The competition will take place over a day and a half, Monday and Tuesday of Automation Fair week, with the winning team announced on Wednesday at 2:30 PM on the Discovery Theater stage.

DMC’s team of experts will be competing against six other teams for honor and glory in a coding challenge that incorporates FactoryTalk Optix, FactoryTalk DataMosaix, FactoryTalk Analytics LogixAI, Studio 5000 Logix Designer, FactoryTalk Historian Site Edition, and Plex.

Good luck to all the teams competing!

Exploring the Expo? Visit us in Booth #752

Stop by our booth to connect with our team, explore our latest automation solutions, and see our custom-built “Drop Bot” in action.

Test your skills against Drop Bot in a life-sized game of Connect Four or challenge your colleagues to a game.

Will we see you at the show? Let us know!

Connect with Us

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What are your coordinates when you stop driving?

Parameters
•    Coordinates of Lebanon, Kansas: 39°48′38″N 98°33′22″W
•    Vernal Equinox: September 22, 2018
•    Time of Sunrise: 7:22:43 am
•    Time of Sunset: 7:30:29 pm

Assumptions
•    You experience no change in elevation
•    Your hovercar has enough fuel to last the entire drive
 

Sun trajectory (source)

Bonus Questions
1.    Is there any non-zero constant velocity for which you will end up where you started?
2.    How can your solution be generalized for different Latitudes and times of the year?

Remember to state any additional simplifying assumptions you use in solving this problem! 

Submit responses to geekchallenge@localhost by October 9, 2018.

Check out previous Geek Challenges.

The post Geek Challenge: A Place in the Sun appeared first on DMC, Inc..

This was a unique problem that needed to utilize at least a little bit of computation (to check if something is prime or not). Luckily, prime calculators are a dime a dozen across the interwebs and many computer languages have a “prime check” method built in.

Congrats to our winner, Alex Bruno! A few people commented on the problem being too constrained, giving the amount of included prime numbers made it simple. This was by design! Alex Bruno came up with a way to greatly minimize the amount of prime checks necessary by creating prime-rich building blocks to create a solid foundation. Although 0373373 was not the answer I had in mind, it followed all the rules and was created in a very interesting way.

No one submitted an answer for the 10-digit bonus, which was surprisingly different from the 7-digit case it many ways. Heavy optimization is necessary for that to ensure your number crunching doesn’t run for days.

Thanks for playing! The formal analysis for my solution and Alex’s solution are below.

Alex Bruno’s Solution

A. Overview:

My analysis split into two schools of thought:

1. Brute force it
2. Start with smaller numbers as "building blocks" and put them together until we get a big enough number with the correct number of primes.

Either way, I needed a good way to calculate the number of primes in any given number. I didn't want to sit and do them all by hand. I wanted to be able to type them into something and have it spit out how many. So, I fired up MATLAB, and banged out a script that would take a number of any size and analyze every possible sub-number and tell me how many primes there were. That script was useful in both approaches.

B. Approaches:

1. Brute Force it

What it says on the tin. I took the script and replaced the while with a for, and told it to run every single 7-digit number from 100000 to 9999999. Obviously, this took a while (I let it run in the background while I worked the other method). After about 30 minutes, it spit out:

3733797

This is one of my submissions, and, as I'll explain in a second, is probably the more accurate one.

2. Building blocks

There are a possible 28 sub-numbers in a 7-digit number (calculated using the nth triangle number formula, which is a sort of additive factorial function). If we need 22 primes, that means a scant six of them can be non-primes.
To that end, I needed to find building blocks that would have as many primes as possible. I decided to start with 3-digit building blocks (six possible sub-numbers), and came up with the following:

Five primes out of six (I started with a list of prime numbers less than 1000):

337
353
379
733
773
797

Six primes out of six:

373

Obviously, I needed to use 373 as a building block. In fact, I could use it twice, and only need to add one extra digit. This looked like this:

373 373 _   OR   373 _ 373     OR    _ 373 373

This left me with 30 options to run, which was very simple using my code. The only combination that gives me a full 22 primes is:

0373373

This is because the 0 in the front, while not being prime itself, makes several extra primes (since 37 is prime, so is 037, etc.). Even with losing that digit as a prime, it adds several others.

C. Conclusion:

So, I have two answers; two 7-digit numbers that have 22 prime sub-numbers. While it is absolutely the more inelegant of the approaches, I would submit the first answer (3733797) as the more correct answer because the prompt for this problem dealt with phone numbers, and I'm not sure they can start with a 0, as with my second answer.

I'll argue that I may have gotten to the first answer using building blocks if I had kept going with that method; the brute force answer does include two of the 3-digit building blocks I found (373 and 379).

Unfortunately, I didn't have time to grapple with the 10-digit optional problem. I can't imagine doing that either of the two ways I did, so I have to ask, what is the more elegant way to do this? I'm sure there's some sort of trick, and I would love to know what it is!

My Solution

Prime number calculation is a deeply researched and investigated field of study with many applications and approaches. Because of this, there are some highly efficient methods for calculating primes. We can leverage these approaches to create an efficient computational solution.

As described, this challenge involves finding primes within a number, with the added complexity of multiple possible sub-groupings and prime combinations. Any given n-digit number has (((n+1)*n))/2 possible groupings (one set of n consecutive digits, two sets of n-1 consecutive digits and so on down to n sets of one digit, so the total number of groupings is the sum from 1 to n). For any given number, all combinations must be investigated to obtain the number of primes it contains.

To avoid excessive calculation, two major computational savings can be applied:

1. Use a prime number sieve (such as the Sieve of Eratosthenes) to pre-calculate all prime numbers within the relevant range (all values less than 〖10〗^(n+1)-1).
2. Use memorization to store a record of previously determined number results.

The first optimization requires creating an array mask of prime/not-prime values. Then, whenever a number is encountered, rather than determining if a number is prime, the corresponding index in the array can be accessed rather than re-calculating the primality of the number. This approach requires greater memory but substantially less computation, especially when many prime computations need to be carried out in a small range. There are several further optimizations that can be applied to the prime number sieve which are not discussed here.

The second optimization leverages an observation on the nature of the result for any given number. Consider the graphic provided in the problem statement:

From these graphics, observe that it is possible to get the number of primes in the n-digit case by taking the sum of the n-1 groups, and subtracting the n-2 group shared by both n-1 groups. Alternatively stated, take the sum of the 6-digit number results and subtract the “middle” 5-digit number case (so 9=6+8-5).

Therefore, for any given number with more than two digits, it is not necessary to calculate the number of primes it contains directly from a sum of its groups, but instead can be found by taking a sum of these sub groupings. An important caveat to this approach is that each entry must be considered with leading zeros as a different value than without leading zeros (i.e. in the example given, 07 is unique from 7).

Applying these computational observations along with the additional unstated constraint that the n-digit number cannot start with a zero, the best answer was found to be 3733797 with 22 primes (with a total computation time of approximately five seconds). The graphic representation of this is shown as:

The Bonus Solution

Extending this approach to the 10-digit case requires some code refactoring, because the prime sieve array mask and memorized prior results require data structures with more indices than provided by a single unsigned 32-bit integer. This can be overcome several ways. One possible approach involves creating custom data-structures that act as wrappers on standard data structures and using a 64-bit integer to address locations. Implementing these structures comes at a significant computational and memory access cost, which substantially slows computation time. Consequently, the 10-digit case took approximately 24 hours to solve.

The best 10-digit number (with 38 primes out of a possible 55) was found to be 7000373379. A graphical representation of this number is shown below.

As an additional interesting note, the distribution of the number of primes in a given number of digits is shown below:

Submit your Geek Challenge questions and ideas at geekchallenge@localhost.

The post Geek Challenge Results: Primetime Telephone Numbers appeared first on DMC, Inc..

At first, I interpreted this as "seven prime digits," which is probably what she meant.

But then it got me thinking, 23 has three prime numbers in it (2, 3, and 23), and 373 contains six primes!

The most primes you can pack into a seven-digit number is 22, so what number contains 22 primes? 

Remember: neither 0 nor 1 are prime; and 5003 contains 5 primes (5, 3, 03, 003, and 5003).

Bonus: What if her phone number was ten digits long? How many primes can you pack into a ten digit number?

The number 2765073 contains nine prime numbers

The solution with the most complete analysis will be this month's Geek Challenge winner!

Submit responses to geekchallenge@localhost by June 1, 2018.

Check out any previous Geek Challenges you might have missed!

The post Geek Challenge: Primetime Telephone Numbers appeared first on DMC, Inc..

Michael Deck is this challenge’s winner for his extensive solution that not only solves the proposed problem but also handles more involved aspects not considered by the OP.  Michael’s solution is below.

1 Introduction

The problem is laid out in our previous blog: The Perfect Bracket.

2 Solution

2.1 Main solution

First, we set forth the probability of having a 100% correct bracket in a given year:

₍₁₎

Where p is the constant probability of the favorite winning a particular game, and n is the number of games held in that year’s tournament to determine the NCAA champion. It follows that the probability of having any errors whatsoever in the bracket is:

₍₂₎

We can quickly see that the probability of having imperfect brackets in all 50 independent years is:

₍₃₎

Thus, the probability of having at least one perfect bracket in this set of 50 years is:

₍₄₎

Now to resolve that pesky n. The vast majority of the time, your bracket pool will ask you to make picks for 63 games. If this is the case for you, set (4) equal to 50%, plug in n = 63, and solve for p:

Huzzah! The answer is C.

2.2 General Formula for p

From the above, it is easy to generalize the formula for p in terms of the number of brackets filled out:
Where b is the number of brackets filled out, c is the probability of at least one perfect bracket.

₍₅₎

3 But Wait, There’s More!

3.1 What if my Pool Picks the Play-in Games too?

I noted that there are 63 games in your pool most of the time. But since 2001, there have been extra “play-in” games that happen prior to the official start of Round 1. From 2001-2010, there was a single extra game for a total of 64, and since 2011 there have been four extra games for a total of 67. You can plug those values into (4) above instead to and alternative solutions that take the play-in games into account.

For 2001-2010:

3.2 An Even More General Formula For p

We can modify (5) to account for a different number of play-in games:
By simply replacing 63 with n to represent the total number of games in the bracket.

₍₆₎

4 Hold on, Check the Problem

The wording of the problem is slightly ambiguous. For the preceding solutions, it has been assumed that the questioner was seeking the probability that at least one of the brackets is perfect. That means the cases where any number of your brackets (including getting 2 perfect, 3 perfect, all the way up to all 50) are perfect are included. But what if the questioner intended to ask how large p should be if you want to get exactly one bracket correct in 50 years?

This solution does not initially appear difficult to set up. Refer back to (3). This formula expresses the case where we are imperfect 50 times. If we only have one perfect bracket, we will have 49 imperfect brackets, so we can use (3) in this solution by decreasing 50 to 49.

The probability of a single perfect bracket is expressed in (1). Combining the adjusted (3) with (1), we get:

₍₇₎

This can easily be identified as (almost) an example of the binomial distribution’s probability mass function. However, (7) does not take into account the number of places in which the single perfect bracket may occur — to take this into account, multiply by the number of ways the single perfect bracket can appear (the binomial coefficient):

₍₈₎

Plugging in n = 63 and setting this equal to 0.5:

₍₉₎

Attempting to solve for p algebraically here is impossible per the Abel-Ruffini theorem — this is effectively a 50th-order polynomial. With a spreadsheet, it is easy to plug in a range of values between 0 and 1 for p⁶³ and see that the probability of exactly one bracket out of 50 being perfect (probably) peaks out around 37.16% when p⁶³  ≈ 0:0200. So, if the text of the problem is taken literally to mean that we seek the value of p for which the probability of having exactly one perfect bracket is 50%, it does not appear that there is a valid solution.

A Final Reality Check

The NCAA has changed the number of teams/games at least 10 times since the inception of the tournament in 1939. Based on the trend, it is likely that there will be more than 67 games at some point in the future. How do we account for this in our answer?

Given the excessive levels of complexity we are introducing here, it’s time to break out some code. I’ve chosen Python to define a function solve_p() that will accept a list of future annual values for n, a target value for the probability of getting 1 bracket correct, and a value for how granularly to search for a solution between 0 and 1, and return its approximation of p:

def prob_imperfect(p, num_games):
    return (1.0 - p ** num_games)

def find_closest(dict_solutions, target):
    adj_dict = {}
    for key in dict_solutions:
        adj_dict[key] = abs(dict_solutions[key] - target)
    return min(adj_dict, key=adj_dict.get)

def solve_p(n_list=([63] * 50), target=0.5, vals=100000):
    num_brackets = len(n_list)
    candidates = {}
    for i in [float(j) / vals for j in range(0,vals,1)]:
        temp_val = 1.0
        for b in n_list:
            temp_val = temp_val * prob_imperfect(i, b)
        candidates[i] = 1.0 - temp_val
    return find_closest(candidates,target)

def prob_imperfect(p, num_games):
    return (1.0 - p ** num_games)

def find_closest(dict_solutions, target):
    adj_dict = {}
    for key in dict_solutions:
        adj_dict[key] = abs(dict_solutions[key] - target)
    return min(adj_dict, key=adj_dict.get)

def solve_p(n_list=([63] * 50), target=0.5, vals=100000):
    num_brackets = len(n_list)
    candidates = {}
    for i in [float(j) / vals for j in range(0,vals,1)]:
        temp_val = 1.0
        for b in n_list:
            temp_val = temp_val * prob_imperfect(i, b)
        candidates[i] = 1.0 - temp_val
    return find_closest(candidates,target)

The helper function prob_imperfect() calculates the probability of an imperfect bracket from p and the number of games. The other helper function find_closest() finds the candidate solution for p whose probability is closest to the target. The default arguments for solve_p() are the list [63,63,…,63] and the target 0.5, which represent the values for our main solution as provided in section 2.1, so we can test that the function works for the static number-of-games scenario by simply running the function with no arguments:

import dmc_geek as dg
dg.solve_p()
0.93424

import dmc_geek as dg
dg.solve_p()
0.93424

Great! Now we can see how high p needs to be if, say, the NCAA decides to add 1 game per year for the next 134 years (assuming I am in a pool where every game is picked):

nl = list(range(67,201))
dg.solve_p(n_list=nl)
0.95074

nl = list(range(67,201))
dg.solve_p(n_list=nl)
0.95074

Good to know! Now, what if I only need a 10% chance of at least one bracket being correct in those 134 years?

dg.solve_p(n_list=nl,target=0.1)
0.92943

dg.solve_p(n_list=nl,target=0.1)
0.92943

Fantastic. Let’s say I want to waste more time on this and would like a more precise answer. How can I increase the granularity of my search space beyond the default of 100,000 test values for p?

v = 1000000
dg.solve_p(n_list=nl,target=0.1,vals=v)
0.929429

v = 1000000
dg.solve_p(n_list=nl,target=0.1,vals=v)
0.929429

Conclusion

As you can see, that p needs to be pretty high for you to have any greater than – how shall I put this – a snowball’s chance in hell of ever picking a perfect bracket in your lifetime. But hey, you don’t need to be perfect to win your coworkers’ cash – just be a little bit better than everyone else, and eventually you might win your pool. Thanks for reading and good luck with your future brackets!

Michael Deck will be receiving a DMC Pocket Protector! Keep an eye out for the next DMC Geek Challenge. 

Here’s where you can find the Solution .tex file and accompanying Python code.

The post Geek Challenge Results: The Perfect Bracket appeared first on DMC, Inc..

The solution with the most complete analysis will be this month's Geek Challenge winner!

Submit responses to geekchallenge@localhost by May 5, 2017.

Check out any previous Geek Challenges you might have missed!

The post Geek Challenge: The Perfect Bracket appeared first on DMC, Inc..

The results are in! February's Geek Challenge winner is Grant Anderson of DMC. Grant's clever breakdown of the problem is shown below.

Grant's Solution

There are five classifications of locations that satisfy this riddle (at least on the surface):

1. As mentioned in the riddle itself, the North Pole.
2. All locations one-mile north of the one-mile long parallel of latitude in the Northern Hemisphere.
3. All locations one-mile north of the one-mile long parallel of latitude in the Southern Hemisphere.
4. All locations one-mile north of a parallel of latitude which is an integral fraction of a one-mile long parallel of latitude in the Northern Hemisphere.
5. All locations one-mile north of a parallel of latitude which is an integral fraction of a one-mile long parallel of latitude in the Southern Hemisphere.

For ease of calculations, we’ll assume the earth is perfectly spherical, rather than the ellipsoid it truly is. Also, since this isn’t a geography question, we’ll assume that our traveler won’t have to take elevations into consideration and place him at sea-level for his entire trip.

There are 90 degrees of latitude between the equator and each pole (the equator being 0 degrees). The circumference of the Earth at the equator is 24,901.92 miles, which, dividing by 360 degrees, means that each degree of longitude at the equator is 69.172 miles.

However, because we are assuming a perfectly spherical earth, we know that each degree of latitude on the Earth is also 69.172 miles. As our traveler moves closer to the poles and away from the equator, the mileage difference between each degree of longitude shrinks.

We’ll first solve for the two parallels of latitude where 360 degrees of longitude is exactly one mile. Or alternately, 1 degree of longitude is 0.0028 miles.

To calculate at which degree of latitude 1 degree of longitude is equal to 0.0028 miles, we can use the following calculation: 

1 degree of longitude = cosine (latitude in degrees) * mile length of a degree at the equator

So, plugging in our values we have: 

0.0028 = cosine(x) * 69.172

Solving for x gives us: 

x = 89.998 degrees

Northern Hemisphere

In the Northern Hemisphere, our traveler needs to go one mile north of 89.998 degrees. Since we know each degree of latitude is 69.172 miles, we find the traveler will need to start out at the 89.998 + (1 / 69.172) degrees North latitude or 90.012 degrees North latitude.

This means he is actually crossing through the North Pole. The upshot is, we can scratch classifications 2 and 4 from the list above. It’s impossible to do this from the Northern hemisphere other than at the North Pole, assuming that you can’t actually walk north of the North Pole.

Southern Hemisphere

In the Southern Hemisphere, once again the traveler needs to go one-mile north of 89.998 degrees, which puts him at 89.983 degrees. One degree of longitude here equals cosine (89.983) * 69.172 or 0.0205 miles. Multiplying by 360 gives us the length of this latitude as 7.389 miles.

Now, we need to calculate the remaining locations that fall under classification 5, all locations one-mile north of a parallel of latitude which is an integral fraction of a one-mile long parallel of latitude in the Southern Hemisphere.

We now calculate the degree of latitude where 720 degrees of longitude is equal to 1 mile. Using our equation from above we find this is the 89.999-degree south parallel. One mile north of that gives us the 89.984-degree south parallel. And its length is 6.953 miles

Next, we calculate the degree of latitude where 1080 degrees of longitude is equal to 1 mile. This puts our traveler less than a thousandth of a degree south of what we just calculated. At this point, we can assume the width of his hovercar from the previous locations trip will cover this location and all subsequent locations that fit the harmonic series.

So if our traveler starts at the North Pole, he’ll move straight down to the 89.983-degree parallel and circle it. After this, he’ll travel to the 89.984-degree parallel and circle it as well. This trip yields a total distance of:

((90 degrees + 89.984 degrees) * 69.172 miles/degree) + 7.389 miles + 6.953 miles = 12,464.195 miles

Traveling at 100 mph means his trip will take about 125 hours. At over 5 days of travel, he’s going to need those energy drinks.

Additional Thoughts

As Grant pointed out, there are an infinite number of rings near the South Pole made from the points that satisfy this riddle. It would be impossible for our traveler to travel along all of these rings since all have a circumference of at least ~6.28 miles (the circumference of the latitude line 1 mile north of the South Pole). This would make our traveler truly eccentric as he would forever travel around the South Pole without ever reaching it (reaching the South Pole would make him a centric traveler).

Grant gave our traveler some closure by pointing out that eventually, the rings become so close to each other that the width of his hover car covers multiple rings and at some point spans all remaining rings to be traveled around. To examine this, and see how wide his hover car would need to be, let's take a closer look at his path along these rings near the South Pole. Let's zoom in and see how far apart the rings are.

Looking at these distances, we see that if our traveler's hovercar is at least 841 ft wide, he could sweep through all the rings in one loop around. If his hover car is at least 420 ft wide, he could do it in 2 loops, covering the first and second ring in his first loop and all the rest in his next loop. 

If his hover car is just the width of a normal car, we see that he'll have to do many loops around before he can cover all of these rings. However, any distance added traveling around these loops is pretty insignificant compared to the first 12,450 miles he travels from the North Pole to the first ring.

One more interesting thing about the spacing of these rings is that the infinite number of points satisfying this riddle, excluding the North Pole, are contained in a relatively small area on the surface of the globe (they are all between 1 and 1.159 miles away from the South Pole).

This area can be calculated as the following surface integral:

\(A = \int_0^{2\pi} \int_{\frac{1mi}{3959mi}}^{\frac{1.159mi}{3959mi}}(3959mi)^2cos(\theta)d\theta d\phi \approx1.076mi^2 \approx689acres\)

3959 mi = Earth's radius

1 mi = distance of inner-most ring from the South Pole (more correctly, what the ring sizes are approaching as they get smaller)

1.159 mi = distance of outer-most ring from the South Pole

The Earth's surface (both land and water) is approximately 126 billion acres, so the area these points are contained in is only 0.0000005% of the total area on Earth!

Submit your comments to geekchallenge@localhost.

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The results are in! Two people correctly answered December’s geek challenge.

Ken Brey of DMC and Jesse Batsche of DMC both identified the correct percentage as D: 60-65%. Ken supplied an exact solution for the probability as the grid of pool balls becomes infinitely large.

However, Jesse is this month’s winner because he wrote a really cool LabVIEW program to solve the problem!

Ken’s Exact Solution

Where X and Y represent the row and column offset of a ball relative to the cue ball, given 4-way symmetry, we can analyze only 1/4 of the space, such that X>0 and Y>0. Now we can use only natural numbers. Given an infinitely large surface, the probability of selecting a ball on the symmetry lines of X=0 or Y=0, rows is infinitely small, and therefore ignoring these can’t affect the outcome.

For a ball at X,Y to be a clear shot, there must be no smaller similar right triangle with integral sides to the right triangle with sides X and Y. So, for all n smaller than X and Y, X/n and Y/n must not both be integers. 

In other words, X and Y are coprime, or the greatest common divisor (GCD) of X and Y is 1. A simple solution for the probability of this question of the probability of GCD being 1 for an infinite set of integers was figured out by a couple of bright undergrads in 1992. Their answer matches more complicated proofs discovered earlier. The answer is 6/pi² or 60.7927%.

References:

Probability of Two Integers Being CoPrime

Jesse’s LabVIEW Solution

Jesse correctly approximated the percentage by building a LabVIEW program to solve the problem for arbitrarily large grids of pool balls. Jesse’s solver iteratively “takes shots” at balls in “rings of increasing radius” (i.e. starting with balls closest to the cue ball, moving outward). It keeps track of shot angles that it has already taken to hit a prior ball. If it aims at a ball, and it requires exactly the same shot angle as a previous ball, then that ball is deemed un-hittable. In addition to calculating the ratio for any size grid Jesse’s solver includes an awesome animated graph showing the hittable and unhittable balls in the grid as it iterates through them. Green boxes represent hittable balls and boxes left red represent unhittable balls.

The results are then shown in a more condensed grid. Shown below are the results for a 103×103 grid.

Congrats to our winner, Jesse, who receives a retro logo DMC pocket protector!

Submit your comments to geekchallenge@localhost.

Check out all the past challenges here!

The post Geek Challenge Results: Infinipool appeared first on DMC, Inc..

To describe the challenge, let’s look at a 3×5 grid of pool balls with the cue ball positioned in the center. We want to know what the odds are that we can hit a ball chosen at random with the cue ball (without jumping or curving around other balls).

Looking at the possible cue ball paths, we see we can hit any ball except the 6 or the 9 because the 7 and 8 balls get in the way. Since there are 14 possible balls to pick from and we can hit 12 of them, there is a 6/7 chance we can hit a ball chosen at random from the 3×5 grid.

For the real challenge, we want to know our odds for an infinitely large grid of pool balls with one cue ball positioned like before. 

Since this grid is substantially larger, the radius of the all the balls is now 0, so that a ball does not interfere with the cue ball path unless the cue ball hits that ball dead center. Because this is a much harder shot, we have also been granted the ability to hit the cue ball infinitely far and with perfect accuracy.

What is the probability that we could hit a ball chosen at random from the infinite grid without another ball being in the way?

A: 0-5%

B: 15-20%

C: 35-40%

D: 60-65%

E: 95-100%

Bonus points for the exact solution.

Please submit your responses to: geekchallenge@localhost.

Check out previous Geek Challenges here!

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To give proper context to the extreme efficiency of the Combinations solution to this problem, I will first demonstrate the more common Series method.  Actually, I’ll let Andrea Gotti of Ingersoll explain:

N : number of points unevenly distributed on the circumference. No more than two chords intersect at a common point inside the circle.
I : number of internal chords intersections. Starting from the images above one can observe as follows:

These numbers can be conveniently collected by using the following table:

The index is used just to keep a compact notation. The number of intersections, expressed as a function, is the sum of all the elements in the first rows.

Therefore, the original problem has become:

The formula can now be expressed as a function of N:

So, substituting to arrive at the final answer:

Wow, that’s a ton of work, and is what the majority of people who answered this question actually went through. A few people, however, found an easier way.  Here is a winning solution from Devon Fritz:

Every set of four perimeter points produces an intersection point between that set of four points.  

So, for each set of unique 4 points on the perimeter, a unique intersection point is formed.  This can be written mathematically as a combination formula for N>3: _NC₄

The formula for combinations can be expressed with factorial:

And substituting the required perimeter points:

OBSERVATION #1: By moving from one row to the next one below, the number in the first column increases by 1, the one in the second by 2, the one in the third by 3, and so on. This allows one to state the problem by using the following recursive formula:

For example:

Consider the term: 

The original problem can now be expressed via this simplified recursive formula:

For example:

OBSERVATION #2:

This allows one to reformulate the problem as follows:

Let's now consider the term: 

And thus, our winners were selected for representing the results of three jam-packed pages of math with just 3 letters: _NC₄

So, congratulations again to John Jacobsma, Devon Fritz, Sudeep Gowrishankar and Adnaan Velji for winning this Geek Challenge!

Submit your comments to geekchallenge@localhost.

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