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Integer Factorization: Trial Division Algorithm

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15 Dec 2023CPOL6 min read 39.1K   446   11   24
Small review of Trial Division algorithm
In this article, we will do a small review of the trial division algorithm. We will look at the classic variants of trial division including brute force, basic version, getting rid of even factors, the wheel and list of primes. Finally, we will take a look at a benchmark.

Background

This article is part of a set of articles. Each article highlights an aspect around Integer Factorization.

xPrompt.zip contains the binary of the xHarbour Scripting Engine for Windows. This is the app needed to run the prg code.

Usage:

• Copy prg file in same directory than xPrompt
• Launch xPrompt
• Type "do FileName.prg" to run the code

xHarbour is freeware and can be downloaded at xHarbour.org[^]

xHarbour is part of the xBase family languages: xBase - Wikipedia[^]

Flowcharts

The flowcharts are related to this article : TopDown Analyze Method with Tree Graphs as Support[^]

If image is too small, open in a new tab and zoom, it is a svg file.

Introduction

When starting to play with Integer Factorization, trying all possible factors is the first idea, that algorithm is named Trial Division.

The algorithm has two purposes - finding a prime factor or finding if an integer is a prime by not by finding a prime factor.

Since the algorithm is about finding a factor, the worst case is when the integer to factorize is a prime.

Trial Division: The Classic Variants

It looks obvious that the most efficient method is to check only prime numbers, but handling the list of primes is a problem by itself. The classical variants exist as solutions to avoid that problem. As methods get more sophisticated, they remove more useless non prime numbers, thus being more efficient.

Brute Force

Some newbies are testing every single number below n. That's overkill. The inefficiency remains hidden as long the integer to factorize is not a prime.

The maximum cost for n is n-2 divisions. Complexity is O(n).

C++
// Trial Division: Brute Force 1
// Check all numbers until Cand - 1
long long TD_BF1(long long Cand) {
Count = 0;
long long Top = Cand - 1;
for (long long Div = 2; Div <= Top; Div++) {
Count++;
if (Cand % Div == 0) {
return Div;
}
}
return Cand;
}
dBase
//  Trial Division Brute Force 1
// May 2019
function TD_BF1(Prod)
Local D, Top
Top= Prod-1
for D= 2 to Top
if Prod % D = 0
return D
endif
next
return Prod

In TrialDiv.cpp and TrialDiv.prg in TrialDiv.zip.

TD_BF1.graphml and TD_BF1.svg in TrialDiv.zip.

Testing all numbers until n/2 is better, but is also overkill. Just like with previous method, the inefficiency remains hidden as long the integer to factorize is not a prime.

The maximum cost for n is n / 2 divisions. Complexity is O(n).

C++
// Trial Division: Brute Force 2
// Check all numbers until Cand / 2
long long TD_BF2(long long Cand) {
Count = 0;
long long Top = Cand / 2;
for (long long Div = 2; Div <= Top; Div++) {
Count++;
if (Cand % Div == 0) {
return Div;
}
}
return Cand;
}
dBase
//  Trial Division Brute Force 2
// May 2019
function TD_BF2(Prod)
Local D, Top
Top= Prod/2
for D= 2 to Top
if Prod % D = 0
return D
endif
next
return Prod

In TrialDiv.cpp and TrialDiv.prg in TrialDiv.zip.

TD_BF2.graphml and TD_BF2.svg in TrialDiv.zip.

Basic Version: Square Root

A little analysis shows that there is no factor to check after the square root.

The maximum cost for n is √n divisions. Complexity id O(√n).

C++
// Trial Division: Square Root
// Check all numbers until Square Root
long long TD_SR(long long Cand) {
Count = 0;
long long Top = sqrt(Cand);
for (long long Div = 2; Div <= Top; Div++) {
Count++;
if (Cand % Div == 0) {
return Div;
}
}
return Cand;
}
dBase
//  Trial Division Square Root
// May 2019
function TD_SR(Prod)
Local D, Top
Top= int(sqrt(Prod))
for D= 2 to Top
if Prod % D = 0
return D
endif
next
return Prod

In TrialDiv.cpp and TrialDiv.prg in TrialDiv.zip.

TD_SR.graphml and TD_SR.svg in TrialDiv.zip.

Get Rid of Even Factors

With further analysis, one can see that there is no even factor beyond 2.

The maximum cost for n is (√n) * 1 / 2 => (√n) * 0.50 divisions. Complexity of is O(√n).

C++
// Trial Division: Square Root + Even Factors
// Check all numbers until Square Root and Skip Even Factors
long long TD_SREF(long long Cand) {
Count = 0;
// check 2 the only Even Prime
Count++;
if (Cand % 2 == 0) {
return 2;
}
long long Top = sqrt(Cand);
for (long long Div = 3; Div <= Top; Div += 2) {
Count++;
if (Cand % Div == 0) {
return Div;
}
}
return Cand;
}
dBase
//  Trial Division Square Root + Even Factors
// May 2019
function TD_SREF(Prod)
Local D, Top
if Prod % 2 = 0
return 2
endif
Top= int(sqrt(Prod))
for D= 3 to Top step 2
if Prod % D = 0
return D
endif
next
return Prod

In TrialDiv.cpp and TrialDiv.prg in TrialDiv.zip.

TD_SREF.graphml and TD_SREF.svg in TrialDiv.zip.

The Wheel

The wheel is an extension of the previous optimization. One builds the wheel from small primes, say 2 and 3, the size of the wheel is 2 * 3 = 6. When one writes a list of numbers in a 6 columns table and removes 2 and 3, one can see that primes are only in the first column and in the fifth column. That is the wheel, we only have to check numbers of columns 1 and 5.

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

The maximum cost for n is (√n) * (1 / 2) * (2 / 3) => (√n) * (1 / 3) => (√n) * 0.33 divisions. Complexity is O(√n).

dBase
//  Trial Division Square Root + Wheel
// May 2019
<pre lang="c++">
// Trial Division: Square Root + Wheel
// Check all numbers until Square Root and Skip useless factors with a wheel
long long TD_SRW(long long Cand) {
long long SPrimes[] = { 2, 3, 0 };
long long Wheel[] = { 4, 2, 0 };
long long Div;

Count = 0;
long long Top = sqrt(Cand);
// Check small primes
for (long long Ind = 0; SPrimes[Ind] != 0; Ind++) {
Div = SPrimes[Ind]; // for debug purpose
if (SPrimes[Ind] > Top) {
return Cand;
}
Count++;
if (Cand % SPrimes[Ind] == 0) {
return SPrimes[Ind];
}
}
// Start the Wheel
Div = 1;
while (Div <= Top) {
for (long long Ind = 0; Wheel[Ind] != 0; Ind++) {
Div += Wheel[Ind];
if (Div > Top) {
break;
}
Count++;
if (Cand % Div == 0) {
return Div;
}
}
}
return Cand;
}
dBase
function TD_SRW(Prod)
local D, Top, SPrimes, Wheel, W
// Check small primes
SPrimes= {2, 3}
Wheel= {4, 2}
for each D in SPrimes
if Prod % D = 0
return D
endif
next
// Start the wheel
D= 1
Top= int(sqrt(Prod))
while D <= Top
for each W in wheel
D += W
if Prod % D = 0
return D
endif
next
enddo
return Prod

In TrialDiv.cpp and TrialDiv.prg in TrialDiv.zip.

TD_SRW.graphml and TD_SRW.svg in TrialDiv.zip.

The wheel can include more primes. With 2, 3 and 5, the size of wheel is 30.

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

The maximum cost for n is (√n) * (1 / 2) * (2 / 3) * (4 / 5) => (√n) * (4 / 15) => (√n) * 0.27 divisions. Complexity is O(√n).

C++
long long SPrimes[] = { 2, 3, 5, 0 };
long long Wheel[] = { 6, 4, 2, 4, 2, 4, 6, 2, 0 };
dBase
// only those 2 lines need change
SPrimes= {2, 3, 5}
Wheel= {6, 4, 2, 4, 2, 4, 6, 2}

Or get even bigger. With 2, 3, 5 and 7, the size of wheel is 210.

The maximum cost for n is (√n) * (1 / 2) * (2 / 3) * (4 / 5) * (6 / 7) => (√n) * (24 / 105) => (√n) * 0.23 divisions. Complexity is O(√n).

The List of Primes

Basically, it is a wheel variant, where the list of small primes exceeds the ones needed for the wheel, and after the end of list, we fall back on the wheel. The advantage over the simple wheel is that it avoids testing non prime factors not filtered by the wheel. As long as we are in the list of primes, the workload is optimum.

Just have to take care about setting the wheel correctly at the end of prime list.

The maximum cost for n is slightly better than the wheel, the list primes just save non prime divisors that are in the wheel. So the longer the list of primes, the better it gets. Complexity is O(√n).

C++
// Trial Division: Square Root + Prime list + Wheel
// Check all numbers until Square Root, start with a list of primes
// and Skip useless factors with a wheel
long long TD_SRPW(long long Cand) {
long long SPrimes[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,
47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113,
127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191,
193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263,
269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347,
349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421,
431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499,
503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593,
599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661,
673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757,
761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853,
857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941,
947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021,
1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093,
1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181,
1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259,
1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321,
1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433,
1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493,
1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579,
1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657,
1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741,
1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831,
1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913,
1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003,
2011, 2017, 2027, 2029, 2039, 2053, 2063, 2069, 2081, 2083, 2087,
2089, 2099, 2111, 2113, 2129, 2131, 2137, 2141, 2143, 2153, 2161,
2179, 2203, 2207, 2213, 2221, 2237, 2239, 2243, 2251, 2267, 2269,
2273, 2281, 2287, 2293, 2297, 2309, 2311, 2333, 2339, 2341, 2347,
2351, 2357, 2371, 2377, 2381, 2383, 2389, 2393, 2399, 2411, 2417,
2423, 2437, 2441, 2447, 2459, 2467, 2473, 2477, 2503, 2521, 2531,
2539, 2543, 2549, 2551, 2557, 2579, 2591, 2593, 2609, 2617, 2621,
2633, 2647, 2657, 2659, 2663, 2671, 2677, 2683, 2687, 2689, 2693,
2699, 2707, 2711, 2713, 2719, 2729, 2731, 2741, 2749, 2753, 2767,
2777, 2789, 2791, 2797, 2801, 2803, 2819, 2833, 2837, 2843, 2851,
2857, 2861, 2879, 2887, 2897, 2903, 2909, 2917, 2927, 2939, 2953,
2957, 2963, 2969, 2971, 2999, 0 };
long long Wheel[] = { 6, 4, 2, 4, 2, 4, 6, 2, 0 };
long long Div;

Count = 0;
long long Top = sqrt(Cand);
// Check small primes
for (long long Ind = 0; SPrimes[Ind] != 0; Ind++) {
Div = SPrimes[Ind]; // for debug purpose
if (SPrimes[Ind] > Top) {
return Cand;
}
Count++;
if (Cand % SPrimes[Ind] == 0) {
return SPrimes[Ind];
}
}
// Start the Wheel
Div = 3001;
while (Div <= Top) {
for (long long Ind = 0; Wheel[Ind] != 0; Ind++) {
if (Div > Top) {
return Cand;
}
Count++;
if (Cand % Div == 0) {
return Div;
}
Div += Wheel[Ind];
}
}
return Cand;
}
dBase
//  Trial Division Square Root + Wheel + list of primes
// with a large list of primes
// May 2019
function TD_SRW2(Prod)
local D, Top, SPrimes, Wheel, W
// Check small primes
SPrimes= {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67,
71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139,
149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211,
223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293}
Wheel= {6, 4, 2, 4, 2, 4, 6, 2}

for each D in SPrimes
if Prod % D = 0
return D
endif
next
// Start the wheel
D= 301
Top= int(sqrt(Prod))
while D <= Top
for each W in wheel
D += W
if Prod % D = 0
return D
endif
next
enddo
return Prod

In TrialDiv.cpp and TrialDiv.prg in TrialDiv.zip.

Benchmarks

Benchmarks are here to highlight efficiency of variants.

As divisions/modulos are what cost time, counting them is enough for the benchmark.

Benchmark 1

This first benchmark is here to highlight the inefficiency of Brute Force variant.

C++
long long Test[] = { 11, 31, 53, 71, 97, 0, 101, 1009, 2003, 3001, 4001,
5003, 6007, 7001, 8009, 9001, 10007, 20011, 30011, 40009, 49999,
1000003, 4000021, 9000011, 0 };
cout << "Comparaison Variantes avec Brute Force" << endl;
cout << "Number\tTD_BF1\tTD_BF2\tTD_SR\tTD_SREF\tTD_SRW\tTD_SRPW" << endl;
for (Ind = 0; Test[Ind] != 0; Ind++) {
cout << Test[Ind] << "\t";
TD_BF1(Test[Ind]);
cout << Count << "\t";
TD_BF2(Test[Ind]);
cout << Count << "\t";
TD_SR(Test[Ind]);
cout << Count << "\t";
TD_SREF(Test[Ind]);
cout << Count << "\t";
TD_SRW(Test[Ind]);
cout << Count << "\t";
TD_SRPW(Test[Ind]);
cout << Count << endl;
}
cout << endl;
dBase
Test = { 11, 31, 53, 71, 97, 101, 1009, 2003, 3001, 4001, 5003, 6007, 7001, 8009, 9001, 10007, 20011, 30011, 40009, 49999, 1000003, 4000021, 9000011 }
? "Comparaison Variantes avec Brute Force"
? "Number","TD_BF1","TD_BF2","TD_SR","TD_SREF","TD_SRW","TD_SRPW"
for Ind = 1 to 5
? Test[Ind]
TD_BF1(Test[Ind])
?? Count
TD_BF2(Test[Ind])
?? Count
TD_SR(Test[Ind])
?? Count
TD_SREF(Test[Ind])
?? Count
TD_SRW(Test[Ind])
?? Count
TD_SRPW(Test[Ind])
?? Count
next
?
C++
Number  TD_BF1  TD_BF2  TD_SR   TD_SREF TD_SRW  TD_SRPW
11      9       4       2       2       2       2
31      29      14      4       3       3       3
53      51      25      6       4       4       4
71      69      34      7       4       4       4
97      95      47      8       5       4       4

Benchmark 2

This extended benchmark is for variants other than Brute Force.

C++
cout << "Comparaison Variantes sans Brute Force" << endl;
cout << "Number\tTD_SR\tTD_SREF\tTD_SRW\tTD_SRPW\tDelta" << endl;
// test after first 0
for (Ind++; Test[Ind] != 0; Ind++) {
cout << Test[Ind] << "\t";
TD_SR(Test[Ind]);
cout << Count << "\t";
TD_SREF(Test[Ind]);
cout << Count << "\t";
TD_SRW(Test[Ind]);
int C1 = Count;
cout << Count << "\t";
TD_SRPW(Test[Ind]);
cout << Count << "\t";
cout << C1 - Count << endl;
}
cout << endl;
dBase
? "Number","TD_SR","TD_SREF","TD_SRW","TD_SRPW"
for Ind = 6 to len(Test)
? Test[Ind]
TD_SR(Test[Ind])
?? Count
TD_SREF(Test[Ind])
?? Count
TD_SRW(Test[Ind])
?? Count
TD_SRPW(Test[Ind])
?? Count
next
?
C++
Comparison Variants sans Brute Force
Number	TD_SR	TD_SREF	TD_SRW	TD_SRPW	Delta
101		9		5		4		4		0
1009	30		16		11		11		0
2003	43		22		14		14		0
3001	53		27		17		16		1
4001	62		32		19		18		1
5003	69		35		20		19		1
6007	76		39		23		21		2
7001	82		42		25		23		2
8009	88		45		26		24		2
9001	93		47		27		24		3
10007	99		50		28		25		3
20011	140		71		40		34		6
30011	172		87		49		40		9
40009	199		100		56		46		10
49999	222		112		62		48		14
1000003	999		500		268		168		100
4000021	1800	901		483		279		204
9000011	2999	1500	802		430		372

Delta shows how the list of primes improves things, and it gets better with long list of primes.

Checking Correctness of Variants

Rgis code checks correctness of code by comparing results of variants.

C++
cout << "Vérification Variantes" << endl;
for (long long Cand = 3; Cand < 10000000; Cand += 10) {
long long d1 = TD_SREF(Cand);
long long d2 = TD_SRPW(Cand);
if (d1 != d2) {
cout << Cand << "\t" << d1 << "\t" << d2 << endl;
}
}

Points of Interest

Trial Division being brute force, one can see that there is brute force and brute force.

History

• 1st April, 2019: First version
• 27th November, 2020: Second version
• 20th December, 2020: Third version: some cleaning, moved download to top
• 25th December, 2020: Fourth version: some cleaning and corrections
• 31st January, 2021: Corrections
• 16th September, 2023: Added flowcharts and corrections.
• 15th December, 2020: Typos

Written By
Database Developer
France
I am a professional programmer.
Problem analyse is certainly what I am best at.
My main programming expertise is in the xBase languages (dBase, Clipper, FoxPro, Harbour, xHarbour), then VBA for Excel and advanced Excel WorkBooks.

I also have knowledge on C/C++, d language, HTML, SVG, XML, XSLT, Javascript, PHP, BASICs, Python, COBOL, Assembly.
My personal interests goes to algorithm optimization and puzzles.

 First Prev Next
 Time? Emil Steen20-Dec-23 11:58 Emil Steen 20-Dec-23 11:58
 From what I can see it only count the number of divisions made. What I remember from testing this many years ago is that the complexity with a "wheel list" made it slower.
 Re: Time? Patrice T22-Dec-23 8:13 Patrice T 22-Dec-23 8:13
 Re: Time? Emil Steen22-Dec-23 12:42 Emil Steen 22-Dec-23 12:42
 Weird notation YDaoust28-Sep-23 23:41 YDaoust 28-Sep-23 23:41
 Re: Weird notation Patrice T29-Sep-23 0:41 Patrice T 29-Sep-23 0:41
 Re: Weird notation YDaoust29-Sep-23 2:05 YDaoust 29-Sep-23 2:05
 Re: Weird notation Patrice T29-Sep-23 3:40 Patrice T 29-Sep-23 3:40
 Re: Weird notation YDaoust29-Sep-23 3:48 YDaoust 29-Sep-23 3:48
 Re: Weird notation Patrice T29-Sep-23 4:00 Patrice T 29-Sep-23 4:00
 Re: Weird notation YDaoust29-Sep-23 4:04 YDaoust 29-Sep-23 4:04
 My vote of 5 Ștefan-Mihai MOGA30-Jan-22 7:56 Ștefan-Mihai MOGA 30-Jan-22 7:56
 Re: My vote of 5 Patrice T30-Jan-22 12:20 Patrice T 30-Jan-22 12:20
 A Suggestion Rick York21-Dec-20 11:51 Rick York 21-Dec-20 11:51
 Re: A Suggestion Patrice T21-Dec-20 13:38 Patrice T 21-Dec-20 13:38
 Re: A Suggestion Rick York21-Dec-20 20:04 Rick York 21-Dec-20 20:04
 Re: A Suggestion Patrice T22-Dec-20 11:32 Patrice T 22-Dec-20 11:32
 Nice Work Rick York20-Dec-20 14:59 Rick York 20-Dec-20 14:59
 Re: Nice Work Patrice T20-Dec-20 15:02 Patrice T 20-Dec-20 15:02
 Re: Nice Work Rick York20-Dec-20 16:48 Rick York 20-Dec-20 16:48
 Re: Nice Work Patrice T20-Dec-20 17:58 Patrice T 20-Dec-20 17:58
 Re: Nice Work Rick York20-Dec-20 18:16 Rick York 20-Dec-20 18:16
 Re: Nice Work Patrice T20-Dec-20 18:22 Patrice T 20-Dec-20 18:22
 Nice work! ssa-ed1-Dec-20 11:09 ssa-ed 1-Dec-20 11:09
 Re: Nice work! Patrice T20-Dec-20 0:47 Patrice T 20-Dec-20 0:47
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