To recap the puzzle:

A man walks into a casino and spots an interesting new game. He is presented with two buttons, either of which he can push. One of the buttons will win him a certain amount of money and the other will win him twice the amount, although he doesn’t know what the amount is. So he pushes one of the buttons at random and is surprised and very pleased to learn he just won $1 million. Suddenly, the machine presents him with two new buttons. He can push one to keep his original choice, keeping the $1 million. Or he can switch his choice to the button he didn’t push. What would you do and why?

When the man is shown how much he won when he pushed the first button randomly, that revealed some information: either the two rewards are $500K and $1 mil or they are $1 mil and $2 mil, but we don’t know which. Given that information, we can devise a strategy to determine whether we should switch our choice.

Suppose we draw a random number from anywhere from the real line using some probability distribution and we take that random number as a proxy for the amount behind the other button. For example, if we randomly selected $750K we would not switch, since $1 mil is bigger. But if we randomly selected $1.5 mil, we would switch, since $1.5 mil is larger than $1 mil. If we happen to select the random number and it falls in the interval of the true rewards, we will guess whether to switch correctly with certainty. And if we draw that number outside the interval, then we will guess correctly half the time.

Let’s formalize that. Suppose we have a pair of prizes (X, 2X) and let S be the event that we switch correctly using our strategy. We choose our random number from Z that comes from some distribution. Then,

But,

So,

which is greater than 50%.

We need to select our underlying probability distribution cleverly then. Suppose we choose a normal distribution with mean $1 mil and standard deviation $500K. That choice should guarantee good probability mass around the intervals (500K,1 mil) and (1 mil, 2 mil).

If it turns out that we are in the interval (500K, 1 mil), then the probability that Z is in (500K, 1 mil) is about 0.34 and the probability of choosing correctly is 1/2 + 1/2(0.34) = 67%. However, if the interval happened to be (1 mil, 2 mil), then the probability that Z is in (1 mil, 2 mil) is 0.48 and the probability of choosing correctly is 1/2 + 1/2(0.48)= 74%.

A man walks into a casino and spots an interesting new game. He is presented with two buttons, either of which he can push. One of the buttons will win him a certain amount of money and the other will win him twice the amount, although he doesn’t know what the amount is. So he pushes one of the buttons at random and is surprised and very pleased to learn he just won $1 million. Suddenly, the machine presents him with two new buttons. He can push one to keep his original choice, keeping the $1 million. Or he can switch his choice to the button he didn’t push. What would you do and why?

It might seem a strange question to ask whether all functions are computable. We are used to seeing functions written down in mathematical language, making them automatically computable. For example, the quadratic function is obviously computable:

We can put any x into this function and calculate the answer. Or to take another example, the sine function is computable since we can write it as an infinite series:

We can put any x into this function and calculate sin(x) to any precision we like by including as many terms as we need in the infinites series. Surprisingly, however, there are an infinite number of functions that we can’t compute. The reason is that there are more functions than there are algorithms to compute them.

What does it mean to compute a function? Intuitively, it means that we can work out the answer on pencil and paper or we can punch numbers into a caculator to compute the function. More generally, a function is computable if we can write a computer program to compute it. For example, one python code to compute the quadratic function would be

def quadratic_function(x):     
    return x**2 + 5*x + 2

How Many Possible Computer Programs Are There?

Computer programs are strings of characters of finite, but arbitrarily large length. For example, the python program above could be represented as the string of characters:

def\space\quadratic_function(x):\enter\\tab\\return x**2 + 5*x + 2

where \space\ is the space character,\enter\ is a carriage return, and \tab\ is a tab. All computer programs can be written as strings of characters of finite length. How many possible computer programs are there? It would seem that there are an infinite number of compute programs, but we’ll need to be more precise about what we mean by infinity.

Countable Infinity

A set is countably infinite if it can be put in order and counted by the infinite positive integers. This definition immediately leads to surprising consequences. The set of even numbers is countably infinite since the even numbers can be put in order and counted by the positive integers.

Thus, it turns out that a subset of a countably infinite set is also countably infinite. When we are talking about infinite numbers, a subset of an infinite set has the same number of elements as the set itself.

What about the rational numbers? They are countably infinite as well, since we can put them in order and count them by the positive integers:

In this diagram, we order the rational numbers by following the arrows, excluding the rational numbers circled in green, since we already counted them previously. 1/1 would be the first element, counted as one, 1/2 would be second, counted as two, 2/1 would be the third element, counted as three, and so on.

Uncountable Infinity

What about the real numbers? Let’s assume we can order them by putting them into a one-to-one correspondence with the natural numbers as in the diagram below, in which we have circled the diagonal element in each real number.

Now let’s construct a new real number with the property that its first digit is different from the first digit of the first real number, its second digit is different from the second digit of the second real number, continuing in that fashion:

This new real number can’t be on the list. As a consequence, there are more real numbers than there are integers. Even if you think you’ve written down an infinite set of real numbers, there are always more. The set of real numbers isn’t countable—it’s uncountably infinite.

Is The Set of Possible Computer Programs Countably or Uncountably Infinite?

The set of all possible computer programs is countably infinite since we can order them. Every computer program is a sequence of character symbols of finite length. To order them, we can consider all legitimate computer programs that have one character, all legtimate computer programs that are two characters in length, etc. We can order them by placing the one character programs first (counting only those that are legitimate, if any, in the programming language we are using), then the two character programs next, and so on. Within the programs of character length N, we can also order them further by defining rules that order sequences of characters: if the first letter is “a”, it comes before any program that begins with any other letter or symbol, including “A” with the rule that lower case letters are counted before upper case letters. If both programs begin with “a,” then we proceed to the next character and apply the same ordering rules. In this way, we can associate every legitimate program with a positive integer. The set of all possible legitimate programs is countably infinite.

How Many Functions Are There?

The number of functions is uncountably infinite. To see this, let’s define a family of functions as follows: for every real number, the function is

where n is a positive integer and d_n is the nth digit in the infinite decimal expansion of the real number. We can define one such function for every real number, implying that this subset of all the functions is uncountably infinite. Since there are more functions than there are algorithms to compute them, there must be an infinite number of functions that are not computable.

Implications

Most real numbers are not computable, since there are an uncoutably infinite number of real numbers and a countably infinite number of computer programs. If a real number is not computable, that means that there is no algorithm to compute its value to arbitrary precision. The computable reals are countable, however, because there are as many of them as there are algorithms.

Chaitin’s Constant

Chaitin’s constant is the probability that a randomly generated computer program will halt. These probabilities are non-computable. They can be approximated, but not to arbitrary precision.

Zeros of Functions

Since most real numbers are non-computable, it should not be surprising to realize that there an infinite number of functions with some real zeros that cannot be computed.

A father tells his children, a boy and a girl, to play in the backyard without getting dirty, but they play in the mud anyway. Both children get mud on their foreheads. The father says, “at least one of you has mud on his or her forehead, which I asked you not to do. I’m going to take away desert tonight, but I’ll give you a chance to get your desert back.” He tells them he will ask the following question twice: do you know if you have mud on your forehead? Each child must answer honestly and simultaneously each time the question is asked. Each child can see the other’s forehead but not his or her own. If both children can correctly tell whether or not they have mud on their foreheads, they can have desert that night. What should the children do?

Solution

When the father asks the question the first time, “do you know if you have mud on your forehead,” both children will answer “no.” Each child sees the mud on the forehead of the other one. Since they know at least one of them has mud on his forehead, neither child can conclude whether he also has mud on his own forehead. They must answer “no” the first time the question is asked.

However, after the question is asked and answered, the girl will immediately conclude she must have mud on her forehead since the boy answered “no.” The boy will also conclude that he must have mud on his forehead for the same reason. So, when the father asks the same question again, they both answer “yes” and enjoy desert that night.

A father tells his children, a boy and a girl, to play in the backyard without getting dirty, but they play in the mud anyway. Both children get mud on their foreheads. The father says, “at least one of you has mud on his or her forehead, which I asked you not to do. I’m going to take away desert tonight, but I’ll give you a chance to get your desert back.” He tells them he will ask the following question twice: do you know if you have mud on your forehead? Each child must answer honestly and simultaneously each time the question is asked. Each child can see the other’s forehead but not his or her own. If both children can correctly tell whether or not they have mud on their foreheads, they can have desert that night. What should the children do?

The solution is here.

This post was contributed by Jingu Chong, who is one of our sponsors.

A nobleman is throwing a party in 24 hours, and he wants to serve wine from 1000 casks in his cellar. Unfortunately, an assassin has poisoned one of these casks. The poison takes 23 hours to take effect, and it shows no symptoms before causing death. The nobleman has servants who are “willing” to taste the bottles.

What is the least number of servants the nobleman needs to correctly identify the poisoned bottle, and what is his strategy?

The solution is here

This post was contributed by Jingu Chong, who is one of our sponsors.

A nobleman is throwing a party in 24 hours, and he wants to serve wine from 1000 casks in his cellar. Unfortunately, an assassin has poisoned one of these casks. The poison takes 23 hours to take effect, and it shows no symptoms before causing death. The nobleman has servants who are “willing” to taste the bottles.

What is the least number of servants the nobleman needs to correctly identify the poisoned bottle, and what is his strategy?

Solution:
The nobleman needs 10 servants. He numbers each servant 0-9, and these represent the respective powers of 2. He numbers each bottle 1-1000. He then labels each bottle by converting the decimal number on each bottle to binary. For example, Bottle 7 becomes 0000000111 and bottle 762 becomes 1011111010. The servants drink from each bottle which has a 1 in the location of their number. So, servants 0, 1, and 2 all drink from Bottle 7 and servants 1,3,4,5,6,7, and 9 all drink from bottle number 762. The servants do this until every bottle has been tested by the appropriate servants.

The nobleman can then look at which servants died. If, for example, Servants 0, 3, 6, and 8 died, the nobleman knows that the poisoned bottle is Bottle 329 (1 + 8 + 64 + 256).

Puzzle 1

Suppose there are two trolls at this fork in the road. One always tells the truth and the other always lies but you don’t know which is which. You have only one gold coin to pay one of them to answer a question. What question will you ask one of them to be 100% sure of which fork to take?

Answer

You simply select one of the trolls randomly and ask, “If I asked the other troll which way I should take, what would he say?” If you selected the lying troll, he would know that the other troll would have told the truth, so the lying troll will lie about the answer. The correct answer, then, is the opposite of what the lying troll would say. On the other hand, if you selected the truthful troll, he would truthfully tell you what the lying troll would say, a lie, and you should therefore take the opposite of his answer. In either case, you should always choose the road that is the opposite of the answer you receive, guaranteeing with certainty that you select the correct road.

Puzzle 2

You are lost in a country in which one third of the inhabitants always lie. The other inhabitants tell the truth, but only three quarters of the time. Desperate to get out of the country, you come to a fork in the road guarded by a troll. One way leads out and the other to certain death, but you don’t know which is which. You can ask the troll which way is out if you pay him a gold coin. You have four gold coins.

1. You pay the troll a gold coin and he advises you to go right. Does that answer help you to decide?

2. You pay the troll another gold coin and he advises you to go right again. Does that answer help you to decide?

3. You pay the troll another gold coin and he advises you to go right yet again. Does that answer help you to decide?

4. You pay the troll your last gold coin and he advises you to go left this time. You must make a final decision. What should you do?

5. What is the chance you get out of this country alive after you’ve paid your last gold coin?

Answer

The variation is good bit harder, requiring some probability calculations. Let’s define some events first. S_n is the event that the troll gives the same answer n times. T is the event that the answer is true. D is the event that the troll is dishonest but not a total liar, i.e., he lies one fourth of the time. The probability we want is P(T|S_n), the conditional probability the response is true given that we have been given the same response n times. By the properties of conditional probability,

P(S_n )= [(3/4)ⁿ +(1/4)ⁿ]*(2/3)+1/3, since 2/3 of the time n truthful answers will be given with probability 3/4 or n false answers will be given with probability 1/4, or 1/3 of the time n false answers will be given with probability 1.

The troll’s first answer is “go right.” What is the probability that answer is truthful? P(S₁ )= [(3/4)¹ +(1/4)¹]*(2/3)+1/3 =1. Therefore, P(T|S₁) = (3/4) * (2/3)/1 =1/2. Unfortunately, the first answer doesn’t improve the odds from 50-50.

The troll’s second answer is “go right.” The probability it is truthful is P(T|S₂) = (3/4)² * (2/3)/ P(S₂ ). But P(S₂ )= [(3/4)² +(1/4)²]*(2/3)+1/3 = (10/16)*(2/3)+1/3 = 18/24 = 3/4. Therefore, P(T|S₂) = (3/4)² * (2/3)/(3/4) = (3/4)*(2/3) = 1/2. The second answer doesn’t help either—it’s still 50-50 odds.

The troll’s third answer is “go right.” The probability it is truthful is P(T|S₃) = (3/4)³ * (2/3)/ P(S₃ ). However, P(S₃ )= [(3/4)³ +(1/4)³]*(2/3)+1/3 = (7/16)*(2/3)+1/3 = 5/8. So, P(T|S₃) = (3/4)³ * (2/3)/(5/8) = 9/20. The probability that the third answer is true is slightly under a half. That helps a little, but not much.

However, the fact that the troll changed his answer on the fourth question helps tremendously, since it shows that he’s dishonest but not a total liar. We now know that he either gave three truthful answers and one false one, or he gave three false answers and one truthful one. Let R be the event that going right is a truthful answer and A be the event that he gave three truthful answers and one false one or three false answers and one truthful one. We want to calculate P(R|A).

When the troll changed his answer, we became 90% certain that going right is the correct road to take, even though he advised us to go left the fourth time we asked.

What would happen if the troll never changed his answer? You could never be certain the troll is a liar based on that, but you would become more and more convinced he’s a liar if he kept giving you the same answer. How many gold coins would you need to become at least 95% sure that his answer is a lie? You would need to find n such that P(T|S_n) <= 5%. If n = 13, P(T|S₁₃) = 4.5%, so you’d need 13 gold coins.

Suppose there are two trolls at this fork in the road. One always tells the truth and the other always lies but you don’t know which is which. You have only one gold coin to pay one of them to answer a question. What question will you ask one of them to be 100% sure of which fork to take?

Here’s a much harder variation: You are lost in a country in which one third of the inhabitants always lie. The other inhabitants tell the truth, but only three quarters of the time. Desperate to get out of the country, you come to a fork in the road guarded by a troll. One way leads out and the other to certain death, but you don’t know which is which. You can ask the troll which way is out if you pay him a gold coin. You have four gold coins.

1. You pay the troll a gold coin and he advises you to go right. Does that answer help you to decide?

2. You pay the troll another gold coin and he advises you to go right again. Does that answer help you to decide?

3. You pay the troll another gold coin and he advises you to go right yet again. Does that answer help you to decide?

4. You pay the troll your last gold coin and he advises you to go left this time. You must make a final decision. What should you do?

5. What is the chance you get out of this country alive after you’ve paid your last gold coin?

Solutions can be found here.

In this post, we’ll see how the modern number system elegantly solves the problem (discussed in this post ) of how to write numbers as big as we please and how to calculate using pen and paper. The modern way we write numbers and do arithmetic is one of the most important inventions in the history of ideas. Indian mathematicians around the 7^th century AD saw that it was possible to write numbers using only ten symbols, a creative solution to a problem that had vexed mathematicians for thousands of years. We often think that mathematics is discovered, but how we chose to write down mathematical symbols was invented. There is no rule that says we must use ten symbols for numbers. Indian mathematicians invented that solution. When Arab mathematicians adopted the Indian numbers, they realized that they could easily write big numbers and that they could do arithmetic with pen and paper. That simple insight eventually led to an explosion of mathematical progress.

Modern numbers originated from the Indian number system

The Indian numerals, developed in the eight century A.D, were founded on two simple, but incredibly powerful ideas. First, you only need to group by ten using ten fundamental symbols, including the number zero. And second, you don’t need any symbols beyond these ten, since you can indicate other groupings such as the one hundred grouping or one thousand grouping by the position of each digit. So, you don’t need an L for 50 or a C for 100 as you would have in the Roman system. You also don’t need cuneiform symbols for one and ten as the Babylonians used for sexagesimals.

The Indian numbers, 1-9 and zero, originally looked like this.

Because these digits had to be hand copied into scholarly manuscripts, they would naturally change a little bit over time. Eventually, as they were copied over and over, they changed a bit after a few hundred years, with the changes indicated in red:

The horizontal line representing the number 1 eventually became vertical, probably because it saved space to write it that way. The other changes are very slight, and probably came about because it’s easier to make the symbol in one smooth motion without lifting the pen. So, the number for 2 likely came from drawing the top horizontal line from left to right, then moving the pen down diagonally to the left, and then making the bottom horizontal line. The modern three likely came about the same way: the three horizontal lines were drawn without lifting the pen. The numbers 4 and 5 are variations. Five came about by adding the curly cue to the bottom of the original 5, possibly as a flourish. The modern eight and nine seem to have developed from writing the number in one continuous circular motion.

During the Arab conquests, the Indian numerals were imported into Islamic science and mathematics and were used extensively by scholars. Arab mathematicians also developed the techniques that schoolchildren use today to do arithmetic by hand using paper and pen. But the Indian numbers never caught on with the general public in the Islamic world of the Middle Ages.

Europe eventually adopted the Indian-Arab number system

At the turn of the thirteenth century, the Indian number system was introduced into Italy by Leonardo of Pisa, also known as Fibonacci. Fibonacci was an Italian merchant and scholar who traveled extensively for business and for mathematical research. While visiting what is today Algeria, Fibonacci learned of the Indian numbering system and the arithmetic Arab mathematicians had developed. Fibonacci thought the Indian-Arab number system so superior to the Roman system and arithmetic with pen and paper so much better than the abacus that he resolved to replace the Roman numbers with the Indian-Arab numbers.

In 1202 AD, Fibonacci published a book, Liber Abaci, Latin for “The Book of Calculation,” although what it means more precisely is the book of calculation without the abacus. Fibonacci’s goal was to write a book that, while as logically rigorous as Euclid’s Elements, would also be an accessible and practical training manual for merchants and scholars. Fibonacci understood the value of marketing, even in mathematics, and so he filled the book with examples from the worlds of finance and trading.

Fibonacci began the first chapter with “The nine Indian figures are 9,8,7,6,5,4,3,2,1. With these 9 figures and sign 0, which the Arabs call zephyr, any number whatsoever is written, as is demonstrated below.” Fibonacci then explained how to do addition, subtraction, multiplication and division by hand, without the abacus, much as we learn to do it in elementary school today. The genius of the Indian system is that the positions of the numbers indicates whether they should be grouped by ones, tens, hundreds, thousands, or something bigger. Arithmetic now became possible without an abacus because we add, subtract, multiply, or divide numbers one position at a time. If we get a result that is too big or too small in the position we are focusing on, we carry the result over to the next position or borrow from the next position and then keep going. Because Arabic is read right to left, Arab arithmetic was done right to left, a custom we still maintain today.

For example, consider adding 45 and 76. 45 in the Indian-Arab notation means 4 tens plus 5 ones and 76 means 7 tens plus 6 ones. In the number 76, 6 is in the ones place and 7 is in the tens place. In the number 45, 5 is in the ones place and four is in the tens place. To add 45 and 76, we first add the 5 and 6 in the ones place. Moving right to left, we get 11, so we keep the 1 in the ones place and carry over ten, which means we carry over 1 ten into the tens place. In the second stage, we add the 1 ten carried over plus the 4 tens plus the 7 tens to get 12 tens. Then, we keep two tens from 12 in the tens place and carry over 1 hundred to the hundreds place. Overall, we have finished with 1 in the hundreds place, 2 in the tens place, and 1 in the ones place. Thus, 45 + 76 = 121. This simple calculation, that’s so easy using pen and paper, is very difficult to perform using Roman numbers, since there are no positions in the numbers indicating how counting should be grouped.

The Indian-Arab numbers introduced the concept of a ‘base’

Nowadays, we call this numbering system a base ten numbering system. The reason we use the term ‘base’ is that the Indian-Arab system is very flexible. It works with any grouping or ‘base,’ which is another word for number grouping. For example, in a base 2 system, there are two numbers, 0 and 1, and we indicate the value of the number by the positions of 0 or 1 in the number. In a base 2 number, reading from right to left, we have a ones place, a twos place, a fours place, an eights place, and so on. Each place is double the previous place. Base 2 numbers are also called binary numbers.

Base 2 or binary numbers

To illustrate how binary numbers work, let’s write 45 in base 2. It would be 101101 or 1 in the thirty-twos place plus 0 in the sixteens place plus 1 in the eights plus 1 in the fours place plus 0 in the twos place plus 1 one in the ones place, or 32 + 8 + 4 +1 = 45.

Arithmetic can be done in binary the same way as in base ten. Binary numbers are essential for computer arithmetic since numbers and calculations on computers are done in base 2. The beauty of Indian-Arab numbers is that we can use any base that is convenient. Computer scientists sometimes use base 16, which are called hexadecimal numbers. They work just like base 10 numbers, except they require 16 basic symbols, 0-9, and the letters a through f.

Indian-Arab numbers eventually replaced Roman numerals

Although Fibonacci’s book began to be used by Italian merchants almost immediately to train apprentices, it took 300 years for the Indian-Arab system to finally prevail over the Roman numbers. Many people resisted the new Indian-Arab numbers and the new arithmetic. Roman numbers had been used for over a thousand years without problems. Abacuses worked just fine. If it’s not broken, why fix it? Nevertheless, the advantages of the new numbers gradually won out and Roman numbers disappeared by and large by 1500 AD, although they are still used for limited purposes today, such as for numbering book chapters.

Decimals based on the Indian-Arab numbers replaced the sexagesimal fraction system

Despite his advocacy for a place-value numbering system using a base of ten, Fibonacci didn’t go to the final step to realize that fractions could also be written in base ten. In our modern numbering system, 0.345 is the same as , essentially equivalent to the sexagesimal system except that ten is used as the base rather than 60. Ironically, Fibonacci thought that if sexagesimal numbers weren’t broken, why fix them? Mathematicians from the time of the Greeks through the1500s in Europe still used them. The Egyptian astronomer Ptolemy as well as Copernicus were able to perform their calculations with acceptable accuracy using sexagesimals, so why change the system?

Just as the advantages of using the Indian base ten place-value system won out for numbers for everyday use, they also won out for fractions. The French mathematician Francoise Viete argued to use decimals exclusively by the latter part of the 16^th century. A little later, the mathematician Simon Stevin wrote a manual for how to do arithmetic with decimals, just as Fibonacci had done for natural numbers. People learned to do decimal calculations from Stevins’ book. By the beginning of the 17^th century the modern base ten numbering system was in place.

Some lessons from the history of numbers

Why did it take over 3000 years, from the time of the Babylonians to the end of the Middle Ages, for the modern numbering system to be developed? There are some important lessons here that we see over and over in the history of mathematics.

First, simple but incredibly powerful ideas, such as the base ten numbering system, were often overlooked by even the greatest of mathematicians. Archimedes missed it.

Second, even when these powerful concepts are discovered, they may not go anywhere on their own. Base ten numbers were developed in the eighth century in India but they didn’t instantly take over. Base ten numbers might have never taken over if the Arab conquests hadn’t brought the idea into the Arab world, where it moved further into Italy. But it took 800 years for that to happen!

Third, somebody needs to understand the significance of an idea, explain it, and push it forward. In this case, that somebody was Fibonacci. Today, to the extent that Fibonacci is known at all he’s generally recognized by mathematicians for a sequence of numbers he is famous for, the Fibonacci sequence. He’s also known among botanists and some gardeners since Fibonacci sequences show up in how plants form and develop. Fibonacci should be remembered for championing perhaps the most important innovation in numbers in history. But even Fibonacci didn’t see that fractions could also be included in a base ten numbering system, which took almost another 100 years to develop. This episode reminds us of another lesson we see over and over in the history of mathematics.

It’s very difficult to know in advance how important a mathematical discovery will turn out to be. Mathematicians usually work on a problem for its own sake. Sometimes the answer becomes vitally useful and sometimes it doesn’t. The proponents of Roman numbers in the late Middle Ages were wrong, but they weren’t terribly wrong. Base ten numbers were better, but they were not that much better than Roman numbers plus the abacus for most everyday purposes.

What the advocates of Roman numbers couldn’t see was that once numbers were severed from the abacus, new calculations that could never have been accomplished using Roman numerals and the abacus became possible. Fibonacci probably didn’t understand fully the profound significance of what he was advocating either. Base ten numbers catapulted mathematics dramatically forward in a startling way that most people at that time could not have foreseen.

Crick’s Comma Free Genetic Code: Crick suggested the "adaptor hypothesis," which posits that amino acids do not interact directly with mRNA but are carried by small molecules that recognize specific codons. (Today, the adaptor molecules would be called tRNAs.) The codons are then thought to be non-overlapping triplets of bases. This code removes the restriction in the Gamow schemes that allowed only certain combinations of amino acids to be adjacent. However, care must be taken to ensure that each triplet uniquely identifies one amino acid. For example, suppose that somewhere in an mRNA there is the partial sequence ... UGUCGUAAG.... (Note that in RNA uracil replaces the thymine of DNA, and so the code is written with U rather than T.) The intended reading frame is ... UGU, CGU, AAG..., but the RNA molecule has no spaces or commas to indicate codon boundaries. The sequence could equally well be read as ... UG, UCG, UAA, G ... or ... U, GUC, GUA, AG.... Each of these alternatives would have a different meaning. Furthermore, adaptor molecules that attached to the messenger RNA in different reading frames might interfere with one another and prevent any protein from being produced at all. He suggested that adaptor molecules exist for only a subset of the 64 codons, with the result that only that subset would be meaningful; the rest of the triplets would be "nonsense codons." The trick is to construct a code in such a way that when any two meaningful codons are put next to each other, the frame-shifted overlap codons are always nonsense. For example, if CGU and AAG are sense codons, then GUA and UAA must be nonsense, because they appear inside the concatenated sequence CGUAAG. Similarly, AGC and GCG are ruled out by the sequence AAGCGU. The codons AAA, CCC, GGG and UUU cannot appear in any comma-free code, since they cannot combine with themselves without generating reading-frame ambiguity. If all the out-of-frame triplets are nonsense, then there is only one possible way to read the message. A code with this property is said to be comma-free, since messages remain unambiguous even when words are run togetherwithoutcommasorspaces. Show that such a comma free code also codes for 20 amino acids. One way to solve this problem is to write a Matlab code for it.

To solve this problem, start by assuming that a particular codon is allowed. Thus, if ACT is a possible codon then to eliminate ambiguity, ACTACT should be uniquely readable as two ACTs. This eliminates CTA and TAC as possible codons because they would result in an ambiguity in the reading frame. Now choose a different possible codon – say CGT and eliminate ambiguous codons.... and so on. Note that not all starting choices will lead to 20 codons. They often end up with fewer than 20. The point is that the maximum number of codons starting with some sequence of choices will result in 20 codons

Solution: The first thing to notice is that AAA, CCC, GGG and TTT cannot be codons in the comma free code because they would each cause reading frame ambiguity (i.e. in a sequence AAAAAAAA, the reading frame is ambiguous). It is also easy to see that if XYZ is an allowed codon, then its cyclic permutations YZX, ZXY cannot be codons. For example, if both XYZ and YZX were allowed, then the sequence XYZXYZ would have a reading frame ambiguity: XYZXYZ and XYZXYZ. Thus, for any allowed codon, its three cyclic permutations are disallowed. Hence the 60 possible codons that remain after discarding AAA, CCC, GGG, TTT should divide into 20 distinct classes. In principle, it should be possible to choose one codon from each class to avoid ambiguity of the reading frame.

Although this seems plausible, the actual exercise of devising such a code is not trivial.

Below is a Matlab program that iteratively finds the codons. The nucleotides have been replaced by numbers: A==1, C==2, G==3, T==4.

When you run the code, it asks you for an input codon. The number of codons it finds which have no reading frame ambiguity (and so can be used in a comma-free-code) depends on the input codon. Starting with some of the codons does not produce twenty codons. For example, starting with 112 ends up finding only 9 codons:

[112,311,411,321,421,312,412,322,422]. However, starting with codon 123 generates a complete set:

C: [123,211,311,411,122,322,422,213,313,413,323,423,433,214,314,414,124,324,424,434]

Other comma free codes can be obtained by choosing different start codons for example, 231 and 312.

The second code below will check that your list of codons has no reading frame ambiguity. The code writes the output array C into a local file called C.csv.

Code to find the codons:

______________________________

% Code to find the codons for Crick's comma free code. The number found

% depends on starting codon (line 16) - the initial value 123 gives all 20

clc;clear all;close all;

C00 = zeros(1,60);

ic = 0;

for i=1:4

for j=1:4

for k=1:4

if abs(i-j)+abs(i-k)+abs(k-j)>0

ic = ic+1;

C00(ic) = i+j*10+k*100;

end

end

end

end

C(1)=input(' give initial codon \n')

for ij=1:20

C0=C00;

n = length(C);

iw = zeros(n,1);

nc = 0;

for i=1:60

for j=1:n

if C0(i)==C(j)

nc = nc+1;

iw(nc) = i;

end

end

end

C0(iw(:))=[];

% now try to add codons one by one and check

m = length(C0);

n=n+1;

err = 1;

kk = 0;

while (err == 1)

kk = kk+1;

if kk>m

C(n)=[];

n=n-1;

for ij=1:n

fprintf('Codon %i = %i \n',ij,C(ij))

end

fprintf('*********\n')

fprintf(' %i codons found \n',n)

break

end

jerr = 0;

C(n) = C0(kk); % add one more codon from C0

% check that this still works

for i=1:n

c3 = mod(C(i),10);

c2 = mod((C(i)-c3)/10,10);

c1 = (C(i)-c2*10-c3)/100;

for j=1:n

d3 = mod(C(j),10);

d2 = mod((C(j)-d3)/10,10);

d1 = (C(j)-d2*10-d3)/100;

I1 = d2*100+d3*10+c1;

I2 = d3*100+c1*10+c2;

I3 = c2*100+c3*10+d1;

I4 = c3*100+d1*10+d2;

for k=1:n

if (I1==C(k))||(I2==C(k))||(I3==C(k))||(I4==C(k))

jerr = 1;

% fprintf(' error %i %i %i %i %i \n',I1,I2,I3,I4,C(k))

end

end

end

end

if jerr == 0

err = 0;

end

end

end

csvwrite('./C.csv',C)

________________________

Code to check the codons:

clc;clear all;close all;

C = csvread('./C.csv');

n=length(C);ierr=0;

for i=1:n

c3 = mod(C(i),10);

c2 = mod((C(i)-c3)/10,10);

c1 = (C(i)-c2*10-c3)/100;

for j=1:n

d3 = mod(C(j),10);

d2 = mod((C(j)-d3)/10,10);

d1 = (C(j)-d2*10-d3)/100;

I1 = d2*100+d3*10+c1;

I2 = d3*100+c1*10+c2;

I3 = c2*100+c3*10+d1;

I4 = c3*100+d1*10+d2;

for k=1:n

if (I1==C(k))||(I2==C(k))

fprintf(' error 1 %i %i %i %i %i \n',I1,I2,I3,I4,C(k))

ierr=1;

end

if (I3==C(k))||(I4==C(k))

fprintf(' error 2 %i %i %i %i %i \n',I1,I2,I3,I4,C(k))

ierr=1;

end

end

end

end

if ierr==0

for ij=1:n

fprintf('Codon %i = %i \n',ij,C(ij))

end

fprintf(' codons have no reading frame ambiguity \n')

end

Gamow’s order free overlapping code: Gamow suggested yet another overlapping-triplet code with an even simpler description: Each codon is defined entirely by its base composition, ignoring the order of the bases within the codon. In other words, any reordering of the three nucleotides does not change the codon. Thus ACT, ATC, CAT, CTA, TAC and TCA are equivalent and specify the same amino acid. Show that the number of codon families in this scheme again turns out to be exactly 20.

Solution: If the codon is defined only by its nucleotide base composition, then with 4 bases A, C, G, T, one possible combination are the ones in the set, where we eliminate all permutations for each codon.

S = {AAA, CCC, GGG, TTT, AAC, AAG, AAT, CCA, CCG, CCT, GGA,

GGC, GGT, TTA, TTC, TTG, ACG, ACT, AGT, CGT}.

We see that the set S has exactly 20 members. A permutation of nucleotides within each of these codons represents the same amino acid.