def can_be_interwoven(s, t, u):
'''Return True if and only if the motifs t and u can be interwoven into the string s.'''
nt, nu = len(t), len(u)
# Allocation + initial condition at i=0. First and last column correspond to the empty motifs.
# At step i in the outer loop (x=s[i]), m[j,k] = (s[0..i] ENDS with an interweaving of
# t[0..j], u[0..k] can be interwoven); m_old holds m[i-1, :, :].
m, m_old = ro.zeros_bool_array(nt + 1, nu + 1), ro.zeros_bool_array(nt + 1, nu + 1)
m[0][0] = -1; # m_max = m[-1][-1]
# Dynamic programming - prefixes of s
for x in s:
ra.copy_list(m, m_old)
# Initial condition - first row (j=0)
for k, uk in enumerate(u, 1):
m[0][k] = m_old[0][k - 1] if x == uk else 0
# Initial condition - first column (k=0)
for j, tj in enumerate(t, 1):
m[j][0] = m_old[j - 1][0] if x == tj else 0
# Dynamic programming - prefixes of t,u in s[0..i] vs. in s[0..i-1].
for j, tj in enumerate(t, 1):
for k, uk in enumerate(u, 1):
mjk = 0
if x == tj: mjk |= m_old[j - 1][ k]
if x == uk: mjk |= m_old[j][k - 1]
m[j][k] = mjk
# Shortcut: since it's a boolean result and we are OR-ing, once it's True it
# must stay True, so no need to complete the entire outer loop over s. Had we
# needed an integer result, we'd have to do the maximization over all s prefixes
if m[-1][-1]: return True
#m_max |= m[-1][-1]
#return m_max
return False
def itwv_matrix(s, t):
'''Return the boolean matrix M.'''
# By symmetry, we only need to call can_be_interwoven() for M's upper triangular part
n = len(t)
m = ro.zeros_bool_array(n, n)
for i, ti in enumerate(t):
for j in xrange(i, n):
m[i][j] = can_be_interwoven(s, ti, t[j])
m[j][i] = m[i][j]
return m