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      "body": "In my introduction I said that I would also present some of my programming and today we will get started with that. The program I want to talk about is a simple football (soccer) simulation that I have been occasionally working on in the past half year or so and appropriately named simplefootball. Rather than look at lots of code (which is in C++ by the way) I would like to focus on the ideas and concepts that went into it.\n\nFor the first part we will look how the result of a game will be generated. I actually got the idea for the entire project when I read about this method somewhere. The trick is to break down the somewhat continuous game into (a finite number of) single blocks that will be processed one after the other. Maybe the most obvious way to do this is to use the 90 minutes of the game as such blocks. Looking at a single minute there will be a certain chance for each team to score in this minute. This will, of course, depend on the offensive and defensive qualities of the teams, their formation, how fit the players still are and tons of other factors but we will worry about that later. One thing this will not replicate is that the chance might be influenced by what happened in the minute before, for example if one team got a penalty it is nearly certain that they will score in the next minute. This is not really a big problem though. If you want, you can think of a penalty as “80% of a goal” and by this it is more or less included in the chance to score.\n\nTo obtain the chance of one team scoring in a single minute we will first take the teams offensive strength (based on the factors mentioned above and maybe some more) and divide it by the opponents defensive strength (depending on similar factors). In a fair match this value will be around 1 = 100%. Since it is rather unrealistic for both teams to score almost every minute we will divide this by a global value I call MINPERGOAL which says how many minutes it takes on average to score for any team. If it were 90 we would expect the “average” game to end exactly 1:1. I am not exactly sure what this should be set to but for the most part it will be between 70 and 80, resulting in 2 to 3 goals per game on average (keep in mind that there are two teams). All this results in a chance of a little over 1% for the team to score in a given minute (in a fair match).\n\nNow that this is figured out we will just do this simulation of one minute ninety times, add up all goals scored for either team and by this get a fairly reasonable simulated result of a football game.\nOne last technicality is that if both teams should score in one minute then no goal is awarded at all since I found it unrealistic for two goals to occur in one minute and this is very unlikely anyway.\n\nIncidentally I rigged up a version of simplefootball to play two identical teams against each other multiple times prior to the world cup so now we can look at the results with MINPERGOAL set to 100:\n![009_sf1_1.JPG](https://cdn.steemitimages.com/DQmZb3dEUW3KCz4QiSKp4XP4tSpdLGmvje19GnwD1dk5KTY/009_sf1_1.JPG)\nand how they change when it is set to 60:\n![009_sf1_2.JPG](https://cdn.steemitimages.com/DQmXSLzChMKtgsvnPGQ1sysmTzhEAJjVC3EPb9oUWJ82Uma/009_sf1_2.JPG)",
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      "body": "Congratulations @jifka! You have completed the following achievement on the Steem blockchain and have been rewarded with new badge(s) :\n\n<table><tr><td>https://steemitimages.com/60x70/http://steemitboard.com/@jifka/votes.png?201901251914</td><td>You made more than 50 upvotes. Your next target is to reach 100 upvotes.</td></tr>\n</table>\n\n<sub>_[Click here to view your Board](https://steemitboard.com/@jifka)_</sub>\n<sub>_If you no longer want to receive notifications, reply to this comment with the word_ `STOP`</sub>\n\n\nTo support your work, I also upvoted your post!\n\n\n> Support [SteemitBoard's project](https://steemit.com/@steemitboard)! **[Vote for its witness](https://v2.steemconnect.com/sign/account-witness-vote?witness=steemitboard&approve=1)** and **get one more award**!",
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      "body": "Today we will come back to complex numbers. As discussed before, a complex number z is represented as a + b * i with a and b real numbers and i the imaginary unit (the square root of -1). Since this is uniquely defined by two real numbers, we can think of the complex numbers as a 2D plane, a giving the position in the horizontal direction and b vertical just like any point on a world map is described by its latitude and longitude.\n\nWhen you think about maps/locations on a smaller scale, you will often make or hear statements like “The train station is two kilometers north of my house” or “Vienna is three hundred kilometers south-east of Prague”. You might realize that if we pick a starting point, we can describe any other point on the map/plane like this, by giving a direction and distance from the starting point. For complex numbers the starting point is 0 (0+0*i) and we can alternatively write any complex number as a pair (r,p) where r is the (positive) distance (r for radius) and p (which is supposed to be the Greek letter phi) is an angle between 0 and 360 degrees (technically, we have to allow the angle to be zero but never 360°; the reason for this will become evident later). This is known as polar coordinates.\n\nNow we need a way to convert the two notations a + b * i and (r,p) into each other. As you can see in the picture, a, b and r form a triangle with a right angle in the lower right. This allows us to use the sine and cosine function: sin(p) = b/r as well as cos(p) = a/r (this is just the definition). This directly yields a = r * cos(p) and b = r * sin(p).\nThe other way around, the Pythagorean theorem gives r^2 = a^2 + b^2 and p can be then be calculated using the inverse of the sine or cosine function.\n\n![008_polar_pic1.JPG](https://cdn.steemitimages.com/DQmRGQGiAN5XYA3gpJsHhoyuLzopByYVHgvS7sykZdvVJb2/008_polar_pic1.JPG)\n\nNow what is all this good for? As mentioned in my first post about complex numbers, any equation like z^n = w can be solved there so obviously taking powers of complex numbers is kind of important. Sadly, taking high powers of sums like a + b * i becomes very difficult. This gets way easier in polar coordinates. In fact there is a very general property: if z has polar coordinates (r,p) and w has (s,q) then the polar coordinates of z * w are (r * s,p+q).\n\nShowing this isn't too hard but it takes some work we don't want to do here. You can make some sense of this property when realizing that positive (real) numbers will always have an angle p of zero and negative numbers an angle of 180°. Multiplying two negatives will bring the angle to 360°, essentially the same as zero (since 360° is a full circle, these describe the same direction; the same will apply for any two angles with a difference divisible by 360). Using induction we can extend this property to general products with any number of factors w * x * y * z * … where the new radius will always be the product of the old ones and the new angle the sum of the angles going in, looping back to zero if it exceeds 360°.\n\nNow we are interested in the special case z = w (= x = y = …). For z = (r,p) this means z^n = (r^n,n * p). In particular, this helps us to solve an equation like z^n = w. If we know w has polar coordinates (s,q), finding the polar coordinates (r,p) of z boils down to r^n = s and n * p = q. Both of this can be easily solved (remember that s is positive).\nIt also becomes evident here that there might be more than one solution. This is well known already for real numbers, e.g. (-2)^2 = 2^2 = 4. Here the reason is that an angle of zero is essentially the same as 360° and could in this case be reached by 2 * 0° (in 2^2) or 2 * 180° (in (-2)^2). As noted before we can argue that p+360° “=” p for any angle p. In fact, there are always n solutions z to z^n = w, one that never reached 360°, one that did so once (and was reset to zero by us), one that did twice, … and finally one that looped around n-1 times. Looping around n or more times is not possible since the initial angle was smaller than 360°.",
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      "author": "jifka",
      "body": "Last time we looked at the geometric mean in contrast to the “normal” arithmetic mean. There is even a third way to determine the “middle” of two numbers, the harmonic mean. The harmonic mean of two numbers a and b is 2/(1/a + 1/b). In other words it is the inverse of the arithmetic mean of the inverse numbers 1/a and 1/b. It might seem like this should be the same as the arithmetic mean but we will look at an example to see that it isn’t.\n\nAs one easily sees the arithmetic mean of one and a million is about half a million. Note that the mean of two and a million or three and a million or fifty and a million will still be very close to half a million (at least in a geometric sense). On the other hand, doubling the million instead of the one will also very nearly double the mean. We have seen this property already. For the harmonic mean we look at the inverse numbers, one and 1/1.000.000 (~0). Their mean is about ½, the inverse and thus the harmonic mean is two, compared to the  arithmetic mean of half a million. Furthermore, adding a few more millions to the second number barely does something to the mean because the inverse was already basically zero. Doubling the one however will mean that its inverse is now ½, resulting in a harmonic mean of four, double what we had before.\n\nIn this example it becomes clear that for the arithmetic mean the bigger number is more important and, if it is much bigger, can determine the mean almost on its own. For the harmonic mean however, the small number is all that (really) matters. In the geometric mean both numbers are equally important, as one easily confirms, quadrupling one of them will double the geometric mean. Particularly in the case of one and a million the geometric mean is a thousand, between the harmonic and arithmetic one. This is no coincidence. As the discussion above suggests, the harmonic mean will always be smaller than the geometric mean which is again smaller than the arithmetic mean. In fact, they can only be the same when the two numbers are identical. \n\nFinally, I want to point out that we compared the means in a geometric sense where, by construction, the geometric mean was right in the middle while the arithmetic mean was much bigger and the harmonic mean smaller. Similarly, in the usual sense of distance (the difference between the numbers) the arithmetic mean is the exact middle and the other two are way smaller. As you might imagine, there is also a “harmonic” way to measure distance where the geometric and arithmetic mean appear too big. Similarly to calculating the mean this is the difference of the inverse numbers (1/a – 1/b).",
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      "title": "The harmonic mean"
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