Getting a specific digit from a ratio expansion in any base (nth digit of x/y)

Question

Is there an algorithm that can calculate the digits of a repeating-decimal ratio without starting at the beginning?

I'm looking for a solution that doesn't use arbitrarily sized integers, since this should work for cases where the decimal expansion may be arbitrarily long.

For example, 33/59 expands to a repeating decimal with 58 digits. If I wanted to verify that, how could I calculate the digits starting at the 58th place?

Edited - with the ratio 2124679 / 2147483647, how to get the hundred digits in the 2147484600th through 2147484700th places.

Tessie

"I'm looking for a solution that doesn't use arbitrarily sized integers" -- my guess is that is not possible without the use of arbitrary-precision integer arithmetic – Streeter

correction: you need integer arithmetic that can handle division and remainder with numbers up to 10x the denominator. – Streeter

Answer 1

OK, 3rd try's a charm :)

I can't believe I forgot about modular exponentiation.

So to steal/summarize from my 2nd answer, the nth digit of x/y is the 1st digit of (10^n-1x mod y)/y = floor(10 * (10^n-1x mod y) / y) mod 10.

The part that takes all the time is the 10^n-1 mod y, but we can do that with fast (O(log n)) modular exponentiation. With this in place, it's not worth trying to do the cycle-finding algorithm.

However, you do need the ability to do (a * b mod y) where a and b are numbers that may be as large as y. (if y requires 32 bits, then you need to do 32x32 multiply and then 64-bit % 32-bit modulus, or you need an algorithm that circumvents this limitation. See my listing that follows, since I ran into this limitation with Javascript.)

So here's a new version.

function abmody(a,b,y)
{
  var x = 0;
  // binary fun here
  while (a > 0)
  {
    if (a & 1)
      x = (x + b) % y;
    b = (2 * b) % y;
    a >>>= 1;
  }
  return x;
}

function digits2(x,y,n1,n2)
{
  // the nth digit of x/y = floor(10 * (10^(n-1)*x mod y) / y) mod 10.
  var m = n1-1;
  var A = 1, B = 10;
  while (m > 0)
  {
    // loop invariant: 10^(n1-1) = A*(B^m) mod y

    if (m & 1)
    {
      // A = (A * B) % y    but javascript doesn't have enough sig. digits
      A = abmody(A,B,y);
    }
    // B = (B * B) % y    but javascript doesn't have enough sig. digits
    B = abmody(B,B,y);
    m >>>= 1;
  }

  x = x %  y;
  // A = (A * x) % y;
  A = abmody(A,x,y);

  var answer = "";
  for (var i = n1; i <= n2; ++i)
  {
    var digit = Math.floor(10*A/y)%10;
    answer += digit;
    A = (A * 10) % y;
  }
  return answer;
}

(You'll note that the structures of abmody() and the modular exponentiation are the same; both are based on Russian peasant multiplication.) And results:

js>digits2(2124679,214748367,214748300,214748400)
20513882650385881630475914166090026658968726872786883636698387559799232373208220950057329190307649696
js>digits2(122222,990000,100,110)
65656565656
js>digits2(1,7,1,7)
1428571
js>digits2(1,7,601,607)
1428571
js>digits2(2124679,2147483647,2147484600,2147484700)
04837181235122113132440537741612893408915444001981729642479554583541841517920532039329657349423345806

Streeter

Answer 2

edit: (I'm leaving post here for posterity. But please don't upvote it anymore: it may be theoretically useful but it's not really practical. I have posted another answer which is much more useful from a practical point of view, doesn't require any factoring, and doesn't require the use of bignums.)

---

@Daniel Bruckner has the right approach, I think. (with a few additional twists required)

Maybe there's a simpler method, but the following will always work:

Let's use the examples q = x/y = 33/57820 and 44/65 in addition to 33/59, for reasons that may become clear shortly.

Step 1: Factor the denominator (specifically factor out 2's and 5's)

Write q = x/y = x/(2^a₂5^a₅z). Factors of 2 and 5 in the denominator do not cause repeated decimals. So the remaining factor z is coprime to 10. In fact, the next step requires factoring z, so you might as well factor the whole thing.

Calculate a₁₀ = max(a₂, a₅) which is the smallest exponent of 10 that is a multiple of the factors of 2 and 5 in y.

In our example 57820 = 2 * 2 * 5 * 7 * 7 * 59, so a₂ = 2, a₅ = 1, a₁₀ = 2, z = 7 * 7 * 59 = 2891.

In our example 33/59, 59 is a prime and contains no factors of 2 or 5, so a₂ = a₅ = a₁₀ = 0.

In our example 44/65, 65 = 5*13, and a₂ = 0, a₅ = a₁₀ = 1.

Just for reference I found a good online factoring calculator here. (even does totients which is important for the next step)

Step 2: Use Euler's Theorem or Carmichael's Theorem.

What we want is a number n such that 10ⁿ - 1 is divisible by z, or in other words, 10ⁿ ≡ 1 mod z. Euler's function φ(z) and Carmichael's function λ(z) will both give you valid values for n, with λ(z) giving you the smaller number and φ(z) being perhaps a little easier to calculate. This isn't too hard, it just means factoring z and doing a little math.

φ(2891) = 7 * 6 * 58 = 2436

λ(2891) = lcm(7*6, 58) = 1218

This means that 10²⁴³⁶ ≡ 10¹²¹⁸ ≡ 1 (mod 2891).

For the simpler fraction 33/59, φ(59) = λ(59) = 58, so 10⁵⁸ ≡ 1 (mod 59).

For 44/65 = 44/(5*13), φ(13) = λ(13) = 12.

So what? Well, the period of the repeating decimal must divide both φ(z) and λ(z), so they effectively give you upper bounds on the period of the repeating decimal.

Step 3: More number crunching

Let's use n = λ(z). If we subtract Q' = 10^a₁₀x/y from Q'' = 10^{(a₁₀ + n)}x/y, we get:

m = 10^a₁₀(10ⁿ - 1)x/y

which is an integer because 10^a₁₀ is a multiple of the factors of 2 and 5 of y, and 10ⁿ-1 is a multiple of the remaining factors of y.

What we've done here is to shift left the original number q by a₁₀ places to get Q', and shift left q by a₁₀ + n places to get Q'', which are repeating decimals, but the difference between them is an integer we can calculate.

Then we can rewrite x/y as m / 10^a₁₀ / (10ⁿ - 1).

Consider the example q = 44/65 = 44/(5*13)

a₁₀ = 1, and λ(13) = 12, so Q' = 10¹q and Q'' = 10¹²⁺¹q.

m = Q'' - Q' = (10¹² - 1) * 10¹ * (44/65) = 153846153846*44 = 6769230769224

so q = 6769230769224 / 10 / (10¹² - 1).

The other fractions 33/57820 and 33/59 lead to larger fractions.

Step 4: Find the nonrepeating and repeating decimal parts.

Notice that for k between 1 and 9, k/9 = 0.kkkkkkkkkkkkk...

Similarly note that a 2-digit number kl between 1 and 99, k/99 = 0.klklklklklkl...

This generalizes: for k-digit patterns abc...ij, this number abc...ij/(10^k-1) = 0.abc...ijabc...ijabc...ij...

If you follow the pattern, you'll see that what we have to do is to take this (potentially) huge integer m we got in the previous step, and write it as m = s*(10ⁿ-1) + r, where 1 ≤ r < 10ⁿ-1.

This leads to the final answer:

• s is the non-repeating part
• r is the repeating part (zero-padded on the left if necessary to ensure that it is n digits)
• with a₁₀ = 0, the decimal point is between the nonrepeating and repeating part; if a₁₀ > 0 then it is located a₁₀ places to the left of the junction between s and r.

For 44/65, we get 6769230769224 = 6 * (10¹²-1) + 769230769230

s = 6, r = 769230769230, and 44/65 = 0.6769230769230 where the underline here designates the repeated part.

You can make the numbers smaller by finding the smallest value of n in step 2, by starting with the Carmichael function λ(z) and seeing if any of its factors lead to values of n such that 10ⁿ ≡ 1 (mod z).

update: For the curious, the Python interpeter seems to be the easiest way to calculate with bignums. (pow(x,y) calculates x^y, and // and % are integer division and remainder, respectively.) Here's an example:

>>> N = pow(10,12)-1
>>> m = N*pow(10,1)*44//65
>>> m
6769230769224
>>> r=m%N
>>> r
769230769230
>>> s=m//N
>>> s
6
>>> 44/65
0.67692307692307696

>>> N = pow(10,58)-1
>>> m=N*33//59
>>> m
5593220338983050847457627118644067796610169491525423728813
>>> r=m%N
>>> r
5593220338983050847457627118644067796610169491525423728813
>>> s=m//N
>>> s
0
>>> 33/59
0.55932203389830504

>>> N = pow(10,1218)-1
>>> m = N*pow(10,2)*33//57820
>>> m
57073676928398478035281909373919059149083362158422691110342442061570390868211691
45624351435489450017295053614666205465236942234520927014873746108612936700103770
32168799723279142165340712556208924247665167762020062262193012798339674852992044
27533725354548599100657212037357315807679003804911795226565202352127291594603943
27222414389484607402282947077135939121411276374956762365963334486336907644413697
68246281563472846765824974057419578000691802144586648218609477689380837080594949
84434451746800415081286751988931165686613628502248356969906606710480802490487720
51193358699411968177101349014181943964026288481494292632307160152196471809062608
09408509166378415773088896575579384296091317883085437564856451054998270494638533
37945347630577654790729851262538913870632998962296783120027672085783465928744379
10757523348322379799377378069872016603251470079557246627464545140089934278796264
26841923209961950882047734347976478727084053960567277758561051539259771705292286
40608785887236250432376340366655136630923555863023175371843652715323417502594258
04219993081978554133517813905223106191629194050501556554825319958491871324801106
88343133863714977516430300933932895191975095122794880664130058803182289865098581
80560359737115185
>>> r=m%N
>>> r
57073676928398478035281909373919059149083362158422691110342442061570390868211691
45624351435489450017295053614666205465236942234520927014873746108612936700103770
32168799723279142165340712556208924247665167762020062262193012798339674852992044
27533725354548599100657212037357315807679003804911795226565202352127291594603943
27222414389484607402282947077135939121411276374956762365963334486336907644413697
68246281563472846765824974057419578000691802144586648218609477689380837080594949
84434451746800415081286751988931165686613628502248356969906606710480802490487720
51193358699411968177101349014181943964026288481494292632307160152196471809062608
09408509166378415773088896575579384296091317883085437564856451054998270494638533
37945347630577654790729851262538913870632998962296783120027672085783465928744379
10757523348322379799377378069872016603251470079557246627464545140089934278796264
26841923209961950882047734347976478727084053960567277758561051539259771705292286
40608785887236250432376340366655136630923555863023175371843652715323417502594258
04219993081978554133517813905223106191629194050501556554825319958491871324801106
88343133863714977516430300933932895191975095122794880664130058803182289865098581
80560359737115185
>>> s=m//N
>>> s
0
>>> 33/57820
0.00057073676928398479

with the overloaded Python % string operator usable for zero-padding, to see the full set of repeated digits:

>>> "%01218d" % r
'0570736769283984780352819093739190591490833621584226911103424420615703908682116
91456243514354894500172950536146662054652369422345209270148737461086129367001037
70321687997232791421653407125562089242476651677620200622621930127983396748529920
44275337253545485991006572120373573158076790038049117952265652023521272915946039
43272224143894846074022829470771359391214112763749567623659633344863369076444136
97682462815634728467658249740574195780006918021445866482186094776893808370805949
49844344517468004150812867519889311656866136285022483569699066067104808024904877
20511933586994119681771013490141819439640262884814942926323071601521964718090626
08094085091663784157730888965755793842960913178830854375648564510549982704946385
33379453476305776547907298512625389138706329989622967831200276720857834659287443
79107575233483223797993773780698720166032514700795572466274645451400899342787962
64268419232099619508820477343479764787270840539605672777585610515392597717052922
86406087858872362504323763403666551366309235558630231753718436527153234175025942
58042199930819785541335178139052231061916291940505015565548253199584918713248011
06883431338637149775164303009339328951919750951227948806641300588031822898650985
8180560359737115185'

Streeter

I'm still trying to wrap my head around the math here. It seems that doing the big number math is more complex than just doing the "long division" to get the digits and then finding repeats with cycle detection or duplicate remainders. The long division method gives one of 'denominator' remainders, in some sequence, I hope there's a clever way to shortcut and get straight to the desired remainder. – Tessie

Except that cycle detection is never going to be foolproof from looking at digits alone: Is 0.12345123451234512345 really enough to say that the repeat digits are 12345? If you keep going and you get 0.123451234512345123456123451234512345123456 then you get a different answer. – Streeter

But you do have a point about long division: there's less state involved. (the 33/57820 example is a good illustration of this!) I might try to combine long division with the Carmichael function which is guaranteed to give you a multiple of the cycle length. – Streeter

Still, you have to know where the nonrepeating part ends. – Streeter

@JasonS When the remainder (not the digits!) forms a cycle you have found the full recurring decimal portion. – Mireielle

@Mireielle oh, that's true. – Streeter

Answer 3

As a general technique, rational fractions have a non-repeating part followed by a repeating part, like this:

nnn.xxxxxxxxrrrrrr

xxxxxxxx is the nonrepeating part and rrrrrr is the repeating part.

1. Determine the length of the nonrepeating part.
2. If the digit in question is in the nonrepeating part, then calculate it directly using division.
3. If the digit in question is in the repeating part, calculate its position within the repeating sequence (you now know the lengths of everything), and pick out the correct digit.

The above is a rough outline and would need more precision to implement in an actual algorithm, but it should get you started.

Statesman

Doesn't step 1 involve calculating the digits to 2x the repeat point? Step 2 would require division of large fractional integers? – Tessie

Step 1 can be done without actually calculating the digits themselves. To do that, find the smallest number that when multiplied by the denominator, gives a number that is of the form 999000, ie. some number of 9s followed by 0s. The number of 0s is the length of the nonrepeating part, and the number of 9s is the length of the repeating part. – Statesman

Answer 4

AHA! caffiend: your comment to my other (longer) answer (specifically "duplicate remainders") leads me to a very simple solution that is O(n) where n = the sum of the lengths of the nonrepeating + repeating parts, and requires only integer math with numbers between 0 and 10*y where y is the denominator.

Here's a Javascript function to get the nth digit to the right of the decimal point for the rational number x/y:

function digit(x,y,n) 
{ 
   if (n == 0) 
      return Math.floor(x/y)%10; 
   return digit(10*(x%y),y,n-1);
}

It's recursive rather than iterative, and is not smart enough to detect cycles (the 10000th digit of 1/3 is obviously 3, but this keeps on going until it reaches the 10000th iteration), but it works at least until the stack runs out of memory.

Basically this works because of two facts:

• the nth digit of x/y is the (n-1)th digit of 10x/y (example: the 6th digit of 1/7 is the 5th digit of 10/7 is the 4th digit of 100/7 etc.)
• the nth digit of x/y is the nth digit of (x%y)/y (example: the 5th digit of 10/7 is also the 5th digit of 3/7)

We can tweak this to be an iterative routine and combine it with Floyd's cycle-finding algorithm (which I learned as the "rho" method from a Martin Gardner column) to get something that shortcuts this approach.

Here's a javascript function that computes a solution with this approach:

function digit(x,y,n,returnstruct)
{
  function kernel(x,y) { return 10*(x%y); }

  var period = 0;
  var x1 = x;
  var x2 = x;
  var i = 0;
  while (n > 0)
  {
    n--;
    i++;
    x1 = kernel(x1,y); // iterate once
    x2 = kernel(x2,y);
    x2 = kernel(x2,y); // iterate twice  

    // have both 1x and 2x iterations reached the same state?
    if (x1 == x2)
    {
      period = i;
      n = n % period;
      i = 0; 
      // start again in case the nonrepeating part gave us a
      // multiple of the period rather than the period itself
    }
  }
  var answer=Math.floor(x1/y);
  if (returnstruct)
    return {period: period, digit: answer, 
      toString: function() 
      { 
        return 'period='+this.period+',digit='+this.digit;
      }};
  else
    return answer;
}

And an example of running the nth digit of 1/700:

js>1/700
0.0014285714285714286
js>n=10000000
10000000
js>rs=digit(1,700,n,true)
period=6,digit=4
js>n%6
4
js>rs=digit(1,700,4,true)
period=0,digit=4

Same thing for 33/59:

js>33/59
0.559322033898305
js>rs=digit(33,59,3,true)
period=0,digit=9
js>rs=digit(33,59,61,true)
period=58,digit=9
js>rs=digit(33,59,61+58,true)
period=58,digit=9

And 122222/990000 (long nonrepeating part):

js>122222/990000
0.12345656565656565
js>digit(122222,990000,5,true)
period=0,digit=5
js>digit(122222,990000,7,true)
period=6,digit=5
js>digit(122222,990000,9,true)
period=2,digit=5
js>digit(122222,990000,9999,true)
period=2,digit=5
js>digit(122222,990000,10000,true)
period=2,digit=6

Here's another function that finds a stretch of digits:

// find digits n1 through n2 of x/y
function digits(x,y,n1,n2,returnstruct)
{
  function kernel(x,y) { return 10*(x%y); }

  var period = 0;
  var x1 = x;
  var x2 = x;
  var i = 0;
  var answer='';
  while (n2 >= 0)
  {
    // time to print out digits?
    if (n1 <= 0) 
      answer = answer + Math.floor(x1/y);

    n1--,n2--;
    i++;
    x1 = kernel(x1,y); // iterate once
    x2 = kernel(x2,y);
    x2 = kernel(x2,y); // iterate twice  

    // have both 1x and 2x iterations reached the same state?
    if (x1 == x2)
    {
      period = i;
      if (n1 > period)
      {
        var jumpahead = n1 - (n1 % period);
        n1 -= jumpahead, n2 -= jumpahead;
      }
      i = 0; 
      // start again in case the nonrepeating part gave us a
      // multiple of the period rather than the period itself
    }    
  }
  if (returnstruct)
    return {period: period, digits: answer, 
      toString: function() 
      { 
        return 'period='+this.period+',digits='+this.digits;
      }};
  else
    return answer;
}

I've included the results for your answer (assuming that Javascript #'s didn't overflow):

js>digit(1,7,1,7,true)
period=6,digits=1428571
js>digit(1,7,601,607,true)
period=6,digits=1428571
js>1/7
0.14285714285714285
js>digit(2124679,214748367,214748300,214748400,true)
period=1759780,digits=20513882650385881630475914166090026658968726872786883636698387559799232373208220950057329190307649696
js>digit(122222,990000,100,110,true)
period=2,digits=65656565656

Streeter

O(N) still seems brutal for a denominator like 214748367 or more I can think of optimizations such as using a larger word size to reduce the constant factor, but then I think it's more difficult to track the intermediate remainders. – Tessie

It's not the O(N) where N is the denominator, but O(N) where N is the period. – Streeter

oops, I meant N = the period + the nonrepeating length. – Streeter

With both sides of the ratio prime, isn't the period the same as the denominator? I suppose that 2124679/214748367 won't run out of memory, just take lots of patience? For prime denominators, any fraction results in the same sequence, just rotated? – Tessie

if the denominator is prime, the period = denominator - 1. But your for your tougher example, 214748367 = 3*89*191*4211 and I get a period of 1759780 when I run my script. – Streeter

Yeah, omitted a digit - the thought was 2^31 - 1, so 2124679/2147483647. How long does the simple script take for 1759780? – Tessie

Several seconds, it's JSDB Javascript based on Mozilla's Spidermonkey (though JSDB is faster than Rhino). I would expect C++ to be quite a bit faster. It's O(N) and the loop operations here are pretty simple. – Streeter

But you'd need 64-bit math to handle 2^31-1, since you have to multiply by 10 before doing a remainder. – Streeter

True if you have a monolithic implementation of ab mod c, or you use an algorithm which does the same w/o requiring larger intermediate products. Otherwise you need to do ab first, then mod c. – Streeter

Answer 5

Ad hoc I have no good idea. Maybe continued fractions can help. I am going to think a bit about it ...

UPDATE

From Fermat's little theorem and because 39 is prime the following holds. (= indicates congruence)

10^39 = 10 (39)

Because 10 is coprime to 39.

10^(39 - 1) = 1 (39)

10^38 - 1 = 0 (39)

[to be continued tomorow]

I was to tiered to recognize that 39 is not prime ... ^^ I am going to update and the answer in the next days and present the whole idea. Thanks for noting that 39 is not prime.

The short answer for a/b with a < b and an assumed period length p ...

• calculate k = (10^p - 1) / b and verify that it is an integer, else a/b has not a period of p
• calculate c = k * a
• convert c to its decimal represenation and left pad it with zeros to a total length of p
• the i-th digit after the decimal point is the (i mod p)-th digit of the paded decimal representation (i = 0 is the first digit after the decimal point - we are developers)

Example

a = 3
b = 7
p = 6

k = (10^6 - 1) / 7
  = 142,857

c = 142,857 * 3
  = 428,571

Padding is not required and we conclude.

3     ______
- = 0.428571
7

Verdha

but it is coprime to 10 and that's what counts. (but you're right, the answer should be edited to reflect this) – Streeter

oops, no it's not sufficient, Fermat's Little Theorem only works on prime exponents. – Streeter

Uuuuppps! It was 04:00 AM when I answered the question and I was quite tired. This might explain why I believed that 39 is prime ... ;) I am going to fix that over the weekend. – Ecstatics

a=1 b=6 p=1, but k=(10^p-1)/b=9/6=1.5 not integral. will your solution only work for prime denominator? – Cassandracassandre