Re: Seaching in Pi a sequence. Looking for a faster method

• To: mathgroup at smc.vnet.net
• Subject: [mg119659] Re: Seaching in Pi a sequence. Looking for a faster method
• From: rafapa <rafapa at us.es>
• Date: Thu, 16 Jun 2011 04:02:01 -0400 (EDT)
• References: <ita4fv\$m7m\$1@smc.vnet.net>

```On Jun 15, 1:19 pm, Dana DeLouis <dana.... at gmail.com> wrote:
> Hi.  Just something a little different.
> Given piesimo[10^7, 9, 7], You are looking for a specific digit (9) repeated a number of times (7).
>
> Here's what we have so far.
>
> Our String:
> pi=StringDrop[ToString[N[Pi,10^7]],2];
>
> Function:
> piesimoS[m_String]:=First/@StringPosition[pi,m]
>
> Check Timing:
>
> piesimoS["9999999"]//Timing
> {0.2513,{1722776,3389380,4313727,5466169}}
>
> This idea finds all sequences of the digit 9 repeated 7 or more times.
>
> DigitSequence[pi,9,7]//Timing
> {0.08725,{{7,1722776},{7,3389380},{7,4313727},{7,5466169}}}
>
> It appears to be about 3 times faster.
> That code returned the length of the sequence, and the starting position.
>
> (* s - String, d - digit to check, start - Minimum length *)
>
> DigitSequence[s_,d_,start_]:=Module[
> {re,z,t},
> z=StringReplace["n{s,}",{"n"->ToString[d],"s"->ToString[start]}];
> re=RegularExpression[z];
> t=StringPosition[s,re,Overlaps->False];
> t=t/.{x_Integer,y_Integer}:>{y-x+1,x};
> SortBy[SortBy[t,Last],First]
> ]
>
> If you only wanted a specific length, then change rule to "n{s,s}"
>
> If you didn't know how many consecutive 9's there are in a string, then this finds 6 or more:
>
> DigitSequence[pi,9,6]//Timing
> {0.08989,
> {{6,762},{6,193034},{6,1985813},{6,2878443},{6,3062881},
> {6,3529731},{6,6951812},{6,7298585},{6,8498459},{7,1722776},
> {7,3389380},{7,4313727},{7,5466169}}}
>
> So, the digit 9 occurs at most 7 times in a row.
>
> This is more in line with your specific length example.
> This does not look at overlap:
>
> DigitSequence2[s_,d_,length_]:=Module[
> {z,t},
> z=StringReplace["n{s,s}",{"n"->ToString[d],"s"->ToString[length]}];
> t=StringPosition[s,RegularExpression[z],Overlaps->False];
> t/.{x_Integer,y_Integer}:>x
> ]
>
> DigitSequence2[pi,9,7]//Timing
> {0.08325,{1722776,3389380,4313727,5466169}}
>
> = = = = = = = = = =
> HTH  : >)
> Dana DeLouis
> \$Version
> 8.0 for Mac OS X x86 (64-bit) (November 6, 2010)
>
> On Jun 10, 6:38 am, Guillermo Sanchez <guillermo.sanc... at hotmail.com> wrote:
>
>
>
>
>
>
>
> > Dear Group
>
> > I have developed this function
>
> > piesimo[n_, m_, r_] := Module[{a}, a = Split[RealDigits[Pi - 3, 10, n]
> > [[1]]]; Part[Accumulate[Length /@ a], Flatten[Position[a, Table[m,
> > {r}]]] - 1] + 1]
>
> > n is the digits of Pi, after 3, where to search a sequence of m digit
> > r times consecutives.
> > For instance:
>
> > piesimo[10^7, 9, 7]
>
> > Gives that the sequence 9999999 happens in positions:
>
> > {1722776, 3389380, 4313727, 5466169}
>
> > I know that in this group I will find  faster methods. Any idea?
>
> > Guillermo

Well,
it looks like the increased speed is due to the use of
RegularExpression. I modified piesimoS is:

piesimoSR[m_String] :=
First /@ StringPosition[pi, RegularExpression[m], Overlaps -> False]

DigitSequence[pi, 9, 7] // Timing

{0.076988, {{7, 1722776}, {7, 3389380}, {7, 4313727}, {7,
5466169}}}

piesimoS["9999999"] // Timing

{0.204969, {1722776, 3389380, 4313727, 5466169}}

piesimoSR["9999999"] // Timing

{0.072989, {1722776, 3389380, 4313727, 5466169}}

```

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