P-EVALUE

P-values    The number of random HSPs with score >= S is described by a Poisson distribution [10,11]. This means that the probability of finding exactly a HSPs with score >=S is given by




where E is the E-value of S given by equation (1) above. Specifically the chance of finding zero HSPs with score >=S is e-E, so the probability of finding at least one such HSP is




This is the P-value associated with the score S. For example, if one expects to find three HSPs with score >= S, the probability of finding at least one is 0.95. The BLAST programs report E-value rather than P-values because it is easier to understand the difference between, for example, E-value of 5 and 10 than P-values of 0.993 and 0.99995. However, when E < 0.01, P-values and E-value are nearly identical.

 

 

改正下面代码错误:using Org.BouncyCastle.Math; using System; using System.Collections.Generic; using System.ComponentModel; using System.Data; using System.Drawing; using System.Linq; using System.Text; using System.Threading.Tasks; using System.Windows.Forms; namespace WindowsFormsApp1 { public partial class Form7 : Form { public Form7() { InitializeComponent(); } private void btnEncrypt_Click(object sender, EventArgs e) { // 生成公钥和私钥 BigInteger p = BigInteger.Parse("857504083339712752489993810777"); BigInteger q = BigInteger.Parse("1029224947942998075080348647219"); BigInteger n = p * q; BigInteger phi = (p - 1) * (q - 1); BigInteger e = 65537; BigInteger d = ModInverse(e, phi); // 加密数据 BigInteger message = BigInteger.Parse(txtData.Text); BigInteger encrypted = ModPow(message, e, n); // 解密数据 BigInteger decrypted = ModPow(encrypted, d, n); // 显示结果 txtEncrypted.Text = encrypted.ToString(); txtDecrypted.Text = decrypted.ToString(); } // 求模反元素 public static BigInteger ModInverse(BigInteger a, BigInteger m) { BigInteger m0 = m; BigInteger y = 0, x = 1; if (m == 1) return 0; while (a > 1) { BigInteger q = a / m; BigInteger t = m; m = a % m; a = t; t = y; y = x - q * y; x = t; } if (x < 0) x += m0; return x; } // 模幂运算 public static BigInteger ModPow(BigInteger b, BigInteger e, BigInteger m) { BigInteger result = 1; while (e > 0) { if ((e & 1) == 1) result = (result * b) % m; e >>= 1; b = (b * b) % m; } return result; } } }
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06-11
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