Algorithm Topic 1

1. Greedy algorithm, change problem

#Greedy Algorithm Money= [100,50,20,5,1]defChange_money (x): change=[0,0,0,0,0] forI,minchEnumerate (Money): Change[i]= x//Money[i] x= x%Money[i]ifX >0:Print("still left%s"%x)return ChangePrint(Change_money (356.2))

2. Greedy algorithm: Digital stitching problem

3. Dynamic planning: Longest ascending subsequence

3. Dynamic planning: The longest common string

def fib (n):
"Fibonacci Series"
f = [All]
For I in range (2, n+1):
F.append (F[-1]+f[-2])
Print (f)
return F[n]

FIB (5)

def LIS (x):
' Longest common subsequence '
F = [0 for _ in range (len (x))]
p = [-1 for _ in range (len (x))]
# initialization
F[0] = 1
P[0] = 1
for k in range (1, Len (F)):
Max_loc =-1
Max_num = 0
# inner loop represents the maximum value of f[i] for all locations less than x[k] in f[0:k]
For I in range (0, K):
If x[i] < X[k]:
If f[i] > max_num:
Max_loc = i
Max_num = F[i]
F[k] = max_num + 1
P[k] = Max_loc

max_i = 0
For I in range (1,len (F)):
If f[i] > f[max_i]:
Max_i = i

Lis = []
i = Max_i
While I >= 0:
Lis.append (X[i])
i = P[i]
Lis.reverse ()
return LIS

Print (LIS ([9,7,2,8,3,5,2]))

def LCS (x, y):
# The longest common subsequence (two-dimensional)
F = [[0 for _ in range (len (y) +1)] ' for ' in range (len (x) +1)]
p = [[0 for _ in range (len (y) +1)] for _ in range (len (x) +1)]
For I in range (1, len (x) +1):
P[i][0] = 2
For j in range (1, len (y) +1):
P[0][J] = 1

# 0 Oblique 1 Transverse j-1 2 vertical i-1
For I in range (1, len (x) +1):
For j in range (1, len (y) +1):
If x[i-1] = = Y[j-1]:
F[I][J] = f[i-1][j-1]+1
P[I][J] = 0
Else
#F [I][j] = max (F[i-1][j], f[i][j-1])
If F[I-1][J] > f[i][j-1]:
F[I][J] = F[i-1][j]
P[I][J] = 2
Else
F[I][J] = f[i][j-1]
P[I][J] = 1

LCS = []
i = Len (x)
j = Len (y)
While I > 0 or J > 0:
If p[i][j] = = 0:
Lcs.append (X[i-1])
I-= 1
J-= 1
Elif P[i][j] = = 1:
J-= 1
Else
I-= 1
Lcs.reverse ()
return LCS
#return F[i][j]

Print (LCS ("Abbcbde", "dbbcdb"))

def lcss_bf (x, y):
# the longest common string (violence)
m = Len (x)
n = Len (y)
Max_len = 0
Max_str = ""
for k in range (0, min (m,n)):
For I in range (0, m-k+1):
For j in range (0, n-k+1):
If x[i:i+k] = = Y[j:j+k]:
If k > Max_len:
Max_len = k
MAX_STR = X[i:i+k]
Return MAX_STR

Print (LCSS_BF ("Abbcbde", "dbbcdb"))

def lcss (x, y):
# the longest common string
F = [[0 for _ in range (len (y) +1)] ' for ' in range (len (x) +1)]
p = [[0 for _ in range (len (y) +1)] for _ in range (len (x) +1)]
# 0 does not match 1 matches
For I in range (1, len (x) +1):
For j in range (1, len (y) +1):
If x[i-1] = = Y[j-1]:
F[I][J] = f[i-1][j-1]+1
P[I][J] = 1
Else
F[I][J] = 0
P[I][J] = 0
Max_val = 0
max_i = 0
Max_j = 0
For I in range (1, len (x) +1):
For j in range (1, len (y) +1):
If F[I][J] > Max_val:
Max_val = F[i][j]
Max_i = i
Max_j = J
#tracback
LCSS = []
i = Max_i
j = Max_j
While p[i][j] = = 1:
Lcss.append (X[i-1])
I-= 1
J-= 1

Lcss.reverse ()
Return LCSS

Print (LCSs ("Abbcbde", "dbbcdb"))

Algorithm Topic 1