 So, thank you for introducing me. I am Azumay Koseyamada from NPTSECA Platform Laboratories. Today I present a paper, Continental Volume Finding Arounding. This is a joint work with Yusafati and K.Tekstawa, who are my colleagues. This is the outline of my presentation. Firstly, I begin with the background. So, this is a volume finding problem. Volume finding problem is one of the most important problems in Pyptop. And applications can be seen everywhere in Pyptop. And the curriculum of the function F is defined to be a pair of the elements in the domain of function, which are mapped to the same value by 8. And in our paper, we focus on a generalized program, large volume finding program. In this program, our goal is to find a large volume of F. And as a large volume, the definition of large volume is the couple of elements in the domain of F, which are mapped to the same value by F. This picture illustrates the example of three volumes. So we can consider a similarity, four volumes, five volumes, six volumes, and the general volumes. And I said that the curriculum finding problem is one of the most important problems in Pyptop. And large volume is also an important problem in Pyptop, and especially for symmetric T, Pyptography. So, and completely to find large volumes, are used to evaluate the safety of Pyptographic scenes. For example, lower volume is used to improve safety. And upper volume is used in general attacks. And so after seeing these applications, I can say that large volume will be important for post-contact safety. And in our paper, we focus on the case that the domain size is much larger than the domain size. And the target function is run, which are technical settings in Pyptop. And so next, in cluster computations, type 1 to find large volumes are known. So the problem is well-studied and well-understood. So cluster type 1 of queries to find a net portion of a random function at least, order L to the power L minus 1 over L. So if L is true, then the bound becomes L to the power 1 over 2, which is known as first label. And if L is 3, the bound becomes L to the power 2 over 3, and so on. And I know here that this bound is for analysis, IDF is a random function. And the complete issue is on the size of the domain. So after seeing this result, my question is, so how about quantum algorithms? Is there a better bound than the classical bound? This is my question. And from now on, I move to quantum computation. And so we use global algorithm. So now I explain a multi-study theory such using global algorithm. So let's consider this problem. Given a function with L, and then find x such that L no x is in L. So let's consider how many queries do we need to find such x? Then let P equals to the 590 log L inverse of L. Then in the classical setting, we need all the L-logic queries to find such x. But in the classical setting, using global algorithm, or using generalizations, we can find such x with all the square roots of L-logic queries. So these are the problems. And next, I move to systematic decision of knowledge. So actually, there is a trillion quantum upper bound to find the multi-quadrons. So let's consider a given function with uniform random here I consider L is greater than or equal to L times L, and L is a small constant. And so what is a trillion complete easy to find the unneeded model? So actually, we can find the unneeded model with order L to the power 1 over 2 queries. This can be done by running a previous such using global hand type. So I can say that this one, order L to the power 1 over 2 is a trillion upper bound of conduct queries for small data. And so no trillion quantum algorithms should give a better one than this one. And so next, I explain the case of the two-quadron. For the two-quadron program, there is typhoon. A typhoon is known. So quantum type bound of queries to find the two-quadron of random constant L is order L to the power 1 over 3. And these results are given by Changbin. And so this bound L to the power 1 over 3 is lower than L to the power 1 over 2. And thus I can say that this bound is non-trivial problem. And so how can we find a question with this quantum complication? OK, next I explain how to find two-quadron of a random function F. But to construct such quantum algorithm, now here I need the worst-case algorithm to find the morphology. So we can find an algorithm of any function F with this complication. Here, this algorithm is worst-case algorithm. I mean, F is a signal to any function. And this complication depends on the size of domain. So for example, if L is 2, then the bound becomes L to the power 2 over 3. And if L is 3, then the bound becomes L to the power 5 over 7, and so on. And so if L is much rather than L, then the complication becomes worse than two-quadron. However, for the case L is 2, then there is a way to get a better amount using these worst-case algorithms. So next, I explain how to construct quantum algorithm for a random function F using this worst-case algorithm. So let's consider random function F. Firstly, take a subset of the size of the power 1 over 2 uniformly at random. Then there is a 2-quadron in x with constant programming. And next, apply worst-case algorithm we described, I described before, on the restricted function F. F is now restricted on the subset X. Then we can find a 2-quadron with all the identity of X to the power 2 over 3 queries. And that's since the identity of X is now all the L to the power 1 over 2, that we can find a 2-quadron with L to the power 1 over 2 to the power 2 over 3, which is equal to L to the power 1 over 3 quantum queries. So now this bound is well done, the trivial bound. And so I can say that this quantum algorithm is non-trivial quantum algorithm. And next, so this is the case of 2-quadron, normal quantum. And next, I extend this technique for the case L is greater than or equal to 3. So this is the quantum algorithm for 2-quadron. And for general case, I modify like this. So eventually, I get this bound, order L to the power L minus 1 over L to the power 1 minus 2 to the power L minus 2 over L minus 1 quantum queries, which is very complex. And so, OK, so is it well done for the L to the power 1 or 2? This is the question. And actually, so we get the bound like this. If L is 2 over 3, then this bound is better than the trivial bound. So it's OK. The bound is non-trivial. But however, for the case L is greater than or equal to 4, then the bound becomes worse than the trivial bound. So it's a problem. And new quantum algorithm is needed for the case L is greater than or equal to 4. OK, so these are the systematization of knowledge. And next, I go to new quantum algorithm and our main results. So this is our main result. We propose a quantum algorithm to find this mass variance. And so suppose M is greater than or equal to L times Z. For arbitrary F, an algorithm can be found with this complexity. And so please pay attention to the right hand side and right side of this equation. This bound is better than the trivial bound. So I can say our quantum algorithm is non-trivial quantum algorithm to find mass variance. And so this graph shows the result of our quantum algorithm. If L is 2, then R1 matches the previous one. And if L is 3, then R1 improves the previous one. And if L is greater than or equal to 4, then the R1 is non-trivial. These are our results. And next, I explain our idea of quantum algorithm. So our idea is to combine BHT algorithm with recursion. So the BHT algorithm is a quantum algorithm proposed by Brassat, Oya, and that, which is a quantum algorithm to find two equations with order n to the power of 3 quantum arrays. So now I explain how this quantum algorithm works. So firstly, this BHT quantum algorithm takes some sets, capital X, oversize order n to the power of 3 quantum arrays. And calculate epsilon x for all small x in capital X. And store their x, epsilon x, in a list of n. And next, for the second set, find a small x prime, which is in domain, but not in capital X, that the variation of x prime is in using the global search. So this is the BHT algorithm. And for the complete HT analysis, the step one requires order n to the power of n more sequence, because we calculate epsilon x for all small x in capital X. And for the second step, we use global search. And the quantum complete HT is order the square root of n over the cardinality of n. And this is equal to order n to the power of 1, 3. So summing up this complete HT, we ordain the desired one. This is the BHT algorithm. And so our idea is to combine this algorithm with recursive. So let's check the example of 3 quantum. So from 2.0 to 3.0. So firstly, I correct order n to the power of 1 over 9, make 2.0, and store these 4.0s in a list of n. And second step, extend one of the 2.0s in n to 3.0 using the global search. This is almost the same as the BHT algorithm. And for a complete HT analysis, the first step requires order n to the power of 1 over 9 queries. Since here, I call quantum algorithm to find the 2.0s for n to the power of 1 over 9 times. So the complete HT runs this times this. This is a very complete HT to find the 2.0. So the quantum very complete HT for step 1, and for the second step, as a severe BHT algorithm, the quantum very complete HT runs like this. So the very complete HT runs, I mean, order n to the power of 1 over 9. And summing up these complete HTs, we get order n to the power of 1 over 9. And this is the example of 3 quantum. And similarly, using the quantum algorithm to find 3 quantum, we can find 4 quantum like this. And the quantum very complete HT to find 4 quantum is order n to the power starting over 27 queries. So these are examples of 3 quantum and 4 quantum. And generalize these techniques. We obtain the generalized quantum algorithm to find L-quotions. So let me denote our quantum algorithm as n-quals, not L-A. So firstly, our quantum algorithm collects order n to the power of 1 over 3 to the power of n minus 1, many n minus 1 quotions by running n-quals f of n minus 1. This is our quantum algorithm to find l minus 1 quotions. And this is them in a sphere. So in the step 1, we recursively call our quantum algorithm to find l minus 1 quotions. This is the first step. And the next step, in the second step, extend one of these l minus 1 quotions in L to l minus 1, using the flow approach. Similarly, as the BST algorithm. So firstly, I collect l minus 1 quotions, many n-quals. And this is the first step. And next step, in the second step, I extend one of these l minus 1 quotions to l-quotions. This is our idea of the quantum algorithm. And so next step, for the completion analysis, the first step requires this quantum query completion. So in the first step, we call the quantum algorithm to find l minus 1 quotions. l to the power 1 over c to the power l minus 1 times. And this is the 20-complete ST of our algorithm to find l minus 1 quotions. And the completion becomes like this. And for the second step, the completion becomes like this. And summing up these two completenesses, we obtain the desired problem. OK, this is our main result, our main quantum algorithm. And actually, we obtained this term, suppose l is greater than y equals to l times n. Then for the function l, our quantum algorithm can find l of the node f with x 15 quantum query completion at most like this. So to prove this term, we gave rigorous mathematical proof and analysis. But the technique is somewhat complex, and I don't describe the proof here. So for more details, please read our paper. And so I move to summary. We propose a new quantum algorithm that finds my quotient with non-trivial quantum query completion ST. And our one improves freedom one for the case l is greater than or equal to 3. This is our test. And the further result direction is that there are three further result directions. One is, so is there a better equivalent? This is the first direction. And the second direction is, what are the requirements? So in our paper, we focus on only an upper bound of query completion. So we are also interested in the lower bound of complexity. And finally, what can we see about time complexity? In our paper, we focus on only quantum query complexity and not time complexity. And so actually, for normal two-quantum algorithms, the time complexity to find two-quantum algorithms becomes sometimes a quantum machine. And so our quantum algorithm actually runs with similar time complexity as quantum query complexity. But we need highly developed quantum hardware, and we need many, many treatments. So but however, I realize that this new band gives new insight to us about the first quantum setility of the cryptos schemes. So that's all. Thank you for your attention. So I can add the first openly search question. And this is about the space complexity, memory complexity. Because in the classical algorithms, there is a big difference between two conditions and three conditions. Two conditions using the raw method, you don't need any memory. For three conditions, you need a lot of memory. And if you are using the end of the one-half quantum algorithm, calculating intercess, again, you don't need any memory regardless of what is the multi-collision. But in your algorithm, you need a lot of memory. So is it clear what is the trade-off? For example, if I only allow you a small amount of memory, what would be the best running time in the quantum algorithm for finding three collisions, four collisions, et cetera? It is a very interesting question. And so I think I have to study this topic. But I have not studied enough. So I can say I'm not seeing anything here. OK. And you have an open-flow room? Can you show me? Are there anything known about raw about other than two collisions, please? Yeah. So actually, there is no raw other than two collisions. So, yeah, so please find a raw one. Find a method proposed by Xiaoyun Wang and grow an algorithm to speed up the method to find a clean such as force Xiaowan. Ah, yes. I think there is something that we can do. But I have not studied so much. So please, please break Xiaowan. The answer is definitely yes. The answer is definitely yes, because what Xiaoyun Wang did was to help in the first few rounds of Xiaowan, something which is very smart, and beyond it is to do a search for a solution to a random search for them. So in the second part of Wang's algorithm, you can just apply it over here, and this will definitely speed up a good time complexity. So you can combine them, but in a better way. Thank you very much. Any others? OK, thanks, speakers.