Rot13 for numbers

问题:

EDIT: Now a Major Motion Blog Post at http://messymatters.com/sealedbids

The idea of rot13 is to obscure text, for example to prevent spoilers. It's not meant to be cryptographically secure but to simply make sure that only people who are sure they want to read it will read it.

I'd like to do something similar for numbers, for an application involving sealed bids. Roughly I want to send someone my number and trust them to pick their own number, uninfluenced by mine, but then they should be able to reveal mine (purely client-side) when they're ready. They should not require further input from me or any third party.

(Added: Note the assumption that the recipient is being trusted not to cheat.)

It's not as simple as rot13 because certain numbers, like 1 and 2, will recur often enough that you might remember that, say, 34.2 is really 1.

Here's what I'm looking for specifically:

A function seal() that maps a real number to a real number (or a string). It should not be deterministic -- seal(7) should not map to the same thing every time. But the corresponding function unseal() should be deterministic -- unseal(seal(x)) should equal x for all x. I don't want seal or unseal to call any webservices or even get the system time (because I don't want to assume synchronized clocks). (Added: It's fine to assume that all bids will be less than some maximum, known to everyone, say a million.)

Sanity check:

> seal(7) 482.2382   # some random-seeming number or string. > seal(7) 71.9217    # a completely different random-seeming number or string. > unseal(seal(7)) 7          # we always recover the original number by unsealing. 

回答1:

You can pack your number as a 4 byte float together with another random float into a double and send that. The client then just has to pick up the first four bytes. In python:

import struct, random def seal(f):    return struct.unpack("d",struct.pack("ff", f, random.random() ))[0] def unseal(f):    return struct.unpack("ff",struct.pack("d", f))[0]  >>> unseal( seal( 3)) 3.0 >>> seal(3) 4.4533985422978706e-009 >>> seal(3) 9.0767582382536571e-010 

回答2:

Here's a solution inspired by Svante's answer.

M = 9999  # Upper bound on bid. seal(x) = M * randInt(9,99) + x unseal(x) = x % M 

Sanity check:

> seal(7) 716017 > seal(7) 518497 > unseal(seal(7)) 7 

This needs tweaking to allow negative bids though:

M = 9999  # Numbers between -M/2 and M/2 can be sealed. seal(x) = M * randInt(9,99) + x unseal(x) =    m = x % M;    if m > M/2 return m - M else return m 

A nice thing about this solution is how trivial it is for the recipient to decode -- just mod by 9999 (and if that's 5000 or more then it was a negative bid so subtract another 9999). It's also nice that the obscured bid will be at most 6 digits long. (This is plenty security for what I have in mind -- if the bids can possibly exceed $5k then I'd use a more secure method. Though of course the max bid in this method can be set as high as you want.)

Instructions for Lay Folk

Pick a number between 9 and 99 and multiply it by 9999, then add your bid. This will yield a 5 or 6-digit number that encodes your bid. To unseal it, divide by 9999, subtract the part to the left of the decimal point, then multiply by 9999. (This is known to children and mathematicians as "finding the remainder when dividing by 9999" or "mod'ing by 9999", respectively.)

This works for nonnegative bids less than 9999 (if that's not enough, use 99999 or as many digits as you want). If you want to allow negative bids, then the magic 9999 number needs to be twice the biggest possible bid. And when decoding, if the result is greater than half of 9999, ie, 5000 or more, then subtract 9999 to get the actual (negative) bid.

Again, note that this is on the honor system: there's nothing technically preventing you from unsealing the other person's number as soon as you see it.

回答3:

If you're relying on honesty of the user and only dealing with integer bids, a simple XOR operation with a random number should be all you need, an example in C#:

static Random rng = new Random();  static string EncodeBid(int bid) {     int i = rng.Next();     return String.Format("{0}:{1}", i, bid ^ i); }  static int DecodeBid(string encodedBid) {     string[] d = encodedBid.Split(":".ToCharArray());     return Convert.ToInt32(d[0]) ^ Convert.ToInt32(d[1]); } 

Use:

int bid = 500; string encodedBid = EncodeBid(bid); // encodedBid is something like 54017514:4017054 and will be different each time int decodedBid = DecodeBid(encodedBid); // decodedBid is 500 

Converting the decode process to a client side construct should be simple enough.

回答4:

Is there a maximum bid? If so, you could do this:

Let max-bid be the maximum bid and a-bid the bid you want to encode. Multiply max-bid by a rather large random number (if you want to use base64 encoding in the last step, max-rand should be (2^24/max-bid)-1, and min-rand perhaps half of that), then add a-bid. Encode this, e.g. through base64.

The recipient then just has to decode and find the remainder modulo max-bid.

回答5:

What you want to do (a Commitment scheme) is impossible to do client-side-only. The best you could do is encrypt with a shared key.

If the client doesn't need your cooperation to reveal the number, they can just modify the program to reveal the number. You might as well have just sent it and not displayed it.

To do it properly, you could send a secure hash of your bid + a random salt. That commits you to your bid. The other client can commit to their bid in the same way. Then you each share your bid and salt.

[edit] Since you trust the other client:

Sender: Let M be your message K = random 4-byte key C1 = M xor hash(K) //hash optional: hides patterns in M xor K //(you can repeat or truncate hash(K) as necessary to cover the message) //(could also xor with output of a PRNG instead) C2 = K append M //they need to know K to reveal the message send C2 //(convert bytes to hex representation if needed)  Receiver: receive C2 K = C2[:4] C1 = C2[4:] M = C1 xor hash(K) 

回答6:

Are you aware that you need a larger 'sealed' set of numbers than your original, if you want that to work?

So you need to restrict your real numbers somehow, or store extra info that you don't show.

回答7:

One simple way is to write a message like:

"my bid is: $14.23: aduigfurjwjnfdjfugfojdjkdskdfdhfddfuiodrnfnghfifyis"

All that junk is randomly-generated, and different every time.

Send the other person the SHA256 hash of the message. Have them send you the hash of their bid. Then, once you both have the hashes, send the full message, and confirm that their bid corresponds to the hash they gave you.

This gives rather stronger guarantees than you need - it's actually not possible from them to work out your bid before you send them your full message. However, there is no unseal() function as you describe.

This simple scheme has various weaknesses that a full zero-knowledge scheme would not have. For example, if they fake you out by sending you a random number instead of a hash, then they can work out your bid without revealing their own. But you didn't ask for bullet-proof. This prevents both accidental and (I think) undetectable cheating, and uses only a commonly-available command line utility, plus a random number generator (dice will do).

If, as you say, you want them to be able to recover your bid without any further input from you, and you are willing to trust them only to do it after posting their bid, then just encrypt using any old symmetric cipher (gpg --symmetric, perhaps) and the key, "rot13". This will prevent accidental cheating, but allow undetectable cheating.

回答8:

One idea that poped into my mind was to maybe base your algorithm on the mathematics used for secure key sharing.

If you want to give two persons, Bob and Alice, half a key each so that only when combining them they will be able to open whatever the key locks, how do you do that? The solution to this comes from mathematics. Say you have two points A (-2,2) and B (2,0) in a x/y coordinate system.

               |        A       +                |                C                | ---+---+---+---|---+---B---+---+---+---                |                +                |                + 

If you draw a straight line between them it will cross the y axis at exactly one single point, C (0,1). If you only know one of the points A or B it is impossible to tell where it will cross. Thus you can let the points A and B be the shared keys which when combined will reveal the y-value of the crossing point (i.e. 1 in this example) and this value is then typically used as a real key for something.

For your bidding application you could let seal() and unseal() swap the y-value between the C and B points (deterministic) but have the A point vary from time to time.

This way seal(y-value of point B) will give completely different results depending on point A, but unseal(seal(y-value of point B)) should return the y-value of B which is what you ask for.

PS It is not required to have A and B on different sides of the y-axis, but is much simpler conceptually to think of it this way (and I recommend implementing it that way as well).

With this straight line you can then share keys between several persons so that only two of them are needed to unlock whatever. It is possible to use curve types other then straight lines to create other key sharing properties (i.e. 3 out of 3 keys are required etc).

回答9:

Pseudo code:

encode:

value = 2000 key = random(0..255); // our key is only 2 bytes  // 'sealing it' value = value XOR 2000;  // add key sealed = (value << 16) | key 

decode:

key = sealed & 0xFF unsealed = key XOR (sealed >> 16) 

Would that work?

回答10:

Since it seems that you are assuming that the other person doesn't want to know your bid until after they've placed their own, and can be trusted not to cheat, you could try a variable rotation scheme:

from random import randint  def seal(input):     r = randint(0, 50)     obfuscate = [str(r)] + [ str(ord(c) + r) for c in '%s' % input ]     return ':'.join(obfuscate)  def unseal(input):     tmp = input.split(':')     r = int(tmp.pop(0))     deobfuscate = [ chr(int(c) - r) for c in tmp ]     return ''.join(deobfuscate)  # I suppose you would put your bid in here, for 100 dollars tmp = seal('$100.00') # --> '1:37:50:49:49:47:49:49' (output varies) print unseal(tmp) # --> '$100.00' 

At some point (I think we may have already passed it) this becomes silly, and because it is so easy, you should just use simple encryption, where the message recipient always knows the key - the person's username, perhaps.

回答11:

If the bids are fairly large numbers, how about a bitwise XOR with some predetermined random-ish number? XORing again will then retrieve the original value.
You can change the number as often as you like, as long as both client and server know it.

回答12:

You could set a different base (like 16, 17, 18, etc.) and keep track of which base you've "sealed" the bid with...

Of course, this presumes large numbers (> the base you're using, at least). If they were decimal, you could drop the point (for example, 27.04 becomes 2704, which you then translate to base 29...)

You'd probably want to use base 17 to 36 (only because some people might recognize hex and be able to translate it in their head...)

This way, you would have numbers like G4 or Z3 or KW (depending on the numbers you're sealing)...

回答13:

Here's a cheap way to piggyback off rot13:

Assume we have a function gibberish() that generates something like "fdjk alqef lwwqisvz" and a function words(x) that converts a number x to words, eg, words(42) returns "forty two" (no hyphens).

Then define

seal(x) = rot13(gibberish() + words(x) + gibberish()) 

and

unseal(x) = rot13(x) 

Of course the output of unseal is not an actual number and is only useful to a human, but that might be ok. You could make it a little more sophisticated with words-to-number function that would also just throw away all the gibberish words (defined as anything that's not one of the number words -- there are less than a hundred of those, I think).

Sanity check:

> seal(7) fhrlls hqufw huqfha frira afsb ht ahuqw ajaijzji > seal(7) qbua adfshua hqgya ubiwi ahp wqwia qhu frira wge > unseal(seal(7)) sueyyf udhsj seven ahkua snsfo ug nuhdj nwnvwmwv 

I know this is silly but it's a way to do it "by hand" if all you have is rot13 available.

文章来源: Rot13 for numbers