Make: More Electronics (2014)
Chapter 36. Experiment 36: The One-Person Paranormal Paradigm
Here’s the plan. There will be a single LED behind a screen. An electronic circuit will switch the LED on or off, and will then prompt the player to guess which state it is in—by using his psychic powers, if he is fortunate enough to possess them.
The player will press a righthand button if he thinks the LED is on, or a lefthand button if he thinks the LED is off. The circuit will tell him whether the guess was correct or incorrect, and the cycle will repeat.
I think we now have all the necessary knowledge to build this circuit. What we don’t have is a way to assess the results. How many correct guesses do you think a player needs to make, compared with his incorrect guesses, to achieve a result which seems too unlikely to be merely a matter of chance? I will explain how this can be assessed, after the circuit has been constructed.
The Last Logic Diagrams
In the previous experiment, I included a logic diagram for the XNOR Randomizer, wired to deliver a 0 or 1 output. I have adjusted this slightly, as shown in Figure 36-1, to make it compatible with the Telepathy Test circuit that I have in mind.
Figure 36-1. The logic diagram for part 1 of the single-person Telepathy Test. The input and outputs labelled A, B, C, and N will connect with part 2.
§ I’ve substituted a seven-input NAND for the seven-input AND gate. Instead of a normally low output that goes high when it detects the disallowed state of 1111111, it has a normally high output that will go low. This will be easier to use in the next part of the logic.
§ I added an XNOR gate beside the binary output. When the output is high, the XNOR output will be low, and vice versa. In other words, it functions like an inverter. So why didn’t I use an inverter? Because I had one XNOR gate to spare in the quad two-input chip that will process feedback for the shift register. Might as well use the spare gate instead of adding an extra inverter chip.
Figure 36-2 shows the schematic for part 1 of the circuit. It’s very similar to the test schematic in the previous experiment. The most visible change is the addition of the NAND gate. The input and outputs identified by letters A, B, C, and N will connect with part 2 of the circuit.
Figure 36-2. Part 1 of the schematic for the single-person Telepathy Test.
Looking at Part 2
Part 2 of the logic diagram is shown in Figure 36-3. In this diagram, timers are colored pink for easy identification, and there are four input/output lines labelled to connect with the ones in Figure 36-1.
Figure 36-3. Part 2 of the logic diagram for the single-person Telepathy Test will process user input, provide feedback, and seed the linear-feedback shift register.
The function of part 2 can be summarized like this:
1. The player sees a “ready” signal.
2. The player presses button A if she thinks the hidden LED is on, or button B if she thinks it’s off.
3. The “ready” signal goes off for about a second. During that interval, the circuit gives a “hit” or “miss” indication, and increments a “hit” counter or a “miss” counter.
4. The “ready” signal comes on again, and the player repeats from step 2.
Behind the scenes, it’s going to be a little more complicated.
Since accuracy is essential, there must be no risk of contact bounce in the input buttons. Therefore, I am routing each input through a one-second timer in the same configuration that I have used before, to create a clean pulse that is not retriggered if someone’s finger holds down a button for too long.
The outputs from the one-second timers must now be compared with the outputs from the previous logic diagram. The shift-register output is on the righthand side and will be high when the hidden LED is on. The XNOR output is on the lefthand side will be high when the hidden LED is off.
There are two ways to make a correct guess:
§ The hidden LED is on, AND the player presses the “LED is on” Guess button
§ OR the hidden LED is off, AND the player presses the “LED is off” Guess button
Likewise there are two ways to make an incorrect guess:
§ The hidden LED is on, AND the player presses the “LED is off” Guess button
§ OR the hidden LED is off, AND the player presses the “LED is on” Guess button
The two pairs of AND gates, each feeding into an OR gate, take care of this logic, in much the same way that I used two pairs of ANDs, each feeding into an OR gate, in the original two-player version of the game.
Two LEDs (red and green circles) will tell the player immediately whether her guess is right or wrong, and two (optional) counters can log the number of hits and misses. I’ll deal with the counters later.
The rest of the circuit waits until either the input timer on the left OR the input timer on the right is triggered. Therefore, the timer outputs are linked in an OR gate at the center. The output from this OR has to do a couple of things.
The Ready Signal
The OR output is tapped by a wire that runs to the left, where it goes into another OR gate, which controls the ready signal. What’s this all about? Well, I want the ready signal to be normally on, but suppressed under two circumstances:
§ For one second after the player has pressed a Guess button
§ OR during the initial “random seeding” of the shift register, when the circuit is not yet ready
By connecting the yellow ready-signal LED to the positive side of the power supply, as shown, it will satisfy these conditions.
The rest of the time, the lefthand OR gate will have two low inputs, so the ready signal will light up as it sinks current into the gate. Remember, a 74HC00 series logic chip can sink as much current (20mA maximum) as it can source. You can use it either way.
I could have used a NOR gate in this location, with the ready-signal LED conventionally grounded, but that would mean using one more chip just to get the NOR. Since I had one OR gate to spare, I took advantage of it.
In Experiment 35 I mentioned the need to random seed the linear shift register (see Chapter 35). I’m choosing to have the player do this manually so that only one fast timer is needed instead of a slow one driving a fast one. When the Random-seed button at bottom-left is not being pressed, it applies a low state to the reset pin of the fast astable timer. Remember that a low state on the reset pin stops the timer. When the Random-seed button is pressed, its high state frees the timer to run.
The output from the fast timer passes through an XOR gate at bottom-right and sends pulses back to the clock input of the shift register.
Before beginning a new trial, a player must remember to push the Random-seed button for an arbitrary interval. Since this is a serious tool for paranormal research (!) I’m going to trust the player to remember the random-seed ritual.
Two More XORs
Now for the tricky part. Going back to the OR gate in the center, its output is connected through a capacitor to an XOR gate below it. The capacitor has a much smaller value than any of those which we have used in previous experiments, because a logic chip has such a sensitive input. In my experience, a 68pF capacitor will deliver a pulse that is short enough to trigger the XOR gate without creating multiple “false positives.” If you test the circuit and find that it is behaving unpredictably, you can try a higher or lower value for the capacitor. I included 47pF and 100pF alternatives in the shopping lists, to help you in this respect.
§ Remember that 1,000pF = 1nF.
The righthand input of the XOR gate is normally high, because of the pullup resistor, and the left-hand input of the AND gate is also normally high, because it is connected with the seven-input NAND gate in the previous section of the circuit, and the NAND has a normally high input, until it detects the disallowed state of 1111111.
When an XOR gate has two high inputs, it creates a low output. Therefore, the output of the first XOR gate is normally low.
The output feeds into the second XOR gate, which has a low left input while the astable timer is not running. So, the second XOR gate has two low inputs, which cause it to have a low output, which connects with the clock of the shift register.
What happens when the player presses a button? Nothing, until the end of the one-second timer pulse. When the pulse ends, the output from a one-second timer goes from high to low. The 68pF capacitor passes that transition as a brief low pulse.
During that brief pulse, the first XOR gate has a low input on the right, while it still has a high input on the left. Consequently, its output goes high. This causes the second XOR gate to have one high input and one low input, so its output goes high and advances the clock of the shift register.
The low pulse from the capacitor is very brief. When it ends, the pullup resistor takes over again, and the two XOR gates return to their initial state. The shift register has been clocked to create a new value, and the ready signal lights up, inviting the player to make another guess.
This is relatively simple, but why do I need two XOR gates to deal with it? Because there are actually three situations where the shift register has to be clocked:
1. When the player has pressed a button. I just described how that will advance the shift register to its next value.
2. When the fast-running timer is cycling the shift register initially.
3. When the shift register reaches the disallowed state of 11111110.
The two XOR gates deal with all the situations where these events can happen simultaneously, or almost simultaneously. This is the difficult part of the circuit to understand.
Timing is Everything
Figure 36-4 shows how the gates will work under normal circumstances when the player presses a button that causes one of the one-second timers to send a pulse. At the end of the pulse, as the timer output goes low, the coupling capacitor passes the transition along, and the XOR logic sends a high pulse to the shift register to advance it to the next state.
Figure 36-4. In response to a low transition on the coupling capacitor, the dual XOR gates send a high clock pulse to the shift register.
But what if the shift register advances to 11111110, the disallowed state? Figure 36-5 shows the sequence of events. The NAND gate in part 1 of the circuit responds to the seven high inputs by taking its output low. But the low pulse from the coupling capacitor hasn’t ended yet, so the clock output goes low.
Figure 36-5. The dual XOR gates process a low signal from the NAND gate in part 1 of the circuit, indicating that the shift register has reached the 11111110 disallowed state.
As soon as the low pulse ends, the pullup resistor on the first XOR gate takes its input high. This causes a high clock output, advancing the shift register to the next state.
The NAND output returns to its high state, because the new state is not disallowed. The clock output goes low, and the circuit is back in equilibrium.
A similar series of events occurs when the shift register is being seeded by the fast astable timer. While the timer is running, if it advances the shift register to the disallowed state, the pair of XOR gates cause the shift register to skip past it. This will happen even if the timer stops when the disallowed state is reached. The shift register will still move on past it.
The system only fails if two events events happen almost simultaneously. Even in this case, the worst that can occur is that the circuit will lodge in the disallowed state until the player presses a button. This will trigger the shift register, because any change in the XOR gates toggles the output to the clock.
§ When feedback is introduced to a logical system, the consequences can be complicated. In a computer, a system clock makes sure that all the chips stay in sync with each other (more or less), and this helps to prevent the kinds of problems that I have been describing. The computer system is “synchronous.”
§ The single-person Telepathy Test is asynchronous, with all the interesting consequences that this entails.
Making Every Guess Count
How are you going to evaluate your success rate in this experiment? If you don’t feel like keeping score manually, and you do have a microcontroller, you can apply the output from each successful guess to one input pin and from each unsuccessful guess to another input pin and run a little program that counts the inputs and displays a running total on an LCD screen.
What if you don’t have a microcontroller, but you want to total your guesses automatically? You could buy yourself a couple of off-the-shelf “event counters.” Just go to eBay and search for “digital counter” or “digital totalizer.” These useful little devices cost maybe $8 each from suppliers in China. You should find that they will make provision for a wide variety of power supplies and inputs. All you need to do is link the counter with your circuit via a common ground and attach the output from your circuit to the counter input through a capacitor (to prevent any flow of DC current).
On the other hand—making marks on paper is not such a bad option. While you’re sitting at a desk, pressing one button or the other, it’s easy to use a pen with your free hand.
Schematic Part 2
Figure 36-6 shows how to put together components that will perform the logical operations in Figure 36-3. You’ll notice that I had to squeeze the components together to fit them into the available space. Still, there are only six chips, which should easily fit on a breadboard, and the connections are relatively simple.
Remember that each of the red, yellow, green, and blue circles represents an LED, and be sure that the yellow one, which is the Ready prompt, goes between an OR gate and the positive side of the power supply, not the negative side. The other three LEDs are conventionally grounded. Naturally you should add series resistors if your LEDs require them.
Points A, B, C, and N are to be connected with points A, B, C, and N in part 1 of the circuit, on an adjacent breadboard.
Figure 36-7 and Figure 36-8 show the breadboarded versions of the circuit. Additional LEDs were included during the testing process.
Figure 36-6. The second part of the schematic for the single-person Telepathy Test.
Figure 36-7. Part 1 of the single-person Telepathy Test, breadboarded.
Figure 36-8. Part 2 of the single-person Telepathy Test, breadboarded.
Testing the Tester
Initially, slow down the fast-running asynchronous timer at the bottom of the second breadboard by using a 10µF capacitor on its pin 6. Also, add low-current LEDs to the outputs of the shift register on the first breadboard. If you use LEDs containing their own series resistors designed for a 12VDC supply voltage, they will draw so little current that the shift register should still communicate successfully with the NAND and XNOR chips.
Hold down the Random-seed button, and you will see the LEDs attached to the shift register cycling through the linear-feedback sequence that is now familiar to you. If this doesn’t happen, you made a wiring error somewhere.
Test the “Guess LED is OFF” pushbutton and “Guess LED is ON” pushbutton, and if each of them generates a pulse lasting about one second, you’re ready for the really interesting part of the test, which is to see if the circuit skips the disallowed state.
When you first apply power, the 100µF capacitor on the first breadboard should prevent fluctuations from loading random values into the shift register. The only problem is that if you disconnect power and then reconnect it again, the capacitor may have kept the chips energized sufficiently for them to retain their previous states. Eventually the stored charge will leak away, but this may take a while. To speed up the process, try shorting out the 100µF capacitor (but make absolutely certain the power is disconnected before you do this).
Assuming that you start with the shift register containing eight low states, you can press the Random-seed button quickly, twenty-four times (still with the 10µF capacitor making it run slowly) to get it to the state immediately preceding the one that is disallowed. (SeeFigure 35-9 to refresh your memory.)
Now if you press the Random-seed button, or either of the Guess buttons, you should find that the shift register skips the 26th state (the disallowed one) and goes straight to the 27th state. This works reliably on the version of the circuit that I built. If you have any problems, your first step would be to try different values for the 68pF capacitor.
Once you have the circuit working, you come to the interesting part: interpreting the results.
How Unlikely Is ESP?
Suppose you run a trial consisting of a thousand guesses. On average, around 500 of them should be correct. But let’s suppose you get 510 right, or 520, or 530. How far from the median do you have to go, for the result to look as if it is not just a matter of chance?
This is a complicated question, so I’ll start by simplifying it. If you make just four guesses in succession, and we use letter Y to represent a correct guess and N to represent a wrong guess, there can be sixteen possible sequences, all of which are equally likely: NNNN, NNNY, NNYN, NNYY, NYNN, NYNY, NYYN, NYYY, YNNN, YNNY, YNYN, YNYY, YYNN, YYNY, YYYN, YYYY.
We don’t care what sequence your guesses are in. We only care about the total number that are correct. This means we need to group the correct and incorrect guesses, regardless of their sequence. Like this:
§ NNNN : 1 way to get 0 guesses correct.
§ NNNY, NNYN, NYNN, YNNN : 4 ways to get 1 guess correct.
§ NNYY, NYYN, YYNN, NYNY, YNYN, YNNY : 6 ways to get 2 guesses correct.
§ YYYN, YYNY, YNYY, NYYY : 4 ways to get 3 guesses correct.
§ YYYY : 1 way to get 4 guesses correct.
Because there are four ways of getting three guesses correct, but the total number of ways to make guesses is sixteen, the chance of making three correct guesses out of four (in any sequence) is 4/16, or 25%.
But wait—what if you get all four out of four guesses correct? That’s even better! So really we should rephrase the question. What are your chances of getting three guesses correct, or even more?
Because we have added another alternative, the chance actually improves to 5/16, or a little over 31%.
If you try to extend this system to calculate the odds for five, six, or more guesses, you find that the number of Y/N combinations gets very big, very quickly. There’s a way to look up the number of combinations, though. You can see it in Figure 36-9. Here’s how to use this array of numbers:
Figure 36-9. Pascal’s Triangle can be used to figure the odds of making any number of correct guesses in a series where a correct and incorrect guess are equally likely.
The second number in each row is the number of guesses in a trial. For example, the bottom row describes a trial where you make eight guesses.
The numbers in each row tell you how many ways there are (that is, the number of permutations) to make correct guesses, beginning with zero correct guesses (only one way to do this, represented by number 1, at the left end of a row) and ending with all correct guesses (only one way to do this, represented by number 1, at the right end of a row). Between these extremes, the numbers in a row in the triangle tell you how many ways there are to make 1, 2, 3 … n correct guesses, where n is the total number of guesses (right or wrong). So, for example, the bottom row shows that if you allow yourself eight guesses, and you want to know how many ways there are to get four of them right, the answer is 70. Just count along the bottom row like this: 0, 1, 2, 3, 4 … and you get to number 70:
§ Remember, the first number in any row (always 1) is the number of ways to get no guesses correct. The second number is the number of ways of getting one correct guess—and so on, ending with the number of ways to get all the guesses correct (always 1).
§ The white number in the righthand column is the total of all the black numbers in that row of the triangle. In other words, it is the total number of different sequences of right and wrong guesses. Notice that this number doubles with each new row of the triangle.
Now you can calculate the odds. Going back to the previous example, if you want to know the odds of making four correct guesses out of eight, you take 70 as the number of ways to make those four correct guesses (in any sequence), and 256 as the total number of permutations, so your odds of getting exactly four correct guesses are 70/256.
But what if you want to know the odds for a range of guesses? In the bottom row, in a series of eight guesses, what are the odds of getting six right, or more? you would add 28 + 8 + 1 to get 37/256. That’s about 14%. You would only expect this to happen, by pure chance, about one time in seven.
Powers of the Triangle
If you have studied much mathematics, you will recognize Figure 36-9 as Pascal’s Triangle. I don’t have room to go into it in detail here, but the interesting aspect of it is that every number in the triangle (other than 1) can be found by adding the two numbers immediately above it.
In theory, if you want a trial in which you make 1,000 guesses, you just need to extend Pascal’s Triangle downward until you get to a line where the second number is 1,000. Now you will see the odds for making any number of correct guesses, from none out of 1,000 to all out of 1,000.
The only problem is, the numbers in Pascal’s Triangle get very large, very quickly. Even a typical computer language is inadequate to handle them. Suppose you have a language that can handle quad integers, meaning whole numbers expressed with 32 binary digits. In decimal notation, a quad integer can have a value of (plus or minus) more than two trillion. But this is only enough to calculate the first 32 lines of Pascal’s Triangle. The size of the numbers in a triangle containing 1,000 lines will present a challenge.
John Walker’s Probabilities
Fortunately, I don’t need to imagine it, because a smart guy named John Walker has already done it for me. Moreover, he has put the results online.
Walker was the founder of Autodesk, which sold the first serious computer-aided design software using MS-DOS. He also happens to have an interest in paranormal phenomena, and pursues that interest in his spare time.
On the page that he created at http://www.fourmilab.ch/rpkp/experiments/bincentre.html, you will find a probability table showing how likely it is to make various numbers of correct guesses out of a total of 1,024. For instance, the odds of getting exactly half of the guesses right (512 out of 1,024) is around 2.5%.
Why isn’t that number higher? Because you are almost as likely to make 511 or 513 correct guesses. What really matters, once again, is the range.
For instance, what are the odds of getting 562 correct guesses—or more? That is, 50 more than the median—or better? Walker anticipated the need to answer this kind of question, so his table shows the cumulative odds, meaning the chance of a certain number of guesses—or better. For 562 or more correct guesses, he lists the chance as 0.000981032. To change this to a percentage, just multiply by 100, and you see that the odds are about 0.098%. This means you could only expect to achieve it on 1 out of 1,019 trials (according to the table).
Let’s suppose that actually happened. Should you think that you must have some paranormal powers? Hmmm, I don’t know about that. Achieving a one-in-a-thousand success is unusual, but by definition, just by chance alone, it will happen about one time in a thousand! By the same logic, I would have to think that if someone happens to win a good amount of money in a casino, he must be psychic, too.
However, the odds in Walker’s table diminish very quickly as the number of correct guesses increases. For instance, it shows that if you achieved 600 or more correct guesses out of 1,024, the odds of that happening would be 1 in 47,491,007. In other words, you’d have to make almost 50 million trials to expect one in which you made 600 or more correct guesses, just by chance alone. If you can achieve this, I’d be impressed—although, you know, 1 in 50 million is probably about the same as the odds of winning a lottery jackpot. Does that mean I should conclude that jackpot winners are all psychic?
You see the problem. After all the trouble we took to establish an evenly weighted random-number generator, it’s still difficult to know how seriously to take any deviation from the norm. Even if you made all of the 1,024 guesses correctly, that could still be a matter of chance—although it is vanishingly unlikely.
Still, testing yourself can be fun. Perhaps 1,024 sounds like a large number, but if you make a guess every three or four seconds, the whole process should only take about an hour. You may never be able to prove beyond doubt that you have psychic powers—but on the other hand, what if your score is about average? That would still be a useful result, because it is a strong indication that you don’t have paranormal abilities!
This raises an interesting question. Which would you prefer to be: psychic, or not psychic?
After thinking about this, I’ve decided that I would rather live in a world where paranormal powers don’t exist. This is because I prefer to believe that there is a rational explanation for everything.
I’m a big fan of rationality. After all, rational thought processes were behind every valid theory in the history of science. Rational investigations confirmed all those theories, and the ultimate rational discipline of mathematics enabled engineers to apply them. Every bridge, building, car, aircraft, spacecraft, and computer throughout modern history has relied on mathematics for its construction.
When I look at the amazing and wonderful achievements that have been realized through the rational capabilities of the brain, I tend to think that even if some psychic powers exist, they would not be quite so impressive after all.