Is there an objective, ideal chance, or is it the result of our ignorance?
In September, several problems were published , with the help of which we studied random processes in everyday objects - locks for bicycles or puzzles. Let us now look at solutions to these problems.
Riddle 1: random combinations
The task was as follows:
Consider a simple code lock for a bicycle, similar to the above image. It has three rotating disks, each of which shows 10 digits in order. When these three discs are rotated so as to give the desired combination - 924 - the lock opens. When you want to close it, you need to mix the numbers so that they are far away from the given combination. But what does “far” mean in this context? If you move the disk as much as possible by 5 positions, you will set the number 479. However, it will be easy for an attacker to accidentally stumble upon this position if he simply rotates all five disks at the same time and sees if the lock opens. Imagine that a cracker has time to test five different combinations. In each case, our potential thief tries our castle after one of the following actions (and in case of failure, returns the castle to its original configuration):
- Rotate one drive at a random number of positions.
- Rotate two discs simultaneously at a random number of positions.
- Rotate all three disks at the same time at a random number of positions.
- Rotate two discs at different angles.
- Rotate all three discs differently.
Our riddle is the following: if the lock unlock code is 924, which set of mixed numbers will be the most stable for random attempts to open the lock, and how many such sets exist? What is the probability of detecting code?
The first formulation of the problem turned out to be somewhat ambiguous, because at first I did not indicate that after each step the thief turns the lock to its original position. One of the readers analyzed this problem, provided that the "random number" in the first three cases is not equal to zero, and the "different" rotation angles in options 4 and 5 are not necessarily equal. However, another reader pointed out that if you accept the last assumption, and turn the lock disks so that two disks are rotated by one angle, and the third by the other - as, for example, in combination 036 - then the thief will not be able to open the lock, since none of the options does not work out such a combination.
The solution to the problem takes into account that in steps 4 and 5, the disks can be rotated at different angles. We also assume that in the first three options the thief can turn the selected discs a full turn, i.e. by 10 (or 0) digits, and return them to their original state. Having specified this, we calculate the probability of each of the actions of the thief. Note that any action taken by a thief to obtain a certain combination is potentially reversible - for this you need to perform a reverse rotation that complements the first and has the same probability. Therefore, the probability that a random rotation of the left disk will lead us from combination 924 to combination 624 is 1 out of 10 chance - as is the probability that a random rotation will lead us back from 624 to 924. And this is true regardless of whether we rotate we accidentally have one drive, two or three. Therefore, in order to calculate how many combinations a thief will need to sort out in order to select the desired one, if he performs a certain action, we can start with our given combination 924 and then calculate how many three-digit combinations we can get from it.
- Starting from number 924, and turning one dial, you can get three-digit combinations of the form x24, 9x4 and 92x, where x is any of 10 digits. There are 10 such combinations each. However, including the same combination of 924 in the second and third options would be unnecessary, so in reality we get 10 + 9 + 9 = 28 different combinations. And if we accidentally turned the numbers of the castle to close it, we got one of these 28 combinations, then the thief will have a 1/28 chance to open the castle.
- Turning the two disks together gives us possible combinations of the form 9 ##, # 2 # and ## 4, where the # signs indicate the difference between the digits of the resulting combination and the initial digits (and this difference will be the same for both disks). There are also 10 pieces of each, and excluding 924 from the second and third forms, we also get 28 combinations and a 1/28 chance of success.
- The rotation of all three discs allows you to get 10 combinations - 035,146, 257, 368, 479, 580, 691, 702, 813 and 924 - and a 1/10 chance of success.
- A random rotation of two discs, not necessarily at the same angles, gives access to all combinations starting with 9 (from 900 and above), all combinations with 2 in the middle, and all combinations ending with 4. Each of the types can be 100 pieces. However, in 9xx combinations, 10 combinations have already been counted, ending in 4, and 10 variants of the x2x combination; in addition, nine other combinations ending in 4 have already been counted in x2x combinations. Therefore, the total number of combinations will be 300 - 10 - 19 = 271 for this step, and it will have a 1/271 chance of success.
- Rotating all three discs at a random angle gives us all three-digit combinations, and a 1/1000 chance of success.
We have two sets of "safe" numbers, the most resistant to hacking attempts. They cannot be obtained by the first four methods, but you can stumble only in the fifth method, where the probability of success is 1/1000. The first persistent combination can be obtained by turning each of the three discs at a different angle so that none of them remains in its original position. Such positions will be 9 Ă— 8 Ă— 7 = 504. Another set of stable combinations can be obtained by turning two disks by one nonzero angle, and the third one by another nonzero angle. This is another 3 x 9 x 8 = 216 combinations, and a total of 720 is obtained. Therefore, 720 combinations are safer than the others.
Riddle 2: from randomness to order in riddles
The task was as follows:
Suppose we solve a puzzle consisting of hexagonal pieces - like honeycombs. The picture of the puzzle is a winding vine. Since the pattern is repeating and self-similar, it cannot be guaranteed that two adjacent pieces fit physically to each other, even if they fit the picture. Suppose that three others can go to each edge of a given piece. Therefore, when two pieces fit each other in the picture, the probability that they fit physically will be 33.33%. However, if you can find another piece that fits both of these, that is, one that has one common edge with each of these two, then your confidence in success will increase. Let's try to evaluate how much it grows.
- You have found three pieces that seem to fit together at first glance, without the obvious displacement of the liana pattern on their adjoining edges. What is the measure of your confidence in the correct selection of pieces?
- You have found a central hexagonal piece, and six surrounding it, and in the picture they seem to coincide. What is the measure of your confidence in the correct selection of pieces?
The larger the groups of pieces become, the stronger your confidence in the correct assembly. It is reasonable to assume that three isolated groups, in which there are a total of seven connected pieces, are not comparable with the only surrounded hexagon described above.
The third part of this puzzle has corrections, and is an attempt to quantify the above difference. Is it possible to come up with a measure of the degree of completion of a partially solved puzzle? This method should allow you to assign a number from 0 to 100 to any partially assembled puzzle of 10x10 hexagons. This number should indicate the degree of completion, roughly correlating with the proportion of the current state of the puzzle in relation to the finished version.
The reader answered the first two questions as follows:
- For three pieces arranged in a triangle, the answer will be p = (2/3) 3 , since there are three faces that can be removed, and the probability of deleting each of them is 2/3. This gives us 1 - p = 0.7037, that is, confidence in 70.37%.
- Six pieces may not coincide 6 + 6 = 12 faces, which gives us 1 - p = 1 - (2/3) 12 = 0.9923 or a confidence of 99.23%.
Using such confidence data, we can choose a simple metric based on the sum of the confidence values ​​for the finished parts of the puzzle so that the fully completed puzzle gives 100% confidence. It is done like this. Take all completed groups of two or more connected pieces. Add up the amount of confidence for each of the individual pieces. That is, for a group of three pieces with a common vertex, we get 3 × 0.7037 = 2.11%, and for a full hexagon we get 7 × 0.9923 = 6.95%. A partially completed puzzle of three groups of three pieces and one hexagon will give you 6.95 + 2.11 + 2.11 + 2.11, or 13.3%. On the other hand, if you have two full hexagons, your total will be 6.95 + 6.95 = 13.9%, even though in this case you used two pieces less.
The reader developed this idea further , and proposed a measure that uses logarithms and is associated with the concept of entropy - a natural measure of disorder and randomness. Its measure for a 10 Ă— 10 grid is n - 100 Ă— (log m) / (log 100), where m is the number of alternative layouts, and n is the total number of pieces placed on the field.
Riddle 3: is perfect coincidence possible?
Today, the prevailing opinion is that quantum physics is based on inherent nature, objective and ideal randomness. I encouraged readers to share their views on this philosophical riddle by joining either the Einstein (E) team or the Bohr (B) team. Team B accepts the objective randomness of the quantum world, and Team E considers physical randomness a logical impossibility, revealing our ignorance of deterministic casual phenomena occurring on subplank scales. Readers' voices were divided approximately equally [as in our vote / approx. transl.].
A reader with the nickname RRG described my motivation to offer such a discussion:
In quantum mechanics, if we consider the standard two-gap experiment , we cannot predict in which place a particular particle will appear on the screen, but we can predict the probability of its getting to one or another place. And these probabilities can be extremely accurate and reliable. This reliability and accuracy of probabilities is a clear sign of the presence of some kind of hidden process.
What is happening is similar to thermodynamics. We can very accurately measure the temperature in a room without knowing what exactly each of the air molecules does. Like probabilities in quantum physics, temperature manifests itself on the basis of a deeper physical level.
That is how I reasoned! Why does a certain particle passing through a double slit hit, say, the upper left part of the screen, and not the lower right? A certain causal chain (possibly mass-energy fluctuations at the level of quantum gravity) should have led to the choice of a certain place in a certain case. If so, then quantum randomness is not an ideal, objective and magical part of the Universe, but a consequence of our ignorance of the principles of physics underlying it - just like a classical randomness.
As the reader wrote Mark Thomas, the probability space defined by the Planck mass-energy can be huge. It can be large enough to achieve indicators close to perfect randomness in the Kolmogorov sense (thanks to another reader for a link with explanations regarding Kolmogorov complexity and randomness). But in this case, the Schrödinger equation will be an approximation, and it cannot be interpreted as something untouchable, and cannot be used as the basis for the now popular “multi-world interpretation” based on considerations of mathematical simplicity. The latter approach is advocated by physicist Sean Carroll .
Reader Rob McChern commented on this passage of mine: “If you knew all the forces acting on a flipped coin or a dice, if you had sufficient computing power, you could predict the result” as follows:
This statement is incorrect. You also need to know all the initial conditions associated with this experiment. And here lies the problem. In any difficult situation, the information content of the initial conditions is much larger than the information content of all the forces or laws of nature. Accordingly, it is much more difficult (and often even impossible in principle) to get all the necessary information about the initial conditions than to get an accurate knowledge of all laws.
I agree that an ideal knowledge of the initial conditions cannot be obtained with infinite accuracy. But I think that most physicists will agree that it is possible to obtain knowledge regarding the tossing of a coin in a room with sufficient accuracy and to predict the result in most cases. Of course, this will not be possible if a hurricane suddenly flies into the window and organizes chaos. It is possible that the aforementioned fluctuations of mass-energy on the Planck scales are the hurricanes that constantly wreak havoc, which is the true cause of quantum randomness. But even in this case, in principle, a causal chain must exist. Team E will simply say that we do not know all the details.
Reader Abhinav Deshpande gave a beautiful, balanced, comprehensive and evidence-backed description of the current state of affairs in this area, as well as links to very interesting articles. He correctly states: “I do not think that the founder of the theory of relativity was well disposed towards nonlocality (even if nonlocality does not permit the transmission of information faster than light).” But we must remember that Bell's theorem was proved ten years after Einstein’s death. And in the face of convincing experimental evidence of Bell's inequalities, Team E had no choice but to tweak Einstein's initial opinion and accept the fact of non-locality and “frightening long-range action”. This means that the existence of superluminal or superspace connections between the components of an entangled quantum object is possible, even if the external transmission of information is limited by the speed of light according to the theory of relativity, and nonlocality never gives a visible leak.
Somehow I came across such a vivid picture: imagine a lake with an opaque surface. A huge wooden elephant upside down floats in it, almost the size of the whole lake, and its legs stick out outside in the four corners of the lake like columns, and its body is hidden under water and is not visible. First you can decide that the four columns are independent objects. However, then you see that their movements are perfectly correlated with each other - they are confused. In the same way, entangled particles form a single entity that can extend to the entire Universe, and its internal connections can be superlight or superspace. An interesting idea connected with this, known as ER = EPR, is a mysterious hypothesis put forward by brilliant theoretical physicists, Juan Maldasena and Leonard Sasskind . The idea is that entangled particles (EPR) are connected by a wormhole, the Einstein-Rosen Bridge (ER). Initially, it was proposed in the context of studying black holes, but perhaps it works for all entangled particles. As Bohm’s theory shows, determinism and quantum mechanics can coexist and deny locality with internal superluminal connections, without the need for objective randomness.