Talk:Karnaugh map

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"This image actually shows two Karnaugh maps..." Oh, does it? Well I wish it didn't! It's quite unreadable. How is the person who has come here expected to learn if you throw them in at the deep end? How about having three images, with this being the third that shows how you have combined the first two? — Preceding unsigned comment added by 203.241.147.20 (talk) 05:56, 8 December 2023 (UTC)[reply]

Hah. It hasn't been fixed in 10 years. That's absolutely hillarious! Qsimanelix (talk) 15:53, 10 October 2024 (UTC)[reply]

I very much agree with "unsigned". The image which shaded in both 0's and 1's would be confusing to someone less familiar with Karnaugh maps. I used https://aashpointo.github.io/KmapWebsite/ to create the graphic. Then I added boxes without shading. Also, now the shading doesn't combine. Blue and green make a beautiful purple, but combining two colors adds to the confusion.

I used the original functions at the bottom and the boxes match. For copyright purposes this should be considered a visual modification of the previous picture.

Feel free to add a 2nd image to explain the PoS formula. I'm preserving the original caption is anyone desires to create a map for a PoS example. "This image actually shows two Karnaugh maps: for the function ƒ, using minterms (colored rectangles) and for its complement, using maxterms (gray rectangles). In the image, E() signifies a sum of minterms, denoted in the article as ∑ m i {\displaystyle \sum m_{i}} ." — Preceding unsigned comment added by Nickalh (talk • contribs) 17:00, 11 February 2025 (UTC)[reply]

Is really the same? Need a little topic here. See the redirectioning. — Preceding unsigned comment added by 201.68.255.23 (talk) 08:17, 11 June 2012 (UTC)[reply]

The current example illustrations of K maps on this page are atrociously har to read due to the coloration system used. Such simple examples are made so awfully difficult to make out. I propose replacing these with normal, standard K map illustrations that use colored outlines and no shading.

152.3.68.83 (talk) 19:26, 5 March 2013 (UTC)[reply]

Agreed. I'll make new diagrams and change the page to use them, because I can't imagine it bothering anyone. If someone is bothered, they will just revert the change. Qsimanelix (talk) 15:50, 10 October 2024 (UTC)[reply]

Although there is now a reference to the Veitch diagram, this is a little misleading. The Karnaugh Map is a special case of a Veitch Diagram. Although the orig;linal form of the Veitch diagram doesn't enjoy any sort of uses now.

This also conflicts with the idea that the Karnaugh map was 'invented' in 1950 - it is more properly an extension of existing work. Anyone have any thoughts on how to resolve this? NVeitch

Additionally, The link to Veitch_chart is circular because it redirects back to this section. I'm not familiar with the preferred way to resolve. spartlow —Preceding undated comment added 20:24, 7 January 2019 (UTC)[reply]

Hm the C,D-row is upside-down. It should go as (0,0)-(0,1)-(1,1)-(1,0). --BL

Perhaps we should mention that the rows and columns are ordered according to a gray code, or otherwise explain the importance of having precisely one input value flip between adjacent boxes. Bovlb 00:37, 2004 Mar 11 (UTC)

Done. Fresheneesz

I do understand that if you don't use Gray code, Karnaugh maps no longer work the way they do. What's the significance, reason and advantage behind using Gray code in Karnaugh maps?

ICE77 (talk) 22:22, 28 December 2010 (UTC)[reply]

Yeah the 'why' would be nice, though, the K-maps still work no matter which order you choose. But, you will get more min/max-terms if you don't go with the gray code which then can be reduced. If you use the K-map with the gray code scheme you probably don't have to reduce your terms at all. The final result, however, will still be the same.

Note: Might say it in general. Just tried it on a few examples. 178.9.117.214 (talk) 14:34, 30 January 2012 (UTC)[reply]

Should there be mentionted that a K-diagram is wrapped around?

So the top left and top right connect and the same goes for bottom left and bottom right.

Kind regards

I second this. Fresheneesz 08:50, 8 March 2006 (UTC)[reply]

There is a line that reads "just as m0, m8, m2, and m10 can be combined into a four-by-four group."(accessed 21/jun/2010). I think that this should read 'a two-by-two group' because there are only two squares on each side making a total of four squares in the group when wrapping the four corners together. Note, I could be really wrong, because I don't understand any of this and there is likely to be more to it than I get. ElTimbalino (talk) 12:16, 21 June 2010 (UTC)[reply]

Should there be a mention that the cells need to be cirlced in a rectangular shape?

Done (not by me tho) Fresheneesz 08:52, 8 March 2006 (UTC)[reply]

Not wishing to be pedantic, but is this strictly true? When doing them by hand any rough shape that groups the right things will do - mine usually come out ellipsoid. The shape of the "circling" is not relevant to the use of the K-map, it's the grouping that matters.Graham 11:32, 8 March 2006 (UTC)[reply]

Well, not only would that be confusing, but what we mean when we say that is that the circled numbers must trace a rectangular shape. Fresheneesz 19:39, 23 March 2006 (UTC)[reply]

What is the correct pronunciation of Karnaugh's name? Stern 00:00, 1 Dec 2004 (UTC)

Same as Carnot, or CAR-NO. Graham 00:08, 1 Dec 2004 (UTC)

What's the purpose of the second diagram? I don't understand it, it appears to duplicate the first with a less clear labelling, and the text doesn't refer to it. I think it should go. Graham 00:29, 21 Dec 2004 (UTC)

It is an alternative syntax, the one we were taught at the university. I actually find it much more clear than the first one: in this one you can see directly where A, B, C and D are ON and where they are OFF just by looking at the line/row. It should actually be rotated 90 degrees so that it could be drawn more directly from the truth table of the function (and so should the already existing map). I'll create an improved version and add a description of it to the article soon. --ZeroOne 17:57, 21 Dec 2004 (UTC)

OK, you can see the improved version here. Refresh the image if it doesn't differ from the one you saw - it is updated:

The little numbers at the corner of each square are the row numbers of the truth table of the function. It is otherwise from 0 to 15, left to right, up to down but the two last columns and two last rows are flipped. There's nothing strange in that, the existing map has the same feature, too. The area "A{", two last rows, is where A is 1. The area "}B", two middle rows, is where B is 1. The area "C{", two last columns, is where C is 1 and the area "D{", two middle columns, is where D is 1. What do you think now? --ZeroOne 18:41, 21 Dec 2004 (UTC)

The updated version does not seem to be showing in the article, even after refresh. It is just an identical copy of the first image, as is the third. Arlo Barnes (talk) 18:43, 23 September 2010 (UTC)[reply]

Weeellll.... to be honest I've been using Karnaugh maps for years and have never seen this form. I'm trying to figure out exactly what it is portraying just by looking at it, and I'm afraid I still don't get it. Maybe I'm just too used to the "old" form. Seems to me the need for the row numbers to be put in each square to help make sense of it mean it might not be as effective as the "old" type - with that one, I can see at a glance what the terms need to be and can simply write them down without any interpretation - but again I'm willing to put that down to familiarity. The flipping of the columns you mention is indeed a feature of the normal map - that's because the bit ordering needs to follow a gray code - something that the article fails to mention, but probably should. Maybe the explanatory text you plan to write will help the lightbulb go on, but the diagram on its own doesn't work for me - but maybe others here will have more to say about it? Graham 22:39, 21 Dec 2004 (UTC)

You don't need those row numbers, you learn them quickly and can always deduce them again should you forget them. Actually the map I'm showing is no different from your map: it just emphasizes the ones better. That is, the rows and columns where a certain variable is 1 are "grouped" with the {'s. The squares which do not belong to a given group belong to the negation (where the variable is 0) of that group. I found a nice picture from the teaching material of my university: [1]. It combines both of the maps in the same picture (because they actually are the same map). Maybe *that* will explain it to you too? :) To add, to me the current map looks almost like if it had 8 rows and 8 columns. The amount of variables shown is too big to handle effectively. --ZeroOne 23:22, 21 Dec 2004 (UTC)

Ah, OK, I get it now. The image on your uni site adds the crucial "lightbulb" detail for me - the AB/CD labelling in the top corner. That said I'm not really convinced that this method is really any better than the normal one - perhaps I was expecting the advantage to be much greater than what is really just a small change to the way the columns and rows are labelled. I'd like to hear some other opinions on this though. By the way, the appearance of 8 columns rather than 4 is not something I'd really found before, since it's clear to me that it is PAIRS of variables that are being considered by each column. Graham 23:47, 21 Dec 2004 (UTC)p

Image:Karnaugh_map_showing_minterms.png

I made the above picture to specify how to find minterms from a karnaugh map. I didn't put this on the main page, because I'm not sure if this is univerasal. Comments please. Fresheneesz 04:57, 6 March 2006 (UTC)[reply]

I have made an SVG image Image:K-map minterms.svg. Cburnett 06:06, 22 December 2006 (UTC)[reply]

This article should mention the possibility and how-to of finding xor terms when minimizing. Fresheneesz 01:56, 19 October 2006 (UTC)[reply]

A karnaugh map doesn't attempt to get the minimum form, it attempts to get the minimum form for a two stage implementation using a particular gate in each stage. Yes there may be tricks that allow one to occasionally spot that there is another way that might save a bit when working in discrete logic but they aren't the main purpose of the maps. Plugwash 16:04, 20 October 2006 (UTC)[reply]

Yeah. K-maps are nice because they always do the computation in 2 steps. It is also just intuitive, regardless of what function is being used. You can also use 3-D K-maps if you have the materials for it, but that doesn't work too great on paper. Qsimanelix (talk) 16:02, 10 October 2024 (UTC)[reply]

Are you talking about Reed-Muller logic ? --68.0.120.35 14:58, 7 December 2006 (UTC)[reply]

I created a few SVG images for this article and, in the process, discovered the listed minterms were not reflected correctly in the images. Please, give the example a thorough review if you have a chance. I did confirm what I could with [2]. Cburnett 08:08, 22 December 2006 (UTC)[reply]

Under Race hazards, first bullet point - shouldn't that read when C is 1, D is 0 and not "when C and D are both 0"? I could be wrong - it's only been 30 years since I have looked at these. Great graphic by the way.--Larry In Cincinnati 03:31, 7 January 2007 (UTC)[reply]

Correct and I've fixed it. Thanks for the complements on the graphics. :) Cburnett 04:01, 7 January 2007 (UTC)[reply]

I feel it appropriate to raise the concern here that illustrations created "de novo" run the risk of being factually incorrect, and fail the encyclopedic requirement of a sound reference source. You can avoid this by using a trusted reference as the source of the illustration's details, even if you create a new rendering to meet copyright requirements. Does anyone else have the same concern? Cburnett seems to have edited the illustration several times to correct errors. This could be avoided by use of an authoritative source material/example. JoGusto (talk) 14:15, 10 August 2013 (UTC)[reply]

Should these be mentioned?

What is a Karnaugh map? Neither in your article, nor in the general literature on the subject, will you find a definition (with one exception). Given a function f : { 0 , 1 } n ↦ { 0 , 1 } {\displaystyle f:\{0,1\}^{n}\mapsto \{0,1\}} , we call its 2 n {\displaystyle 2^{n}} 'arguments' ( x 1 , … , x n ) {\displaystyle (x_{1},\dots ,x_{n})} — each an element of { 0 , 1 } n {\displaystyle \{0,1\}^{n}} — an input combination or input event. We can now define:

A Karnaugh map for n input variables x 1 , … , x n {\displaystyle x_{1},\dots ,x_{n}} is a Venn diagram whose universe consists of all 2 n {\displaystyle 2^{n}} input events ( x 1 , … , x n ) {\displaystyle (x_{1},\dots ,x_{n})} . The sets defined and pictured on the universe are n non-disjoint sets X 1 , … , X n {\displaystyle {\mathcal {X}}_{1},\dots ,{\mathcal {X}}_{n}} , any such set X i {\displaystyle {\mathcal {X}}_{i}} containing all input events whose i-th input variable x i {\displaystyle x_{i}} is 1, i.e., ( x 1 , … , x i , … , x n ) ∈ X i ⇔ x i = 1 {\displaystyle (x_{1},\dots ,x_{i},\dots ,x_{n})\in {\mathcal {X}}_{i}\Leftrightarrow x_{i}=1} .

Karnaugh maps and their usage are covered quite extensively in chapters 6, 7 and 21 (Reduced K-maps) of

Shimon P. Vingron:'Switching Theory. Insight through Predicate Logic' Springer-Verlag, 2004, ISBN 3-540-40343-4.

I have not put this, and other aspects touched on in the above reference, in the main article so as not to intrude on your work: I leave it to you whether you want to make use of the above comment.

S.P.Vingron, 81.217.16.172 14:16, 28 August 2007 (UTC)[reply]

Why must the blue inverse term be restricted by A ¯ {\displaystyle {\overline {A}}} ? If we really don't care about the m15 case, it can be treated as 1 for the non-inverse case and 0 for the inverse case; while this makes the non-inverse and inverse cases non-equivalent for all inputs, they are still equivalent for all inputs that we care about.

However, it doesn't really matter, since treating the don't-care as 1 allows the inverse function to be simplified to F = ( A + B ) ( A + C ) ( A + D ¯ ) {\displaystyle F=\left(A+B\right)\left(A+C\right)\left(A+{\overline {D}}\right)} . I can't take credit for this second bit though, an applet gave it to me when I was checking the first paragraph. Anomie 17:07, 14 October 2007 (UTC)[reply]

Shouldn't the ( A + D ¯ ) {\displaystyle \left(A+{\overline {D}}\right)} term be added to the inverse case to remove a hazard when moving from m7 to either m3 or m5? Anomie 17:07, 14 October 2007 (UTC)[reply]

The section title "Elimination" seems unhelpful and obtuse for a section which deals with race hazards and how to eliminate them. The origin of a race hazard as the result of the reduction, how to fix the race hazard, when is the race hazard important (purely combinatoric logic) and when might it be ignored (registered, clocked "sequential" circuits). JoGusto (talk) 14:28, 10 August 2013 (UTC)[reply]

For the green encircling, besides A and B, C also maintains the same state of 1. So, shouldn't the second term be A B ¯ C {\displaystyle A{\overline {B}}C} and not A B ¯ {\displaystyle A{\overline {B}}} ?

Sepia tone 04:40, 25 October 2007 (UTC)[reply]

I don't know which image you're looking at, but I can't find one where a green encircling has C remaining unchanged. In Image:K-map 6,8,9,10,11,12,13,14.svg, for example, the green circles all four minterms in the rightmost ( A B ¯ {\displaystyle A{\overline {B}}} ) column, overlapping with the right half of the red group. Anomie 11:20, 25 October 2007 (UTC)[reply]

You are right. My bad. Sepia tone (talk) 02:53, 30 November 2007 (UTC)[reply]

Can we have examples for 5 and 6 variables?

For 5 variables A,B,C,D,E: We can make a 4 variable frame A,B,C,D, then fill in each entry with 1,0,x(don't care),E,e(Ebar). Then put in the circles, noting that E is not 0 and is not 1.

Truth table

ABCDE Out
00000 0
00001 1
00010 0
00011 1
00100 0
00101 1
00110 0
00111 1
01000 0
01001 1
01010 0
01011 1
01100 0
01101 1
01110 0
01111 1
10000 0
10001 0
10010 0
10011 0
10100 1
10101 1
10110 1
10111 1
11000 1
11001 1
11010 0
11011 0
11100 0
11101 0
11110 1
11111 1

Map

  |0110|C
  |0011|D
--|----|
00|EEEE|
10|0110|
11|1010|
01|EEEE|
--|----|
AB|

Out=aE+AbC+ACD+ABcd

(lower case is not)

No, usually you would use Boolean Algebra when simplifying five, six, or more variable circuits. It's certainly possible, but do you really want to have thirty-two squares to fill out? ChyranandChloe (talk) 02:29, 14 June 2008 (UTC)[reply]

It is possible to work with 5 variables any beyond. For an odd number like 5 you would create two 2x2 maps. One for E = 0 and one for E = 1. However, like ChyranandChloe said it isn't wise to work with a huge K-map unless you have to but even then I'd rather look for a software solution. 178.9.117.214 (talk) 14:44, 30 January 2012 (UTC)[reply]

I agree... but you would probably use an automated tool based on the Quine-McCluskey minimization algorithm. Perhaps it should be mentioned. JoGusto (talk) 14:31, 10 August 2013 (UTC)[reply]

If you think that the above suggestion is about 32 entries or an additional dimension, then you have miss-read it. It is describing how to add an extra variable without increasing the number of dimensions. We learnt this at university circa 1992-1995. It may have another name. Look at the example it has 5 variable, but only 2 dimensions. From what I remember it allows 6 variable with 2 dimensions. Ctrl-alt-delor (talk) 12:21, 01 October 2019 (UTC)[reply]

You can use 4×4×2 cuboid for 5 variables. 220.228.194.112 (talk) 03:34, 11 December 2020 (UTC)[reply]

Shouldn't entries such as K-Map link to this page? Thanks

129.7.108.128 (talk) 18:28, 6 February 2008 (UTC)[reply]

K-map does, K-Map doesn't because "map" isn't capitalized. But if you put "K-Map" in the search box then you hit this article just fine. Cburnett (talk) 19:03, 6 February 2008 (UTC)[reply]

I think it would helpful if we could include a unsimplified digital electronic circuit design and a simplified circuit design in addition to the boolean algebra - it makes it difficult to apply if you don't have the circuit to refer to. ChyranandChloe (talk) 00:01, 30 May 2008 (UTC)[reply]

Good point! Can you do it? yoyo (talk) 05:40, 9 June 2010 (UTC)[reply]

What the heck is the "axion law"? This term is used at least twice in this article, and links to the boolean logic page, in which the word "axion" never appears. linas (talk) 21:34, 4 June 2008 (UTC)[reply]

It is "axiom laws" (an axion is a hypothetical elementary partial - in physics). An Axiom is a postulate, a assumption that is accepted and known to be true. The way that is is used in the article (and in boolean algebra) is that we automatically accept that if you have A and NOT A they cancel each other out; it's like adding − 1 {\displaystyle -1} and 1 {\displaystyle 1} , you get 0 {\displaystyle 0} . They go to a lot of trouble explaining it in the Boolean algebra page:

"...Two Boolean laws having no numeric counterpart are the laws characterizing logical negation, namely x∧¬x = 0..."

∧ = AND; ¬ = NOT by the way. The edit fixes are made, thanks. ChyranandChloe (talk) 01:55, 6 June 2008 (UTC)[reply]

In the example diagram it is not obvious that the brown section is the overlapping of the red and green sections. This got me confused for quite a while! I suggest that using eg striped red and green for the overlapping section, or make the outlines easier to see.--Czar Kirk (talk) 04:30, 13 June 2008 (UTC)[reply]

I suggest we stop coloring our K-maps or at least the inverses. I'll make an effort to redo them, but I'm still learning how to use Inkscape. ChyranandChloe (talk) 02:29, 14 June 2008 (UTC)[reply]

Yeah, the current colouring scheme is illegible. I'd suggest using multiple maps to avoid the tangle. Dcoetzee 02:34, 14 June 2008 (UTC)[reply]

I am trying to learn about Karnaugh maps, I saw something a lot time ago during the first semester in school.

Diagrams in the style suggested by ZeroOne seems more clear to me.

As far as I remember, instead of using Gray codes, the arrangement was done by drawing the variables in the way that is shown in ZeroOne's diagram. If I am not wrong it is a way to achieve Gray codes always, if one follows that mnemonic rule.

It was not clear to me, why the lower 2 cells at the bottom right corner together with the 2 cells at the top right were discarded. It is a 2^2=4 set of cells filled with 1s.

I think this article may be substantially improved if this examples are worked, drawing a figure for each step, using diagrams like the one suggested by ZeroOne, and also adding a more formal definition of what a Karnaugh map is as suggested above, complemented by a formal proof that the method works.

The part about glitches detection is important if it can be explained in more detail would be great.

Thank you! —Preceding unsigned comment added by Elias (talk • contribs) 06:20, 3 January 2009 (UTC)[reply]

Different schools teach in different ways, when I first learned it, rather than gray code it was A'B' A'B AB A'B' with "skip the third". But you're right, if we could get step by step examples would it would substantially simplify the the learning process. This article has been essentially neglected, it was actually first written in Wikiversity and transplanted to Wikipedia, so improving it would not be hard. ChyranandChloe (talk) 06:32, 3 January 2009 (UTC)[reply]

The yellow square and teal square are both redundant groupings. The actual minimal POS and SOP equations should only have 3 terms each, not 4. 75.151.246.133 (talk) 22:11, 2 May 2010 (UTC)[reply]

Agree about the redundant groups in the top diagram. The equations should be F = AC' + AB' + BCD' and F = (A'+C)(A'+B)(B'+C'+D). I see they're written that way when the diagram is repeated further down the page. I don't want to edit it, however, since the diagram shows the redundant groups (AD' is shown on the diagram as yellow). The diagram should be updated along with the equations.

Another point: I think it's confusing to label both equations F = ... when they represent the complements of each other. Perhaps F = ... and F' = ... would be better. Alternatively, F = ... and G = ..., to at least indicate the two equations don't represent the same cells in the map. TeWaitere (talk) 06:37, 16 April 2024 (UTC)[reply]

I just discovered that geneticists describe Mendelian inheritance using a diagram called a Punnett square which strikes me as being both structurally and functionally very similar to a Karnaugh map. Is it worth mentioning this correspondence anywhere? (Perhaps just a See Also link?) —Steve Summit (talk) 03:33, 30 June 2009 (UTC)[reply]

I can see the "structurally similar" here (inputs along the borders) but not the "functionally similar" --- the analogy actually seems quite superficial, as any Cartesian product of mappings can be represented as somewhat along these lines. So some further explanation about how the Punnett square is functionally similar is required. The P square is not used (afaik) to find minimal representations or prime implicants, whatever the analogs in genetics might be. (I am both a geneticist and a logician, so I think I would have spotted any such analogy, but then I do not wish to claim outright that it isn't there just because I failed to observe it.) 2A01:CB0C:CD:D800:9C38:3BDB:AF80:1542 (talk) 15:56, 11 May 2021 (UTC)[reply]

A Karnaugh map is not the same thing as a Veitch diagram. Veitch's diagram is used by virtually no one. I have the originals of both papers. The 2nd drawing on the right is a virtual reproduction of drawings from both papers. Notice that the Veitch diagram does not use the "Gray (en)coding" or the encoding from the "vertices of a hypercube".

I'll have change this paragraph eventually unless someone persuades me to do otherwise. Bill Wvbailey (talk) 18:20, 22 December 2009 (UTC)[reply]

I just went ahead and changed the lead paragraph. If anyone doubts the truth of the above I'll just photograph or pdf the relevant drawings from both papers to make the point. BTW: I was able to get the Karnaugh paper off the web, but I had to buy the damn Veitch paper; someone with access to an academic library can probably get it from the ACM. It is not a particularly satisfactory paper when compared to the Karnaugh paper. Bill Wvbailey (talk) 18:38, 22 December 2009 (UTC)[reply]

I am by no means a subject matter expert, but the second function in the example does not look right to me. Can someone confirm that it is indeed correct? Specifically, I am referring to this:

• f ( A , B , C , D , E ) = ( A ¯ B C D E ¯ ) + ( A B ¯ C ¯ D ¯ E ¯ ) + ( A B ¯ C ¯ D E ) + ( A B ¯ C D E ¯ ) + ( A B ¯ C D E ) + ( A B C ¯ D ¯ , E ¯ ) + ( A B C ¯ D E ) + ( A B C D E ¯ ) {\displaystyle f(A,B,C,D,E)=({\overline {A}}BCD{\overline {E}})+(A{\overline {B}}\,{\overline {C}}\,{\overline {D}}{\overline {E}})+(A{\overline {B}}\,{\overline {C}}{D}E)+(A{\overline {B}}CD{\overline {E}})+(A{\overline {B}}CDE)+(AB{\overline {C}}\,{\overline {D}},{\overline {E}})+(AB{\overline {C}}DE)+(ABCD{\overline {E}})}

It doesn't seem to correspond with the previous example or what comes after.

I searched through the edit history and found a prior revision of that example that makes a lot more sense to me. Here is the revision where it changed from something I understand to something I don't.

207.182.200.34 (talk) 15:18, 24 August 2011 (UTC)[reply]

The example is thoroughly corrupted. It has 5 variables: A, B, C, D, E. This requires it to have 2⁵ = 32 rows, 5 "input" columns labelled as A, B, C, D, E with all 0 & 1 patterns below them and one output column. The example has none of the above. A four-variable example would have 16 "input" rows and 4 input columns labelled A, B, C, D; one output column. I rolled the version back to the uncorrupted truth table, but the formulas underneath the drawings look corrupted, plus unlike a real K-map, the minterms (the 16 squares) don't have the numbers that correspond to the rows inside them. From upper left to lower right the minterm boxes are usually numbered as follows (base 10 with the binary equivalents below this matrix). Thus with the boxes numbered it's easy to go back and forth from truth-table to map:

0, 1, 3, 2

4, 5, 7, 6

12, 13, 15, 14

8, 9, 11, 10

0000, 0001, 0011, 0010

0100, 0101, 0111, 0110

1100, 1101, 1111, 1110

1000, 1001, 1011, 1010

If you look closely at the binary versions you'll see that as you go left to right or right to left, or up or down one square at a time (but not diagonally), only one "1" or "0" changes at a time. This includes the left-right edge wraparound and the top-bottom edge wrap-around. Other number-schemes are allowed but the "only-one-variable change" cannot be violated. That's the key to the Karnaugh map.

See the proper box-numbering scheme above (circa 2004 -- but this is a different example, but the numbering scheme is correct), and the 8-square version directly above.

Hope this helps. BillWvbailey (talk) 20:58, 24 August 2011 (UTC)[reply]

It definitely helps. Thanks. 207.182.200.34 (talk) 18:13, 25 August 2011 (UTC)[reply]

I went back to my sources to see their cell-numbering schemes. To a man/woman they numbered them in a different way than what I showed above. This includes M. Karnaugh himself in his very-most original paper. Here is how they all do it:

0, 4, 12, 8

1, 5, 13, 9

3, 7, 15, 11

2, 6, 14, 10

0000, 0100, 1100, 1000

0001, 0101, 1101, 1001

0011, 0111, 1111, 1011

0010, 0110, 1110, 1010

In words: the "classical" numbering scheme shown above is "my" numbering scheme mirror-imaged and rotated counter-clockwise. This is just to set the record straight. BillWvbailey (talk) 00:48, 28 August 2011 (UTC)[reply]

I reverted the recent edits made in good faith by 108.41.142.178; here is why:

• According to the Karnaugh map methodology, the "green group" in the picture in section "Karnaugh map#Solution" should be as large as possible in order to obtain the simplest possible expression realizing the function f. Since the whole rightmost column contains 1s, the green group should be at least that large (it can't be larger, on the other hand). That is, it should be AB' rather than AB'C only ("'" denoting complement here).
• The inner caption of the picture says "F=AC'+AB'+BCD'" rather than "F=AC'+AB'C+BCD'". The text should be consistent with that.
• Looking very closely at the square ABCD=1001, some green can be seen at its bottom, probably intended to indicate that the green area is not restricted to the two fields below it.

In order to avoid repetitions of 108.41.142.178's misunderstanding of the picture, the extensions of the red, green, and blue field should be made more clear in an improved version of the picture. There could be more emphasis on the borders and less on the areas; and the border lines of the red and green field should not coincide exactly, such that both are visible. If the partitioning of the 0 area is not discussed in the article (I didn't check that), it should be removed from the picture. - Jochen Burghardt (talk) 16:39, 30 September 2013 (UTC)[reply]

From the article: "For the red grouping: [...] C does not change. It is always 0, so its complement, NOT-C, should be included. Thus, \overline{C} should be included."

This is true in the bottom picture, but not the top picture (File:Karnaugh.svg). That picture shows "CD" as "10" or "11" in the red group, in which case C is always 1. Can anyone fix either the image or description? 108.202.199.178 (talk) 15:38, 18 February 2015 (UTC)[reply]

Just before the sub-section labeled "Karnaugh map", it may be helpful to actually write out the complete (16-term/16-factor) function for both sum-of-products and product-of-sums. I figured this out on my own after carefully reading the math and the definition of minterm/maxterm (from the linked Canonical_normal_form), but the SoP/PoS notion is a lot simpler than either article makes it appear.

I think having the complete function would also really help to illustrate how the solution below (of 3 terms) is vastly simpler than the trivial (canonical?) version. However, if showing the entire function is thought out of scope, is there perhaps a better/more-straightforward description that could be linked at this point in the article?--preferably one that clearly explains how one might construct the full form, but not as overly complicated as the Canonical_normal_form article. 108.202.199.178 (talk) 16:14, 18 February 2015 (UTC)[reply]

Are Karnaugh Map solutions to truth tables an example of a heuristic ? They will usually produce a good solution, even if they don't always find the best solution? Paul Murray (talk) 03:56, 18 June 2015 (UTC)[reply]