However, despite repeated attempts to alter the default setup, or to create a new one, or to alter the "Throwback" format, MacTerm refused to accept the font change, and in all cases in remained at Monaco. ![]() In each case, whenever I clicked on the field to change the font, it would launch Apple's font chooser, where I could select a different font and font size. However, regardless of whether I tried to change the default format, or set up a new format specifically geared towards BBSing, the app would not let me change the font. I actually have three different fonts on my machine which meet this requirement. I encountered several problems while trying to set up this app for BBSing purposes, and also when attempting to log on to my BBS.īeing as a true PC-ANSI-based BBS uses the CP437 character set - i.e., extended ASCII or high ASCII glyphs - to draw ANSI graphics, I tried to change the default font in the app from Monaco to a font which actually contains the CP437 symbols, in the translation section of the app's preferences. I use an iMac running Yosemite 10.10.3, and my BBS uses the Hermes II BBS software. In the picture (created using latex) you can see the expression – it is the same in it's minimal DNF and minimal CNF – and the sum of minterms equivalent to it.Being an old school BBSer - I run my own BBS at telnet 202.128.4.177 and BBS website at - I downloaded this app to see how well it works as a telnet client for BBSing. Just any product is not a minterm, so the original expression could be in the form of both the product of sum and the sum of products, but not the valid sum-of-minterms. = (¬a⋅¬b⋅¬c)+(¬a⋅¬b⋅c)+(¬a⋅b⋅¬c)+(¬a⋅b⋅c)+(a⋅¬b⋅¬c)+(a⋅b⋅¬c)+(a⋅b⋅c)Īs you can see, even if these two boolean expressions are equivalent to each other, the original one (on the left side of the equation) is not written as the sum-of-minterms expression (on the right side of the equation). However, this is not a valid sum of minterms, because there is none: f(a,b,c) = ∏(5) = M5 = (¬a + b + ¬c)įor the original expression to be also the sum of minterms, it would need to mark out every single true/one cell in your K-map separately like this: f(a,b,c) = ∑(0,1,2,3,4,6,7) = m0 + m1 + m2 + m3 + m4 + m6 + m7 = ![]() It is also the same boolean expression as the original one, so that is a valid product-of-maxterms expression too. It is a valid product-of-maxterms expression, even if there is only one. ![]() There is only one maxterm present in the truth table (and your K-map) and the only maxterm determining the function's output as logical 0. The following truth-table corresponds to the given function: index | a | b | c || f(a,b,c) | term matching the row/K-map cell In a truth table a maxterm or a minterm matches only one row. ![]() In a K-map a minterm or a maxterm marks out only one cell. A cluster of literals in a boolean expression forms a minterm or a maxterm only, if there are all literals (variables of the given function or their negation) included in it.Ī minterm is a product of all literals of a function, a maxterm is a sum of all literals of a function.
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