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I have a confusion in mind that since in hashing we transform a large string into a smaller fixed size string, so its unlikely that any large string has a unique hash. There had to be collisions and repetition since the number of bits are reduced and mathematically it seems unlikely that smaller number of bits represent a large number of bits with uniqueness.

For guaranteed uniqueness its only possible if the input bits is equal to output bits.

Not having a possibility to find a pre-image is a different thing, my question is about the existence of a pre-image, whether it should exist or not? I think it should always exist.

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    The title of your questions (about collision resistance, i.e. ability to create two images with the same hash) does not align with the topic of the question (do multiple images exist at all). To answer the last: as you already determined correctly multiple images resulting in the same hash must be possible just because there is a limited number of hashes and an unlimited number of images. – Steffen Ullrich May 15 '16 at 6:07
  • yes! question updated – Firdous May 15 '16 at 10:56
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As you already determined correctly multiple images resulting in the same hash must be possible. The argumentation is simple: there is a limited number of hashes and an unlimited number of images which means that there must be multiple images resulting in the same hash.

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    The name for this argument is the "pigeonhole principle". – Lie Ryan May 15 '16 at 11:44
  • @LieRyan Indeed - though in real life the birthday paradox might be more important – Hagen von Eitzen May 15 '16 at 11:47
  • @HagenvonEitzen: Birthday paradox and pigeonhole principle are completely different things and birthday paradox is unrelated to this question (don't only look at the title of the question) while pigeonhole is the correct answer. – Steffen Ullrich May 15 '16 at 11:49

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