I'm new to boolean expressions.
I've been given the task to simplify
F(w,x,y,z) = xy’ + x’z’ + wxz + wx’y by using K map.
I've done it and the result is wx’+w’y’+xyz.
Now I have to "Write it in a standard SOP form. You need to provide the steps through which you get the standard SOP".
And i have no idea how to do it. I thought result after k map is sop.
Yes, you already have it in SOP form. But the second question is about Standard (aka canonical) SOP form. That's much simpler to find than having to use K-maps (but it's often long), it's just the sum of minterms.
I think your solution does not cover all ones. These Karnaugh maps show the original expression, the simplified version (minimal SOP) and the canonical SOP, where every product contains all literals (all given variables or their negation).
The original expression is
F(w,x,y,z) = x·¬y + ¬x·¬z + w·x·z + w·¬x·y
– there are two fours and two pairs circled in the corresponding (first one) K-map.
The original expression simplified using K-map (shown in the second one):
F(w,x,y,z) = x·¬y + ¬x·¬z + w·y·z
is different than yours, but you can check for example with wolframalpha online tool, that it is the simplified original expression.
It is also the minimal DNF, but not a sum of minterms (where the output is equal to 1), because there are not all variables in every product of the sum.
The third K-map shows ten minterms circled. They form the canonical DNF:
F(w,x,y,z) = m0 + m2 + m4 + m5 + m8 + m10 + m11 + m12 + m13 + m15 =
= ¬w·¬x·¬y·¬z + ¬w·¬x·y·¬z + ¬w·x·¬y·¬z + ¬w·x·¬y·z + w·¬x·¬y·¬z
+ w·¬x·y·¬z + w·¬x·y·z + w·x·¬y·¬z + w·x·¬y·z + w·x·y·z
I checked your simplified expression, but there are not all ones covered (even if there were some useful do not care states (marked X)). Maybe you made a typo. Or could there be a typo in the original expression?
We can implement the K-Map algorithm in python for 4 variables, as shown below. The function accepts the Boolean function in SOP (sum of products) form and the names of the variables and returns a simplified reduced representation. Basically you need to create rectangular groups containing total terms in power of two like 8, 4, 2 and try to cover as many elements as you can in one group (we need to cover all the ones).
For example, the function can be represented F(w,x,y,z) = xy’ + x’z’ + wxz + wx’y in SOP form as f(w,x,y,z)=∑(0,2,4,5,8,10,11,12,13,15), as can be seen from the below table:
As can be seen from the output of the next code snippet, the program outputs the simplified form x¬y + ¬x¬z + wyz, where negation of a boolean variable x is represented as ¬x in the code.
from collections import defaultdict
from itertools import permutations, product
def kv_map(sop, vars):
sop = set(sop)
not_covered = sop.copy()
sop_covered = set([])
mts = [] # minterms
# check for minterms with 1 variable
all_3 = [''.join(x) for x in product('01', repeat=3)]
for i in range(4):
for v_i in [0,1]:
if len(not_covered) == 0: continue
mt = ('' if v_i else '¬') + vars[i]
s = [x[:i]+str(v_i)+x[i:] for x in all_3]
sop1 = set(map(lambda x: int(x,2), s))
if len(sop1 & sop) == 8 and len(sop_covered & sop1) < 8: # if not already covered
mts.append(mt)
sop_covered |= sop1
not_covered = not_covered - sop1
if len(not_covered) == 0:
return mts
# check for minterms with 2 variables
all_2 = [''.join(x) for x in product('01', repeat=2)]
for i in range(4):
for j in range(i+1, 4):
for v_i in [0,1]:
for v_j in [0,1]:
if len(not_covered) == 0: continue
mt = ('' if v_i else '¬') + vars[i] + ('' if v_j else '¬') + vars[j]
s = [x[:i]+str(v_i)+x[i:] for x in all_2]
s = [x[:j]+str(v_j)+x[j:] for x in s]
sop1 = set(map(lambda x: int(x,2), s))
if len(sop1 & sop) == 4 and len(sop_covered & sop1) < 4: # if not already covered
mts.append(mt)
sop_covered |= sop1
not_covered = not_covered - sop1
if len(not_covered) == 0:
return mts
# check for minterms with 3 variables similarly (code omitted)
# ... ... ...
return mts
mts = kv_map([0,2,4,5,8,10,11,12,13,15], ['w', 'x', 'y', 'z'])
mts
# ['x¬y', '¬x¬z', 'wyz']
The following animation shows how the above code (greedily) simplifies the Boolean function given in SOP form (the basic goal is to cover all the 1s with minimum number of power-2 blocks). Since the algorithm is greedy it may get stuck to some local minimum, that we need to be careful about.
Related
I have just started learning dynamic programming and was able to do some of the basic problems, such as fibbonaci, the knapsack and a few more problems. Coming across the problem
below, I got stuck and do not know how to proceed forward. What confuses me is what would be the base case in this case, and the overlapping problems. Not knowing
this prevents me from developing a relation. They are not as apparent in this example as they were in the previous ones I have solved thus far.
Suppose we are given some string origString, a string toMatch and some number maxNum greater than or equal to 0. How can we count in how many ways it is possible to take maxNum number of nonempty and nonoverlapping substrings of the string origString to make up the string toMatch?
Example:
If origString = "ppkpke", and toMatch = "ppke"
maxNum = 1: countWays("ppkpke", "ppke", 1) will give 0 because toMatch is not a substring of origString.
maxNum = 2: countWays("ppkpke", "ppke", 2) will give 4 because 4 different combinations of 2 substring made up of "ppkpke" can make "ppke".
Those strings are "ppk" & "e", "pp" & "ke" , "p" & "pke" (excluding "p") and "p" & "pke" (excluding "k")
As an initial word of caution, I’d say that although my solution happens to match the expected output for the tiny test set, it is very likely wrong. It’s up to you to double-check it on other examples you may have etc.
The algorithm walks the longer string and tries to spread the shorter string over it. The incremental state of the algorithm consists of tuples of 3 elements:
long string coordinate i (origString[i] == toMatch[j])
short string coordinate j (origString[i] == toMatch[j])
number of ways we made it into that^^^ state
Then we just walk along the strings over and over again, using stored, previously discovered state, and sum up the total number(s) of ways each state was achieved — in the typical dynamic programming fashion.
For a state to count as a solution, j must be at the end of the short string and the number of iterations of the dynamic algorithm must be equivalent to the number of substrings we wanted at that point (because each iteration added one substring).
It is not entirely clear to me from the assignment whether maxNum actually means something like “exactNum”, i.e. exactly that many substrings, or whether we should sum across all lower or equal numbers of substrings. So the function returns a dictionary like { #substrings : #decompositions }, so that the output can be adjusted as needed.
#!/usr/bin/env python
def countWays(origString, toMatch, maxNum):
origLen = len(origString)
matchLen = len(toMatch)
state = {}
for i in range(origLen):
for j in range(matchLen):
o = i + j
if origString[o] != toMatch[j]:
break
state[(o, j)] = 1
sums = {}
for n in range(1, maxNum):
if not state:
break
nextState = {}
for istart, jstart in state:
prev = state[(istart, jstart)]
for i in range(istart + 1, origLen):
for j in range(jstart + 1, matchLen):
o = i + j - jstart - 1
if origString[o] != toMatch[j]:
break
nextState[(o, j)] = prev + nextState.get((o, j), 0)
sums[n] = sum(state[(i, j)] for i, j in state if j == matchLen - 1)
state = nextState
sums[maxNum] = sum(state[(i, j)] for i, j in state if j == matchLen - 1)
return sums
result = countWays(origString='ppkpke', toMatch='ppke', maxNum=5)
print('for an exact number of substrings:', result)
print(' for up to a number of substrings:', {
n: s for n, s in ((m, sum(result[k] for k in range(1, m + 1)))
for m in range(1, 1 + max(result.keys())))})
This^^^ code is a quick and ugly hack and nothing more. There is a huge room for improvement, including (but not limited to) the use of generator functions (yield), the use of #memoize etc. Here’s some output:
for an exact number of substrings: {1: 0, 2: 4, 3: 8, 4: 4, 5: 0}
for up to a number of substrings: {1: 0, 2: 4, 3: 12, 4: 16, 5: 16}
It would be an interesting (and nicely challenging) exercise to store a bit more of the dynamic state (e.g. to keep it for each n) and then reconstruct and pretty-print (efficiently) the exact string (de)compositions that were counted.
Here is a recursive solution.
Compares the first character of source and target, and if they're equal, choose to either take it (advancing by 1 char in both strings) or not take it (advancing by 1 char in source but not in target). The value of k is decremented everytime a new substring is created; there is an additional variable continued which is True if we're in the middle of building a substring, and False otherwise.
def countWays(source, target, k, continued=False):
if len(target) == 0:
return (k == 0)
elif (k == 0 and not continued) or len(source) == 0:
return 0
elif source[0] == target[0]:
if continued:
return countWays(source[1:], target[1:], k, True) + countWays(source[1:], target[1:], k-1, True) + countWays(source[1:], target, k, False)
else:
return countWays(source[1:], target[1:], k-1, True) + countWays(source[1:], target, k, False)
else:
return countWays(source[1:], target, k, False)
print(countWays('ppkpke', 'ppke', 1))
# 0
print(countWays('ppkpke', 'ppke', 2))
# 4
print(countWays('ppkpke', 'ppke', 3))
# 8
print(countWays('ppkpke', 'ppke', 4))
# 4
print(countWays('ppkpke', 'ppke', 5))
# 0
I'm working on the problem stated in the question statement. I know my solution is correct (ran the program) but I'm curious as to whether or not I'm analyzing my code (below) correctly.
def parens(num)
return ["()"] if num == 1
paren_arr = []
parens(num-1).each do |paren|
paren_arr << paren + "()" unless "()#{paren}" == "#{paren}()"
paren_arr << "()#{paren}"
paren_arr << "(#{paren})"
end
paren_arr
end
parens(3), as an example, will output the following:
["()()()", "(()())", "(())()", "()(())", "((()))"]
Here's my analysis:
Every f(n) value is roughly 3 times as many elements as f(n-1). So:
f(n) = 3 * f(n-1) = 3 * 3 * f(n-2) ~ (3^n) time cost.
By a similar analysis, the stack will be occupied by f(1)..f(n) and so the space complexity should be 3^n.
I'm not sure if this analysis for either time or space is correct. On the one hand, the stack only holds n function calls, but each of these calls returns an array ~3 times as big as the call before it. Does this factor into space cost? And is my time analysis correct?
As others have mentioned, your solution is not correct.
My favourite solution to this problem generates all the valid combinations by repeatedly incrementing the current string to the lexically next valid combination.
"Lexically next" breaks down into a few rules that make it pretty easy:
The first difference in the string changes a '(' to a ')'. Otherwise the next string would be lexically before the current one.
The first difference is as far to the right as possible. Otherwise there would be smaller increments.
The part after the first difference is lexically minimal, again because otherwise there would be smaller increments. In this case that means that all the '('s come before all the ')'.
So all you have to do is find the rightmost '(' that can be changed to a ')', flip it, and then append the correct number of '('s and ')'s.
I don't know Ruby, but in Python it looks like this:
current="(((())))"
while True:
print(current)
opens=0
closes=0
pos=0
for i in range(len(current)-1,-1,-1):
if current[i]==')':
closes+=1
else:
opens+=1
if closes > opens:
pos=i
break
if pos<1:
break
current = current[:pos]+ ")" + "("*opens + ")"*(closes-1)
Output:
(((())))
((()()))
((())())
((()))()
(()(()))
(()()())
(()())()
(())(())
(())()()
()((()))
()(()())
()(())()
()()(())
()()()()
Solutions like this turn out to be easy and fast for many types of "generate all the combinations" problems.
Recursive reasoning makes a simple solution. If the number of left parens remaining to emit is positive, emit one and recur. If the number of right parens remaining to emit is greater than the number of left, emit and recur. The base case is when all parens, both left and right, have been emitted. Print.
def parens(l, r = l, s = "")
if l > 0 then parens(l - 1, r, s + "(") end
if r > l then parens(l, r - 1, s + ")") end
if l + r == 0 then print "#{s}\n" end
end
As others have said, the Catalan numbers give the number of strings that will be printed.
While this Ruby implementation doesn't achieve it, a lower level language (like C) would make it easy to use a single string buffer: O(n) space. Due to substring copies, this one is O(n^2). But since the run time and output length are O(n!), O(n) space inefficiency doesn't mean much.
I found Tom Davis' article, "Catalan Numbers," very helpful in explaining one recursive method for defining the Catalan Numbers. I'll try to explain it myself (in part, to see how much of it I've understood) as it may be applied to finding the set of all unique arrangements of N matched parentheses (e.g., 1 (); 2 ()(), (()); etc. ).
For N > 1 let (A)B represent one arrangement of N matched parentheses, where A and B each have only balanced sets of parentheses. Then we know that if A contains k matched sets, B must have the other N - k - 1, where 0 <= k <= N - 1.
In the following example, a dot means the group has zero sets of parentheses:
C_0 => .
C_1 => (.)
To enumerate C_2, we arrange C_1 as AB in all ways and place the second parentheses around A:
. () = AB = C_0C_1 => (.)()
() . = AB = C_1C_0 => (()) .
Now for C_3, we have three partitions for N - 1, each with its own combinations: C_0C_2, C_1C_1, C_2C_0
C_0C_2 = AB = . ()() and . (()) => ()()(), ()(())
C_1C_1 = AB = ()() => (())()
C_2C_0 = AB = ()() . and (()) . => (()()), ((()))
We can code this method by keeping a set for each N and iterating over the combinations for each partition. We'll keep the individual arrangements as bits: 0 for left and 1 for right (this appears backwards when cast as a binary string).
def catalan
Enumerator.new do |y|
# the zero here represents none rather than left
s = [[0],[2]]
y << [0]
y << [2]
i = 2
while true
s[i] = []
(0..i - 1).each do |k|
as = s[k]
bs = s[i - k - 1]
as.each do |a|
bs.each do |b|
if a != 0
s[i] << ((b << (2*k + 2)) | (1 << (2*k + 1)) | (a << 1))
else
s[i] << (2 | (b << 2))
end
end # bs
end # as
end # k
y.yield(s[i])
i = i + 1
end # i
end # enumerator
end
catalan.take(4)
# => [[0], [2], [10, 12], [42, 50, 44, 52, 56]]
The yielder is lazy: although the list is infinite, we can generate as little as we like (using .take for example):
catalan.take(4).last.map{|x| x.to_s(2)}
# => ["101010", "110010", "101100", "110100", "111000"]
The former generation obliges us to keep all previous sets in order to issue the next. Alternatively, we can build a requested set through a more organic type, meandering recursion. This next version yields each arrangement to the block, so we can type:
catalan(4){
|x| (0..7).reduce(""){
|y,i| if x[i] == 0 then y + "(" else y + ")" end
}
}.take(14)
# => ["(((())))", "((()()))", "((())())", "((()))()", "(()(()))", "(()()())",
# "(()())()", "(())(())", "(())()()", "()((()))", "()(()())", "()(())()",
# "()()(())", "()()()()"]
Direct generation:
def catalan(n)
Enumerator.new do |y|
s = [[0,0,0]]
until s.empty?
left,right,result = s.pop
if left + right == 2 * n
y << yield(result)
end
if right < left
s << [left, right + 1, result | 1 << (left + right)]
end
if left < n
s << [left + 1, right, result]
end
end
end
end
SO,
The problem
I have two integers, which are in first case, positive, and in second case - just any integers. I need to create a map function F from them to some another integer value, which will be:
Result should be integer value. For first case (x>0, y>0), positive integer value
Symmetric. That means F(x, y) = F(y, x)
Unique. That means F(x0, y0) = F(x1, y1) <=> (x0 = x1 ^ y0 = y1) V (y0 = x1 ^ x0 = y1)
My approach
At first glance, for positive integers we could use expression like F(x, y) = x2 + y2, but that will fail - for example, 892 + 232 = 132 + 912 As for second (common) case - that's even more complicated.
Use-case
That may be useful when dealing with some things, which supposed to be order-independent and need to be unique. For example, if we want to find cartesian product of many arrays and we want result to be unique independent of order, i.e. <x,z,y> is equal to <x,y,z>. It may be done with:
function decartProductPair($one, $two, $unique=false)
{
$result = [];
for($i=0; $i<count($one); $i++)
{
for($j=0; $j<count($two); $j++)
{
if($unique)
{
if($i!=$j)
{
$result[$i*$i+$j*$j]=array_merge((array)$one[$i],(array)$two[$j]);
// ^
// |
// +----//this is the place where F(i,j) is needed
}
}
else
{
$result[]=array_merge((array)$one[$i], (array)$two[$j]);
}
}
}
return array_values($result);
}
Another use-case is to properly group sender and receiver in some SQL table, so that different senders/receivers will be differed while they should stay symmetric. Something like:
SELECT
COUNT(1) AS message_count,
sender,
receiver
FROM
test
GROUP BY
-- this is the place where F(sender, receiver) is needed:
sender*sender + receiver*receiver
(By posting samples I wanted to show that issue is certainly related to programming)
The question
As mentioned, the question is - what can be used as F? I want as simple F as it's possible. Keep in mind two cases:
Integer x>0, y>0. F(x,y) > 0
Any integer x, y and so any integer F(x,y) as a result
May be F isn't just an expression - but some algorithm to find desired result for any x,y (so tagging with algorithm too). However, expression is better because it's more like that it will be able to use that expression in SQL or PHP or whatever. Feel free to edit tagging because I'm not sure if two tags here is enough
Most simple solution: f(x,y) = x^5 + y^5
No positive integer is known which can be written as the sum of two fifth powers in more than one way.
As for now, this is unsolved math problem.
You need a MAX_INTEGER constant, and the result will need to hold MAX_INTEGER**2 (say: be a long, if both are int's). In that case, one such function is:
f(x,y) = min(x,y)*MAX_INTEGER + max(x,y)
But I propose a different solution: use a hash function (say md5) of the string resulting from the concatenation of str(min(x,y)), a separator (say ".") and str(max(x,y)). That is:
f(x,y) = md5(str(min(x,y)) + "." + str(max(x,y)))
It is not unique, but collisions are very rare, and probably OK for most use cases. If still worried about collisions, save the actualy {x,y} along with f(x,y), and check if collisions happened.
Sort input numbers and interleave their bits:
x = 5
y = 3
Step 1. Sorting: 3, 5
Step 2. Mixing bits: 11, 101 -> 1_1_, 1_0_1 -> 11011 = 27
So, F(3, 5) = 27
A compact representation is x*(x+3)/2 + y*(x+1) + (y*(y-1))/2, which comes from an arrangement like this:
x->
y 0 1 3 6 10 15
| 2 4 7 11 16
v 5 8 12 17
9 13 18
14 19
20
According to [Stackoverflow:mapping-two-integers-to-one-in-a-unique-and-deterministic-way][1], if we symmetrize the formula we would have the following:
(x + y) * (x + y + 1) / 2 + min(x, y)
This might just work. For
(x + y) * (x + y + 1) / 2 + x
is unique, then the first formula is also unique.
[1]: Mapping two integers to one, in a unique and deterministic way
Sometimes, you want to use an if/unless modifier but the conditionally executed part includes a local variable that is to be defined within the condition. For example,
a = [1, 2, 3]
n = 3*(x**2) + 4*x + 5 if x = a[2]
m = 6*(y**2) + 7*y + 8 unless (y = a[0]).zero?
will give a parsing error because x,y is read before the if/unless modifier. In order to avoid that, I think it is pretty much common (at least for me) to use and instead of if and or instead of unless:
x = a[2] and n = 3*(x**2) + 4*x + 5
(y = a[0]).zero? or m = 6*(y**2) + 7*y + 8
Besides the fact that it does not raise an error, is there any difference? Is there any side effects in doing this? And, is there a better way?
Alternatives that I would choose (and have used) when this issue arises:
if x=a[2]
n = 3*(x**2) + 4*x + 5
end
if x=a[2] then n = 3*(x**2) + 4*x + 5 end
x=nil
n = 3*(x**2) + 4*x + 5 if x=a[2]
the logical or returns true if one of the sides evaluates to true...
are you sure that the or and unless lines are equal
(could be, that the or line always evaluates to true since m = 6*(y**2) + 7*y + 8 is true)?
EDIT:
nope i was wrong take a look at this doc - seems to be what you are looking for
I just realized that there is a difference between
if x = something then f(x) end (1)
and
x = something and f(x) (2)
That is when something takes the value false. In this case, (1) returns nil, whereas (2) returns false.
Consider the problem in which you have a value of N and you need to calculate how many ways you can sum up to N dollars using [1,2,5,10,20,50,100] Dollar bills.
Consider the classic DP solution:
C = [1,2,5,10,20,50,100]
def comb(p):
if p==0:
return 1
c = 0
for x in C:
if x <= p:
c += comb(p-x)
return c
It does not take into effect the order of the summed parts. For example, comb(4) will yield 5 results: [1,1,1,1],[2,1,1],[1,2,1],[1,1,2],[2,2] whereas there are actually 3 results ([2,1,1],[1,2,1],[1,1,2] are all the same).
What is the DP idiom for calculating this problem? (non-elegant solutions such as generating all possible solutions and removing duplicates are not welcome)
Not sure about any DP idioms, but you could try using Generating Functions.
What we need to find is the coefficient of x^N in
(1 + x + x^2 + ...)(1+x^5 + x^10 + ...)(1+x^10 + x^20 + ...)...(1+x^100 + x^200 + ...)
(number of times 1 appears*1 + number of times 5 appears * 5 + ... )
Which is same as the reciprocal of
(1-x)(1-x^5)(1-x^10)(1-x^20)(1-x^50)(1-x^100).
You can now factorize each in terms of products of roots of unity, split the reciprocal in terms of Partial Fractions (which is a one time step) and find the coefficient of x^N in each (which will be of the form Polynomial/(x-w)) and add them up.
You could do some DP in calculating the roots of unity.
You should not go from begining each time, but at max from were you came from at each depth.
That mean that you have to pass two parameters, start and remaining total.
C = [1,5,10,20,50,100]
def comb(p,start=0):
if p==0:
return 1
c = 0
for i,x in enumerate(C[start:]):
if x <= p:
c += comb(p-x,i+start)
return c
or equivalent (it might be more readable)
C = [1,5,10,20,50,100]
def comb(p,start=0):
if p==0:
return 1
c = 0
for i in range(start,len(C)):
x=C[i]
if x <= p:
c += comb(p-x,i)
return c
Terminology: What you are looking for is the "integer partitions"
into prescibed parts (you should replace "combinations" in the title).
Ignoring the "dynamic programming" part of the question, a routine
for your problem is given in the first section of chapter 16
("Integer partitions", p.339ff) of the fxtbook, online at
http://www.jjj.de/fxt/#fxtbook