Japanese Mathematics Explained: Soroban, Anzan, Wasan and Sangaku

Japanese mathematics has two living halves. The soroban (算盤), a 1 plus 4 bead abacus, and anzan (暗算), calculation on an imagined soroban, are still taught and graded nationally today. Wasan (和算), the native mathematics of the Edo period, gave the world Seki Takakazu's determinants in 1683 and sangaku (算額), geometry problems hung in temples and shrines.

This post works through it by hand. Every problem in our fixed problem set is done on the soroban with the bead state recorded after each step, and every number was checked with Python. Then we follow Seki and his pupil Takebe Katahiro into determinants, the Bernoulli numbers and enri, solve two published temple problems, and explain why wasan faded after the Meiji Restoration.

Key takeaways

• The modern soroban has one bead worth 5 and four beads worth 1 on each rod, so every digit from 0 to 9 has exactly one bead pattern.
• All soroban arithmetic reduces to four moves: direct, five complement, ten complement and a mixed move.
• Anzan runs the same moves on an imagined board. The League for Soroban Education of Japan (Zenshuren) grades soroban from 15th kyu to 10th dan, and its championship scores the individual contest out of 1,500.
• Seki Takakazu's 1683 work is the earliest known written treatment of determinants, about ten years before Leibniz's 1693 letter; historians treat the two as independent. Seki also found the Bernoulli numbers independently of Jacob Bernoulli.
• Enri, the circle method, used doubling polygons, an acceleration trick now called Aitken's method, and Takebe Katahiro's infinite series.
• Wasan declined after the 1872 Education Order made Western mathematics the school subject. The soroban survived; wasan research did not.

The problem set for this series

This is the Japanese post in a series on mathematical traditions. The Indian, Chinese and comparison posts work the same fixed problems:

The soroban diagrams were not typed by hand. A small simulator applied each bead move, named its rule and drew the board, and checked every final value against ordinary arithmetic.

A short timeline of Japanese mathematics

How does the soroban work?

The 1 plus 4 layout

A soroban is a frame of vertical rods crossed by a horizontal beam (梁, hari). Each rod has one upper bead, called the heaven bead or godama (五珠), worth 5, and four lower beads, called earth beads or ichidama (一珠), each worth 1. A bead counts when it touches the beam. So the heaven bead counts when it is pushed down, and earth beads count when they are pushed up.

Each rod holds one digit: 5 if the heaven bead is down, plus the number of earth beads up. With only four earth beads, each digit has exactly one pattern. A twentieth century soroban in the Whipple Museum at Cambridge shows the standard form: 15 rods, one bead above the divider and four below on each. Older boards carried more beads, closer to the Chinese suanpan.

Here is every digit on one board. O is a bead and | is bare rod. A heaven bead in the top row counts nothing; in the second row it is down against the beam (=) and counts 5. Below the beam, beads above the gap touch the beam and count 1 each.

Rods A to J set to 0, 1, 2 ... 9
rod   A B C D E F G H I J
      O O O O O | | | | |
      | | | | | O O O O O
      ===================
      | O O O O | O O O O
      O | O O O O | O O O
      O O | O O O O | O O
      O O O | O O O O | O
      O O O O | O O O O |
value 0 1 2 3 4 5 6 7 8 9

Rod H has the heaven bead down (5) and two earth beads up (2), so it shows 7. Rod A has nothing touching the beam, so 0.

Most boards mark every third rod with a dot, the unit points (定位点, teiiten), so you can place the ones digit and read thousands at a glance. Here we name rods with letters and always say which rod holds the ones digit.

Finger technique

The usual Japanese teaching uses two fingers of the right hand. The thumb pushes earth beads up, which adds. The index finger pushes earth beads down, which subtracts, and moves the heaven bead both ways. Adding 6 to an empty rod is one pinch: the index finger brings the heaven bead down while the thumb lifts one earth bead.

To clear the board, tilt it so the beads slide down, then run the index finger along the beam to push every heaven bead up.

The four moves: direct, five complement, ten complement and mixed

Adding a digit to a rod is always one of four moves. The same is true of subtraction, in reverse.

The pairs to memorise are the complements of 5 (1 and 4, 2 and 3) and of 10 (1 and 9, 2 and 8, 3 and 7, 4 and 6, 5 and 5). The board does the carrying.

Here is a short run that uses three of the moves: start with 4, then add 3, 8 and 6. In this and every later step table, each bead move names the rod, the change and the rule: direct, (five) for a five complement, (ten) for a ten complement and (mixed) for the combination. The last column reads the rods left to right as one string of digits.

Start: 4 on rod C
rod   A B C
      O O O
      | | |
      =====
      | | O
      O O O
      O O O
      O O O
      O O |
value 0 0 4

Final: 4 + 3 + 8 + 6 = 21
rod   A B C
      O O O
      | | |
      =====
      | O O
      O O |
      O | O
      O O O
      O O O
value 0 2 1

The soroban never "knows" a sum. Adding 8 to 7 is not recalled as 15; it is performed as "take 2 here, add 1 there". That is the core idea of the instrument.

Working the problem set on the soroban

How soroban multiplication is laid out

In a common layout, the two factors sit on the left of the board and the product is built on rods to the right. Each digit of the multiplicand is multiplied by each digit of the multiplier, left to right, and the two digit partial product (2 x 3 is written "06") is added into the product rods at a position fixed by the two place values.

The diagrams show only the product rods, with the product's ones digit on the last rod. The "Rods" column names the two rods that receive each partial product.

98 x 97

Four partial products: 9 x 9, 9 x 7, 8 x 9, 8 x 7. Product rods A to G, ones on G.

After step 2
rod   A B C D E F G
      O O O | | O O
      | | | O O | |
      =============
      | | | O O O |
      O O O O O O O
      O O O O | O O
      O O O | O | O
      O O O O O O O
value 0 0 0 8 7 3 0

Final: 98 x 97 = 9506
rod   A B C D E F G
      O O O | | O |
      | | | O O | O
      =============
      | | | O | | O
      O O O O O O |
      O O O O O O O
      O O O O O O O
      O O O | O O O
value 0 0 0 9 5 0 6

Step 3 is where beginners stall. Rod E shows 7 and we must add 7. The ten complement says "take 3 here, add 1 to the left", but a rod showing 7 has the heaven bead and only two earth beads, so you cannot remove 3 earth beads. The mixed move handles it: add 2 earth beads, lift the heaven bead away (-5), and carry 1 to rod D. The net change on rod E is +2 - 5 = -3, which together with the carry is +7. Step 4 then shows a carry that itself needs a five complement: rod E holds 4 (four earth beads up), and adding 1 means bringing the heaven bead down and dropping all four earth beads.

Count the work: four partial products, ten single rod moves, and never more than one two digit product held in mind. The board holds the running total.

47 x 53

Product rods A to G, ones on G.

Final: 47 x 53 = 2491
rod   A B C D E F G
      O O O O O | O
      | | | | | O |
      =============
      | | | O O O O
      O O O O O O |
      O O O | O O O
      O O O O O O O
      O O O O | | O
value 0 0 0 2 4 9 1

Every move is direct, which makes this the easiest product in the set on the soroban. Other traditions use the difference of squares, 50² - 3² = 2,491; a soroban user does not need the trick.

386 x 524

Nine partial products. Product rods A to H, ones on H. This is the problem where the soroban starts to pay off, because the carries take care of themselves.

After step 6
rod   A B C D E F G H
      O O O | | O O O
      | | | O O | | |
      ===============
      | | O O O O O |
      O O | O O | O O
      O O O O O O | O
      O O O O O O O O
      O O O | | O O O
value 0 0 1 9 9 1 2 0

Final: 386 x 524 = 202264
rod   A B C D E F G H
      O O O O O O | O
      | | | | | | O |
      ===============
      | | O | O O O O
      O O O O O O | O
      O O | O | | O O
      O O O O O O O O
      O O O O O O O |
value 0 0 2 0 2 2 6 4

Step 7 is the interesting one. Adding 30 lands a 3 on rod E, which holds 9. The ten complement takes 7 from E and carries 1 left to D, but D also holds 9, so it becomes 0 and carries again into C. On paper this is where people lose a carry. On the soroban the carry is a bead you move, finished before the next partial product.

85²

Squaring is multiplication with equal factors: 85 x 85. Product rods A to G, ones on G.

Final: 85 x 85 = 7225
rod   A B C D E F G
      O O O | O O |
      | | | O | | O
      =============
      | | | O O O |
      O O O O O O O
      O O O | | | O
      O O O O O O O
      O O O O O O O
value 0 0 0 7 2 2 5

The Vedic shortcut for squares ending in 5 is faster: 8 x 9 = 72, then 25, giving 7,225. On the soroban, two partial products (8 x 5 and 5 x 8) are the same 40, so an experienced operator adds 80 once.

7,429 ÷ 37

Soroban division (商除法, the method of placing quotients) works like long division, but you subtract partial products directly from the dividend on the board instead of writing them underneath.

The dividend 7,429 goes on rods F to I, ones on I. The divisor 37 is set on the far left of the board, outside the diagrams. The placement rule for each quotient digit: compare the divisor with the leading digits of the dividend. If the divisor is equal to or smaller than those digits (37 against 74), the quotient digit goes two rods to the left of the first dividend digit. If the divisor is larger, it goes one rod to the left. Here 37 fits into 74 twice, so 2 goes on rod D.

Start: dividend 7429 on rods F to I (ones on I)
rod   A B C D E F G H I
      O O O O O | O O |
      | | | | | O | | O
      =================
      | | | | | O O O O
      O O O O O O O O O
      O O O O O | O | O
      O O O O O O O O O
      O O O O O O | O |
value 0 0 0 0 0 7 4 2 9

Final: quotient 200 on D to F, remainder 29 on H to I
rod   A B C D E F G H I
      O O O O O O O O |
      | | | | | | | | O
      =================
      | | | O | | | O O
      O O O O O O O O O
      O O O | O O O | O
      O O O O O O O O O
      O O O O O O O O |
value 0 0 0 2 0 0 0 2 9

After subtracting 2 x 37 = 74 from the leading digits, the board holds 29 on rods H and I. That is smaller than 37, so the remaining quotient digits are 0 and the division stops. Because the dividend's ones digit was on rod I and the divisor has two digits, the quotient's ones digit is three rods to the left, on rod F: rods D, E and F read 2, 0, 0, which is 200. The remainder 29 sits on H and I.

Check: 37 x 200 + 29 = 7,400 + 29 = 7,429.

√7,744

Square roots (開平, kaihei) are part of soroban grading. Zenshuren's dan level exam lets candidates choose root extraction (開法, kaihō) as one of their optional events. The method below is one standard soroban procedure, close to pen and paper long square root.

Group the digits in pairs from the ones digit: 77 | 44. The root has one digit per pair, so two digits. The number goes on rods E to H, ones on H. The root is built on rods B and C.

Start: 7744 on rods E to H, grouped 77 | 44
rod   A B C D E F G H
      O O O O | | O O
      | | | | O O | |
      ===============
      | | | | O O O O
      O O O O O O O O
      O O O O | | O O
      O O O O O O O O
      O O O O O O | |
value 0 0 0 0 7 7 4 4

Root digit 8 on B; 77 - 64 leaves 1344 on E to H
rod   A B C D E F G H
      O | O O O O O O
      | O | | | | | |
      ===============
      | O | | O O O O
      O O O O | O O O
      O O O O O O O O
      O | O O O | O O
      O O O O O O | |
value 0 8 0 0 1 3 4 4

Final: root 88 on B and C, remainder 0
rod   A B C D E F G H
      O | | O O O O O
      | O O | | | | |
      ===============
      | O O | | | | |
      O O O O O O O O
      O O O O O O O O
      O | | O O O O O
      O O O O O O O O
value 0 8 8 0 0 0 0 0

Step 1: the largest square not above 77 is 64, so the first root digit is 8. Subtract 64 from the 77 on rods E and F, which leaves 13, so rods E to H now read 1, 3, 4, 4.

Step 2: double the root so far, 2 x 8 = 16, and use it as a trial divisor on 134. 134 ÷ 16 gives 8, so try 8 as the second digit and subtract 16 x 8 = 128 with its ones digit on rod G (it is worth 1,280). Rod G holds 4 and must lose 8, so it borrows 1 from F and adds the complement 2, which needs a five complement.

Step 3: subtract the square of the new digit, 8 x 8 = 64, on rods G and H. The remainder is 0, so 7,744 is a perfect square and its root is 88, read from rods B and C.

Check: 88 x 88 = 7,744.

1/19 as a decimal

Division by 19 never ends; you keep going until you have the places you need. Each step: read the remainder one rod further right (times 10), find how many 19s fit, set that digit and subtract.

After three quotient digits: 0.052 on A to D, remainder 12 on H and I
rod   A B C D E F G H I
      O O | O O O O O O
      | | O | | | | | |
      =================
      | | | O | | | O O
      O O O O O O O | O
      O O O | O O O O |
      O O O O O O O O O
      O O O O O O O O O
value 0 0 5 2 0 0 0 1 2

Eighteen quotient digits on C to T (decimal point between B and C), remainder back to 1
rod   A B C D E F G H I J K L M N O P Q R S T
      O O O | O | O O | | | | O | O | | O O O
      | | | O | O | | O O O O | O | O O | | |
      =======================================
      | | | | O O O O | O O O O O O O O O O O
      O O O O O | O | O O O O O O O | O O O |
      O O O O | O O O O | O O O | O O O O | O
      O O O O O O | O O O | O O O | O | O O O
      O O O O O O O O O O O | | O O O O | O O
value 0 0 0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1

After the 18th digit the remainder is 1 again, so 052631578947368421 repeats forever. The period is 18 because 10 is a primitive root modulo 19; the remainder column visits each of 1 to 18 exactly once.

On paper you write 18 subtraction rows; on the board you overwrite the same few rods 18 times, so a slip leaves no record. The defence is to re-check, which we cover after the linear system.

The linear system, Seki's way

The soroban has no algebra, so the system x + y + z = 6, 2y + 5z = -4, 2x + 5y - z = 27 is solved with an algebraic method and the board does the arithmetic. The method we use is Seki Takakazu's own: determinants. The Dictionary of Scientific Biography notes that for three equations Seki's rule was basically the same as the rule now named after Sarrus: three products along the "down" diagonals are added and three products along the "up" diagonals are subtracted.

Write the coefficients in a table, as Seki's school did on the counting board:

The determinant D of the coefficients. The three added products are 1 x 2 x (-1) = -2, 1 x 5 x 2 = 10 and 1 x 0 x 5 = 0. The three subtracted products are 1 x 2 x 2 = 4, 1 x 5 x 5 = 25 and 1 x 0 x (-1) = 0. So D = (-2 + 10 + 0) - (4 + 25 + 0) = 8 - 29 = -21.

A soroban cannot show a negative number, so keep a plus pile on the left and a minus pile on the right. A negative product joins the opposite pile, so the plus pile is 10 and the minus pile is 4 + 25 + 2 = 31.

Determinant D: plus pile 10 on A to D, minus pile 31 on F to I
rod   A B C D E F G H I
      O O O O O O O O O
      | | | | | | | | |
      =================
      | | O | | | | O O
      O O | O O O O O |
      O O O O O O O O O
      O O O O O O O | O
      O O O O O O O O O
value 0 0 1 0 0 0 0 3 1

At the end you subtract the smaller pile from the larger and remember the sign:

31 - 10 = 21 on H and I; the minus pile was larger, so D = -21
rod   A B C D E F G H I
      O O O O O O O O O
      | | | | | | | | |
      =================
      | | | | | | | O O
      O O O O O O O O |
      O O O O O O O | O
      O O O O O O O O O
      O O O O O O O O O
value 0 0 0 0 0 0 0 2 1

The numerator determinants. Replace one column at a time with the constants. For x the rows become (6, 1, 1), (-4, 2, 5), (27, 5, -1); the added products are -12, 135 and -20 and the subtracted ones 54, 150 and 4, so the plus pile is 135 and the minus pile 54 + 150 + 4 + 12 + 20 = 240.

Dx: plus pile 135 on B to D, minus pile 240 on G to I
rod   A B C D E F G H I
      O O O | O O O O O
      | | | O | | | | |
      =================
      | O O | | | O O |
      O | O O O O O O O
      O O O O O O | O O
      O O | O O O O O O
      O O O O O O O | O
value 0 1 3 5 0 0 2 4 0

So Dx = 135 - 240 = -105. In the same way Dy = 72 - 135 = -63 (plus pile 4 + 60 + 8, minus pile 135) and Dz = 74 - 32 = 42 (plus pile 54 + 20, minus pile 24 + 8).

The answer. x = Dx / D = -105 / -21 = 5, y = Dy / D = -63 / -21 = 3, z = Dz / D = 42 / -21 = -2.

Check in the original equations: 5 + 3 - 2 = 6; 2 x 3 + 5 x (-2) = 6 - 10 = -4; 2 x 5 + 5 x 3 - (-2) = 10 + 15 + 2 = 27. All three hold.

For a 3 x 3 system this costs 24 triple products, more work than elimination. Seki's interest was not speed on small systems; his determinants were a general tool for eliminating unknowns.

Checking answers: casting out nines (九去法)

Soroban students are taught to verify answers, and the oldest cheap check is casting out nines (九去法, kyūkyohō). Replace each number by its remainder after division by 9, which is the repeated sum of its digits. If the arithmetic was right, the same operation on the remainders gives the same remainder as the answer.

The check is fast but not complete. It cannot see two swapped digits (9,506 and 9,056 have the same digit sum) or an error that is a multiple of 9. It is a checksum: it catches most slips for almost no cost.

What is anzan, and how is it graded?

Calculating on an imagined board

Anzan (暗算) literally means "dark calculation", calculation without a visible tool. In the soroban tradition it means something specific: the operator pictures a soroban and moves imagined beads with the same rules shown above. Many learners move their fingers in the air while doing it.

Here is what a flash anzan item looks like. Fifteen three digit numbers flash on a screen one after another and the operator keeps a running total on the mental board. Our example sequence:

The operator never says 7,467 + 237 in words. They picture the board at 7,467, add 2 to the hundreds rod (a five complement), 3 to the tens rod, then 7 to the ones rod, whose carry ripples through the tens into the hundreds, and read 7,704. At the end the picture is:

The mental image at the end: 8358
rod   A B C D E
      O | O | |
      | O | O O
      =========
      | O O | O
      O O O O O
      O O O O O
      O | | O |
      O O O O O
value 0 8 3 5 8

What the research says, and what it does not

Peer reviewed work supports a narrow, useful claim. In a three year randomised controlled trial with 204 elementary school students at a school in Vadodara, India, Barner and colleagues (Child Development, 2016) found that children taught mental abacus improved on arithmetic tasks relative to controls. They did not find that the training changed basic cognitive abilities such as working memory. Instead, children who started with better spatial working memory learned the technique better.

The limits matter. That is one trial, in one school, with one curriculum, and its gains were in arithmetic. It does not support claims that the abacus raises intelligence or general exam performance, and we make none. Older work cited by the same paper, such as Hatano, Miyake and Binks (1977), showed that experts can keep calculating while speaking. That is evidence about experts, not a promise for every learner.

Grades and exams in Japan

Two national bodies run most soroban grading in Japan, and their grades are not interchangeable. The League for Soroban Education of Japan (全国珠算教育連盟, Zenshuren), based in Kyoto, runs soroban, anzan and flash anzan exams. The League of Japan Abacus Associations (日本珠算連盟, Nisshuren) works with chambers of commerce on their abacus exams; its FAQ notes that only a few chambers run the anzan and dan exams. Check which body's grade a school or employer means.

Zenshuren's soroban exam, from its published exam page:

Mitorizan (見取算) is a column of numbers to add and subtract, the soroban equivalent of totalling a ledger. Zenshuren states that its grades from jun 6th kyu up are backed by Japan's education ministry.

Its flash anzan exam has 20 kyu grades (20th up to 1st) and 10 dan grades (shodan to 10th dan). Each question is worth 10 points; 100 points passes, shodan is set at 100, every further 20 points is one dan higher, and 10th dan needs 280 or more.

The All Japan Soroban Championship

Zenshuren's 2026 All Japan Soroban Championship (全日本珠算選手権大会) was set for 8 August 2026 at the Kyoto International Conference Center, with about 600 places. Entry requires a Zenshuren dan grade in both soroban and anzan, or a recommendation from a Zenshuren branch. The published rules:

The individual winner takes the title "soroban Japan number one" (そろばん日本一), and the rules list a Minister of Education cup as planned.

What is wasan?

Wasan (和算) is the mathematics practised in Japan during the Edo period (1603 to 1868). It grew from Chinese texts, notably Zhu Shijie's Suanxue qimeng of 1299 with its counting board algebra, into a culture of schools, secret teachings and published problem books.

Seki Takakazu and determinants before Leibniz

Seki Takakazu (関孝和) was born around 1642 and died on 5 December 1708 in Edo, after a career as an examiner of accounts and later a Shogunate samurai. MacTutor calls him "the first person to study determinants".

In 1683 Seki wrote Kaifukudai no Hō (解伏題之法, roughly "Method of solving the dissimulated problems"). According to MacTutor's history of matrices and determinants, it laid out coefficient tables in the Chinese counting board style and, without any word for "determinant", gave general methods for computing them for 2 x 2, 3 x 3, 4 x 4 and 5 x 5 arrays. Seki used them to eliminate unknowns from equations, not to solve systems of linear equations the way we did above.

The first European appearance was ten years later. In 1693 Leibniz wrote to de l'Hôpital with the condition, in modern terms, that a coefficient array has determinant zero. MacTutor's Seki biography says Leibniz worked independently. The European theory then developed through Cramer, Bezout, Vandermonde and Laplace in the eighteenth century, while Seki's methods stayed inside Japanese schools.

A fair way to state the priority: Seki wrote down determinants first, and MacTutor's biography adds that Seki's version was the more general. It does not mean European mathematics learned from Seki; it did not.

Seki and the Bernoulli numbers

Seki also studied sums of powers, 1^p + 2^p + ... + n^p. The Dictionary of Scientific Biography describes his method for these sums and notes that the coefficients 1/6, 1/30 and 1/42 that appear in them are the Bernoulli numbers. His results appeared in Katsuyō Sanpō (括要算法), published after his death. Most sources, including the summary on MacTutor's Seki page that "he also discovered Bernoulli numbers before Jacob Bernoulli", rest on a publication date of 1712; the DSB's bibliography gives 1709. Jacob Bernoulli's Ars Conjectandi, which introduced the numbers in Europe, was published in Basel in 1713, eight years after his death. Either way, both men's results were published posthumously within a few years of each other, and independently.

The formulas for sums of powers, worked for n = 10 and checked by direct addition:

The last coefficients, 1/6, -1/30 and 1/42, are the Bernoulli numbers B2, B4 and B6 (modern signs). Knowing them, you can write the sum for any power without adding anything up.

Fractions in the table are written as mixed numbers so the arithmetic stays visible: 1,000/3 is 333 1/3, and 10/6 is 1 2/3. In the sixth power line the fractional parts 3/7, -2/3 and 5/21 add up to exactly 0, which is why the total is a whole number.

Enri: the circle principle

Enri (円理) is the wasan study of circles: circumference, arcs and the sphere.

Seki's doubling polygons. According to the Dictionary of Scientific Biography, Seki inscribed regular polygons of 4, 8, 16 sides and so on, up to 2^17 = 131,072 sides, in a circle of diameter 1. Their perimeters approach π from below, and each follows from the last using only square roots:

Even with 131,072 sides the perimeter is correct only to nine decimal places. Seki combined the last three. Each doubling cuts the error by about a factor of four, so you can extrapolate to the limit. In modern notation:

s = c16 + (c16 - c15)(c17 - c16) / ((c16 - c15) - (c17 - c16))

which is exactly what is now called Aitken's delta squared process, published by Alexander Aitken in 1926. The DSB reports Seki's result as s = 3.14159265359, "accurate except for the last figure". With correctly computed perimeters, the same formula gives 3.141592653589793238, matching π to 18 decimal places, because the error of the polygon perimeters is so regular.

One honest note. The DSB article, as reproduced on MacTutor, prints Seki's three perimeters. Its value for 2^15 sides agrees with our recomputation to 17 decimal places, but its values for 2^16 and 2^17 sides differ from the true perimeters in the 11th and 14th decimal places. We cannot tell from the reproduction whether that is a typesetting slip or Seki's own arithmetic, so we quote the perimeters from our recomputation, and Seki's final value from the DSB.

Takebe's series. Seki's pupil Takebe Katahiro (建部賢弘, 1664 to 1739) went further. In Tetsujutsu Sankei (1722) he found an infinite series for the length of a circular arc s in terms of its sagitta k, the height from the middle of the chord to the arc, for a circle of diameter 1:

(s/2)² = k + (1/3)k² + (8/45)k³ + (4/35)k⁴ + (128/1575)k⁵ + (128/2079)k⁶ + ...

MacTutor writes the coefficients as Takebe built them, as running products: 1/3, then (1 x 8)/(3 x 15), then (1 x 8 x 9)/(3 x 15 x 14), and so on. In modern terms the right hand side is the Taylor series of the square of arcsin(√k). MacTutor reports that most historians now credit the method of enri to Takebe rather than Seki, although it was long attributed to the teacher.

Worked example: take a semicircle. Its sagitta equals the radius, k = 1/2, and its arc is half the circumference, s = π/2. So the series should add up to (π/4)², which is 0.61685...

Convergence is slow at k = 1/2. MacTutor describes Takebe's acceleration technique as essentially equivalent to the Romberg algorithm, and says it gave him π correct to 40 decimal places.

What are sangaku?

Sangaku (算額) are wooden tablets with geometry problems painted on them, hung as offerings in Buddhist temples and Shinto shrines during the Edo period. People from many walks of life made them, including samurai, farmers and merchants. A tablet usually gives the figure, the problem, the answer and sometimes a short method, often without proof. The standard English sources are Hidetoshi Fukagawa and Daniel Pedoe's Japanese Temple Geometry Problems (1989) and Fukagawa and Tony Rothman's Sacred Mathematics: Japanese Temple Geometry (Princeton University Press, 2008). We take both problems below from the 1989 collection as catalogued, with solutions, in J. Marshall Unger's A Collection of Sangaku Problems (Ohio State University, 2009), and we checked each solution numerically.

Sangaku 1: three circles on a line (Gunma, 1824)

The problem. Two circles of radius r1 and r2 rest on a straight line and touch each other. A third, smaller circle of radius r3 rests on the same line and touches both. How is r3 related to r1 and r2?

Fukagawa and Pedoe list it as problem 1.1.1, "well known", from a tablet in Gunma prefecture dated 1824. Sources disagree on the tablet. Cut the Knot cites the same figure as Fukagawa and Pedoe's problem 1.1.2, from a Miyagi tablet dated 1892, and suspects a misprint because the book points to a solution published in 1810. We follow Unger's citation, problem 1.1.1, Gunma, 1824.

The solution, step by step.

1. Two touching circles of radii R and r on the line have centres R + r apart, with a height difference of R - r.
2. By Pythagoras, the distance between their touching points on the line is √((R + r)² - (R - r)²) = 2√(Rr).
3. Apply this to all three pairs. Call the touching points on the line A (circle 1), C (circle 3) and B (circle 2). Then AB = 2√(r1 r2), AC = 2√(r1 r3) and CB = 2√(r2 r3).
4. Since C lies between A and B, AC + CB = AB: 2√(r1 r3) + 2√(r2 r3) = 2√(r1 r2).
5. Divide both sides by 2√(r1 r2 r3). The result is the tablet's answer:

1/√r3 = 1/√r1 + 1/√r2

Worked numbers. Let r1 = 36 and r2 = 9. Then 1/√r3 = 1/6 + 1/3 = 1/2, so √r3 = 2 and r3 = 4.

Check with coordinates. The centres sit at (0, 36), (24, 4) and (36, 9). Their distances are √1,600 = 40 = 36 + 4, √169 = 13 = 9 + 4 and √2,025 = 45 = 36 + 9, so every pair touches.

The relation is a special case of Descartes' circle theorem, with the straight line treated as a circle of zero curvature.

Sangaku 2: a circle in a cut square (Miyagi, 1877)

The problem. A square of side a has one diagonal drawn. A line runs from a third corner to the midpoint of the opposite side. A circle is inscribed in the triangle that sits opposite that midpoint. What is its radius?

Fukagawa and Pedoe give it as problem 3.1.3, from a surviving tablet in Miyagi prefecture dated 1877, and the tablet's answer is r = 2a / (3 + √5 + √8). The solution below follows the one Unger credits to Alexander Bogomolny, set out in coordinates.

The solution, step by step.

1. Put the square at A = (0, 0), B = (a, 0), C = (a, a), D = (0, a), with diagonal AC and a line from B to M = (a/2, a), the midpoint of DC.
2. Along AC, y = x. The point a fraction t of the way from B to M is (a - ta/2, ta). Setting y = x gives t = 2/3, so the lines cross at P = (2a/3, 2a/3).
3. The triangle opposite M is ABP. Its base AB = a and its height is the y coordinate of P, 2a/3. Its area is (1/2) x a x (2a/3) = a²/3.
4. Its other sides: AP is two thirds of the diagonal, (2/3) x a√2 = a√8/3. BP is two thirds of BM, and BM = a√5/2, so BP = a√5/3.
5. For any triangle, area = inradius x semiperimeter, or 2 x area = r x perimeter. The perimeter is a + a√8/3 + a√5/3 = a(3 + √8 + √5)/3.
6. So 2a²/3 = r x a(3 + √5 + √8)/3, which gives

r = 2a / (3 + √5 + √8)

matching the tablet.

Worked numbers. For a = 10, r = 20 / (3 + 2.23607 + 2.82843) = 20 / 8.06450 = 2.4800. As an independent check, the incentre of triangle ABP for a = 1 is at (0.59873, 0.24800), and its distance to each side is 0.24800.

The pattern is typical of sangaku: a simple figure, one hidden fact (the crossing point divides both lines 2 to 1, because it is the centroid of a larger triangle), then standard area formulas.

Why did wasan decline after the Meiji Restoration?

Wasan was strong right up to the end of the Edo period. Marion Cousin's study of Meiji era geometry textbooks (Revue d'Histoire des Mathématiques, 2017) gives the turning point clearly: in 1872 the Education Order, the Gakusei (学制), planned the abandonment of wasan and the exclusive teaching of Western mathematics (yōsan, 洋算).

A new state school system. The Gakusei set up a centralised school system on the French model, replacing the patchwork of independent schools of the Edo period, and class teaching replaced the individual tutoring typical of wasan schools. Wasan's institutions, its private schools and licences, had no place in the new structure.

Western mathematics as the language of modernisation. The Meiji government's aim was rapid modernisation, and engineering, the military, navigation and the sciences arrived with Western notation, proofs and calculus. Cousin notes that the first Japanese treatises on yōsan appeared in the 1850s but that very few were published before the 1870s, so a whole new mathematical language for textbooks had to be built, and it was.

A generational handover. Some wasan mathematicians wanted to keep Edo practices, and the debate went on for a time. But by the early twentieth century, Cousin writes, Western mathematics had stabilised in primary and secondary schools, with national curricula and textbooks such as Kikuchi Dairoku's in geometry and Fujisawa Rikitarō's in arithmetic.

What survived. Not everything was lost. Cousin stresses that some elements of traditional teaching outlived the modernisation policy. The most visible survivor is the soroban itself: it stayed in practical use, and soroban schools and the national grading bodies described above still operate today. Wasan's research tradition, the determinants, series and sangaku, became history. Its problems are still studied, but the schools that produced them are gone.

Soroban, anzan, paper and wasan: which wins where?

The founder's question for this series is which method is "more powerful". On our problem set the honest answer depends on the problem.

The soroban wins on speed and reliability of arithmetic, after training. The board holds the running total and carries are physical moves, so working memory holds only one digit pair at a time.

It loses on transparency. Every step overwrites the last, so a slip leaves no trace. Paper keeps the history, which makes it easier to check and to teach from.

It has nothing to say about structure. The soroban cannot see that 47 x 53 is 50² - 9, that 1/19 repeats every 18 digits, or that a system with a zero determinant has no unique solution. Those are ideas, and they came from wasan, from Indian and Chinese mathematics and from Europe.

A practice plan

A twelve week plan for adults, following the order of the Zenshuren syllabus: addition and subtraction, then multiplication, division and anzan. Practise 20 to 30 minutes a day.

For a recognised grade, find a school affiliated with Zenshuren or Nisshuren, since each runs its own exams. For wasan and sangaku, work the problems in Fukagawa and Rothman's book with pencil and paper and check your answers numerically.

Common mistakes

• Not clearing the board. A stray bead from the last problem is the most common cause of a wrong answer.
• Counting beads instead of reading patterns. Learn to read each digit as a shape.
• Losing the unit rod. Decide which rod holds the ones digit before you start; correct digits on the wrong rods are off by a factor of ten.
• Skipping the mixed move. Learners who drill only five and ten complements freeze at 7 + 7 or 6 + 8.
• Overclaiming the benefits. The peer reviewed evidence supports arithmetic gains, not general intelligence gains.
• Trusting every tablet. Many sangaku give only an answer, and Unger records one (Fukagawa and Pedoe 3.1.7) whose printed answer is clearly wrong.
• Mixing up Seki and Takebe. Most historians now credit enri and the arc length series to Takebe.

FAQ

What is the difference between a soroban and an abacus?

A soroban is the Japanese abacus. It has one bead worth 5 above the beam and four beads worth 1 below it on each rod, so every digit has exactly one bead pattern. The Chinese suanpan traditionally has two upper and five lower beads.

How do you add on a soroban?

Each digit is added on its own rod using one of four moves: a direct move when the beads are free, a five complement (add 5, remove the difference), a ten complement (remove the complement to 10 and carry 1 left) or a mixed move that combines both. Addition never needs a memorised sum beyond the complements of 5 and 10.

What are the soroban grades in Japan?

Zenshuren grades soroban from 15th kyu up to 1st kyu, then from jun shodan to 10th dan. Kyu grades need 100 or more out of 150 in each event; dan grades need 80 or more out of 300 in all six events for jun shodan, rising to 280 or more for 10th dan. Nisshuren and the chambers of commerce run a separate set of exams.

Does learning the soroban make children smarter?

There is no good evidence for a general intelligence effect. A three year randomised trial with 204 children (Barner and colleagues, 2016) found arithmetic gains compared with controls, but no change in basic cognitive abilities such as working memory.

Did Seki Takakazu discover determinants before Leibniz?

On the surviving written record, yes. Seki described determinants for arrays up to 5 x 5 in 1683, and Leibniz described a determinant condition in a 1693 letter. MacTutor credits Seki with the earlier record and treats the two developments as independent, so the fair summary is independent discovery, with Seki's account written down first.

Did Seki discover the Bernoulli numbers?

He found them independently in his work on sums of powers, published after his death in Katsuyō Sanpō (usually dated 1712; one reference work gives 1709). Jacob Bernoulli's Ars Conjectandi followed in 1713. Each man worked without knowledge of the other.

What is a sangaku?

A sangaku is a wooden tablet with a geometry problem, usually with its answer, offered at a Japanese temple or shrine during the Edo period. Many survive, and they are collected in Fukagawa and Pedoe's 1989 book and Fukagawa and Rothman's Sacred Mathematics (2008).

Why did traditional Japanese mathematics disappear?

The 1872 Education Order planned to teach only Western mathematics in the new state schools, and Western notation and methods became the language of science and engineering. Wasan schools lost their role, though the soroban stayed in use.

Where Safeguard fits

Safeguard does not do mathematics, and nothing in this post was computed by a Safeguard model; the checks were plain Python. The honest parallel is about method. A soroban operator checks a result with casting out nines, a cheap checksum that catches most slips and misses a few known cases. A software supply chain needs the same habit at a larger scale: every package you install should be identifiable, checkable against a known good version, and traceable to where it came from, because a build, like a soroban, overwrites its working and keeps no record of how it got there unless you make it. Software bills of materials, signatures and reproducible build records are the supply chain's version of a checked calculation. Our post on using SBOMs in incident response shows what that looks like when an advisory lands.

The simulator behind this post's diagrams is a small Python script. Even a script like that runs on an interpreter, and real projects pull in third party libraries that carry their own vulnerabilities; keeping those dependencies on patched versions is the part of the job Safeguard is built for.

Fix it without waiting for upstream

Whatever the vulnerability, most of the risk sits in two gaps. The first is the time between a fix being published and that fix actually running in your production builds. The second, and worse, is the time before any fix exists at all, while a maintainer works out a patch and a release schedule. Waiting through both gaps is a habit, not a requirement. Safeguard gives teams three ways to close them.

1. Autonomous remediation: pull requests, not tickets

When a fixed version exists, Safeguard's Griffin models open a tested, policy gated pull request that moves the affected dependency to the safe version. You can run it on one repository or in bulk across every project that carries the vulnerable version. Remediation covers eight categories, including dependencies, licenses, security posture and findings. Reachability analysis ranks the vulnerable code paths your application actually calls first, so the first pull requests you merge are the ones that matter most. Autonomous remediation ships switched off, with every category set to Manual, so your team decides how much runs on its own and what still needs a human approval.

2. Gold Open Source: the safe version in seconds, free

gold.safeguard.sh is free and needs no login. Search any CVE or package to see affected and fixed versions, CVSS, EPSS, CISA KEV status and a recommended safe version, or use the "Am I affected?" checker. The same data is available through a free read API, README status badges and RSS threat feeds. The safeguard-sh/gold-scan-action GitHub Action can fail a build on a known exploited vulnerability, on any known vulnerability, or never, so a version you have just removed cannot quietly come back in next week's merge.

3. Gold Registry: start from components that are already fixed

registry.safeguard.sh provides hardened, drop in replacements for upstream packages and container images. Every Gold artifact is held continuously to zero exploitable critical or high CVEs, zero malware and a completed license check, with SLSA L3 or higher provenance, a Sigstore signature and a public evidence page you can hand to an auditor. An artifact loses its Gold badge if it regresses. Teams can point their own package directory at the registry so hardened equivalents resolve without changing how developers install anything.

When upstream has no fix yet, or will not ship one, Safeguard first works to get the fix accepted upstream. If the issue is severe or upstream will not move, Safeguard builds a fork, fixes the vulnerability there and publishes a hardened Gold build, so you are not stuck on a maintainer's release schedule. You can also request a hardened, compatibility tested build of a specific package. This fork and patch flow is in early access today, so ask which packages it covers before you plan around it.

One honest limit: when the vulnerable code lives in a vendor appliance, a hosted service or an operating system runtime you do not build yourself, the fix still has to come from that vendor or distribution. The steps above apply to the open source dependencies and container images you build and ship, which is where most teams carry the largest and least tracked share of their exposure.

See how Safeguard would help

These demos show dependency scanning and remediation workflows, not mathematics. You can download Safeguard and follow the CLI docs to try them on your own projects.

This demo is a real CLI session. After installing, safeguard opens the interactive terminal; asked to "fix the lodash CVE in this repo", it reads the finding and the advisory, asks before any command that changes something, edits package.json to lodash 4.17.21, runs npm install and npm test with approval, prints a summary and pushes nothing. The same flow applies to the Python or JavaScript tooling behind a teaching site, a flash anzan trainer or a calculation script.

Click through it at your own pace, or open it full screen.

This demo sets up the Safeguard runner on your own machine, approves it with its device code and scans from the terminal. Pointed at a project folder, the scan lists the packages and versions it depends on and their known vulnerabilities, so you know what to update before you share the project.

Click through it at your own pace, or open it full screen.

Quick check

Sources

• MacTutor: Takakazu Seki biography
• MacTutor: Seki, Dictionary of Scientific Biography article by A. Kobori
• MacTutor: Matrices and determinants
• MacTutor: Takebe Katahiro biography
• MacTutor: Jacob Bernoulli biography
• J. Marshall Unger, A Collection of Sangaku Problems, Ohio State University (2009)
• Cut the Knot: Three Tangent Circles Sangaku
• Marion Cousin, Sur la création d'une nouvelle langue mathématique japonaise, Revue d'Histoire des Mathématiques 23 (2017)
• Barner et al., Learning Mathematics in a Visuospatial Format, Child Development 87(4), 2016 (Harvard DASH)
• Zenshuren (League for Soroban Education of Japan): home
• Zenshuren: soroban exam (珠算検定試験)
• Zenshuren: flash anzan exam
• Zenshuren: 2026 All Japan Soroban Championship rules (PDF)
• Nisshuren (League of Japan Abacus Associations)
• Whipple Museum of the History of Science: soroban, 20th century
• Hidetoshi Fukagawa and Daniel Pedoe, Japanese Temple Geometry Problems (Charles Babbage Research Centre, 1989), cited through Unger
• Hidetoshi Fukagawa and Tony Rothman, Sacred Mathematics: Japanese Temple Geometry (Princeton University Press, 2008)

Last reviewed: 25 October 2026. Exam rules and championship details change each year; check the Zenshuren and Nisshuren pages above for the current versions.