms.
Could not find this approach mentioned here or elsewhere so thought it might be useful for someone.
Your approach is fine. I'm not sure why you would ever need anything faster than what you've described.
Or, this would give you an instantaneous solution: Before you actually need it, calculate what you would need from 1 to some maximum number. You can store the numbers needed at each step. If you have a range like your second example, it would be what's needed for 1 to 300, minus what's needed for 1 to 50.
Now you have a lookup table that can be called at will. Doing up to 10,000 would only take a few MB and, what, a few minutes to compute, once?
This doesn't answer your exact question, but it's interesting to note the distribution of first digits according to Benford's Law. For example, if you choose a set of numbers at random, 30% of them will start with "1", which is somewhat counter-intuitive.
I don't know of any distributions describing subsequent digits, but you might be able to determine this empirically and come up with a simple formula for computing an approximate number of digits required for any range of numbers.
If "better" means "clearer," then I doubt it. If it means "faster," then yes, but I wouldn't use a faster algorithm in place of a clearer one without a compelling need.
#!/usr/bin/ruby1.8
def digits_for_range(min, max, leading_zeros)
bins = [0] * 10
format = [
'%',
('0' if leading_zeros),
max.to_s.size,
'd',
].compact.join
(min..max).each do |i|
s = format % i
for digit in s.scan(/./)
bins[digit.to_i] +=1 unless digit == ' '
end
end
bins
end
p digits_for_range(1, 49, false)
# => [4, 15, 15, 15, 15, 5, 5, 5, 5, 5]
p digits_for_range(1, 49, true)
# => [13, 15, 15, 15, 15, 5, 5, 5, 5, 5]
p digits_for_range(1, 10000, false)
# => [2893, 4001, 4000, 4000, 4000, 4000, 4000, 4000, 4000, 4000]
Ruby 1.8, a language known to be "dog slow," runs the above code in 0.135 seconds. That includes loading the interpreter. Don't give up an obvious algorithm unless you need more speed.
If you need raw speed over many iterations, try a lookup table:
Build an array with 2 dimensions: 10 x max-house-number
int nDigits[10000][10] ; // Don't try this on the stack, kids!
Fill each row with the count of digits required to get to that number from zero.
Hint: Use the previous row as a start:
n=0..9999:
if (n>0) nDigits[n] = nDigits[n-1]
d=0..9:
nDigits[n][d] += countOccurrencesOf(n,d) //
Number of digits "between" two numbers becomes simple subtraction.
For range=51 to 300, take the counts for 300 and subtract the counts for 50.
0's = nDigits[300][0] - nDigits[50][0]
1's = nDigits[300][1] - nDigits[50][1]
2's = nDigits[300][2] - nDigits[50][2]
3's = nDigits[300][3] - nDigits[50][3]
etc.
You can separate each digit (look here for a example), create a histogram with entries from 0..9 (which will count how many digits appeared in a number) and multiply by the number of 'numbers' asked.
But if isn't what you are looking for, can you give a better example?
Edited:
Now I think I got the problem. I think you can reckon this (pseudo C):
int histogram[10];
memset(histogram, 0, sizeof(histogram));
for(i = startNumber; i <= endNumber; ++i)
{
array = separateDigits(i);
for(j = 0; k < array.length; ++j)
{
histogram[k]++;
}
}
Separate digits implements the function in the link.
Each position of the histogram will have the amount of each digit. For example
histogram[0] == total of zeros
histogram[1] == total of ones
...
Regards