Find the a factor of an integer which is nearest to another integer

Question

If I know two integers $n$ and $m$ and $m < n$, how can I find two different integers $x$ and $y$ that are nearest to $m$, and satisfy Mod[n,x]==Mod[n,y]==0 ?

For example,

f[720,8] ==> {8,9}
f[720,20] ==> {18,20}

If $n$ is small, we can brute force to test whether an integer is a factor of $n$, and find the nearest two of the factors:

twoNear[n_Integer, m_Integer] := Module[{ls, sortls},
  ls = Select[Range[1, n], Mod[n, #] == 0 &];
  sortls = Sort[Transpose[{ls - m, ls}], Abs[#1[[1]]] < Abs[#2[[1]]] &];
  Sort@sortls[[1 ;; 2, 2]]
 ]

twoNear[720,8]
twoNear[720,20]
(*{8, 9}*)
(*{18, 20}*)

but how to deal with problems such as n=20!, m=1*^9 , n=40!,m=1*^20 ?

Shouldn't f[720,8] return {8,8}? Or do you want that $x\ne y$? — whuber, Mar 11 '13 at 22:44
Note that this is a knapsack problem (and NP-Hard), so there will be a limit on what can be achieved. Think of $n$ as giving you a bag of objects; there will be $e_i$ objects with values $\log(p_i)$ when $p_i^{e_i}$ is a term in the prime power factorization of $n$. You want to solve two problems: (1) find subsets of objects whose total values are as large as possible, subject to not exceeding $\log(m)$; and (2) find subsets whose totals are as small as possible, subject to not being less than $\log(m)$. You actually need the best and second-best solutions of both. — whuber, Mar 11 '13 at 23:17

score 8 · Accepted Answer · edited Apr 13 '17 at 12:55

My lucky day (night). I get to take my own answer which was of dubious value here and basically repurpose it to give a good response to this question.

We set this up as a mixed linear program by trying to get sums of logs of factors as close to m as possible, subject to coefficients being nonnegative integers that do not exceed the powers of the corresponding factors. There is an added complication in that we need not just the optimal value but also the next best.

So here we go.

nearDivisors[n_, m_] := Module[
  {fax = FactorInteger[n], prec, logs, maxes, len, c, coeffs, c1, c2, 
   c3, diff, epsilon, constraints, vars, min, vals, res1, res2, res3, 
   newproblems, equalities, best},
  prec = Max[20, Log[10., m]];
  logs = Log[N[fax[[All, 1]], prec]];
  maxes = fax[[All, 2]];
  len = Length[fax];
  coeffs = Array[c, len];
  c1 = Table[0 <= c[j] <= maxes[[j]], {j, len}];
  diff = coeffs.logs - Log[m];
  c2 = {diff <= epsilon, -diff <= epsilon};
  c3 = Element[coeffs, Integers];
  constraints = Join[c1, c2, {c3}];
  vars = Join[coeffs, {epsilon}];
  {min, vals} = FindMinimum[{epsilon, constraints}, vars];
  res1 = Round[Exp[coeffs.logs /. vals]];
  vals = coeffs /. vals;
  newproblems = Flatten[Table[
     equalities = Table[c[k] == vals[[k]], {k, j - 1}];
     {Join[constraints, equalities, {c[j] <= vals[[j]] - 1}],
      Join[constraints, equalities, {c[j] >= vals[[j]] + 1}]}
     , {j, len}], 1];
  best = Infinity;
  Do[min = 
    Quiet[FindMinimum[{epsilon, newproblems[[j]]}, vars, 
      WorkingPrecision -> prec]];
   If[Head[min] =!= FindMinimum,
    If[min[[1]] < best,
     best = min[[1]];
     res2 = Round[Exp[coeffs.logs /. min[[2]]]];
     ]
    ], {j, Length[newproblems]}];
  {res1, res2}
  ]

Here are the now-standard benchmark tests. We also check that we used adequate precision so that we can accurately recover the results.

Timing[ndsmall = nearDivisors[20!, 10^9]]

(* Out[967]= {1.840000, {999949860, 1000194048}} *)

20!/ndsmall

(* Out[968]= {2433024000, 2432430000} *)

Timing[ndmiddle = nearDivisors[30!, 10^9]]

(* Out[971]= {20.350000, {999949860, 1000065000}} *)

30!/ndmiddle

(* Out[972]= {265266160257466368000000, 265235619496923758592000} *)

Now we get ambitious.

Timing[ndbig = nearDivisors[40!, 10^20]]

(* Out[969]= {41.020000, {99999622686575390625, 100000079233442099712}} *)

40!/ndbig

(* Out[970]= {8159183618174107243291607040, 8159146367706464256000000000} *)

--- edit ---

Okay, why not...

Timing[ndbigger = nearDivisors[80!, 10^20]]
(* Out[1017]= {2070.830000, {100000004179992913920, 
  100000028588579559040}} *)

80!/ndbigger

(* Out[1018]= \
{715694540546656942785553897344003751254098684983349359903188074751331\
095778689024000000000000000000, \
7156943658557848407002572034351080637013939631522009525004233001820750\
27228721152000000000000000000} *)

--- end edit ---

@sOrce Knapsack questions are always interesting. But reusing an answer just a few minutes later, that's a rare thing. — Daniel Lichtblau, Mar 12 '13 at 21:29

kglr · Answer 2 · 2013-03-11T23:27:09.993

2

Brute-forcing for "small" n:

 nrstDvsrsF = Nearest[Divisors[#1], #2, 2] &
 nrstDvsrsF[720, 8]
 (* {8, 9} *)
 nrstDvsrsF[720, 20]
 (* {20, 18} *)
 AbsoluteTiming[nrstDvsrsF[20!, 1*^9]]
 (* {0.031003, {999949860, 1000194048}} *)
 AbsoluteTiming[nrstDvsrsF[30!, 1*^9]]
 (* {14.320432, {999949860, 1000065000}} *)

Warning: As noted by @whuber in comments, do not try this with n=40!,m=1*^20.

edited Mar 11 '13 at 23:27

answered Mar 11 '13 at 22:47

kglr

394,356
18
477
896

And what happens to RAM usage and computing time when you try the last example in the question? :-) (Actually, that one is barely manageable--why not increase the $40!$ to $80!$, though...) – whuber Mar 11 '13 at 22:53
@whuber @ kguler I can't get an answer for the example n=40!,m=1*^20 before my disk is full, so I have quit the job. – user0501 Mar 11 '13 at 22:56
@whuber, still waiting to see - with fingers crossed that I will not have to restart Windows :) – kglr Mar 11 '13 at 22:56

Find the a factor of an integer which is nearest to another integer

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