EASTNORTHEAST - This is not exactly the hint Jim Sanborn (JS) gave for K4 on the 29th of January this year. He only gave NORTHEAST - which refers to the positions 26-34 of K4's plaintext. Beside BERLIN and CLOCK it is the third revealed plaintext word of K4. However, also this hint does not seem to help much. However, it just so happened, that a member in the yahoo kryptos group had a conversation with Jim Sanborn due to a submitted solution. Sandborn's answer to the question contained again the last clue which surprisingly was EASTNORTHEAST at position 22-34. Jim Sanborns compass rose at CIA There is disagreement if Jim revealed this on purpose or he did it accidentially, but the new extended clue seem to be serious and valid.Interestingly, EASTNORTHEAST is exactly the direction which is illustrated on the compass rose on one of the stones around kryptos, also created by Jim Sanborn. Actually, i dont really kn...
This post is dedicated to give you an implementation (in SAGE) of the FKT-algorithm. I coded this while spending some time on the ideas i explained in this blog post. I searched several hours to find an existing implementation without success. Since it took me a while to code this on my own, i think it is worth to present it, therewith other can use it in their own work. It works well, but if you find an error, please leave a note.
The FKT-algorithm can be used to count the number of perfect matchings in a planar graph. And a graph is planar if and only if it does not have K$_5$ or K$_{3,3}$ as a minor.
A paper that will be presented at the CCC conference this year is called:
########################################################################
# AUTHOR: Dr. Christian Schridde
# E-MAIL: christianschridde [at] googlemail [dot] com
#
# DESCRIPTION: Implementation of the FKT-algorithm
#
# INPUT: Adjacency matrix A of a undirected loop-free planar graph G
# OUTPUT: Skew matrix M, such that PerfMatch(G) = Sqrt(Determinat(M))
########################################################################
def FKT(A):
# make some local copies
B_graph = copy(A);
B_digraph = copy(A);
# FIRST: REMOVE SELF-LOOPS
for i in range(0,B_graph.nrows()):
B_graph[i,i] = 0;
B_digraph[i,i] = 0;
# SECOND: TRANSFORM B_graph TO AN UNDIRECTED GRAPH
G = DiGraph(B_graph);
G = G.to_undirected();
B_graph = G.adjacency_matrix();
# THIRD: COMPUTE A PLANAR EMBEDDING
if (G.is_planar(set_embedding=true, set_pos=true) == false):
print "G is not planar";
return null;
embd = G.get_embedding();
# each face is listed in clockwise order, except the first one which is the outer face
faces = G.trace_faces(embd);
# FOURTH: GET THE SPANNING TREE
T1 = G.min_spanning_tree(algorithm='Prim_fringe');
T1 = Graph(T1);
# FIFTH: ASSIGN AN ARBITRARY DIRECTION TO THE EDGES
DG = DiGraph(B_digraph);
DG = DG.to_undirected();
B_digraph = DG.adjacency_matrix();
for i in range(0,B_digraph.nrows()):
for j in range(0,B_digraph.nrows()):
if (B_digraph[i,j] == 1):
if (B_digraph[j,i] == 1):
r = randint(0,1)
if (r==0):
B_digraph[j,i] = 0;
else:
B_digraph[i,j] = 0;
G = DiGraph(B_digraph);
# SIXTH: FIND FACE THAT HAS ONLY ONE EDGE NOT IN T1
edgesT1 = T1.edges(labels=false);
adj_T1 = T1.adjacency_matrix();
# give the edges in T1 the orientation from B_digraph
for edge in edgesT1:
if (B_digraph[edge[0],edge[1]] == 0):
adj_T1[edge[0],edge[1]] = 0;
else:
adj_T1[edge[1],edge[0]] = 0;
T1 = DiGraph(adj_T1);
edgesT1 = T1.edges(labels=false);
# remove the first face, which is the outer face
# QUESTION: Is it always correct to remove this face?
faces.pop(0);
while (len(faces) > 0):
index = -1;
for face in faces:
countMissingEdges = 0;
missingEdge = 0;
index += 1;
for edge in face:
try:
idx1 = edgesT1.index(edge);
except ValueError:
try:
idx2 = edgesT1.index(reverseEdge(edge));
except ValueError:
countMissingEdges += 1;
missingEdge = edge;
else:
doNothing();
else:
doNothing();
if (countMissingEdges == 1):
# in this face, only one edge is missing.
# Place the missing edge such that the total number
# of clockwise edges of this face is odd
# add this edge to the spanning tree
if (is_odd(numberOfClockwiseEdges(face,edgesT1))):
# insert counterclockwise in adj_T1;
if (isClockwise(missingEdge,face) == false):
adj_T1[missingEdge[0],missingEdge[1]] = 1;
else:
adj_T1[missingEdge[1],missingEdge[0]] = 1;
else:
# insert clockwise in adj_T1
if (isClockwise(missingEdge,face) == true):
adj_T1[missingEdge[0],missingEdge[1]] = 1;
else:
adj_T1[missingEdge[1],missingEdge[0]] = 1;
# rebuild the graph
T1 = DiGraph(adj_T1);
edgesT1 = T1.edges(labels=false);
# remove the face that was found
faceFound = faces.pop(index);
break;
return toSkewSymmetricMatrix(adj_T1);
Also, you need the following helper functions:
############################
# Returns true if the given edge is
# clockwise oriented regarding the given face
def isClockwise(e,face):
try:
face.index(e);
except ValueError:
return false;
else:
return true;
############################
# This is a placeHolder function
def doNothing():
return 0;
############################
# Reverses a given edge
def reverseEdge(edge):
return (edge[1],edge[0]);
############################
# Inputs are the face and the orientedEdges from the spanning-tree T1
# Note, that all edges in face are clockwise (since it is the result
# from the obtained embedding).
# It returns how many of the edges from T1 that are part of the
# given face are actually clockwise.
def numberOfClockwiseEdges(face, edgesT1):
clockwise = 0;
for edge in face:
try:
edgesT1.index(edge);
except ValueError:
doNothing();
else:
clockwise += 1;
return clockwise;
############################
# Transforms a given matrix A to a
# skewSymmetric Matrix
def toSkewSymmetricMatrix(A):
for i in range(0,A.nrows()):
for j in range(0,A.nrows()):
if (A[i,j] == 1):
A[j,i] = -1;
return A;
The FKT-algorithm can be used to count the number of perfect matchings in a planar graph. And a graph is planar if and only if it does not have K$_5$ or K$_{3,3}$ as a minor.
A paper that will be presented at the CCC conference this year is called:
Counting the Number of Perfect Matchings in $K_5$-free Graphs [in polynomial time] Simon Straub (Ulm University), Thomas Thierauf (Aalen University), Fabian Wagner (Ulm University)I did not manage to read it yet, but i am curious to see, how they circumvent the K$_{3,3}$ minor problem.
########################################################################
# AUTHOR: Dr. Christian Schridde
# E-MAIL: christianschridde [at] googlemail [dot] com
#
# DESCRIPTION: Implementation of the FKT-algorithm
#
# INPUT: Adjacency matrix A of a undirected loop-free planar graph G
# OUTPUT: Skew matrix M, such that PerfMatch(G) = Sqrt(Determinat(M))
########################################################################
def FKT(A):
# make some local copies
B_graph = copy(A);
B_digraph = copy(A);
# FIRST: REMOVE SELF-LOOPS
for i in range(0,B_graph.nrows()):
B_graph[i,i] = 0;
B_digraph[i,i] = 0;
# SECOND: TRANSFORM B_graph TO AN UNDIRECTED GRAPH
G = DiGraph(B_graph);
G = G.to_undirected();
B_graph = G.adjacency_matrix();
# THIRD: COMPUTE A PLANAR EMBEDDING
if (G.is_planar(set_embedding=true, set_pos=true) == false):
print "G is not planar";
return null;
embd = G.get_embedding();
# each face is listed in clockwise order, except the first one which is the outer face
faces = G.trace_faces(embd);
# FOURTH: GET THE SPANNING TREE
T1 = G.min_spanning_tree(algorithm='Prim_fringe');
T1 = Graph(T1);
# FIFTH: ASSIGN AN ARBITRARY DIRECTION TO THE EDGES
DG = DiGraph(B_digraph);
DG = DG.to_undirected();
B_digraph = DG.adjacency_matrix();
for i in range(0,B_digraph.nrows()):
for j in range(0,B_digraph.nrows()):
if (B_digraph[i,j] == 1):
if (B_digraph[j,i] == 1):
r = randint(0,1)
if (r==0):
B_digraph[j,i] = 0;
else:
B_digraph[i,j] = 0;
G = DiGraph(B_digraph);
# SIXTH: FIND FACE THAT HAS ONLY ONE EDGE NOT IN T1
edgesT1 = T1.edges(labels=false);
adj_T1 = T1.adjacency_matrix();
# give the edges in T1 the orientation from B_digraph
for edge in edgesT1:
if (B_digraph[edge[0],edge[1]] == 0):
adj_T1[edge[0],edge[1]] = 0;
else:
adj_T1[edge[1],edge[0]] = 0;
T1 = DiGraph(adj_T1);
edgesT1 = T1.edges(labels=false);
# remove the first face, which is the outer face
# QUESTION: Is it always correct to remove this face?
faces.pop(0);
while (len(faces) > 0):
index = -1;
for face in faces:
countMissingEdges = 0;
missingEdge = 0;
index += 1;
for edge in face:
try:
idx1 = edgesT1.index(edge);
except ValueError:
try:
idx2 = edgesT1.index(reverseEdge(edge));
except ValueError:
countMissingEdges += 1;
missingEdge = edge;
else:
doNothing();
else:
doNothing();
if (countMissingEdges == 1):
# in this face, only one edge is missing.
# Place the missing edge such that the total number
# of clockwise edges of this face is odd
# add this edge to the spanning tree
if (is_odd(numberOfClockwiseEdges(face,edgesT1))):
# insert counterclockwise in adj_T1;
if (isClockwise(missingEdge,face) == false):
adj_T1[missingEdge[0],missingEdge[1]] = 1;
else:
adj_T1[missingEdge[1],missingEdge[0]] = 1;
else:
# insert clockwise in adj_T1
if (isClockwise(missingEdge,face) == true):
adj_T1[missingEdge[0],missingEdge[1]] = 1;
else:
adj_T1[missingEdge[1],missingEdge[0]] = 1;
# rebuild the graph
T1 = DiGraph(adj_T1);
edgesT1 = T1.edges(labels=false);
# remove the face that was found
faceFound = faces.pop(index);
break;
return toSkewSymmetricMatrix(adj_T1);
Also, you need the following helper functions:
############################
# Returns true if the given edge is
# clockwise oriented regarding the given face
def isClockwise(e,face):
try:
face.index(e);
except ValueError:
return false;
else:
return true;
############################
# This is a placeHolder function
def doNothing():
return 0;
############################
# Reverses a given edge
def reverseEdge(edge):
return (edge[1],edge[0]);
############################
# Inputs are the face and the orientedEdges from the spanning-tree T1
# Note, that all edges in face are clockwise (since it is the result
# from the obtained embedding).
# It returns how many of the edges from T1 that are part of the
# given face are actually clockwise.
def numberOfClockwiseEdges(face, edgesT1):
clockwise = 0;
for edge in face:
try:
edgesT1.index(edge);
except ValueError:
doNothing();
else:
clockwise += 1;
return clockwise;
############################
# Transforms a given matrix A to a
# skewSymmetric Matrix
def toSkewSymmetricMatrix(A):
for i in range(0,A.nrows()):
for j in range(0,A.nrows()):
if (A[i,j] == 1):
A[j,i] = -1;
return A;
Comments
Post a Comment