# Pi Estimation using Integers

I ran across this problem in a high school math competition:

"You must use the integers $1$ to $9$ and only addition, subtraction, multiplication, division, and exponentiation to approximate the number $\pi$ as accurately as possible. Each integer must be used at most one time. Parenthesis are able to be used."

I tried writing a program in MATLAB to solve this problem but it was very inefficient (brute-force) and took too long.

What is the correct answer, and how would I figure it out?

EXAMPLE:

$$\pi \approx \frac {(6+8+2^3)} 7 = \frac {22} 7 \approx 3.142857$$

• I think this question is too tough for a high school math competition. – Lord Soth Jul 16 '13 at 19:39
• Maybe we could work backwards from $22/7$ or the Ramanujan continued fraction.en.wikipedia.org/wiki/Approximations_of_π#20th_century – Torsten Hĕrculĕ Cärlemän Jul 16 '13 at 19:41
• I can beat 22/7 with $3 + (5+8)/(9^2 + 7 + 4) \approx 3.141304$. But I can't think of a way to solve for the best possible answer outside of an exhaustive search. – MartianInvader Jul 16 '13 at 19:45
• May some digits be written in row, without signs + - * / between them? Or all digits separately? Other words: is this example correct: $\dfrac{2485}{791}\approx 3.14159292$?? – Oleg567 Jul 16 '13 at 21:52
• This contest: web.monroecc.edu/MathPuzzler/March13puzzle ? – Oleg567 Jul 28 '13 at 15:24

Here are three approximations which I came up with, just by sitting down and "playing" with the numbers (no brute-force algorithm needed):

1.) $\pi \approx 3+\dfrac {(5 \times 7)^{(\large8/9)^4}} {2^6+1} = 3.1415926539$

2.) $\pi \approx \left (\dfrac{3^7} {5+8+9} -2\right)^{1/4} = 3.141592652$

3.) $\pi \approx \dfrac{7^3+2 \times 6} {(8+5) \times 9 - 4} = \dfrac {355} {113} = 3.1415929$

I'm sure that there are other solutions out there, these are just the first that I could think of.

• playing with them? – Torsten Hĕrculĕ Cärlemän Jul 16 '13 at 19:51
• ${}{}$Software? – Lord Soth Jul 16 '13 at 19:53
• Srinivasa, is that you? – Daniel R Jul 16 '13 at 21:39
• What do you mean just playing around? How on earth did you come up with the first one? It seems so random, but you must've had a reason for setting it up that way. – Ovi Jul 17 '13 at 2:25
• Withouth software or calculators, how did you took the $(8/9)^4$ power of $35$?. With some pain I could get the fourth root of the second expression, after getting the square root two times, still, I don't get the motivation for doing that.The third one is the only human-like identity... – chubakueno Jul 17 '13 at 23:01

I find formula, which provides $11$ correct digits of $\pi$ (including the leading $3.$):

$$\begin{array}{|c|}\hline\\ {\Large \pi \approx \;}{\huge 3 +} \dfrac {\huge 9^{^{\frac{2}{5\cdot 7} - (1+6)^{-4}}}} {\huge 8} \large = 3.1415926535 \color{Tan}{916691}..., \\ \hline \end{array}\tag{1}$$ error (deviation) is equal to $\Delta = 1.875904... \times 10^{-12}$.

The question is closely related to pandigital approximation of $\pi$
(see http://mathworld.wolfram.com/PiApproximations.html), eq. $(32),(35) - (38)$.

Other solutions with high accuracy (which provide $10$ correct digits):

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx {\Large 1+\dfrac{(6-9^{-5})^{(4+7)^{8^{-2}}}}{3}} & = 3.141592653 \color{Tan}{473482}... & (\Delta=1.163107... \times 10^{-10}); & (2) \end{array} \begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx \dfrac{\huge 12^{ \frac{7-\frac{3^{-5}}{6}}{8 \cdot 9}}}{\Large 4} & = 3.141592653 \color{Tan}{437419}... & (\Delta=1.523732... \times 10^{-10}); & (3) \end{array}

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx {\large 56^{4^{\Large-9^{-3^{-2}}}}}-\dfrac{7-1}{8} & = 3.141592653 \color{Tan}{772701}... & (\Delta=1.829078... \times 10^{-10}); & (4) \end{array} formula $(4)$ looks like "telescope".

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx \Bigl(4 \cdot 92^{3/8}\Bigr)^{\Large\frac{1+\frac{6}{7}}{5}} & = 3.141592653 \color{Tan}{854014}... & (\Delta=2.642208... \times 10^{-10}); & (5) \end{array}

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx 94 \cdot \left(\dfrac{8}{13}\right)^{7} - \dfrac{5^{-6}}{2} & = 3.141592653 \color{Tan}{854369}... & (\Delta=2.645759... \times 10^{-10}); & (6) \end{array}

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx 4-1+\dfrac{\Bigl((6-7^{-3})\cdot 9\Bigr)^{\Large 2^{-5}}}{8} & = 3.141592653 \color{Tan}{863925}... & (\Delta=2.741318... \times 10^{-10}); & (7) \end{array}

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx 3+\dfrac{1}{7+4^{-2}} - \left(6-\dfrac{5}{8}\right)^{-9} & = 3.141592653 \color{Tan}{303244}... & (\Delta=2.865490... \times 10^{-10}); & (8) \end{array}

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx {\large 3+\dfrac{1-9^{\Large-(6-8^{-5})}}{7+4^{-2}}} & = 3.141592653 \color{Tan}{904056}... & (\Delta=3.142632... \times 10^{-10}); & (9) \end{array}

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx \color{gray}{3+\dfrac{\left(5 \cdot 7\right)^{(8/9)^4}}{2^6+1}} & \color{gray}{ = 3.141592653 \color{Tan}{915932}...} & \color{gray}{ (\Delta=3.261394... \times 10^{-10});} & \color{gray}{(10)} \end{array} formula $(10)$ is obtained by user14069, see above (cited just for accuracy comparison).

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx \color{gray}{3+\dfrac{1-(9-8^{-5})^{-6}}{7+4^{-2}}} & \color{gray}{= 3.141592653 \color{Tan}{916501}...} & \color{gray}{(\Delta=3.267085... \times 10^{-10});} & \color{gray}{(11)} \end{array} formula $(11)$ is obtained by Ed Pegg, see source (cited just for accuracy comparison).

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx 3+\dfrac{75248}{9^6-1} & = 3.141592653 \color{Tan}{921421}... & (\Delta=3.316278... \times 10^{-10}); & (12) \end{array}

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx 3+\dfrac{1-9^{-6}}{7+4^{-2}} & = 3.141592653 \color{Tan}{921922}... & (\Delta=3.321291... \times 10^{-10}); & (13) \end{array} formula $(13)$ looks very easy; it contains $7$ digits only.

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx 3+\dfrac{5-4-9^{-6}}{7+\dfrac{1}{2 \cdot 8}} & = 3.141592653 \color{Tan}{921922}... & (\Delta=3.321291... \times 10^{-10}); & (13') \end{array}

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx 3+\dfrac{8\cdot(1-9^{-6})}{57-\dfrac{2}{4}} & = 3.141592653 \color{Tan}{921922}... & (\Delta=3.321291... \times 10^{-10}); & (13'') \end{array}

formulas $(13'), (13'')$ are equal to $(13)$, but use all digits.

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx 3+\dfrac{8}{\Bigl(1+9^{-6}\Bigr)\cdot \Bigl(57-\dfrac{2}{4}\Bigr)} & = 3.141592653 \color{Tan}{922423}... & (\Delta=3.326304... \times 10^{-10}); & (14) \end{array}

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx {\large \dfrac{1+\Biggl(\dfrac{2^{\Large 4^{6/5}}}{9}\Biggr)^3}{8}-7} & = 3.141592653 \color{Tan}{226228}... & (\Delta= 3.635644... \times 10^{-10}); & (15) \end{array}

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx \dfrac{3}{\dfrac{1}{4}-{\Large 7^{- \bigl(2+5^{6^{-8}}\bigr)} }}-9 & = 3.141592653 \color{Tan}{223921}... & (\Delta= 3.658712... \times 10^{-10}); & (16) \end{array}

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx 3+\left(8-1+(2 \cdot 5)^{4/9} \right)^{-6/7} &= 3.141592653 \color{Tan}{169312}... &(\Delta=4.204806...\times 10^{-10}); &(17) \end{array}

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx {\Large 3 \cdot \Bigl(1+\frac{4}{9} \Bigr)^{5^{-8^{6 \cdot 7^{-2}}}}} & = 3.141592653 \color{Tan}{160095}... & (\Delta= 4.296978... \times 10^{-10}); & (18) \end{array}

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx 3+\left(8-1+4^{-2}\right)^{\Large{-7^{\Large{5\cdot 6^{-9}}}}} &= 3.141592653 \color{Tan}{130264}... &(\Delta=4.595292... \times 10^{-10}); &(19) \end{array}

\begin{array}{lrrr} \phantom{WWWWWWWWWWWWW} & & & \\ \pi\approx \dfrac{7}{5}\cdot \Biggl(\Large 1 + 4^{\LARGE 2^{\Large -\frac{8-9^{-6}}{3}}}\Biggr) &= 3.141592653 \color{Tan}{021507}... &(\Delta=5.682853... \times 10^{-10}). &(20) \end{array}

Note: formulas $(3) - (6), (12), (13''), (14)$ contain concatenation of digits. Probably this trick is not allowed in contest/task.

And I'd like to say a few words about famous approximation $\pi \approx \dfrac{355}{113}$.
$\pi \approx \dfrac{2485}{791} = 3.141592\color{Tan}{920353}...$ ($7$ correct digits). It contains $1$ operation only.

And about beauty of fraction $\dfrac{355}{113}$:
what a nice pandigital finite generalized continued fractions for $\pi$: $$\begin{array}{|ccc|} \hline & \Large\pi & \\ & \wr\wr & \\ 2+\cfrac{9}{5+\cfrac{4}{1+\cfrac{3}{7+\cfrac{6}{8\color{#DDDDDD}{\small +0}}}}} & = \dfrac{355}{113} = & 3+\cfrac{1}{6+\cfrac{9}{8+\cfrac{4}{5+\cfrac{7}{2\color{#DDDDDD}{\small +0}}}}} \\ \hline \end{array}$$

These formulas looked intriguing, so I decided to try to compute some of them myself with Maple, using the approximation $$\pi \approx \frac{355}{113} \approx 3.1415929.$$

The algorithm uses dynamic programming where the nodes are the subsets of $\{1,2,3,\ldots,8,9\}$ and the corresponding value is a table of all integer values less than some maximum that can be expressed by a legal formula as specified in the problem definition using the digits in the subset. The recursive step splits the argument into two sets and combines the values obtained for them using the five operators that are given. In the present case I used $360$ as the maximum value because $355<360.$ Finally in order to express a fraction we iterate over all 2-partitions of the nine digits, looking to find a pair where the first subset can express the numerator and the second one the denominator. The approximations are not limited to $\pi$, one may try for any fraction with small numerator and denominator.

Here are some of the formulas that I obtained: $$\frac{1 \times \left(2 \times 6 + {7}^{3}\right)}{5 + 9 \times \left(4 + 8\right)}$$ $$\frac{1 + 5 + 6 + {7}^{3}}{{9}^{2} + 4 \times 8}$$ $$\frac{5 \times \left(8 + 9 \times \left(1 + 6\right)\right)}{{7}^{2} + {4}^{3}}$$ $$\frac{5 \times \left(\left(8 \times 9\right) - 1\right)}{7 - \left(\frac{4 - \left({6}^{3}\right)}{2}\right)}$$ $$\frac{4 + 8 + {7}^{3}}{1 - \left(9 - \left({\left(5 + 6\right)}^{2}\right)\right)}$$ $$\frac{\left(2 + 3\right) \times \left(\left(8 \times 9\right) - 1\right)}{\left(4 \times \left(5 \times 6\right)\right) - 7}$$ $$\frac{5 \times \left(\left(6 \times \left(3 + 9\right)\right) - 1\right)}{\left({\left(4 + 7\right)}^{2}\right) - 8}$$ $$\frac{\left(6 - 1\right) \times \left(8 + 7 \times 9\right)}{{2}^{5} + {3}^{4}}$$

Here are some examples obtained by enabling concatenation: $$\frac{5 \times 71}{2 - \left(3 - \left(6 + 9 \times \left(4 + 8\right)\right)\right)}$$ $$\frac{71 \times \left(2 + 3\right)}{4 + 5 + 6 + 98}$$ $$\frac{5 \times \left(73 - 2\right)}{6 - \left(1 - \left(9 \times \left(4 + 8\right)\right)\right)}$$ $$\frac{5 \times \left(9 + 62\right)}{1 + 8 \times \left(3 + 4 + 7\right)}$$ $$\frac{5 \times \left(7 + {8}^{2}\right)}{1 + 3 \times 6 + 94}$$ $$\frac{\left(7 \times 52\right) - 9}{1 - \left(8 \times \left(4 - \left(3 \times 6\right)\right)\right)}$$ $$\frac{5 \times \left(7 + 64\right)}{1 - \left(9 - \left({\left(3 + 8\right)}^{2}\right)\right)}$$ $$\frac{\left(51 \times \left(3 + 4\right)\right) - 2}{8 + 7 \times \left(6 + 9\right)}$$ $$\frac{31 + 9 \times \left({6}^{2}\right)}{4 \times 7 + 85}$$ $$\frac{\left(9 - 4\right) \times \left(72 - 1\right)}{83 + 5 \times 6}$$ $$\frac{5 \times \left(8 + 64 - 1\right)}{92 + 3 \times 7}$$ $$\frac{\left(9 \times 41\right) - \left(6 + 8\right)}{7 + 2 \times 53}$$ $$\frac{4 + 8 + {7}^{\left(\frac{9}{3}\right)}}{51 + 62}$$ $$\frac{5 + 7 \times \left(\left(6 \times 9\right) - 4\right)}{31 + 82}$$

The following examples have special property which is left to the reader to discover. $$\frac{71 \times \left(2 + 8\right)}{4 + 3 \times \left(9 + 65\right)}$$ $$\frac{71 \times \left(4 + 6\right)}{2 \times \left(3 \times 5 + 98\right)}$$ $$\frac{5 \times \left(61 + {3}^{4}\right)}{{2}^{7} + 98}$$ $$\frac{\left(9 \times \left(84 - 5\right)\right) - 1}{\left(3 \times 76\right) - 2}$$ $$\frac{6 - \left(8 \times \left(4 - 92\right)\right)}{1 + 3 \times 75}$$ $$\frac{\left(4 + 6\right) \times \left(8 + 7 \times 9\right)}{1 + {\left(3 \times 5\right)}^{2}}$$

Last but not least, we have $$\frac{3 \times \left(5 \times 71\right)}{9 + \left(8 \times 42\right) - 6}$$ $$\frac{71 \times \left(6 + 9\right)}{2 + 5 + 4 \times 83}$$ $$\frac{41 + 8 \times \left({2}^{7}\right)}{{3}^{5} + 96}$$ $$\frac{\left(6 + 9\right) \times \left(72 - 1\right)}{\left(8 \times 43\right) - 5}$$ $$\frac{3 \times \left(5 \times \left(\left(8 \times 9\right) - 1\right)\right)}{\left({7}^{\left(\frac{6}{2}\right)}\right) - 4}$$ $$\frac{1 + \left(5 + 9\right) \times \left(82 - 6\right)}{\left({7}^{3}\right) - 4}.$$

It would be interesting to know whether the next best approximation, which according to Wikipedia is $$\frac{52163}{16604} \approx 3.141592387$$ can be represented this way, whether there is another fraction with a smaller denominator that produces more correct places and whether one has to add concatenation to the set of operators for this to occur.

This is the code of the Maple program. Enjoy!

with(combinat,powerset);

maxval := 360;

repr :=
proc(s)
local d, f, dstr, fstr, dtstr, ftstr, res, aprob, bprob, x, y,
leftset, rightset, check, typeset;
option remember;

res := table();

if nops(s) = 1 then
d := op(1,s);
res[d] := [d, sprintf("%d", d), sprintf("%d", d)];

return op(res);
fi;

check :=
proc(val)
if type(res[val], list) then return false; fi;
if type(val, integer) and abs(val)<maxval then return true fi;

return false;
end proc;

typeset :=
proc(dtstr, what, ftstr)
local pard, parf;

if length(dtstr)=1 then
pard := dtstr;
else
pard := sprintf("\\left(%s\\right)", dtstr);
fi;
if length(ftstr)=1 then
parf := ftstr;
else
parf := sprintf("\\left(%s\\right)", ftstr);
fi;

if evalb(what = "+") then
return sprintf("%s + %s", dtstr, ftstr);
elif evalb(what = "-") then
return sprintf("%s - %s", pard, parf);
elif evalb(what = "*") then
return sprintf("%s \\times %s", pard, parf);
elif evalb(what = "/") then
return
sprintf("\\frac{%s}{%s}", dtstr, ftstr);
fi;

return sprintf("{%s}^{%s}", pard, parf);
end proc;

for leftset in powerset(s) do
rightset := s minus leftset;

if nops(leftset)=0 or nops(rightset)=0 then next fi;

aprob := repr(leftset);
bprob := repr(rightset);

if nops(op(op(res))) = 2*maxval-1 then next fi;

for x in op(aprob) do for y in op(bprob) do
d := x[1]; dstr:= x[2]; dtstr := x[3];
f := y[1]; fstr:= y[2]; ftstr := y[3];

if check(d+f) then
res[d+f] := [d+f, sprintf("(%s) + (%s)", dstr, fstr),
typeset(dtstr, "+", ftstr)];
fi;

if check(d-f) then
res[d-f] := [d-f, sprintf("(%s) - (%s)", dstr, fstr),
typeset(dtstr, "-", ftstr)];
fi;

if check(d*f) then
res[d*f] := [d*f, sprintf("(%s) * (%s)", dstr, fstr),
typeset(dtstr, "*", ftstr)];
fi;

if f <> 0 and check(d/f) then
res[d/f] := [d/f, sprintf("(%s) / (%s)", dstr, fstr),
typeset(dtstr, "/", ftstr)];
fi;

if not(d=0 or f=0) and evalf(abs(f*log(d))<10)
and check(d^f) then
res[d^f] := [d^f, sprintf("(%s) ^ (%s)", dstr, fstr),
typeset(dtstr, "^", ftstr)];
fi;

od od;
od;

return op(res);
end;

approx_frac :=
proc(what)
local allset, numerset, denomset, x, y,
xt, yt, xpair, ypair;

allset := {seq(k, k=1..9)};

for numerset in powerset(allset) do
denomset := allset minus numerset;

if nops(numerset)>=1 and nops(denomset)>=1 then
xt := repr(numerset);
yt := repr(denomset);

xpair := xt[numer(what)];
ypair := yt[denom(what)];

if type(xpair, list) and type(ypair, list) then
printf("(%s) / (%s) %a %a\n\\frac{%s}{%s}\n\n",
xpair[2], ypair[2],
xpair[1]/ypair[1], evalf(xpair[1]/ypair[1]),
xpair[3], ypair[3]);
fi;
fi;
od;
end;
• The arguments of the function "repr" are subsets of the set of nine digits and the return value is a hash table that maps values that can be computed to three-tuples consisting of the value, a simple string representation of the formula, and a LaTeX representation of the formula, as I did not want to typeset them manually, this being a source of errors. – Marko Riedel Jul 19 '13 at 3:08
• Obviously the space of possible values can be expanded significantly by working with rationals and irrationals as the intermediate values rather than just integers. – Marko Riedel Jul 19 '13 at 4:11

I modified my program to include rationals with small denominators as intermediate values.

Using the approximation $$\pi\approx \sqrt[4]{\frac{2143}{22}} \approx 3.14159265$$ which also has nine decimal places and is apparently due to Ramanujan (Wikipedia), we get the result $$\left(\left(\frac{{3}^{6}}{7 + \frac{8 - 5}{9}}\right) - 2\right)^{1/4}$$

This is a variation of the expression shown in the OP, but computed with limited human intervention.

If we permit concatenation of digits, we get the following pretty result: $$\left(97 + 3 \times \left(\frac{6}{52 - 8}\right)\right)^{1/4}.$$ Another one of this type is $$\left(3 + 96 - \left(\frac{5}{2 + \frac{8}{7}}\right)\right)^{1/4}.$$

More formulas of this type are $$\left(97 + \frac{\frac{3}{5 - \left(\frac{8}{6}\right)}}{2}\right)^{1/4}$$ $$\left(98 - \left(\frac{\frac{7}{2} + 3}{6 + 5}\right)\right)^{1/4}$$ $$\left(\left(\frac{{\left(8 - 5\right)}^{6}}{\frac{3}{9} + 7}\right) - 2\right)^{1/4}$$

The code for the modified algorithm follows:

with(combinat,powerset);

maxnumer := 2200;
denomset := {1, 2, 3, 5, 11, 22};

maxv_comp_const :=
proc()
local s, d;

s:= {};

for d in denomset do
s := s union {seq(k/d, k=0..maxnumer)};
od;

nops(s);
end proc;

maxcount := maxv_comp_const();

repr :=
proc(s)
local d, f, dstr, fstr, dtstr, ftstr, res, aprob, bprob, x, y,
leftset, rightset, check, typeset;
global maxcount;
option remember;

res := table();

if nops(s) = 1 then
d := op(1,s);
res[d] := [d, sprintf("%d", d), sprintf("%d", d)];

return op(res);
fi;

check :=
proc(val)
global maxnumer, denomset;

if not(type(val, rational)) then return false; fi;

if val<0 then return false; fi;
if type(res[val], list) then return false; fi;

if numer(val)<=maxnumer and denom(val) in denomset then
return true
fi;

return false;
end proc;

typeset :=
proc(dtstr, what, ftstr)
local pard, parf;

if length(dtstr)=1 then
pard := dtstr;
else
pard := sprintf("\\left(%s\\right)", dtstr);
fi;
if length(ftstr)=1 then
parf := ftstr;
else
parf := sprintf("\\left(%s\\right)", ftstr);
fi;

if evalb(what = "+") then
return sprintf("%s + %s", dtstr, ftstr);
elif evalb(what = "-") then
return sprintf("%s - %s", pard, parf);
elif evalb(what = "*") then
return sprintf("%s \\times %s", pard, parf);
elif evalb(what = "/") then
return
sprintf("\\frac{%s}{%s}", dtstr, ftstr);
fi;

return sprintf("{%s}^{%s}", pard, parf);
end proc;

for leftset in powerset(s) do
rightset := s minus leftset;

if nops(leftset)=0 or nops(rightset)=0 then next fi;

aprob := repr(leftset);
bprob := repr(rightset);

if nops(op(op(res))) >= maxcount then next fi;

for x in op(aprob) do for y in op(bprob) do
d := x[1]; dstr:= x[2]; dtstr := x[3];
f := y[1]; fstr:= y[2]; ftstr := y[3];

if check(d+f) then
res[d+f] := [d+f, sprintf("(%s) + (%s)", dstr, fstr),
typeset(dtstr, "+", ftstr)];
fi;

if check(d-f) then
res[d-f] := [d-f, sprintf("(%s) - (%s)", dstr, fstr),
typeset(dtstr, "-", ftstr)];
fi;

if check(d*f) then
res[d*f] := [d*f, sprintf("(%s) * (%s)", dstr, fstr),
typeset(dtstr, "*", ftstr)];
fi;

if f <> 0 and check(d/f) then
res[d/f] := [d/f, sprintf("(%s) / (%s)", dstr, fstr),
typeset(dtstr, "/", ftstr)];
fi;

if not(d=0 or f=0) and evalf(abs(f*log(d))<10)
and check(d^f) then
res[d^f] := [d^f, sprintf("(%s) ^ (%s)", dstr, fstr),
typeset(dtstr, "^", ftstr)];
fi;

od od;
od;

return op(res);
end;

approx_pi :=
proc()
local what, allset, allrepr, numerset, denomset, x, y,
xt, yt, xpair, ypair;

what := 2143/22;
allset := {seq(k, k=1..9)} minus {1,4};
allrepr := repr(allset);

xpair := allrepr[what];
if type(xpair, list) then
printf("(%s)^(1/4) %a %a\n", xpair[2],
xpair[1]^(1/4), evalf(xpair[1]^(1/4)));
printf("\\left(%s\\right)^{1/4}\n\n", xpair[3]);
fi;
end;
• There is a vastly improved version of this program at the following stackexchange link. – Marko Riedel Jul 26 '13 at 3:54

$$\frac{71\times5}{63\times2-9-8+4}=\frac{355}{113}=3.141592\color{#A0A0A0}{92}$$

• You can add $8$ as well by writing $-9-8+4$. – Asaf Karagila Jul 20 '13 at 0:02
• @AsafKaragila: thanks, I was just playing around with $8$. – robjohn Jul 20 '13 at 0:02
• Why red and not, say, underlining? – Asaf Karagila Jul 20 '13 at 8:37
• @AsafKaragila: underlining looks link-like. How about gray for the non-matching digits? – robjohn Jul 20 '13 at 9:36
• Yeah, that's less painful. Thanks! Another option was boxing. $$\boxed{3.141592}92$$ – Asaf Karagila Jul 20 '13 at 9:44

Here are some more formulas using Ramanujan's approximation. $$\left(\left(\frac{\frac{{3}^{8}}{\left(9 + 7\right) - 5}}{6}\right) - 2\right)^{1/4}$$ $$\left(\left(\frac{\frac{{3}^{8}}{9 + 2}}{6}\right) - 7 + 5\right)^{1/4}$$ $$\left(\frac{\left(\left(\frac{{3}^{8}}{9 + 2}\right) - 7\right) - 5}{6}\right)^{1/4}$$

$$\frac{71*5}{2^6+49} =\frac{355}{113} = 3.141592.$$