tag:blogger.com,1999:blog-64556432007364634502024-02-20T16:58:01.614-08:00Dot Star MoneyChrishttp://www.blogger.com/profile/05191219955389953492noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6455643200736463450.post-66552150310773036802016-01-11T21:52:00.000-08:002016-01-16T13:35:04.023-08:00Toughest TopCoder Problems #1: CollectingBonusesTime for a new series! I am very cynical of "competitive programming websites," that is to say I think they hardly showcase talent and instead showcase whose lucky enough to remember what specific bit of knowledge is required per-problem. What I'd like to do then, is take possibly the most well known competitive programming website (<a href="https://www.topcoder.com/">TopCoder</a>), take a look at their "hardest problems," and deconstruct/solve them to both prove my point, and uncover the knowledge required behind them. I am in no way involved in the TopCoder community nor do I care to be, so everything you see here is original work.<br />
<br />
To find what TopCoder would indicate are the worst of the worst, I've gone to their <a href="https://www.topcoder.com/tc?module=ProblemArchive">problem archive</a> and sorted the problems by the lowest solve percentage. There are about two pages of 0% (so nobody got 'em), and then we start to see a trickle in of problems one to a handful of people were able to solve. Keep in mind, "solving" one of these problems means you were able to get it from solution to software in at most 75m.<br />
<br />
Lets start with the very first problem on this list that had a successful submission<span style="font-size: small;">...</span><br />
<div style="text-align: center;">
<br /></div>
<div style="text-align: center;">
<h2>
<span style="font-family: inherit; font-size: large;"><a href="https://community.topcoder.com/stat?c=problem_statement&pm=8764">CollectingBonuses</a></span></h2>
<div>
<br /></div>
</div>
<div style="text-align: left;">
</div>
<h3>
<span style="color: #666666;">The Problem</span> </h3>
<br />
I take it most are capable of reading the problem statement, but to summarize an equivalent form of the problem that is less verbose:<br />
<i></i><br />
<blockquote class="tr_bq">
<span style="color: #999999;">There is a set of \(n\) unique numbers. Let the following sampling process be defined: Choose a number from the set with each number's selection equally likely. Compute the expected value \(P\) of samplings </span><span style="font-family: inherit;"><span style="color: #999999;">such that from the set of all numbers sampled, there are </span><span style="color: #999999; font-family: inherit;">\(k\) </span><span style="color: #999999;">distinct values.</span></span></blockquote>
<br />
<span style="font-family: inherit;">Like all <span style="font-family: inherit;">competitive programming problems, this one is deceptively simple. <span style="font-family: inherit;">At first glance<span style="font-family: inherit;"> it ap<span style="font-family: inherit;">pears you could just cheese it an<span style="font-family: inherit;">d run a Monte-Car<span style="font-family: inherit;">lo simul<span style="font-family: inherit;">ation, yet they <span style="font-family: inherit;">stipulate the expected value be computed accurately to <span style="font-family: inherit;">within a precision o<span style="font-family: inherit;">f </span></span></span></span></span></span></span></span></span></span></span><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;">\(1.0\times10^{-9}\)</span></span><span style="font-family: inherit;">, which would require<span style="font-family: inherit;"> more computation<span style="font-family: inherit;"> time</span> than <span style="font-family: inherit;">TopCoder allows</span></span>. Okay so </span></span></span></span></span></span></span></span></span></span></span>fair enough<span style="font-family: inherit;">, <span style="font-family: inherit;">we'll need to fin<span style="font-family: inherit;">d a way<span style="font-family: inherit;"> to compute this value directly<span style="font-family: inherit;"><span style="font-family: inherit;">.</span></span></span></span></span></span><br />
<br />
<h3>
<span style="color: #666666;">A Formula for the Expected Value</span></h3>
<h4>
<span style="font-family: inherit;"> </span></h4>
<span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;">Anyone with a basic understanding of probability shouldn't have too much difficulty recovering a formula for the expected value. </span></span></span></span></span></span></span><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;">You <span style="font-family: inherit;">might try solving this problem by thinking about how the expected value of samplings changes as we sample towards \(k\) unique<span style="font-family: inherit;"> sampled <span style="font-family: inherit;">numbers</span></span>. <span style="font-family: inherit;">When \(k=1\), the expecte<span style="font-family: inherit;">d value \(P\) <span style="font-family: inherit;">is \(1\) <span style="font-family: inherit;">as our <span style="font-family: inherit;">set of <span style="font-family: inherit;">sampl<span style="font-family: inherit;">ed <span style="font-family: inherit;">numbers </span>has exactly \(k=1\) unique members<span style="font-family: inherit;">: we certainly don't need to sample more than once to get \(1\) unique <span style="font-family: inherit;">number</span>. </span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span><br />
<br />
<span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;">What about when <span style="font-family: inherit;">\(k=2\)? Well <span style="font-family: inherit;">sample <span style="font-family: inherit;">once to get our first unique <span style="font-family: inherit;">number</span>, <span style="font-family: inherit;"><span style="font-family: inherit;">so \(P=1\) so far. Now<span style="font-family: inherit;"> to determine how many sampl<span style="font-family: inherit;">ings we <span style="font-family: inherit;">expect </span>to make to get</span></span> a second unique <span style="font-family: inherit;">number</span>, we have to think about the probability of selecting <span style="font-family: inherit;">a <span style="font-family: inherit;">number </span>that isn't the <span style="font-family: inherit;">one we already selected<span style="font-family: inherit;">. If we have \(n\) different numbers <span style="font-family: inherit;">to choose from, then the probability of picking one we haven<span style="font-family: inherit;">'<span style="font-family: inherit;">t alread<span style="font-family: inherit;">y chos<span style="font-family: inherit;">en<span style="font-family: inherit;"> is<span style="font-family: inherit;">:</span> \[\frac{n-1}{n}\]T<span style="font-family: inherit;"><span style="font-family: inherit;">his can </span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span>be inverted to give us an expected<span style="font-family: inherit;"> value of samples for picking a 2nd unique number<span style="font-family: inherit;"> so: \[\frac{n}{n-1}\]Th<span style="font-family: inherit;"><span style="font-family: inherit;">erefo<span style="font-family: inherit;"><span style="font-family: inherit;">re, \[P=1+\frac{n}{n-1}\]<span style="font-family: inherit;">When \(k=2\).</span></span></span></span></span></span></span><br />
<br />
<span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;">When <span style="font-family: inherit;">\(k=3\) we can f<span style="font-family: inherit;">ollow very similar logic. <span style="font-family: inherit;">We can just use the expected value from whe<span style="font-family: inherit;">n \(k=2\) a<span style="font-family: inherit;">nd <span style="font-family: inherit;">add to it the expected value of samples for picking a 3rd uni<span style="font-family: inherit;">q<span style="font-family: inherit;">ue <span style="font-family: inherit;">number</span>.<span style="font-family: inherit;"> Similar to last time, when <span style="font-family: inherit;">a 3rd <span style="font-family: inherit;">number</span> is sampled, the probability<span style="font-family: inherit;"> of picking one that isn't one of the <span style="font-family: inherit;">num<span style="font-family: inherit;">bers </span></span>we've alread<span style="font-family: inherit;">y selected i<span style="font-family: inherit;">s<span style="font-family: inherit;">:</span> \[\frac{n-2}{n}\]<span style="font-family: inherit;">A</span></span>gain <span style="font-family: inherit;">then <span style="font-family: inherit;">the e<span style="font-family: inherit;">xpected value <span style="font-family: inherit;">of samples we mu<span style="font-family: inherit;">st take is: \[\frac{n}{n-2}\]T<span style="font-family: inherit;">his makes our expecte<span style="font-family: inherit;">d<span style="font-family: inherit;"> value of samples <span style="font-family: inherit;">for \(k=3\)<span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;">: \[P=1+\frac{n}{n-1}+\frac{n}{n-2}\]If a pattern wasn't immediately obvious before,<span style="font-family: inherit;"> <span style="font-family: inherit;">its<span style="font-family: inherit;"> clear that we <span style="font-family: inherit;">are encounte<span style="font-family: inherit;">ring a sum of<span style="font-family: inherit;"> some kind. </span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span><br />
<br />
We can consider the last term of our latest formula for \(P\) as the beginning of a summation, and attempt to generalize the formula above using sigma notation: \[P=n \sum_{i=n-k+1}^{n} \frac{1}{i}\]<br />
<h3>
<span style="color: #666666;">
Computing Large Values of \(P\) </span></h3>
<br />
<span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;">Now anyone who is <span style="font-family: inherit;">famil<span style="font-family: inherit;">iar with competitive<span style="font-family: inherit;"> programming knows that the<span style="font-family: inherit;"> <span style="font-family: inherit;">easy <span style="font-family: inherit;">bit </span></span></span>above is far from the solution to the problem. The problem <span style="font-family: inherit;">further requires that our function compute <span style="font-family: inherit;">ex<span style="font-family: inherit;">pected values for \(n\) and \(k\) as large a<span style="font-family: inherit;">s \(1.0\times10^{18}\). Obviousl<span style="font-family: inherit;">y we can't just grind out the sum stated above as we <span style="font-family: inherit;">firstly<span style="font-family: inherit;">, well... we'd be there for a while... Secondly, if we were somehow able to add that many terms of the sum, the numerical inaccuracies would build up <span style="font-family: inherit;">pretty sever<span style="font-family: inherit;">ely by the time we got to \(1.0\times10^{18}\)<span style="font-family: inherit;">. S<span style="font-family: inherit;">o what do we do?</span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span><br />
<br />
<span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;">As it turns out, the<span style="font-family: inherit;"> previously <span style="font-family: inherit;">mentioned sum is <span style="font-family: inherit;">known as a Harmonic <span style="font-family: inherit;">series. This <span style="font-family: inherit;">mathematical form has been <span style="font-family: inherit;">around for <span style="font-family: inherit;">ages, and <span style="font-family: inherit;">there is a well known relation<span style="font-family: inherit;">ship between a <span style="font-family: inherit;">specific form of Harmonic series and what is known as the <span style="font-family: inherit;">digamma function<span style="font-family: inherit;">:\[\sum_{i=0}^{\infty}\frac{1}{i+x} =-\psi^{0}(x)\]Our strategy here will be to leverage <span style="font-family: inherit;">som<span style="font-family: inherit;">e of the ways</span><span style="font-family: inherit;"> <span style="font-family: inherit;">we can represent</span> \(\psi^{0}(x)\) <span style="font-family: inherit;">to compute values for our <span style="font-family: inherit;">Harmonic series for \(P\). Let us first represent the<span style="font-family: inherit;"> series for \(P\) in terms of \(\psi^{0}(x)\). <span style="font-family: inherit;">(For now, we won't greatly delve in<span style="font-family: inherit;">to the digamma function, just know that <span style="font-family: inherit;">its relation<span style="font-family: inherit;">ship with Harmonic seri<span style="font-family: inherit;">es <span style="font-family: inherit;">lends itself to solving this problem.)</span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span><br />
<br />
<span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;">Our <span style="font-family: inherit;">sum for the <span style="font-family: inherit;">digamma function was infinite<span style="font-family: inherit;">, yet we seek a sum<span style="font-family: inherit;"> that is finite. <span style="font-family: inherit;">We can deal with this by representing the sum for \(P\) in two parts, one part that contains both our finite sum and the unwanted terms that stretch out to infinity, and then a second part with just the <span style="font-family: inherit;">unwanted infinite <span style="font-family: inherit;">terms.<span style="font-family: inherit;"> By subtracting the second part from the first, we can recover <span style="font-family: inherit;">the finite sum for<span style="font-family: inherit;"> \(P\). In other words, by subtracting two infinite sums, we recover a finite sum. Lets start with the first one by pretending that our sum for \(P\) has upper limit \(\infty\) instead of \(k\). Since our sum counts up from \(n-k+1\), we can represent this variation of our sum for \(P\) as:\[\sum_{i=0}^{\infty} \frac{1}{i+(n-k+1)}=-\psi^{0}(n-k+1)\]Now we aren't interested in the terms that go past \(i=n\), so for our second sum which we will subtract from the first, we will use:\[\sum_{i=0}^{\infty} \frac{1}{i+(n+1)}=-\psi^{0}(n+1)\]Therefore, we can combine the two sums in terms of digamma and aesthetically rearrange to recover:\[P=n(\psi^{0}(n+1)-\psi^{0}(n-k+1))\]This leaves only the difficulty of computing values for digamma, which conveniently bears an approximation perfect for our application. </span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span><br />
<br />
<h3>
<span style="color: #666666;">
Computing Values for Digamma</span></h3>
<div>
<br /></div>
<div>
Now it is entirely permissible for us to use summation formula for \(P\) when \(k\) is sufficiently small, but when \(k\) is large, as previously stated, we'll need to use the digamma formula for \(P\). The takeaway here is that we need only use the digamma formation when \(k\) is large, which opens for the possibility to replace \(\psi^{0}(x)\) with an asymptotic approximation assuming we have one <i>and </i>its a very good approximation. I imagine you see where this is going? As it turns out, we have one (sort of)! </div>
<div>
<br /></div>
Let us first examine the exponential of the expression inside the parenthesis in the digamma formation of \(P\):\[{\Large\mathrm{e}^{\psi{0}(n+1)-\psi{0}(n-k+1)}=\mathrm{e}^{\psi{0}(n+1)}\mathrm{e}^{-\psi{0}(n-k+1)}}\]This expression is useful considering there happens to be a very good approximation for \(\Large\mathrm{e}^{\psi{0}(x)}\), namely:\[{\Large\mathrm{e}^{\psi{0}(x)}}\approx x-\frac{1}{2}, \mbox{if }x>1\]Using this approximation, we can write our previously mentioned exponential:\[{\Large\mathrm{e}^{\psi{0}(n+1)}\mathrm{e}^{-\psi{0}(n-k+1)}}=(n+\frac{1}{2})(n-k+\frac{1}{2})\]Woah, now how is that for an improvement?<br />
<br />
Of course, the exponential of the inner parenthesis in the formula for \(P\) isn't very useful, so we'll remove the exponentiation by taking the natural logarithm of the exponential form and bring \(n\) back in to recover an approximation for \(P\):\[P\approx\ln\left((n+\frac{1}{2})(n-k+\frac{1}{2})\right) n\]<span style="color: #666666;"></span><br />
<h3>
<span style="color: #666666;">Conclusion</span></h3>
<br />
Finding the expected value is now simply a matter of checking the size of \(k\), and at some point such that the approximation provides enough digits of precision (which should be well before the number of terms gets out of control), switch to the approximation of the series to compute \(P\) instead of using the series directly. I never got around to coding this solution as I solved the problem in about 40 minutes at 4:00am-ish, but if you can't already tell, the code should be a piece of ice cream cake preferably because that's my favorite kind of cake.<br />
<span style="font-family: inherit;">
<span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"><span style="font-family: inherit;"> </span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span> <br />
<span style="font-family: inherit;"></span>Chrishttp://www.blogger.com/profile/05191219955389953492noreply@blogger.com1tag:blogger.com,1999:blog-6455643200736463450.post-84263805296490281712016-01-11T21:09:00.003-08:002016-01-11T21:09:27.778-08:00What happened to all of the posts?I've been doing a little house cleaning and I wanted to refocus the type of content we throw up here. While all of the old game-development posts are gone for now, we intend to post more of that type of stuff as we go forward. Chrishttp://www.blogger.com/profile/05191219955389953492noreply@blogger.com1