I just finished reading China Mieville's novel Kraken. It was really cool, it did take me a while to get into Mieville's voice, but once I got into the swing of it I really enjoyed it. Here is the blurb in case you are interested:

"In the Darwin Centre at London’s Natural History Museum, Billy Harrow, a cephalopod specialist, is conducting a tour whose climax is meant to be the Centre’s prize specimen of a rare Architeuthis dux--better known as the Giant Squid. But Billy’s tour takes an unexpected turn when the squid suddenly and impossibly vanishes into thin air."

One of the lines in the book struck me as surprisingly familiar. Here is the quote from the book:

"She flicked through a pad by her bed, where she made notes of various summonings. A spaceape, all writhing tentacles, to stimulate her audio nerve directly? Too much attitude."

After thinking about this for a moment, it clicked that this is a reference to a Burial song called Spaceape (from his self titled album). The line goes:

"Living spaceapes, creatures, covered, smothered in writhing tentacles
Stimulating the audio nerve directly"

I couldn't find any reference online to the inclusion of Burial lyrics in Mieville's novels. Okay I thought, that's a cool a easter egg, but it got my thinking, are there any other song lyrics buried in Mieville's books? And if so, is there any way we can scan them automatically?

Collecting Lyrics

The first step was to get a database of song lyrics which we can use to scan the novels for, Unfortunately there is no easy place to find a database of song lyrics, so I was forced to scrape them from a lyrics site.

I used the following free chrome extension web scraper which is very easy to use, and in my experience very reliable:
webscraper.io/

After about 10 minutes of setting it up, and about an hour of leaving it to run. I had managed to scrape most of Burial's lyrics in to csv files.

I also scraped lyrics by Kode9 and Spaceape so I could see if they were referenced anywhere. It's hard to know which artist I should look for, but both of these have been mentioned by Mieville in interviews.

The web scraping add-in has an easy to use GUI. Here is a screenshot of what it looks like to set it up:
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Analysing the text

This is now the hardest part of the problem. We have electronic copies of China Mieville's novels in .txt format, and we have a collection of lyrics in .txt format which we would like to compare them against, how can we programmatically analyse whether Mieville references other Burial lyrics in one of his books?

If we simply attempt to match entire strings, then we have the issue that we might miss out on some references due to small changes in word ordering or punctuation. For example, in the example above using Burial's Spaceape, the wording is slightly different and the tenses of some of the words have been changed, therefore looking for an exact match between lyrics and text will probably not work. If on the other hand we don't match complete strings, but just try to match words, then we will be overwhelmed by small words like 'the' and 'a' which will be used multiple times in both Burial's song lyrics, and in China Mieville's novels.

There are two main approaches I came up with.to solve this problem. My first thought was to match individual words, generating a huge list of matches, and then to count the number of uses of each word in Mieville's novels, and then sorting by the words that match but which are also the most uncommon. For example I would imagine that Spaceape is only ever used once in all of Mieville's novels, giving us information about how unusual this word is. Combined with the fact that this word is also used in a Burial lyric, gives us enough information to assume that there is a high probability of a match, at this point we could investigate manually. 

I ultimately didn't go down this road. Instead, I had the idea to try to adapt plagiarism detection software to this problem. When you think about it, the two problems are actually quite similar. Plagiarism detection is about trying to automatically check two documents for similar phrases, without relying on complete matches.

Open Source Plagiarism Detection

I found the following free-to-use program created by Lou Bloomfield of the University of Virginia which is perfect for what I was trying to do.
plagiarism.bloomfieldmedia.com/wordpress/

It compares two sets of files and then creates a side by side hyperlinked comparison, which can be viewed in chrome, highlighting the parts of the documents where a possible match has been detected. There are various settings you can tweak to specify how close of a match you are interested in. 

I have included a screenshot below of the section where the Spaceape line is detected. There were about 500 matches detected when I ran this, but it only took about a minute to scroll through and check up on the ones that looked significant.
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Results

Ultimately, this analysis felt like a bit of a failure. These were the only lyrics I could find in all of his novels and while  there is always the chance that I need to expand the number of artists I'm looking at, or refine my detection methods I imagine this is all there is. I still thought the process was quite cool so I thought I'd write up what I had done anyway.

If you have any thoughts, let me know by leaving a comment.

When you are playing a drinking game, do you ever catch yourself calculating the probabilities of various events happening? If you are like most people the answer to that is probably "..... um..... no....... why would I ever do that...."

Okay, maybe that's not the kind of thing you think about when playing a drinking game. But I bet you think it would be interesting to know what the odds are anyway? No? Really? Still not that interested? .... Okay... well I find it interesting, so I'm going to write about it anyway. 

Eyes up/Eyes down

I've only played this game a couple of times but it's pretty fun, it requires no equipment, and has quite a lot of drinking. All players sit in a circle, one player calls out "Eye's down" at which point everyone looks down, the same player then says "Eye's up" at which point everyone looks up at a random person in the circle. If the person you are looking at is also looking at you, then you both drink. Pretty simple.

What is the probability that you will drink? Let's denote the event that you have matched someone with $M$. Then:

$$P(M) = \frac{1}{n-1} $$
Where $n$ is the number of players.

​Of course people in real life tend not to pick a random person to look at, and even if they attempt to pick a random person, people have been shown to be terrible at picking randomly. But for the purposes of this analysis, unless we make this assumption, the only alternative would be to play the game hundreds of times, record how often matches are made, and then make an empirical claim. As fun as it sounds to play this game hundreds of times in a row, it would be better for your health to just assume a uniform spread of guesses.

The fact that you have a $ \frac{1}{n-1} $ chance of getting a match means that the more people are playing the game, the less likely each person is to drink. Does this apply to the group as a whole though? What is the probability that someone in the group drinks? If we had a hundred people playing, then each individual would hardly ever drink. In fact you would expect them to drink once every $99$ goes. But would you expect it to also be unlikely that anyone at all drinks?  I spent about an hour trying to calculate this and didn't get very far.

Calculating through conditional probabilities doesn't seem to help, and I couldn't come up with a decent approach to count the permutations of all the possible ways of selecting pairings for elements in a set, such that there are no 2-cycles. In the end I gave up and just wrote a program to calculate the answer stochastically. Monte Carlo analysis really is such a versatile tool for problems like these. Anytime you can formulate the problem easily, but struggle to come up with an analytical solution, Monte Carlo analysis will probably work.

Here is the table I generated of the probability of someone in the group drinking for a group of size n:

There is a definite trend here, if we plot the values on the chart it looks pretty clear that out solution converges to $0.6$. No idea why though! I might come back to this problem another time, but if anyone has any thoughts then please let me know.
​

Odds

So this is not technically a drinking game, but is played quite often by people who are out drinking. The rules for this game are pretty simple, anytime you would like to dare someone to do something you say "odds of you doing this". The dare might be anything, from 'eating a spoon full of chilli powder, downing your drink, or even getting a tattoo (that last one is a real one I heard about from a friend who playing the game while travelling in Eastern Europe). The person you have challenged then gives you a number. They might say $20$, then each of you count down from $3$ and say an integer from $1$ to $20$ (or whatever number they selected). If you both said the same number, then they have to do the dare. Therefore the higher number you pick, the less likely it is that you will have to do the dare. 

What are the odds of you having to do whatever it is you are betting on, based on you selecting the number $n$? This is quite a straightforward problem, the answer is just $\frac{1} {n} $. I was playing this game last week with someone who thought the answer might be $\frac{1} {n^2} $. This would give a very different likelihood of having to do the dare. For example if you selected $20$, then you would have a $5 \%$ chance of having to do the dare, but according to my friend's calculation he thought you would have $0.25 \%$ chance of doing the dare.

Here is a table showing the correct probabilities for various values of $n$.

Dirty Pint Flip

I wouldn't recommend playing this game to be honest. I played it in uni once, but it's a bit grim. All the players sit in a circle with a central pint glass in the middle, the players take it in turn to pour any amount of their own drink into the central pint glass as they would like, they then have to flip a coin and guess whether it will be heads or tails. If they get it right, then the game moves on to the person to their left, if they get it wrong, then they have to down the dirty pint. When I played it some people were drinking wine, some people were drinking beer... it was not a good combination.

What is the probability that you will have to drink? This is quite straight forward, it is simply the probability that you will guess the coin flip correctly. Which is just $\frac{1}{2} $.

What about the question of how much you will drink on average? That is, on average if you have to drink, how many people will have poured their drink into the glass before you drink? We can model this as a geometric distribution and calculate the probabilities of each possible number of people. Let's denote the number of people who have poured in before you, given you are about to drink as $N$, then:

The expected value of the number of drinks is then the sumproduct of these two columns . This gives us a value of $2$. Therefore, if you have to drink, the average number of drinks that will have been poured into the pint is $2$.

Two Dice Nominate

This is another game I've only played a couple of times. And I'm not sure if it has a better name than the one I've given it, if you know of one then let me know. In this game, you sit in a circle with a range of drinks on the table, you take it in turns to nominate a person, a drink, and a number between $2$ and $12$. You then roll two dice and add them together. If you have rolled the correct number, the person you nominated drinks the drink you selected, if however you roll a double, then you drink that drink. If it is both a double and you get it correct, then you both have to drink.

The reason that this game does not have much depth to it is that you can always just pick the most likely number to come up when rolling two dice. Here is a table showing the probability of each value coming up:

Therefore you might as well pick $7$ every single time! Is there anything else we can say about this game? We know that if we roll  a double, then we have to drink the drink. What is the probability of you rolling a double in this game? The probability of this is always just $\frac{1}{6} $. Interesting this is the same probability as rolling the dice such that the sum is $7$. Therefore, if you do pick $7$, there is an equal chance of you and the person you are nominating drinking the drink.
​

Recently I've been reading The Mathematics of Poker (2009, Bill Chen and Jerrod Ankenman) and I came across an interesting idea that I thought I'd write about. For me, understanding how to analyse this situation really gets to the heart of how I think about poker. I'd love to spend more time playing and studying poker but it's such a time-sink and I don't really have the oodles of free time it would require, but every now and again I'll still open up a poker book and read about something, this is one of those interesting topics I was reading about hopefully you find it as interesting as I do.

Calling a shove pre-flop in heads up

The scenario being analysed in the book is a relatively common situation, particularly in online poker where people are more inclined to shove than in real life games. The question is:

How should we analyse whether to call when we have a moderately strong hand against an opponent who has gone all-in pre flop.

Let's set up an example so that we have something concrete to talk about. Say we have pocket Kings pre flop, and our opponent goes all-in, how should we decide whether to call? Obviously without knowing our opponent's hand there is no 100% correct answer. There is  however one very useful way of analysing the situation.

Equity against a range

We need to ask ourselves - what cards would are opponent go all-in with, and how does my current hand fare against that range? i.e. we need to calculate our equity against our opponent's range. We are adding another layer of stochastic uncertainty on the event, instead of trying to guess what hand our opponent has (which is almost impossible) we are trying to guess what kind of hands he might go all-in with (which is very much possible). We then take this extra level of uncertainty and calculate the correct mathematical way to proceed.

On the one extreme, let's suppose that based on our read of how our opponent is playing, we might think that they would only go all-in with very strong hands, in this case just pocket Aces. We would then be a 4:1 underdog if we call with Ks, and we should definitely fold. (In order to calculate this we could use any standard poker calculator like the following)

www.cardplayer.com/poker-tools/odds-calculator/texas-holdem

One the other hand, suppose we know for a fact that our opponent has not looked at their cards at all but has still decided to go all in. In this case we should definitely call. The only cards we will be behind are pocket Aces, which make up a small fraction of the possible hands that our opponent could shove with, and we will be ahead or equal against all other possible hands. Therefore we would have a positive EV when calling.

What if our read on our opponent's range is somewhere in between though?

What we need to do is calculate our equity against each individual hand in our opponent's range, and then calculate the probability of our opponent having a given hand from that range. That is to say, in order to calculate the conditional expectation, we need to calculate the unconditional expectations against each hand and then multiply by the conditional probability of our opponent having that hand, given our belief about our opponent's range. 

Numerical Example

Let's go back to our numerical example, and suppose that we have pocket Kings, and we put our opponent on either Pocket Aces, Kings, Queens, or Jacks. All of these hands are equally likely, so there is a 25% chance of our opponent having each hand. We can look up our equity against each hand (after you've been playing for a while, you naturally start to memorise hand equities anyway)

Our probability of winning is then:

$$P(A) * P(beating A) + P(K)*P(beating K) + P(Q)*P(beating Q) + P(J) * P(beating J)$$
Putting in our values:

$$ 0.25*0.2 + 0.25*0.5 + 0.25*0.8 + 0.25*0.8 = 0.575.$$

​We therefore see we have a positive expectation against this range, and should call.

No one actually thinks like this in real games?

It is a general misconception that professional poker is a game where players are trying to guess exactly what hand their opponent has, are constantly trying to bluff each other, or trying to pick up on subtle tells or signs that their opponent is or isn't bluffing.

The more mundane truth is that poker is ultimately a game of imperfect information, where the better player is the one who can correctly interpret the betting information their opponent is giving them, and can then quickly and accurately make the kind of judgements described above during a game.

Obviously poker players are not carrying out these calculations in their head to multiple decimal places in real time, what they will do though is review their hands after a game, calculate exactly what they should have done, and try to build up an intuition as to what the correct answer is, so that in the middle of a game they can quickly make decisions.

Software to Analyse this situation

Is there an easy software based method way of calculating our equity against a range? After I did a quick google there are a few programs that offer this type of analysis. For example:

combonator.com/
www.power-equilab.com/

More interestingly though, I also found the following open source software, that can be adapted to carry out this type of analysis:

github.com/zekyll/OMPEval

At some point, I might try to use this code to set up a page on this website to let people analyse this situation.

I was thinking about this last week at work when I was coding part of a model involving the parameters of a truncated lognormal distribution. The lognormal distribution definitely feels like it was named the wrong way round.
​
What is a Log-normal Distribution?

In other words, a Log-normal distribution is a distribution such that the log of the distribution is a normal distribution. It is not, as you might think, a distribution which is the log of the normal distribution.

So if $Y \sim N( \mu , {\sigma}^2 ) $ then $Log ( Y ) $ is not a lognormal distribution, instead $ e ^ Y $ is a lognormal distribution.

So to create a lognormal distribution, we don't take the log of the normal distribution, we take the exponential!
​
Why does this matter?

The Reinstatement Premium payable following a loss to an Excess of Loss contract, is related to the ceded loss to the contract by a simple formula. Therefore, it seems reasonable that we should be able to come up with a simple formula relating the price charged for the Excess of Loss contract to the price charged for a Reinstatement Premium Protection (RPP) cover. I was in a meeting last week with two brokers who were trying to do just this. They had come up with an indicative price for an Excess of Loss programme and were trying to use this to price the equivalent RPP cover.

At the time I didn't have an answer for them, and when I did a quick Google, nothing came up. When thinking about it subsequently though, there are a couple of easy approximate methods we can use. Below I discuss three different methods for how you can price an RPP cover, two of which do not require any stochastic modelling assuming you already know the price of the Excess of Loss layers.

Let's quickly review what we mean by all these terms so that we are starting from the same point.

What is a Reinstatement Premium?

If you already understand how a Reinstatement Premium works, then feel free to skip this section.

Most Excess of Loss contracts will have some form of reinstatement premium. This is a payment from the Insurer to the Reinsurer to reinstate the protection in the event of a loss. In the London market, most contracts willl have either $1$, $2$, or $3$ reinstatements and generally these will be payable at $100 \%$.

What is a Reinstatement Premium Protection Cover?

The reinstatement premiums can be quite a large proportion of the overall premium paid for the Excess of Loss reinsurance. This is especially the case for lower level, working layers. Furthermore, the Reinstatement Premium will be payable when the insurer has just suffered a loss. From the point of view of the insurer, this additional payment comes at the worst possible time. The Insurer is being asked to fork over another large premium to the Reinsurer, just after having suffered a loss.

To address some of these concerns, Reinsurers developed a product called a Reinstatement Premium Protection cover (RPP cover). This cover pays the Insurer's Reinstatement Premium for them, which gives the insurer further indemnification in the event of a loss. Here's an example of how it works in practice:

Let's suppose we are considering a $5m$ xs $5m$ Excess of Loss contract, there is one reinstatement at $100 \%$ (written $1$ @ $100 \%$), and the Rate on Line is $25 \%$. The Rate on Line is just the Premium divided by the Limit. So here, the Premium can be found by multiplying the Limit and the RoL:

$$5m* 25 \% = 1.25m$$

So we see that the Insurer will have to pay the Reinsurer $1.25m$ at the start of the contract. Now let's suppose there is a loss of $7m$. The Insurer will recover $2m$ from the Resinsurer, but they will also have to make a payment  to cover the reinstatement premium of: $\frac {2m}  {5m} * (5m * 25 \% ) = 2m * 25 \% = 0.5m$ to reinstate the cover. So the Insurer will actually have to pay out $5.5m$. The RPP cover, if purchased by the insurer, would pay the additional $0.5m$ on behalf of the insurer

Now that we know how it works, how would we price the RPP cover?

Three methods for pricing an RPP cover

Method 1 - Full stochastic model

If we have priced the original Excess of Loss layer ourselves using a Monte Carlo model, then it should be relatively straight forward to price the RPP cover. We can just look at the expected Reinstatements, and apply a suitable loading for profit and expenses. This loading will probably be broadly in line with the loading that is applied to the expected losses to the Excess of Loss layer, but accounting for the fact that the writer of the RPP cover will not receive any form of Reinstatement for their Reinsurance.

What if we do not have a stochastic model set up to price the Excess of Loss layer? What if all we know is the price being charged for the Excess of Loss layer?

Method 2 - Simple formula

This was the situation I was in when I was asked to price the RPP cover last week. The broker had come up with some very approximate pricing for a programme of Excess of Loss layers. This pricing was driven by the burning cost, and other commercial factors rather than any actuarial modelling. The brokers wanted to come up with an approximate price for the RPP cover, just based on the price of the Excess of Loss layer. The two should be related, as they pay out dependant on the same underlying losses. So what can we say?

If we denote the Expected Losses to the layer by $EL$, then the Expected Reinstatement Premium should be:
$$EL * ROL $$
To see this is the case, I used the following reasoning; if we had losses in one year equal to the $EL$ (I'm talking about actual losses, not expected losses here), then the Reinstatement Premium for that year would be the proportion of the layer which had been exhausted $\frac {EL}  {Limit} $ multiplied by the Deposit Premium $Limit * ROL$ i.e.:

$$ RPP = \frac{EL}  {Limit} * Limit * ROL  = EL * ROL$$

Great! So we have our formula right? The issue now is that we don't know what the $EL$ is. We do however know the $ROL$, does this help?

If we let $DP$ denote the deposit premium, which is the amount we initially pay for the Excess of Loss layer and we assume that we are dealing with a working layer, then we can assume that:

Which is our first very simple formula for the price that should be charged for an RPP. Was there anything we missed out though in our analysis? 
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Method 3 - A more complicated formula:

There is one subtlety we glossed over in order to get our simple formula. The writer of the Excess of Loss layer will also receive the Reinstatement Premiums during the course of the contract. The writer of the RPP cover on the other hand, will not receive any reinstatement premiums (or anything equivalent to a reinstatement premium). Therefore, when comparing the Premium charged for an Excess of Loss layer against the Premium charged for the equivalent RPP layer, we should actually consider the total expected Premium for the Excess of Loss Layer rather than just the Deposit Premium.

What will the additional premium be? We already have a formula for the expected Reinstatement premium:

$$EL * ROL $$

Therefore the total expected premium for the Excess of Loss Layer is the Deposit Premium plus the additional Premium:

$$ DP + EL * ROL $$

This total expected premium is charged in exchange for an expected loss of $EL$.

So at this point we know the Total Expected Premium for the Excess of Loss contract, and we can relate the expected loss to the Excess of Loss layer to the Expected Loss to the RPP contract. 

i.e. For an expected loss to the RPP of $EL * ROL$, we would actually expect an equivalent premium for the RPP to be:

$$ RPP =  (DP + EL * ROL) * ROL $$

This formula is already loaded for Profit and Expenses, as it is based on the total premium charged for the Excess of Loss contract. It does however still contain the $EL$ as one of its terms which we do not know.

We have two choices at this point. We can either come up with an assumption for the profit and expense loading (which in this hard market might be as little as only be $5 \% - 10 \%$ ). And then replace $EL$ with a scaled down $DP$:

$$RPP =  \frac{DP} {1.075} * ( 1 + ROL) * ROL $$

Or we could simply replace the $EL$ with the $DP$, which is partially justified by the fact that the $EL$ is only used to multiply the $ROL$, and will therefore have a relatively small impact on the result. Giving us the following formula:

$$RPP =  DP ( 1 + ROL) * ROL $$

Which of the three methods is the best?

The full stochastic model is always going to be the best in my opinion. If we do not have access to one though, then out of the two formulas, the more complicated formula we derived should be more accurate (by which I mean more actuarially correct). If I was doing this in practice, I would probably calculate both, to generate some sort of range, but tend towards the second formula.

That being said, when I compared the prices that the Brokers had come up with, which is based on what they thought they could actually place in the market, against my formulas, I found that the simple version of the formula was actually closer to the Broker's estimate of how much these contacts could be placed for in the market. Since the simple formula always comes out with a lower price than the more complicated formula, this suggests that there is a tendency for RPPs to be under-priced in the market.

This systematic under-pricing may be driven by commercial considerations rather than faulty reasoning on the part of market participants. According to the Broker I was discussing these contracts with, a common reason for placing an RPP is to give a Reinsurer who does not currently have a line on the underlying Excess of Loss layer, but who would like to start writing it, a chance to have an involvement in the same risk, without diminishing the signed lines for the existing markets. So let's say that Reinsurer A writes $100 \%$ of the Excess of Loss contract, and Reinsurer B would like to take a line on the contract. The only way to give them a line on the Excess of Loss contract is to reduce the line that Reinsurer A has. The insurer may not wish to do this though if Reinsurer A is keen to maintain their line. So the Insurer may allow Reinsurer B to write the RPP cover instead, and leave Reinsurer A with $100 \%$ of the Excess of Loss contract. This commercial factor may be one of the reasons that traditionally writers of an RPP would be inclined to give favourable terms relative to the Excess of Loss layer so as to encourage the insurer to allow them on to the main programme and to encourage them to allow them to wrte the RPP cover at all.

Moral Hazard

​One point that is quite interesting to note about how these deals are structured is that RPP covers can have quite a significant moral hazard effect on the Insurer. The existence of Reinstatement Premiums is at least partially a mechanism to prevent moral hazard on the part of the Insurer. To see why this is the case, let's go back to our example of the $5m$ xs $5m$ layer. An insurer who purchases this layer is now exposed to the first $5m$ of any loss. But they are indemnified for the portion of the loss above $5m$, up to a limit of $5m$. If the insurer is presented with two risks which are seeking insurance - one with a total sum insured of $10m$, and another with a total sum insured of $6m$, the net retained exposure is the same for both risks from the point of view of the insurer. By including a reinstatement premium as part of the Excess of Loss layer, an therefore ensuring that the insurer has to make a payment any time a loss ceded to the layer, the reinsurer is ensuring that the insurer keeps their financial incentive to not have losses in this range. 

By purchasing an RPP cover, the insurer is removing their financial interest in losses which are ceded to the layer. There is an interesting conflict of interest in that the RPP cover will almost always be written by a different reinsurer to the Excess of Loss layer. The Reinsurer that is writing the RPP cover is therefore increasing the moral hazard risk whichever Reinsurer has written the Excess of Loss layer. Which will almost always be business written by one of the Reinsurer's competitors! 

Working Layers and unlimited Reinstatements

Another point to note is that this pricing analysis makes a couple of implicit assumptions. The first is that there is a sensible relationship between the expected loss to the layer and the premium charged for the layer. This will normally only be the case for 'working layers'. These are layers to which a reasonable amount of loss activity is expected. If we are dealing with clash or other higher layers, then the pricing of these layers will be more heavily driven by considerations beyond the expected loss to the layer. These might be capital considerations on the part of the Reinsurer, commercial considerations such as 

Another implicit assumption in this analysis is that the reinstatements offered are unlimited,. If this is not the case, then the statement that the expected reinstatement is $EL * ROL$ no longer holds. If we have limited reinstatements (which is the case in practice most of the time) then we would expect the expected reinstatement to be less than or equal to this.
​

Over the Christmas break, I finally got around to playing some Computer Games, which I unfortunately haven't done in ages. One really fun game (if you're like me and love difficult puzzles) is a game called Myst.

It was originally released in 1993 as a point and click puzzle adventure game, but it has subsequently been remade, using the Unity engine, as a full 3d interactive game. I really liked the fact that there is no time limit, no danger of dying, just a cool story, interesting puzzles and relaxing music.

One puzzle that I thought was cool involved aligning gear wheels. If you haven't played the game and don't want spoilers, then stop reading now.

The Gear Puzzle

​We are given the contraption in the image below, and are told that we need to end up with the numbers 2,2,1 (reading from top to bottom). We have three levers we can pull, the left lever moves the top two gears left once, the right lever moves the bottom two gears right once. The lever on the back wall resets the machine. The starting values of the gears are 3,3,3. So how many times do we need to pull each lever in order to end up at 2,2,1?

After playing around with it for a while I began to think that it might not even be possible to get to 2,2,1 using the two levers. Then I thought if that's true, can I prove it as well?

Proof that the puzzle is impossible

Instead of thinking of the value as $3$, let's call it $0$ instead. We can then think of the starting state as a vector:

To see that we are not able to do this with just the basic operations from the starting point we are given, note that the the sum of the first and third elements of our starting vector is equal to the second vector mod $3$, that is, $0 + 0 \equiv 0$ $(mod$ $3)$. For the vector we trying to create, the first and third is not congruent to the second element mod $3$. Instead we have $ 2 + 1 \equiv 0$ $(mod$ $3) $ rather than $2$.

The two operations we are applying to our starting vector preserve this congruence property, as it applies to both the vectors we are adding on as well. Therefore, we can conclude that this property is an invariant of the system, and because our goal does not satisfy this condition, we will never be able to reach:

by just using simple pulls of the lever. We need to realise this in order to force us to look for an alternative way to solve the puzzle. Otherwise we would probably just keep yanking on the levers in the hope that we will eventually get there. Once we realise this and start looking around, we are able to find another way to manipulate the machine, allowing us to make the numbers we need.