Ok, here it is. I currently belive that this publishing makes the
system un-patentable by anyone but me (and I can only patent it in the
US, not in the EC). It is my intent that this algorithm be unfettered
by copyright, liscence, trademark, patent, or any other icky
intelectual property right. So let me state here that the algorith is
in the public domain. I release all copyright to it. There, i hope
that does it.
But if I'm wrong, oh well. I don't think there is much economic
worth in this scheme. But, I would be happy to be proven wrong! I
expect that the odds that this system actually work are pretty long.
But I've been over it too much, and can't see any holes, its time for
others to poke at it. Besides, I like the tase of crow.
j'
This is an -*-outline-*- of my public key crypto system
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* Informal introduction
** Description of the system
*** Key generation
In total secrecy, Andy generates two graphs, one for encoding 1's and
the other for encoding 0's. He then openly publishes these two graphs.
*** Sending a bit
In total secrecy, Beth selects one of the two graphs, and generates a
new graph isomorphic to the selected graph based. Then Beth publicly
sends the new graph to Andy.
*** Recieving a bit
To decrypt which bit Andy recieved, he must determin which graph Beth
selected, and permuted. He must solve one case of the GI problem. To
make this easy, he has hidden trapdoor indentifiers in the published
graphs. Using my special JGI algorithm, and the trapdoor identifiers,
Andy will be able to discover which bit Beth sent.
*** The trapdoor information
To make hiding a trapdoor identifier possible, Andy also publishes a
labeling of the two graphs. For each node and each edge in the
published graphs, Andy associates a labeling string. (He uses 2k-bit
binary numbers as labels.)
When he constructs the graphs, Andy insures that each one has a
Hamiltonian Circuit. The trapdoor information is the labeling of the
Hamiltonian Circuits of the two graphs. Naturally, each graph has a
different Hamiltonian Circuit from the other, with a different labeling.
** Informal security argument
For Eve to be able to determin the bit sent from Beth to Andy, she
must be able to either solve instances of the Graph Isomorphism problem,
or find the trapdoor identifier in the graph that Beth sends to Andy and
also in the two published graphs.
(I will ignore the posibility that Andy's and Beth's 'total secrecy'
is penetrable by Eve. She might have psychic powers, or access to
sophisticated spying technology. If this is the case, too bad for Andy
and Beth.)
*** The Graph Isomorphism problem
Graph Isomorphism (GI) is a problem for which people believe there is
no polynomial time solution. Although GI is belived to be easyer than
problems known to be NP complete. So we belive that Eve has a fairly
hard problem ahead of her, although the problem might not quite fit the
usual definition of intractable.
*** The Hameltonain Circuit problem
Instead Eve could try to discover the trapdoor information. But since
the Hamiltonian Circuit Decision problem is NP complete, and since NP
complete problems are (belived) at least as hard as GI, it doesn't seem
that there is much profit for Eve to try this aproach.
* The formal version
** Key generation
For a particular security parameter k, the published key consists of
an ordered pair of graphs <G0, G1>. G0 is used for sending 0 bits, and
G1 for sending 1 bits.
Both G0 and G1 contain 2^4k nodes, and 2^4k*2^2k==2^6k edges. Each
graph contains a Hameltonian Circuit. Each node, and each edge of each
graph is labled with a member of {0,1}^k (the set of bit strings k bits
long). Each node has exactly 2^2k outgoing edges (and 2^2k incomming?).
To construct a graph, begin with a random set of labled nodes.
Construct the Hameltonian Circuit by adding edges from vi to vj, each
with a random label. Note (one of) the string(s) which is formed by
appending the node and edge lables in order along the Hameltonian
Circuit. This is the trapdoor information which makes the graph
isomorphism problem easy. Next add edges to the graph until each node
has exactly 2^2k outgoing edges, label each edge at random.
(Here is where I should talk about how the GI problem is only rarely
hard, and that the edges labeled at random garantees that we _sometimes_
land in the hard susbset of the GI problem. It would be nice to make a
better construction which always landed in the hard subset of GI. But
this is likely to be a hard research problem. Oh well.)
** Sending a bit
Reciever sends two graphs as described above to the sender. The
sender decides which bit to send -- 1 or 0. The sender then selects a
permutation P of the nodes of the apropriate graph. The sender then
sends the isomorphic graph defined by the permutation P to the reciever.
The reciever uses my GI algorithm to determin which graph was sent.
** Recieving a bit
The reciever runs the folowing algorithm twice in parallel, and the
algorithm to finish first determins which graph was sent. The other
algorithm is terminated (since its result is unnecesary.)
*** Description of the algorithm
The JGI algorithm takes as input a trapdoor string T of labels <tn1,
te1, tn2, te2...> (tni, and tei are strings of binary digits), and a
graph G=<V,E> of |V| nodes. It either halts and accepts the input, or
halts and rejects the input. After initializing, the algorithm will
halt in exactly V iterations of the main loop.
**** Initialization
For each node v in the graph, if the node's label matches the first
label in the trapdoor, create a set sv containing v. Also create a
pair pv of <v, sv>. Finally add the pair pv to the active set.
Remove the first label from the trapdoor string.
**** Main Loop
While the trapdoor string T is not empty and the active set is not
empty, do the Outer Loop. After performing the outer loop, make the
next active set be the active set, and then remove the first two
labels from the trapdoor string.
***** Outer Loop
For each pair pi=<vi, svi> in the active set, do the Inner Loop.
****** Inner Loop
For each edge e=<vj, vk> in E where vi==vj, if T's first label
matches e's label, and if vk is not in svi, and if T's second label
matches vk's label, then add the pair pi'=<vk, svi union {vk}> to the
next active set.
**** Final step
If the trapdoor string is empty, halt and accept. If the active set
is empty, and the trapdoor string is not, halt and reject.
*** Proof of polynomial time and space behavior
(This is a little weak, but I belive it can fly.)
The main loop executes no more than |V| times since the trapdoor
string contains exactly |V| node labels, and each iteration removes
one of them. The important question is how many new pairs are added
to the next active set, for each pair in the active set, by the outer
and inner loops. For one of my graphs, the expected number is (less
than) one. To see this note that the product of number of edge labels
and the number of node labels equals the numbe of edges leaving a
node. However, the test to see if the new vk is already a member of
the old svi reduces this number.
** Proof of security
The evesdropper must solve the GI problem for the subset of graphs
constructed, or must discover the trapdoor information, and use my GI
algorithm. To show how hard this is, I will show that GI of the
subset of graphs generated is (polynomial time) GI complete, and I
will show that discovering the trapdoor information is as hard as the
Hameltonian circuit path discovery problem.
*** The reduction to HP
Now how am I going to do this? Ideas are solicited.
*** The reduction to GI
(All I actually present are the constructions for the reductions. I
don't proove that isomorphism and (where apropriate) hameltonian
posetion is retained. But I am convinced. Just tiered of typeing.)
I will write GI for graph isomorphism, LGI for labeld graph
isomorphism, HLGI to Hameltonian posesing labeled graph isomorphism,
FAHLGI for fixed (at |V|^1/2) arity Hameltonian posesing labeled graph
isomorphism.
The subset of graphs that are generated in the key generation process
are exactly those of the FAHLGI problem. (This is true by
construction.)
**** FAHLGI <= GI <= FAHLGI
I will now prove that FAHLGI <= GI <= FAHLGI. I will prove this by
the chain FAHLGI <= HLGI <= FAHLGI, HLGI <= LGI <= HLGI, LGI <= GI <=
LGI.
***** FAHLGI <= HLGI <= FAHLGI
****** FAHLGI <= HLGI
Obvious: Since FAHLGI is a subset of HLGI, a HLGI algorithm will work
just fine when given graphs from the FAHLGI problem.
****** HLGI <= FAHLGI
Replace each node with a clique of size |V|. Label the nodes in the
clique as the original node, and the edges in the clique 00. For each
ordered pair of nodes <v1,v2> in V^2, add an edge from one of the nodes
in the clique for v1 to one of the nodes in the clique for v2. Label
the new edge 11x if the there is an edge <v1,v2> in E and its label is
x, label the new edge label 10x for some random x, if <v1,v2> is not in
E.
***** HLGI <= LGI <= HLGI
****** HLGI <= LGI
Obvious: Since HLGI is a subset of LGI, a LGI algorithm will work just
fine when given graphs from the HLGI problem.
****** LGI <= HLGI
For each v labeled x, construct v', v'' labeled 0x and 1x resp. For
each v', and each v'', add the edges <v', v''> and <v'', v'> each
labeled 0x for some random x. For each e= <v1, v2> in E labeled x add
e'= <v1', v2'> labeled 1x.
***** LGI <= GI <= LGI
****** LGI <= GI
For each node label add a new node, and an edge from the new node to
each of the nodes so labeled. For each edge, add an intermediate node.
For each label of the edges, construct a new node, and edges from it to
the new edge nodes.
****** GI <= LGI
Obvious construction: give each node and edge the label 0.
