Deadlock Detection - resources

I am not sure how to properly detect deadlocks. My book provides this table for five processes P0 to P4 and three resource types A, B, and C. Resource A has seven instances, resource type B has two instances, and resource type C has six instances. This is the resource-allocation state at time T0:
Allocation Request Available
A B C A B C A B C
P0 0 1 0 0 0 0 0 0 0
P1 2 0 0 2 0 2
P2 3 0 3 0 0 0
P3 2 1 1 1 0 0
P4 0 0 2 0 0 2
It goes on to state that the system is not in a deadlocked state, but does not provide much detail for the problem solving technique. Can anyone break it down for me so that I may follow along with the steps easily and attempt a similar problem?
Here is my attempt at understanding the problem:
P0 has 1 of resource B and requires no resources. P0 relinquishes its resources
A B C A B C A B C
P0 0 0 0 0 0 0 0 1 0
P1 2 0 0 2 0 2
P2 3 0 3 0 0 0
P3 2 1 1 1 0 0
P4 0 0 2 0 0 2
P1 has 2 of resource A and requires 2 of resource A and C. There are not enough resources available. P2 has 3 of resource A and C and requires no resources. P2 relinquishes its resources.
A B C A B C A B C
P0 0 0 0 0 0 0 3 1 3
P1 2 0 0 2 0 2
P2 0 0 0 0 0 0
P3 2 1 1 1 0 0
P4 0 0 2 0 0 2
Hope I'm on the right track thus far.


Deadlocks occur when you once process is in the state "have A, need B" and some other process is in "have B, need A".
if I'm reading your tables right, then this would indicate a deadlock:
Allocation Request Available
A B C A B C A B C
P0 0 1 0 1 0 0 0 0 0
P1 2 0 0 2 1 2
P0 has resource B, and requires resource A, with no more resource A's available
P1 has 2 of resource A, and requires source B, with no more B's available.
The deadlock cannot be resolved until more of resources A or B become available. That'd mean P0 or P1 have to relinquish some of their allocations (or P2+ doing the same).

Related

Pattern identification and sequence detection

I have a dataset 'df' that looks something like this:
MEMBER seen_1 seen_2 seen_3 seen_4 seen_5 seen_6
A 1 0 0 1 0 1
B 1 1 0 0 1 0
C 1 1 1 0 0 1
D 0 0 1 0 0 1
As you can see there are several rows of ones and zeros. Can anyone suggest me a code in python such that I am able to count the number of times '1' occurs continuously before the first occurrence of a 1, 0 and 0 in order. For example, for member A, the first double zero event occurs at seen_2 and seen_3, so the event will be 1. Similarly for the member B, the first double zero event occurs at seen_3 and seen_4 so there are two 1s that occur before this. The resultant table should have a new column 'event' something like this:
MEMBER seen_1 seen_2 seen_3 seen_4 seen_5 seen_6 event
A 1 0 0 1 0 1 1
B 1 1 0 0 1 0 2
C 1 1 1 0 0 1 3
D 0 0 1 0 0 1 1

My approach:
df = df.set_index('MEMBER')
# count 1 on each rows since the last 0
s = (df.stack()
.groupby(['MEMBER', df.eq(0).cumsum(1).stack()])
.cumsum().unstack()
)
# mask of the zeros:
u = s.eq(0)
# look for the first 1 0 0
idx = (~u &
u.shift(-1, axis=1, fill_value=False) &
u.shift(-2, axis=1, fill_value=False) ).idxmax(1)
# look up
df['event'] = s.lookup(idx.index, idx)
Test data:
MEMBER seen_1 seen_2 seen_3 seen_4 seen_5 seen_6
0 A 1 0 1 0 0 1
1 B 1 1 0 0 1 0
2 C 1 1 1 0 0 1
3 D 0 0 1 0 0 1
4 E 1 0 1 1 0 0
Output:
MEMBER seen_1 seen_2 seen_3 seen_4 seen_5 seen_6 event
0 A 1 0 1 0 0 1 1
1 B 1 1 0 0 1 0 2
2 C 1 1 1 0 0 1 3
3 D 0 0 1 0 0 1 1
4 E 1 0 1 1 0 0 2

How to understand synergy in information theory?

In information theory, multivariate mutual information (MMI) could be synergy (negative) or redundancy (positive). To simulate this two cases, assuming three variables X, Y and Z, all of them takes 0 or 1 (binary variable). And we repeat sampling them 12 times.
Case 1:
X = [ 0 0 0 0 0 0 1 1 1 1 1 1 ]
Y = [ 0 0 0 0 1 1 0 0 1 1 1 1 ]
Z = [ 0 0 1 1 1 1 0 0 0 0 1 1 ]
In this case, we assume a mechanism among XYZ taht when both Y and Z are 0 or 1, X takes 0 or 1 respectively. When Y = 0, Z = 1, then X takes 0, and Y = 1, Z = 0, then X takes 1.
The mmi(X,Y,Z) = -0.1699 in this case, indicating a synergy effect among three variable.
Case 2:
X = [ 0 0 0 0 0 0 1 1 1 1 1 1 ]
Y = [ 0 0 0 0 0 1 0 1 1 1 1 1 ]
Z = [ 0 1 1 1 1 1 0 0 0 0 0 1 ]
the machanism in this case is same as above. The difference is there are more samples of XY takes different value and less samples of both XY are 0 or 1.
The mmi(X,Y,Z) = 0.0333, indicating a redundancy.
So far, can I say in these two cases, synergy and redundancy show the similar mechanism (or relationship) among three variables? But how do we understand redundancy and particularly synergy in realistic data?

How is the port address decoded?

According to the image below, it says the output ports are from F0-F7H. My question is how are these ports addresses determine? For example, F0(active low) (Y0) is determine from the inputs A0-A7 inputs? If so, how are these input ports mathematically come to F0?

The 74ALS138 is a 1-8 demultiplexer, this means it takes a number between 0 and 7 and activates one of its eight output lines (active low).
The A, B and C input signals are the three bit encoding the input number (23 = 8) while G1, GA and GB are the enable signals.
To enable the chip G1 must be high and GA and GB must be low, any other combination will disable the chip (all output is high).
To have G1 high we must have the bit 4 of the address high, analogously bit 5, 6 and 7 must be high.
Finally, bit 3 must be low.
This gives an address of the form 1111 0xxx, ranging from 0f0h to 0f7h.
The lowest three bits select the output line.
Regarding the tie between A, B and C and the outputs, you can start with a truth table:
A B C Y0 Y1 Y2 Y3 Y4 Y5 Y6 Y7
0 0 0 0 1 1 1 1 1 1 1
0 0 1 1 0 1 1 1 1 1 1
0 1 0 1 1 0 1 1 1 1 1
0 1 1 1 1 1 0 1 1 1 1
1 0 0 1 1 1 1 0 1 1 1
1 0 1 1 1 1 1 1 0 1 1
1 1 0 1 1 1 1 1 1 0 1
1 1 1 1 1 1 1 1 1 1 0
Each output Yi must be computed independently; since each of them is zero exactly once, there is no need to optimize it with a Karnaugh map and since there are a lot of ones, working with Maxterms is definitively better.
So for example for Y0 the formula is
Y0 = A + B + C
Due to the rules when dealing with maxterms (It's a product of sum, each factor being negated iif the input variable is 1).
The other relations are
Y1 = A + B + C'
Y2 = A + B' + C
Y3 = A + B' + C'
Y4 = A' + B + C
Y5 = A' + B + C'
Y6 = A' + B' + C
Y7 = A' + B' + C'
This doesn't take into account the enable inputs, internally we can have a single enable signal E by taking E = G1 * GA' * GB' then the truth table for Y0 becomes
E A B C Y0
0 0 0 0 1
0 0 0 1 1
0 0 1 0 1
0 0 1 1 1
0 1 0 0 1
0 1 0 1 1
0 1 1 0 1
0 1 1 1 1
1 0 0 0 0
1 0 0 1 1
1 0 1 0 1
1 0 1 1 1
1 1 0 0 1
1 1 0 1 1
1 1 1 0 1
1 1 1 1 1
This just translates to Y0 = E' + A + B + C.
If you consider that X + Y === (X' * Y')' by De Morgan's laws and call (X * Y)' NAND you see that Y0 = NAND(E, A', B', C') which is exactly the implementation in the 74ALS138 datasheet:
Datasheet courtesy of Matt Clark

The table in the data sheet seems quite clear: Y0 is active if A0, A1, A2, and A3 are low and A4, A5, A6, and A7 are high. Y1 is active under all the same conditions except A0 is high.

Thanks for the comments to help me connect the dots in figuring out how the address is map to the input address

how to remove outermost logic?

how to remove outermost logic?
such as
input column D result
And(OR(A,B),C)
output column E binary number
OR(A,B)
A B C result(D)after extract(E)
0 0 0 0 0
0 0 1 0 0
0 1 0 0 1
0 1 1 1 1
1 0 0 0 1
1 0 1 1 1
1 1 0 0 1
1 1 1 1 1
i tried in excel
=IF(NOT(AND(D2,C2))=TRUE,1,0)
but can not remove outermost logic
result after extract
0 0 0 =IF(AND(OR(A2,B2),C2)=TRUE,1,0) =IF(OR(A2,B2)=TRUE,1,0) =IF(NOT(AND(D2,C2))=TRUE,1,0)
0 0 1 =IF(AND(OR(A3,B3),C3)=TRUE,1,0) =IF(OR(A3,B3)=TRUE,1,0) =IF(NOT(AND(D3,C3))=TRUE,1,0)
0 1 0 =IF(AND(OR(A4,B4),C4)=TRUE,1,0) =IF(OR(A4,B4)=TRUE,1,0) =IF(NOT(AND(D4,C4))=TRUE,1,0)
0 1 1 =IF(AND(OR(A5,B5),C5)=TRUE,1,0) =IF(OR(A5,B5)=TRUE,1,0) =IF(NOT(AND(D5,C5))=TRUE,1,0)
1 0 0 =IF(AND(OR(A6,B6),C6)=TRUE,1,0) =IF(OR(A6,B6)=TRUE,1,0) =IF(NOT(AND(D6,C6))=TRUE,1,0)
1 0 1 =IF(AND(OR(A7,B7),C7)=TRUE,1,0) =IF(OR(A7,B7)=TRUE,1,0) =IF(NOT(AND(D7,C7))=TRUE,1,0)
1 1 0 =IF(AND(OR(A8,B8),C8)=TRUE,1,0) =IF(OR(A8,B8)=TRUE,1,0) =IF(NOT(AND(D8,C8))=TRUE,1,0)
1 1 1 =IF(AND(OR(A9,B9),C9)=TRUE,1,0) =IF(OR(A9,B9)=TRUE,1,0) =IF(NOT(AND(D9,C9))=TRUE,1,0)

By "remove the outermost logic", I assume you want to remove the IF function.
One thing to note is that in a formula like =IF(AND(OR(A2,B2),C2)=TRUE,1,0) you never need the =TRUE test. =IF(AND(OR(A2,B2),C2),1,0) will work exactly the same.
There are a couple of ways to convert a boolean (i.e. true/false value) into an integer without the explicit IF. One is --AND(OR(A2,B2),C2). Another is int(AND(OR(A2,B2),C2)).

Transform a Matrix into an Edge List in EXCEL

I have an excel matrix that looks like this:
ID A B C D E F G
XYZ 0 1 0 1 0 1 0
ZVC 1 0 1 0 1 0 1
...
ABC 0 1 0 1 0 1 1
I would like to transform this matrix into three columns:
XYZ A 0
XYZ B 1
XYZ C 0
...
ABC F 1
ABC G 1
What would be an efficient way to do that (possibly without macros)?

I actually found a simple solution to my problem: https://www.youtube.com/watch?v=N3wWQjRWkJc

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