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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T05:40:52+00:00 2026-05-15T05:40:52+00:00

I just spent some time reading https://stackoverflow.com/questions/2768248/is-md5-really-that-bad (I highly recommend!). In it, it talks

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I just spent some time reading https://stackoverflow.com/questions/2768248/is-md5-really-that-bad (I highly recommend!).

In it, it talks about hash collisions. Maybe I’m missing something here, but can’t you just encrypt your password using, say, MD5 and then, say, SHA-1 (or any other, doesn’t matter.) Wouldn’t this increase the processing power required to brute-force the hash and reduce the possibility of collision?

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  1. Editorial Team
    Editorial Team
    2026-05-15T05:40:53+00:00Added an answer on May 15, 2026 at 5:40 am

    You are talking about 2 distinct (although related) problems. First is the likely-hood of a collision, and the second is the ability to run the algorithm on tons of values to find the original value which created the hash.

    1. Collisions. If you run sha1(md5(text)) you first get the hash of md5, then pass that to sha1. Lets assume the sha1 function has a 128-bit output, and the md5 also has 128-bit output. Your chance of collision in the md5 function is 1/2^128. Then your chance of collision in the sha1 is 1/2^128. If either collides then the function overall collides and hence the result is (1/2^128) + (1/2^128) or 1/2^127
    2. Brute forcing. Running sha1(md5(text)) will only double the time it takes to find the original string. This is nothing in terms of security. FOr instance, if you have 128-bits of output space for each algorithm, and it takes 1 hour to brute force, then it will take 2 hours to run the same brute force twice to get the original string. This would be the same as increasing the output space to 129-bits. However, if you want to really make brute forcing impossible, what you have to do is double the output-size (which can be compared to the key size in encryption).
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