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SECCON CTF quals 2013 | Cryptanalysis (Crypto 300pt)
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| a = 1234577 | |
| b = 3213242 | |
| M = 7654319 | |
| def add(A,B): | |
| if A==(0,0): return B | |
| if B==(0,0): return A | |
| x1,y1 = A | |
| x2,y2 = B | |
| if A!=B: | |
| p = (y2-y1)*pow(x2-x1,M-2,M) | |
| else: | |
| p = (3*x1*x1+a)*pow(2*y1,M-2,M) | |
| x3 = p*p-x1-x2 | |
| y3 = p*(x1-x3)-y1 | |
| return (x3%M,y3%M) | |
| base = (5234568, 2287747) | |
| pub = (2366653, 1424308) | |
| crypt = [(5081741, 6744615), (610619, 6218)] | |
| X = (0,0) | |
| for i in range(M): | |
| if X==pub: | |
| secret = i | |
| print "secret:" + str(secret) | |
| break | |
| X = add(X, base) | |
| print i | |
| #secret = 1584718 | |
| tmp =(0,0) | |
| for i in range(1584718): | |
| tmp = add(tmp,crypt[0]) | |
| plain = add(crypt[1], (tmp[0],tmp[1]*-1)) | |
| print plain | |
| # 2171002 + 3549912 = 5720914 |
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