Beside *10^n, is there a faster way to add thousands of zeros to end of a number?



In[230]:= First[Timing[prime*10^100000000]]

Out[230]= 3.625

In[25]:= First[
   Join[IntegerDigits[prime], ConstantArray[0, 100000000]]]]]

Out[25]= 7.547


According to Mr.Wizard answer:

In[46]:= First[Timing[2213*1*^200000000]]

Out[46]= 0.015 (* The real timing on the clock is 5 seconds *)

In[45]:= First[Timing[2213*10^200000000]]

Out[45]= 5.11

In[3]:= First[Timing[ToExpression["2213*1*^200000000"]]]

Out[3]= 5.078

But the strange thing is that by using special 1*^ notation, the reported timing is almost zero while by clock it takes roughly the same time of 10^n. Maybe Timing function doesn't take Mathematica built-in notations into account.!!

Updated 2

Yet The fastest method to beat native *^10 adding zero method is by shifting, even though the improvement is very tiny.

In[16]:= First[Timing[1223*10^100000000]]

Out[16]= 2.625

In[17]:= First[Timing[BitShiftLeft[1223*5^100000000, 100000000]]]

Out[17]= 2.547
  • $\begingroup$ Can you provide the problem example where performance suffers from slow expanding of number by zeroes at its end? $\endgroup$ – Piotr Semenov Nov 24 '12 at 12:35
  • $\begingroup$ In the Karatsuba multiplication where number = A*10^n + B and numbernumber = AA*10^2n + 2AB*10^n + BB. n can be very large. $\endgroup$ – Mohsen Afshin Nov 24 '12 at 13:02
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    $\begingroup$ I think there is no way to improve this thing in Mathematica: 1) obviously, translation to/from list representation of integers makes your code slower. 2) using Compile also will give no effect because you use very big numbers those cannot be fitted by machine numbers. Mathematica is well suited for symbolic computations and prototyping of numerical algorithms but not for great numerical performance. $\endgroup$ – Piotr Semenov Nov 24 '12 at 13:51
  • $\begingroup$ @Piotr Semenov What improved speeds can you obtain, in any other language, when multiplying numbers in the 10^8 digit size range? $\endgroup$ – Daniel Lichtblau Nov 24 '12 at 21:31
  • $\begingroup$ @Daniel Lichtblau I mean the absolute time that program takes to run. For example, I am not ok with 1.5 hours that Mathematica spends doing some stuff. So in this case I prefer writing fast C++ code that takes several seconds of time to do the same stuff. So I claimed that you have to choose more suitable environment to have great performance. Mathematica's usage is limited for applications those are to be as fast as possible. $\endgroup$ – Piotr Semenov Nov 25 '12 at 11:04

My original answer is incorrect -- it is preserved as a record of my own hubris. :^)

Simply, as Rojo points out, the calculation is still being done with 1*^1000, it's just being done at a different time. One may see this by manually observing the time taken for evaluation on an idle machine, or by setting this option which will print the total time taken to evaluate a Cell of code:

SetOptions[$FrontEnd, EvaluationCompletionAction -> "ShowTiming"]

The syntax n*^x is still useful when making definitions as it evaluates before the definition is made. For example, a function defined f1 = # * 1*^100000000 & will be faster in use than f2 = # * 10^100000000 & because the huge coefficient is only calculated once, but it is still calculated.

Logically, since Mathematica still runs on a binary system, a mantissa of two does not have the overhead that a decimal one has:

(* Time: 2.18 seconds *)

(* Time: 0.00 seconds *)

There is a simple solution to this problem and it relies on a specific input form (*^):

prime = Prime @ 813580;





Crucially, 1*^1000 is directly interpreted by Mathematica as the number 101000 while 10^1000 actually calculates this number.

This turns out to be anything but "simple" as I cannot think of a clean way to insert exponents programmatically into the input syntax n*^exp in a manner than retains its benefit.

Hardly nice, but at least as a proof of concept one could bodge this together with $PreRead:

$PreRead = # /. 
    RowBox[{"zeros", "[", exp_, "]"}] :> 
     "1*^" <> ToString@ToExpression@exp &;

prime = Prime @ 813580;
e = 100000000;

prime*zeros[e]; // AbsoluteTiming // First


prime*zeros[e] === prime*1*^100000000


  • 1
    $\begingroup$ Your solution does the magic but how can I use it with variables like First[AbsoluteTiming[prime*1*^ e]]. It gives error $\endgroup$ – Mohsen Afshin Nov 24 '12 at 14:15
  • 3
    $\begingroup$ thanks for your reply but using ToExpression causes Mathametica to calculate the result again. Timing become 2.453 again $\endgroup$ – Mohsen Afshin Nov 24 '12 at 14:24
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    $\begingroup$ @MohsenAfshin I'm not surprised; it is a very limited replacement rule as written. You'll also not be able to use zeros @ x or x // zeros. As I said this will need to be made more robust if it is to actually be used. I'm still bothered that I can't find a better way so I keep looking, rather than improving this, but if I come up empty in a day or two I'll rewrite this to be at least passably practical. $\endgroup$ – Mr.Wizard Nov 24 '12 at 23:21
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    $\begingroup$ I think the problem is that 1*^200000 also has to "do the calculation". It't just that it's done in the parsing stage before any timing function can get to report it. So, there's nothing advantagous in principle about trying to do the math while parsing, and as soon as you move the parsing inside a timing function (by using ToExpression) you get a reality report $\endgroup$ – Rojo Nov 25 '12 at 20:24
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    $\begingroup$ @Rojo FACEPALM -- Oh dear, I'm afraid you're right. Talk about embarrassing, but as always I learn from these things. Now, what do I do with this post? $\endgroup$ – Mr.Wizard Nov 25 '12 at 23:54

You could try timing this method:

FromDigits[Join[{123}, ConstantArray[0, 20]]]

12 300 000 000 000 000 000 000

  • 1
    $\begingroup$ IntegerDigits[123] is an extra that can be easily removed. One can perform just by FromDigits[{123}~Join~ConstantArray[0,20]]. But I think this code is slower than just 123 * 10^n due to translation to/from list representation of integer. $\endgroup$ – Piotr Semenov Nov 24 '12 at 13:38
  • $\begingroup$ It doubles processing time $\endgroup$ – Mohsen Afshin Nov 24 '12 at 13:40
  • $\begingroup$ The time consuming phase is FromDigits not the joining phase. $\endgroup$ – Mohsen Afshin Nov 24 '12 at 13:46

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