The deletion method allows you to check whether a given solution to the transport problem is a reference solution.
Let an admissible solution to the transport problem, which has m+n-1 non-zero coordinates, be written in the table. For this solution to be a reference solution, the condition vectors corresponding to the positive coordinates must be linearly independent. To do this, the cells of the table occupied by the solution must be arranged so that it is impossible to form a cycle from them.
A row or column of a table with one occupied cell cannot be included in any cycle, since a cycle has two and only two cells in each row or column. Therefore, you can first cross out either all rows of the table containing one occupied cell each, or all columns containing one occupied cell each, then return to the columns (rows) and continue crossing them out. If, as a result of deleting, all rows and columns are crossed out, it means that from the occupied cells of the table it is impossible to select a part that forms a cycle, and the system of corresponding vectors-conditions is linearly independent, and the solution is a reference one. If, after deleting, some cells remain, then these cells form a cycle, the system of corresponding vectors-conditions is linearly dependent, and the solution is not a reference one.
Below are examples of “crossed out” (reference) and “non-crossed out” (non-support) solutions:
;
“crossed out” “not crossed out”
6. Methods for constructing the initial reference solution. Northwest corner method.
There are a number of methods for constructing an initial reference solution, the simplest of which is the northwest corner method. IN this method The stocks of the next supplier are used to supply the requests of the next consumers until they are completely exhausted, after which the stocks of the next supplier by number are used.
Filling out the transport task table starts from the left top corner and consists of a number of similar steps. At each step, based on the stocks of the next supplier and the requests of the next consumer, only one cell is filled and, accordingly, one supplier or consumer is excluded from consideration. This is done this way:
It is customary to enter zero shipments into the table only when they fall into cell (i,j) to be filled in. If transportation is required to be placed in the next cell of the table (i,j), and the i-th supplier or j-th consumer has zero inventories or requests, then a transportation equal to zero (basic zero) is placed in the cell, and after that, as usual, the relevant supplier or consumer is excluded from consideration. Thus, only basic zeros are entered into the table, the remaining cells with zero transportation remain empty.
To avoid errors, after constructing the initial reference solution, it is necessary to check that the number of occupied cells is equal to m+n-1 and the condition vectors corresponding to these cells are linearly independent.
Theorem4. The solution to the transport problem, constructed by the northwest corner method, is the reference one.
Proof. The number of table cells occupied by the reference solution should be equal to N=m+n-1. At each step of constructing a solution using the northwest corner method, one cell is filled in and one row (supplier) or one column (consumer) of the problem table is excluded from consideration. After m+n-2 steps, m+n-2 cells will be occupied in the table. At the same time, one row and one column will remain uncrossed, with only one unoccupied cell. When this last cell is filled, the number of occupied cells will be m+n-2+1=m+n-1.
Let us check that the vectors corresponding to the cells occupied by the reference solution are linearly independent. Let's use the deletion method. All occupied cells can be crossed out if you do this in the order in which they are filled.
It must be borne in mind that the northwest corner method does not take into account the cost of transportation, so the reference solution constructed by this method may be far from optimal.
In order for the linear programming transport problem to have a solution, it is necessary and sufficient that the total inventories of suppliers equal the total demands of consumers, i.e. the task must be with the right balance.
Theorem 38.2 Property of the system of constraints of the transport problem
The rank of the system of vectors-conditions of the transport problem is equal to N=m+n-1 (m - suppliers, n-consumers)
Reference solution to the transport problem
The reference solution of a transport problem is any feasible solution for which the condition vectors corresponding to the positive coordinates are linearly independent.
Due to the fact that the rank of the system of vectors-conditions of the transport problem is equal to m+n - 1, the reference solution cannot have more than m+n-1 non-zero coordinates. The number of non-zero coordinates of a non-degenerate reference solution equals m+n-1, and for a degenerate reference solution it is less than m+n-1
CycleCycle such a sequence of cells in the transport problem table is called (i 1 , j 1),(i 1 , j 2),(i 2 , j 2),...,(i k , j 1), in which there are two and only two adjacent cells arranged in one row or column, with the first and last cells also in the same row or column.
The cycle is depicted as a table of the transport problem in the form of a closed broken line. In a cycle, any cell is a corner cell in which a polyline link rotates 90 degrees. The simplest cycles are shown in Figure 38.1
Theorem 38.3An admissible solution to the transport problem X=(x ij) is a reference solution if and only if no cycle can be formed from the occupied cells of the table.
Cross-out method
The deletion method allows you to check whether a given solution to the transport problem is a reference solution.
Let an admissible solution to the transport problem, which has m+n-1 non-zero coordinates, be written in a table. For this solution to be a reference solution, the condition vectors corresponding to the positive coordinates, as well as the base zeros, must be linearly independent. To do this, the cells of the table occupied by the solution must be arranged so that it is impossible to form a cycle from them.
A table row or column with one occupied cell cannot be included in any cycle, since a cycle has two and only two cells in each row or column. Therefore, to first cross out either all rows of the table containing one occupied cell each, or all columns containing one occupied cell each, then return to the columns (rows) and continue crossing out.
If, as a result of deleting, all rows and columns are crossed out, it means that from the occupied cells of the table it is impossible to select a part that forms a cycle, and the system of corresponding vectors-conditions is linearly independent, and the solution is a reference one.
If, after deleting, some cells remain, then these cells form a cycle, the system of corresponding vectors-conditions is linearly dependent, and the solution is not a reference one.
Examples of “crossed out” (reference) and “not crossed out” (non-reference solutions):
Cross-out logic:
- Cross out all columns that have only one occupied cell (5 0 0), (0 9 0)
- Cross out all lines that have only one occupied cell (0 15), (2 0)
- Repeat cycle (7) (1)
Methods for constructing an initial reference solution
Northwest Angle Method
There are a number of methods for constructing an initial reference solution, the simplest of which is the northwest corner method.
In this method, the inventories of the next numbered supplier are used to supply the requests of the next numbered consumers until they are completely exhausted, after which the inventories of the next supplier number are used.
Filling out the transport task table starts from the upper left corner, which is why it is called the northwest corner method.
The method consists of a number of similar steps, at each of which, based on the stocks of the next supplier and the requests of the next consumer, only one cell is filled in and, accordingly, one supplier or one consumer is excluded from consideration.
Example 38.1Create a support solution using the northwest corner method.
1. We distribute the stocks of the 1st supplier.
If the reserves of the first supplier are greater than the requests of the first consumer, then write in cell (1,1) the amount of the request of the first consumer and move on to the second consumer. If the reserves of the first supplier are less than the requests of the first consumer, then we write in cell (1,1) the amount of the reserves of the first supplier, exclude the first supplier from consideration and move on to the second supplier.
Example: since its reserves a 1 =100 are less than the requests of the first consumer b 1 =100, then in cell (1,1) we write down transportation x 11 =100 and exclude the supplier from consideration.
We determine the remaining unsatisfied requests of the 1st consumer b 1 = 150-100 = 50.
2.We distribute the stocks of the 2nd supplier.
Since its reserves a 2 = 250 are greater than the remaining unsatisfied requests of the 1st consumer b 1 =50, then in cell (2,1) we write down transportation x 21 =50 and exclude the 1st consumer from consideration.
We determine the remaining inventories of the 2nd supplier a 2 = a 2 - b 1 = 250-50 = 200. Since the remaining inventories of the 2nd supplier are equal to the demands of the 2nd consumer, we write x 22 = 200 in cell (2,2) and exclude at our discretion either the 2nd supplier or the 2nd consumer. In our example, we excluded the 2nd supplier.
We calculate the remaining unsatisfied requests of the second consumer b 2 =b 2 -a 2 =200-200=0.
150 | 200 | 100 | 100 | ||
100 | 100 | |
|||
250 | 50 |
200 |
250-50=200 200-200=0 | ||
200 | |||||
150-100-50=0 |
3. We distribute the stocks of the 3rd supplier.
Important! In the previous step, we had the choice to exclude the supplier or the consumer. Since we excluded the supplier, the requests of the 2nd consumer still remained (albeit equal to zero).
We must write the remaining requests equal to zero in cell (3,2)
This is due to the fact that if transportation is required to be placed in the next cell of the table (i, j), and the supplier with number i or consumer with number j has zero inventories or requests, then transportation equal to zero (basic zero) is placed in the cell, and either the relevant supplier or the consumer is then excluded from consideration.
Thus, only basic zeros are entered into the table, the remaining cells with zero transportation remain empty.
To avoid errors, after constructing the initial reference solution, it is necessary to check that the number of occupied cells is equal to m+n-1 (the base zero is also considered an occupied cell), and the condition vectors corresponding to these cells are linearly independent.
Since in the previous step we excluded the second supplier from consideration, we write x 32 =0 in cell (3.2) and exclude the second consumer.
Supplier 3's inventories have not changed. In cell (3.3) we write x 33 =100 and exclude the third consumer. In cell (3,4) we write x 34 =100. Due to the fact that our task is with the right balance, the stocks of all suppliers are exhausted and the demands of all consumers are satisfied completely and simultaneously.
Reference solution | ||||
150 | 200 | 100 | 100 | |
100 | 100 | |||
250 | 50 | 200 | ||
200 | 0 | 100 | 100 |
4. We check the correctness of the construction of the reference solution.
The number of occupied cells should be equal to N=m(suppliers)+m(consumers) - 1=3+4 - 1=6.
Using the cross-out method, we make sure that the solution found is “cross-out” (the basic zero is marked with an asterisk).
Consequently, the condition vectors corresponding to the occupied cells are linearly independent and the constructed solution is indeed a reference one.
Minimum Cost Method
Method minimum cost is simple and allows you to construct a reference solution that is quite close to the optimal one, since it uses the cost matrix of the transport problem C=(c ij).
Like the northwest corner method, it consists of a number of similar steps, at each of which only one cell of the table is filled in, corresponding to the minimum cost:
and only one row (supplier) or one column (consumer) is excluded from consideration. The next cell corresponding to is filled in according to the same rules as in the northwest corner method. A supplier is excluded from consideration if its cargo inventory is fully used. The consumer is excluded from consideration if his requests are fully satisfied. At each step, either one supplier or one consumer is eliminated. Moreover, if the supplier has not yet been excluded, but its inventories are equal to zero, then at the step when this supplier is required to deliver cargo, a base zero is entered in the corresponding cell of the table and only then the supplier is excluded from consideration. Same with the consumer.
Using the minimum cost method, construct an initial reference solution to the transport problem.
1. Let’s write down the cost matrix separately to make it more convenient to choose the minimum costs.
2. Among the elements of the cost matrix, select the lowest cost C 11 =1, mark it with a circle. This cost occurs when transporting cargo from 1 supplier to 1 consumer. In the appropriate box we write down the maximum possible volume of transportation:
x 11 = min (a 1; b 1) = min (60; 40) =40 those. the minimum between the stocks of the 1st supplier and the requests of the 1st consumer.
2.1. We reduce the inventories of the 1st supplier by 40.
2.2. We exclude the 1st consumer from consideration, since his requests are fully satisfied. In matrix C we cross out the 1st column.
3. In the remaining part of the matrix C, the minimum cost is the cost C 14 =2. The maximum possible transportation that can be carried out from the 1st supplier to the 4th consumer is equal to x 14 = min (a 1 "; b 4 ) = min (20; 60) = 20, where a 1 with a prime is the remaining inventory of the first supplier.
3.1. The supplies of the 1st supplier are exhausted, so we exclude it from consideration.
3.2. We reduce the requests of the 4th consumer by 20.
4. In the remaining part of the matrix C, the minimum cost is C 24 =C 32 =3. Fill in one of the two cells of the table (2.4) or (3.2). Let's write it in a cage x 24 = min (a 2; b 4) = min (80; 40) =40 .
4.1. The 4th consumer's requests have been satisfied. We exclude it from consideration by crossing out the 4th column in matrix C.
4.2. We reduce the inventory of the 2nd supplier 80-40=40.
5. In the remaining part of the matrix C, the minimum cost is C 32 =3. Let's write transportation in cell (3,2) of the table x 32 = min (a 3; b 2) = min (100; 60) =60.
5.1. Let's exclude the 2nd consumer from consideration. We exclude the 2nd column from matrix C.
5.2. Let's reduce the inventories of the 3rd supplier 100-60=40
6. In the remaining part of the matrix C, the minimum cost is C 33 =6. Let's write transportation in cell (3,3) of the table x 33 = min (a 3 "; b 3 ) = min (40; 80) =40
6.1. Let us exclude the 3rd supplier from consideration, and the 3rd row from matrix C.
6.2. We determine the remaining requests of the 3rd consumer 80-40=40.
7. The only element left in the matrix C is C 23 =8. We write in the cell of the table (2.3) transportation X 23 =40.
8. We check the correctness of the construction of the reference solution.
The number of occupied cells in the table is N=m+n - 1=3+4 -1.
Using the deletion method, we check the linear independence of the condition vectors corresponding to the positive coordinates of the solution. The order of deletion is shown in the X matrix:
Conclusion: The solution by the minimum cost method (Table 38.3) is “crossed out” and therefore reference.
Hello Srgy!
Psht et chttl Vshy rssylk, ktru n nhdt all plzny... D t t prktk-t nt. I n smm for z dvn zntrsvn vzmzhnstyu svta skrtchtn. For me this is always a dream. I was on this tm rzgvry with brtm. n skzl slsch: sl honor chn fast,t n spvsh all nfrmcyu plntsnn brbtt. Skrst chtnya prktchsk straight prprtsnln skrst thinking. to be honest - to be honest - to be honest. Tue brtn, k szhlnyu, n actvt. Spsby sskstng vyshn skrst chtnya - this fktsya.
And here is the original
Hello Sergey!
This is written by a reader of your newsletter, which he finds very useful... But there is no practice. In fact, I have been interested in the possibility of mastering speed reading for a long time. But for some reason it always seemed like a dream to me. I had conversations on this topic with my brother. He said the following: if you read very quickly, you don’t have time to fully process all the information. The speed of reading is almost directly proportional to the speed of thinking. Increase your thinking speed and your reading speed will also increase. But the opposite, unfortunately, does not apply. Methods to artificially increase reading speed are fiction.
Even after the text has been shortened by 50% by eliminating some letters, it is still readable.
Not every word (every letter) carries an information load. Some words can be perceived as hieroglyphs.
To increase your reading speed, just start reading through the word. You might argue that at school you were taught to read carefully and think about every word. Perhaps this reading rule is still relevant and has not become obsolete, like the recommendations that when reading you must run your finger along the lines or read the text aloud (from reading textbooks of the last century).
There are developments in software implementation of the method. If anyone is interested in creating an advisor, please write.Here is a description of the method.
Money management is based on a Martingale modification - Labouchere,
also known as the “strike-out method”. This method is not as extreme as a regular martingale.
What is the principle of transaction management?At the dawn of casinos, for playing games of equal value (for example, red - black), a method of doubling the bet when losing was invented. I won’t go into detail, but this method, although mathematically certainly allows you to win, has negative features. The bets increase exponentially and sooner or later, you will either win, or face the lack of the necessary amount in your pocket for the next doubling of the bet, or with a limit maximum bet on the gaming table.
Let me remind you that the mathematical probability of winning when playing classic roulette is 49%. 1% is ZERO, this is the advantage of the casino.
The deletion method is as follows. We divide our deposit into 100 parts.
1% of the deposit is one contract.We start the game with 1 contract. We take paper and pen and write down the bets in a column one below the other.
-1
We add 1 more contract to the lost one. The next bid is 2 contracts. For example, we won. Write it down in a column
-1
+2
In total, we won 1 contract. We cross out everything and start again. The next bid is 1 contract.Let's look at a more interesting series.
For example, we lost the first bet. Write it down on paper
-1
We add 1 more contract to the lost one. The next bid is 2 contracts. For example, we lost. Write it down in a column
-1
-2
Now to the first bet in the column (-1), add the last bet (-2). Total 3 contracts. Let's say we lost. We write it down in a column.
-1
-2
-3
Now to the first bet in the column (-1), add the last bet (-3). Total 4 contracts. Let's say we lose again. Write it down in a column
-1
-2
-3
-4
Now to the first bet in the column (-1), add the last bet (-4). Total 5 contracts. Let's say we lose again. Write it down in a column
-1
-2
-3
-4
-5
Five losses in a row. It happens... The next bid is 6 Contracts.
For example, we won. We write it down in a column.
-1
-2
-3
-4
-5
+6
The 6 contracts that we won compensated for the loss of -1 and – 5 contracts! Now, cross out -1, -5 and +6.
Left:
-2
-3
-4
Now to the first bet in the column (-2), add the last bet (-4). Total 6 contracts. The next bid is 6 Contracts. Let's say we win again. Write it down in a column
-2
-3
-4
+6
The 6 contracts that we won compensated for the loss of -2 and – 4 contracts! Now, cross out -2, -4 and +6.
-3 contracts left. Since there is nothing else in the column, we add 1.
The next bid is 4 contracts. If we win, then we cross out everything, remain in the black by 1 contract and start the series again.We had such a series
-1
-2
-3
-4
-5
+6
+6
+4Three profitable trades compensated for 5 losing ones.
I advise you to practice on paper several times until the principle becomes automatic.So, pay attention! In order for the system to function and win, it is necessary to have a number of profitable transactions above 33% -40% percent!!!
If anyone is in doubt, write your own long series. You can practice at any online casino that has a test game for virtual money. Divide your deposit into 100 parts. Bet only on red or only on black. Keep in mind that such a method of play may be considered dishonest by the casino, and the casino computer will, after some time, begin to give you series of the opposite color of 10-20-30 in a row, of course, we will no longer talk about any 33-40 percent ratio and you will lose.But the principle remains UNCHANGED, 33% of winnings compensate for 66% of losses.
Thus, using such money management in practical Forex trading, we need a trading system that has a 50% probability of winning, and the ratio of possible profit to possible loss is greater than or equal to 1,
those. Profit factor >=1.
chemical properties of strokes
text and spots covering these strokes. The difference in properties makes it easier to identify flooded texts. If the properties of the substance of the spot and the strokes are close or identical, solving the problems becomes extremely difficult.
The construction of a general methodology for identifying flooded texts presents certain difficulties due to the variety of objects encountered.
During the study, the expert first of all needs to find out:
What is the nature of the material and type of writing device used to produce the document;
What is the nature of the stain material covering the text;
What technical methods are advisable to use to restore the content of the document;
In what order should they be applied?
When preparing documents, use various materials letters. They are also, as a rule, used when applying strokes and spots covering the text. These include ink, ballpoint pen pastes and felt-tip pens, stamp inks, ink, pencils, carbon papers and tapes. Their reflectivity (brightness, color) in the visible, ultraviolet and infrared zones of the spectrum is determined by their chemical composition.
The most common among writing materials are organic-based inks, which include one or more dyes, the mixture of which determines their color: black, violet, blue, green, red, etc. Black, blue, violet, red stamps are made on this basis paints. Their spectral properties are similar to the corresponding brands of ink.
Unlike ink strokes, strokes made in a document with paste, ink, or graphite pencils have a peculiar reflective ability. Their individual areas can glare under directional lighting.
How to determine the nature of dyes? The nature of coloring substances can be determined by studying the brightness and color (spectral) properties of strokes in the visible, UV, and IR ranges. (The properties of the writing material used to make the main text of the document are determined by studying the strokes that are not covered by the stain.)
For any combination of coloring substances, it is necessary, first of all, to inspect the document from different viewing angles, both in reflected light and in transmission. In cases where the eye detects differences in the optical density or color of the strokes and the spot covering them, it is possible to determine the content of the filled-in records. If it was not possible to visually solve the task of establishing the content of the text, you must use various options color discrimination photography.
The most accurate are instrumental methods. According to spectrophotometric evaluation data (spectral reflectance curves of dyes), the spectral regions where maximum differences in the brightness of the separated elements of the document are observed are found. Effective for color discrimination in many cases are visual analysis of the color properties of dyes using color atlases, the color triangle, and experimental selection of an effective illumination zone by observing an object through various brands of filters. It must be remembered that filters based on the color of the required stroke transmit rays, and absorb the opposite ones, thereby achieving the desired contrast.
Digital photography has significantly expanded the capabilities of expert research due to the high spectral sensitivity of digital sensors to various zones of the spectrum. In addition, with spectrozonal photography, as with color discrimination photography, in order to achieve maximum color contrast, it is possible to edit images by using digital computer programs such as Photoshop. For example, one of the functions of such programs is Hue/Saturation, which allows you to widely change the color gamut of an object and the saturation of shades. By shifting the color shades of the image from the original to the end of the spectral range, both in one direction and the other on the shade scale in the range from -180° to +180°, the area where color contrast differentiated dyes is maximum.
When studying texts written with black coloring substances (inks and pastes) and smeared with substances of the same color, the following methods are used.
Changing luminance contrast under special lighting conditions. If the difference in brightness is due to an unequal type of reflection (directional, diffuse), for example, when the coloring matter of the strokes of a filled text has a specific shine compared to the material of the masking spot, or vice versa, shooting with bright-field lighting is used. Shooting in these lighting conditions gives good results upon identifying records executed graphite pencil and covered in black ink. The direction angle of the illuminator rays is selected experimentally.
Differences in densities in areas of the document where the strokes are covered by a stain and areas under the stain, free from strokes, are established by shooting in transmitted rays. To reduce the density of the paper, it is recommended to moisten it with pure gasoline before shooting. Shooting can be done with or without filters. This way you can identify records covered in blood, etc.
In cases where the writing materials used to produce a document are opaque to IR rays - graphite pencil, black ink, black ink on copy paper, black printing ink, as well as dyes with salts of heavy metals - iron, chromium, copper, etc. - are transparent - aniline dyes, and the dye of the masking spot is transparent, use the method of research in reflected IR rays, for example, the VC-30 device, in which only the strokes of the recordings will be observed. Texts written in graphite pencil and crossed out (smeared over) with colored pencils (not ink) can also be detected in reflected IR rays.
IR luminescence photography is one of the effective methods used to identify records filled with a substance similar to the coloring substance of the text. If positive results were not achieved when studying the front side of the document, it is advisable to conduct a study of its back side. Positive results are obtained when shooting IR luminescence on the back of a document if the recordings are made with ink containing methylene blue and brilliant green dyes, since they have high penetrating ability in the paper environment.
Techniques for changing luminance contrast known in the digital photographic process - enhancing, weakening, contrast equalizing, adding and subtracting images (photographic masking), filtering details - can also lead to a positive result. They have proven to be very effective for this purpose in the digital photographic process. software changes in brightness and contrast.
These include graphic editors such as Adobe Photoshop Brightness/Contrast should be included - the simplest means and the least accurate; Levels - a more complex tool that includes several ways to control tonality and gives good results; Curves - capable of changing brightness certain levels without affecting the others.
The diffuse-copying method (DCM) is used to identify records made with writing materials containing organic dyes that are discolored by the action of an alkaline solution of sodium hydrosulfite and some insoluble dyes (for example, ballpoint pen paste) filled with water-insoluble substances (for example, ink) and not discolored in the specified solution.
To establish the effectiveness of DCM, it is recommended to carry out a preliminary analysis (test), the essence of which is to copy a small area of text strokes and spots onto moistened photographic paper, followed by treating it with an alkaline solution of sodium hydrosulfite. If only text strokes are discolored, all recommended operations can be carried out in full (copying over the entire filled area, highlighting, developing, etc.).
Wet copying is used when the dye of the detected records has greater copying power compared to the dye of the spot. For copying, use fixed photographic paper or polyvinyl chloride film, moistened, respectively, with distilled water (sometimes acidified with acetic acid) or an organic solvent. If the coloring matter of the spot is copied better than the coloring matter of the text, then with repeated copying you can gradually remove part of the dye of the spot. As a result of these actions, the text becomes visible. If the substance of the streaks is insoluble in water, then use a PVC film, which is moistened with organic solvents (dimethylformamide, dichlorhexane, benzene, chlorobenzene, acetone, alcohol).
Solvents that more vigorously dissolve the coloring matter of the strokes of the text being revealed are preferable. To do this, using drop reactions, a solvent can be selected directly for the substance of the streaks (outside the spot) and the substance of the spot. PVC film (or fixed photographic paper) is moistened with this solvent. Excess solvent is removed with filter paper, and then the film is applied to the area of the document being examined.
Often copied strokes are barely visible. If they cannot be identified by further photography in order to enhance the contrast, then the print is studied in filtered UV rays. This may reveal a difference in the luminescence of the text strokes and the spot. Positive results are often obtained when studying a copy using infrared luminescence. In this modification, the wet copying method is called adsorption-luminescence.
If you have a computer, you can use a flatbed scanner to display the image of faintly visible copied strokes on the monitor screen and, using the Photoshop program, enhance the contrast and make them clearly readable on the screen.
Mechanical removal of stain substance. If the stain is formed by large particles of a substance, identification of records is possible using mechanical action on the substance of the stain, for example, plasticine, rubber, or an elastic band. Lightening stains formed by insoluble substances can be done with fixed photographic paper.
In particular, if the text is crossed out with a graphite pencil, then the use of rubber gives good results. The rubber should be slightly moistened beforehand. Then, after countertyping, the layer with adhering graphite is cut off from the rubber. Copying continues until the identified text becomes visible.
It is advisable to wash off if the dye of the detected strokes does not dissolve in water or organic solvents, or is less soluble than the dye of the spot.
The list of methods used to identify flooded texts does not end there. Currently, a number of so-called private methods have been proposed that are effective for solving individual issues.
Identification of crossed out entries. Crossed-out entries can be identified using methods recommended for studying flooded and smeared texts. This mainly applies to the situation when the text strokes are completely invisible from under the strikethrough strokes. If the crossing out is done incompletely or with a different color dye, it is possible to use other additional techniques.
The method of photographic exclusion (subtractive masking) is as follows. First, a photograph is taken of the document with the entries crossed out. natural light. Then, with the same position of the object and the camera, color separation photography is carried out in order to obtain an image in which the detected recordings would be eliminated or significantly weakened. When photographing, the following can be changed: the direction of illumination of the object, the spectral composition of the light or shutter speed. The resulting image is inverted in a computer graphics editor, transforming from positive to negative. The converted image is combined with the image obtained in the visible part of the spectrum. The combination is carried out in the Adobe Photoshop graphic editor, using various image blending modes, which are set in the palette dialog box. Of all the possible image overlay modes (multiplication, lightening, addition, subtraction, difference, etc.), the H (normal) mode is suitable for photographic masking, which leads to a complete replacement of the brightness values of the background image with the brightness of the overlay.
When combining two positive images, use the blending mode P (difference), the effect of which is to subtract one brightness value from another and then store the absolute value in the total channel, or I (exclusion).
In case of failure, it is advisable to use the method of deciphering crossed out entries. Decoding is carried out on enlarged digital photographs. One of the technical methods of decoding is retouching the strikeout lines. Retouching is performed in a digital graphics editor.