Let a be input value, k1 and k2 be two keys. Then

 X=(a+k1) mod m;

Can i decrypt back? Because i am not getting exact answer while decrypting. How to rotate bits in verilog? Can you please help with code if this possible?


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  • This question does not appear to be about electronics design within the scope defined in the help center.
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  • \$\begingroup\$ Welcome to crypto.stackexchange - This appears to be a question for the electrical engineering stackexchange. I will migrate this there for you. The relation to cryptography is not obvious and does not seem important to the question, so I recommend editing it to ask about arithmetic and bit shifts without mentioning cryptography. \$\endgroup\$ – Ella Rose May 15 at 17:39
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    \$\begingroup\$ Despite comments to the contrary, this is not primarily a question about electrical engineering. This question should not have been migrated. \$\endgroup\$ – duskwuff May 15 at 19:08

I might be wrong, but I do not think that you will be able to decrypt it. You are setting the first X to a remainder so there is no way to guarantee that you get the original value back. For example, assuming k1 is a positive value, and we set m = 2, then X has 2 possible values, 0 and 1. If the first X is 0, and a = 1, than k1 can be any odd number

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  • \$\begingroup\$ Actually i was able to decrypt back when i was using only modular addition and bit wise exor opeartion. After adding shift, only few values are decrypting back. Others not! \$\endgroup\$ – SHRIVATHSA TUNGA May 16 at 0:43

I leave the decryption question to somebody else, but how to rotate a vector in Verilog:

reg [7:0] rotate_me;
always @( posedge clk) 
   rotate_me <= {rotate_me[6:0],rotate_me[7]};
 // or in the opposite direction:
   rotate_me <= {rotate_me[0],rotate_me[7:1]};

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