Do you want a mathematician or a computer?

Recently I have been applying to a couple of jobs for next year, and have beenquite fascinated by the anxiety other people express over the whole applicationprocess. It seems everyone is so intent on being perfect for the job they’reapplying for, that they don’t seem to notice that if they have to ‘change’ whothey are to get the job, the job probably won’t suit them. So they doctor uptheir C.V’s and buy different clothes and attend the interviews readyingthemselves to answer in such a way that their weaknesses become strengths….

The best one I’ve heard from this is: “what is your biggest weakness?”
“I’m dedicated to my work, so I often work straight through a lunch break”.

Yeah right.
Anyways, a Game theorist wrote a lot more on this here :

an awesome blog to read, and the articles make you think.

My favourite part of interviews are the questions that makeyou think. Asking what I would do in a given situation is boring, but asking mehow to solve an abstract idea – that’s fun.
My girlfriend had an interview a few months ago, and one ofthe questions was:

using square tiles that have an area of 1 unit squared, how can you lay thetiles such that the number of tiles used is twice the number of tiles in theperimeter?

the way people approach this is fascinating. Most of us will start writing outequations, so perimeter of a rectangle is 2 x (width+length) and the area iswidth x length, so, using w,l to denote width, length respectively,

which isn’t easily solved. It would probably be best to start throwing integersinto the equation, and see what comes out.
So if we define w as 8, we get 4+l/2=l4=l/2 8=l so a square of 8 by 8 works! Right?
We forgot to notice that thecorner tiles are counted twice. The perimeter calculation should actually be2*(w+l-1) (but the area still stays the same).
4+4(l-1)/w =l

again, try values of w

Which gives a perimeter of 28tiles and an area of 56, which works.
But to do this in an interviewwhere people are stressed out….that’s a bit more tricky. You’ll forget todouble the perimeter. You’ll forget the repeated tiles. And you won’t noticethat if you tile around the equator, you simply need to place 4 rows of tiles,one above the other, to satisfy the requirements…
There are quite a few questionslike this.  
Here’s another one from the blog Ilisted before:

the solution is awesome.

An alternative solution is tocreate a gold cutting device that has 6 knives. Then you just need one cut.
Now, from my limited experience ofinterviews and applications, it seems that most companies claiming to belooking for mathematicians are actually looking for a computer or a softwaredeveloper (which is good news to software developers!). Almost all graduatejobs base part of the application on marks attained at university.  Interestingly, those marks are made up ofexams, tests and assignments. Assignments seldom give more than 20%  towards the final mark. Roughly 40% of theassignments are based on an ability to think like a mathematician.  The rest is rote learning/memorisation or useof a method.  Unfortunately, the examsand tests are also based on this.  This meansthat most courses have at most 10% of the course based on the ability to think,and 90% on being able to memorise proofs and theorems and methods. 
This is all well and good…except for a little thing called the internet. Throughthe use of google, wolframalpha and Wikipedia, that 90% of the course can bedone by someone with no previous knowledge of the course.  I would also guess that they can do it moreaccurately, and faster.
So to base your applicants ontheir university scores seems to me to be a waste of time. If that is all youwanted, you’d be better off getting a software developer or simply using thosethree sites – this way will be much cheaper! If you ever come across a problemyou can’t solve using those sites…well, then there isn’t a high chance yourgraduate with straight A’s could have solved it either…

But enough of that. It’s fairly obvious I have crap marks and think I’m abetter mathematician than those marks show, but that’s my opinion. I likesolving problems, and this is a taster of the next post:

what do the numbers 1089 and 495 have in common?


truthful guests of patterns

Let me introduce to some logic games/ideas…
There is an island with only two tribes on it. Tribe number 1 is the “truth tellers”, tribe number 2 is the “liars”. The truth tellers can only tell the truth, the liars can only lie. You’re visiting the island and come across three of the inhabitants. You ask the first what tribe he is from, but he responds in a voice so low and guttural you cannot make out what he said. You ask the second one what the first one said and he states “He said he is a truth teller”. the final inhabitant looks solemnly at the second, then, turning to you says “he lied to you”.
Which tribe was the third speaker from?
I suggest you think about this first before reading on, it’s quite a fun intellectual exercise to do, and requires nothing more than 5 mins of contemplation.
The trick with this question is to look first not at the third speaker, but one of the other two. Now, we don’t know what the first speaker said, so maybe we should see if we can find out what he said.
Imagine the first speaker was a truth teller. He would then tell us he is a Truth teller. But if he was a liar, he’d still tell us he was a truth teller! This tells us that the second speaker is a truth teller, which tells us that the last speaker is a liar.
A slightly more complicated version of this story: the truth tellers love to eat outsiders (ie: you) and the liars love to help outsiders get off the island safely. Both tribes respect the other, and so if you are with a liar, the truth tellers will not try and abduct you and if you are with a truth teller the liars will not try and save you. Inhabitants will never take an outsider, the outsider must always choose to go with them, regardless of the tribe. There are no differences between the tribes other than their honesty and what they do with outsiders. The only way to get off the island is with the liars, and if an outsider doesn’t choose a tribe s/he will starve to death…
So. Imagine you’re on the island and you run into 3 inhabitants. In as few questions as possible, how do you determine who you go with?
I like the idea of asking them questions about themselves (it’s more personal that way…and requires some well thought out questions). I’ll give them names for illustration: Andy, Bob and Carlise. Then you could ask Bob “if I asked Andy which tribe he was from, what would he say?”, asking Carlise the same question but about Bob, then asking Andy about Carlise. Whichever tribe they mention is the tribe they’re from. A simpler way is to ask them whether it’s raining or not!
There are many variations on these – I think there is a book on them, but I can’t seem to find it. Some of the variations involve different conditions for the inhabitants (they all tell the truth on certain days of the week) language issues, etc.
Now, as stated, these are logic games/puzzles. Some people may not realise it ties into mathematics – but it does! All these games can be broken down into symbols and letters and linked together, giving you a set of tools to solve the problems with. If you want to know more about mathematical logic, study it, look it up online, or ask me!
I would now like to go on a small rant if I may about something that bugs me about people who don’t know much about maths…why do people always expect that because you study maths you are going to be a whiz at mental arithmetic? There are people who are amazing at mental arithmetic, and there are others who quite enjoy working out the patterns of arithmetic ( indeed, I’m one of the latter. For example, take any three digit number where the 1st and 3rd digit differ by at least 2. Reverse the number. Subtract the smaller number from the bigger one. Take your new number, and reverse it. Add those two numbers together. And your answer is 1089. Always. And those patterns are very interesting to analyse as they make cool math tricks) but not every mathematician likes or is good at simple calculations – it’s why we have calculators, although it is definitely better for your brain if you spend some time calculating it yourself! In truth, expecting all mathematicians to be able to calculate quickly and easily in their head is like expecting an engineer to build a boeing 747 from a junk site, or a chef to make a wedding cake with ease, or an artist to take stunning photos, all musicians to be able to sing opera. Whilst there are some that can, it does not mean that all of them can.
Moving on 🙂
Using the logic we played with before, we can attack almost any problem and solve it. However, it is probably best to sort out a method of approach first!
First, find the simplest version of the problem.
See how it behaves.
Slowly complicate the problem toward the original.
look for patterns.
prove the patterns.
(if you read the blog about Fibonacci’s rabbits, it ties into the idea of simplifying the problem first).
there are many mini steps that can fit in there, and some steps that could be there but aren’t, and some that are in front of others that should be behind them….but this is just a general guide line.
So now for the next problem.
Mr and Mrs Senzigic were having a party. They invited 4 other couples around, and had a very good time. At about midnight, Mr Senzigic gathered everyone around and asked them all how many hands they had shaken. To his amazement, everyone had shaken a different amount of hands (even himself), ie, one had shaken 0, one 1, one 2…all the way to 8. Nobody had shook the same person twice, and nobody had shaken their spouses hand. How many hands did Mrs Senzigic shake?
On first look, this problem seems impossible. How, without knowing more detail, can you even begin to hope to solve to this??? The only thing you can deduce is that Mrs Senzigic shook the same number of hands as someone else.
So lets take it down to a simple state of affairs, with only two couples, the Hosts, and The Guests. HM will refer to Mr Host, HF will be Mrs Host, GM Mr Guest, GF Mrs Guest.
Let’s imagine that HM – being the nice man he is – shakes every hands except his wife and his own. This means that his wife has shook none, but both of the guests have shook one. if one of the guests shakes hands again, they will be on two shakes, but they can only get there if they shake HF, which would mean she has shaken two hands as well. This would cause a problem, and you would get a similar situation if you started by assuming HF had shaken everybody’s hand. The only option left the is that one of the Guests shakes two hands, so lets assume GF shakes both HM and HF hands, then they each have one, GF has two, and GM has none.
if you look at the same situation with 2 Guest couples, you find that Mr Guest 1 would shake 4 hands, his wife none, Mrs Guest 2 three hands, Mr Guest 2 one hand, and the Hosts would shake the 2 hands each.
The proof for this is very simple. Start with the last couple to arrive. They walk in, walk around the room greeting everybody, one of them shaking everybody’s hand (8 shakes), the other shaking none(0 shakes). The same for the couple before them, they would have walked in and one of them would have shaken everybody’s hand who was there (6 shakes for now) and their partner shaking none. When the couple arrived after them, the earlier couple would have both been shaken by the shaker in the new couple, meaning that the earlier couple would now have shook 7 and 1 hands respectively. The same pattern continues for the first two couples to arrive. Because of this, it ends up that every couple shakes a total of 8 hands. S0 8 and 0, 7 and 1, 6 and 2, 5 and 3…4 and 4?! but that would mean that there was a repeat in the numbers!!
and then you remember that that Mrs Senzigic shook the same amount of hands as one other person…therefore she must be in the couple of four and four, therefore Mrs Senziga shook four hands.
Hope you followed that, if not, let me know :).
Now for the final problem – and this was in an assignment for a fairly difficult paper that I’m not doing – yet.
Imagine a square, 1×1 units.

 Divide each of these sides into three new sides.
 On the middle piece of each sides, draw a new square with three new sides.

Repeat this pattern for each side.

What is the total area of this shape?
Have fun! Email me if you want the solution.