拿去用,常見幾種加密算法的Python實(shí)現(xiàn)

作者 | 我被狗咬了
來源 | Python亂燉

生活中我們經(jīng)常會(huì)遇到一些加密算法，今天我們就聊聊這些加密算法的Python實(shí)現(xiàn)。部分常用的加密方法基本都有對(duì)應(yīng)的Python庫，基本不再需要我們用代碼實(shí)現(xiàn)具體算法。

一、MD5加密

import?hashlib
m?=?hashlib.md5()
m.update(str.encode("utf8"))
print(m.hexdigest())

二、SHA1加密

import hashlib
sha1?=?hashlib.sha1()
data?=?'2333333'
sha1.update(data.encode('utf-8'))
sha1_data?=?sha1.hexdigest()
print(sha1_data)

三、HMAC加密

import?hmac
import?hashlib
#?第一個(gè)參數(shù)是密鑰key，第二個(gè)參數(shù)是待加密的字符串，第三個(gè)參數(shù)是hash函數(shù)
mac?=?hmac.new('key','hello',hashlib.md5)
mac.digest()??#?字符串的ascii格式
mac.hexdigest()??#?加密后字符串的十六進(jìn)制格式

四、DES加密

import?binascii
from?pyDes?import?des,?CBC,?PAD_PKCS5
#?需要安裝?pip?install?pyDes

def?des_encrypt(secret_key,?s):
????iv?=?secret_key
????k?=?des(secret_key,?CBC,?iv,?pad=None,?padmode=PAD_PKCS5)
????en?=?k.encrypt(s,?padmode=PAD_PKCS5)
????return?binascii.b2a_hex(en)

def?des_decrypt(secret_key,?s):
????iv?=?secret_key
????k?=?des(secret_key,?CBC,?iv,?pad=None,?padmode=PAD_PKCS5)
????de?=?k.decrypt(binascii.a2b_hex(s),?padmode=PAD_PKCS5)
????return?de

secret_str?=?des_encrypt('12345678',?'I?love?YOU~')
print(secret_str)
clear_str?=?des_decrypt('12345678',?secret_str)
print(clear_str)

五、AES加密

import?base64
from?Crypto.Cipher?import?AES

'''
AES對(duì)稱加密算法
'''
#?需要補(bǔ)位，str不是16的倍數(shù)那就補(bǔ)足為16的倍數(shù)
def?add_to_16(value):
????while?len(value)?%?16?!=?0:
????????value?+=?'\0'
????return?str.encode(value)??#?返回bytes
#?加密方法
def?encrypt(key,?text):
????aes?=?AES.new(add_to_16(key),?AES.MODE_ECB)??#?初始化加密器
????encrypt_aes?=?aes.encrypt(add_to_16(text))??#?先進(jìn)行aes加密
????encrypted_text?=?str(base64.encodebytes(encrypt_aes),?encoding='utf-8')??#?執(zhí)行加密并轉(zhuǎn)碼返回bytes
????return?encrypted_text
#?解密方法
def?decrypt(key,?text):
????aes?=?AES.new(add_to_16(key),?AES.MODE_ECB)??#?初始化加密器
????base64_decrypted?=?base64.decodebytes(text.encode(encoding='utf-8'))??#?優(yōu)先逆向解密base64成bytes
????decrypted_text?=?str(aes.decrypt(base64_decrypted),?encoding='utf-8').replace('\0',?'')??#?執(zhí)行解密密并轉(zhuǎn)碼返回str
????return?decrypted_text

六、RSA加密

#?-*-?coding:?UTF-8?-*-
#?reference?codes:?https://www.jianshu.com/p/7a4645691c68

import?base64
import?rsa
from?rsa?import?common

#?使用?rsa庫進(jìn)行RSA簽名和加解密
class?RsaUtil(object):
????PUBLIC_KEY_PATH?=?'xxxxpublic_key.pem'??#?公鑰
????PRIVATE_KEY_PATH?=?'xxxxxprivate_key.pem'??#?私鑰

????#?初始化key
????def?__init__(self,
?????????????????company_pub_file=PUBLIC_KEY_PATH,
?????????????????company_pri_file=PRIVATE_KEY_PATH):

????????if?company_pub_file:
????????????self.company_public_key?=?rsa.PublicKey.load_pkcs1_openssl_pem(open(company_pub_file).read())
????????if?company_pri_file:
????????????self.company_private_key?=?rsa.PrivateKey.load_pkcs1(open(company_pri_file).read())

????def?get_max_length(self,?rsa_key,?encrypt=True):
????????"""加密內(nèi)容過長時(shí)?需要分段加密?換算每一段的長度.
????????????:param?rsa_key:?鑰匙.
????????????:param?encrypt:?是否是加密.
????????"""
????????blocksize?=?common.byte_size(rsa_key.n)
????????reserve_size?=?11??#?預(yù)留位為11
????????if?not?encrypt:??#?解密時(shí)不需要考慮預(yù)留位
????????????reserve_size?=?0
????????maxlength?=?blocksize?-?reserve_size
????????return?maxlength

????#?加密?支付方公鑰
????def?encrypt_by_public_key(self,?message):
????????"""使用公鑰加密.
????????????:param?message:?需要加密的內(nèi)容.
????????????加密之后需要對(duì)接過進(jìn)行base64轉(zhuǎn)碼
????????"""
????????encrypt_result?=?b''
????????max_length?=?self.get_max_length(self.company_public_key)
????????while?message:
????????????input?=?message[:max_length]
????????????message?=?message[max_length:]
????????????out?=?rsa.encrypt(input,?self.company_public_key)
????????????encrypt_result?+=?out
????????encrypt_result?=?base64.b64encode(encrypt_result)
????????return?encrypt_result

????def?decrypt_by_private_key(self,?message):
????????"""使用私鑰解密.
????????????:param?message:?需要加密的內(nèi)容.
????????????解密之后的內(nèi)容直接是字符串，不需要在進(jìn)行轉(zhuǎn)義
????????"""
????????decrypt_result?=?b""

????????max_length?=?self.get_max_length(self.company_private_key,?False)
????????decrypt_message?=?base64.b64decode(message)
????????while?decrypt_message:
????????????input?=?decrypt_message[:max_length]
????????????decrypt_message?=?decrypt_message[max_length:]
????????????out?=?rsa.decrypt(input,?self.company_private_key)
????????????decrypt_result?+=?out
????????return?decrypt_result

????#?簽名?商戶私鑰?base64轉(zhuǎn)碼
????def?sign_by_private_key(self,?data):
????????"""私鑰簽名.
????????????:param?data:?需要簽名的內(nèi)容.
????????????使用SHA-1?方法進(jìn)行簽名（也可以使用MD5）
????????????簽名之后，需要轉(zhuǎn)義后輸出
????????"""
????????signature?=?rsa.sign(str(data),?priv_key=self.company_private_key,?hash='SHA-1')
????????return?base64.b64encode(signature)

????def?verify_by_public_key(self,?message,?signature):
????????"""公鑰驗(yàn)簽.
????????????:param?message:?驗(yàn)簽的內(nèi)容.
????????????:param?signature:?對(duì)驗(yàn)簽內(nèi)容簽名的值（簽名之后，會(huì)進(jìn)行b64encode轉(zhuǎn)碼，所以驗(yàn)簽前也需轉(zhuǎn)碼）.
????????"""
????????signature?=?base64.b64decode(signature)
????????return?rsa.verify(message,?signature,?self.company_public_key)

七、ECC加密

#?-*-?coding:utf-8?*-
#?author:?DYBOY
#?reference?codes:?https://blog.dyboy.cn/websecurity/121.html
#?description:?ECC橢圓曲線加密算法實(shí)現(xiàn)
"""
????考慮K=kG ，其中K、G為橢圓曲線Ep(a,b)上的點(diǎn)，n為G的階（nG=O∞?），k為小于n的整數(shù)。
????則給定k和G，根據(jù)加法法則，計(jì)算K很容易但反過來，給定K和G，求k就非常困難。
????因?yàn)閷?shí)際使用中的ECC原則上把p取得相當(dāng)大，n也相當(dāng)大，要把n個(gè)解點(diǎn)逐一算出來列成上表是不可能的。
????這就是橢圓曲線加密算法的數(shù)學(xué)依據(jù)
????點(diǎn)G稱為基點(diǎn)（base?point）
????k（k????K為公開密鑰（public?key)
"""

def?get_inverse(mu,?p):
????"""
????獲取y的負(fù)元
????"""
????for?i?in?range(1,?p):
????????if?(i*mu)%p?==?1:
????????????return?i
????return?-1

def?get_gcd(zi,?mu):
????"""
????獲取最大公約數(shù)
????"""
????if?mu:
????????return?get_gcd(mu,?zi%mu)
????else:
????????return?zi

def?get_np(x1,?y1,?x2,?y2,?a,?p):
????"""
????獲取n*p，每次+p，直到求解階數(shù)np=-p
????"""
????flag?=?1??#?定義符號(hào)位（+/-）

????#?如果?p=q??k=(3x2+a)/2y1mod?p
????if?x1?==?x2?and?y1?==?y2:
????????zi?=?3?*?(x1?**?2)?+?a??#?計(jì)算分子??????【求導(dǎo)】
????????mu?=?2?*?y1????#?計(jì)算分母

????#?若P≠Q(mào)，則k=(y2-y1)/(x2-x1)?mod?p
????else:
????????zi?=?y2?-?y1
????????mu?=?x2?-?x1
????????if?zi*?mu?0:
????????????flag?=?0????????#?符號(hào)0為-（負(fù)數(shù)）
????????????zi?=?abs(zi)
????????????mu?=?abs(mu)

????#?將分子和分母化為最簡
????gcd_value?=?get_gcd(zi,?mu)?????#?最大公約數(shù)
????zi?=?zi?//?gcd_value????????????#?整除
????mu?=?mu?//?gcd_value
????#?求分母的逆元??逆元：??a ∈G ，ョb∈G 使得 ab = ba = e
????#?P(x,y)的負(fù)元是?(x,-y?mod?p)=?(x,p-y)?，有P+(-P)=?O∞
????inverse_value?=?get_inverse(mu,?p)
????k?=?(zi?*?inverse_value)

????if?flag?==?0:???????????????????#?斜率負(fù)數(shù)?flag==0
????????k?=?-k
????k?=?k?%?p
????#?計(jì)算x3,y3?P+Q
????"""
????????x3≡k2-x1-x2(mod?p)
????????y3≡k(x1-x3)-y1(mod?p)
????"""
????x3?=?(k?**?2?-?x1?-?x2)?%?p
????y3?=?(k?*?(x1?-?x3)?-?y1)?%?p
????return?x3,y3

def?get_rank(x0,?y0,?a,?b,?p):
????"""
????獲取橢圓曲線的階
????"""
????x1?=?x0?????????????#-p的x坐標(biāo)
????y1?=?(-1*y0)%p??????#-p的y坐標(biāo)
????tempX?=?x0
????tempY?=?y0
????n?=?1
????while?True:
????????n?+=?1
????????#?求p+q的和，得到n*p，直到求出階
????????p_x,p_y?=?get_np(tempX,?tempY,?x0,?y0,?a,?p)
????????#?如果?==?-p,那么階數(shù)+1，返回
????????if?p_x?==?x1?and?p_y?==?y1:
????????????return?n+1
????????tempX?=?p_x
????????tempY?=?p_y

def?get_param(x0,?a,?b,?p):
????"""
????計(jì)算p與-p
????"""
????y0?=?-1
????for?i?in?range(p):
????????#?滿足取模約束條件，橢圓曲線Ep(a,b)，p為質(zhì)數(shù)，x,y∈[0,p-1]
????????if?i**2%p?==?(x0**3?+?a*x0?+?b)%p:
????????????y0?=?i
????????????break

????#?如果y0沒有，返回false
????if?y0?==?-1:
????????return?False

????#?計(jì)算-y（負(fù)數(shù)取模）
????x1?=?x0
????y1?=?(-1*y0)?%?p
????return?x0,y0,x1,y1

def?get_graph(a,?b,?p):
????"""
????輸出橢圓曲線散點(diǎn)圖
????"""
????x_y?=?[]
????#?初始化二維數(shù)組
????for?i?in?range(p):
????????x_y.append(['-'?for?i?in?range(p)])

????for?i?in?range(p):
????????val?=get_param(i,?a,?b,?p)??#?橢圓曲線上的點(diǎn)
????????if(val?!=?False):
????????????x0,y0,x1,y1?=?val
????????????x_y[x0][y0]?=?1
????????????x_y[x1][y1]?=?1

????print("橢圓曲線的散列圖為：")
????for?i?in?range(p):??????????????#?i=?0->?p-1
????????temp?=?p-1-i????????#?倒序

????????#?格式化輸出1/2位數(shù)，y坐標(biāo)軸
????????if?temp?>=?10:
????????????print(temp,?end="?")
????????else:
????????????print(temp,?end="??")

????????#?輸出具體坐標(biāo)的值，一行
????????for?j?in?range(p):
????????????print(x_y[j][temp],?end="??")
????????print("")???#換行

????#?輸出?x?坐標(biāo)軸
????print("??",?end="")
????for?i?in?range(p):
????????if?i?>=10:
????????????print(i,?end="?")
????????else:
????????????print(i,?end="??")
????print('\n')

def?get_ng(G_x,?G_y,?key,?a,?p):
????"""
????計(jì)算nG
????"""
????temp_x?=?G_x
????temp_y?=?G_y
????while?key?!=?1:
????????temp_x,temp_y?=?get_np(temp_x,temp_y,?G_x,?G_y,?a,?p)
????????key?-=?1
????return?temp_x,temp_y

def?ecc_main():
????while?True:
????????a?=?int(input("請(qǐng)輸入橢圓曲線參數(shù)a(a>0)的值："))
????????b?=?int(input("請(qǐng)輸入橢圓曲線參數(shù)b(b>0)的值："))
????????p?=?int(input("請(qǐng)輸入橢圓曲線參數(shù)p(p為素?cái)?shù))的值："))???#用作模運(yùn)算

????????#?條件滿足判斷
????????if?(4*(a**3)+27*(b**2))%p?==?0:
????????????print("您輸入的參數(shù)有誤，請(qǐng)重新輸入！?。n")
????????else:
????????????break

????#?輸出橢圓曲線散點(diǎn)圖
????get_graph(a,?b,?p)

????#?選點(diǎn)作為G點(diǎn)
????print("user1：在如上坐標(biāo)系中選一個(gè)值為G的坐標(biāo)")
????G_x?=?int(input("user1：請(qǐng)輸入選取的x坐標(biāo)值："))
????G_y?=?int(input("user1：請(qǐng)輸入選取的y坐標(biāo)值："))

????#?獲取橢圓曲線的階
????n?=?get_rank(G_x,?G_y,?a,?b,?p)

????#?user1生成私鑰，小key
????key?=?int(input("user1：請(qǐng)輸入私鑰小key（<{}）：".format(n)))

????#?user1生成公鑰，大KEY
????KEY_x,kEY_y?=?get_ng(G_x,?G_y,?key,?a,?p)

????#?user2階段
????#?user2拿到user1的公鑰KEY，Ep(a,b)階n，加密需要加密的明文數(shù)據(jù)
????#?加密準(zhǔn)備
????k?=?int(input("user2：請(qǐng)輸入一個(gè)整數(shù)k（<{}）用于求kG和kQ：".format(n)))
????k_G_x,k_G_y?=?get_ng(G_x,?G_y,?k,?a,?p)?????????????????????????#?kG
????k_Q_x,k_Q_y?=?get_ng(KEY_x,?kEY_y,?k,?a,?p)?????????????????????#?kQ

????#?加密
????plain_text?=?input("user2：請(qǐng)輸入需要加密的字符串:")
????plain_text?=?plain_text.strip()
????#plain_text = int(input("user1：請(qǐng)輸入需要加密的密文："))
????c?=?[]
????print("密文為：",end="")
????for?char?in?plain_text:
????????intchar?=?ord(char)
????????cipher_text?=?intchar*k_Q_x
????????c.append([k_G_x,?k_G_y,?cipher_text])
????????print("({},{}),{}".format(k_G_x,?k_G_y,?cipher_text),end="-")


????#?user1階段
????#?拿到user2加密的數(shù)據(jù)進(jìn)行解密
????#?知道?k_G_x,k_G_y，key情況下，求解k_Q_x,k_Q_y是容易的，然后plain_text?=?cipher_text/k_Q_x
????print("\nuser1解密得到明文：",end="")
????for?charArr?in?c:
????????decrypto_text_x,decrypto_text_y?=?get_ng(charArr[0],?charArr[1],?key,?a,?p)
????????print(chr(charArr[2]//decrypto_text_x),end="")

if?__name__?==?"__main__":
????print("*************ECC橢圓曲線加密*************")
????ecc_main()