Time consumption for certificate verification

Question

Hello everyone,
I tried to track the time consumption for certificate verification. I generated the certificate by myself using X509V1CertificateGenerator in org.bouncycastle.asn1.x509. The signature algorithm is "SHA1withRSA". I generated two different certificates using different public exponent and tracked the time for certificate verification respectively. The first time, the public exponent I used to generate a certificate is 3; the second time, the public exponent I used to generate a certificate is 65537.
In theory, the operation for signature verification is C=M^e(mod n), in another word, the time consumption should depend on the length of e. More specifically, Time complexity of C=M^e(mod n) should be linear with the length of exponent e, under the same n.
The method I used for tracking the time is:
//record the time before verification
long time1 = (new Date()).getTime();
//certificate verification
aCert.verify(publickey)/*Here, 'aCert' is a certificate to be verified. 'publickey' is a public key which I retrieved from an already trusted certificate, and in this case, the first time, e=3, the second time, e=65537.*/
//record the time after verification
long time2 = (new Date()).getTime();
//the time consumption is:
long period = time2 - time1;But, after I execute using different public exponent 3 and 65537, the time consumption is almost the same, not as it supposed to be.(in theory, due to the different length of 3 and 65537 in binary form(one is 2-bit, the other is 17-bit), the time consumtion should be 17/2 times different). Actualy, in my case, the time for verification using e=3 is 551 ms; and the time for e=65537 is 531 ms.
I am just wondering, how to explain this bizarre thing?
Any help is every welcome! Thanks a lot.

Answer 1

You can check the source code of the oddModPow() method in java/math/BigInteger.java (check src.zip). There are some optimizations in it, so try to verify if they can make the running time of the case "exponent = 3" nearly equal to the case "exponent = 65537". In the first case, 3 = 11 in binary (two bits 1); in the second case, 65537 = 10000000000000001 in binary (two bits 1 as well). Maybe it can be a hint for you.