Olfaction

Like know you these plastic packs you crack and they get cold? I read that it is often this spontaneous endothermic reaction taking place. NH4NO3 (s) –> NH4+ (aq) + NO3- (aq) So, with well-suited electrochemical chamber, I should be able to create current until the reaction reaches equilibrium? Oh wait, because of ΔG° = ΔH° - TΔS° , this reaction will only be spontaneous at high temperatures. So isn’t it simply, that if we create a cold spot in a warmer environment, there is going to be a flow of molecule to fill the hole and therefore kinetic energy?

Oh thanks, that’s the kind of answer I was expecting. That would explain why the Nernst-Equation allows me to convert redox potential into Gibbs Energy and not into Enthalpy… ΔG = -z F ΔE Where G ist the molar Gibbs Free Energy, z the number of exchanged electrons and E ist the standard redox potential…

But why is it called Gibb's free energy, if there is no way to use it? Is there maybe a way to create some gradient and do some diffusion that could be harnessed into work, altough no heat is being produced?

Hello, could someone please help me with this? Some reactions are spontaneous (with dG < 0) although they are endothermic (dH > 0). My question is: Is it possible to extract work from such a reaction? The process I understand to produce work is to heat something in a contained space and transfer the heat pressure into kinetic energy with a turbine. (Sorry I know this is not very formal). For example, this reaction is a spontaneous endothermic reaction. NH4NO3 (s) –> NH4+ (aq) + NO3- (aq) Is there a way I can get my turbine to work with this reaction? Is the Gibbs Free Energy that comes from the raise in Entropy of any use? And how so, if I get not heat? Thank you for you help. I am quite confused.

No my point is that ''TMC'' in my exemple has nothing to do with halting at all. '' We define a function C(n,m) that outputs 1 if the n-th TM outputs a 3 on input m and 0 otherwise. We are interested in the restriction C(n, n). Note that C is definitely a function, we just don't yet know if it's computable '' And using this proof we showed it cannot exist. I actually tried to find a TM that would be ''obviously'' computable, in order to illustrate my problem about the proof. Now you may argue that TMC does has to do with halting, as to know whether or not the n-th TM outputs a 3 or not, it has to know if it halts. So I came with an other exemple: Say we have a TM that simply counts the number of steps on the tape of other TMs. Call it TMC. Say we apply TMD on the output of TMC. TMD is a time keeper TM, that outputs a ‘’0’’ whenever the output of TMC exceeds say 1000 steps and a ‘’1’’ otherwise. Is TMD x TMC actually computable? Let’s first assume so, to be concise. Let’s rename TMD x TMC = TME. Let’s undergo again the proof of the non-computability of TMH, but now applying to TME. Here it goes: We enumerate all the TMs (there are lots of ways to do this, details are unimportant). We define a function E(n,m) that outputs 1 if the n-th TM exceeds 1000 steps on input m and 0 otherwise. We are interested in the restriction E(n, n). Note that E is definitely a function; we just don't yet know if it's computable. Assume there is a TM, call it TME, that computes E(n,n). Define another TM, called TMY, as follows. TMY(k) first runs TME(k) to determine E(k,k). If E(k,k) = 1, meaning that the k-th TM exceeds 1000 steps on k, then TMY(k) goes into an infinite loop that contains more than 1000 steps. If E(k,k) = 0, meaning that the k-th TM exceeds 1000 steps on k, then TMY(k) = 1. Now if TMy is in fact a Turing machine, it's on our list somewhere, say it's the y-th TM. What is TMY(y)? Well if E(y,y) = 1, meaning that TME exceeds 1000 steps on y, then TMY goes into an infinite loop that contains more than 1000 steps on y. But if E(x,x) = 0, meaning that TME exceeds 1000 steps on y, then TMY(y) = 1. So TMY exceeds 1000 steps on y if and only if it doesn't exceeds 1000 steps. That's a contradiction, showing that TMY can't exist. Therefore E is not in fact computable by a Turing machine. TME can't exist. thank you for reading Now if someone could show with other means, that TME actually is computable for any TM and and any input k, I think my point would be done.

@wtf In order to go further, I will use your own description of the proof to address my question again. Say we take a very simple algorithm instead of H(n,m). One you will not reproach me of ''assuming it is intuitively computable''. Say all what TMH(n,m) does is: if the n-th TM when applied the input m, outputs the number 3, then it outputs 1, and otherwise outputs 0. We name this new TM ‘TMC’. And here goes your proof again with some TMC instead of TMH: We enumerate all the TMs (there are lots of ways to do this, details are unimportant). We define a function C(n,m) that outputs 1 if the n-th TM outputs a 3 on input m and 0 otherwise. We are interested in the restriction C(n, n). Note that C is definitely a function, we just don't yet know if it's computable. Assume there is a TM, call it TMC, that computes C(n,n). Define another TM, called TMx, as follows. TMx(k) first runs TMC(k) to determine C(k,k). If C(k,k) = 1, meaning that the k-th TM outputs a 3 on k, then TMx(k) outputs a 6. If C(k,k) = 0, meaning that the k-th TM fails to output a 3 on k, then TMx(k) = 3. Now if TMx is in fact a Turing machine, it's on our list somewhere, say it's the x-th TM. What is TMx(x)? Well if C(x,x) = 1, meaning that TMC outputs a 3 on x, then TMx outputs a 6 on x. But if C(x,x) = 0, meaning that TMH fails output a 3 on x, then TMx(x) = 3. So TMx outputs a 3 on x if and only if it doesn't output a 3. That's a contradiction, showing that TMx can't exist. Therefore C is not in fact computable by a Turing machine. TMC can't exist. So how come using this proof we can show that any TM of the type X(n,m) = 1 or 0 doesn’t exist? This proof recalls me a lot of Russell’s paradox…

I respect your criticism, and thank you for trying to help me. but I kind of feel you did'nt read my explanation of the proof in my last post as you are refering to things I said in my first post again here? And because you said so explicitly you wont go through it? But to answer your first question yes, the way I put it here is pretty much word for word how it is in Penrose's book. Anyway, Thank you for providing me another version of the proof, I will go through it.

Thank you for your support. < The fact that you refer to the proof of the noncomputability of the halting problem as the "diagonal slash" makes me think you don't understand the proof. I could be wrong but you are vague where precision is required. Yes I knew the way I would express things will be problematic. That is why I put the link to a detail of the proof, and I thought this would avoid me expressing things wrong. I will try to express my understanding of the proof in my own words, as it helps to show whether I understand it or not. And then go on again with my question in simpler terms and avoiding alternative explanation. This will be a lot based on Penrose’s 1989 The Emperor's New Mind, chapter: the insolubility of Hilbert's Problem; We have Tn(m), where ‘n’ is the number of the Turing machine (TM), and (m) is the input. Whenever Tn(m) doesn’t end its calculation, we express as: Tn(m) = □. Now, we ‘assume’ we are provided with a TM ‘H’ that can tells us whether Tn(m) ends its calculation or not. Here what TMH does: H (n;m) = 0 if Tn(m) = □, and H (n;m) = 1 if Tn(m) = stops. We first imagine an infinite array, which lists all the outputs of all possible TMs acting on all the possible different inputs. The nth row of the array displays the output of the nth TM, as applied to the various inputs m = 0,1,2,3,4…: n\m … 0 1 2 3 4 5 6 7 … … □ □ □ □ □ □ □ □ 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2 0 2 0 2 0 2 0 2 0 3 0 □ 0 □ 0 □ 0 □ 0 4 0 □ 1 □ 2 □ 3 □ 4 5 1 1 1 □ □ 1 1 1 1 Note: all the outputs are fictive and were chosen for their explicative meaningfulness: i.e. the ns are not real TM machine numbers, and the output are not accurate if applied to inputs ms. We then apply to the whole table: Tn(m) x H (n;m) Which gives as all the actual inputs as from TMns exept that when this TM wasn’t to stop, the □ is replaced by a ‘0’ (hence the purpose of TMH) n\m … 0 1 2 3 4 5 6 7 … … 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2 0 2 0 2 0 2 0 2 0 3 0 0 0 0 0 0 0 0 0 4 0 0 1 0 2 0 3 0 4 5 1 1 1 0 0 1 1 1 1 Here comes the Cantors diagonal slash in play, watch it picks up the numbers in fat and display new sequence. n\m … 0 1 2 3 4 5 6 7 … … 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2 0 2 0 2 0 2 0 2 0 3 0 0 0 0 0 0 0 0 0 4 0 0 1 0 2 0 3 0 4 5 1 1 1 0 0 1 1 1 1 Says the machine that does the diagonal slash is termed TMD we would have: D(n;m) x Tn(m) x H(n;m) = 0, 1, 2, 0, 0, 1, … To which we add one 1+ (D(n;m) x Tn(m) x H(n;m)) = 1, 2, 3, 1, 1, 2, … It is important to note at this point that according to Penrose, if we assume TMH exists, all the subsequent operations are perfectly computable. The paradox arises, as if our array is to list all the outputs of all possible TMs acting on all the possible different inputs, then among of the outputs from the array should also figure our new sequence: 1, 2, 3, 1, 1, 2, … Yet it cannot be so, for our new sequence differs from the first row in the first entry, from the second row in the second entry and so on. From this contradiction, one had proven that the TMH cannot exist. End of the proof. Here now my question: Say we repeat the same proof, but instead of using TMH, we use TMH*. TMH*, is opposite to TMH, is well defined. TMH* checks if another TM goes longer than 1000 steps, outputting a 1 if it stops before then and a 0 otherwise. As an argument that this TMH* indeed exists its coding may look like that in Turing notation if Tn1 does 1. 00 ► 00R, 2. 01 ►131R Which in text reads as: 1. If you are in the internal state ‘0’ and see a 0 on the tape, then stay to internal state 0, leave a 0 on the tape, and move to right (R). The coding TMH* (n1,m) would look like: 1. 00 ► 010R, 2. 01 ►130111R 3. Whenever you are in the internal state nn(output of the Tn1) followed by thousand ‘1’ then display 0 and stop. This is clearly just a sketch but I’m sure any computer scientist would agree that such a ‘step counter’ TM exists? Let’s then take our array again, and then make Tn(m) x H* (n;m). It would look like previously, except with more 0s, as the criterion for what is a non-halting TM is much more restrictive. n\m … 0 1 2 3 4 5 6 7 … … 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2 0 2 0 2 0 2 0 0 0 3 0 0 0 0 0 0 0 0 0 4 0 0 1 0 2 0 3 0 4 5 1 1 0 0 0 1 0 1 0 Again we go, D(n;m) x Tn(m) x H*(n;m) = 0, 1, 2, 0, 0, 0, … And 1+ (D(n;m) x Tn(m) x H(n;m)) = 1, 2, 3, 1, 1, 1, … Again, the new sequence differs from the other sequences in the original array (the first row in the first entry, from the second row in the second entry and so on). So we come to the conclusion TMH* also wouldn’t exist? I hope now it is clearer and one may spot where I get wrong. Thank you

Uncool -> Yes This is exactlly what I mean.

Well Thank you for the terminology tips. The word ''paradoxe'' is used here to express that if we accepted that the halting testing TM was existing, the maths get paradoxal. if you assume TMH does'nt exists, then effectively there is no paradoxe. I would appreciate some insights about how, when applying the same proof to a TMH* (that just defines non-halting for a TM as exceeding a fixed amount of steps), one does'nt come with the conclusion that TMH* doesnt exist. Thank you

Hallo, Maybe can someone please help me here; I've been reading that there is no general Turing Machine (TM) for specifying whether or not a given TM will end its calculation. One could prove this claim by applying Cantor's diagonal argument to a plot exhibiting the output of all possible TMs when applied an infinite set of inputs. See Penrose 1989 The Emperor's New Mind, chapter: the insolubility of Hilbert's Problem; or https://www.decodedscience.org/turing-machines-and-the-halting-problem/12672/3 The proof implies a hypothetic foolproof TM termed H (TMH), that indeed always gives a computable output to the question of whether a given TM halts or not (i.e. ''1'' if it does, ''0'' if it does not). The use of Cantor's diagonal slash applied to the output of TMH is then showed to lead to paradoxes, and one concludes the TMH cannot exist. All is well explained in the link. My problem starts here, Let say that instead of TMH, we have a TM that simply displays a ''0'' whenever a TM exceeds a fixed number of steps. Let’s term this Turing Machine TMH*. Then we go through the same processes, using the diagonal slash, and it seems to me, we come to the same conclusion, that TMH* couldn’t exist! However, it seems from common sense that TMH* is perfectly computable, predictable and deterministic and therefore can exist as a genuine Turing Machine. Isn’t the use of the Diagonal Argument to argue against the existance of a general TMH inappropriate? Doesn’t the ''paradoxal result'' instead tells us that two distinct computable processes can lead to two different sets of computable numbers, at yet those both set be infinite? I.e. there is distinct ''phases'' of infinite sets, and two infinite sets may be distinct from each other, as well regarding to their sizes? Indeed like Cantor's initial use of the diagonal slash… That would mean, the TM doing the diagonal slash, say TMD, is not contained among the initial set of TMs, because it belongs to a distinct infinite set? But then, all that has nothing to do with resolving the Halting Problem. Where am I wrong? Thank you for your help. P.S. Sorry if some terms or expressions are misused here, I've made effort, but I am new to the field. I would be happy to try to clarify if needed.