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Research article

Driver identification and fatigue detection algorithm based on deep learning


  • Received: 07 December 2022 Revised: 03 February 2023 Accepted: 12 February 2023 Published: 27 February 2023
  • In order to avoid traffic accidents caused by driver fatigue, smoking and talking on the phone, it is necessary to design an effective fatigue detection algorithm. Firstly, this paper studies the detection algorithms of driver fatigue at home and abroad, and analyzes the advantages and disadvantages of the existing algorithms. Secondly, a face recognition module is introduced to crop and align the acquired faces and input them into the Facenet network model for feature extraction, thus completing the identification of drivers. Thirdly, a new driver fatigue detection algorithm based on deep learning is designed based on Single Shot MultiBox Detector (SSD) algorithm, and the additional layer network structure of SSD is redesigned by using the idea of reverse residual. By adding the detection of drivers' smoking and making phone calls, adjusting the size and number of prior boxes of SSD algorithm, improving FPN network and SE network, the identification and verification of drivers can be realized. The experimental results showed that the number of parameters decreased from 96.62 MB to 18.24 MB. The average accuracy rate increased from 89.88% to 95.69%. The projected number of frames per second increased from 51.69 to 71.86. When the confidence threshold was set to 0.5, the recall rate of closed eyes increased from 46.69% to 65.87%, that of yawning increased from 59.72% to 82.72%, and that of smoking increased from 65.87% to 83.09%. These results show that the improved network model has better feature extraction ability for small targets.

    Citation: Yuhua Ma, Ye Tao, Yuandan Gong, Wenhua Cui, Bo Wang. Driver identification and fatigue detection algorithm based on deep learning[J]. Mathematical Biosciences and Engineering, 2023, 20(5): 8162-8189. doi: 10.3934/mbe.2023355

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  • In order to avoid traffic accidents caused by driver fatigue, smoking and talking on the phone, it is necessary to design an effective fatigue detection algorithm. Firstly, this paper studies the detection algorithms of driver fatigue at home and abroad, and analyzes the advantages and disadvantages of the existing algorithms. Secondly, a face recognition module is introduced to crop and align the acquired faces and input them into the Facenet network model for feature extraction, thus completing the identification of drivers. Thirdly, a new driver fatigue detection algorithm based on deep learning is designed based on Single Shot MultiBox Detector (SSD) algorithm, and the additional layer network structure of SSD is redesigned by using the idea of reverse residual. By adding the detection of drivers' smoking and making phone calls, adjusting the size and number of prior boxes of SSD algorithm, improving FPN network and SE network, the identification and verification of drivers can be realized. The experimental results showed that the number of parameters decreased from 96.62 MB to 18.24 MB. The average accuracy rate increased from 89.88% to 95.69%. The projected number of frames per second increased from 51.69 to 71.86. When the confidence threshold was set to 0.5, the recall rate of closed eyes increased from 46.69% to 65.87%, that of yawning increased from 59.72% to 82.72%, and that of smoking increased from 65.87% to 83.09%. These results show that the improved network model has better feature extraction ability for small targets.



    In 1997, Van Hamme [19,(H.2)] proved the following supercongruence: for any prime p3(mod4),

    (p1)/2k=0(12)3kk!30(modp2), (1.1)

    where (a)n=a(a+1)(a+n1) is the rising factorial. It is easy to see that (1.1) is also true when the sum is over k from 0 to p1, since (1/2)k/k!0(modp) for (p1)/2<kp1. Nowadays various generalizations of (1.1) can be found in [8,10,11,12,13,14,16,17]. For example, Liu [12] proved that, for any prime p3(mod4) and positive integer m,

    mp1k=0(12)3kk!30(modp2). (1.2)

    The first purpose of this paper is to prove the following q-analogue of (1.2), which was originally conjectured by the author and Zudilin [10,Conjecture 2].

    Theorem 1.1. Let m and n be positive integers with n3(mod4). Then

    mn1k=0(1+q4k+1)(q2;q4)3k(1+q)(q4;q4)3kqk0(modΦn(q)2), (1.3)
    [5pt]mn+(n1)/2k=0(1+q4k+1)(q2;q4)3k(1+q)(q4;q4)3kqk0(modΦn(q)2). (1.4)

    Here and in what follows, the q-shifted factorial is defined by (a;q)0=1 and (a;q)n=(1a)(1aq)(1aqn1) for n1, and the n-th cyclotomic polynomial Φn(q) is defined as

    Φn(q)=1kngcd(n,k)=1(qζk),

    where ζ is an n-th primitive root of unity. Moreover, the q-integer is given by [n]=[n]q=1+q++qn1.

    The m=1 case of (1.3) was first conjectured by the author and Zudilin [9,Conjecture 4.13] and has already been proved by themselves in a recent paper [11]. For some other recent progress on q-congruences, the reader may consult [2,3,4,5,6,7,8,10,15].

    In 2016, Swisher [18,(H.3) with r=2] conjectured that, for primes p3(mod4) and p>3,

    (p21)/2k=0(12)3kk!3p2(modp5), (1.5)

    The second purpose of this paper is to prove the following q-congruences related to (1.5) modulo p4.

    Theorem 1.2. Let n3(mod4) be a positive integer. Modulo Φn(q)2Φn2(q)2, we have

    (n21)/2k=0(1+q4k+1)(q2;q4)3k(1+q)(q4;q4)3kqk[n2]q2(q3;q4)(n21)/2(q5;q4)(n21)/2q(1n2)/2, (1.6)
    [5pt]n21k=0(1+q4k+1)(q2;q4)3k(1+q)(q4;q4)3kqk[n2]q2(q3;q4)(n21)/2(q5;q4)(n21)/2q(1n2)/2. (1.7)

    Let n=p3(mod4) be a prime and take q1 in Theorem 1.2. Then Φp(1)=Φp2(1)=p, and

    limq1(q3;q4)(p21)/2(q5;q4)(p21)/2=(p21)/2k=14k14k+1=(34)(p21)/2(54)(p21)/2.

    Therefore, we obtain the following conclusion.

    Corollary 1. Let p3(mod4) be a prime. Then

    (p21)/2k=0(12)3kk!3p2(34)(p21)/2(54)(p21)/2(modp4), (1.8)
    [5pt]p21k=0(12)3kk!3p2(34)(p21)/2(54)(p21)/2(modp4). (1.9)

    Comparing (1.5) and (1.8), we would like to propose the following conjecture, which was recently confirmed by Wang and Pan [20].

    Conjecture 1. Let p3(mod4) be a prime and r a positive integer. Then

    (p2r1)/2k=14k14k+11(modp2). (1.10)

    Note that the r=1 case is equivalent to say that (1.5) is true modulo p4.

    We need to use Watson's terminating 8ϕ7 transformation formula (see [1,Appendix (Ⅲ.18)] and [1,Section 2]):

    8ϕ7[a,qa12,qa12,b,c,d,e,qna12,a12,aq/b,aq/c,aq/d,aq/e,aqn+1;q,a2qn+2bcde]=(aq;q)n(aq/de;q)n(aq/d;q)n(aq/e;q)n4ϕ3[aq/bc, d, e, qnaq/b,aq/c,deqn/a;q,q], (2.1)

    where the basic hypergeometric r+1ϕr series with r+1 upper parameters a1,,ar+1, r lower parameters b1,,br, base q and argument z is defined as

    r+1ϕr[a1,a2,,ar+1b1,,br;q,z]:=k=0(a1;q)k(a2;q)k(ar+1;q)k(q;q)k(b1;q)k(br;q)kzk.

    The left-hand side of (1.4) with m0 can be written as the following terminating 8ϕ7 series:

    8ϕ7[q2,q5,q5,q2,q,q2,q4+(4m+2)n,q2(4m+2)nq,q,q4,q5,q4,q2(4m+2)n,q4+(4m+2)n;q4,q]. (2.2)

    By Watson's transformation formula (2.1) with qq4, a=b=d=q2, c=q, e=q4+(4m+2)n, and nmn+(n1)/2, we see that (2.2) is equal to

    (q6;q4)mn+(n1)/2(q(4m+2)n;q4)mn+(n1)/2(q4;q4)mn+(n1)/2(q2(4m+2)n;q4)mn+(n1)/2×4ϕ3[q3, q2,q4+(4m+2)n, q2(4m+2)nq4,q5,q6;q4,q4]. (2.3)

    It is not difficult to see that there are exactly m+1 factors of the form 1qan (a is an integer) among the mn+(n1)/2 factors of (q6;q4)mn+(n1)/2. So are (q(4m+2)n;q4)mn+(n1)/2. But there are only m factors of the form 1qan (a is an integer) in each of (q4;q4)mn+(n1)/2 and (q2(4m+2)n;q4)mn+(n1)/2. Since Φn(q) is a factor of 1qN if and only if n divides N, we conclude that the fraction before the 4ϕ3 series is congruent to 0 modulo Φn(q)2. Moreover, for any integer x, let fn(x) be the least non-negative integer k such that (qx;q4)k0 modulo Φn(q). Since n3(mod4), we have fn(2)=(n+1)/2, fn(3)=(n+1)/4, fn(4)=n, fn(5)=(3n1)/4, and fn(6)=(n1)/2. It follows that the denominator of the reduced form of the k-th summand

    (q3;q4)k(q2;q4)k(q4+(4m+2)n;q4)k(q2(4m+2)n;q4)k(q4;q4)2k(q5;q4)k(q6;q4)kq4k

    in the 4ϕ3 summation is always relatively prime to Φn(q) for any non-negative integer k. This proves that (2.3) (i.e. (2.2)) is congruent to 0 modulo Φn(q)2, thus establishing (1.4) for m0.

    It is easy to see that (q2;q4)3k/(q4;q4)3k is congruent to 0 modulo Φn(q)3 for mn+(n1)/2<k(m+1)n1. Therefore, the q-congruence (1.3) with mm+1 follows from (1.4).

    The author and Zudilin [11,Theorem 1.1] proved that, for any positive odd integer n,

    (n1)/2k=0(1+q4k+1)(q2;q4)3k(1+q)(q4;q4)3kqk[n]q2(q3;q4)(n1)/2(q5;q4)(n1)/2q(1n)/2(modΦn(q)2), (3.1)

    which is also true when the sum on the left-hand side of (3.1) is over k from 0 to n1. Replacing n by n2 in (3.1) and its equivalent form, we see that the q-congruences (1.6) and (1.7) hold modulo Φn2(q)2.

    It is easy to see that, for n3(mod4),

    [n2]q2(q3;q4)(n21)/2(q5;q4)(n21)/2q(1n2)/20(modΦn(q)2)

    because [n2]q2=(1qn2)/(1q2) is divisible by Φn(q), and (q3;q4)(n21)/2 contains (n+1)/2 factors of the form 1qan (a is an integer), while (q5;q4)(n21)/2 only has (n1)/2 such factors. Meanwhile, by Theorem 1.1, the left-hand sides of (1.6) and (1.7) are both congruent to 0 modulo Φn(q)2 since (n21)/2=(n1)n/2+(n1)/2. This proves that the q-congruences (1.6) and (1.7) also hold modulo Φn(q)2. Since the polynomials Φn(q) and Φn2(q) are relatively prime, we finish the proof of the theorem.

    Swisher's (H.3) conjecture also indicates that, for positive integer r and primes p3(mod4) with p>3, we have

    (p2r1)/2k=0(12)3kk!3p2r(modp2r+3). (4.1)

    Motivated by (4.1), we shall give the following generalization of Theorem 1.2.

    Theorem 4.1. Let n and r be positive integers with n3(mod4). Then, modulo Φn2r(q)2rj=1Φn2j1(q)2, we have

    (n2r1)/2k=0(1+q4k+1)(q2;q4)3k(1+q)(q4;q4)3kqk[n2r]q2(q3;q4)(n2r1)/2(q5;q4)(n2r1)/2q(1n2r)/2, (4.2)
    [5pt]n2r1k=0(1+q4k+1)(q2;q4)3k(1+q)(q4;q4)3kqk[n2r]q2(q3;q4)(n2r1)/2(q5;q4)(n2r1)/2q(1n2r)/2. (4.3)

    Proof. Replacing n by n2r in (3.1) and its equivalent form, we see that (4.2) and (4.3) are true modulo Φn2r(q)2. Similarly as before, we can show that

    [n2r]q2(q3;q4)(n2r1)/2(q5;q4)(n2r1)/2q(1n2r)/20(modrj=1Φn2j1(q)2).

    Further, by Theorem 1.1, we can easily deduce that the left-hand sides of (4.2) and (4.3) are also congruent to 0 modulo rj=1Φn2j1(q)2.

    Letting n=p3(mod4) be a prime and taking q1 in Theorem 4.1, we are led to the following result.

    Corollary 2. Let p3(mod4) be a prime and let r1. Then

    (p2r1)/2k=0(12)3kk!3p2r(34)(p2r1)/2(54)(p2r1)/2(modp2r+2), (4.4)
    [5pt]p2r1k=0(12)3kk!3p2r(34)(p2r1)/2(54)(p2r1)/2(modp2r+2). (4.5)

    In light of (1.10), the supercongruence (4.4) implies that (4.1) holds modulo p2r+2 for any odd prime p.

    It is known that q-analogues of supercongruences are usually not unique. See, for example, [2]. The author and Zudilin [10,Conjecture 1] also gave another q-analogue of (1.2), which still remains open.

    Conjecture 2 (Guo and Zudilin). Let m and n be positive integers with n3(mod4). Then

    mn1k=0(q;q2)2k(q2;q4)k(q2;q2)2k(q4;q4)kq2k0(modΦn(q)2),mn+(n1)/2k=0(q;q2)2k(q2;q4)k(q2;q2)2k(q4;q4)kq2k0(modΦn(q)2). (4.6)

    The author and Zudilin [10,Theorem 2] themselves have proved (4.6) for the m=1 case. Motivated by Conjecture 2, we would like to give the following new conjectural q-analogues of (1.8) and (1.9).

    Conjecture 3. Let n3(mod4) be a positive integer. Modulo Φn(q)2Φn2(q)2, we have

    (n21)/2k=0(q;q2)2k(q2;q4)k(q2;q2)2k(q4;q4)kq2k[n2](q3;q4)(n21)/2(q5;q4)(n21)/2,n21k=0(q;q2)2k(q2;q4)k(q2;q2)2k(q4;q4)kq2k[n2](q3;q4)(n21)/2(q5;q4)(n21)/2.

    There are similar such new q-analogues of (4.4) and (4.5). We omit them here and leave space for the reader's imagination.

    The author is grateful to the two anonymous referees for their careful readings of this paper.



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