street vendor: how are the metric TSP and the Eulerian cycle equivalent?

On paper: https://arxiv.org/pdf/1908.00227.pdf states that

In the problem of metric TSP, which we study here, distances satisfy the inequality of the triangle. Therefore, the problem is equivalent to finding a closed Eulerian walk of minimal cost.

They do not appear to be equivalent at all, since for a complete chart of size 5, with all costs of edge 1, a TSP metric solution would be cost (and length) 5, while a Eulerian cycle would be cost (and length) ) 10. Am I misunderstood?

wp query: continue or interrupt the while cycle

I'm trying to find a way to keep it while the loop goes for posts in another section in boot columns or is there any way to "break it" and then keep it where it stayed? The first image shows the while cycle that shows 5 posts the way I want. The image below shows how I want the following 6 publications to remain vertical. Now I'm showing the posts, but it starts from the first post and just duplicates them. enter the description of the image hereenter the description of the image here

  
& # 39; send & # 39 ;, & # 39; posts_per_page & # 39; => 5, ); $ blogposts = new WP_Query ($ args); $ i = 0; while ($ blogposts-> have_posts ()) { $ blogposts-> the_post (); yes ($ i < 2) : ?>
<a href = "">
<img class = "card-img" src = "https://wordpress.stackexchange.com/"alt =" Card image ">
- -
& # 39; send & # 39 ;, & # 39; posts_per_page & # 39; => 6, ); $ blogposts = new WP_Query ($ args); while ($ blogposts-> have_posts ()) { $ blogposts-> the_post (); ?>
<a href = ""> <img class = "card-3-img" src = "" class = "card-img-top h-100" alt = "...">

- -

double cycle for

It is possible to make a for in python with two variables that are traversed at the same time but on different sides in the same list

MacBook Pro is constantly in kernel panic, freezing or entering the restart cycle

2018 MacBook Pro 15 "a1990 model shuts down on its own. The fans run at full speed and turns off. Sometimes I can log in for a minute at most, other times it goes into an infinite restart cycle. The recovery mode causes an error ( -1008F) by holding down D for the diagnosis and it says that there is no hardware problem. I have tried all the different boot commands with mixed results, since I have said that sometimes I can enter the computer and everything I had previously opened It is there and accessible, but shortly after it will shut down. (Panico kernel?)

I could get this out on one of the reboots

panic(cpu 4 caller 0xffffff80190689ca): Kernel trap at 0xffffff81e85a3520, type 14=page fault, registers:
CR0: 0x0000000080010033, CR2: 0xffffff81e85a3520, CR3: 0x00000001ef8de011, CR4: 0x00000000003626e0
RAX: 0x0000000000000000, RBX: 0xffffff804b0f4000, RCX: 0x0000000000000000, RDX: 0x0000000000000017
RSP: 0xffffff81e85a3520, RBP: 0xffffff81e85a3520, RSI: 0x0000000000000017, RDI: 0xffffff804b0f4000
R8:  0x0000000000000000, R9:  0xffffff7f9c5dbd28, R10: 0xffffff7f9c5dbe6c, R11: 0x0000000010000003
R12: 0xffffff804b155660, R13: 0x0000000000000016, R14: 0x000000000000001f, R15: 0xffffff804bf4935b
RFL: 0x0000000000010246, RIP: 0xffffff81e85a3520, CS:  0x0000000000000008, SS:  0x0000000000000010
Fault CR2: 0xffffff81e85a3520, Error code: 0x0000000000000011, Fault CPU: 0x4 Kernel NX fault, PL: 0, VF: 2

Backtrace (CPU 4), Frame : Return Address
0xffffff81e85a2f80 : 0xffffff8018f3f98b 
0xffffff81e85a2fd0 : 0xffffff8019076c15 
0xffffff81e85a3010 : 0xffffff801906861e 
0xffffff81e85a3060 : 0xffffff8018ee6a40 
0xffffff81e85a3080 : 0xffffff8018f3f077 
0xffffff81e85a3180 : 0xffffff8018f3f45b 
0xffffff81e85a31d0 : 0xffffff80196d2d89 
0xffffff81e85a3240 : 0xffffff80190689ca 
0xffffff81e85a33c0 : 0xffffff80190686c8 
0xffffff81e85a3410 : 0xffffff8018ee6a40 
0xffffff81e85a3430 : 0xffffff81e85a3520 
0xffffff81e85a3520 : 0xffffff7f9c5dbefa 
0xffffff81e85a3560 : 0xffffff7f9c603a44 
0xffffff81e85a35e0 : 0xffffff7f9c60af0b 
0xffffff81e85a36b0 : 0xffffff7f9c5f85b1 
0xffffff81e85a36e0 : 0xffffff7f9c5db152 
0xffffff81e85a3750 : 0xffffff7f9c5db071 
0xffffff81e85a37b0 : 0xffffff7f9c594f2a 
0xffffff81e85a3870 : 0xffffff7f9c594633 
0xffffff81e85a3900 : 0xffffff80191cb82e 
0xffffff81e85a3980 : 0xffffff80191c0fac 
0xffffff81e85a3b80 : 0xffffff80191c14bf 
0xffffff81e85a3bd0 : 0xffffff80191ae7cd 
0xffffff81e85a3f10 : 0xffffff80191aef7f 
0xffffff81e85a3f40 : 0xffffff801959b999 
0xffffff81e85a3fa0 : 0xffffff8018ee7206 
      Kernel Extensions in backtrace:
         com.apple.filesystems.apfs(1412.11.7)(B26D7DE6-BE2D-3E44-B334-1346AEC6C3BB)@0xffffff7f9c55a000->0xffffff7f9c67efff
            dependency: com.apple.kec.corecrypto(1.0)(263BCEB0-E4C5-3540-9E03-CC1F0A4D5BDF)@0xffffff7f99e32000
            dependency: com.apple.driver.AppleEffaceableStorage(1.0)(D4D0276C-28F7-3878-AC53-86A1134225E0)@0xffffff7f99d8e000
            dependency: com.apple.iokit.IOStorageFamily(2.1)(CAC103D2-4533-3A81-8190-D0133B4F8626)@0xffffff7f99b75000

BSD process name corresponding to current thread: xpcproxy
Boot args: chunklist-security-epoch=0 -chunklist-no-rev2-dev

Mac OS version:
19A603

Kernel version:
Darwin Kernel Version 19.0.0: Wed Sep 25 20:18:50 PDT 2019; root:xnu-6153.11.26~2/RELEASE_X86_64
Kernel UUID: 70EDD61F-86EE-3E1B-873F-98D909B78160
Kernel slide:     0x0000000018c00000
Kernel text base: 0xffffff8018e00000
__HIB  text base: 0xffffff8018d00000
System model name: MacBookPro15,1 (Mac-937A206F2EE63C01)
System shutdown begun: NO

System uptime in nanoseconds: 4668631807
last loaded kext at 1396725744: >usb.IOUSBHostHIDDevice 1.2 (addr 0xffffff7f99aef000, size 49152)
loaded kexts:
>usb.realtek8153patcher 5.0.0
>BCMWLANFirmware4355.Hashstore  1
>BCMWLANFirmware4364.Hashstore  1
>BCMWLANFirmware4377.Hashstore  1
>!AFileSystemDriver 3.0.1
@filesystems.hfs.kext   522.0.9
@BootCache  40
@!AFSCompression.!AFSCompressionTypeDataless 1.0.0d1
@!AFSCompression.!AFSCompressionTypeZlib    1.0.0
>!AVirtIO   1.0
>!ABCMWLANBusInterfacePCIe  1
@filesystems.apfs   1412.11.7
@private.KextAudit  1.0
>!ASmartBatteryManager  161.0.0
>!AACPIButtons  6.1
>!ASMBIOS   2.1
>!AACPIEC   6.1
>!AAPIC 1.7
$!AImage4   1
@nke.applicationfirewall    302
$TMSafetyNet    8
@!ASystemPolicy 2.0.0
|EndpointSecurity   1
>usb.IOUSBHostHIDDevice 1.2
>usb.cdc.ecm    5.0.0
>usb.cdc.ncm    5.0.0
>usb.cdc    5.0.0
>usb.networking 5.0.0
>usb.!UHostCompositeDevice  1.2
|IOSurface  269.6
@filesystems.hfs.encodings.kext 1
>!ABCMWLANCore  1.0.0
>mDNSOffloadUserClient  1.0.1b8
>IOImageLoader  1.0.0
|IOSerial!F 11
|IO80211!FV2    1200.12.2b1
>corecapture    1.0.4
|IOSkywalk!F    1
>!AXsanScheme   3
>!AThunderboltNHI   5.5.8
|IOThunderbolt!F    7.4.5
>usb.!UVHCIBCE  1.2
>usb.!UVHCI 1.2
>usb.!UVHCICommonBCE    1.0
>usb.!UVHCICommon   1.0
>!AEffaceableNOR    1.0
|IOBufferCopy!C 1.1.0
|IOBufferCopyEngine!F   1
|IONVMe!F   2.1.0
>usb.!UHostPacketFilter 1.0
|IOUSB!F    900.4.2
>usb.!UXHCIPCI  1.2
>usb.!UXHCI 1.2
>!AEFINVRAM 2.1
>!AEFIRuntime   2.1
>!ASMCRTC   1.0
|IOSMBus!F  1.1
|IOHID!F    2.0.0
$quarantine 4
$sandbox    300.0
@kext.!AMatch   1.0.0d1
>!AKeyStore 2
>!UTDM  489.11.2
|IOSCSIBlockCommandsDevice  422.0.2
>!ACredentialManager    1.0
>KernelRelayHost    1
>!ASEPManager   1.0.1
>IOSlaveProcessor   1
>!AFDEKeyStore  28.30
>!AEffaceable!S 1.0
>!AMobileFileIntegrity  1.0.5
@kext.CoreTrust 1
|CoreAnalytics!F    1
|IOTimeSync!F   800.14
|IONetworking!F 3.4
>DiskImages 493.0.0
|IO!B!F 7.0.0f8
|IO!BPacketLogger   7.0.0f8
|IOUSBMass!SDriver  157.11.2
|IOSCSIArchitectureModel!F  422.0.2
|IO!S!F 2.1
|IOUSBHost!F    1.2
>usb.!UCommon   1.0
>!UHostMergeProperties  1.2
>!ABusPower!C   1.0
|IOReport!F 47
>!AACPIPlatform 6.1
>!ASMC  3.1.9
>watchdog   1
|IOPCI!F    2.9
|IOACPI!F   1.4
@kec.pthread    1
@kec.Libm   1
@kec.corecrypto 1.0!

enter the description of the image here

Click for full size

Algorithms: If I have an MST and add any edge to create a cycle, will removing the heaviest edge of that cycle result in an MST?

Let's say I have an MST, $ T $. I choose an edge that is not in $ T $ and change your weight, and add it to $ T $ to create a cycle Will removing the heaviest edge of that cycle result in an MST?

MST means minimum expansion tree of a chart. I found these two publications:

and I follow both to the case where $ w_ {old}> w $ Y $ e notin T $. Both say that removing the heavier edge will guarantee an MST, but I don't see how to prove it. The cycle property only says that if it has an MST, it cannot have an edge, which is the heaviest edge in a cycle of the original graph $ G $; You are NOT saying that IF you have a tree that does not contain an edge that turns out to be the heaviest edge of any cycle in the original graphic $ G $, you are an MST.

To ask the most explicit question in terms of the problem I was trying to solve, I will copy a part of the first link:

If your weight was reduced, add it to the original MST. This will create a cycle. Scan the cycle, looking for the heaviest edge (this could select the original edge again). Remove this edge.

I don't understand why this guarantees that we find an MST. Sure, we get an expansion tree, but why removing this heavier edge produces a MINIMUM expansion tree?

Mobilizer – How to cycle through the dynamically generated response keys?

In MEL, how to cycle through the dynamically generated response keys to verify that it has been verified or not. I have shared a block of sample code. I want to scroll through the answer keys. Here the answers are dynamically generating.


            
                Pink
            
            
                Red
            
            
                Violet
            
        .......
        
                Cyan
            

            Select the colors you prefer 

         

please suggest me the best approach here

Thank you

java: set the Boolean flag to true if a variable with a given value appears twice in each cycle

How to write this with clean code rules:

orderLines is a list of JSON mapped objects in a given ORDER.
I want to check if the SECTION appears the second time in my for loop, then I have to set my indicator section Second appeared to true because in each SI I have to invoke another method (setList ()) with one of the two parameters depending on my indicator .

How to do that? With a counter? When a counter = 2, does the flag change to true? What is the most professional approach in that case?

 boolean sectionAppearedSecondTime = false;
    for (OrderLine orderLine : orderLines) {

                    if (orderLine.getBase()) {
                        //do something
setList(list1 OR list2 depending on the flag)
                    } else if (orderLine.getSection()){
                        if(orderLine.getSection() appeared the second time, then){
    sectionAppearedSecondTime = true;
setList(list1 OR list2 depending on the flag)
    }
                    } else if (orderLine.getSpecialty())) {
                        //do something
setList(list1 OR list2 depending on the flag)
                    } else if ...
                    }

linear programming – LP – m restrictions given for 2 variables find the maximum cycle radius

Dice $ m $ restrictions for 2 variables $ x_1, x_2 $ :

$ d_ix_1 + e_ix_2 leq b_i $ for $ i = 1, … m $

you need to create a linear program that finds the maximum radius of a cycle so that all the points within the cycle are in the feasible range of the previous restrictions.

so I know the formula for the distance between some $ (x, y) $ and something $ ax + for + c = 0 $

then i tried

  • $ Maximum $ $ R $ S t
  • $ d_ix_1 + e_ix_2 leq b_i $ for each $ i $
  • $ R leq | d_ix_1 + e_ix_2 – b_i | / { sqrt {{e_i} ^ 2 + {d_i} ^ 2}} $ for each $ i $
  • $ R geq 0 $

I know that the standard linear program does not get the absolute value function, but obviously we can only have 2 inequalities to get rid of it.

What you think? and how I can eventually the relevant $ (x_0, y_0) $ that will be the center of the cycle that $ R $ is your radio going to be specific $ x_1, x_2 $ Will that make R maximum, I guess?

Update to handle the final cycle of Windows Server 2008

As Microsoft has terminated support for Windows Server 2008, I am forced to look for an alternative.

What I need is:
1. History of reliability.
2. Windows 2016 as an option (I don't want to load an ISO).
3. At least 2 virtual cores.
4. At least 4 GB of RAM.
5. At least 80 GB of SSD storage.
6. 3 TB + bandwidth.
7. An automatic backup service that allows me to restore the server (from the previous day, week, month).
8. Access to RDP / console.

co.combinatorics – Number of expansion trees in the third power of the cycle

Next page

https://oeis.org/search?q=1%2C1%2C2%2C4%2C11%2C16&language=english&go=Search

gives the number of expansion trees in the third power of the cycle, for example,

$ 1, 1, 2, 4, 11, 16, 49, 72, 214, 319, 947, 1408, ldots $.

Is it not clear to me that the definition of "expansion trees"? It is the number of expansion trees in the third cycle. $ C_3 ^ 3 $ $ 2 $? I think the number is $ 3 $ or $ 1 $ until isomorphic.

Could you show me the definition of "expansion trees" on this page?