I'm a beginner to logic circuits and I'm trying to construct a truth table for a LED dice circuit.
I've got 7 outputs in my table, 1 for each LED, but I can't figure out how many inputs I need.
I've been told that the formula below gives the number of inputs, but I don't know what Y is. Can anyone confirm that the formula is correct, and tell me what Y is so I can work this out? Thanks
n = log(Y + 1) / log(2)
There is no telling how many inputs you must have. You could have as many as you wish. But there is a minimum.
Every input line can be thought of as a digit in a binary number. So in order to identify 7 different numbers we would need at least three binary digits (000 to 111). So the formula would be the ceil(log2(Y)) where Y in the number of output lines.
A great example of such a circuit would be a demultiplexer. You will notice the number of selector bits in the DEMUX is ceil(log2(Y)) the number of output lines.
Related
I'm trying to solve the Hidden Sequence problem on Code Chef, but I don't fully understand the explanation. I especially don't understand what's the use of Y in the triplets.
We know that there is a hidden sequence A1,A2,…,AN which contains only integers between 1 and K inclusive. We have acquired M triplets (X1,Y1,Z1),(X2,Y2,Z2),…,(XM,YM,ZM). A very reliable source has given us intel that for each valid i, the Yi-th occurrence of the integer Xi in the sequence A is AZi, i.e. AZi=Xi and there are Yi−1 indices j<Zi such that Aj=Xi
Find any sequence A consistent with this information or determine that no such sequence exists.
Could anyone explain it?
The hidden array may have duplicate values, like [1,2,3,1,2,3,1,1].
One of the triplets could be X=3, Y=2, Z=6.
This tells us that the second 3 is at position 6. So:
X = the value
Y = the occurrence of that value X (whether it is the first, second, third, ... occurrence)
Z = the position of the Y-th occurrence of that value X
This may be the only info you will get about the value 3. So you might not get a triplet saying X=3, Y=1, Z=3, which would tell you where the first 3 is positioned. Instead your algorithm must derive that from the other triplets.
The algorithm should somehow stay aware that in the first 5 positions there must be a 3. From other triplets it will know similar things. At a certain point this bulk of information will not allow that 3 to occur at just any of those 5 positions, as some positions will be needed for other values. This will narrow down the possibilities, until maybe only a few are left over, or maybe none.
Hope this explains what Y is about.
A few days ago I came across such a problem at the contest my uni was holding:
Given the history of guesses in a mastermind game using digits instead
of colors in a form of pairs (x, y) where x is the guess and y is how
many digits were placed correctly, guess the correct number. Each
input is guaranteed to have a solution.
Example for a 5-place game:
(90342, 2)
(70794, 0)
(39458, 2)
(34109, 1)
(51545, 2)
(12531, 1)
Should yield:
39542
Create an algorithm to correctly guess the result in an n-place
mastermind given the history.
So the only idea I had was to keep the probability of each digit being correct based on the correct shots in a given guess and then try to generate the most possible number, then the next one and so on - so for example we'd have 9 being 40% possible for the first place (cause the first guess has 2/5=40% correct), 7 being impossible and so on. Then we do the same for other places in the number and finally generate a number with the highest probability to test it against all the guesses.
The problem with this approach, though, is that generating the next possible number, and the next, and so on (as we probably won't score a home run in the first try) is really non-trivial (or at least I don't see an easy way of implementing this) and since this contest had something like a 90 minute timeframe and this wasn't the only problem, I don't think something so elaborate was the anticipated approach.
So how could one do it easier?
An approach that comes to mind is to write a routine that can generally filter an enumeration of combinations based on a particular try and its score.
So for your example, you would initially pick one of the most constrained tries (one of the ones with a score of 2) as a filter and then enumerate all combinations that satisfy it.
The output from that enumeration is then used as input to a filter run for the next unprocessed try, and so on, until the list of tries is exhausted.
The candidate try that comes out of the final enumeration is the solution.
Probability does not apply here. In this case a number is either right or wrong. There is no "partially right".
For 5 digits you can just test all 100,000 possible numbers against the given history and throw out the ones where the matches are incorrect. This approach becomes impractical for larger numbers at some point. You will be left with a list of numbers that meet the criteria. If there is exactly one in the list, then you have solved it.
python code, where matches counts the matching digits of its 2 parameters:
for k in range(0,100000):
if matches(k,90342)==2 and matches(k,70794)==0 and matches(k,39458)==2 and matches(k,34109)==1 and matches(k,51545)==2 and matches(k,12531):
print k
prints:
39542
I was asked the following question in an interview. I don't have any idea about how to solve this. Any suggestions?
Given the start and an ending integer as user input,
generate all integers with the following property.
Example:
123 , 1+2 = 3 , valid number
121224 12+12 = 24 , valid number
1235 1+2 = 3 , 2+3 = 5 , valid number
125 1+2 <5 , invalid number
A couple of ways to accomplish this are:
Test every number in the input range to see if it qualifies.
Generate only those numbers that qualify. Use a nested loop for the two starting values, append the sum of the loop indexes to the loop indexes to come up with the qualifying number. Exit the inner loop when the appended number is past the upper limit.
The second method might be more computationally efficient, but the first method is simpler to write and maintain and is O(n).
I don't know what the interviewer is looking for, but I would suspect the ability to communicate is more important than the answer.
The naive way to solve this problem is by iterating the numbers in the set range, parsing the numbers into a digit sequence and then testing the sequence according to the rule. There is an optimization in that the problem essentially asks you to find fibonnaci numbers so you can use two variables or registers and add them sequentially.
It is unclear from your question whether the component numbers have to have the same number of digits. If not, then you will have to generate all the combinations of the component number arrangements.
I'm trying to solve this problem on SPOJ : http://www.spoj.pl/problems/EDIT/
I'm trying to get a decent recursive description of the algorithm, but I'm failing as my thoughts keep spinning in circles! Can you guys help me out with this one? I'll try to describe what approach I'm trying to solve this.
Basically I want to solve a problem of size j-i where i is the starting index and j is the ending index. Now, there should be two cases. If j-i is even then both the starting and the ending letters have to be the same case, and they have to be the opposite case when j-i is odd. I also want to reduce the problem of a lower size (j-i-1 or j-i-2), but I feel that if I know a solution to a smaller problem, then constructing a solution of a just bigger problem should also take into account the starting and ending letter cases of the smaller problem. This is exactly where I'm getting confused. Can you guys put my thoughts on the right track?
I think recursion is not the best way to go with this problem. It can be solved quite fast if we take a different approach!
Let us consider binary strings. Say an uppercase char is 1 and a lowercase one is 0. For example
AaAaB -> 10101
ABaa -> 1100
a -> 0
a "correct" alternating chain is either 10101010.. or 010101010..
We call the minimum number of substitutions required to change one string into the other the Hamming distance between the strings. What we have to find is the minimum Hamming distance between the input binary string and one of the two alternating chains of the same length.
It's not difficult: we XOR each string and then count the number of 1s. (link). For example, let's consider the following string: ABaa.
We convert it in binary:
ABaa -> 1100
We generate the only two alternating chains of length 4:
1010
0101
We XOR them with the input:
1100 XOR 1010 = 0101
1100 XOR 0101 = 1010
We count the 1s in the results and take the minimum. In this case, it's 2.
I coded this procedure in Java with some minor optimization (buffered I/O, no real need to generate the alternating chains) and it got accepted: (0.60 seconds one).
Given any string s of length n, there are only two possible "alternating chain".
This 2 variants can be defined sequentially by settings the first letter state (if first is upper then second is lower, third is upper...).
A simple linear algorithm would be to make 2 simple assumptions about the first letter:
First letter is UpperCase
First letter is LowerCase
For each assumption, run a simple edit distance algorithm and you are done.
You can do it recursively, but you'll need to pass and return a lot of state information between functions, which I think is not worthwhile when this problem can be solved by a simple loop.
As the others say, there are two possible "desired result" strings: one starts with an uppercase letter (let's call it result_U) and one starts with a lowercase letter (result_L). We want the smaller of EditDistance(input, result_U) and EditDistance(input, result_L).
Also observe that, to calculate EditDistance(input, result_U), we do not need to generate result_U, we just need to scan input 1 character at a time, and each character that is not the expected case will need 1 edit to make it the correct case, i.e. adds 1 to the edit distance. Ditto for EditDistance(input, result_L).
Also, we can combine the two loops so that we scan input only once. In fact, this can be done while reading each input string.
A naive approach would look like this:
Pseudocode:
EditDistance_U = 0
EditDistance_L = 0
Read a character
To arrive at result_U, does this character need editing?
Yes => EditDistance_U += 1
No => Do nothing
To arrive at result_L, does this character need editing?
Yes => EditDistance_L += 1
No => Do nothing
Loop until end of string
EditDistance = min(EditDistance_U, EditDistance_L)
There are obvious optimizations that can be done to the above also, but I'll leave it to you.
Hint 1: Do we really need 2 conditionals in the loop? How are they related to each other?
Hint 2: What is EditDistance_U + EditDistance_L?
I was trying to solve this problem on SPOJ, in which I have to find how many numbers are there in a range whose digits sum up to a prime. This range can be very big, (upper bound of 10^8 is given). The naive solution timed out, I just looped over the entire range and checked the required condition. I cant seem find a pattern or a formula too. Could someone please give a direction to proceed in??
Thanks in advance...
Here are some tips:
try to write a function that finds how many numbers in a given range have a given sum of the digits. Easiest way to implement this is to write a function that returns the number of numbers with a given sum of digits up to a given value a(call this f(sum,a)) and then the number of such numbers in the range a to b will be f(sum,b) - f(sum, a - 1)
Pay attention that the sum of the digits itself will not be too high - up to 8 * 9 < 100 so the number of prime sums to check is really small
Hope this helps.
I (seriously) doubt whether this 'opposite' approach will be any faster than #izomorphius's suggestion, but it might prompt some thoughts about improving the performance of your program:
1) Get the list of primes in the range 2..71 (you can omit 1 and 72 from any consideration since neither is prime).
2) Enumerate the integer partitions of each of the prime numbers in the list. Here's some Python code. You'd want to modify this so as not to generate partitions which were invalid, such as those containing numbers larger than 9.
3) For each of those partitions, pad out with 0s to make a set of 8 digits, then enumerate all the permutations of the padded set.
Now you have the list of numbers you require.
Generate the primes using the sieve of Eratosthenes up to the maximum sum (9 + 9...). Put them in a Hash table. Then you could likely loop quickly through 10^8 numbers and add up their sums. There might be more efficient methods, but this should be quick enough.