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Generate distinctly different RGB colors in graphs - c#

When generating graphs and showing different sets of data it usually a good idea to difference the sets by color. So one line is red and the next is green and so on. The problem is then that when the number of datasets is unknown one needs to randomly generate these colors and often they end up very close to each other (green, light green for example).
Any ideas on how this could be solved and how it would be possibler to generate distinctly different colors?
I'd be great if any examples (feel free to just discuss the problem and solution without examples if you find that easier) were in C# and RGB based colors.

You have three colour channels 0 to 255 R, G and B.
First go through
0, 0, 255
0, 255, 0
255, 0, 0
Then go through
0, 255, 255
255, 0, 255
255, 255, 0
Then divide by 2 => 128 and start again:
0, 0, 128
0, 128, 0
128, 0, 0
0, 128, 128
128, 0, 128
128, 128, 0
Divide by 2 => 64
Next time add 64 to 128 => 192
follow the pattern.
Straightforward to program and gives you fairly distinct colours.
EDIT: Request for code sample
Also - adding in the additional pattern as below if gray is an acceptable colour:
255, 255, 255
128, 128, 128
There are a number of ways you can handle generating these in code.
The Easy Way
If you can guarantee that you will never need more than a fixed number of colours, just generate an array of colours following this pattern and use those:
static string[] ColourValues = new string[] {
"FF0000", "00FF00", "0000FF", "FFFF00", "FF00FF", "00FFFF", "000000",
"800000", "008000", "000080", "808000", "800080", "008080", "808080",
"C00000", "00C000", "0000C0", "C0C000", "C000C0", "00C0C0", "C0C0C0",
"400000", "004000", "000040", "404000", "400040", "004040", "404040",
"200000", "002000", "000020", "202000", "200020", "002020", "202020",
"600000", "006000", "000060", "606000", "600060", "006060", "606060",
"A00000", "00A000", "0000A0", "A0A000", "A000A0", "00A0A0", "A0A0A0",
"E00000", "00E000", "0000E0", "E0E000", "E000E0", "00E0E0", "E0E0E0",
};
The Hard Way
If you don't know how many colours you are going to need, the code below will generate up to 896 colours using this pattern. (896 = 256 * 7 / 2) 256 is the colour space per channel, we have 7 patterns and we stop before we get to colours separated by only 1 colour value.
I've probably made harder work of this code than I needed to. First, there is an intensity generator which starts at 255, then generates the values as per the pattern described above. The pattern generator just loops through the seven colour patterns.
using System;
class Program {
static void Main(string[] args) {
ColourGenerator generator = new ColourGenerator();
for (int i = 0; i < 896; i++) {
Console.WriteLine(string.Format("{0}: {1}", i, generator.NextColour()));
}
}
}
public class ColourGenerator {
private int index = 0;
private IntensityGenerator intensityGenerator = new IntensityGenerator();
public string NextColour() {
string colour = string.Format(PatternGenerator.NextPattern(index),
intensityGenerator.NextIntensity(index));
index++;
return colour;
}
}
public class PatternGenerator {
public static string NextPattern(int index) {
switch (index % 7) {
case 0: return "{0}0000";
case 1: return "00{0}00";
case 2: return "0000{0}";
case 3: return "{0}{0}00";
case 4: return "{0}00{0}";
case 5: return "00{0}{0}";
case 6: return "{0}{0}{0}";
default: throw new Exception("Math error");
}
}
}
public class IntensityGenerator {
private IntensityValueWalker walker;
private int current;
public string NextIntensity(int index) {
if (index == 0) {
current = 255;
}
else if (index % 7 == 0) {
if (walker == null) {
walker = new IntensityValueWalker();
}
else {
walker.MoveNext();
}
current = walker.Current.Value;
}
string currentText = current.ToString("X");
if (currentText.Length == 1) currentText = "0" + currentText;
return currentText;
}
}
public class IntensityValue {
private IntensityValue mChildA;
private IntensityValue mChildB;
public IntensityValue(IntensityValue parent, int value, int level) {
if (level > 7) throw new Exception("There are no more colours left");
Value = value;
Parent = parent;
Level = level;
}
public int Level { get; set; }
public int Value { get; set; }
public IntensityValue Parent { get; set; }
public IntensityValue ChildA {
get {
return mChildA ?? (mChildA = new IntensityValue(this, this.Value - (1<<(7-Level)), Level+1));
}
}
public IntensityValue ChildB {
get {
return mChildB ?? (mChildB = new IntensityValue(this, Value + (1<<(7-Level)), Level+1));
}
}
}
public class IntensityValueWalker {
public IntensityValueWalker() {
Current = new IntensityValue(null, 1<<7, 1);
}
public IntensityValue Current { get; set; }
public void MoveNext() {
if (Current.Parent == null) {
Current = Current.ChildA;
}
else if (Current.Parent.ChildA == Current) {
Current = Current.Parent.ChildB;
}
else {
int levelsUp = 1;
Current = Current.Parent;
while (Current.Parent != null && Current == Current.Parent.ChildB) {
Current = Current.Parent;
levelsUp++;
}
if (Current.Parent != null) {
Current = Current.Parent.ChildB;
}
else {
levelsUp++;
}
for (int i = 0; i < levelsUp; i++) {
Current = Current.ChildA;
}
}
}
}

To implement a variation list where by your colors go, 255 then use all possibilities of that up, then add 0 and all RGB patterns with those two values. Then add 128 and all RGB combinations with those. Then 64. Then 192. Etc.
In Java,
public Color getColor(int i) {
return new Color(getRGB(i));
}
public int getRGB(int index) {
int[] p = getPattern(index);
return getElement(p[0]) << 16 | getElement(p[1]) << 8 | getElement(p[2]);
}
public int getElement(int index) {
int value = index - 1;
int v = 0;
for (int i = 0; i < 8; i++) {
v = v | (value & 1);
v <<= 1;
value >>= 1;
}
v >>= 1;
return v & 0xFF;
}
public int[] getPattern(int index) {
int n = (int)Math.cbrt(index);
index -= (n*n*n);
int[] p = new int[3];
Arrays.fill(p,n);
if (index == 0) {
return p;
}
index--;
int v = index % 3;
index = index / 3;
if (index < n) {
p[v] = index % n;
return p;
}
index -= n;
p[v ] = index / n;
p[++v % 3] = index % n;
return p;
}
This will produce patterns of that type infinitely (2^24) into the future. However, after a hundred or so spots you likely won't see much of a difference between a color with 0 or 32 in the blue's place.
You might be better off normalizing this into a different color space. LAB color space for example with the L,A,B values normalized and converted. So the distinctness of the color is pushed through something more akin to the human eye.
getElement() reverses the endian of an 8 bit number, and starts counting from -1 rather than 0 (masking with 255). So it goes 255,0,127,192,64,... as the number grows it moves less and less significant bits, subdividing the number.
getPattern() determines what the most significant element in the pattern should be (it's the cube root). Then proceeds to break down the 3N²+3N+1 different patterns that involve that most significant element.
This algorithm will produce (first 128 values):
#FFFFFF
#000000
#FF0000
#00FF00
#0000FF
#FFFF00
#00FFFF
#FF00FF
#808080
#FF8080
#80FF80
#8080FF
#008080
#800080
#808000
#FFFF80
#80FFFF
#FF80FF
#FF0080
#80FF00
#0080FF
#00FF80
#8000FF
#FF8000
#000080
#800000
#008000
#404040
#FF4040
#40FF40
#4040FF
#004040
#400040
#404000
#804040
#408040
#404080
#FFFF40
#40FFFF
#FF40FF
#FF0040
#40FF00
#0040FF
#FF8040
#40FF80
#8040FF
#00FF40
#4000FF
#FF4000
#000040
#400000
#004000
#008040
#400080
#804000
#80FF40
#4080FF
#FF4080
#800040
#408000
#004080
#808040
#408080
#804080
#C0C0C0
#FFC0C0
#C0FFC0
#C0C0FF
#00C0C0
#C000C0
#C0C000
#80C0C0
#C080C0
#C0C080
#40C0C0
#C040C0
#C0C040
#FFFFC0
#C0FFFF
#FFC0FF
#FF00C0
#C0FF00
#00C0FF
#FF80C0
#C0FF80
#80C0FF
#FF40C0
#C0FF40
#40C0FF
#00FFC0
#C000FF
#FFC000
#0000C0
#C00000
#00C000
#0080C0
#C00080
#80C000
#0040C0
#C00040
#40C000
#80FFC0
#C080FF
#FFC080
#8000C0
#C08000
#00C080
#8080C0
#C08080
#80C080
#8040C0
#C08040
#40C080
#40FFC0
#C040FF
#FFC040
#4000C0
#C04000
#00C040
#4080C0
#C04080
#80C040
#4040C0
#C04040
#40C040
#202020
#FF2020
#20FF20
Read left to right, top to bottom. 729 colors (9³). So all the patterns up to n = 9. You'll notice the speed at which they start to clash. There's only so many WRGBCYMK variations. And this solution, while clever basically only does different shades of primary colors.
Much of the clashing is due to green and how similar most greens look to most people. The demand that each be maximally different at start rather than just different enough to not be the same color. And basic flaws in the idea resulting in primary colors patterns, and identical hues.
Using CIELab2000 Color Space and Distance Routine to randomly select and try 10k different colors and find the maximally-distant minimum-distance from previous colors, (pretty much the definition of the request) avoids clashing longer than the above solution:
Which could be just called a static list for the Easy Way. It took an hour and a half to generate 729 entries:
#9BC4E5
#310106
#04640D
#FEFB0A
#FB5514
#E115C0
#00587F
#0BC582
#FEB8C8
#9E8317
#01190F
#847D81
#58018B
#B70639
#703B01
#F7F1DF
#118B8A
#4AFEFA
#FCB164
#796EE6
#000D2C
#53495F
#F95475
#61FC03
#5D9608
#DE98FD
#98A088
#4F584E
#248AD0
#5C5300
#9F6551
#BCFEC6
#932C70
#2B1B04
#B5AFC4
#D4C67A
#AE7AA1
#C2A393
#0232FD
#6A3A35
#BA6801
#168E5C
#16C0D0
#C62100
#014347
#233809
#42083B
#82785D
#023087
#B7DAD2
#196956
#8C41BB
#ECEDFE
#2B2D32
#94C661
#F8907D
#895E6B
#788E95
#FB6AB8
#576094
#DB1474
#8489AE
#860E04
#FBC206
#6EAB9B
#F2CDFE
#645341
#760035
#647A41
#496E76
#E3F894
#F9D7CD
#876128
#A1A711
#01FB92
#FD0F31
#BE8485
#C660FB
#120104
#D48958
#05AEE8
#C3C1BE
#9F98F8
#1167D9
#D19012
#B7D802
#826392
#5E7A6A
#B29869
#1D0051
#8BE7FC
#76E0C1
#BACFA7
#11BA09
#462C36
#65407D
#491803
#F5D2A8
#03422C
#72A46E
#128EAC
#47545E
#B95C69
#A14D12
#C4C8FA
#372A55
#3F3610
#D3A2C6
#719FFA
#0D841A
#4C5B32
#9DB3B7
#B14F8F
#747103
#9F816D
#D26A5B
#8B934B
#F98500
#002935
#D7F3FE
#FCB899
#1C0720
#6B5F61
#F98A9D
#9B72C2
#A6919D
#2C3729
#D7C70B
#9F9992
#EFFBD0
#FDE2F1
#923A52
#5140A7
#BC14FD
#6D706C
#0007C4
#C6A62F
#000C14
#904431
#600013
#1C1B08
#693955
#5E7C99
#6C6E82
#D0AFB3
#493B36
#AC93CE
#C4BA9C
#09C4B8
#69A5B8
#374869
#F868ED
#E70850
#C04841
#C36333
#700366
#8A7A93
#52351D
#B503A2
#D17190
#A0F086
#7B41FC
#0EA64F
#017499
#08A882
#7300CD
#A9B074
#4E6301
#AB7E41
#547FF4
#134DAC
#FDEC87
#056164
#FE12A0
#C264BA
#939DAD
#0BCDFA
#277442
#1BDE4A
#826958
#977678
#BAFCE8
#7D8475
#8CCF95
#726638
#FEA8EB
#EAFEF0
#6B9279
#C2FE4B
#304041
#1EA6A7
#022403
#062A47
#054B17
#F4C673
#02FEC7
#9DBAA8
#775551
#835536
#565BCC
#80D7D2
#7AD607
#696F54
#87089A
#664B19
#242235
#7DB00D
#BFC7D6
#D5A97E
#433F31
#311A18
#FDB2AB
#D586C9
#7A5FB1
#32544A
#EFE3AF
#859D96
#2B8570
#8B282D
#E16A07
#4B0125
#021083
#114558
#F707F9
#C78571
#7FB9BC
#FC7F4B
#8D4A92
#6B3119
#884F74
#994E4F
#9DA9D3
#867B40
#CED5C4
#1CA2FE
#D9C5B4
#FEAA00
#507B01
#A7D0DB
#53858D
#588F4A
#FBEEEC
#FC93C1
#D7CCD4
#3E4A02
#C8B1E2
#7A8B62
#9A5AE2
#896C04
#B1121C
#402D7D
#858701
#D498A6
#B484EF
#5C474C
#067881
#C0F9FC
#726075
#8D3101
#6C93B2
#A26B3F
#AA6582
#4F4C4F
#5A563D
#E83005
#32492D
#FC7272
#B9C457
#552A5B
#B50464
#616E79
#DCE2E4
#CF8028
#0AE2F0
#4F1E24
#FD5E46
#4B694E
#C5DEFC
#5DC262
#022D26
#7776B8
#FD9F66
#B049B8
#988F73
#BE385A
#2B2126
#54805A
#141B55
#67C09B
#456989
#DDC1D9
#166175
#C1E29C
#A397B5
#2E2922
#ABDBBE
#B4A6A8
#A06B07
#A99949
#0A0618
#B14E2E
#60557D
#D4A556
#82A752
#4A005B
#3C404F
#6E6657
#7E8BD5
#1275B8
#D79E92
#230735
#661849
#7A8391
#FE0F7B
#B0B6A9
#629591
#D05591
#97B68A
#97939A
#035E38
#53E19E
#DFD7F9
#02436C
#525A72
#059A0E
#3E736C
#AC8E87
#D10C92
#B9906E
#66BDFD
#C0ABFD
#0734BC
#341224
#8AAAC1
#0E0B03
#414522
#6A2F3E
#2D9A8A
#4568FD
#FDE6D2
#FEE007
#9A003C
#AC8190
#DCDD58
#B7903D
#1F2927
#9B02E6
#827A71
#878B8A
#8F724F
#AC4B70
#37233B
#385559
#F347C7
#9DB4FE
#D57179
#DE505A
#37F7DD
#503500
#1C2401
#DD0323
#00A4BA
#955602
#FA5B94
#AA766C
#B8E067
#6A807E
#4D2E27
#73BED7
#D7BC8A
#614539
#526861
#716D96
#829A17
#210109
#436C2D
#784955
#987BAB
#8F0152
#0452FA
#B67757
#A1659F
#D4F8D8
#48416F
#DEBAAF
#A5A9AA
#8C6B83
#403740
#70872B
#D9744D
#151E2C
#5C5E5E
#B47C02
#F4CBD0
#E49D7D
#DD9954
#B0A18B
#2B5308
#EDFD64
#9D72FC
#2A3351
#68496C
#C94801
#EED05E
#826F6D
#E0D6BB
#5B6DB4
#662F98
#0C97CA
#C1CA89
#755A03
#DFA619
#CD70A8
#BBC9C7
#F6BCE3
#A16462
#01D0AA
#87C6B3
#E7B2FA
#D85379
#643AD5
#D18AAE
#13FD5E
#B3E3FD
#C977DB
#C1A7BB
#9286CB
#A19B6A
#8FFED7
#6B1F17
#DF503A
#10DDD7
#9A8457
#60672F
#7D327D
#DD8782
#59AC42
#82FDB8
#FC8AE7
#909F6F
#B691AE
#B811CD
#BCB24E
#CB4BD9
#2B2304
#AA9501
#5D5096
#403221
#F9FAB4
#3990FC
#70DE7F
#95857F
#84A385
#50996F
#797B53
#7B6142
#81D5FE
#9CC428
#0B0438
#3E2005
#4B7C91
#523854
#005EA9
#F0C7AD
#ACB799
#FAC08E
#502239
#BFAB6A
#2B3C48
#0EB5D8
#8A5647
#49AF74
#067AE9
#F19509
#554628
#4426A4
#7352C9
#3F4287
#8B655E
#B480BF
#9BA74C
#5F514C
#CC9BDC
#BA7942
#1C4138
#3C3C3A
#29B09C
#02923F
#701D2B
#36577C
#3F00EA
#3D959E
#440601
#8AEFF3
#6D442A
#BEB1A8
#A11C02
#8383FE
#A73839
#DBDE8A
#0283B3
#888597
#32592E
#F5FDFA
#01191B
#AC707A
#B6BD03
#027B59
#7B4F08
#957737
#83727D
#035543
#6F7E64
#C39999
#52847A
#925AAC
#77CEDA
#516369
#E0D7D0
#FCDD97
#555424
#96E6B6
#85BB74
#5E2074
#BD5E48
#9BEE53
#1A351E
#3148CD
#71575F
#69A6D0
#391A62
#E79EA0
#1C0F03
#1B1636
#D20C39
#765396
#7402FE
#447F3E
#CFD0A8
#3A2600
#685AFC
#A4B3C6
#534302
#9AA097
#FD5154
#9B0085
#403956
#80A1A7
#6E7A9A
#605E6A
#86F0E2
#5A2B01
#7E3D43
#ED823B
#32331B
#424837
#40755E
#524F48
#B75807
#B40080
#5B8CA1
#FDCFE5
#CCFEAC
#755847
#CAB296
#C0D6E3
#2D7100
#D5E4DE
#362823
#69C63C
#AC3801
#163132
#4750A6
#61B8B2
#FCC4B5
#DEBA2E
#FE0449
#737930
#8470AB
#687D87
#D7B760
#6AAB86
#8398B8
#B7B6BF
#92C4A1
#B6084F
#853B5E
#D0BCBA
#92826D
#C6DDC6
#BE5F5A
#280021
#435743
#874514
#63675A
#E97963
#8F9C9E
#985262
#909081
#023508
#DDADBF
#D78493
#363900
#5B0120
#603C47
#C3955D
#AC61CB
#FD7BA7
#716C74
#8D895B
#071001
#82B4F2
#B6BBD8
#71887A
#8B9FE3
#997158
#65A6AB
#2E3067
#321301
#FEECCB
#3B5E72
#C8FE85
#A1DCDF
#CB49A6
#B1C5E4
#3E5EB0
#88AEA7
#04504C
#975232
#6786B9
#068797
#9A98C4
#A1C3C2
#1C3967
#DBEA07
#789658
#E7E7C6
#A6C886
#957F89
#752E62
#171518
#A75648
#01D26F
#0F535D
#047E76
#C54754
#5D6E88
#AB9483
#803B99
#FA9C48
#4A8A22
#654A5C
#965F86
#9D0CBB
#A0E8A0
#D3DBFA
#FD908F
#AEAB85
#A13B89
#F1B350
#066898
#948A42
#C8BEDE
#19252C
#7046AA
#E1EEFC
#3E6557
#CD3F26
#2B1925
#DDAD94
#C0B109
#37DFFE
#039676
#907468
#9E86A5
#3A1B49
#BEE5B7
#C29501
#9E3645
#DC580A
#645631
#444B4B
#FD1A63
#DDE5AE
#887800
#36006F
#3A6260
#784637
#FEA0B7
#A3E0D2
#6D6316
#5F7172
#B99EC7
#777A7E
#E0FEFD
#E16DC5
#01344B
#F8F8FC
#9F9FB5
#182617
#FE3D21
#7D0017
#822F21
#EFD9DC
#6E68C4
#35473E
#007523
#767667
#A6825D
#83DC5F
#227285
#A95E34
#526172
#979730
#756F6D
#716259
#E8B2B5
#B6C9BB
#9078DA
#4F326E
#B2387B
#888C6F
#314B5F
#E5B678
#38A3C6
#586148
#5C515B
#CDCCE1
#C8977F
Using brute force to (testing all 16,777,216 RGB colors through CIELab Delta2000 / Starting with black) produces a series. Which starts to clash at around 26 but could make it to 30 or 40 with visual inspection and manual dropping (which can't be done with a computer). So doing the absolute maximum one can programmatically only makes a couple dozen distinct colors. A discrete list is your best bet. You will get more discrete colors with a list than you would programmatically. The easy way is the best solution, start mixing and matching with other ways to alter your data than color.
#000000
#00FF00
#0000FF
#FF0000
#01FFFE
#FFA6FE
#FFDB66
#006401
#010067
#95003A
#007DB5
#FF00F6
#FFEEE8
#774D00
#90FB92
#0076FF
#D5FF00
#FF937E
#6A826C
#FF029D
#FE8900
#7A4782
#7E2DD2
#85A900
#FF0056
#A42400
#00AE7E
#683D3B
#BDC6FF
#263400
#BDD393
#00B917
#9E008E
#001544
#C28C9F
#FF74A3
#01D0FF
#004754
#E56FFE
#788231
#0E4CA1
#91D0CB
#BE9970
#968AE8
#BB8800
#43002C
#DEFF74
#00FFC6
#FFE502
#620E00
#008F9C
#98FF52
#7544B1
#B500FF
#00FF78
#FF6E41
#005F39
#6B6882
#5FAD4E
#A75740
#A5FFD2
#FFB167
#009BFF
#E85EBE
Update:
I continued this for about a month so, at 1024 brute force.
public static final String[] indexcolors = new String[]{
"#000000", "#FFFF00", "#1CE6FF", "#FF34FF", "#FF4A46", "#008941", "#006FA6", "#A30059",
"#FFDBE5", "#7A4900", "#0000A6", "#63FFAC", "#B79762", "#004D43", "#8FB0FF", "#997D87",
"#5A0007", "#809693", "#FEFFE6", "#1B4400", "#4FC601", "#3B5DFF", "#4A3B53", "#FF2F80",
"#61615A", "#BA0900", "#6B7900", "#00C2A0", "#FFAA92", "#FF90C9", "#B903AA", "#D16100",
"#DDEFFF", "#000035", "#7B4F4B", "#A1C299", "#300018", "#0AA6D8", "#013349", "#00846F",
"#372101", "#FFB500", "#C2FFED", "#A079BF", "#CC0744", "#C0B9B2", "#C2FF99", "#001E09",
"#00489C", "#6F0062", "#0CBD66", "#EEC3FF", "#456D75", "#B77B68", "#7A87A1", "#788D66",
"#885578", "#FAD09F", "#FF8A9A", "#D157A0", "#BEC459", "#456648", "#0086ED", "#886F4C",
"#34362D", "#B4A8BD", "#00A6AA", "#452C2C", "#636375", "#A3C8C9", "#FF913F", "#938A81",
"#575329", "#00FECF", "#B05B6F", "#8CD0FF", "#3B9700", "#04F757", "#C8A1A1", "#1E6E00",
"#7900D7", "#A77500", "#6367A9", "#A05837", "#6B002C", "#772600", "#D790FF", "#9B9700",
"#549E79", "#FFF69F", "#201625", "#72418F", "#BC23FF", "#99ADC0", "#3A2465", "#922329",
"#5B4534", "#FDE8DC", "#404E55", "#0089A3", "#CB7E98", "#A4E804", "#324E72", "#6A3A4C",
"#83AB58", "#001C1E", "#D1F7CE", "#004B28", "#C8D0F6", "#A3A489", "#806C66", "#222800",
"#BF5650", "#E83000", "#66796D", "#DA007C", "#FF1A59", "#8ADBB4", "#1E0200", "#5B4E51",
"#C895C5", "#320033", "#FF6832", "#66E1D3", "#CFCDAC", "#D0AC94", "#7ED379", "#012C58",
"#7A7BFF", "#D68E01", "#353339", "#78AFA1", "#FEB2C6", "#75797C", "#837393", "#943A4D",
"#B5F4FF", "#D2DCD5", "#9556BD", "#6A714A", "#001325", "#02525F", "#0AA3F7", "#E98176",
"#DBD5DD", "#5EBCD1", "#3D4F44", "#7E6405", "#02684E", "#962B75", "#8D8546", "#9695C5",
"#E773CE", "#D86A78", "#3E89BE", "#CA834E", "#518A87", "#5B113C", "#55813B", "#E704C4",
"#00005F", "#A97399", "#4B8160", "#59738A", "#FF5DA7", "#F7C9BF", "#643127", "#513A01",
"#6B94AA", "#51A058", "#A45B02", "#1D1702", "#E20027", "#E7AB63", "#4C6001", "#9C6966",
"#64547B", "#97979E", "#006A66", "#391406", "#F4D749", "#0045D2", "#006C31", "#DDB6D0",
"#7C6571", "#9FB2A4", "#00D891", "#15A08A", "#BC65E9", "#FFFFFE", "#C6DC99", "#203B3C",
"#671190", "#6B3A64", "#F5E1FF", "#FFA0F2", "#CCAA35", "#374527", "#8BB400", "#797868",
"#C6005A", "#3B000A", "#C86240", "#29607C", "#402334", "#7D5A44", "#CCB87C", "#B88183",
"#AA5199", "#B5D6C3", "#A38469", "#9F94F0", "#A74571", "#B894A6", "#71BB8C", "#00B433",
"#789EC9", "#6D80BA", "#953F00", "#5EFF03", "#E4FFFC", "#1BE177", "#BCB1E5", "#76912F",
"#003109", "#0060CD", "#D20096", "#895563", "#29201D", "#5B3213", "#A76F42", "#89412E",
"#1A3A2A", "#494B5A", "#A88C85", "#F4ABAA", "#A3F3AB", "#00C6C8", "#EA8B66", "#958A9F",
"#BDC9D2", "#9FA064", "#BE4700", "#658188", "#83A485", "#453C23", "#47675D", "#3A3F00",
"#061203", "#DFFB71", "#868E7E", "#98D058", "#6C8F7D", "#D7BFC2", "#3C3E6E", "#D83D66",
"#2F5D9B", "#6C5E46", "#D25B88", "#5B656C", "#00B57F", "#545C46", "#866097", "#365D25",
"#252F99", "#00CCFF", "#674E60", "#FC009C", "#92896B", "#1E2324", "#DEC9B2", "#9D4948",
"#85ABB4", "#342142", "#D09685", "#A4ACAC", "#00FFFF", "#AE9C86", "#742A33", "#0E72C5",
"#AFD8EC", "#C064B9", "#91028C", "#FEEDBF", "#FFB789", "#9CB8E4", "#AFFFD1", "#2A364C",
"#4F4A43", "#647095", "#34BBFF", "#807781", "#920003", "#B3A5A7", "#018615", "#F1FFC8",
"#976F5C", "#FF3BC1", "#FF5F6B", "#077D84", "#F56D93", "#5771DA", "#4E1E2A", "#830055",
"#02D346", "#BE452D", "#00905E", "#BE0028", "#6E96E3", "#007699", "#FEC96D", "#9C6A7D",
"#3FA1B8", "#893DE3", "#79B4D6", "#7FD4D9", "#6751BB", "#B28D2D", "#E27A05", "#DD9CB8",
"#AABC7A", "#980034", "#561A02", "#8F7F00", "#635000", "#CD7DAE", "#8A5E2D", "#FFB3E1",
"#6B6466", "#C6D300", "#0100E2", "#88EC69", "#8FCCBE", "#21001C", "#511F4D", "#E3F6E3",
"#FF8EB1", "#6B4F29", "#A37F46", "#6A5950", "#1F2A1A", "#04784D", "#101835", "#E6E0D0",
"#FF74FE", "#00A45F", "#8F5DF8", "#4B0059", "#412F23", "#D8939E", "#DB9D72", "#604143",
"#B5BACE", "#989EB7", "#D2C4DB", "#A587AF", "#77D796", "#7F8C94", "#FF9B03", "#555196",
"#31DDAE", "#74B671", "#802647", "#2A373F", "#014A68", "#696628", "#4C7B6D", "#002C27",
"#7A4522", "#3B5859", "#E5D381", "#FFF3FF", "#679FA0", "#261300", "#2C5742", "#9131AF",
"#AF5D88", "#C7706A", "#61AB1F", "#8CF2D4", "#C5D9B8", "#9FFFFB", "#BF45CC", "#493941",
"#863B60", "#B90076", "#003177", "#C582D2", "#C1B394", "#602B70", "#887868", "#BABFB0",
"#030012", "#D1ACFE", "#7FDEFE", "#4B5C71", "#A3A097", "#E66D53", "#637B5D", "#92BEA5",
"#00F8B3", "#BEDDFF", "#3DB5A7", "#DD3248", "#B6E4DE", "#427745", "#598C5A", "#B94C59",
"#8181D5", "#94888B", "#FED6BD", "#536D31", "#6EFF92", "#E4E8FF", "#20E200", "#FFD0F2",
"#4C83A1", "#BD7322", "#915C4E", "#8C4787", "#025117", "#A2AA45", "#2D1B21", "#A9DDB0",
"#FF4F78", "#528500", "#009A2E", "#17FCE4", "#71555A", "#525D82", "#00195A", "#967874",
"#555558", "#0B212C", "#1E202B", "#EFBFC4", "#6F9755", "#6F7586", "#501D1D", "#372D00",
"#741D16", "#5EB393", "#B5B400", "#DD4A38", "#363DFF", "#AD6552", "#6635AF", "#836BBA",
"#98AA7F", "#464836", "#322C3E", "#7CB9BA", "#5B6965", "#707D3D", "#7A001D", "#6E4636",
"#443A38", "#AE81FF", "#489079", "#897334", "#009087", "#DA713C", "#361618", "#FF6F01",
"#006679", "#370E77", "#4B3A83", "#C9E2E6", "#C44170", "#FF4526", "#73BE54", "#C4DF72",
"#ADFF60", "#00447D", "#DCCEC9", "#BD9479", "#656E5B", "#EC5200", "#FF6EC2", "#7A617E",
"#DDAEA2", "#77837F", "#A53327", "#608EFF", "#B599D7", "#A50149", "#4E0025", "#C9B1A9",
"#03919A", "#1B2A25", "#E500F1", "#982E0B", "#B67180", "#E05859", "#006039", "#578F9B",
"#305230", "#CE934C", "#B3C2BE", "#C0BAC0", "#B506D3", "#170C10", "#4C534F", "#224451",
"#3E4141", "#78726D", "#B6602B", "#200441", "#DDB588", "#497200", "#C5AAB6", "#033C61",
"#71B2F5", "#A9E088", "#4979B0", "#A2C3DF", "#784149", "#2D2B17", "#3E0E2F", "#57344C",
"#0091BE", "#E451D1", "#4B4B6A", "#5C011A", "#7C8060", "#FF9491", "#4C325D", "#005C8B",
"#E5FDA4", "#68D1B6", "#032641", "#140023", "#8683A9", "#CFFF00", "#A72C3E", "#34475A",
"#B1BB9A", "#B4A04F", "#8D918E", "#A168A6", "#813D3A", "#425218", "#DA8386", "#776133",
"#563930", "#8498AE", "#90C1D3", "#B5666B", "#9B585E", "#856465", "#AD7C90", "#E2BC00",
"#E3AAE0", "#B2C2FE", "#FD0039", "#009B75", "#FFF46D", "#E87EAC", "#DFE3E6", "#848590",
"#AA9297", "#83A193", "#577977", "#3E7158", "#C64289", "#EA0072", "#C4A8CB", "#55C899",
"#E78FCF", "#004547", "#F6E2E3", "#966716", "#378FDB", "#435E6A", "#DA0004", "#1B000F",
"#5B9C8F", "#6E2B52", "#011115", "#E3E8C4", "#AE3B85", "#EA1CA9", "#FF9E6B", "#457D8B",
"#92678B", "#00CDBB", "#9CCC04", "#002E38", "#96C57F", "#CFF6B4", "#492818", "#766E52",
"#20370E", "#E3D19F", "#2E3C30", "#B2EACE", "#F3BDA4", "#A24E3D", "#976FD9", "#8C9FA8",
"#7C2B73", "#4E5F37", "#5D5462", "#90956F", "#6AA776", "#DBCBF6", "#DA71FF", "#987C95",
"#52323C", "#BB3C42", "#584D39", "#4FC15F", "#A2B9C1", "#79DB21", "#1D5958", "#BD744E",
"#160B00", "#20221A", "#6B8295", "#00E0E4", "#102401", "#1B782A", "#DAA9B5", "#B0415D",
"#859253", "#97A094", "#06E3C4", "#47688C", "#7C6755", "#075C00", "#7560D5", "#7D9F00",
"#C36D96", "#4D913E", "#5F4276", "#FCE4C8", "#303052", "#4F381B", "#E5A532", "#706690",
"#AA9A92", "#237363", "#73013E", "#FF9079", "#A79A74", "#029BDB", "#FF0169", "#C7D2E7",
"#CA8869", "#80FFCD", "#BB1F69", "#90B0AB", "#7D74A9", "#FCC7DB", "#99375B", "#00AB4D",
"#ABAED1", "#BE9D91", "#E6E5A7", "#332C22", "#DD587B", "#F5FFF7", "#5D3033", "#6D3800",
"#FF0020", "#B57BB3", "#D7FFE6", "#C535A9", "#260009", "#6A8781", "#A8ABB4", "#D45262",
"#794B61", "#4621B2", "#8DA4DB", "#C7C890", "#6FE9AD", "#A243A7", "#B2B081", "#181B00",
"#286154", "#4CA43B", "#6A9573", "#A8441D", "#5C727B", "#738671", "#D0CFCB", "#897B77",
"#1F3F22", "#4145A7", "#DA9894", "#A1757A", "#63243C", "#ADAAFF", "#00CDE2", "#DDBC62",
"#698EB1", "#208462", "#00B7E0", "#614A44", "#9BBB57", "#7A5C54", "#857A50", "#766B7E",
"#014833", "#FF8347", "#7A8EBA", "#274740", "#946444", "#EBD8E6", "#646241", "#373917",
"#6AD450", "#81817B", "#D499E3", "#979440", "#011A12", "#526554", "#B5885C", "#A499A5",
"#03AD89", "#B3008B", "#E3C4B5", "#96531F", "#867175", "#74569E", "#617D9F", "#E70452",
"#067EAF", "#A697B6", "#B787A8", "#9CFF93", "#311D19", "#3A9459", "#6E746E", "#B0C5AE",
"#84EDF7", "#ED3488", "#754C78", "#384644", "#C7847B", "#00B6C5", "#7FA670", "#C1AF9E",
"#2A7FFF", "#72A58C", "#FFC07F", "#9DEBDD", "#D97C8E", "#7E7C93", "#62E674", "#B5639E",
"#FFA861", "#C2A580", "#8D9C83", "#B70546", "#372B2E", "#0098FF", "#985975", "#20204C",
"#FF6C60", "#445083", "#8502AA", "#72361F", "#9676A3", "#484449", "#CED6C2", "#3B164A",
"#CCA763", "#2C7F77", "#02227B", "#A37E6F", "#CDE6DC", "#CDFFFB", "#BE811A", "#F77183",
"#EDE6E2", "#CDC6B4", "#FFE09E", "#3A7271", "#FF7B59", "#4E4E01", "#4AC684", "#8BC891",
"#BC8A96", "#CF6353", "#DCDE5C", "#5EAADD", "#F6A0AD", "#E269AA", "#A3DAE4", "#436E83",
"#002E17", "#ECFBFF", "#A1C2B6", "#50003F", "#71695B", "#67C4BB", "#536EFF", "#5D5A48",
"#890039", "#969381", "#371521", "#5E4665", "#AA62C3", "#8D6F81", "#2C6135", "#410601",
"#564620", "#E69034", "#6DA6BD", "#E58E56", "#E3A68B", "#48B176", "#D27D67", "#B5B268",
"#7F8427", "#FF84E6", "#435740", "#EAE408", "#F4F5FF", "#325800", "#4B6BA5", "#ADCEFF",
"#9B8ACC", "#885138", "#5875C1", "#7E7311", "#FEA5CA", "#9F8B5B", "#A55B54", "#89006A",
"#AF756F", "#2A2000", "#576E4A", "#7F9EFF", "#7499A1", "#FFB550", "#00011E", "#D1511C",
"#688151", "#BC908A", "#78C8EB", "#8502FF", "#483D30", "#C42221", "#5EA7FF", "#785715",
"#0CEA91", "#FFFAED", "#B3AF9D", "#3E3D52", "#5A9BC2", "#9C2F90", "#8D5700", "#ADD79C",
"#00768B", "#337D00", "#C59700", "#3156DC", "#944575", "#ECFFDC", "#D24CB2", "#97703C",
"#4C257F", "#9E0366", "#88FFEC", "#B56481", "#396D2B", "#56735F", "#988376", "#9BB195",
"#A9795C", "#E4C5D3", "#9F4F67", "#1E2B39", "#664327", "#AFCE78", "#322EDF", "#86B487",
"#C23000", "#ABE86B", "#96656D", "#250E35", "#A60019", "#0080CF", "#CAEFFF", "#323F61",
"#A449DC", "#6A9D3B", "#FF5AE4", "#636A01", "#D16CDA", "#736060", "#FFBAAD", "#D369B4",
"#FFDED6", "#6C6D74", "#927D5E", "#845D70", "#5B62C1", "#2F4A36", "#E45F35", "#FF3B53",
"#AC84DD", "#762988", "#70EC98", "#408543", "#2C3533", "#2E182D", "#323925", "#19181B",
"#2F2E2C", "#023C32", "#9B9EE2", "#58AFAD", "#5C424D", "#7AC5A6", "#685D75", "#B9BCBD",
"#834357", "#1A7B42", "#2E57AA", "#E55199", "#316E47", "#CD00C5", "#6A004D", "#7FBBEC",
"#F35691", "#D7C54A", "#62ACB7", "#CBA1BC", "#A28A9A", "#6C3F3B", "#FFE47D", "#DCBAE3",
"#5F816D", "#3A404A", "#7DBF32", "#E6ECDC", "#852C19", "#285366", "#B8CB9C", "#0E0D00",
"#4B5D56", "#6B543F", "#E27172", "#0568EC", "#2EB500", "#D21656", "#EFAFFF", "#682021",
"#2D2011", "#DA4CFF", "#70968E", "#FF7B7D", "#4A1930", "#E8C282", "#E7DBBC", "#A68486",
"#1F263C", "#36574E", "#52CE79", "#ADAAA9", "#8A9F45", "#6542D2", "#00FB8C", "#5D697B",
"#CCD27F", "#94A5A1", "#790229", "#E383E6", "#7EA4C1", "#4E4452", "#4B2C00", "#620B70",
"#314C1E", "#874AA6", "#E30091", "#66460A", "#EB9A8B", "#EAC3A3", "#98EAB3", "#AB9180",
"#B8552F", "#1A2B2F", "#94DDC5", "#9D8C76", "#9C8333", "#94A9C9", "#392935", "#8C675E",
"#CCE93A", "#917100", "#01400B", "#449896", "#1CA370", "#E08DA7", "#8B4A4E", "#667776",
"#4692AD", "#67BDA8", "#69255C", "#D3BFFF", "#4A5132", "#7E9285", "#77733C", "#E7A0CC",
"#51A288", "#2C656A", "#4D5C5E", "#C9403A", "#DDD7F3", "#005844", "#B4A200", "#488F69",
"#858182", "#D4E9B9", "#3D7397", "#CAE8CE", "#D60034", "#AA6746", "#9E5585", "#BA6200"
};

I have put up a page online for procedurally generating visually distinct colors:
http://phrogz.net/css/distinct-colors.html
Unlike other answers here that evenly step across RGB or HSV space (where there is a nonlinear relationship between the axis values and the perceptual differences), my page uses the standard CMI(I:c) color distance algorithm to prevent two colors from being too visually close.
The final tab of the page allows you to sort the values in several ways, and then interleave them (ordered shuffle) so that you get very distinct colors placed next to one another.
As of this writing, it only works well in Chrome and Safari, with a shim for Firefox; it uses HTML5 range input sliders in the interface, which IE9 and Firefox do not yet support natively.

I think the HSV (or HSL) space has more opportunities here. If you don't mind the extra conversion, it's pretty easy to go through all the colors by just rotating the Hue value. If that's not enough, you can change the Saturation/Value/Lightness values and go through the rotation again. Or, you can always shift the Hue values or change your "stepping" angle and rotate more times.

There's a flaw in the previous RGB solutions. They don't take advantage of the whole color space since they use a color value and 0 for the channels:
#006600
#330000
#FF00FF
Instead they should be using all the possible color values to generate mixed colors that can have up to 3 different values across the color channels:
#336600
#FF0066
#33FF66
Using the full color space you can generate more distinct colors. For example, if you have 4 values per channel, then 4*4*4=64 colors can be generated. With the other scheme, only 4*7+1=29 colors can be generated.
If you want N colors, then the number of values per channel required is: ceil(cube_root(N))
With that, you can then determine the possible (0-255 range) values (python):
max = 255
segs = int(num**(Decimal("1.0")/3))
step = int(max/segs)
p = [(i*step) for i in xrange(segs)]
values = [max]
values.extend(p)
Then you can iterate over the RGB colors (this is not recommended):
total = 0
for red in values:
for green in values:
for blue in values:
if total <= N:
print color(red, green, blue)
total += 1
Nested loops will work, but are not recommended since it will favor the blue channel and the resulting colors will not have enough red (N will most likely be less than the number of all possible color values).
You can create a better algorithm for the loops where each channel is treated equally and more distinct color values are favored over small ones.
I have a solution, but didn't want to post it since it isn't the easiest to understand or efficient. But, you can view the solution if you really want to.
Here is a sample of 64 generated colors: 64 colors

I needed the same functionality, in a simple form.
What I needed was to generate as unique as possible colors from an an increasing index value.
Here is the code, in C# (Any other language implementation should be very similar)
The mechanism is very simple
A pattern of color_writers get generated from indexA values from 0 to 7.
For indices < 8, those colors are = color_writer[indexA] * 255.
For indices between 8 and 15, those colors are = color_writer[indexA] * 255 + (color_writer[indexA+1]) * 127
For indices between 16 and 23, those colors are = color_writer[indexA] * 255 + (color_writer[indexA+1]) * 127 + (color_writer[indexA+2]) * 63
And so on:
private System.Drawing.Color GetRandColor(int index)
{
byte red = 0;
byte green = 0;
byte blue = 0;
for (int t = 0; t <= index / 8; t++)
{
int index_a = (index+t) % 8;
int index_b = index_a / 2;
//Color writers, take on values of 0 and 1
int color_red = index_a % 2;
int color_blue = index_b % 2;
int color_green = ((index_b + 1) % 3) % 2;
int add = 255 / (t + 1);
red = (byte)(red+color_red * add);
green = (byte)(green + color_green * add);
blue = (byte)(blue + color_blue * add);
}
Color color = Color.FromArgb(red, green, blue);
return color;
}
Note: To avoid generating bright and hard to see colors (in this example: yellow on white background) you can modify it with a recursive loop:
int skip_index = 0;
private System.Drawing.Color GetRandColor(int index)
{
index += skip_index;
byte red = 0;
byte green = 0;
byte blue = 0;
for (int t = 0; t <= index / 8; t++)
{
int index_a = (index+t) % 8;
int index_b = index_a / 2;
//Color writers, take on values of 0 and 1
int color_red = index_a % 2;
int color_blue = index_b % 2;
int color_green = ((index_b + 1) % 3) % 2;
int add = 255 / (t + 1);
red = (byte)(red + color_red * add);
green = (byte)(green + color_green * add);
blue = (byte)(blue + color_blue * add);
}
if(red > 200 && green > 200)
{
skip_index++;
return GetRandColor(index);
}
Color color = Color.FromArgb(red, green, blue);
return color;
}

In case someone needs to generate random medium to high dark color for white foreground in C#, here is the code.
[DllImport("shlwapi.dll")]
public static extern int ColorHLSToRGB(int H, int L, int S);
public static string GetRandomDarkColor()
{
int h = 0, s = 0, l = 0;
h = (RandomObject.Next(1, 2) % 2 == 0) ? RandomObject.Next(0, 180) : iApp.RandomObject.Next(181, 360);
s = RandomObject.Next(90, 160);
l = RandomObject.Next(80, 130);
return System.Drawing.ColorTranslator.FromWin32(ColorHLSToRGB(h, l, s)).ToHex();
}
private static string ToHex(this System.Drawing.Color c)
{
return "#" + c.R.ToString("X2") + c.G.ToString("X2") + c.B.ToString("X2");
}
You can replace RandomObject with your own Random class object.

I would start with a set brightness 100% and go around primary colors first:
FF0000, 00FF00, 0000FF
then the combinations
FFFF00, FF00FF, 00FFFF
next for example halve the brightness and do same round. There's not too many really clearly distinct colors, after these I would start to vary the line width and do dotted/dashed lines etc.

I implemented this algorithm in a shorter way
void ColorValue::SetColorValue( double r, double g, double b, ColorType myType )
{
this->c[0] = r;
this->c[1] = g;
this->c[2] = b;
this->type = myType;
}
DistinctColorGenerator::DistinctColorGenerator()
{
mFactor = 255;
mColorsGenerated = 0;
mpColorCycle = new ColorValue[6];
mpColorCycle[0].SetColorValue( 1.0, 0.0, 0.0, TYPE_RGB);
mpColorCycle[1].SetColorValue( 0.0, 1.0, 0.0, TYPE_RGB);
mpColorCycle[2].SetColorValue( 0.0, 0.0, 1.0, TYPE_RGB);
mpColorCycle[3].SetColorValue( 1.0, 1.0, 0.0, TYPE_RGB);
mpColorCycle[4].SetColorValue( 1.0, 0.0, 1.0, TYPE_RGB);
mpColorCycle[5].SetColorValue( 0.0, 1.0, 1.0, TYPE_RGB);
}
//----------------------------------------------------------
ColorValue DistinctColorGenerator::GenerateNewColor()
{
int innerCycleNr = mColorsGenerated % 6;
int outerCycleNr = mColorsGenerated / 6;
int cycleSize = pow( 2, (int)(log((double)(outerCycleNr)) / log( 2.0 ) ) );
int insideCycleCounter = outerCycleNr % cyclesize;
if ( outerCycleNr == 0)
{
mFactor = 255;
}
else
{
mFactor = ( 256 / ( 2 * cycleSize ) ) + ( insideCycleCounter * ( 256 / cycleSize ) );
}
ColorValue newColor = mpColorCycle[innerCycleNr] * mFactor;
mColorsGenerated++;
return newColor;
}

You could also think of the color space as all combinations of three numbers from 0 to 255, inclusive. That's the base-255 representation of a number between 0 and 255^3, forced to have three decimal places (add zeros on to the end if need be.)
So to generate x number of colors, you'd calculate x evenly spaced percentages, 0 to 100. Get numbers by multiplying those percentages by 255^3, convert those numbers to base 255, and add zeros as previously mentioned.
Base conversion algorithm, for reference (in pseudocode that's quite close to C#):
int num = (number to convert);
int baseConvert = (desired base, 255 in this case);
(array of ints) nums = new (array of ints);
int x = num;
double digits = Math.Log(num, baseConvert); //or ln(num) / ln(baseConvert)
int numDigits = (digits - Math.Ceiling(digits) == 0 ? (int)(digits + 1) : (int)Math.Ceiling(digits)); //go up one if it turns out even
for (int i = 0; i < numDigits; i++)
{
int toAdd = ((int)Math.Floor(x / Math.Pow((double)convertBase, (double)(numDigits - i - 1))));
//Formula for 0th digit: d = num / (convertBase^(numDigits - 1))
//Then subtract (d * convertBase^(numDigits - 1)) from the num and continue
nums.Add(toAdd);
x -= toAdd * (int)Math.Pow((double)convertBase, (double)(numDigits - i - 1));
}
return nums;
You might also have to do something to bring the range in a little bit, to avoid having white and black, if you want. Those numbers aren't actually a smooth color scale, but they'll generate separate colors if you don't have too many.
This question has more on base conversion in .NET.

for getting nth colour. Just this kind of code would be enough. This i have use in my opencv clustering problem. This will create different colours as col changes.
for(int col=1;col<CLUSTER_COUNT+1;col++){
switch(col%6)
{
case 1:cout<<Scalar(0,0,(int)(255/(int)(col/6+1)))<<endl;break;
case 2:cout<<Scalar(0,(int)(255/(int)(col/6+1)),0)<<endl;break;
case 3:cout<<Scalar((int)(255/(int)(col/6+1)),0,0)<<endl;break;
case 4:cout<<Scalar(0,(int)(255/(int)(col/6+1)),(int)(255/(int)(col/6+1)))<<endl;break;
case 5:cout<<Scalar((int)(255/(int)(col/6+1)),0,(int)(255/(int)(col/6+1)))<<endl;break;
case 0:cout<<Scalar((int)(255/(int)(col/6)),(int)(255/(int)(col/6)),0)<<endl;break;
}
}

You could get a random set of your 3 255 values and check it against the last set of 3 values, making sure they are each at least X away from the old values before using them.
OLD: 190, 120, 100
NEW: 180, 200, 30
If X = 20, then the new set would be regenerated again.

Related

I recently read an article explaining how to generate a weighted random number, and there's a piece of the code that I don't understand:
int r = ((int)(rand.Next() * (323567)) % prefix[n - 1]) + 1;
Why is rand.Next being multiplied by a constant 323567? Would the code work without this constant?
Below is the full code for reference, and you can find the full article here: https://www.geeksforgeeks.org/random-number-generator-in-arbitrary-probability-distribution-fashion/
Any help is appreciated, thank you!!
// C# program to generate random numbers
// according to given frequency distribution
using System;
class GFG{
// Utility function to find ceiling
// of r in arr[l..h]
static int findCeil(int[] arr, int r,
int l, int h)
{
int mid;
while (l < h)
{
// Same as mid = (l+h)/2
mid = l + ((h - l) >> 1);
if (r > arr[mid])
l = mid + 1;
else
h = mid;
}
return (arr[l] >= r) ? l : -1;
}
// The main function that returns a random number
// from arr[] according to distribution array
// defined by freq[]. n is size of arrays.
static int myRand(int[] arr, int[] freq, int n)
{
// Create and fill prefix array
int[] prefix = new int[n];
int i;
prefix[0] = freq[0];
for(i = 1; i < n; ++i)
prefix[i] = prefix[i - 1] + freq[i];
// prefix[n-1] is sum of all frequencies.
// Generate a random number with
// value from 1 to this sum
Random rand = new Random();
int r = ((int)(rand.Next() * (323567)) % prefix[n - 1]) + 1; // <--- RNG * Constant
// Find index of ceiling of r in prefix array
int indexc = findCeil(prefix, r, 0, n - 1);
return arr[indexc];
}
// Driver Code
static void Main()
{
int[] arr = { 1, 2, 3, 4 };
int[] freq = { 10, 5, 20, 100 };
int i, n = arr.Length;
// Let us generate 10 random numbers
// according to given distribution
for(i = 0; i < 5; i++)
Console.WriteLine(myRand(arr, freq, n));
}
}
UPDATE:
I ran this code to check it:
int[] intArray = new int[] { 1, 2, 3, 4, 5 };
int[] weights = new int[] { 5, 20, 20, 40, 15 };
List<int> results = new List<int>();
for (int i = 0; i < 100000; i++)
{
results.Add(WeightedRNG.GetRand(intArray, weights, intArray.Length));
}
for (int i = 0; i < intArray.Length; i++)
{
int itemsFound = results.Where(x => x == intArray[i]).Count();
Console.WriteLine($"{intArray[i]} was returned {itemsFound} times.");
}
And here are the results with the constant:
1 was returned 5096 times.
2 was returned 19902 times.
3 was returned 20086 times.
4 was returned 40012 times.
5 was returned 14904 times.
And without the constant...
1 was returned 100000 times.
2 was returned 0 times.
3 was returned 0 times.
4 was returned 0 times.
5 was returned 0 times.
It completely breaks without it.

The constant does serve a purpose in some environments, but I don't believe this code is correct for C#.
To explain, let's look at the arguments to the function. The first sign something is off is passing n as an argument instead of inferring it from the arrays. The second sign is it's poor practice in C# to deal with paired arrays rather than something like a 2D array or sequence of single objects (such as a Tuple). But those are just indicators something is odd, and not evidence of any bugs.
So let's put that on hold for a moment and explain why a constant might matter by looking a small example.
Say you have three numbers (1, 2, and 3) with weights 3, 2, and 2. This function first builds up a prefix array, where each item includes the chances of finding the number for that index and all previous numbers.
We end up with a result like this: (3, 5, 7). Now we can use the last value and take a random number from 1 to 7. Values 1-3 result in 1, values 4 and 5 result in 2, and values 6 and 7 result in 3.
To find this random number the code now calls rand.Next(), and this is where I think the error comes in. In many platforms, the Next() function returns a double between 0 and 1. That's too small to use to lookup your weighted value, so you then multiply by a prime constant related the platform's epsilon value to ensure you have a reasonably large result that will cover the entire possible range of desired weights (in this case, 1-7) and then some. Now you take the remainder (%) vs your max weight (7), and map it via the prefix array to get the final result.
So the first error is, in C#, .Next() does not return a double. It is documented to return a non-negative random integer between 0 and int.MaxValue. Multiply that by 323567 and you're asking for integer overflow exceptions. Another sign of this mistake is the cast to int: the result of this function is already an int. And let's not even talk the meaningless extra parentheses around (323567).
There is also another, more subtle, error.
Let's the say the result of the (int)(rand.Next() * 323567) expression is 10. Now we take this value and get the remainder when dividing by our max value (%7). The problem here is we have two chances to roll a 1, 2, or 3 (the extra chance is if the original was 8, 9, or 10), and only once chance for the remaining weights (4-7). So we have introduced unintended bias into the system, completely throwing off the expected weights. (This is a simplification. The actual number space is not 1-10; it's 0 * 323567 - 0.999999 * 323567. But the potential for bias still exists as long that max value is not evenly divisible by the max weight value).
It's possible the constant was chosen because it has properties to minimize this effect, but again: it was based on a completely different kind of .Next() function.
Therefore the rand.Next() line should probably be changed to look like this:
int r = rand.Next(prefix[n - 1]) +1;
or this:
int r = ((int)(rand.NextDouble() * (323567 * prefix[n - 1])) % prefix[n - 1]) + 1;
But, given the other errors, I'd be wary of using this code at all.
For fun, here's an example running several different options:
https://dotnetfiddle.net/Y5qhRm
The original random method (using NextDouble() and a bare constant) doesn't fare as badly as I'd expect.

I'm new to c#. I have a task to make a type of minesweeper, but which immediately opens a solution.
static void Main(string[] args)
{
Console.Write("Enter the width of the field: ");
int q = Convert.ToInt32(Console.ReadLine());
Console.Write("Enter the length of the field: ");
int w = Convert.ToInt32(Console.ReadLine());
Console.Write("Enter the number of bombs: ");
int c = Convert.ToInt32(Console.ReadLine());
Random rand = new Random();
var charArray = new char[q, w];
var intArray = new int[q, w];
for (int i = 0; i < q; i++)
{
for (int j = 0; j < w; j++)
{
intArray[i, j] = rand.Next(2);
charArray[i, j] = intArray[i, j] == 0 ? '_' : '*';
Console.Write(charArray[i, j]);
}
Console.WriteLine();
}
}
}
}
Two arrays should be output. Everything should be closed on the first one, that is, there should be only the characters: _ and *
0 - these are places without mines, I replaced them with a symbol _
1 - these are places with mines, I replaced them with an asterisk symbol, but they do not accept the number of mines entered by the user. And it is necessary that there are as many "*" characters as there are mines.
And in the second array there should be an already open solution of the game. That is, the cells next to which there are mines should take a number meaning the number of mines next to this cell.
Please help me..
Compiling the current code

Random random = new Random();
while(c > 0)
{
var rq = random.Next(q);
var rw = random.Next(w);
if(intArray[rq,rw] == 0)
{
intArray[rq, rw] = 1;
c--;
}
}

I would suggest dividing the problem in smaller manageable chunks. For instance, you can place the bombs in a initial step, and on a second step build the solution. You can build the solution at the same time you place the bombs, although for clarity you can do it after.
Naming of variables is also important. If you prefer using single letter variable names, I believe that's fine for the problem limits, however I would use meaningful letters easier to remember. eg: W and H for the width and height of the board, and B for the total number of bombs.
The first part of the problem then can be described as placing B bombs in a WxH board. So instead of having nested for statements that enumerate WxH times, it's better to have a while loop that repeats the bomb placing logic as long as you have remaining bombs.
Once you generate a new random location on the board, you have to check you haven't placed a bomb there already. You can have an auxiliary function HasBomb that checks that:
bool HasBomb(char[,] charArray, int x, int y)
{
return charArray[x,y] == '*';
}
I'll leave error checking out, this function being private can rely on the caller sending valid coordinates.
Then the bomb placing procedure can be something like:
int remainingBombs = B;
while (remainingBombs > 0)
{
int x = rand.Next(W);
int y = rand.Next(H);
if (!HasBomb(charArray, x, y)
{
charArray[x,y] = '*';
remainingBombs--;
}
}
At this point you may figure out another concern. If the number B of bombs to place is larger than the available positions on the board WxH, then you wont be able to place the bombs on the board. You'll have to check for that restriction when requesting the values for W, H and B.
Then in order to create the array with the number of bombs next to each position, you'll need some way to check for all the neighbouring positions to a given one. If the position is in the middle of the board it has 8 neighbour positions, if it's on an edge it has 5, and if it's on a corner it has 3. Having a helper function return all the valid neighbour positions can be handy.
IEnumerable<(int X, int Y)> NeighbourPositions(int x, int y, int W, int H)
{
bool leftEdge = x == 0;
bool topEdge = y == 0;
bool rightEdge = x == W - 1;
bool bottomEdge = y == H - 1;
if (!leftEdge && !topEdge)
yield return (x-1, y-1);
if (!topEdge)
yield return (x, y-1);
if (!rightEdge && !topEdge)
yield return (x+1, y-1);
if (!leftEdge)
yield return (x-1, y);
if (!rightEdge)
yield return (x+1, y);
if (!leftEdge && !bottomEdge)
yield return (x-1, y+1);
if (!bottomEdge)
yield return (x, y+1);
if (!rightEdge && !bottomEdge)
yield return (x+1, y+1)
}
This function uses Iterators and touples. If you feel those concepts are too complex as you said are new to C#, you can make the function return a list with coordinates instead.
Now the only thing left is to iterate over the whole intArray and increment the value on each position for each neighbour bomb you find.
for (int x = 0; x < W; x++)
{
for (int y = 0; y < H; y++)
{
foreach (var n in NeighbourPositions(x, y, W, H))
{
if (HasBomb(charArray, n.X, n.Y))
intArray[x,y]++;
}
}
}

The answers here are mostly about generating random x and random y put in loop and trying to put the mine into empty cells. It is ok solution, but if you think of it, it is not that sufficient. Every time you try to find a new random cell, there is chance that cell is already a mine. This is pretty much alright, if you don't put too many mines into your field, but if you put some greater number of mines, this event can occur quite often. This means that the loop might take longer than usually. Or, theoretically, if you wanted to put 999 mines into 1000 cell field, it would be really hard for the loop to fill all the necessary cells, especially for the last mine. Now, I am not saying that the solutions here are bad, I think, it's really alright solution for many people. But if someone wanted a little bit efficient solution, I have tried to crate it.
Solution
In this solution, you iterate each cell and try to calculate a probability of the mine to be placed there. I have come up with this easy formula, which calculates the probability:
Every time you try to place a mine into one cell, you calculate this formula and compare it to random generated number.
bool isMine = random.NextDouble() < calcProbability();

This question already has answers here:
Calculate median in c#
(12 answers)
Closed 5 years ago.
So I'm struggling to get this piece to output a median value with 4 values.
The output produces a value one above the actual middle value and I cannot seem to get it to output a decimal even when I change 2 to 2.0. I can get it to output a value with 3 numbers just haven't achieved it with 4.
Console.Write("Median Value: ");
var items = new[]{num1, num2, num3, num4 };
Array.Sort(items);
Console.WriteLine(items[items.Length/2]);
This work is an extension task in my computing class so I may have very well taken a completely wrong approach to this task.
Thanks in advance

If you look at the explanations in Wikipedia, it's quite simple:
Easy explanation of the sample median
In individual series (if number of observation is very low) first one must arrange all the observations in order.
Then count(n) is the total number of observation in given data.
If n is odd then Median (M) = value of ((n + 1)/2)th item term.
If n is even then Median (M) = value of [(n/2)th item term + (n/2 + 1)th item term]/2.
How does that translate to code ?
using System;
namespace ConsoleApp1
{
internal static class Program
{
private static void Main(string[] args)
{
var array = new[] {1, 2, 3, 4};
Array.Sort(array);
var n = array.Length;
double median;
var isOdd = n % 2 != 0;
if (isOdd)
{
median = array[(n + 1) / 2 - 1];
}
else
{
median = (array[n / 2 - 1] + array[n / 2]) / 2.0d;
}
Console.WriteLine(median);
}
}
}
Note that you have to subtract one when getting the value of an element in the array since array indices are zero-based.

public decimal GetMedian(int[] array)
{
int[] tempArray = array;
int count = tempArray.Length;
Array.Sort(tempArray);
decimal medianValue = 0;
if (count % 2 == 0)
{
// count is even, need to get the middle two elements, add them together, then divide by 2
int middleElement1 = tempArray[(count / 2) - 1];
int middleElement2 = tempArray[(count / 2)];
medianValue = (middleElement1 + middleElement2) / 2;
}
else
{
// count is odd, simply get the middle element.
medianValue = tempArray[(count / 2)];
}
return medianValue;
}
Your implementation works for odd sized collections. But when dealing with even sized collections, you find the middle pair of numbers, and then find the value that is half way (simple arithmetical average) between them. This is easily done by adding them together and dividing by two.
The method call based on your code:
Console.Write("Median Value: ");
int[] items = new int[] {num1, num2, num3, num4};
var median = GetMedian(items);
Console.WriteLine(median);
See it running on Ideone.
How to find the median value
Add a median method to a list (source)

I have an enum defined like that:
[Flags]
public enum Orientation
{
North = 1,
North_East = 2,
East = 4,
South_East = 8,
South = 16,
South_West = 32,
West = 64,
North_West = 128
}
Is there a generic way to tell if exactly one flag is set, multiple or none?
I don't care for what the value of the enum is, I just want to know how many flags are set.
This is not a duplicate for counting the number of bits. If I initialize the enum like this and count the number of bits I get 10. But the number of set flags would be 8.
GeographicOrientation go = (GeographicOrientation) 1023;

You can determine this with a simple bit trick after converting the value to int:
int v = (int)enumValue;
If v == 0, then no flags are set
Otherwise, if ((v-1)&v) == 0, then exactly one flag is set
Otherwise, multiple flags are set.
The only tricky one is #2. Here is an explanation: consider a binary number with exactly one bit set to 1. Then subtracting 1 would make the following change:
0000010000000
- 1
-------------
0000001111111
All zeros following the lone 1 become 1s, 1 becomes zero, and the rest of the bits remain the same. AND-ing v and v-1 produces zero, because there is no pair of 1s in the same position between the two numbers.
When there are two or more 1s, the bits to the left of the lone 1 will remain unchanged. Therefore, at least one position will have a 1 in the result of bitwise AND.

you can use this code :
var item = Orientation.North | Orientation.South;
int i = 0;
foreach (Orientation e in Enum.GetValues(typeof(Orientation)))
if(item.HasFlag(e))
i++;
Console.WriteLine(i);

var test = Orientation.North;
var flagCount = GetFlagCount(test);
public int GetFlagCount(Enum testValue)
{
return Enum.GetValues(testValue.GetType()).Cast<Enum>().Count(testValue.HasFlag);
}

If you are looking for the "shortest" way:
Orientation o = Orientation.East | Orientation.West; // o.ToString() = "East, West"
var c = o.ToString().Split().Count();
or even shorter:
var c = (o + "").Split().Count();
Update
To support values above 255, you can use any of those ugly hacks:
Orientation o = (Orientation) 1023;
var c = ((Orientation)(byte)o + "").Split().Count();
c = ((Orientation)((int)o & 255) + "").Split().Count();
or just define the enum as byte:
[Flags]
public enum Orientation : byte
{
North = 1,
North_East = 2,
East = 4,
South_East = 8,
South = 16,
South_West = 32,
West = 64,
North_West = 128
}
Update 2
I personally wouldn't use the string method in production code especially when just a bit count is needed.
Anyway, I just thought of another hack just for fun. Base 2 log will return a whole number when one bit is set, -Infinity when 0, and anything else when more than one bit is set. For Example
Math.Log(0, 2 ) // -Infinity
Math.Log(0, 64) // 6.0
Math.Log(0, 65) // 6.0223678130284544
So, (byte)go != 0 can be used to check if any flags are set, and then Math.Log((byte)go, 2) % 1 == 0 to check if only one flag is set.
But, dasblinkenlight's solution seems like the best.

Off the top of my head:
var map = 1;
var count = 0;
while (map <= North_West)
{
if( ((int)Value & map) > 0)
count += 1;
map = map << 1; //left shift, a.k.a. multiply by 2
}

This is not a duplicate for counting the number of bits. If I initialize the enum like this and count the number of bits I get 10. But the number of set flags would be 8.
Please consider this statement directly from MSDN:
Flags enumerations are used for masking bit fields and doing bitwise comparisons.
If you are designing or using the enum in some manner that does NOT allow counting the number of flags using bitwise operations, then you are designing or using the enum improperly.
[Flags]
public enum MyEnum
{
Foo = 1,
Bar = 2,
Bizz = 4,
Buzz = 8,
Boo = 16
}
var foo = MyEnum.Foo | MyEnum.Foo | MyEnum.Bizz;
// foo == MyEnum.Foo | MyEnum.Bizz because setting the same flag twice does nothing
var bar = (MyEnum)27 // Some invalid combination
// bar == 27. Why would you do this instead of MyEnum.Boo | MyEnum.Buzz | MyEnum.Bar | MyEnum.Foo
If you design your enum properly, you can count the flag, and optionally short circuit if there is more than one flag set since continuing to count would be pointless.
var foo = MyEnum.Foo | MyEnum.Bizz;
int fooValue = (int)foo;
int numberOfFlags = 0;
while (fooValue > 0 && numberOfFlags < 2) // Stop counting if more than one flag since we don't need exact count
{
if ((fooValue & 1) == 1)
{
numberOfFlags++;
}
fooValue >>= 1;
}
Console.WriteLine(numberOfFlags);

I have this code that finds numbers in a given range that contain only 3 and 5 and are polynoms(symetrical, 3553 for example). The problem is that the numbers are between 1 and 10^18, so there are cases in which I have to work with big numbers, and using BigInteger makes the program way too slow, so is there a way to fix this ? Here's my code:
namespace Lucky_numbers
{
class Program
{
static void Main(string[] args)
{
string startString = Console.ReadLine();
string finishString = Console.ReadLine();
BigInteger start = BigInteger.Parse(startString);
BigInteger finish = BigInteger.Parse(finishString);
int numbersFound = 0;
for (BigInteger i = start; i <= finish; i++)
{
if (Lucky(i.ToString()))
{
if (Polyndrome(i.ToString()))
{
numbersFound++;
}
}
}
}
static bool Lucky(string number)
{
if (number.Contains("1") || number.Contains("2") || number.Contains("4") || number.Contains("6") || number.Contains("7") || number.Contains("8") || number.Contains("9") || number.Contains("0"))
{
return false;
}
else
{
return true;
}
}
static bool Polyndrome(string number)
{
bool symetrical = true;
int middle = number.Length / 2;
int rightIndex = number.Length - 1;
for (int leftIndex = 0; leftIndex <= middle; leftIndex++)
{
if (number[leftIndex] != number[rightIndex])
{
symetrical = false;
break;
}
rightIndex--;
}
return symetrical;
}
}
}
Edit: Turns out it's not BigInteger, it's my shitty implementation.

You could use ulong:
Size: Unsigned 64-bit integer
Range: 0 to 18,446,744,073,709,551,615
But I would guess that BigInteger is not a problem here. I think you should create algorithm for palindrome creation instead of brute-force increment+check solution.
Bonus
Here is a palyndrome generator I wrote in 5 minutes. I think it will be much faster than your approach. Could you test it and tell how much faster it is? I'm curious about that.
public class PalyndromeGenerator
{
private List<string> _results;
private bool _isGenerated;
private int _length;
private char[] _characters;
private int _middle;
private char[] _currentItem;
public PalyndromeGenerator(int length, params char[] characters)
{
if (length <= 0)
throw new ArgumentException("length");
if (characters == null)
throw new ArgumentNullException("characters");
if (characters.Length == 0)
throw new ArgumentException("characters");
_length = length;
_characters = characters;
}
public List<string> Results
{
get
{
if (!_isGenerated)
throw new InvalidOperationException();
return _results.ToList();
}
}
public void Generate()
{
_middle = (int)Math.Ceiling(_length / 2.0) - 1;
_results = new List<string>((int)Math.Pow(_characters.Length, _middle + 1));
_currentItem = new char[_length];
GeneratePosition(0);
_isGenerated = true;
}
private void GeneratePosition(int position)
{
if(position == _middle)
{
for (int i = 0; i < _characters.Length; i++)
{
_currentItem[position] = _characters[i];
_currentItem[_length - position - 1] = _characters[i];
_results.Add(new string(_currentItem));
}
}
else
{
for(int i = 0; i < _characters.Length; i++)
{
_currentItem[position] = _characters[i];
_currentItem[_length - position - 1] = _characters[i];
GeneratePosition(position + 1);
}
}
}
}
Usage:
var generator = new PalyndromeGenerator(6, '3', '5');
generator.Generate();
var items = generator.Results.Select(x => ulong.Parse(x)).ToList();

Strange riddle, but can be simplified if I understand the requirement.
I would first map these numbers to binary as there is only two possible
"lucky" digits, then generate the numbers by counting in binary until
I have completed nine bits. Reflect it for the full number, then
convert 0 to 3 and 1 to 5.
Example 1101
Reflect it = 10111101 --> 53555535
Do this from 0 all the way to 111111111

Declare start and finish to be static inside the class.
Change the method Lucky to:
static bool Lucky(string number)
{
return !(number.Contains("1") || number.Contains("2") || number.Contains("4") || number.Contains("6") || number.Contains("7") || number.Contains("8") || number.Contains("9") || number.Contains("0"));
}
Also, you can use Parallel library to parallelize the computation.
Instead of using a regular for loop, you could use a Parallel.For.

Look at the problem a different way - how many strings of up to 9 characters (using only '3' and '5') can you make? for each string you have 2 palindromes (one repeating the last character, one not) that you can make.
e.g.
3 -> 33
5 ->, 55
33 -> 333, 3333
35 -> 353, 3553
53 -> 535, 5335
...

The only suggestion I have is to use a 3rd party library like intx, or some unmanaged code. The intx author reports that it can work faster than BigInteger in some situations: "System.Numerics.BigInteger class was introduced in .NET 4.0 so I was interested in performance of this solution. I did some tests (grab test code from GitHub) and it appears that BigInteger has performance in general comparable with IntX on standard operations but starts losing when FHT comes into play (when multiplying really big integers, for example)."

Since the number has to be symmetrical, you only need to check the first half of the number. You don't need to check 18 digits, you only have to check to 9 digits and then swap the order of the characters and add them to the back as a string.

One thing I can think of is if you are only going to count integers that are containing 3 or 5 you don't need to traverse the entire list of numbers between your beginning & ending range.
Instead, look at your character set as either '3' or '5'. Then you can simply go through the allowed permutations of half of the number itself, leaving the other half to be completed to successfully create a polyndrome.
There are some rules to this method which would help, such as :
if the starting number's left-most digit was greater than 5 there is no need to attempt for that specific number of digits.
if both numbers fall on the same amount of digits but left-most digits do not traverse / include 5 or 3, no need to process.
Developing some set of rules such as this may help other than attempting to check every possible permutation.
So, for example, your Lucky function would become something more along the lines of :
static bool Lucky(string number)
{
if((number[0] != '3') && (number[0] != '5'))
{
return false;
} //and you could continue this for the entire string
...
}