I'm working on trying to identify a winning hand in mahjong, does anyone have any tips or a working formula?
The game of mahjong is like gin rummy with 13 cards, sets of 4,3 or flushes of 3. A winning hand includes the stolen or given card, which totals 14 cards. A winning hand is sets of:
3 sets of 4 plus 1 pair
4 sets of 3 plus 1 pair.
2 sets of 4 plus 2 sets of 3, no pair.
Flushes are in sets of 3, but all one suite is a bonus.
All pairs is also an option.
There are also points associated with this.
Any guidance or help would be much appreciated.
To follow up, and add info, this may help others, and gould help us to answer our own question,
I am aiming at Hong Kong Mahjong Rules.
When I work through an algorithm, the one I am investigating is the removal of streets, and if fail, remove the set.
works when I have a b3333 b4 b5 b666
check b345 - remove
check b345 fail, b333 is a set, remove
check b678, fail, b666 is a set.
none left, Win
but when I have a b222 b33 b44 b5. I need to skip and restart.
check b234 ok, remove
check b234 ok, remove
check b234 fail, single b2 (Failed hand)
restart, skip 1.
check b345 ok, remove
end, return to start
check b234 ok remove
check b234 fail, b22 is pair.
none left win.
but this method is troublesome, because there could be a double state to retry.
b222 b33 b44 b555 b66 b77
check b234 ok, remove
check b234 ok, remove
check b234 fail, resart, skip 1.
check b345 ok, remove
check b345 ok, remove
check b567 ok, remove
check b678 fail, no 7, skip one rety
any ideas?
I have implemented a successful mahjong game in Android. The algorithm I used was as follows:
1) Count the occurrence of each tile into an array of 34
2) Remove every possible pair
3) Recursively check the remaining tiles for triplets
Hope that helps
Related
Is it possible to indicate monitor position in Weston/Wayland?
I have two monitors and been testing the Weston compositor, but I have been unable to indicate which monitor should be the main one (or which one should show the "left part" of the screen).
Checking the weston.ini docs (http://manpages.ubuntu.com/manpages/xenial/en/man5/weston.ini.5.html) I found info about setting resolution, scaling and transform/rotation, but nothing about the position of the monitors.
I have been interested into the same some weeks ago. I send the question to the wayland IRC channel.
You can have a look at:
https://people.freedesktop.org/~cbrill/dri-log/index.php?channel=wayland&highlight_names=&date=2017-12-19
https://people.freedesktop.org/~cbrill/dri-log/index.php?channel=wayland&highlight_names=&date=2017-12-20
Here is the relevant part:
22:31 maggu2810: Is there any change to change the display / output position in weston? I am using three outputs but don't know how to configure which one if left / right of the other one.
22:33 maggu2810: ... chance to change ...
22:35 maggu2810: the output sections in weston.ini currently contains the name, the mode and the scale. I didn't find any "position" option or anything that looks like an option to modify the "order" of the displays.
07:07 pq: maggu2810, yes, not implemented yet, there were indeed some WIP patches in 2016
Perhaps you will be able to post a feature request in their mailing list. I don't think they will look for feature requests on SO ;)
I have a quick question about the details of running a model in JAGS and BUGS.
Say I run a model with n.burnin=5000, n.iter=5000 and thin=2. Does this mean that the program will:
Run 5,000 iterations, and discard results; and then
Run another 10,000 iterations, only keeping every second result?
If I save these simulations as a CODA object, are all 10,000 saved, or only the thinned 5,000? I'm just trying to understand which set of iterations are used to make the ACF plot?
With JAGS, n.burnin=5000, n.iter=5000 and thin=2, means you keep nothing. You run 5000, discard the first 5000 of these 5000 and then only keep a half of the remaining values of the chain (keep 1 value and discard the next one ..).
Use for example n.burnin=2000, n.iter=7000, thin=50, n.chains=5 : so you have (7000-2000)/50 * 5 = 500 values.
Could you be more specific which software you're talking about? It looks like you're referring to the arguments of the function bugs() in the R2WinBUGS package (except that the argument is called n.thin not thin). Looking at help(bugs) it just says n.burnin is the "number of iterations to discard at the beginning". Which doesn't specifically answer your question, but looking at the source for bugs.script() in that package suggests to me that it would run 5000 iterations burn in, as you suspected. You could send a suggestion to the maintainers of that package to clarify their documentation.
In your example, bugs() would then run 0 further iterations after the burn-in. Here the documentation is clearer - n.iter is the total number of iterations including the burn-in.
For your second question, the CODA output from WinBUGS (and any software which calls WinBUGS or OpenBUGS) will only include the thinned sample.
I got this error on a CJ1W-CT021 card. It happen all of a sudden after its been running the program for some time. How i found it was by going to the IO Table and Unit Set up. Clicked on parameters for that card and found two settings in red.
Output Control Mode and And/Or Counter Output Patterns. This was there reading
Output Control Mode = 0x40 No Applicable Set Data
And/Or Counter Output Patterns = 0x64 No Applicable Set Data
no idea on how or why these would change they should of been
Output Control Mode = Range Mode
And/Or Counter Output Patterns = Logically Or
I have added some new code, but nothing big or really even used as i had the outputs of the new rungs jumped out. One thing i thought might cause this is every cycle of the program it was checking the value of an encoder connected to this card. Maybe checking it too offten? Anyhow if anyone has any idea what these do or how they would change please post.
Thanks
Glen
EDIT.. I wanted to add the bits i used, dont think any are part of this cards internal io but i may be wrong?
Work bits 66.01 - 66.06 , 60.02 - 60.07 , 160.12, 160.01 - 160.04, 161.02, 161.03
and
Data Bits (D)20720, 20500, 20600, 20000, 20590, 20040
I would check section 4-1 through 4-2-4 of the CT021 manual - make sure you aren't writing to reserved memory locations used for configuration data of the CT021 unit.
EDIT:
1) Check Page 26 of the above manual to see the location of the machine switch settings. The bottom dial sets the '1's digit and the top dial sets the '10's digit (ie machine number can be 0-99);
2) Per page 94, D-Memory is allocated from D20000 + (N X 100) (400 Words) where N is equal to the machine number.
I would guess that your machine number is set to 0 (ie: both dials at '0'), 5, or 6. In the case of machine number '0', this would make the reserved DM range D20000 -> D20399. In this case (see pages 97, 105) D20000 would contain configuration data for Output Control Mode (bits 00-07) and Counter Output Patterns (bits 08-15). It looks like you are writing 0x6440 to D20000 (or D20500, D20600 for machine number 5 or 6, respectively) and are corrupting the configuration data.
If your machine number is 0 then stay away from D20000-D20399 unless you are directly trying to modify the counter's configuration state (ie: don't use them in your program!).
If the machine number is 1 then likewise for D20100-D20499, etc. If you have multiple counters they can overlap ranges so they should always be set with machine numbers which are 4 apart from each other.
During the reading of PLC documentation (Omron CP1L PLC and CX-Programmer), there are some missing explanation. For example, it defines "Flag" as "a bit that serves as an interface between in*struction", does that mean flag is some sort of conditional Power Flow?
It gets more confusing with the terms "Differential Up/Down", "Carry Flag"? What are flags and what do they do in the ladder logic? Are they a simple or instruction to use or just a concept that I don't really need to program in ladder?
[EDITED]
Where to add / modify / delete the flag in an instruction? I open up the edit but flag isn't there.
Ok, this is a better question.
PLCs are like any program - data is stored as different types. Think of flags as interchangeable with the term "bit", "boolean", etc. They are very important.
If you have CX-Programmer, a much better place to get information is the Instruction Reference (Help --> Instruction Reference --> yourPLC). These show time diagrams of most of the instructions and how each of the parameters and flags operate.
For example, a basic timer (TIM) works by assigning it a value. If you use a BCD type 100ms timer and assign its SV (setpoint value) a BCD value of 300 then you have created a timer with a 30 second limit (300 x 100ms). When the timer turns on it will begin counting and the PV (process value) will start from 300 and count down. When the value reaches zero the timer's flag turns ON to indicate that it has expired. If the timer's number is, say T100 then you can use T100 as a contact in another rung of logic - it will be true when the timer's execution conditions are TRUE and the timer has expired.
Differentials (UP/DOWN) are special flags which are true for only one PLC scan (ie: they are true for one execution cycle only) when their input conditions change from FALSE to TRUE (ie:OFF to ON) for UP differentials, and TRUE to FALSE (ie:ON to OFF) for DOWN differentials. You would use differentials in cases where you wanted to perform an action the moment a given condition changes.
Flags can be used for almost anything. You can use them as general booleans in your own programs, they can be parts of certain operations (ie: the CY (carry) flag is used on arithmetic operations which result in a carry - other flags are used to indicate overflows or div/0 errors, etc).
Edit again : (to answer extended question).
A basic timer's completion flag is a contact with its number. Say I have a 100ms timer, T100, which turns on when contact 10.00 is on:
10.00 ___
|-----| |---------------------------------------|TIM|
|100|
| |
|#20|
|___|
Now, once 10.00 has been ON for two seconds, the timer will elapse and the flag for timer 100, T100, will turn ON. If I had another rung where
| T100 W15.00
|-----| |-----------------------------------( )
Then work bit W15.00 would be turned on when the timer elapsed and would remain on so long as the timer's input condition remained satisfied (ie: so long as 10.00 remains ON). Flags work in different ways for different things, however. Each operation can use them in different ways.
The example from the Omron Instruction Reference (Help -> Instruction Reference -> [select PLC]) looks like this :
Well very good explanation with example and the flags value can be found in memory area it is pure binary either 0 or 1, as i was read the documentation work-bit memory location changes as per timer type e.g TIM/TIMX or TIMH or TIMHX, both are BCD timers but unit for the timer changes.
The Operating System Concepts 6th edition present one trival algorithm to implementate ciritical section.
do{
while (turn != i);
critical section
trun = j;
remainder section
} while(1);
Note,Pi is the process with identifier i,Pj is the process with identifier j.To simlify the question,the book limit the i,j to 0 and 1,the two processes constriant enviroment.
Question1 is,dose this algorithm voilate the Progress requirement which is one of the three requirements to ciritical section solution?
In my opinion,when Pi is in its remainder section,it cannot participate in the decision on whether Pj can enter the critical section.Then it is bound to the requirement.
Or my understanding of progress requirement is totally wrong.So because if Pi retired from the remainder section,It could not get into the cirtical section immediate,this alg violate the rule.
Question2,
If turn == 0 and P1 is ready to enter
itscritical section,P1 cannot do
so,even thought P0 may be in its
remainder section
Whats the meaning of this statement?As far as I could think,I could not understand why turn == 0 and p0 be in its remainder section could be exist concurrently...
So is this statement wrong?
Suppose that turn = 0 initially. P0 does its critical section and sets turn = 1. Now, P1 must execute its critical section before P0 may execute its one again. But just because both threads have a critical section doesn't mean that they're going to want to alternate their use of it in this way — in fact, P1 may never take its turn. (And in the general case, you can't determine this at compile time.)
So basically the problem is that the threads are forced to alternate their turns, even if one of them actually doesn't want to enter its critical section for an indefinitely long time.
By the way, your answer to Question 1 is correct. The algorithm doesn't fail the Progress condition, it fails the Bounded Waiting condition.