JPS5921424B2 - How to unlock using a card - Google Patents

Description

[Detailed Description of the Invention] The present invention provides an unlocking method using a card, which is particularly suitable for hotels where there are many rooms that need to be locked and the person unlocking the locks changes frequently. Regarding the unlocking method used.

The hotel has many rooms that need to be locked.

The door of each room is usually equipped with a mechanical lock, such as a tumbler lock, which can be opened and closed with a key.

With a lock that opens and closes with a key, the key must be handed over to use the room and then taken back after use, which requires a lot of effort to manage the keys.

Furthermore, if the key is lost, the lock itself needs to be replaced as a crime prevention measure against those who find it, which takes time and money.

Therefore, we installed an automatic unlocking device that unlocks the lock using an electrical signal, and installed a card reader, a storage device that stores a secret number, and a processing unit on the door, and locked the door using a card with the secret number recorded on it. A solution has been devised.

This device compares the secret number read from the card by the card reader with the secret number stored in the storage device, and unlocks when the two numbers match.

Then, each time the user of the same room changes, a card with a new secret number recorded is handed to the user.

The secret number is changed in a central control unit that centrally controls a large number of rooms, and the new secret number that has been changed is transmitted each time from the central control unit to the storage device of the corresponding room.

Therefore, with this method, communication cables must be installed between the central control unit and the lock/unlock devices in each room, which increases equipment costs and requires maintenance and inspection at appropriate intervals. Must have.

Therefore, there is a need for an unlocking method using a card that does not require communication cables between the central control unit and the lock/unlock devices in each room.

On the other hand, in the above unlocking device, the secret number is changed when the user of the room changes, so if there is no next user of the room, the secret number stored in the above storage device will be replaced by the previous user's card. The number remains the same.

In hotels, etc., rooms often remain vacant for a day or two or even several days, and in such cases, the previous user has to pay for the room until the next user starts using it. It becomes possible to use it without any problems.

This invention was made in view of the above circumstances, and the secret number can be changed every time the user changes, so there is no need to take special security measures against loss or abandonment of the card. Furthermore, there is no need to install a cable between the central control unit and the system can be realized at low cost.In addition, it is possible to unlock the room using a card only for the scheduled period of stay, and after the period of stay has elapsed. provides an unlocking method in which the same card cannot be used to unlock the same room unless a new card is issued.

Hereinafter, a case will be described in which the present invention is applied to doors of a large number of rooms in a hotel with reference to the drawings.

FIGS. 1 to 4 show an overview of the process from issuing a card to unlocking a door using this card.

A card issuing machine 1 is installed at the front desk of a hotel.

When the cards 2 and 4 having the magnetic stripe 3 are inserted into the card issuing machine 1, required card data is magnetically recorded on the magnetic stripe 3 according to the procedure described later.

There are two types of cards: card 2, which is handed to a hotel user, and is issued when the user starts using a room, for example, when checking in at the hotel.

By using this user card 2, it is possible to unlock only the room to be used within the scheduled period of stay.

The other one is maid card 4 used by hotel maids.
It is.

The maid uses the maid card to unlock each room in order to clean each room once a day.

User card 2 contains card data such as hotel code H, room number N1, secret number G, and two types of auxiliary numbers A.
, B, a designation code C for selecting an auxiliary number, and the number of days of stay are recorded.

In Figure 2, these numbers or codes G, A, B
, C is appended with a suffix 2° indicating the number of times the card is issued to the same unlocking device.

When the card 2 is issued, this card 2 is handed over to the user who uses the corresponding room.

The door 5 of each room is provided with an unlocking device, and this unlocking device includes a card reader 6.

A card reader 6 is mounted on the outside of the door 5.

A fixed knob 7 is also provided on the outside of the door 5, and each room number N is displayed at the top.

The card reader 6 is of a manual scanning type, and when the user inserts the card 2 into the card reader 6 and scans it,
Door 5 is unlocked.

Of course, an automatic scanning type card reader can also be used.

The maid card 4 records a hotel code H and a maid secret number GO.

The hotel maid can also insert this card 4 into the card reader 6 to unlock the door.

FIG. 5 shows the appearance of the locking device provided for each door of each room in the hotel.

This locking device has an automatic locking function, and when the door 5 is closed, the trigger bolt 9 is activated, the latch 8 is fixed, and the door is automatically locked.

When the card 2 or 4 is inserted into the card reader 6 as described above, the unlocking nozzle 69 (FIG. 9), which will be described later, is energized and the indicator light 13 is turned on.

Then, when the button 11 of the knob 1 is pressed, the lunch is released and the lock is unlocked.

This lock device is equipped with an auxiliary mechanism that unlocks by mechanical operation using a key, and the lock is unlocked by inserting the key into the keyhole 12 and turning it.

Further, an inner bundle (not shown) is provided on the inside of the door 5, and the door can be freely unlocked using this bundle.

FIG. 6 shows the internal configuration of the card issuing machine 1.

The card issuing machine 1 includes a numeric keypad 15, an input bag 7a register 16, a random number generator 11, a display 24,
An encoder 25, a clock 26 and its control circuit 27, an arithmetic circuit 20, arithmetic registers 21 and 22 (expressed as Ra 1 * Ra 2 for convenience), and a 7-lag 23 (Fl) for storing arithmetic results. ), a register group 30 for storing various data, a RAM 40 for storing data,
It is composed of a register 39 (indicated by AD) for specifying the address of the RAM 40.

The numeric keypad 15 is also provided with keys for items representing the date and room class.

Random number generator 17 includes an oscillator 18 and a counter 19.

The register group 30 includes seven registers 31 to 31 that respectively store seven types of card data.

The RAM 40 also includes storage areas 41 to 43 for information regarding the grade of each room, and storage area 4 for information regarding each room.
4 to 58, a storage area 60 for hotel codes H, and a storage area 61 for the number of rooms M.

In this example, the rooms are graded, with first class, second class, and third class rooms.

The room numbers for 1st class rooms are 1-10, the room numbers for 2nd class rooms are 11-293, etc., and the room numbers are 30 or less.

In the areas 44 to 58, the room number N1 secret number 01 designation code C and the length of stay are stored for each room.

In the area 41, the first address of the storage area for information about the first class room, in this example, the address of the area 44 is stored.

Similarly, in areas 42 and 43, the leading addresses of storage areas for information on degrees Celsius for rooms 2, 3, etc., in this example, the addresses of areas 50 and 52 are stored, respectively.

Now, add the subscript l to the secret number G1 auxiliary numbers A, B and designation code C of the card previously issued for a certain room, and get G1, AI, and B1, respectively.

Represented by C1, the secret number G1 of the card issued this time for the same room, the auxiliary numbers A, B, and the designation code C are appended with the suffix 2 as G2 and A2, respectively.

B2. It is represented by C2.

The card issuing machine 1 creates the current secret number G2 from the previous secret number G1 and designation code C1, and either the current auxiliary number A2 or B2.

Here, the designation code C1 and the auxiliary number A2. B2 is a random number and is read out from the random number generator 17.

Auxiliary numbers A and B may be numbers of the same rank, but 3, 4
Digits are preferred.

The designation code C1 was read out when creating the secret number 01 last time, and when creating the secret number 02 this time, the designation code C2 is read out for the next time L -1<, and the designation code C is either 1 or O. One value is taken, and auxiliary number A or B is selected depending on that value.

In this embodiment, when the designation code C is 1, the auxiliary number A is selected, and when the designation code C is O, the auxiliary number B is selected.

There are many calculation methods to create this secret number G2.

For example, G2=G1-+lA2 or B2)
...(υG2=GIX(A2 or B2)
...(2) G2=mXG1+nX(A2 or B 2
) - (3) G2 - (C1) + (A2 or B2)
... (4) etc. can be mentioned.

Here, m and n are constants, and the RAM of the card issuing machine 1
40 and a RAM 90 of the unlocking device 10, which will be described later.

In this way, the current secret number G2 is the previous secret number G
1 so that the required relationship exists between the two.

In this embodiment, equation (1) is used.

Let us take as an example the case where a card for room number N is issued.

RAM400 Storage area 54.5 for room number N
5.56 stores the room number N1, the previous secret number G1, and the designation code C1.

Further, an O is written in the storage area for the length of stay for the empty room.

The room with room number N is empty, and O is stored in area 51.

Note that consecutive numbers are assigned to the addresses of each area 41 to 61.

Referring to FIG. 7, when a numerical value indicating the grade of the room is entered using the numeric keypad 15, this is read into the buffer register 16 (step 101), and if there is a key indicating "etc." on the numeric keypad 15 (step 102). ), step 10
Put the numerical value indicating the grade read in step 1 into register Ra1 (
Step 103), and input the reference address into register Ra2 (step 104).

The reference address is the address of the area (the path shown) immediately before the area 41.

If the address of area 41 is 41, then address 40 is placed in register Ra2.

Next, the contents of both registers Ra1 and Ra2 are added and placed in the address or register AD (step 105).
.

As a result of this calculation, when the 1st class room is designated, the reference address is incremented by 1, so area 41 is designated, and when the 2.3rd class room is designated, areas 42 and 43 are designated accordingly.

Specify one of areas 41 to 43 with the contents of the address register AD corresponding to the input grade, read out the first address for each grade stored in that area, and store it in the address register AD. (Step 106) and is also stored in register Ra2 (Step 107).

If the third class room is specified, the address of area 52 is read from area 43.

Then, the area 52 is designated with this address, and the room number 30 stored in the area 52 is read out and stored in the register 32 (step 108).

Next, 1 is placed in register Ra1 (step 109), the contents of both registers Ra 1 and Ra 2 are added, the result is placed in register Ra2 (step 110), and then the contents of register Ra2 are sent to address register AD. to designate the next area (path shown) after area 52 (step 111).

Since the previous secret cauldron MCI of room number 30 is stored in the area next to area 52, it is read out and stored in register 33 (step 112).

Register Ral contains 1 (step 109)
, register Ra2 contains the address of the area next to area 52, and by adding these, the area in which the designation code C1 is stored is designated.

The sum of the contents of both registers Ra1 and Ra2 is put into register Ra2 (step 113), and
Send the contents of to the address register AD (step 1
14), the designation code C1 is read out and stored in the register 34 (step 115).

Similarly, the contents of both registers Ral and Ra2 are added (step 116), thereby specifying the area where the number of days of stay is stored (step 117).

The number of days the worker will stay is read out and stored in the register 37 (step 118).

Then, the number of days of stay is read out and placed in register Ral (step 119), O is placed in register Ra2 (step 120), and both registers Ral are read out.

Subtract the contents of Ra2.

If the result of this subtraction is 0 (the number of days of stay D=0), the flag F1 is set to 1, and if it is other than 0, the 7 lag F1 is set to 0 (step 121).

7 If the flag F1 is 1, the number of days of stay is 0 and the room is empty, so proceed to step 131. If the flag F1 is 0, the room is not empty, so another room is searched. Then proceed to step 123 (step 122
).

In step 123, the contents of address register AD are moved to register Ra2, and the final address is transferred to register Ra1.
(step 124), and both registers Ra1. Ra
Subtract the contents of 2 (Status 7'125).

The final address is the final address of an area that contains information regarding each room for each grade. For example, in the case of 1st grade, the number of days of stay of room number 9 is stored and is the address of the area.

If the address of the area whose length of stay was read in step 118 is the final address, the subtraction result in step 125 is O, and in this case, 72 lag F1 is set as 1, otherwise, 7 lag F1 is set as 1. Cent to 0.

If the flag F1 is 1 (step 126), it means that the inspection of all the rooms for each class has been completed, and as a result, there are no empty rooms, so the display 24 displays "No rooms" and processing is continued. Finish (step 130).

If flag F1 is O (step 126, since the search for the eight rooms has not yet been completed,
By putting 1 into register Ra1 (step 121) and adding the contents of both registers Ra 1 and Ra 2, the address of the area where the room number to be searched for next is stored is created, and this addition result is stored in the register. Ra2 (step 128), and sends it to address register AD to specify the area to be searched (step 1
29).

Then, returning to step 108, the room number N, secret number G1, and designation code C1 are read out and stored in each register 32, 33, and 34, and the number of days of stay is read out and checked to see if it is O.

In this way, the number of days of stay in each room is checked, and if, for example, the number of days of stay in the room with room number N is 0, this room number N is displayed on the display 24 (step 131).

Then, the count value of the counter 19 at that time is read out from the random number generator 17 as an auxiliary number A2, and stored in the register 35 (step 132).

Next, it waits for the number of days of stay and "day" to be entered using the numeric keypad 15 (steps 133 and 134), and if the day is entered, the number of days of stay entered manually is stored in the register 31 (steps 133 and 134).
Step 135).

As in step 132, a random number is read from the random number generator 11 and stored in the register 36 as the auxiliary number B2 (step 136).

Next, the previous designation code C1 is read from the register 34 and placed in the register Ral (step 137), 0 is placed in the register Ra2 (step 138), and the arithmetic circuit 2
Both registers Ra1. By subtracting the contents of Ra2, it is determined whether the previous specified code C1 is 1 or 0, and this result is sent to 7 lag F1 (step 1
39).

If the designation code C1 is 1, the subtraction result is 1, and in this case, O is set in the flag F1.

Conversely, if the designated code C1 is O, the subtraction result is O, and in this case, 1 is added to the 7 lag F1.

Checking the state of 7-lag F1 (step 140), 7-lag F1
If 1 is 0, the auxiliary number A2 is read from the register 35 and placed in the register Ra2 (step 7'141), and if the flag F1 is 1, the auxiliary number B2 is read from the register 36 and placed in the register Ra2 (step 142). , reads the previous secret number G1 from the register 33 and stores it in the register Ra.
Enter IK (step 143).

Then, both registers Ra1. The contents of Ra2 are added and the result is stored in the register 33 as the current secret number G2 (step 144).

By the processing of steps 140 to 144, the specified code C1
It will be understood that the current secret number G2 is calculated using the auxiliary number A2 or B2 selected by the user and the previous secret number G1.

Since the last digit of the count value of the counter 19 of the random number generator 17 is 1 or 0, read this and enter the current designation code C.
2 and stored in the register 34 (step 145).

Using the address when the number of days of stay was read in step 118, the number of days of stay in the register 31 is stored in the area of the number of days of stay of room number N in the RAM 40 (57 in this case).
Step 146) of updating the number of days of stay in the RAM 40 by storing the number of days of stay in the RAM 40.

The contents of address register AD specifying area 57 are put into register Ral (step 14γ), and register R
Put 1 in a2 (step 148), and set both registers Ra.
The contents of 1 t Ra 2 are subtracted and the subtraction result is sent to address register AD (step 149) to designate area 56.

Then, the designation code C2 of the register 34 is stored in the area 56 (step 7''150).

Similarly, the contents of address register AD are transferred to register R.
a1 (step 151), and both registers Ra 1
, Ra 2 and use this subtraction result as address.
Put it in register AD, specify area 55 (Stella 7
"152), and the current secret number G2 of the register 33 is stored in the area 55 (step 153).

As described above, the secret number G1 and the designation code C1 of the storage areas 55 and 56 are changed to G2. Changed to C2.

The hotel code H is read from the area 60 of the RAM 40 and placed in the register 31 (step 154), and finally, from the register group 30, the hotel code H1, room number N1, secret number G2, designation code C2, auxiliary number A2, . B2 and the length of stay are read and transferred to the encoder 25, and the card 2 is read out and transferred to the encoder 25.
(step 155).

In the above procedure, the room number N1, secret number G, and designation code C1 are read out from the RAM 40 and stored in each register 32, 33, and 34 for each room search (steps 108, 112, and 115). First, only the number of days of stay is searched sequentially, and after finding the room D=O, each number N related to that room is searched.

It is also possible to read out Gl and CI.

In the above example, when you finish using the room, at the front desk,
The number of days of stay in the corresponding room stored in the RAM 40 is set to O, but the number of days of stay in all the areas in the RAM 40 (if the number of days of stay is O) is displayed at 10 am, for example, due to the 1 change signal from the clock 26. The stored number of days of stay may be updated every day by subtracting 1 from (excluding).

Referring to FIG. 8, when a 1 change signal is output from the clock 26 (step 160), the total number of rooms M
Step 161: Store in area 61 of R, AM40
), the address of the area (area 41 in this example) in which the number of days of stay in the first room is stored in the RAM 40 is entered into the address register AD (step 162).

Then, by this address register AD, area 4 is
1 is specified and the number of days of stay in that area is read out and stored in register Ra1 (step 163).

Next, put 0 in register Ra2 (step 164),
The contents of both registers Ra1 and Ra2 are subtracted, and if the result is 0, the 7 lag F1 is set to 1, and if the result is other than 0, the flag F1 is set to 0 (step 165).

If the 7-lag F1 is 1 (step 166), the read length of stay is D'lJ''-O, so the process moves to step 170. If the 7-lag F1 is O, 1 is subtracted from the length of stay. Proceed to step 167 to subtract.

In step 167, 1 is first put into register Ra2, and the contents of register Ra 1 e Ra 2 are subtracted (step 168), and this subtraction! The result D-1 is stored in the original area 47 (step 169).

In order to check whether the subtraction process has ended for all the days of stay, the number of rooms M is read from the area 61 and placed in the register Ra1 (step 170), and
2 is set to 1 (step 111), and both registers Ra
1 s Ra 2 is subtracted and the result is stored in register Ral (step 172).

In this way, the number of rooms M is calculated for each subtraction process for one length of stay.
Subtract 1 from.

Therefore, the number of rooms M is the number of days of stay that has not yet been subtracted.

Register R to check whether this number of rooms M is 0.
Put O in a2, subtract the contents of both registers Ral and Ra2, and if the subtraction result is O, put 7 lag F1 into 1, otherwise put 7 lag F1 into O (
Step 114).

Then, looking at the state of flag F1 (step 175),
If the flag F1 is 1, it means that the processing has been completed for all areas of the length of stay, and the processing ends.

If flag F1 is O, the contents of address register AD are transferred to register Ra in order to move to the next area for the number of days of stay.
1 (step 176), 4 is placed in register Ra2 (step 111), the contents of both registers Ral and Ra2 are subtracted, and the result is placed in address register AD (step 118).

Thereafter, the process moves to step 163, where the number of days of stay in the next area is read out and similar subtraction processing is performed one after another.

FIG. 9 shows an unlocking device 10 provided on the door 5. As shown in FIG.

The unlocking device 10 includes a card reader 6, a clock 65, a control circuit 66 including an instruction register, a duder, an interrupt control circuit, etc., a timer 61, and a limp 70 ring 68 (FF
(represented by
0 and calculation registers 71 and 72 (Rbl, Rb
2), a flag 73 (represented by F2) for storing calculation results, a register group 80 for storing various data, and a RAM 90 for storing data.

In this embodiment, the unlocked state is maintained for a predetermined time T after unlocking, and after this time T has elapsed, the locked state is returned again.

A timer 67 is used to measure the constant time T.

Register group 80 is composed of seven registers 81-87.

The RAM 90 also includes an area 91 for storing the hotel code H, an area 92 for storing the room number N, an area 93 for storing the secret number G, an area 94 for storing the designation code C, and an area 95 for storing the length of stay. , an area 96 for storing the set time T of the timer 61, and an area 97 for storing the maid's secret number GO.

The unlocking device 10 reads the data on the card 2 or 4 and determines that the secret number G in the card data has one of the following relationships with the secret numbers G and Go stored in the storage areas 93 and 97 of the RAM 90. If the number of days of stay in the storage area 95 of the RAM 90 is not O, the lock is released.

The predetermined relationship 1 regarding the secret number is a case where the secret number G read by the card reader 6 matches the secret number G in the storage area 93.

Part 2 is that the secret number G read by the card reader 6 is the (1st ) is the case.

That is, the secret number and auxiliary number in the card data are G2, A2, and B2, respectively, and the storage area 9 is
3's secret number is G1, and the designated code of storage area 94 is 0.
1, these secret numbers Gl.

G2 is the auxiliary number A selected by the designation code C1
2 or B2 to release the lock if the relationship of equation (1) is satisfied.

In this case, since the user of room number N has changed, the secret number G2 and designation code C2 in the card data are changed to Gl.

They are stored in storage areas 93 and 94 of the RAM 90, respectively, instead of C1.

Part 3 is when maid card 4 is used,
Maid card 4's maid secret number GO is R, AM9
This is a case where the number matches the secret number GO stored in area 91 of 0.

Now, storage areas 91, 92, 96°97 of RAM 90
are respectively hotel code H1 room number N1 timer 6
It is assumed that the set time T of 7 and the secret number GO for the maid are stored in advance.

Further, the knob 70 and the knob 68 have been reset.

Referring to FIG. 10, when card 2 or 4 is inserted into card reader 6, card 2 or 4 is read and the read card data is read into register group 80 (step 181).

Then, the hotel code H in the storage area 91 of the RAM 90 is placed in the register Rbl (step 182), and the hotel code H is placed in the register Rb2 in the register 81 (step 183), thereby checking the hotel code.

The contents of both registers Rbl and Rb2 are subtracted, and if the result is 0, the 7 lag F2 is set to 1, and if the subtraction result is other than O, the flag F2 is set to O (step 18
4).

Then, looking at the state of 7-lag F2, if 7-lag F2 is 1, then the hotel code H in the card data is correct, and the next step is to check the room number N in step 186.

If the flag F2 is O, it means that the hotel code H is incorrect and an unauthorized card is being used, so the process ends without unlocking.

In step 186, as in the case of checking hotel code H, room number N in memory area 92 of RAM 90 is checked.
is placed in register Rbl, and the next register 820 room number N is placed in register Rb2 (step 187). Both registers Rb1. Subtract the contents of Rb2 and the result is O
If so, set flag F2 to 1 (step 188
).

Then, check the state of flag F2 (step 189).

If it is 1, the room number N is assumed to be correct and the process moves to step 190 to check the secret number G.

If the flag F2 is 0, the process ends and the lock is not unlocked.

The current secret number G2, designation code C2, auxiliary numbers A2 and B2, and the number of days of stay are stored in the registers 83 to 87, respectively, and the memory areas 93 and 9 of the RAM 90 are stored in the registers 83 to 87.
4.95 stores the previous secret number G1, designation code C1, and number of days of stay.

In step 190, the secret numbers G and G1 stored in the storage area 93 are read and stored in the register Rb1, and then the secret number G2 or GO of the register 83 is read out and stored in the register Rb1.
2 (step 191), and both registers Rbl and Rb
The contents of 2 are subtracted, and if the subtraction result is O, the flag F2 is set to 1 (step 192).

If the subtraction result is not 0, 7 lags F2 are sent to O.

Then, looking at the state of 7 lag F2 (step 193),
If it is l, the process advances to step 194; if it is 0, the process advances to step 203.

If the secret number in area 93 of RAM 90 is G1 and the secret number in register 83 is 02, these secret numbers do not match, and the answer at step 193 is NO.

This is a case where the user of room number N changes and a new card is issued.

Therefore, next, it is checked whether the numbers G1 and G2 are related by equation (1).

In step 203, the designation code C1 stored in the area 94 of the RAM 90 is read out and placed in the register Rbl, and then O is placed in the register Rb2 (step 204).
), subtract the contents of both registers Rbl and Rb2, and if the result is 0, set 7 lag F2 to 1, and the subtraction result is 1.
If so, set flag F2 to O (step 205
).

Then, if the state of the 7-lag F2 is 0, the process proceeds to step 207 to select the auxiliary number A2, and if it is 1, the process proceeds to step 209 to select the auxiliary number B2 (step 206).

In either case, first enter the secret number G1 of Elijah 93 into register Rb1 (steps 207 and 209), and
If lag F2 is 00, put the auxiliary number A2 of register 85 into register R, b2 (step 208), and set 7 lag F2.
If 2 is 1, the auxiliary number B2 of the register 86 is placed in the register Rb2 (step 210).

Then, the secret number G1 of register Rbl and the register R
The auxiliary number A2f or B2 of b2 is added to the auxiliary number A2f or B2, and the result is stored in the register Rb1 (step 211).

Read secret number G2 from register 83 and register Rb
2 (step 212), and both registers Rbl and Rb
If the content of 2 is subtracted and the result is 0, 7 lag F2 is placed in 1, and if the result is 0, 7 lag F2 is placed in 0 (step 213).

The contents of the register Rbl are the sum of the secret number G1 and the auxiliary number A2 or B2, and the contents of the register Rb2 are the secret number G2. Therefore, the above subtraction result is 0, which means that the above secret numbers G1 and G2 This means that the equation (1) is satisfied.

Therefore, in this case, the process proceeds to unlocking (step 214).

Step 2130 If the subtraction result is not 0, both secret numbers G
l, G2 is not in the relationship of (LX).

Therefore, if the flag F2 is 0 (step 21
4) Do not unlock (finish the process).

In this way, if the secret number stored in the RAM 90 and the secret number recorded on the card 2 do not have a certain relationship (for example, equation (0)), the lock cannot be unlocked.

If the 7lag F2 is 1 in step 214, the current secret number G2 stored in the register 83 is stored in the RAM 90.
Write the secret number in area 93 of (Step 2)
15).

Further, the designation code C2 of the register 84 is written into the storage area 94 of the RAM 90 to update the designation code and prepare for the next issued card (step 216).

Furthermore, the number of days of stay in the register 87 is written into the area 95 of the RAM 90 (step 217).

After the above processing, the set time T stored in the area 96 of the RAM 90 is read out and stored in the timer 67 (step 198), and the
is set (step 199).

When flip 70 Tsubu FF is sent, Nrenoid 69
works, so it is unlocked.

Step 200), the timer 67 is counted by -I from the content of the timer 67 until the content of the timer 61 becomes 0.
When the content becomes 0 (step 201), the 7-lip flop FF is reset (step 202).

In this way, the door is unlocked only for a certain period of time T during which the flip 70 knob FF is inserted.

Note that the time measurement process in step 200 is performed using register Rb1°f.
Subtraction processing may be performed using il and B2.

Since the secret number in the area 93 of the RAM 90 has been updated from 01 to G2 in step 215, if you try to unlock it again using the same card, the contents of the register 83 and the contents of the storage area 93 will match (step 215). 193
(Yes)

In this case, the process advances to step 194 to check the length of stay.

First, the number of days of stay of area 95 in RAM 90 is entered into register Rb1, and then 0 is entered into register Rb2 (step 195).

Both registers Rb1. Subtract the contents of Rb2 and if the result is 0, cent 72gF2 to 1,
If it is other than 0, 72gF2 is set to 0 (step 196).

7lag F2 being 1 means that the number of days of stay 1)f is Q, so in this case the process ends without unlocking.

If the length of stay is other than 0 (F2=O), step 1
Proceed to 98 and use the Flip 70 knob FF to unlock it for a certain period of time.

Since the data of the room number N is not recorded in the maid card 4 (the content is O), the determination at step 189 is NO.

In this case, the process moves to step 218 to check the maid's secret number GO, and first enters the maid's secret number GO in area 91 of the RAM 90 into registers R, b1, and then enters the maid's secret number GO in register 83. Register Rb2
(step 219).

Then, the contents of both registers R, bl, and Rb2 are subtracted, and if the result is O, flag F2 is set to 1, and O
In other cases, the flag F2 is set to O (step 220).

When the flag F2 is 00, the secret numbers in the area 97 and the register 83 do not match, so the process ends without unlocking (step 211).

If flag F2 is 1, the number of days of stay in area 95 of RAM 90 is entered in register Rb1 (step 222);
0 is placed in register Rb2 (step 223), and the contents of both registers Rbl and Rb2 are subtracted to see if the number of days of stay is 0 (step 224).

If the result of this subtraction is O, add 1 to flag F2,
If the value is other than 0, 0 is placed in the flag F2.

If flag F2 is 00, the number of days of stay is not 0, so proceed to step 226, put 1 in register Rb2,
Subtract 1 from the number of days of stay, which is the content of register Rb1, and put the result in register Rb1 (step 227).
.

Then, at ℃, the contents D-1 of this register Rbl are stored in RAM90.
Enter the area 95 and update the length of stay (Step 2)
28).

Hotel maids unlock each room once a day to clean the room.

Therefore, the remaining number of days of the user's stay stored in the RAM 90 is decremented by one each time the maid unlocks the room.

After subtracting 1 from the number of days of stay of area 95 in RAM 90 (step 228), and if the number of days of stay is O (
In step 225), the process moves to step 198 and an unlocking process is executed.

The daily update (subtraction of 1) of the number of days of stay stored in the area 95 of the RAM 90 is performed using the 1 output from the clock 65.
This can also be done using a change signal.

An example of this process is shown in FIG. When the clock 65 outputs a daily change signal once a day (step 231,
Using the same procedure as steps 222 to 228, only when the number of days of stay is not O, 1 is subtracted from the number of days of stay and the result is entered in the area 95, and the number of days of stay is updated (steps 232 to 238).

If there are multiple maids for a large number of rooms, the rooms each maid is responsible for cleaning are determined in advance, and each room can be unlocked only by the maid in charge.

Furthermore, in case the maid in charge changes or the maid card is lost, it is preferable to perform the secret number update process for the maid card in the same way as for the user card.

In the above example, the room number N identifies whether the card is for a user or a maid, but an identification code may be recorded on the magnetic stripe of the card to distinguish between a user and a maid. .

In this case, a code indicating the user and a code indicating the maid are stored in the RAM 90 in advance.

Then, after checking the hotel code H, it is compared with the identification code or maid code read from the card, and the process is performed according to the result.

Furthermore, although the length of stay of the user is 21 days,
It is also possible to measure the time in units of one hour, and to reduce the stored content related to the stay period in the storage device every predetermined number of unit times, for example, 6 hours or 12 hours.

Although there are two auxiliary numbers, A and B', three or more auxiliary numbers may be prepared and one or more of these auxiliary numbers may be designated by the designation code C.

Therefore, in this case, the designation code C takes a value of 3 or more instead of a binary value of 0 or 1.

Even when the card issuing machine 1 creates the current secret number G2, when the unlocking device 10 checks whether the previous secret number G1 and the current secret number G2 are in the relationship shown in (LX). Although the previous designation code C1 is used to select one of the auxiliary numbers A2 and B2, it goes without saying that the current designation code C2 may also be used.

As described above, in the unlocking method according to the present invention, when issuing a card, the previous secret number and the auxiliary number selected by the designated code from among the plurality of auxiliary numbers are used to keep the same number as the previous secret number. A secret number has been created, and this secret number, multiple auxiliary numbers, and a designated code are recorded on the card.

When unlocking each room, the data recorded on the card is read, and it is determined whether the read secret number matches the secret number stored in the storage device based on the above certain relationship. If the secret number of the card and the secret number of the storage device match, and if there is a certain relationship as described above, the lock is unlocked. When a certain relationship exists, the secret number of the card is stored in the storage device and the secret number is updated.

Therefore, the secret number of the card can be changed every time the user changes, and once the secret number is updated, it can no longer be unlocked using the previous card.
There is no need to take special security measures for cards used before the last time.

In addition, the changed secret number must not be transmitted from the central controller to the unlocking device (it can be determined whether the secret number on the card is legitimate or not, so the central controller and each unlocking device can It is not necessary to connect the equipment with a cable, and the equipment L5 can be installed at an extremely low cost.

Furthermore, since multiple auxiliary numbers are prepared and selected by specified codes, the above-mentioned certain relationships are very easy to understand (
Since the auxiliary number that is not involved in the calculation of the secret number is also recorded on the card, it is extremely effective in terms of crime prevention.

It is possible to unlock the same room using the card only during the scheduled stay period, and after the first half of the stay, the same room can be unlocked using the same card unless a new card is issued. cannot be locked.

Therefore, after the scheduled stay period has passed, the next user of the room will not be able to use the room (even if the room is vacant, this will prevent the situation where the previous user uses the room without paying). be able to.

[Brief explanation of drawings]

Figures 1 to 4 are explanatory diagrams showing an overview of the process from issuing a card to unlocking the door using the card, Figure 5 is a perspective view showing the external appearance of the locking device and unlocking device on the door, and Figure 6 is FIG. 7 is a block diagram showing the internal configuration of the card issuing machine. FIG. 7 is a flow chart showing the operation of the card issuing machine. FIG. 8 is a flow chart showing the operation when updating the length of stay using a clock. The figure is a block diagram showing the internal configuration of the unlocking device, FIG. 10 is a 70- chart showing the operation of the unlocking device, and FIG. 11 is a 70- chart showing the operation when updating the length of stay using a clock. It is a chart. 6...- Card league, 10... Unlocking device, 70''... Arithmetic circuit, 90...R
A.M.

Claims (1)

[Claims] 1. Unlocking using a card on which a secret number, a plurality of auxiliary numbers, a designated code for selecting at least one of the plurality of auxiliary numbers, and a period of stay are recorded. A method comprising: a card reader for reading data recorded on a card; a storage device for storing at least a secret number and a length of stay; and a correspondence between the secret number read from the card and the secret number stored in the storage device. comprising an arithmetic device that checks the relationship, and when the secret number of the card and the secret number of the storage device are in a certain relationship using an auxiliary number selected by a designated code from among the plurality of auxiliary numbers; At the same time as unlocking, the secret number read from the card and the period of stay are stored in the storage device,
Each time a unit period elapses, the stay period stored in the storage device is reduced by 1 per unit period, and the secret number of the card and the secret number of the storage device match 1 and are stored in the storage device. An unlocking method using a card that unlocks the door even if the length of stay exceeds a predetermined number. 2. The unlocking method using a card according to claim 1, wherein the period of stay is the number of days of stay, and the number of days of stay stored in the storage device is decreased by 1 every time one day passes. 3. An unlocking method using a card according to claim 1, which includes a clock device and reduces the length of stay stored in the storage device based on the output of the clock device. 4 The above period of stay is the number of days of stay, and the method includes periodically unlocking the door once a day using a forced unlocking card on which a secret number for forced unlocking that is different from the above card is recorded. A secret number for forced unlocking is stored in the storage device in advance, and when the secret number for forced unlocking of the card matches the secret number for forced unlocking of the storage device, the card is unlocked and the secret number for forced unlocking is stored in the storage device. An unlocking method using a card according to claim 1, which reduces the number of days of stay stored in the device by one.